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Magnéli oxides as promising n-type thermoelectrics

  • The discovery of a large thermopower in cobalt oxides in 1997 lead to a surge of interest in oxides for thermoelectric application. Whereas conversion efficiencies of p-type oxides can compete with non-oxide materials, n-type oxides show significantly lower thermoelectric performances. In this context so-called Magnéli oxides have recently gained attention as promising n-type thermoelectrics. A combination of crystallographic shear and intrinsic disorder lead to relatively low thermal conductivities and metallic-like electrical conductivities in Magnéli oxides. Current peak-zT values of 0.3 around 1100 K for titanium and tungsten Magnéli oxides are encouraging for future research. Here, we put Magnéli oxides into context of n-type oxide thermoelectrics and give a perspective where future research can bring us.

    Citation: Gregor Kieslich, Wolfgang Tremel. Magnéli oxides as promising n-type thermoelectrics[J]. AIMS Materials Science, 2014, 1(4): 184-190. doi: 10.3934/matersci.2014.4.184

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  • The discovery of a large thermopower in cobalt oxides in 1997 lead to a surge of interest in oxides for thermoelectric application. Whereas conversion efficiencies of p-type oxides can compete with non-oxide materials, n-type oxides show significantly lower thermoelectric performances. In this context so-called Magnéli oxides have recently gained attention as promising n-type thermoelectrics. A combination of crystallographic shear and intrinsic disorder lead to relatively low thermal conductivities and metallic-like electrical conductivities in Magnéli oxides. Current peak-zT values of 0.3 around 1100 K for titanium and tungsten Magnéli oxides are encouraging for future research. Here, we put Magnéli oxides into context of n-type oxide thermoelectrics and give a perspective where future research can bring us.


    The algebraic classification (up to isomorphism) of algebras of dimension $ n $ from a certain variety defined by a certain family of polynomial identities is a classic problem in the theory of non-associative algebras. There are many results related to the algebraic classification of small-dimensional algebras in many varieties of non-associative algebras [11,12,2,3,4,6,13,9,16]. So, algebraic classifications of $ 2 $-dimensional algebras [16,19], $ 3 $-dimensional evolution algebras [1], $ 3 $-dimensional anticommutative algebras [17], $ 4 $-dimensional division algebras [5,7], $ 4 $-dimensional nilpotent algebras [13] and $ 6 $-dimensional anticommutative nilpotent algebras [12] have been given. In the present paper, we give the algebraic classification of $ 5 $-dimensional nilpotent commutative algebras. The variety of commutative algebras is defined by the following identity: $ xy = yx. $ It contains commutative $ \mathfrak{CD} $-algebras, Jordan algebras, mock-Lie algebras and commutative associative algebras as subvarieties. On the other hand, it is a principal part in the varieties of weakly associative algebras and flexible algebras.

    The algebraic study of central extensions of associative and non-associative algebras has been an important topic for years (see, for example, [10,20] and references therein). Our method for classifying nilpotent commutative algebras is based on the calculation of central extensions of nilpotent algebras of smaller dimensions from the same variety (first, this method has been developed by Skjelbred and Sund for Lie algebra case in [20]) and the classifications of all complex $ 5 $-dimensional nilpotent commutative (non-Jordan) $ \frak{CD} $-algebras [11]; nilpotent Jordan (non-associative) algebras [9]; and nilpotent associative commutative algebras [18].

    Throughout this paper, we use the notations and methods well written in [10], which we have adapted for the commutative case with some modifications. Further in this section we give some important definitions.

    Let $ ({\bf A}, \cdot) $ be a complex commutative algebra and $ \mathbb V $ be a complex vector space. The $ \mathbb C $-linear space $ {\rm Z^{2}}\left( \bf A,\mathbb V \right) $ is defined as the set of all bilinear maps $ \theta \colon {\bf A} \times {\bf A} \longrightarrow {\mathbb V} $ such that $ \theta(x,y) = \theta(y,x) . $ These elements will be called cocycles. For a linear map $ f $ from $ \bf A $ to $ \mathbb V $, if we define $ \delta f\colon {\bf A} \times {\bf A} \longrightarrow {\mathbb V} $ by $ \delta f (x,y ) = f(xy ) $, then $ \delta f\in {\rm Z^{2}}\left( {\bf A},{\mathbb V} \right) $. We define $ {\rm B^{2}}\left({\bf A},{\mathbb V}\right) = \left\{ \theta = \delta f\ : f\in {\rm Hom}\left( {\bf A},{\mathbb V}\right) \right\} $. We define the second cohomology space $ {\rm H^{2}}\left( {\bf A},{\mathbb V}\right) $ as the quotient space $ {\rm Z^{2}} \left( {\bf A},{\mathbb V}\right) \big/{\rm B^{2}}\left( {\bf A},{\mathbb V}\right) $.

    Let $ \operatorname{Aut}({\bf A}) $ be the automorphism group of $ {\bf A} $ and let $ \phi \in \operatorname{Aut}({\bf A}) $. For $ \theta \in {\rm Z^{2}}\left( {\bf A},{\mathbb V}\right) $ define the action of the group $ \operatorname{Aut}({\bf A}) $ on $ {\rm Z^{2}}\left( {\bf A},{\mathbb V}\right) $ by

    $\phi \theta (x,y) = \theta \left( \phi \left( x\right) ,\phi \left( y\right) \right) $.

    It is easy to verify that $ {\rm B^{2}}\left( {\bf A},{\mathbb V}\right) $ is invariant under the action of $ \operatorname{Aut}({\bf A}). $ So, we have an induced action of $ \operatorname{Aut}({\bf A}) $ on $ {\rm H^{2}}\left( {\bf A},{\mathbb V}\right) $.

    Let $ \bf A $ be a commutative algebra of dimension $ m $ over $ \mathbb C $ and $ {\mathbb V} $ be a $ \mathbb C $-vector space of dimension $ k $. For the bilinear map $ \theta $, define on the linear space $ {\bf A}_{\theta } = {\bf A}\oplus {\mathbb V} $ the bilinear product " $ \left[ -,-\right] _{{\bf A}_{\theta }} $" by $ \left[ x+x^{\prime },y+y^{\prime }\right] _{{\bf A}_{\theta }} = xy +\theta(x,y) $ for all $ x,y\in {\bf A},x^{\prime },y^{\prime }\in {\mathbb V} $. The algebra $ {\bf A}_{\theta } $ is called a $ k $-dimensional central extension of $ {\bf A} $ by $ {\mathbb V} $. One can easily check that $ {\bf A_{\theta}} $ is a commutative algebra if and only if $ \theta \in {\rm Z^2}({\bf A}, {\mathbb V}) $.

    Call the set $ \operatorname{Ann}(\theta) = \left\{ x\in {\bf A}:\theta \left( x, {\bf A} \right) = 0\right\} $ the annihilator of $ \theta $. We recall that the annihilator of an algebra $ {\bf A} $ is defined as the ideal $ \operatorname{Ann}( {\bf A} ) = \left\{ x\in {\bf A}: x{\bf A} = 0\right\} $. Observe that $ \operatorname{Ann}\left( {\bf A}_{\theta }\right) = (\operatorname{Ann}(\theta) \cap\operatorname{Ann}({\bf A})) \oplus {\mathbb V} $.

    The following result shows that every algebra with a non-zero annihilator is a central extension of a smaller-dimensional algebra.

    Lemma 1.1. Let $ {\bf A} $ be an $ n $-dimensional commutative algebra such that

    $\dim (\operatorname{Ann}({\bf A})) = m\neq0$.

    Then there exists, up to isomorphism, a unique $ (n-m) $-dimensional commutative algebra $ {\bf A}' $ and a bilinear map $ \theta \in {\rm Z^2}({\bf A'}, {\mathbb V}) $ with $ \operatorname{Ann}({\bf A'})\cap\operatorname{Ann}(\theta) = 0 $, where $ \mathbb V $ is a vector space of dimension m, such that $ {\bf A} \cong {{\bf A}'}_{\theta} $ and $ {\bf A}/\operatorname{Ann}({\bf A})\cong {\bf A}' $.

    Proof. Let $ {\bf A}' $ be a linear complement of $ \operatorname{Ann}({\bf A}) $ in $ {\bf A} $. Define a linear map $ P \colon {\bf A} \longrightarrow {\bf A}' $ by $ P(x+v) = x $ for $ x\in {\bf A}' $ and $ v\in\operatorname{Ann}({\bf A}) $, and define a multiplication on $ {\bf A}' $ by $ [x, y]_{{\bf A}'} = P(x y) $ for $ x, y \in {\bf A}' $. For $ x, y \in {\bf A} $, we have

    $ P(xy) = P((x-P(x)+P(x))(y- P(y)+P(y))) = P(P(x) P(y)) = [P(x), P(y)]_{{\bf A}'}. $

    Since $ P $ is a homomorphism, $ P({\bf A}) = {\bf A}' $ and $ {\bf A}' $ is a commutative algebra and also $ {\bf A}/\operatorname{Ann}({\bf A})\cong {\bf A}' $, which gives us the uniqueness. Now, define the map $ \theta \colon {\bf A}' \times {\bf A}' \longrightarrow\operatorname{Ann}({\bf A}) $ by $ \theta(x,y) = xy- [x,y]_{{\bf A}'} $. Thus, $ {\bf A}'_{\theta} $ is $ {\bf A} $ and therefore $ \theta \in {\rm Z^2}({\bf A'}, {\mathbb V}) $ and $ \operatorname{Ann}({\bf A'})\cap\operatorname{Ann}(\theta) = 0 $.

    Definition 1.2. Let $ {\bf A} $ be an algebra and $ I $ be a subspace of $ \operatorname{Ann}({\bf A}) $. If $ {\bf A} = {\bf A}_0 \oplus I $ then $ I $ is called an annihilator component of $ {\bf A} $. A central extension of an algebra $ \bf A $ without annihilator component is called a non-split central extension.

    Our task is to find all central extensions of an algebra $ \bf A $ by a space $ {\mathbb V} $. In order to solve the isomorphism problem we need to study the action of $ \operatorname{Aut}({\bf A}) $ on $ {\rm H^{2}}\left( {\bf A},{\mathbb V} \right) $. To do that, let us fix a basis $ e_{1},\ldots ,e_{s} $ of $ {\mathbb V} $, and $ \theta \in {\rm Z^{2}}\left( {\bf A},{\mathbb V}\right) $. Then $ \theta $ can be uniquely written as $ \theta \left( x,y\right) = \sum\limits_{i = 1}^{s} \theta _{i}\left( x,y\right) e_{i} $, where $ \theta _{i}\in {\rm Z^{2}}\left( {\bf A},\mathbb C\right) $. Moreover, $ \operatorname{Ann}(\theta) = \operatorname{Ann}(\theta _{1})\cap\operatorname{Ann}(\theta _{2})\cap\ldots \cap\operatorname{Ann}(\theta _{s}) $. Furthermore, $ \theta \in {\rm B^{2}}\left( {\bf A},{\mathbb V}\right) $ if and only if all $ \theta _{i}\in {\rm B^{2}}\left( {\bf A}, \mathbb C\right) $. It is not difficult to prove (see [10,Lemma 13]) that given a commutative algebra $ {\bf A}_{\theta} $, if we write as above $ \theta \left( x,y\right) = \sum\limits_{i = 1}^{s} \theta_{i}\left( x,y\right) e_{i}\in {\rm Z^{2}}\left( {\bf A},{\mathbb V}\right) $ and $ \operatorname{Ann}(\theta)\cap \operatorname{Ann}\left( {\bf A}\right) = 0 $, then $ {\bf A}_{\theta } $ has an annihilator component if and only if $ \left[ \theta _{1}\right] ,\left[ \theta _{2}\right] ,\ldots ,\left[ \theta _{s}\right] $ are linearly dependent in $ {\rm H^{2}}\left( {\bf A},\mathbb C\right) $.

    Let $ {\mathbb V} $ be a finite-dimensional vector space over $ \mathbb C $. The Grassmannian $ G_{k}\left( {\mathbb V}\right) $ is the set of all $ k $-dimensional linear subspaces of $ {\mathbb V} $. Let $ G_{s}\left( {\rm H^{2}}\left( {\bf A},\mathbb C\right) \right) $ be the Grassmannian of subspaces of dimension $ s $ in $ {\rm H^{2}}\left( {\bf A},\mathbb C\right) $. There is a natural action of $ \operatorname{Aut}({\bf A}) $ on $ G_{s}\left( {\rm H^{2}}\left( {\bf A},\mathbb C\right) \right) $. Let $ \phi \in \operatorname{Aut}({\bf A}) $. For $ W = \langle \left[ \theta _{1}\right] ,\left[ \theta _{2}\right] ,\dots,\left[ \theta _{s} \right] \rangle \in G_{s}\left( {\rm H^{2}}\left( {\bf A},\mathbb C \right) \right) $ define $ \phi W = \langle \left[ \phi \theta _{1}\right] ,\left[ \phi \theta _{2}\right] ,\dots,\left[ \phi \theta _{s}\right] \rangle $. We denote the orbit of $ W\in G_{s}\left( {\rm H^{2}}\left( {\bf A},\mathbb C\right) \right) $ under the action of $ \operatorname{Aut}({\bf A}) $ by $ \operatorname{Orb}(W) $. Given

    $ W_{1} = \langle \left[ \theta _{1}\right] ,\left[ \theta _{2}\right] ,\dots, \left[ \theta _{s}\right] \rangle ,W_{2} = \langle \left[ \vartheta _{1}\right] ,\left[ \vartheta _{2}\right] ,\dots,\left[ \vartheta _{s}\right] \rangle \in G_{s}\left( {\rm H^{2}}\left( {\bf A},\mathbb C\right) \right), $

    we easily have that if $ W_{1} = W_{2} $, then $ \bigcap\limits_{i = 1}^{s}\operatorname{Ann}(\theta _{i})\cap \operatorname{Ann}\left( {\bf A}\right) = \bigcap\limits_{i = 1}^{s} \operatorname{Ann}(\vartheta _{i})\cap\operatorname{Ann}( {\bf A}) $, and therefore we can introduce the set

    $ {\bf T}_{s}({\bf A}) = \left\{ W = \langle \left[ \theta _{1}\right] , \dots,\left[ \theta _{s}\right] \rangle \in G_{s}\left( {\rm H^{2}}\left( {\bf A},\mathbb C\right) \right) : \bigcap\limits_{i = 1}^{s}\operatorname{Ann}(\theta _{i})\cap\operatorname{Ann}({\bf A}) = 0\right\}, $

    which is stable under the action of $ \operatorname{Aut}({\bf A}) $.

    Now, let $ {\mathbb V} $ be an $ s $-dimensional linear space and let us denote by $ {\bf E}\left( {\bf A},{\mathbb V}\right) $ the set of all non-split $ s $-dimensional central extensions of $ {\bf A} $ by $ {\mathbb V} $. By above, we can write

    $ {\bf E}\left( {\bf A},{\mathbb V}\right) = \left\{ {\bf A}_{\theta }:\theta \left( x,y\right) = \sum\limits_{i = 1}^{s}\theta _{i}\left( x,y\right) e_{i} \ \ {\rm{and}} \ \ \langle \left[ \theta _{1}\right] ,\left[ \theta _{2}\right] ,\dots, \left[ \theta _{s}\right] \rangle \in {\bf T}_{s}({\bf A}) \right\} . $

    We also have the following result, which can be proved as in [10,Lemma 17].

    Lemma 1.3. Let $ {\bf A}_{\theta },{\bf A}_{\vartheta }\in {\bf E}\left( {\bf A},{\mathbb V}\right) $. Suppose that $ \theta \left( x,y\right) = \sum\limits_{i = 1}^{s}\theta _{i}\left( x,y\right) e_{i} $ and $ \vartheta \left( x,y\right) = \sum\limits_{i = 1}^{s} \vartheta _{i}\left( x,y\right) e_{i} $. Then the commutative algebras $ {\bf A}_{\theta } $ and $ {\bf A}_{\vartheta } $ are isomorphic if and only if

    $ \operatorname{Orb}\langle \left[ \theta _{1}\right] , \left[ \theta _{2}\right] ,\dots,\left[ \theta _{s}\right] \rangle = \operatorname{Orb}\langle \left[ \vartheta _{1}\right] ,\left[ \vartheta _{2}\right] ,\dots,\left[ \vartheta _{s}\right] \rangle. $

    This shows that there exists a one-to-one correspondence between the set of $ \operatorname{Aut}({\bf A}) $-orbits on $ {\bf T}_{s}\left( {\bf A}\right) $ and the set of isomorphism classes of $ {\bf E}\left( {\bf A},{\mathbb V}\right) $. Consequently we have a procedure that allows us, given a commutative algebra $ {\bf A}' $ of dimension $ n-s $, to construct all non-split central extensions of $ {\bf A}' $. This procedure is:

    $ 1.\ $For a given commutative algebra $ {\bf A}' $ of dimension $ n-s $, determine $ {\rm H^{2}}( {\bf A}',\mathbb {C}) $, $ \operatorname{Ann}({\bf A}') $ and $ \operatorname{Aut}({\bf A}') $.

    $ 2.\ $Determine the set of $ \operatorname{Aut}({\bf A}') $-orbits on $ {\bf T}_{s}({\bf A}') $.

    $ 3.\ $For each orbit, construct the commutative algebra associated with a representative of it.

    The idea of the definition of a $ \mathfrak{CD} $-algebra comes from the following property of Jordan and Lie algebras: the commutator of any pair of multiplication operators is a derivation. This gives three identities of degree four, which reduce to only one identity of degree four in the commutative or anticommutative case. Namely, a commutative algebra is a commutative $ \frak{CD} $-algebra ($ \frak{CCD} $-algebra) if it satisfies the following identity:

    $ ((xy)a)b + ((xb)a)y + x((yb)a) = ((xy)b)a + ((xa)b)y+ x((ya)b). $

    The above described method gives all commutative ($ \mathfrak{CCD} $- and non-$ \mathfrak{CCD} $-) algebras. But we are interested in developing this method in such a way that it only gives non-$ \mathfrak{CCD} $ commutative algebras, because the classification of all $ \mathfrak{CCD} $-algebras is done in [11]. Clearly, any central extension of a commutative non-$ \mathfrak{CCD} $-algebra is a non-$ \mathfrak{CCD} $-algebra. But a $ \mathfrak{CCD} $-algebra may have extensions which are not $ \mathfrak{CCD} $-algebras. More precisely, let $ \mathfrak{D} $ be a $ \mathfrak{CCD} $-algebra and $ \theta \in {\rm Z_\mathfrak{C}^2}(\mathfrak{D}, {\mathbb C}). $ Then $ {\mathfrak{D}}_{\theta } $ is a $ \mathfrak{CCD} $-algebra if and only if

    $ \theta(x,y) = \theta(y,x), $
    $ \theta((xy)a,b)+\theta((xb)a,y)+\theta(x,(yb)a) = \theta((xy)b,a)+\theta((xa)b,y)+\theta(x,(ya)b). $

    for all $ x,y,a,b\in {\mathfrak{D}}. $ Define the subspace $ {\rm Z_\mathfrak{D}^2}({\mathfrak{D}},{\mathbb C}) $ of $ {\rm Z_\mathfrak{C}^2}({\bf \mathfrak{D}},{\mathbb C}) $ by

    $ Z2D(D,C)={θZ2C(D,C):θ(x,y)=θ(y,x),θ((xy)a,b)+θ((xb)a,y)+θ(x,(yb)a)=θ((xy)b,a)+θ((xa)b,y)+θ(x,(ya)b) for all x,y,a,bD}. $

    Observe that $ {\rm B^2}({ \mathfrak{D}},{\mathbb C})\subseteq{\rm Z_\mathfrak{D}^2}({\mathfrak{D}},{\mathbb C}). $ Let $ {\rm H_\mathfrak{D}^2}({\mathfrak{D}},{\mathbb C}) = {\rm Z_\mathfrak{D}^2}({\mathfrak{D}},{\mathbb C}) \big/{\rm B^2}({\mathfrak{D}},{\mathbb C}). $ Then $ {\rm H_\mathfrak{D}^2}({\mathfrak{D}},{\mathbb C}) $ is a subspace of $ {\rm H_\mathfrak{C}^2}({\mathfrak{D}},{\mathbb C}). $ Define

    $ {\bf R}_{s}({\mathfrak{D}}) = \left\{ {\bf W}\in {\bf T}_{s}({\mathfrak{D}}) :{\bf W}\in G_{s}({\rm H_\mathfrak{D}^2}({\mathfrak{D}},{\mathbb C}) ) \right\}, $
    $ {\bf U}_{s}({\mathfrak{D}}) = \left\{ {\bf W}\in {\bf T}_{s}({\mathfrak{D}}) :{\bf W}\notin G_{s}({\rm H_\mathfrak{D}^2}({\mathfrak{D}},{\mathbb C}) ) \right\}. $

    Then $ {\bf T}_{s}({\mathfrak{D}}) = {\bf R}_{s}( {\mathfrak{D}}) $$ {\bf U}_{s}( {\mathfrak{D}}). $ The sets $ {\bf R}_{s}({\mathfrak{D}}) $ and $ {\bf U}_{s}({\mathfrak{D}}) $ are stable under the action of $ \operatorname{Aut}({\mathfrak{D}}). $ Thus, the commutative algebras corresponding to the representatives of $ \operatorname{Aut}({\mathfrak{D}}) $ -orbits on $ {\bf R}_{s}({\mathfrak{D}}) $ are $ \mathfrak{CCD} $-algebras, while those corresponding to the representatives of $ \operatorname{Aut}({\mathfrak{D}} ) $-orbits on $ {\bf U}_{s}({\mathfrak{D}}) $ are not $ \mathfrak{CCD} $-algebras. Hence, we may construct all non-split commutative non-$ \mathfrak{CCD} $-algebras $ \bf{A} $ of dimension $ n $ with $ s $-dimensional annihilator from a given commutative algebra $ \bf{A} ^{\prime } $ of dimension $ n-s $ in the following way:

    $1.\ $If $ \bf{A}^{\prime } $ is non-$ \mathfrak{CCD} $, then apply the procedure.

    $2.\ $Otherwise, do the following:

    ${\rm{(a)}} \ $Determine $ {\bf U}_{s}(\bf{A}^{\prime }) $ and $ \operatorname{Aut}(\bf{A}^{\prime }). $

    $ {\rm{(b)}}\ $Determine the set of $ \operatorname{Aut}(\bf{A}^{\prime }) $-orbits on $ {\bf U }_{s}(\bf{A}^{\prime }). $

    ${\rm{(c)}} \ $For each orbit, construct the commutative algebra corresponding to one of its representatives.

    Let us introduce the following notations. Let $ {\bf A} $ be a nilpotent algebra with a basis $ e_{1},e_{2}, \ldots, e_{n}. $ Then by $ \Delta_{ij} $ we will denote the bilinear form $ \Delta_{ij}:{\bf A}\times {\bf A}\longrightarrow \mathbb C $ with $ \Delta_{ij}(e_{l},e_{m}) = \delta_{il}\delta_{jm}, $ if $ i\leq j $ and $ l\leq m. $ The set $ \left\{ \Delta_{ij}:1\leq i\leq j\leq n\right\} $ is a basis for the linear space of bilinear forms on $ {\bf A}, $ so every $ \theta \in {\rm Z^2}({\bf A},\bf \mathbb V ) $ can be uniquely written as $ \theta = \sum\limits_{1\leq i\leq j\leq n} c_{ij}\Delta _{{i}{j}} $, where $ c_{ij}\in \mathbb C $. Let us fix complex number $ \eta_k $ ($ \eta_k^k = -1, $ $ \eta_k^l\neq1 $ for $ 0<l<k $). For denote our algebras, we will use the following notations:

    $ NΞjjth5dimensional family ofcommutative nonCCDalgebras with parametrs Ξ.Nijjth idimensional nonCCDalgebra.Nijjth idimensional CCDalgebra. $

    Remark 1. All families of algebras from our final list do not have intersections, but inside some families of algebras there are isomorphic algebras. All isomorphisms between algebras from a certain family of algebras constucted from the representative $ \nabla(\Sigma) $ are given in the list of distinct orbit representations. The notation $ \langle \nabla(\Xi) \rangle^{O(\Xi_1) = O(\Xi_2)} $ represents that the elements $ \langle \nabla(\Xi_1) \rangle $ and $ \langle \nabla(\Xi_2) \rangle $ have the same orbit.

    Thanks to [8] we have the complete classification of complex $ 4 $-dimensional nilpotent commutative algebras. It will be re-written by some different way for separating $ \mathfrak{CCD} $- and non-$ \mathfrak{CCD} $-algebras.

    $ N301,N401:e1e1=e2H2C=H2DN302,N402:e1e1=e2e1e2=e3H2CH2DN303,N403:e1e2=e3H2C=H2DN304,N404:e1e1=e2e2e2=e3H2CH2DN405:e1e1=e2e1e3=e4H2C=H2DN406:e1e1=e2e3e3=e4H2C=H2DN407:e1e1=e4e2e3=e4H2C=H2DN408:e1e1=e2e1e2=e3e2e2=e4H2CH2DN409:e1e1=e2e2e3=e4H2CH2DN410:e1e1=e2e1e2=e4e3e3=e4H2CH2DN411:e1e1=e2e1e3=e4e2e2=e4H2CH2DN412:e1e1=e2e2e2=e4e3e3=e4H2CH2DN413(λ):e1e1=e2e1e2=e3e1e3=e4e2e2=λe4H2CH2DN414:e1e2=e3e1e3=e4H2CH2DN415:e1e2=e3e1e3=e4e2e2=e4H2CH2DN416:e1e2=e3e1e3=e4e2e3=e4H2CH2DN417:e1e2=e3e3e3=e4H2CH2DN418:e1e1=e4e1e2=e3e3e3=e4H2CH2DN419:e1e1=e4e1e2=e3e2e2=e4e3e3=e4H2CH2DN401:e1e1=e2e1e2=e3e2e3=e4N402:e1e1=e2e1e2=e3e1e3=e4e2e3=e4N403:e1e1=e2e1e2=e3e3e3=e4N404:e1e1=e2e1e2=e3e2e2=e4e3e3=e4N405:e1e1=e2e1e3=e4e2e2=e3N406:e1e1=e2e1e2=e4e1e3=e4e2e2=e3N407:e1e1=e2e2e2=e3e2e3=e4N408:e1e1=e2e1e3=e4e2e2=e3e2e3=e4N409:e1e1=e2e2e2=e3e3e3=e4N410:e1e1=e2e2e2=e3e1e2=e4e3e3=e4N411(λ):e1e1=e2e1e2=λe4e2e2=e3e2e3=e4e3e3=e4 $

    Here we will collect all information about $ \mathbf{N}_{02}^{3*}: $

    $ \begin{array}{|l|l|l|l|} \hline \rm{ }  & \rm{ } & \rm{Cohomology} & \rm{Automorphisms} \\ \hline {\mathbf{N}}^{3*}_{02} & \begin{array}{l}e_1e_1 = e_2 \\ e_1e_2 = e_3 \end{array} & H2D(N302)=[Δ13],[Δ22],H2C(N302)=H2D(N302)[Δ23],[Δ33] & \phi = (x00yx20z2xyx3)\\ \hline \end{array} $

    Let us use the following notations:

    $ 1=[Δ13],2=[Δ22],3=[Δ23],4=[Δ33]. $

    Take $ \theta = \sum\limits_{i = 1}^{4}\alpha_i\nabla_i\in\mathrm{H}^2_{\mathfrak{C}}(\mathbf{N}^{3*}_{02}) . $ Since

    $ \phi^T(00α10α2α3α1α3α4)\phi = (ααα1αα2α3α1α3α4), $

    we have

    $ \alpha_1^* = (\alpha_1x+\alpha_3y+\alpha_4z)x^3, \quad \alpha_2^* = (\alpha_2x^2+4\alpha_3xy+4\alpha_4y^2)x^2 ,\\ \alpha_3^* = (\alpha_3x+2\alpha_4y)x^4 , \qquad \alpha_4^* = \alpha_4x^6. $

    We are interested only in $ (\alpha_3,\alpha_4)\neq(0,0) $ and consider the vector space generated by the following two cocycles:

    $ \theta_1 = \alpha_1\nabla_1+\alpha_2\nabla_2+\alpha_3\nabla_3+\alpha_4\nabla_4 \ \ \rm{and} \ \ \theta_2 = \beta_1\nabla_1+\beta_2\nabla_2+\beta_3\nabla_3. $

    Thus, we have

    $ α1=(α1x+α3y+α4z)x3,β1=(β1x+β3y)x3,α2=(α2x2+4α3xy+4α4y2)x2,β2=(β2x+4β3y)x3,α3=(α3x+2α4y)x4,β3=β3x5.α4=α4x6. $

    Consider the following cases.

    $ 1.\ $$ \alpha_4\neq0, $ then:

    $ {\rm{(a)}} \ $$ \beta_3 = 0, \beta_2\neq 0, \beta_1 = 0 , $ then by choosing $ x = 2 \alpha_4^2, $ $ y = -\alpha_3 \alpha_4, $ $ z = \alpha_3^2-2 \alpha_1 \alpha_4, $ we have the representatives $ \langle \nabla_4, \nabla_2 \rangle; $

    ${\rm{(b)}}\ $ $ \beta_3 = 0, \beta_2\neq 0, \beta_1\neq 0 , $ then by choosing

    $x = 2 \alpha_4^2 \beta_2, $ $y = -\alpha_3 \alpha_4 \beta_2,$ $z = \alpha_3^2 (-2 \beta_1+\beta_2)+2 \alpha_4 (\alpha_2 \beta_1-\alpha_1 \beta_2),$

    we have the representatives $ \langle \nabla_4, \nabla_1 +\alpha \nabla_2 \rangle_{\alpha\neq 0}; $

    $ {\rm{(c)}} \ $$ \beta_3 = 0, \beta_2 = 0, \beta_1\neq 0 , $ then by choosing $ y = -\frac{x \alpha_3}{2 \alpha_4}, $ we have two representatives $ \langle \nabla_4, \nabla_1 \rangle $ and $ \langle \nabla_2+\nabla_4, \nabla_1 \rangle, $ depending on $ \alpha_3^2 = \alpha_2 \alpha_4 $ or not. The first representative will be joint with the family from the case (1b);

    ${\rm{(d)}} \ $$ \beta_3\neq0, 4\alpha_2\beta_3^2 = 4\beta_2\alpha_3\beta_3-\beta_2^2\alpha_4, \beta_2 = 4\beta_1, $ then by choosing

    $ x = 4\beta_3\alpha_4, y = -\beta_2\alpha_4, z = \beta_2\alpha_3-4\alpha_1\beta_3,$

    we have the representative $ \langle \nabla_4, \nabla_3 \rangle; $

    $ {\rm{(e)}} \ $$ \beta_3\neq0, 4\alpha_2\beta_3^2 = 4\beta_2\alpha_3\beta_3-\beta_2^2\alpha_4, \beta_2\neq4\beta_1, $ then by choosing

    $ x = \frac{4\beta_1-\beta_2}{4\beta_3}, y = \frac{\beta_2^2-4\beta_1\beta_2}{16\beta_3^2}, z = \frac{(4\beta_1-\beta_2)(8\beta_1\alpha_3\beta_3-4\beta_1\beta_2\alpha_4-8\alpha_1\beta_3^3+\beta_2^2\alpha_4)}{32\beta_3^3\alpha_4},$

    we have the representative $ \langle \nabla_4, \nabla_1+\nabla_3 \rangle; $

    $ {\rm{(f)}} \ $$ \beta_3\neq0, 4\alpha_2\beta_3^2\neq4\beta_2\alpha_3\beta_3-\beta_2^2\alpha_4, $ then by choosing

    $ x = \sqrt{\frac{4\alpha_2\beta_3^2-4\beta_2\alpha_3\beta_3+\beta_2^2\alpha_4}{4\beta_3^2\alpha_4}},$ $ y = -\frac{ \beta_2\sqrt{\alpha_4 \beta_2^2-4 \alpha_3 \beta_2 \beta_3+4 \alpha_2 \beta_3^2}}{8\beta_3^2\sqrt{\alpha_4}}, $ $z = \frac{(8\beta_1\alpha_3\beta_3-4\beta_1\beta_2\alpha_4-8\alpha_1\beta_3^3+\beta_2^2\alpha_4)\sqrt{4\alpha_2\beta_3^2-4\beta_2\alpha_3\beta_3+\beta_2^2\alpha_4}}{16\beta_3^3\alpha_4\sqrt{\alpha_4}},$

    we have the family of representatives $ \langle \nabla_2+\nabla_4, \alpha\nabla_1+\nabla_3 \rangle. $

    $2.\ \alpha_4 = 0, \alpha_3\neq0, $ then we may suppose that $ \beta_3 = 0 $ and

    $ {\rm{(a)}}\ $ if $ \beta_1\neq0, \beta_2 = 4\beta_1, \alpha_2 = 4\alpha_1, $ then by choosing $ x = \alpha_3, y = -\alpha_1, z = 0, $ we have the representative $ \langle \nabla_3, \nabla_1+4\nabla_2 \rangle; $

    $ {\rm{(b)}}\ $if $ \beta_1\neq0, \beta_2 = 4\beta_1, \alpha_2\neq4\alpha_1, $ then by choosing $ x = \frac{\alpha_2-4\alpha_1}{\alpha_3}, y = \frac{4\alpha^2_1-\alpha_1\alpha_2}{\alpha_3^2}, z = 0, $ we have the representative $ \langle -24(\nabla_2+\nabla_3), \nabla_1+4\nabla_2 \rangle; $

    $ {\rm{(c)}}\ $if $ \beta_1\neq0, \beta_2\neq4\beta_1 , $ then by choosing $ x = \alpha_3(\beta_2-4\beta_1), y = \beta_1\alpha_2-\alpha_1\beta_2, z = 0, $ we have the family of representatives $ \langle \nabla_3, \nabla_1+\alpha\nabla_2 \rangle_{\alpha\neq4}, $ which will be jointed with the case (2a);

    $ {\rm{(d)}}\ $if $ \beta_1 = 0, $ then we have the representative $ \langle -3\nabla_3, \nabla_2 \rangle $.

    Summarizing, we have the following distinct orbits:

    $ \langle \nabla_1, \nabla_2+\nabla_4 \rangle, \, \langle \nabla_1+4\nabla_2, -24(\nabla_2+\nabla_3) \rangle, \, \langle \nabla_1+\lambda\nabla_2, \nabla_3 \rangle, \, \langle \nabla_1 +\lambda \nabla_2, \nabla_4\rangle, \\ \langle \alpha\nabla_1+\nabla_3, \nabla_2+\nabla_4 \rangle, \, \langle \nabla_1+\nabla_3, \nabla_4 \rangle, \, \langle \nabla_2, -3 \nabla_3 \rangle, \, \langle \nabla_2, \nabla_4 \rangle, \, \langle \nabla_3, \nabla_4 \rangle. $

    Note that the algebras constructed from the orbits $ \langle \nabla_1+4\nabla_2, -24(\nabla_2+\nabla_3) \rangle, $ $ \langle \nabla_1 + \lambda \nabla_2, \nabla_3 \rangle, $ $ \langle \nabla_1 +\alpha \nabla_2, \nabla_4\rangle, $ $ \langle \nabla_2, -3 \nabla_3 \rangle $ and $ \langle \nabla_2, \nabla_4 \rangle $ are parts of some families of algebras which found below. Hence, we have the following new algebras:

    $ N12:e1e1=e2e1e2=e3e1e3=e4e2e2=e5e3e3=e5N4168:e1e1=e2e1e2=e3e1e3=e4e2e2=4e424e5e2e3=24e5Nλ,0170:e1e1=e2e1e2=e3e1e3=e4e2e2=λe4e2e3=e5Nλ,0184:e1e1=e2e1e2=e3e1e3=e4e2e2=λe4e3e3=e5Nα13:e1e1=e2e1e2=e3e1e3=αe4e2e2=e5e2e3=e4e3e3=e5N14:e1e1=e2e1e2=e3e1e3=e4e2e3=e4e3e3=e5N176:e1e1=e2e1e2=e3e2e2=e4e2e3=3e5N080:e1e1=e2e1e2=e3e2e2=e4e3e3=e5N15:e1e1=e2e1e2=e3e2e3=e4e3e3=e5 $

    Here we will collect all information about $ \mathbf{N}_{04}^{3*}: $

    $ \begin{array}{|l|l|l|l|}    \hline  {\mathbf{N}}^{3*}_{04} &  \begin{array}{l}e_1e_1 = e_2 \\ e_2e_2 = e_3  \end{array} & H2D(N304)=[Δ12],H2C(N304)=H2D(N304)[Δ13],[Δ23],[Δ33]& \phi = (x000x20z0x4)\\ \hline \end{array} $

    Let us use the following notations:

    $ 1=[Δ12],2=[Δ13],3=[Δ23],4=[Δ33]. $

    Take $ \theta = \sum\limits_{i = 1}^{4}\alpha_i\nabla_i\in\mathrm{H}^2_{\mathfrak{C}}(\mathbf{N}^{3*}_{04}) . $ Since

    $ \phi^T(0α1α2α10α3α2α3α4)\phi = (αα1α2α1αα3α2α3α4), $

    we have

    $ α1=(α1x+α3z)x2,α2=(α2x+α4z)x4,α3=α3x5,α4=α4x8. $

    Consider the following cases:

    $ 1.\ $$ \alpha_4\neq0, $ then consider the vector space generated by the following two cocycles:

    $ \theta_1 = \alpha_1\nabla_1+\alpha_2\nabla_2+\alpha_3\nabla_3+\alpha_4\nabla_4 \ \ \rm{and} \ \ \theta_2 = \beta_1\nabla_1+\beta_2\nabla_2+\beta_3\nabla_3. $

    Thus, we have

    $ α1=(α1x+α3z)x2,β1=(β1x+β3z)x2,α2=(α2x+α4z)x4,β2=β2x5,α3=α3x6,β3=β3x6.α4=α4x8. $

    Then we consider the following subcases:

    $ {\rm{(a)}}\ $$ \beta_3 = 0, \alpha_3 = 0, $ then we have:

    ${\rm{(i)\ }} $if $ \beta_1 = 0, \alpha_1 = 0, $ then we have the representative $ \langle \nabla_4,\nabla_2 \rangle; $

    ${\rm{(ii)}}\ $if $ \beta_1 = 0, \alpha_1\neq0, $ then by choosing $ x = \sqrt[5]{ \alpha_1{\alpha_4}^{-1}}, $ we have the representative $ \langle \nabla_1+\nabla_4,\nabla_2 \rangle; $

    ${\rm{(iii)}}\ $if $ \beta_1\neq0, \beta_2 = 0, $ then by choosing $ x = 1 $ and $ z = - \alpha_2 \alpha_4^{-1}, $ we have the representative $ \langle \nabla_4,\nabla_1 \rangle; $

    $ {\rm{(iv)}}\ $ if $ \beta_1\neq0, \beta_2\neq0, $ then by choosing $ x = \sqrt{{\beta_1}{\beta_2}^{-1}} $ and $ z = \frac{\alpha_1\beta_2-\beta_1\alpha_2}{\alpha_4 \sqrt{\beta_1\beta_2}}, $ we have the representative $ \langle \nabla_4,\nabla_1+\nabla_2 \rangle. $

    ${\rm{(b)}}\ \beta_3 = 0, \alpha_3\neq0, $ then we have:

    ${\rm{(i)}}\ $if $ \beta_2 = 0, $ then by choosing $ x = \sqrt{{\alpha_3}{\alpha_4}^{-1}} $ and $ z = -{\alpha_2}\sqrt{{\alpha_3} \alpha_4^{-3}}, $ we have the representative $ \langle \nabla_3+\nabla_4,\nabla_1 \rangle; $

    ${\rm{(ii)}}\ $if $ \beta_2\neq0, \beta_1 = 0, $ then by choosing

    $ x = \sqrt{{\alpha_3}{\alpha_4}^{-1}}\ {\rm{and}}\ z = -{\alpha_1}\sqrt{{\alpha_3}^{-1} \alpha_4^{-1}},$

    we have the representative $ \langle \nabla_3+\nabla_4,\nabla_2 \rangle; $

    $ {\rm{(iii)}}\ $if $ \beta_2\neq0, \beta_1\neq0, \beta_2\alpha_3 = \beta_1\alpha_4, $ then by choosing $ x = \sqrt{{\alpha_3}{\alpha_4}^{-1}} $ and $ z = 0, $ we have the family of representatives $ \langle \alpha\nabla_1+\nabla_3+\nabla_4,\nabla_1+\nabla_2 \rangle; $

    $ {\rm{(iv)}}\ $if $ \beta_2\neq0, \beta_1\neq0, \beta_2\alpha_3\neq\beta_1\alpha_4, $ then by choosing

    $ x = \sqrt{\frac{\beta_1}{\beta_2}} {\text{ and }} z = \frac{(\alpha_1\beta_2-\beta_1\alpha_2)\sqrt{\beta_1}}{(\beta_1\alpha_4-\beta_2\alpha_3)\sqrt{\beta_2}},$

    we have the family of representatives $ \langle \alpha\nabla_3+\nabla_4,\nabla_1+\nabla_2 \rangle_{\alpha\neq0,1}, $ which will be jointed with the case (1(a)iv).

    ${\rm{(c)}}\ \beta_3\neq0, \alpha_3 = 0, $ then we have:

    $ {\rm{(i)}}\ $if $ \beta_2 = 0, \alpha_1 \neq0, $ then by choosing $ x = \sqrt[5]{ \alpha_1 \alpha_4^{-1}} $ and $ z = -\alpha_2 \sqrt[5]{\alpha_1 \alpha_4^{-6}}, $ we have the family of representatives $ \langle \nabla_1+\nabla_4,\alpha\nabla_1+\nabla_3 \rangle; $

    $ {\rm{(ii)}}\ $if $ \beta_2 = 0, \alpha_1 = 0, $ then by choosing $ z = -\frac{\alpha_2x}{\alpha_4}, $ we have two representatives $ \langle \nabla_4,\nabla_3 \rangle $ or $ \langle \nabla_4,\nabla_1+\nabla_3 \rangle $ depending on whether $ \beta_1\alpha_4 = \alpha_2 \beta_3 $ or not;

    $ {\rm{(iii)}}\ $if $ \beta_2\neq0, $ then by choosing $ x = {\frac{\beta_2}{\beta_3}} $ and $ z = -\frac{\alpha_2\beta_2}{\beta_3\alpha_4}, $ we have the family of representatives $ \langle \alpha\nabla_1+\nabla_4,\beta\nabla_1+\nabla_2+\nabla_3 \rangle. $

    2. $ \alpha_4 = 0, \alpha_3\neq0 $, then we may suppose that $ \beta_3 = 0. $ Thus, we have

    $ α1=(α1x+α3z)x2,β1=β1x3,α2=α2x5,β2=β2x5,α3=α3x6, $

    and consider the following subcases:

    $ {\rm{(a)}}\ $$ \beta_2 = 0, $ then we have two representatives $ \langle \nabla_3,\nabla_1 \rangle $ or $ \langle \nabla_2+\nabla_3,\nabla_1 \rangle, $ depending on whether $ \alpha_2 = 0 $ or not;

    $ {\rm{(b)}}\ $$ \beta_2\neq0, \alpha_2 = 0, $ then by choosing $ z = -\frac{\alpha_1x}{\alpha_3}, $ we have two representatives $ \langle \nabla_3,\nabla_2 \rangle $ or $ \langle \nabla_3,\nabla_1+\nabla_2 \rangle, $ depending on whether $ \beta_1 = 0 $ or not.

    3. $ \alpha_4 = 0, \alpha_3 = 0, \beta_3 = 0, \beta_2 = 0, \alpha_2\neq0, $ then we have the representative $ \langle \nabla_2,\nabla_1 \rangle. $

    Summarizing, we have the following distinct orbits:

    $ \langle \nabla_1, \nabla_2 \rangle, $ $ \langle \nabla_1, \nabla_2+\nabla_3 \rangle,$ $ \langle \nabla_1, \nabla_3 \rangle,$ $ \langle \nabla_1, \nabla_3+\nabla_4 \rangle, $ $ \langle \nabla_1, \nabla_4 \rangle,$ $\langle \nabla_1+\nabla_2, \alpha\nabla_1+\nabla_3+\nabla_4 \rangle^{O(\alpha) = O(-\alpha)}, $ $ \langle \nabla_1+\nabla_2,\nabla_3 \rangle, $ $ \langle \nabla_1+\nabla_2, \alpha \nabla_3 + \nabla_4 \rangle_{\alpha\neq 1},$ $ \langle \beta\nabla_1+\nabla_2+\nabla_3, \alpha \nabla_1 + \nabla_4 \rangle, $ $ \langle \alpha\nabla_1+\nabla_3, \nabla_1+\nabla_4 \rangle^{O(\alpha) = O(-\eta_3 \alpha) = O(\eta_3^2 \alpha)}, $ $ \langle \nabla_1+\nabla_3, \nabla_4 \rangle,$ $\langle \nabla_1+\nabla_4, \nabla_2 \rangle,$ $ \langle \nabla_2, \nabla_3 \rangle, $ $ \langle \nabla_2, \nabla_3+\nabla_4 \rangle, $ $ \langle \nabla_2, \nabla_4 \rangle,$ $ \langle \nabla_3, \nabla_4 \rangle.$

    Note that, the orbit $ \langle \nabla_1, \nabla_2 \rangle $ after a change of the basis of the constructed algebra gives a part of the family $ {\mathbf{N}}_{79}^{\alpha}, $ which will be found below. Hence, we have the following new algebras:

    $ N076:e1e1=e2e1e2=e3e1e4=e5e2e2=e4N16:e1e1=e2e1e2=e4e1e3=e5e2e2=e3e2e3=e5N17:e1e1=e2e1e2=e4e2e2=e3e2e3=e5N18:e1e1=e2e1e2=e4e2e2=e3e2e3=e5e3e3=e5N19:e1e1=e2e1e2=e4e2e2=e3e3e3=e5Nα20:e1e1=e2e1e2=e4+αe5e1e3=e4e2e2=e3e2e3=e5e3e3=e5N21:e1e1=e2e1e2=e4e1e3=e4e2e2=e3e2e3=e5Nα122:e1e1=e2e1e2=e4e1e3=e4e2e2=e3e2e3=αe5e3e3=e5Nα,β23:e1e1=e2e1e2=βe4+αe5e1e3=e4e2e2=e3e2e3=e4e3e3=e5Nα24:e1e1=e2e1e2=αe4+e5e2e2=e3e2e3=e4e3e3=e5N25:e1e1=e2e1e3=e4e2e2=e3e2e3=e4e3e3=e5N26:e1e1=e2e1e2=e4e1e3=e5e2e2=e3e3e3=e4N27:e1e1=e2e1e3=e4e2e2=e3e2e3=e5N28:e1e1=e2e1e3=e4e2e2=e3e2e3=e5e3e3=e5N29:e1e1=e2e1e3=e4e2e2=e3e3e3=e5N30:e1e1=e2e2e2=e3e2e3=e4e3e3=e5 $

    Here we will collect all information about $ {\mathbf N}_{02}^{4*}: $

    $ \begin{array}{|l|l|l|l|}   \hline {\mathbf{N}}^{4*}_{02} & \begin{array}{l} e_1e_1 = e_2\\ e_1e_2 = e_3 \end{array} & H2D(N402)=[Δ13],[Δ22],[Δ14],[Δ24],[Δ44]H2C(N402)=H2D(N402)[Δ23],[Δ33],[Δ34] & \phi = (x000qx200w2xqx3re00t)\\ \hline \end{array} $

    Let us use the following notations:

    $ 1=[Δ13],2=[Δ14],3=[Δ22],4=[Δ23],5=[Δ24],6=[Δ33],7=[Δ34],8=[Δ44]. $

    Take $ \theta = \sum\limits_{i = 1}^{8}\alpha_i\nabla_i\in\mathrm{H}^2_{\mathfrak{C}}(\mathbf{N}^{4*}_{02}) . $ Since

    $ \phi^T(00α1α20α3α4α5α1α4α6α7α2α5α7α8)\phi = (ααα1α2αα3α4α5α1α4α6α7α2α5α7α8), $

    we have

    $ α1=(α1x+α4q+α6w+α7e)x3,α2=(α1x+α4q+α6w+α7e)r+(α2x+α5q+α7w+α8e)t,α3=(α3x2+4α4xq+4α6q2)x2,α4=(α4x+2α6q)x4,α5=(α4r+α5t)x2+2(α6r+α7t)xq,α6=α6x6,α7=(α6r+α7t)x3,α8=α6r2+2α7rt+α8t2. $

    We interested in $ (\alpha_2,\alpha_5,\alpha_7,\alpha_8)\neq(0,0,0,0) $ and $ (\alpha_4,\alpha_6,\alpha_7)\neq(0,0,0) . $ Let us consider the following cases:

    $ 1.\ $$ \alpha_6 = 0, \alpha_7 = 0, $ then $ \alpha_4\neq0 $ and we have the following subcases:

    $ {\rm{(a)}}\ $$ \alpha_8 = 0, \alpha_2\alpha_4-\alpha_1\alpha_5 = 0, $ then we have a split extension;

    $ {\rm{(b)}}\ $$ \alpha_8 = 0, \alpha_2\alpha_4-\alpha_1\alpha_5\neq0, \alpha_3 = 4\alpha_1, $ then by choosing

    $ x = \sqrt[4]{\alpha_2\alpha_4-\alpha_1\alpha_5}, t = \alpha_4^2, r = -\alpha_4\alpha_5, q = -\frac{\alpha_1\sqrt[4]{\alpha_2\alpha_4-\alpha_1\alpha_5}}{\alpha_4}, $

    we have the representative $ \langle \nabla_2+\nabla_4 \rangle; $

    $ {\rm{(c)}}\ $$ \alpha_8 = 0, \alpha_2\alpha_4-\alpha_1\alpha_5\neq0, \alpha_3\neq4\alpha_1, $ then by choosing

    $ x = \frac{\alpha_3-4\alpha_1}{\alpha_4}, t = \frac{(\alpha_3-4\alpha_1)^4}{\alpha_4^2(\alpha_2\alpha_4-\alpha_1\alpha_5)}, r = \frac{\alpha_5(\alpha_3-4\alpha_1)^4}{\alpha_4^3(\alpha_1\alpha_5-\alpha_2\alpha_4)}, q = \frac{4\alpha^2_1-\alpha_1\alpha_3}{\alpha^2_4},$

    we have the representative $ \langle \nabla_2+\nabla_3+\nabla_4 \rangle; $

    $ {\rm{(d)}}\ $$ \alpha_8\neq0, \alpha_3 = 4\alpha_1, $ then by choosing

    $ x = \alpha_4\alpha_8, t = \alpha^3_4\alpha^2_8, q = -\alpha_1\alpha_8, r = -\alpha_4^2\alpha_5\alpha_8^2, e = \alpha_1\alpha_5-\alpha_2\alpha_4, $

    we have the representative $ \langle \nabla_4+\nabla_8 \rangle; $

    $ {\rm{(e)}}\ $$ \alpha_8\neq0, \alpha_3\neq4\alpha_1, $ then by choosing

    $x=α34α1α4,t=(α34α1)5α24α8,q=4α21α1α3α24,r=α5(α34α1)5α34α8,e=(4α1α3)(α2α4α1α5)α24α8,$

    we have the representative $ \langle \nabla_3+\nabla_4+\nabla_8 \rangle. $

    2. $ \alpha_6 = 0, \alpha_7\neq0, $ then we have the following subcases:

    $ {\rm{(a)}}\ $$ \alpha_4 = 0, \alpha_3 = 0, $ then by choosing

    $x=2α27,q=α5α7,e=2α1α7,w=α25+2α1α82α2α7,t=2α7,r=α8,$

    we have the representative $ \langle \nabla_7 \rangle; $

    ${\rm{(b)}}\ $$ \alpha_4 = 0, \alpha_3\neq0, $ then by choosing

    $ x = 1,$ $q = -\frac{\alpha_5}{2\alpha_7},$ $ e = - \frac{\alpha_1}{\alpha_7},$ $w = \frac{\alpha_5^2+2\alpha_1\alpha_8-2\alpha_2\alpha_7}{2 \alpha_7^2},$ $t = \frac{\alpha_3}{\alpha_7},$ $ r = -\frac{\alpha_3\alpha_8}{2\alpha_7^2}, $

    we have the representative $ \langle \nabla_3+\nabla_7 \rangle; $

    $ {\rm{(c)}}\ $$ \alpha_4\neq0, \alpha_3\alpha_7^2-2\alpha_4\alpha_5\alpha_7+\alpha_4^2\alpha_8 = 0, $ then by choosing

    $ x=α7,t=α4,e=α34α14α7,r=α4α82α7,q=α3α74α4,w=4α1α4α84α2α4α7+α3(α5α7α4α8)4α4α37,$

    we have the representative $ \langle \nabla_4+\nabla_7 \rangle; $

    ${\rm{(d)}}\ $$ \alpha_4\neq0, \alpha_3\alpha_7^2-2\alpha_4\alpha_5\alpha_7+\alpha_4^2\alpha_8\neq0, $ then by choosing

    $x=α32α4+α5α7α4α82α27,q=α3(α3α272α4α5α7+α24α8)8α24α27,w=(α3α272α4α5α7+α24α8)(4α2α4α74α1α4α8+α3(α5α7+α4α8))8α24α47,e=(4α1α3)(α3α272α4α5α7+α24α8)8α4α37,t=(α3α272α4α5α7+α24α8)24α4α57,r=α8(α3α272α4α5α7+α24α8)28α4α67,$

    we have the representative $ \langle \nabla_4+ \nabla_5+\nabla_7 \rangle. $

    3. $ \alpha_6\neq0, $ then we have the following subcases:

    $ {\rm{(a)}}\ $$ \alpha_6\alpha_8-\alpha_7^2 = 0, \alpha_5\alpha_6-\alpha_4\alpha_7 = 0, \alpha_2\alpha_6-\alpha_1\alpha_7 = 0, $ then we have a split extension;

    ${\rm{(b)}}\ $$ \alpha_6\alpha_8-\alpha_7^2 = 0, \alpha_5\alpha_6-\alpha_4\alpha_7 = 0, \alpha_2\alpha_6-\alpha_1\alpha_7\neq0, \alpha_3\alpha_6-\alpha^2_4 = 0, $ then by choosing

    $x = 1, t = \frac{\alpha_6^2}{\alpha_2\alpha_6-\alpha_1\alpha_7}, q = -\frac{\alpha_4}{2\alpha_6}, r = \frac{\alpha_6\alpha_7}{\alpha_1\alpha_7-\alpha_2\alpha_6}, e = 0, w = \frac{\alpha^2_4-2\alpha_1\alpha_6}{\alpha_6},$

    we have the representative $ \langle \nabla_2+\nabla_6 \rangle; $

    $ {\rm{(c)}}\ $$ \alpha_6\alpha_8-\alpha_7^2 = 0, \alpha_5\alpha_6-\alpha_4\alpha_7 = 0, \alpha_2\alpha_6-\alpha_1\alpha_7\neq0, \alpha_3\alpha_6-\alpha^2_4\neq0, $ then by choosing

    $x=α3α6α24α26,t=(α3α6α24)5α36(α2α6α1α7),q=α4α3α6α242α26,r=(α3α6α24)5α7α46(α2α6α1α7),e=0,$

    and $ w = \frac{(\alpha_4^2-2 \alpha_1 \alpha_6) \sqrt{\alpha_3 \alpha_6-\alpha_4^2}}{2 \alpha_6^3}, $ we have the representative $ \langle \nabla_2+\nabla_3+\nabla_6 \rangle; $

    ${\rm{(d)}}\ $$ \alpha_6\alpha_8-\alpha_7^2 = 0, \alpha_5\alpha_6-\alpha_4\alpha_7\neq0, 2\alpha_6(\alpha_2\alpha_6-\alpha_1\alpha_7) = \alpha_4(\alpha_5\alpha_6-\alpha_4\alpha_7), $ then by choosing

    $ t = \frac{\alpha_6^2}{\alpha_5\alpha_6-\alpha_4\alpha_7}x^4,$ $q = -\frac{\alpha_4}{2\alpha_6}x,$ $r = \frac{\alpha_6\alpha_7}{\alpha_4\alpha_7-\alpha_5\alpha_6}x^4,$ $e = 0,$ $w = \frac{\alpha^2_4-2\alpha_1\alpha_6}{\alpha_6}x,$

    we have the representatives $ \langle \nabla_5+\nabla_6 \rangle $ and $ \langle \nabla_3+\nabla_5+\nabla_6 \rangle $ depending on whether $ \alpha_3\alpha_6 = \alpha_4^2 $ or not;

    $ {\rm{(e)}}\ $$ \alpha_6\alpha_8-\alpha_7^2 = 0, \alpha_5\alpha_6-\alpha_4\alpha_7\neq0, 2\alpha_6(\alpha_2\alpha_6-\alpha_1\alpha_7)\neq\alpha_4(\alpha_5\alpha_6-\alpha_4\alpha_7), $ then by choosing $ x = \frac{2\alpha_6(\alpha_2\alpha_6-\alpha_1\alpha_7)-\alpha_4(\alpha_5\alpha_6-\alpha_4\alpha_7)}{2\alpha_6^2(\alpha_5\alpha_6-\alpha_4\alpha_7)}, $ $ t = \frac{\alpha_6^2}{\alpha_5\alpha_6-\alpha_4\alpha_7}x^4, $ $ q = -\frac{\alpha_4}{2\alpha_6}x, $ $ r = \frac{\alpha_6\alpha_7}{\alpha_4\alpha_7-\alpha_5\alpha_6}x^4, $ $ e = 0, $ $ w = \frac{\alpha^2_4-2\alpha_1\alpha_6}{\alpha_6}x, $ we have the representative $ \langle \nabla_2+\alpha\nabla_3+\nabla_5+\nabla_6 \rangle; $

    $ {\rm{(f)}}\ $$ \alpha_6\alpha_8-\alpha_7^2\neq0, \alpha_5\alpha_6-\alpha_4\alpha_7 = 0, $ then by choosing

    $t=α6x3α6α8α27,q=α4x2α6,r=α7x3α6α8α27,e=(α1α7α2α6)xα6α8α27,w=(α242α26+α1α8α2α7α27α6α8)x,$

    we have the representatives $ \langle \nabla_6+\nabla_8 \rangle $ and $ \langle \nabla_3+\nabla_6+\nabla_8 \rangle $ depending on whether $ \alpha_3\alpha_6 = \alpha_4^2 $ or not.

    $ {\rm{(g)}}\ $$ \alpha_6\alpha_8-\alpha_7^2\neq0, \alpha_5\alpha_6-\alpha_4\alpha_7\neq0, $ then by choosing

    $x=α5α6α4α7α26(α6α8α27),t=(α5α6α4α7)3α26(α27α6α8)2,q=α4(α4α7α5α6)2α6α26(α6α8α27),r=α7(α4α7α5α6)3α36(α27α6α8)2,e=α6(α5α6α4α7)(α4α5α6α24α7+2α6(α2α6+α1α7))2α36(α6α8α27)3,w=α6(α5α6α4α7)(α24α8α4α5α7+2α6(α2α7α1α8))2α36(α6α8α27)3,$

    we have the representative $ \langle \alpha\nabla_3+\nabla_5+\nabla_6+\nabla_8 \rangle. $

    Summarizing, we have the following distinct orbits

    $2+3+4,2+α3+5+6,2+3+6,2+4,2+6,3+4+8,3+5+6,α3+5+6+8,3+6+8,3+7,4+5+7,4+7,4+8,5+6,6+8,7,$

    which gives the following new algebras:

    $ N31:e1e1=e2e1e2=e3e1e4=e5e2e2=e5e2e3=e5Nα32:e1e1=e2e1e2=e3e1e4=e5e2e2=αe5e2e4=e5e3e3=e5N33:e1e1=e2e1e2=e3e1e4=e5e2e2=e5e3e3=e5N34:e1e1=e2e1e2=e3e1e4=e5e2e3=e5N35:e1e1=e2e1e2=e3e1e4=e5e3e3=e5N36:e1e1=e2e1e2=e3e2e2=e5e2e3=e5e4e4=e5N37:e1e1=e2e1e2=e3e2e2=e5e2e4=e5e3e3=e5Nα38:e1e1=e2e1e2=e3e2e2=αe5e2e4=e5e3e3=e5e4e4=e5N39:e1e1=e2e1e2=e3e2e2=e5e3e3=e5e4e4=e5N40:e1e1=e2e1e2=e3e2e2=e5e3e4=e5N41:e1e1=e2e1e2=e3e2e3=e5e2e4=e5e3e4=e5N42:e1e1=e2e1e2=e3e2e3=e5e3e4=e5N43:e1e1=e2e1e2=e3e2e3=e5e4e4=e5N44:e1e1=e2e1e2=e3e2e4=e5e3e3=e5N45:e1e1=e2e1e2=e3e3e3=e5e4e4=e5N46:e1e1=e2e1e2=e3e3e4=e5 $

    Here we will collect all information about $ {\mathbf N}_{04}^{4*}: $

    $ \begin{array}{|l|l|l|l|}   \hline {\mathbf{N}}^{4*}_{04} & \begin{array}{l} e_1e_1 = e_2 \\ e_2e_2 = e_3 \end{array} & H2D(N404)=[Δ12],[Δ14],[Δ24],[Δ44],H2C(N404)=H2D(N404)[Δ13],[Δ23],[Δ33],[Δ34] & \phi = (x0000x200y0x4rz00t)\\ \hline \end{array} $

    Let us use the following notations:

    $ 1=[Δ12],2=[Δ13],3=[Δ14],4=[Δ23],5=[Δ24],6=[Δ33],7=[Δ34],8=[Δ44]. $

    Take $ \theta = \sum\limits_{i = 1}^{8}\alpha_i\nabla_i\in\mathrm{H}^2_{\mathfrak{C}}(\mathbf{N}^{4*}_{04}) . $ Since

    $ \phi^T(0α1α2α3α10α4α5α2α4α6α7α3α5α7α8)\phi = (αα1α2α3α1αα4α5α2α4α6α7α3α5α7α8), $

    we have

    $ α1=(α1x+α4y+α5z)x2,α2=(α2x+α6y+α7z)x4,α3=(α2x+α6y+α7z)r+(α3x+α7y+α8z)t,α4=α4x6,α5=(α4r+α5t)x2,α6=α6x8,α7=(α6r+α7t)x4,α8=α6r2+2α7rt+α8t2. $

    We interested in $ (\alpha_3,\alpha_5,\alpha_7,\alpha_8)\neq(0,0,0,0) $ and $ (\alpha_2,\alpha_4,\alpha_6,\alpha_7)\neq(0,0,0,0) . $ Let us consider the following cases:

    $1.\ $$ \alpha_6 = 0, \alpha_7 = 0, \alpha_4 = 0, $ then $ \alpha_2\neq0 $ and we have the following cases:

    $ {\rm{(a)}}\ $if $ \alpha_8 = 0, \alpha_5 = 0, $ then by choosing $ t = 1 $ and $ r = -\frac{\alpha_3}{\alpha_2}, $ we have a split extension;

    $ {\rm{(b)}}\ $if $ \alpha_8 = 0, \alpha_5\neq0, $ then by choosing

    $ x = \alpha_2\alpha_5,$ $t = \alpha_2^4\alpha_5^2,$ $z = -\alpha_1\alpha_2$, $r = -\alpha_2^3\alpha_3\alpha_5^2,$ $y = 0,$

    we have the representative $ \langle \nabla_2+\nabla_5 \rangle; $

    $ {\rm{(c)}}\ $if $ \alpha_8\neq0, \alpha_5 = 0, \alpha_1 = 0, $ then by choosing

    $x = \alpha_3,$ $t = {\sqrt{\alpha_2}}\alpha_8^2,$ $z = -{\alpha_3},$ $r = 0,$ $y = 0,$

    we have the representative $ \langle \nabla_2+\nabla_8 \rangle; $

    ${\rm{(d)}}\ $if $ \alpha_8\neq0, \alpha_5 = 0, \alpha_1\neq0, $ then by choosing

    $ x = \sqrt{{\alpha_1}{\alpha_2}^{-1}},$ $t = \sqrt[4]{\alpha_1^5\alpha_2^{-3}}\sqrt{\alpha_8^{-1}},$ $z = -\sqrt{\alpha_1\alpha_2^{-1}} \alpha_3\alpha_8^{-1},$ $r = 0,$ $y = 0,$

    we have the representative $ \langle \nabla_1+\nabla_2+\nabla_8 \rangle; $

    $ {\rm{(e)}}\ $if $ \alpha_8\neq0, \alpha_5\neq0, $ then by choosing

    $ x = {\frac{\alpha^2_5}{\alpha_2\alpha_8}}, t = {\frac{\alpha_5^5}{\alpha_2^2\alpha^3_8}}, z = -{\frac{\alpha_1\alpha_5}{\alpha_2\alpha_8}}, r = \frac{\alpha_5^4(\alpha_1\alpha_8-\alpha_3\alpha_5)}{\alpha_2^3\alpha_8^3}, y = 0,$

    we have the representative $ \langle \nabla_2+\nabla_5+\nabla_8 \rangle. $

    2. $ \alpha_6 = 0, \alpha_7 = 0, \alpha_4\neq0, $ then by choosing $ r = -\frac{\alpha_5}{\alpha_4}t, y = -\frac{\alpha_1x+\alpha_5z}{\alpha_4}, $ we have $ \alpha^*_1 = \alpha^*_5 = 0 . $ Now we can suppose that $ \alpha_1 = 0, \alpha_5 = 0, $ and we have the following subcases:

    ${\rm{(a)}}\ $if $ \alpha_8 = 0, \alpha_3 = 0, $ then we have a split extension;

    $ {\rm{(b)}}\ $if $ \alpha_8 = 0, \alpha_3\neq0, \alpha_2 = 0, $ then by choosing $ x = \alpha_3, $ $ y = 0, $ $ z = 0, $ $ r = 0, $ $ t = \alpha_3^4, $ we have the representative $ \langle \nabla_3+\nabla_4 \rangle; $

    ${\rm{(c)}}\ $if $ \alpha_8 = 0, \alpha_3\neq0, \alpha_2\neq0, $ then by choosing $ x = \frac{\alpha_2}{\alpha_4}, $ $ y = 0, $ $ z = 0, $ $ r = 0, t = \frac{\alpha_2^5}{\alpha_3\alpha_4^4}, $ we have the representative $ \langle \nabla_2+\nabla_3+\nabla_4 \rangle; $

    ${\rm{(d)}}\ $if $ \alpha_8\neq0, \alpha_2 = 0, $ then by choosing $ x = 1, y = 0, z = -\frac{\alpha_3}{\alpha_8}, r = 0, t = \sqrt{\frac{\alpha_4}{\alpha_8}}, $ we have the representative $ \langle \nabla_4+\nabla_8 \rangle; $

    $ {\rm{(e)}}\ $if $ \alpha_8\neq0, \alpha_2\neq0, $ then by choosing $ x = \frac{\alpha_3}{\alpha_4}, y = 0, z = -\frac{\alpha_2\alpha_3}{\alpha_4\alpha_8}, r = 0, t = \frac{\alpha_2^3}{\sqrt{\alpha_4^5\alpha_8}}, $ we have the representative $ \langle \nabla_2+\nabla_4+\nabla_8 \rangle. $

    3. $ \alpha_6 = 0, \alpha_7\neq0, $ then by choosing $ r = -\frac{\alpha_8t}{2\alpha_7}, y = -\frac{(\alpha_2\alpha_8-\alpha_3\alpha_7)x}{\alpha_7^2}, z = -\frac{\alpha_2x}{\alpha_7}, $ we have $ \alpha^*_2 = \alpha_3^* = \alpha^*_8 = 0. $ Now we can suppose that $ \alpha_2 = 0, $ $ \alpha_3 = 0, $ $ \alpha_8 = 0, $ and consider the following cases:

    ${\rm{(a)}}\ $if $ \alpha_4 = 0, \alpha_5 = 0, \alpha_1 = 0, $ then we have the representative $ \langle \nabla_7 \rangle; $

    ${\rm{(b)}}\ $if $ \alpha_4 = 0, \alpha_5 = 0, \alpha_1\neq0, $ then by choosing $ x = \frac{1}{\alpha_7}, t = \alpha_1, y = 0, z = 0, r = 0, $ we have the representative $ \langle \nabla_1+\nabla_7 \rangle; $

    ${\rm{(c)}}\ $if $ \alpha_4 = 0, \alpha_5\neq0, \alpha_1 = 0, $ then by choosing $ x = \sqrt{\frac{\alpha_5}{\alpha_7}}, t = 1, y = 0, z = 0, r = 0, $ we have the representative $ \langle \nabla_5+\nabla_7 \rangle; $

    $ {\rm{(d)}}\ $if $ \alpha_4 = 0, \alpha_5\neq0, \alpha_1\neq0, $ then by choosing $ x = \sqrt{\frac{\alpha_5}{\alpha_7}}, t = \sqrt{\frac{\alpha_1^2}{\alpha_5\alpha_7}}, y = 0, z = 0, r = 0, $ we have the representative $ \langle \nabla_1+\nabla_5+\nabla_7 \rangle; $

    $ {\rm{(e)}}\ $if $ \alpha_4\neq0, \alpha_5 = 0, \alpha_1 = 0, $ then by choosing $ x = \sqrt{\alpha_7}, t = \alpha_4, y = 0, z = 0, r = 0, $ we have the representative $ \langle \nabla_4+\nabla_7 \rangle; $

    ${\rm{(f)}}\ $if $ \alpha_4\neq0, \alpha_5 = 0, \alpha_1\neq0, $ then by choosing $ x = \sqrt[3]{\frac{\alpha_1}{\alpha_4}}, t = \sqrt[3]{\frac{\alpha_1\alpha_4^2}{\alpha^3_7}}, y = 0, z = 0, r = 0, $ we have the representative $ \langle \nabla_1+\nabla_4+\nabla_7 \rangle; $

    ${\rm{(g)}}\ $if $ \alpha_4\neq0, \alpha_5\neq0, $ then by choosing $ x = \sqrt{\frac{\alpha_5}{\alpha_7}}, t = \sqrt{\frac{\alpha_4\alpha_5}{\alpha^2_7}}, y = 0, z = 0, r = 0, $ we have the representative $ \langle \alpha\nabla_1+\nabla_4+\nabla_5+\nabla_7 \rangle. $

    4. $ \alpha_6\neq0, $ then by choosing $ r = -\frac{\alpha_7t}{\alpha_6}, y = -\frac{\alpha_2x+\alpha_7z}{\alpha_6}, $ we have $ \alpha^*_2 = \alpha_7^* = 0 . $ Now we can suppose that $ \alpha_2 = 0, \alpha_7 = 0, $ and we have:

    $ {\rm{(a)}}\ $if $ \alpha_8 = 0, \alpha_5 = 0, $ then $ \alpha_3\neq0 $ and we have the following subcases:

    $ {\rm{(i)}}\ $$ \alpha_4 = 0, \alpha_1 = 0, $ then by choosing $ x = \alpha_3, t = \alpha_3^6\alpha_6, y = 0, z = 0, r = 0, $ we have the representative $ \langle \nabla_3+\nabla_6 \rangle; $

    ${\rm{(ii)}}\ $$ \alpha_4 = 0, \alpha_1\neq0, $ then by choosing $ x = \sqrt[5]{\frac{\alpha_1}{\alpha_6}}, t = \sqrt[5]{\frac{\alpha_1^7}{\alpha_3^5\alpha^2_6}}, y = 0, z = 0, r = 0, $ we have the representative $ \langle \nabla_1+\nabla_3+\nabla_6 \rangle; $

    ${\rm{(iii)}}\ $$ \alpha_4\neq0, $ then by choosing $ x = \sqrt{\frac{\alpha_4}{\alpha_6}}, t = \sqrt[5]{\frac{\alpha_4^7}{\alpha_3^2\alpha^5_6}}, y = 0, z = 0, r = 0, $ we have the representative $ \langle \alpha\nabla_1+\nabla_3+\nabla_4+\nabla_6 \rangle. $

    (b) $ \alpha_8 = 0, \alpha_5\neq0, $ then we have the following subcases:

    ${\rm{(i)}} \ $$ \alpha_4 = 0, \alpha_3 = 0, $ then by choosing $ x = \sqrt[6]{\frac{\alpha_5}{\alpha_6}}, t = 1, z = -\frac{\alpha_1}{\sqrt[6]{\alpha_5^5\alpha_6}},y = 0, r = 0, $ we have the representative $ \langle \nabla_5+\nabla_6 \rangle; $

    ${\rm{(ii)}}\ $$ \alpha_4 = 0, \alpha_3\neq0, $ then by choosing $ x = \frac{\alpha_3}{\alpha_5}, t = \alpha_3^6\alpha_5^{-7}\alpha_6, z = -\frac{\alpha_1\alpha_3}{\alpha_5^2}, y = 0, r = 0, $ we have the representative $ \langle \nabla_3+\nabla_5+\nabla_6 \rangle; $

    ${\rm{(iii)}}\ $$ \alpha_4\neq0, $ then by choosing $ x = \sqrt{\frac{\alpha_4}{\alpha_6}}, t = \frac{\alpha_4^3}{\alpha_5\alpha^2_6}, z = -\alpha_1\alpha_5^{-1}\sqrt{\alpha_4\alpha_6^{-1}}, y = 0, r = 0, $ we have the representative $ \langle \alpha\nabla_3+\nabla_4+\nabla_5+\nabla_6 \rangle. $

    (c) $ \alpha_8\neq0, $ then we have the following subcases:

    ${\rm{(i)}}\ $$ \alpha_5 = 0, \alpha_4 = 0, \alpha_1 = 0, $ then by choosing $ x = 1, t = \sqrt{\frac{\alpha_6}{\alpha_8}}, z = -\frac{\alpha_3}{\alpha_8},y = 0, r = 0, $ we have the representative $ \langle \nabla_6+\nabla_8 \rangle; $

    ${\rm{(ii)}}\ $$ \alpha_5 = 0, \alpha_4 = 0, \alpha_1\neq0, $ then by choosing

    $x=5α1α16,t=10α81α36α58,z=α3α185α1α16,y=0,r=0,$

    we have the representative $ \langle \nabla_1+\nabla_6+\nabla_8 \rangle; $

    ${\rm{(iii)}}\ $$ \alpha_5 = 0, \alpha_4\neq0, $ then by choosing

    $x = \sqrt{\alpha_4 \alpha_6^{-1}},$ $t = \alpha_4^2 \sqrt{ \alpha_6^{-3} \alpha_8^{-1}},$ $z = -\alpha_3{\alpha_8^{-1}}\sqrt{\alpha_4 \alpha_6^{-1}},$ $y = 0,$ $r = 0,$

    we have the representative $ \langle \alpha\nabla_1+\nabla_4+\nabla_6+\nabla_8 \rangle; $

    $ {\rm{(iv)}}\ $$ \alpha_5\neq0, $ then by choosing $ x = \sqrt[4]{\frac{\alpha^2_5}{\alpha_6\alpha_8}}, $ $ t = \frac{\alpha_5^2}{\sqrt{\alpha_6\alpha_8^3}}, $ $ z = -\frac{\alpha_3\sqrt{\alpha_5}}{\sqrt[4]{\alpha_6\alpha_8^5}}, $ $ y = 0, $ $ r = 0, $ we have the representative $ \langle \alpha\nabla_1+\beta\nabla_4+\nabla_5+\nabla_6+\nabla_8 \rangle. $

    Summarizing, we have the following distinct orbits:

    $1+2+8,1+3+6,α1+3+4+6O(α)=O(α),α1+β4+5+6+8O(α,β)=O(α,β)=O(±iα,β),α1+4+6+8O(α)=O(α),1+4+7,α1+4+5+7O(α)=O(α),1+5+7,1+6+8,1+7,2+3+4,2+4+8,2+5,2+5+8,2+8,3+4,α3+4+5+6O(α)=O(α),3+5+6,3+6,4+7,4+8,5+6,5+7,6+8,7.$

    Hence, we have the following new algebras:

    $ N47:e1e1=e2e1e2=e5e1e3=e5e2e2=e3e4e4=e5N48:e1e1=e2e1e2=e5e1e4=e5e2e2=e3e3e3=e5Nα49:e1e1=e2e1e2=αe5e1e4=e5e2e2=e3e2e3=e5e3e3=e5Nα,β50:e1e1=e2e1e2=αe5e2e2=e3e2e3=βe5e2e4=e5e3e3=e5e4e4=e5Nα51:e1e1=e2e1e2=αe5e2e2=e3e2e3=e5e3e3=e5e4e4=e5N52:e1e1=e2e1e2=e5e2e2=e3e2e3=e5e3e4=e5Nα53:e1e1=e2e1e2=αe5e2e2=e3e2e3=e5e2e4=e5e3e4=e5N54:e1e1=e2e1e2=e5e2e2=e3e2e4=e5e3e4=e5N55:e1e1=e2e1e2=e5e2e2=e3e3e3=e5e4e4=e5N56:e1e1=e2e1e2=e5e2e2=e3e3e4=e5N57:e1e1=e2e1e3=e5e1e4=e5e2e2=e3e2e3=e5N58:e1e1=e2e1e3=e5e2e2=e3e2e3=e5e4e4=e5N59:e1e1=e2e1e3=e5e2e2=e3e2e4=e5N60:e1e1=e2e1e3=e5e2e2=e3e2e4=e5e4e4=e5N61:e1e1=e2e1e3=e5e2e2=e3e4e4=e5N62:e1e1=e2e1e4=e5e2e2=e3e2e3=e5Nα63:e1e1=e2e1e4=αe5e2e2=e3e2e3=e5e2e4=e5e3e3=e5N64:e1e1=e2e1e4=e5e2e2=e3e2e4=e5e3e3=e5N65:e1e1=e2e1e4=e5e2e2=e3e3e3=e5N66:e1e1=e2e2e2=e3e2e3=e5e3e4=e5N67:e1e1=e2e2e2=e3e2e3=e5e4e4=e5N68:e1e1=e2e2e2=e3e2e4=e5e3e3=e5N69:e1e1=e2e2e2=e3e2e4=e5e3e4=e5N70:e1e1=e2e2e2=e3e3e3=e5e4e4=e5N71:e1e1=e2e2e2=e3e3e4=e5 $

    Here we will collect all information about $ {\mathbf N}_{08}^{4*}: $

    $ \begin{array}{|l|l|l|l|}   \hline {\mathbf{N}}^{4*}_{08} & \begin{array}{l} e_1e_1 = e_2 \\ e_1e_2 = e_3 \\ e_2e_2 = e_4 \end{array} & H2D(N408)=[Δ13],[Δ14]+3[Δ23]H2C(N408)=H2D(N408)[Δ14],[Δ24],[Δ33],[Δ34],[Δ44] & \phi = (x000yx200z2xyx30ty2x2yx4)\\ \hline \end{array} $

    Let us use the following notations:

    $ 1=[Δ13],2=[Δ14]+3[Δ23],3=[Δ14],4=[Δ24],5=[Δ33],6=[Δ34],7=[Δ44]. $

    Take $ \theta = \sum\limits_{i = 1}^{7}\alpha_i\nabla_i\in\mathrm{H}^2_{\mathfrak{C}}(\mathbf{N}^{4*}_{08}) . $ Since

    $ \phi^T(00α1α2+α3003α2α4α13α2α5α6α2+α3α4α6α7)\phi = (ααα1α2+α3αα3α2α4α13α2α5α6α2+α3α4α6α7), $

    we have

    $ α1=(α1x+3α2y+α5z+α6t)x3+((α2+α3)x+α4y+α6z+α7t)x2y,α2=13(3α2x3+(α4+2α5)x2y+3α6xy2+α7y3)x2,α3=((α2+α3)x+α4y+α6z+α7t)x413(3α2x3+(α4+2α5)x2y+3α6xy2+α7y3)x2,α4=(α4x2+2α6xy+α7y2)x4,α5=(α5x2+2α6xy+α7y2)x4,α6=(α6x+α7y)x6,α7=α7x8. $

    We are interested in $ (\alpha_3,\alpha_4, \alpha_5, \alpha_6,\alpha_7)\neq(0,0,0,0,0), $ $ (\alpha_2+\alpha_3,\alpha_4, \alpha_6, \alpha_7)\neq(0,0,0,0) $ and $ (\alpha_1,\alpha_2, \alpha_5, \alpha_6)\neq(0,0,0,0). $ Let us consider the following cases:

    $ 1.\ $$ \alpha_7 = 0, \alpha_6 = 0, \alpha_5 = 0, \alpha_4 = 0, $ then $ \alpha_3\neq0, $ $ \alpha_2+\alpha_3 \neq 0 $ and $ (\alpha_1,\alpha_2)\neq(0,0). $

    ${\rm{(a)}}\ $ if $ \alpha_2\neq-\frac{\alpha_3}{4}, $ then by choosing $ x = 4\alpha_2+\alpha_3, y = -\alpha_1, $ we have the representative $ \langle \alpha\nabla_2+\nabla_3 \rangle_{\alpha\neq0,-\frac{1}{4},-1}; $

    ${\rm{(b)}}\ $if $ \alpha_2 = -\frac{\alpha_3}{4}, $ then we have the representatives

    $ \langle -\frac{1}{4}\nabla_2+\nabla_3 \rangle {\text{ and }} \langle \nabla_1-\frac{1}{4}\nabla_2+\nabla_3 \rangle $

    depending on $ \alpha_1 = 0 $ or not.

    2. $ \alpha_7 = 0, \alpha_6 = 0, \alpha_5 = 0, \alpha_4\neq0, $ then by choosing $ y = -\frac {3\alpha_2}{\alpha_4}x, $ we have $ \alpha_2^* = 0. $ This we can suppose $ \alpha_2 = 0, $ which implies $ \alpha_1\neq 0 $ and choosing $ x = \sqrt {\alpha_1 \alpha_4^{-1}}, $ we have the representative $ \langle \nabla_1+\alpha\nabla_3+\nabla_4 \rangle. $

    3. $ \alpha_7 = 0, \alpha_6 = 0, \alpha_5\neq0. $

    ${\rm{(a)}}\ $ if $ \alpha_4 = 0, $ then $ \alpha_2\neq-\alpha_3 $ and choosing

    $ x = \frac{\alpha_2+\alpha_3}{\alpha_5},$ $y = \frac{3\alpha_2\alpha_3+3\alpha_3^2}{2\alpha_5^2},$ $z = -\frac{(\alpha_2+\alpha_3)(2\alpha_1\alpha_5+12\alpha_2\alpha_3+3\alpha_3^2)}{4\alpha_5^3},$

    we have the representative $ \langle \nabla_2+\nabla_5 \rangle; $

    ${\rm{(b)}}\ $if $ \alpha_4\neq0, \alpha_4\neq\alpha_5, 2(\alpha_2\alpha_5-\alpha_2\alpha_4+\alpha_3\alpha_5)+\alpha_3\alpha_4 = 0, $ then by choosing

    $ x = 2(\alpha_4-\alpha_5), y = 3\alpha_3, z = 0, t = 0, $

    we have the representative $ \langle \alpha\nabla_4+\nabla_5 \rangle_{\alpha\neq0,1}; $

    ${\rm{(c)}}\ $ if $ \alpha_4\neq0, \alpha_4\neq\alpha_5, 2(\alpha_2\alpha_5-\alpha_2\alpha_4+\alpha_3\alpha_5)+\alpha_3\alpha_4\neq0, $ then by choosing

    $ x = \frac{2(\alpha_2\alpha_5-\alpha_2\alpha_4+\alpha_3\alpha_5)+\alpha_3\alpha_4}{2(\alpha_5^2-\alpha_4\alpha_5)},y = \frac{3\alpha_3(2(\alpha_2\alpha_5-\alpha_2\alpha_4+\alpha_3\alpha_5)+\alpha_3\alpha_4)}{2\alpha_5(\alpha_5-\alpha_4)^2}, z = \\ -\frac{(2 \alpha_2 (\alpha_4-\alpha_5)-\alpha_3 (\alpha_4+2 \alpha_5)) (4 \alpha_1 (\alpha_4-\alpha_5)^2-24 \alpha_2 \alpha_3 (\alpha_4-\alpha_5)+3 \alpha_3^2 (\alpha_4+2 \alpha_5))}{8 (\alpha_4-\alpha_5)^3 \alpha_5^2}, t = 0, $

    we have the family of representatives $ \langle \nabla_2+\alpha\nabla_4+\nabla_5 \rangle_{\alpha\neq0,1} $;

    ${\rm{(d)}}\ $if $ \alpha_4\neq0, \alpha_4 = \alpha_5, $ then by choosing $ y = -\frac{\alpha_2x}{\alpha_5} $ and $ z = \frac{(\alpha_3\alpha_5-\alpha_1\alpha_5+3\alpha_2^2)x}{\alpha_5^2}, $ we have the representatives $ \langle \nabla_4+\nabla_5 \rangle $ and $ \langle \nabla_3+\nabla_4+\nabla_5 \rangle $ depending on whether $ \alpha_3 = 0 $ or not. Note that $ \langle \nabla_4+\nabla_5 \rangle = \langle \nabla_2+\nabla_4+\nabla_5 \rangle $ and it will be jointed with the family from the case (3c).

    4. if $ \alpha_7 = 0, \alpha_6\neq0, $ then by choosing $ x = 1, $ $ y = \frac{ \sqrt{( \alpha_4 + 2 \alpha_5)^2 - 36 \alpha_2 \alpha_6}- \alpha_4 - 2 \alpha_5 }{6 \alpha_6}, $

    $z = y^2 - \frac{ \alpha_3}{\alpha_6} + \frac{2 y ( \alpha_5-\alpha_4)}{3 \alpha_6}{\text{ and }}t = -\frac{x^2 \alpha_1 + x y (4 \alpha_2 + \alpha_3) + x z \alpha_5 + y (y \alpha_4 + z \alpha_6)}{ \alpha_6)}, $

    we have $ \alpha_1^* = \alpha_2^* = \alpha_3^* = 0. $ Now we can suppose that $ \alpha_1 = 0, \alpha_2 = 0, \alpha_3 = 0, $ and we have the following cases:

    $ {\rm{(a)}}\ $if $ \alpha_4 = 0, \alpha_5 = 0, $ then by choosing $ x = 1, y = 0, z = 0, t = 0, $ we have the representative $ \langle \nabla_6 \rangle; $

    ${\rm{(b)}}\ $ if $ \alpha_4 = 0, \alpha_5\neq0, $ then by choosing $ x = -\frac{4 \alpha_5}{3 \alpha_6}, y = \frac{8 \alpha_5^2}{9 \alpha_6^2}, z = 0, t = 0, $ we have the representative $ \langle \nabla_4+\frac{1}{4}\nabla_5+\nabla_6 \rangle; $

    ${\rm{(c)}}\ $if $ \alpha_4\neq0, $ then by choosing $ x = \frac{\alpha_4}{\alpha_6}, y = 0, z = 0, t = 0, $ we have the family of representatives $ \langle \nabla_4+\alpha\nabla_5+\nabla_6 \rangle, $ which will be jointed with the representative from the case (4b).

    5. if $ \alpha_7\neq0, $ then by choosing $ x = 1, $ $ y = -\frac{\alpha_6}{\alpha_7}, $ $ t = \frac{ 2 \alpha_6^3+2 (\alpha_4-\alpha_5) \alpha_6 \alpha_7-3 \alpha_3 \alpha_7^2 }{3\alpha_7^3} $ and $ z = 0, $ we have $ \alpha_3^* = 0, \alpha_6^* = 0. $ Now we can suppose that $ \alpha_3 = 0, \alpha_6 = 0, $ and we have the following cases:

    ${\rm{(a)}}\ $if $ \alpha_5\neq0, $ then by choosing $ x = \sqrt{\alpha_5\alpha_7^{-1}}, y = 0, z = \sqrt{\alpha_1^2 \alpha_5^{-1}\alpha_7^{-1}}, t = 0, $ we have the family of representatives $ \langle \alpha\nabla_2+\beta\nabla_4+\nabla_5+\nabla_7 \rangle; $

    ${\rm{(b)}}\ $if $ \alpha_5 = 0, \alpha_2 = 0, $ then $ \alpha_1\neq 0 $ and we have the family of representatives $ \langle \nabla_1+\alpha\nabla_4+\nabla_7 \rangle; $

    ${\rm{(c)}}\ $if $ \alpha_5 = 0, \alpha_2\neq0, $ then by choosing $ x = \sqrt[3]{\alpha_2\alpha_7^{-1}}, y = 0, z = 0, t = 0, $ we have the family of representatives $ \langle \alpha\nabla_1+ \nabla_2+\beta\nabla_4+\nabla_7 \rangle. $

    Summarizing all cases we have the following distinct orbits

    $1142+3,α1+2+β4+7O(α,β)=O(η3α,η23β)=O(η23α,η3β),1+α3+4O(α)=O(α),1+α4+7O(α)=O(α),α2+3α0,1,2+α4+5,α2+β4+5+7O(α,β)=O(α,β),3+4+5,α4+5α0,1,4+α5+6,6, $

    which gives the following new algebras:

    $ N72:e1e1=e2e1e2=e3e1e3=e5e1e4=34e5e2e2=e4e2e3=34e5Nα,β73:e1e1=e2e1e2=e3e1e3=αe5e1e4=e5e2e2=e4e2e3=3e5e2e4=βe5e4e4=e5Nα74:e1e1=e2e1e2=e3e1e3=e5e1e4=αe5e2e2=e4e2e4=e5Nα75:e1e1=e2e1e2=e3e1e3=e5e2e2=e4e2e4=αe5e4e4=e5Nα0,176:e1e1=e2e1e2=e3e1e4=(1+α)e5e2e2=e4e2e3=3αe5Nα77:e1e1=e2e1e2=e3e1e4=e5e2e2=e4e2e3=3e5e2e4=αe5e3e3=e5Nα,β78:e1e1=e2e1e2=e3e1e4=αe5e2e2=e4e2e3=3αe5e2e4=βe5e3e3=e5e4e4=e5N79:e1e1=e2e1e2=e3e1e4=e5e2e2=e4e2e4=e5e3e3=e5Nα0,180:e1e1=e2e1e2=e3e2e2=e4e2e4=αe5e3e3=e5Nα81:e1e1=e2e1e2=e3e2e2=e4e2e4=e5e3e3=αe5e3e4=e5N82:e1e1=e2e1e2=e3e2e2=e4e3e4=e5 $

    Here we will collect all information about $ {\mathbf N}_{09}^{4*}: $

    $ \begin{array}{|l|l|l|l|}   \hline {\mathbf{N}}^{4*}_{09} & \begin{array}{l} e_1e_1 = e_2 \\ e_2e_3 = e_4 \end{array} & H2D(N409)=[Δ12],[Δ13],[Δ22],[Δ33]H2C(N409)=H2D(N409)[Δ14],[Δ24],[Δ34],[Δ44]& \phi = (x0000x20000r0t0sx2r)\\ \hline \end{array} $

    Let us use the following notations:

    $ 1=[Δ12],2=[Δ13],3=[Δ14],4=[Δ22],5=[Δ24],6=[Δ33],7=[Δ34],8=[Δ44]. $

    Take $ \theta = \sum\limits_{i = 1}^{8}\alpha_i\nabla_i\in\mathrm{H}^2_{\mathfrak{C}}(\mathbf{N}^{4*}_{09}) . $ Since

    $ \phi^T(0α1α2α3α1α40α5α20α6α7α3α5α7α8)\phi = (αα1α2α3α1α4αα5α1αα6α7α3α5α7α8), $

    we have

    $ α1=(α1x+α5t)x2,α2=(α2x+α7t)r+(α3x+α8t)s,α3=(α3x+α8t)x2r,α4=α4x4,α5=α5x4r,α6=(α6r+α7s)r+(α7r+α8s)s,α7=(α7r+α8s)x2r,α8=α8r2x4. $

    We are interested in $ (\alpha_3,\alpha_5,\alpha_7,\alpha_8)\neq(0,0,0,0) $. Let us consider the following cases:

    $ 1.\ $$ \alpha_8 = 0, \alpha_7 = 0, \alpha_5 = 0, $ then $ \alpha_3\neq0 $ and we have

    ${\rm{(a)}}\ $if $ \alpha_1 = 0, \alpha_4 = 0, \alpha_6 = 0, $ then by choosing $ x = 1, r = \alpha_3, s = -\alpha_2, t = 0, $ we have the representative $ \langle \nabla_3 \rangle; $

    ${\rm{(b)}}\ $if $ \alpha_1 = 0, \alpha_4 = 0, \alpha_6\neq0, $ then by choosing $ x = \alpha_6, r = \alpha_3\alpha_6^2, s = -\alpha_2\alpha_6^2, t = 0, $ we have the representative $ \langle \nabla_3+\nabla_6 \rangle; $

    ${\rm{(c)}}\ $if $ \alpha_1 = 0, \alpha_4\neq0, \alpha_6 = 0, $ then by choosing $ x = \alpha_3^2, r = \alpha_3\alpha_4, s = -\alpha_2\alpha_4, t = 0, $ we have the representative $ \langle \nabla_3+\nabla_4 \rangle; $

    $ {\rm{(d)}}\ $if $ \alpha_1 = 0, \alpha_4\neq0, \alpha_6\neq0, $ then by choosing $ x = {\alpha_3}^{-1}\sqrt{\alpha_4\alpha_6}, $ $ r = {\alpha_3}^{-2}{\sqrt{\alpha_4^3\alpha_6}}, $ $ s = -\alpha_2{\alpha_3}^{-3}\sqrt{\alpha_4^3\alpha_6}, $ $ t = 0, $ we have the representative $ \langle \nabla_3+\nabla_4+\nabla_6 \rangle; $

    ${\rm{(e)}}\ $if $ \alpha_1\neq0, \alpha_4 = 0, \alpha_6 = 0, $ then by choosing $ x = 1, $ $ r = {\alpha_1}{\alpha_3}^{-1}, $ $ s = -{\alpha_1\alpha_2}{\alpha^{-2}_3}, $ $ t = 0, $ we have the representative $ \langle \nabla_1+\nabla_3 \rangle; $

    $ {\rm{(f)}}\ $if $ \alpha_1\neq0, \alpha_4 = 0, \alpha_6\neq0, $ then by choosing $ x = \sqrt[3]{\alpha_1\alpha_6 \alpha_3^{-2}}, $ $ r = \alpha_1\alpha_3^{-1}, $ $ s = -\alpha_1\alpha_2\alpha^{-2}_3, $ $ t = 0, $ we have the representative $ \langle \nabla_1+\nabla_3+\nabla_6 \rangle; $

    ${\rm{(g)}}\ $if $ \alpha_1\neq0, \alpha_4\neq0, $ then by choosing

    $ x = \alpha_1 \alpha_4^{-1},$ $r = \alpha_1\alpha_3^{-1},$ $s = -\alpha_1\alpha_2 \alpha^{-2}_3,$ $ t = 0,$

    we have the family of representatives $ \langle \nabla_1+\nabla_3+\nabla_4+\alpha\nabla_6 \rangle. $

    2. $ \alpha_8 = 0, \alpha_7 = 0, \alpha_5\neq0 $ and we have

    ${\rm{(a)}}\ $if $ \alpha_3 = 0, \alpha_2 = 0, \alpha_4 = 0, \alpha_6 = 0, $ then by choosing $ r = 1, x = \alpha_5, t = -\alpha_1, s = 0, $ we have the representative $ \langle \nabla_5 \rangle; $

    $ {\rm{(b)}}\ $if $ \alpha_3 = 0, \alpha_2 = 0, \alpha_4 = 0, \alpha_6\neq0, $ then by choosing $ x = \alpha_5\alpha_6, $ $ r = \alpha_5^5\alpha_6^3, s = 0, t = -\alpha_1\alpha_6, $ we have the representative $ \langle \nabla_5+\nabla_6 \rangle; $

    ${\rm{(c)}}\ $if $ \alpha_3 = 0, \alpha_2 = 0, \alpha_4\neq0, \alpha_6 = 0, $ then by choosing $ x = 1, r = \alpha_4{\alpha_5}^{-1}, t = -{\alpha_1}{\alpha_5}^{-1}, s = 0, $ we have the representative $ \langle \nabla_4+\nabla_5 \rangle; $

    $ {\rm{(d)}}\ $if $ \alpha_3 = 0, \alpha_2 = 0, \alpha_4\neq0, \alpha_6\neq0, $ then by choosing

    $x = {\sqrt[4]{\alpha_4\alpha_6\alpha_5^{-2}}},$ $r = {\alpha_4}{\alpha_5}^{-1},$ $t = -{\alpha_1\sqrt[4]{\alpha_4\alpha_6 \alpha_5^{-6}}}, s = 0,$

    we have the representative $ \langle \nabla_4+\nabla_5+\nabla_6 \rangle; $

    $ {\rm{(e)}}\ $if $ \alpha_3 = 0, \alpha_2\neq0, \alpha_4 = 0, \alpha_6 = 0, $ then by choosing

    $ r = 1, x = \sqrt[3]{{\alpha_2}{\alpha_5}^{-1}}, t = -\alpha_1\sqrt[3]{\alpha_2\alpha_5^{-4}}, s = 0,$

    we have the representative $ \langle \nabla_2+\nabla_5 \rangle; $

    ${\rm{(f)}}\ $if $ \alpha_3 = 0, \alpha_2\neq0, \alpha_4 = 0, \alpha_6\neq0, $ then by choosing

    $ x = \sqrt[3]{{\alpha_2}{\alpha_5}^{-1}},$ $r = \alpha_6^{-1}\sqrt[3]{\alpha_2^{4}\alpha_5^{-1}},$ $t = -\alpha_1\sqrt[3]{\alpha_2\alpha_5^{-4}}, s = 0,$

    we have the representative $ \langle \nabla_2+\nabla_5+\nabla_6 \rangle; $

    ${\rm{(g)}}\ $if $ \alpha_3 = 0, \alpha_2\neq0, \alpha_4\neq0, $ then by choosing

    $ x = \sqrt[3]{\alpha_2\alpha_5^{-1}}, r = {\alpha_4}{\alpha_5}^{-1}, t = -\alpha_1\sqrt[3]{\alpha_2 \alpha_5^{-4}}, s = 0,$

    we have the family of representatives $ \langle \nabla_2+\nabla_4+\nabla_5+\alpha\nabla_6 \rangle; $

    ${\rm{(h)}}\ $if $ \alpha_3\neq0, \alpha_4 = 0, \alpha_6 = 0, $ then by choosing $ x = {\alpha_3}{\alpha_5}^{-1}, r = \alpha_3, s = -\alpha_2, t = -{\alpha_1\alpha_3}{\alpha_5^{-2}}, $ we have the representative $ \langle \nabla_3+\nabla_5 \rangle; $

    ${\rm{(i)}}\ $if $ \alpha_3\neq0, \alpha_4 = 0, \alpha_6\neq0, $ then by choosing

    $ x = {\alpha_3}{\alpha_5}^{-1}, r = {\alpha_3^4}{\alpha_5^{-3}\alpha_6^{-1}}, s = -{\alpha_2\alpha_3^3}{\alpha_5^{-3}\alpha_6^{-1}}, t = -{\alpha_1\alpha_3}{\alpha_5^{-2}},$

    we have the representative $ \langle \nabla_3+\nabla_5+\nabla_6 \rangle; $

    $ {\rm{(j)}}\ $if $ \alpha_3\neq0, \alpha_4\neq0, $ then by choosing $ x = {\alpha_3}{\alpha_5}^{-1}, r = {\alpha_4}{\alpha_5}^{-1}, s = -{\alpha_2\alpha_4}{\alpha_3^{-1}\alpha_5^{-1}}, t = -{\alpha_1\alpha_3}{\alpha_5^{-2}}, $ we have the family of representatives $ \langle \nabla_3+\nabla_4+\nabla_5+\alpha\nabla_6 \rangle. $

    3. $ \alpha_8 = 0, \alpha_7\neq0, $ then by choosing $ x = 2 \alpha_7^2, $ $ t = \alpha_3 \alpha_6-2 \alpha_2 \alpha_7, $ $ s = -\alpha_6, $ $ r = 2 \alpha_7, $ we have $ \alpha_2^* = \alpha_6^* = 0. $ Now we can suppose that $ \alpha_2 = 0 $ and $ \alpha_6 = 0, $ then for $ s = 0 $ and $ t = 0, $ we have:

    ${\rm{(a)}}\ $if $ \alpha_1 = 0, \alpha_3 = 0, \alpha_4 = 0, \alpha_5 = 0, $ then by choosing $ r = 1, x = 1, $ we have the representative $ \langle \nabla_7 \rangle; $

    ${\rm{(b)}}\ $if $ \alpha_1 = 0, \alpha_3 = 0, \alpha_4 = 0, \alpha_5\neq0, $ then by choosing $ x = \alpha_7, r = \alpha_5\alpha_7, $ we have the representative $ \langle \nabla_5+\nabla_7 \rangle; $

    ${\rm{(c)}}\ $if $ \alpha_1 = 0, \alpha_3 = 0, \alpha_4\neq0, \alpha_5 = 0, $ then by choosing $ x = \sqrt{\alpha_7}, r = \sqrt{\alpha_4}, $ we have the representative $ \langle \nabla_4+\nabla_7 \rangle; $

    ${\rm{(d)}}\ $if $ \alpha_1 = 0, \alpha_3 = 0, \alpha_4\neq0, \alpha_5\neq0, $ then by choosing $ x = \alpha_4\sqrt{\alpha_7 \alpha_5^{-3}}, r = {\alpha_4}{\alpha_5}^{-1}, $ we have the representative $ \langle \nabla_4+\nabla_5+\nabla_7 \rangle; $

    ${\rm{(e)}}\ $if $ \alpha_1 = 0, \alpha_3\neq0, \alpha_5 = 0, $ then by choosing $ r = \alpha_3, x = \alpha_5, $ we have the family of representatives $ \langle \nabla_3+\alpha\nabla_4+\nabla_7 \rangle; $

    $ {\rm{(f)}}\ $if $ \alpha_1 = 0, \alpha_3\neq0, \alpha_5\neq0, $ then by choosing $ x = {\alpha_3}{\alpha_5}^{-1}, r = {\alpha_3^2}{\alpha_5^{-1}\alpha_7^{-1}}, $ we have the family of representatives $ \langle \nabla_3+\alpha\nabla_4+\nabla_5+\nabla_7 \rangle; $

    ${\rm{(g)}}\ $if $ \alpha_1\neq0, \alpha_3 = 0, \alpha_4 = 0, \alpha_5 = 0, $ then by choosing $ x = \alpha_1\alpha_7, r = \alpha_1, $ we have the representative $ \langle \nabla_1+\nabla_7 \rangle; $

    $ {\rm{(h)}}\ $if $ \alpha_1\neq0, \alpha_3 = 0, \alpha_4 = 0, \alpha_5\neq0, $ then by choosing $ x = \sqrt[3]{{\alpha_1\alpha_7}{\alpha^{-2}_5}}, r = \sqrt[3]{{\alpha_1^2}{\alpha_5^{-1}\alpha_7^{-1}}}, $ we have the representative $ \langle \nabla_1+\nabla_5+\nabla_7 \rangle; $

    $ {\rm{(i)}}\ $if $ \alpha_1\neq0, \alpha_3 = 0, \alpha_4\neq0, $ then by choosing $ x = {\alpha_1}{\alpha_4}^{-1}, r = {\alpha_1}{\sqrt{\alpha_4^{-1}\alpha_7^{-1}}}, $ we have the family of representatives $ \langle \nabla_1+\nabla_4+\alpha\nabla_5+\nabla_7 \rangle; $

    ${\rm{(j)}}\ $if $ \alpha_1\neq0, \alpha_3\neq0, $ then by choosing $ x = {\alpha_1\alpha_7}{\alpha_3^{-2}}, r = {\alpha_1}{\alpha_3}^{-1}, $ we have the family of representatives $ \langle \nabla_1+\nabla_3+\alpha\nabla_4+\beta\nabla_5+\nabla_7 \rangle. $

    4. $ \alpha_8\neq0, $ then by choosing $ x = \alpha_8, t = - \alpha_3, s = - \alpha_7, r = \alpha_8, $ we have $ \alpha_3^* = \alpha_7^* = 0. $ Now we can suppose that $ \alpha_3 = 0 $ and $ \alpha_7 = 0, $ then for $ s = 0 $ and $ t = 0, $ we have

    ${\rm{(a)}}\ $if $ \alpha_1 = 0, \alpha_2 = 0, \alpha_4 = 0, \alpha_6 = 0, $ then we have the representatives $ \langle \nabla_8 \rangle $ and $ \langle \nabla_5+\nabla_8 \rangle, $ depending on whether $ \alpha_5 = 0 $ or not;

    ${\rm{(b)}}\ $if $ \alpha_1 = 0, \alpha_2 = 0, \alpha_4 = 0, \alpha_6\neq0, $ then we have the representatives $ \langle \nabla_6+\nabla_8 \rangle $ and $ \langle \nabla_5+\nabla_6+\nabla_8 \rangle, $ depending on whether $ \alpha_5 = 0 $ or not;

    $ {\rm{(c)}}\ $if $ \alpha_1 = 0, \alpha_2 = 0, \alpha_4\neq0, \alpha_6 = 0, $ then we have the representatives $ \langle \nabla_4+\nabla_8 \rangle $ and $ \langle \nabla_4+\nabla_5+\nabla_8 \rangle, $ depending on whether $ \alpha_5 = 0 $ or not;

    $ {\rm{(d)}}\ $if $ \alpha_1 = 0, \alpha_2 = 0, \alpha_4\neq0, \alpha_6\neq0, $ then by choosing $ x = \sqrt[4]{{\alpha_6}{\alpha_8}^{-1}}, r = \sqrt{{\alpha_4}{\alpha_8}^{-1}} , $ we have the representative $ \langle \nabla_4+\alpha\nabla_5+\nabla_6+\nabla_8 \rangle; $

    ${\rm{(e)}}\ $if $ \alpha_1 = 0, \alpha_2\neq0, \alpha_4 = 0, \alpha_5 = 0, $ then we have the representatives $ \langle \nabla_2+\nabla_8 \rangle $ and $ \langle \nabla_2+\nabla_6+\nabla_8 \rangle, $ depending on whether $ \alpha_6 = 0 $ or not;

    $ {\rm{(f)}}\ $if $ \alpha_1 = 0, \alpha_2\neq0, \alpha_4 = 0, \alpha_5\neq0, $ then by choosing $ x = \sqrt[3]{{\alpha_2}{\alpha_5}^{-1}}, r = {\alpha_5}{\alpha_8}^{-1}, $ we have the representative $ \langle \nabla_2+\nabla_5+\alpha\nabla_6+\nabla_8 \rangle; $

    ${\rm{(g)}}\ $if $ \alpha_1 = 0, \alpha_2\neq0, \alpha_4\neq0, $ then by choosing $ x = \sqrt[6]{{\alpha_2^2 \alpha_4^{-1}\alpha_8^{-1}}}, $ $ r = \sqrt{{\alpha_4}{\alpha_8}^{-1}}, $ we have the representative $ \langle \nabla_2+\nabla_4+\alpha\nabla_5+\beta\nabla_6+\nabla_8 \rangle; $

    $ {\rm{(h)}}\ $if $ \alpha_1\neq0, \alpha_2 = 0, \alpha_4 = 0, \alpha_5 = 0, $ then we have the representatives $ \langle \nabla_1+\nabla_8 \rangle $ and $ \langle \nabla_1+\nabla_6+\nabla_8 \rangle $ depending on whether $ \alpha_6 = 0 $ or not;

    ${\rm{(i)}}\ $if $ \alpha_1\neq0, \alpha_2 = 0, \alpha_4 = 0, \alpha_5\neq0, $ then by choosing $ x = {\alpha_1\alpha_8}{\alpha_5^{-2}}, r = {\alpha_5}{\alpha_8}^{-1}, $ we have the representative $ \langle \nabla_1+\nabla_5+\alpha\nabla_6+\nabla_8 \rangle; $

    ${\rm{(j)}}\ $if $ \alpha_1\neq0, \alpha_2 = 0, \alpha_4\neq0, $ then by choosing $ x = {\alpha_1}{\alpha_4}^{-1}, r = \sqrt{{\alpha_4}{\alpha_8}^{-1}}, $ we have the representative $ \langle \nabla_1+\nabla_4+\alpha\nabla_5+\beta\nabla_6+\nabla_8 \rangle; $

    $ {\rm{(k)}}\ $if $ \alpha_1\neq0, \alpha_2\neq0, $ then by choosing $ x = \sqrt[5]{{\alpha_2^2}{\alpha_1^{-1}\alpha_8^{-1}}}, r = \sqrt[5]{{\alpha_1^4}{\alpha_2^{-3}\alpha_8^{-1}}}, $ we have the representative $ \langle \nabla_1+\nabla_2+\alpha\nabla_4+\beta\nabla_5+\gamma\nabla_6+\nabla_8 \rangle. $

    Summarizing, we have the following distinct orbits:

    $ 1+2+α4+β5+γ6+8O(α,β,γ)=O(η5α,η35β,η5γ)=O(η25α,η5β,η25γ)=O(η35α,η45β,η35γ)=O(η45α,η25β,η45γ),1+3,1+3+α4+β5+7,1+3+4+α6,1+3+6,1+4+α5+β6+8O(α,β)=O(α,β),1+4+α5+7O(α,β)=O(α,β),1+5+α6+8,1+5+7,1+6+8,1+7,1+8,2+4+5+α6O(α)=O(η3α)=O(η23α),2+4+α5+β6+8,2+5,2+5+6,2+5+α6+8O(α,β)=O(α,β)=O(α,η23β)=O(α,η23β)=O(α,η3β)=O(α,η3β),2+6+8,2+8,3,3+4,3+4+5+α6,3+α4+5+7,3+4+6,3+5,3+5+6,3+6,4+5,4+5+6,4+α5+6+8O(α)=O(α),4+5+7,4+5+8,4+7,4+8,5,5+6,5+6+8,5+7,5+8,6+8,7,8,$

    which gives the following new algebras:

    $ Nα,β,γ83:e1e1=e2e1e2=e5e1e3=e5e2e2=αe5e2e3=e4e2e4=βe5e3e3=γe5e4e4=e5N84:e1e1=e2e1e2=e5e1e4=e5e2e3=e4Nα,β85:e1e1=e2e1e2=e5e1e4=e5e2e2=αe5e2e3=e4e2e4=βe5e3e4=e5Nα86:e1e1=e2e1e2=e5e1e4=e5e2e2=e5e2e3=e4e3e3=αe5N87:e1e1=e2e1e2=e5e1e4=e5e2e3=e4e3e3=e5Nα,β88:e1e1=e2e1e2=e5e2e2=e5e2e3=e4e2e4=αe5e3e3=βe5e4e4=e5Nα89:e1e1=e2e1e2=e5e2e2=e5e2e3=e4e2e4=αe5e3e4=e5Nα90:e1e1=e2e1e2=e5e2e3=e4e2e4=e5e3e3=αe5e4e4=e5N91:e1e1=e2e1e2=e5e2e3=e4e2e4=e5e3e4=e5N92:e1e1=e2e1e2=e5e2e3=e4e3e3=e5e4e4=e5N93:e1e1=e2e1e2=e5e2e3=e4e3e4=e5N94:e1e1=e2e1e2=e5e2e3=e4e4e4=e5Nα95:e1e1=e2e1e3=e5e2e2=e5e2e3=e4e2e4=e5e3e3=αe5Nα,β96:e1e1=e2e1e3=e5e2e2=e5e2e3=e4e2e4=αe5e3e3=βe5e4e4=e5N97:e1e1=e2e1e3=e5e2e3=e4e2e4=e5N98:e1e1=e2e1e3=e5e2e3=e4e2e4=e5e3e3=e5Nα99:e1e1=e2e1e3=e5e2e3=e4e2e4=e5e3e3=αe5e4e4=e5N100:e1e1=e2e1e3=e5e2e3=e4e3e3=e5e4e4=e5N101:e1e1=e2e1e3=e5e2e3=e4e4e4=e5N102:e1e1=e2e1e4=e5e2e3=e4N103:e1e1=e2e1e4=e5e2e2=e5e2e3=e4Nα104:e1e1=e2e1e4=e5e2e2=e5e2e3=e4e2e4=e5e3e3=αe5Nα105:e1e1=e2e1e4=e5e2e2=αe5e2e3=e4e2e4=e5e3e4=e5N106:e1e1=e2e1e4=e5e2e2=e5e2e3=e4e3e3=e5N107:e1e1=e2e1e4=e5e2e3=e4e2e4=e5N108:e1e1=e2e1e4=e5e2e3=e4e2e4=e5e3e3=e5N109:e1e1=e2e1e4=e5e2e3=e4e3e3=e5N110:e1e1=e2e2e2=e5e2e3=e4e2e4=e5N111:e1e1=e2e2e2=e5e2e3=e4e2e4=e5e3e3=e5Nα112:e1e1=e2e2e2=e5e2e3=e4e2e4=αe5e3e3=e5e4e4=e5N113:e1e1=e2e2e2=e5e2e3=e4e2e4=e5e3e4=e5N114:e1e1=e2e2e2=e5e2e3=e4e2e4=e5e4e4=e5N115:e1e1=e2e2e2=e5e2e3=e4e3e4=e5N116:e1e1=e2e2e2=e5e2e3=e4e4e4=e5N117:e1e1=e2e2e3=e4e2e4=e5N118:e1e1=e2e2e3=e4e2e4=e5e3e3=e5N119:e1e1=e2e2e3=e4e2e4=e5e3e3=e5e4e4=e5N120:e1e1=e2e2e3=e4e2e4=e5e3e4=e5N121:e1e1=e2e2e3=e4e2e4=e5e4e4=e5N122:e1e1=e2e2e3=e4e3e3=e5e4e4=e5N123:e1e1=e2e2e3=e4e3e4=e5N124:e1e1=e2e2e3=e4e4e4=e5 $

    Here we will collect all information about $ {\mathbf N}_{10}^{4*}: $

    $ \begin{array}{|l|l|l|l|}   \hline {\mathbf{N}}^{4*}_{10} & \begin{array}{l} e_1e_1 = e_2 \\ e_1e_2 = e_4 \\ e_3e_3 = e_4 \end{array} & \begin{array}{l} \mathrm{H}^2_{\mathfrak{D}}(\mathbf{N}^{4*}_{10}) = \\  \Big\langle \begin{array}{c}  [\Delta_{13}],[\Delta_{14}],[\Delta_{22}], \\   [\Delta_{23}],[\Delta_{33}] \end{array} \Big\rangle\\ \mathrm{H}^2_{\mathfrak{C}}(\mathbf{N}^{4*}_{10}) = \mathrm{H}^2_{\mathfrak{D}}(\mathbf{N}^{4*}_{10})\oplus\\ {\langle [\Delta_{24}], [\Delta_{34}], [\Delta_{44}] \rangle} \end{array} & ϕ=(x000yx2zrx0z0r0tz2+2xysx3),r2=x3\\ \hline \end{array} $

    Let us use the following notations:

    $ 1=[Δ13],2=[Δ14],3=[Δ22],4=[Δ23],5=[Δ24],6=[Δ33],7=[Δ34],8=[Δ44]. $

    Take $ \theta = \sum\limits_{i = 1}^{8}\alpha_i\nabla_i\in\mathrm{H}^2_{\mathfrak{C}}(\mathbf{N}^{4*}_{10}) . $ Since

    $ \phi^T(00α1α20α3α4α5α1α4α6α7α2α5α7α8)\phi = (ααα1α2αα3α4α5α1α4α6+αα7α2α5α7α8), $

    we have

    $ α1=(α3y+α4z+α5t)zrx+(α1x+α4y+α6z+α7t)r+(α2x+α5y+α7z+α8t)s,α2=(α2x+α5y+α7z+α8t)x3,α3=α3x4+2α5x2(z2+2xy)+α8(z2+2xy)2,α4=(α3x2+α5(z2+2xy))zrx+(α4x2+α7(z2+2xy))r+(α5x2+α8(z2+2xy))s,α5=(α5x2+α8(z2+2xy))x3,α6=(α4rα3zrx+α5s)zrx+(α6rα4zrx+α7s)r+(α7rα5zrx+α8s)s(α3y+α4z+α5t)x2(α2x+α5y+α7z+α8t)(z2+2xy),α7=(α7rα5zrx+α8s)x3,α8=α8x6. $

    We are interested in $ (\alpha_5,\alpha_7,\alpha_8)\neq(0,0,0) . $ Let us consider the following cases:

    $ 1.\ $$ \alpha_8 = 0, \alpha_5 = 0, $ then $ \alpha_7\neq0. $ Now by choosing

    $y=α22+α2α3+α4α72α27x,z=α2α7x,s=3α22α3+α2(α23+6α4α7)+α7(α3α4+2α6α7)4α37x3,t=α27(α242α1α7)+α32α3+α22(α23+3α4α7)+2α2α7(α3α4+α6α7)2α47x,$

    we have $ \alpha^*_1 = 0, \alpha^*_2 = 0, \alpha^*_4 = 0, \alpha^*_6 = 0. $ Then we have the representatives $ \langle \nabla_7 \rangle $ or $ \langle \nabla_3+\nabla_7 \rangle $ depending on whether $ \alpha_3 = 0 $ or not.

    $ 2.\ $$ \alpha_8 = 0, \alpha_5\neq0, $ then by choosing

    $y=α2α5+α27α25x,z=α7α5x,s=α3α7α4α5α25x3,t=α2α3α5+α25α6+3α4α5α72α3α27α35x,$

    we have $ \alpha^*_2 = \alpha^*_4 = \alpha^*_6 = 0 $ and $ \alpha_7^* = 0. $ Therefore, we can suppose that $ \alpha_2 = 0, \alpha_4 = 0, \alpha_6 = 0, \alpha_7 = 0, $ and have the following cases:

    ${\rm{(a)}}\ $if $ \alpha_1 = 0, \alpha_3 = 0, $ then we have the representative $ \langle \nabla_5 \rangle; $

    ${\rm{(b)}}\ $if $ \alpha_1 = 0, \alpha_3\neq0, $ then by choosing $ x = \frac{\alpha_3}{\alpha_5}, r^2 = x^3, $ we have the representative $ \langle \nabla_3+\nabla_5 \rangle; $

    ${\rm{(c)}}\ $if $ \alpha_1\neq0, $ then by choosing $ x = \sqrt[5]{\frac{\alpha^2_1}{\alpha^2_5}}, r^2 = x^3, $ we have the family of representatives $ \langle \nabla_1+\alpha\nabla_3+\nabla_5 \rangle. $

    3. $ \alpha_8\neq0, $ then by choosing $ y = -\frac{\alpha_5x^2+\alpha_8z^2}{2\alpha_8x}, s = \frac{\sqrt{x}(\alpha_5z-\alpha_7x)}{\alpha_8}, t = -\frac{\alpha_2x+\alpha_5y+\alpha_7z}{\alpha_8}, $ we have $ \alpha^*_2 = \alpha^*_5 = \alpha^*_7 = 0 . $ Therefore, we can suppose that $ \alpha_2 = 0, \alpha_5 = 0, \alpha_7 = 0, $ and have the following cases:

    ${\rm{(a)}}\ $if $ \alpha_3 = 0, \alpha_4 = 0, \alpha_6 = 0, $ then we have the representative $ \langle \nabla_8 \rangle $ and $ \langle \nabla_1+\nabla_8 \rangle $ depending on whether $ \alpha_1 = 0 $ or not.

    ${\rm{(b)}}\ $if $ \alpha_3 = 0, \alpha_4 = 0, \alpha_6\neq0, $ then by choosing $ x = \sqrt[3]{\frac{\alpha_6}{\alpha_8}}, r^2 = x^3, z = -\frac{\alpha_1}{\sqrt[3]{\alpha^2_6\alpha_8}}, $ we have the representative $ \langle \nabla_6+\nabla_8 \rangle; $

    $ {\rm{(c)}}\ $if $ \alpha_3 = 0, \alpha_4\neq0, $ then by choosing $ x = \sqrt[5]{\frac{\alpha^2_4}{\alpha^2_8}}, r^2 = x^3, z = \frac{\alpha_6}{3\sqrt[5]{\alpha_4^3\alpha^2_8}}, $ we have the representative $ \langle \alpha\nabla_1+\nabla_4+\nabla_8 \rangle; $

    ${\rm{(d)}}\ $if $ \alpha_3\neq0, $ then by choosing $ x = \sqrt{\frac{\alpha_3}{\alpha_8}}, r^2 = x^3, z = \frac{\alpha_4}{\sqrt{\alpha_3\alpha_8}}, $ we have the representative $ \langle \alpha\nabla_1+\nabla_3+\beta\nabla_6+\nabla_8 \rangle. $

    Summarizing, we have the following distinct orbits:

    $1+α3+5O(α)=O(η45α)=O(η35α)=O(η25α)=O(η5α),α1+3+β6+8O(α,β)=O(α,β)=O(η3α,η23β)=O(η3α,η23β)=O(η23α,η3β)=O(η23α,η3β),α1+4+8O(α)=O(α)=O(η45α)=O(η45α)=O(η35α)=O(η35α)=O(η25α)=O(η25α)=O(η5α)=O(η5α),1+8,3+5,3+7,5,6+8,7,8,$

    which gives the following new algebras:

    $ Nα125:e1e1=e2e1e2=e4e1e3=e5e2e2=αe5e2e4=e5e3e3=e4Nα,β126:e1e1=e2e1e2=e4e1e3=αe5e2e2=e5e3e3=e4+βe5e4e4=e5Nα127:e1e1=e2e1e2=e4e1e3=αe5e2e3=e5e3e3=e4e4e4=e5N128:e1e1=e2e1e2=e4e1e3=e5e3e3=e4e4e4=e5N129:e1e1=e2e1e2=e4e2e2=e5e2e4=e5e3e3=e4N130:e1e1=e2e1e2=e4e2e2=e5e3e3=e4e3e4=e5N131:e1e1=e2e1e2=e4e2e4=e5e3e3=e4N132:e1e1=e2e1e2=e4e3e3=e4+e5e4e4=e5N133:e1e1=e2e1e2=e4e3e3=e4e3e4=e5N134:e1e1=e2e1e2=e4e3e3=e4e4e4=e5 $

    Here we will collect all information about $ {\mathbf N}_{11}^{4*}: $

    $ \begin{array}{|l|l|l|l|}   \hline {\mathbf{N}}^{4*}_{11} & \begin{array}{l} e_1e_1 = e_2 \\ e_1e_3 = e_4 \\ e_2e_2 = e_4 \end{array} & H2D(N411)=[Δ12],[Δ22],[Δ23],[Δ33]H2C(N411)=H2D(N411)[Δ14],[Δ24],[Δ34],[Δ44] & \phi = (x0000x200z0x30t2xzsx4)\\ \hline \end{array} $

    Let us use the following notations:

    $ 1=[Δ12],2=[Δ14],3=[Δ22],4=[Δ23],5=[Δ24],6=[Δ33],7=[Δ34],8=[Δ44]. $

    Take $ \theta = \sum\limits_{i = 1}^{8}\alpha_i\nabla_i\in\mathrm{H}^2_{\mathfrak{C}}(\mathbf{N}^{4*}_{11}) . $ Since

    $ \phi^T(0α10α2α1α3α4α50α4α6α7α2α5α7α8)\phi = (αα1αα2α1α3+αα4α5αα4α6α7α2α5α7α8), $

    we have

    $ α1=(α1x+α4z+α5t)x2+2(α2x+α7z+α8t)xz,α2=(α2x+α7z+α8t)x4,α3=(α3x2+4α5xz+4α8z2)x2(α6z+α7t)x3(α2x+α7z+α8t)s,α4=(α4x+2α7z)x4+(α5x+2α8z)xs,α5=(α5x+2α8z)x5,α6=α6x6+2α7x3s+α8s2,α7=(α7x3+α8s)x4,α8=α8x8. $

    We are interested in $ (\alpha_2,\alpha_5,\alpha_7,\alpha_8)\neq(0,0,0,0) . $ Let us consider the following cases:

    $1.\ $$ \alpha_8 = 0, \alpha_7 = 0, \alpha_5 = 0 , $ then $ \alpha_2\neq0 $ and we have

    ${\rm{(a)}}\ $ if $ \alpha_4 = 0, \alpha_6 = 0, $ then by choosing $ x = 2\alpha_2, z = -\alpha_1, s = 8\alpha_2^2\alpha_3, t = 0, $ we have the representative $ \langle \nabla_2 \rangle; $

    $ {\rm{(b)}}\ $ if $ \alpha_4 = 0, \alpha_6\neq0, $ then by choosing $ x = \frac{\alpha_2}{\alpha_6}, z = -\frac{\alpha_1}{2\alpha_6}, s = \frac{\alpha_1\alpha_2\alpha_6+2\alpha_2^2\alpha_3}{2\alpha_6^3}, t = 0, $ we have the representative $ \langle \nabla_2+\nabla_6 \rangle; $

    ${\rm{(c)}}\ $if $ \alpha_4\neq0, \alpha_4 = -2\alpha_2, \alpha_1\neq0, $ then by choosing

    $ x = \sqrt{\frac{\alpha_1}{\alpha_2}}, z = 0, s = \frac{\alpha_1\alpha_3\sqrt{\alpha_1}}{\alpha_2^2\sqrt{\alpha_2}}, t = 0,$

    we have the family of representatives $ \langle \nabla_1+\nabla_2-2\nabla_4+\alpha\nabla_6 \rangle; $

    $ {\rm{(d)}}\ $ if $ \alpha_4\neq0, \alpha_4 = -2\alpha_2, \alpha_1 = 0, \alpha_6\neq 0, $ then by choosing $ x = \alpha_2\alpha_6^{-1}, $ $ z = 0, $ $ s = \alpha_2^2\alpha_3\alpha_6^{-3}, $ $ t = 0, $ we have the representative $ \langle \nabla_2-2\nabla_4 +\nabla_6 \rangle; $

    $ {\rm{(e)}}\ $ if $ \alpha_4\neq0, \alpha_4 = -2\alpha_2, \alpha_1 = 0,\alpha_6 = 0, $ then by choosing $ x = \alpha_2, $ $ z = 0, $ $ s = \alpha_2^2\alpha_3, $ $ t = 0, $ we have the representative $ \langle \nabla_2-2\nabla_4 \rangle; $

    $ {\rm{(f)}}\ $if $ \alpha_4\neq0, \alpha_4\neq-2\alpha_2, \alpha_6 = 0, $ then by choosing

    $ x = \alpha_4+2\alpha_2, z = -\alpha_1, s = \frac{\alpha_3(\alpha_4+2\alpha_2)^3}{\alpha_2}, t = 0,$

    we have the the family of representatives $ \langle \nabla_2+\alpha\nabla_4 \rangle_{\alpha\neq0,-2}, $ which will be jointed with the cases (1a) and (1b);

    ${\rm{(g)}}\ $if $ \alpha_4\neq0, \alpha_4\neq-2\alpha_2, \alpha_6\neq0, $ then by choosing

    $ x = \frac{\alpha_2}{\alpha_6}, z = -\frac{\alpha_1\alpha_2}{\alpha_6(\alpha_4+2\alpha_2)}, s = \frac{\alpha_1\alpha^2_2\alpha_6+2\alpha_2^3\alpha_3+\alpha_2^2\alpha_3\alpha_4}{\alpha_6^3(\alpha_4+2\alpha_2)}, t = 0,$

    we have the family of representatives $ \langle \nabla_2+\alpha\nabla_4+\nabla_6 \rangle_{\alpha\neq0,-2}, $ which will be jointed with the cases (1b) and (1d).

    2. $ \alpha_8 = 0, \alpha_7 = 0, \alpha_5\neq0 , $ then we have

    $ {\rm{(a)}}\ $ if $ \alpha_6 = 0, \alpha_2 = 0, $ then by choosing

    $ x = 4\alpha_5, z = -\alpha_3, s = -64\alpha_4\alpha_5^2, t = \frac{\alpha_3\alpha_4-4\alpha_1\alpha_5}{\alpha_5} $

    we have the representative $ \langle \nabla_5 \rangle; $

    ${\rm{(b)}}\ $ if $ \alpha_6 = 0, \alpha_2\neq0, $ then by choosing

    $x=α2α5,z=α22α4+α2α3α54α35,s=α32α4α45,t=(α2α4+2α22)(α2α4+α3α5)4α1α2α254α45,$

    we have the representative $ \langle \nabla_2+\nabla_5 \rangle; $

    ${\rm{(c)}}\ $ if $ \alpha_6\neq0, \alpha_6 = 4\alpha_5, \alpha_3 = 0, \alpha_2 = 0, $ then by choosing $ x = \alpha_5, z = 0, s = -\alpha_4\alpha_5^2, t = -\alpha_1, $ we have the representative $ \langle \nabla_5+4\nabla_6 \rangle; $

    $ {\rm{(d)}}\ $ if $ \alpha_6\neq0, \alpha_6 = 4\alpha_5, \alpha_2\alpha_4+\alpha_3\alpha_5 = 0, \alpha_2\neq0, $ then by choosing

    $ x = \frac{\alpha_2}{\alpha_5}, z = 0, s = -\frac{\alpha_2^3\alpha_4}{\alpha_5^4}, t = -\frac{\alpha_1\alpha_2}{\alpha_5^2}, $

    we have the representative $ \langle \nabla_2+\nabla_5+4\nabla_6 \rangle; $

    ${\rm{(e)}}\ $if $ \alpha_6\neq0, \alpha_6 = 4\alpha_5, \alpha_2\alpha_4+\alpha_3\alpha_5\neq0, $ then by choosing

    $ x = \frac{\sqrt{\alpha_2\alpha_4+\alpha_3\alpha_5}}{\alpha_5}, z = 0,$ $s = -\frac{\alpha_4\sqrt{(\alpha_2\alpha_4+\alpha_3\alpha_5)^3}}{\alpha^4_5},$ $t = -\frac{\alpha_1\sqrt{\alpha_2\alpha_4+\alpha_3\alpha_5}}{\alpha^2_5}, $

    we have the family of representatives $ \langle \alpha\nabla_2+\nabla_3+\nabla_5+4\nabla_6 \rangle; $

    ${\rm{(f)}}\ $if $ \alpha_6\neq0, \alpha_6\neq4\alpha_5, \alpha_2 = 0, $ then by choosing

    $ x = \alpha_6-4\alpha_5, z = \alpha_3,$ $s = \frac{\alpha_4(4\alpha_5-\alpha_6)^3}{\alpha_5},$ $t = \frac{4\alpha_1\alpha_5-\alpha_1\alpha_6-\alpha_3\alpha_4}{\alpha_5},$

    we have the family of representatives $ \langle \nabla_5+\alpha\nabla_6 \rangle_{\alpha\neq0,4}, $ which will be jointed with the cases (2a) and (2c);

    $ {\rm{(g)}}\ $if $ \alpha_6\neq0, \alpha_6\neq4\alpha_5, \alpha_2\neq0, $ then by choosing

    $ x=α2α5,z=α2(α2α4+α3α5)α25α64α25,s=α4α32α45,t=α2(2α22α4+α3α4α5+α2(α24+2α3α5)α1α5(4α5α6))α35(4α5α6), $

    we have the family of representatives $ \langle \nabla_2+\nabla_5+\alpha\nabla_6 \rangle_{(\alpha\neq0,4)}, $ which will be jointed with the cases (2a) and (2c).

    3. $ \alpha_8 = 0, \alpha_7\neq0 , $ then by choosing $ z = -\frac{\alpha_2}{\alpha_7}x, $ $ s = -\frac{\alpha_6}{2\alpha_7}x^3, $ $ t = \frac{ \alpha_2 (\alpha_6-4 \alpha_5)+\alpha_3 \alpha_7}{\alpha^2_7}x, $ we have $ \alpha_2^* = \alpha_3^* = \alpha_6^* = 0 . $ Now we can suppose that $ \alpha_2 = 0, \alpha_3 = 0, \alpha_6 = 0 $ and have the following subcases:

    ${\rm{(a)}}\ $if $ \alpha_1 = 0, \alpha_4 = 0, \alpha_5 = 0, $ then we have the representative $ \langle \nabla_7 \rangle; $

    ${\rm{(b)}}\ $if $ \alpha_1 = 0, \alpha_4 = 0, \alpha_5\neq0, $ then by choosing $ x = \frac{\alpha_5}{\alpha_7}, z = 0, s = 0, t = 0, $ we have the representative $ \langle \nabla_5+\nabla_7 \rangle; $

    ${\rm{(c)}}\ $if $ \alpha_1 = 0, \alpha_4\neq0, $ then by choosing $ x = \sqrt{\frac{\alpha_4}{\alpha_7}}, z = 0, s = 0, t = 0, $ we have the family of representatives $ \langle \nabla_4+\alpha\nabla_5+\nabla_7 \rangle; $

    $ {\rm{(d)}}\ $if $ \alpha_1\neq0, $ then by choosing $ x = \sqrt[4]{\frac{\alpha_1}{\alpha_7}}, z = 0, s = 0, t = 0, $ we have the family of representatives $ \langle \nabla_1+\alpha\nabla_4+\beta\nabla_5+\nabla_7 \rangle. $

    4. $ \alpha_8\neq0 , $ then by choosing $ z = -\frac{\alpha_5}{2\alpha_8}x, s = -\frac{\alpha_7}{\alpha_8}x^3, t = \frac{\alpha_5\alpha_7-2\alpha_2\alpha_8}{2\alpha^2_8}x, $ we have $ \alpha_2^* = \alpha_5^* = \alpha_7^* = 0 . $ Now we can suppose that $ \alpha_2 = 0, \alpha_5 = 0, \alpha_7 = 0 $ and have the following subcases:

    ${\rm{(a)}}\ $if $ \alpha_1 = 0, \alpha_3 = 0, \alpha_4 = 0, \alpha_6 = 0, $ then we have the representative $ \langle \nabla_8 \rangle; $

    ${\rm{(b)}}\ $if $ \alpha_1 = 0, \alpha_3 = 0, \alpha_4 = 0, \alpha_6\neq0, $ then by choosing $ x = \sqrt{\frac{\alpha_6}{\alpha_8}}, z = 0, s = 0, t = 0, $ we have the representative $ \langle \nabla_6+\nabla_8 \rangle; $

    ${\rm{(c)}}\ $if $ \alpha_1 = 0, \alpha_3 = 0, \alpha_4\neq0, $ then by choosing $ x = \sqrt[3]{\frac{\alpha_4}{\alpha_8}}, z = 0, s = 0, t = 0, $ we have the family of representatives $ \langle \nabla_4+\alpha\nabla_6+\nabla_8 \rangle; $

    ${\rm{(d)}}\ $if $ \alpha_1 = 0, \alpha_3\neq0, $ then by choosing $ x = \sqrt[4]{\frac{\alpha_3}{\alpha_8}}, z = 0, s = 0, t = 0, $ we have the family of representatives $ \langle \nabla_3+\alpha\nabla_4+\beta\nabla_6+\nabla_8 \rangle; $

    ${\rm{(e)}}\ $if $ \alpha_1\neq0, $ then by choosing $ x = \sqrt[5]{\frac{\alpha_1}{\alpha_8}}, z = 0, s = 0, t = 0, $ we have the family of representatives $ \langle \nabla_1+\alpha\nabla_3+\beta\nabla_4+\gamma\nabla_6+\nabla_8 \rangle. $

    Summarizing, we have the following distinct orbits:

    $1+224+α6O(α)=O(α),1+α3+β4+γ6+8O(α,β,γ)=O(η5α,η25β,η35γ)=O(η25α,η45β,η5γ)=O(η35α,η5β,η45γ)=O(η45α,η35β,η25γ),1+α4+β5+7O(α,β)=O(α,β)=O(α,iβ)=O(α,iβ),α2+3+5+46O(α)=O(α),2+α4,2+α4+6,2+5+α6,3+α4+β6+8,4+α5+7,4+α6+8,5+α6,5+7,6+8,7,8,$

    which gives the following new algebras:

    $ Nα135:e1e1=e2e1e2=e5e1e3=e4e1e4=e5e2e2=e4e2e3=2e5e3e3=αe5Nα,β,γ136:e1e1=e2e1e2=e5e1e3=e4e2e2=e4+αe5e2e3=βe5e3e3=γe5e4e4=e5Nα,β137:e1e1=e2e1e2=e5e1e3=e4e2e2=e4e2e3=αe5e2e4=βe5e3e4=e5Nα138:e1e1=e2e1e3=e4e1e4=αe5e2e2=e4+e5e2e4=e5e3e3=4e5Nα139:e1e1=e2e1e3=e4e1e4=e5e2e2=e4e2e3=αe5Nα140:e1e1=e2e1e3=e4e1e4=e5e2e2=e4e2e3=αe5e3e3=e5Nα141:e1e1=e2e1e3=e4e1e4=e5e2e2=e4e2e4=e5e3e3=αe5Nα,β142:e1e1=e2e1e3=e4e2e2=e4+e5e2e3=αe5e3e3=βe5e4e4=e5Nα143:e1e1=e2e1e3=e4e2e2=e4e2e3=e5e2e4=αe5e3e4=e5Nα144:e1e1=e2e1e3=e4e2e2=e4e2e3=e5e3e3=αe5e4e4=e5Nα145:e1e1=e2e1e3=e4e2e2=e4e2e4=e5e3e3=αe5N146:e1e1=e2e1e3=e4e2e2=e4e2e4=e5e3e4=e5N147:e1e1=e2e1e3=e4e2e2=e4e3e3=e5e4e4=e5N148:e1e1=e2e1e3=e4e2e2=e4e3e4=e5N149:e1e1=e2e1e3=e4e2e2=e4e4e4=e5 $

    Here we will collect all information about $ {\mathbf N}_{12}^{4*}: $

    $ \begin{array}{|l|l|l|l|}   \hline {\mathbf{N}}^{4*}_{12} & \begin{array}{l} e_1e_1 = e_2 \\ e_2e_2 = e_4 \\ e_3e_3 = e_4 \end{array} & H2D(N412)=[Δ12],[Δ13],[Δ23],[Δ33]H2C(N412)=H2D(N412)[Δ14],[Δ24],[Δ34],[Δ44] & \phi_{\pm} = (x0000x20000±x20t0sx4)\\ \hline \end{array} $

    Let us use the following notations:

    $ 1=[Δ12],2=[Δ13],3=[Δ14],4=[Δ23],5=[Δ24],6=[Δ33],7=[Δ34],8=[Δ44]. $

    Take $ \theta = \sum\limits_{i = 1}^{8}\alpha_i\nabla_i\in\mathrm{H}^2_{\mathfrak{C}}(\mathbf{N}^{4*}_{12}) . $ Since

    $ \phi_{\pm}^T(0α1α2α3α10α4α5α2α4α6α7α3α5α7α8)\phi_{\pm} = (αα1α2α3α10α4α5α2α4α6α7α3α5α7α8), $

    we have

    $ α1=(α1x+α5t)x2,α2=(α3x+α8t)s±(α2x+α7t)x2,α3=(α3x+α8t)x4,α4=(α5s±α4x2)x2,α5=α5x6,α6=α6x4±2α7sx2+α8s2,α7=(α8s±α7x2)x4,α8=α8x8. $

    We will consider only the action of $ \phi_+ $ for find representatives and after that we will see that the set of our representatives gives distinct orbits under action of $ \phi_+ $ and $ \phi_-. $ We are interested in $ (\alpha_3,\alpha_5,\alpha_7,\alpha_8)\neq(0,0,0,0) . $ Let us consider the following cases:

    $ 1.\ $$ \alpha_8 = 0, \alpha_5 = 0, \alpha_7 = 0, $ then $ \alpha_3\neq0 $ and we have the following subcases:

    $ {\rm{(a)}}\ $if $ \alpha_1 = 0, \alpha_4 = 0, \alpha_6 = 0, $ then by choosing $ x = \alpha_3, s = -\alpha_2\alpha_3, t = 0, $ we have the representative $ \langle \nabla_3 \rangle; $

    ${\rm{(b)}}\ $if $ \alpha_1 = 0, \alpha_4 = 0, \alpha_6\neq0, $ then by choosing $ x = \frac{\alpha_6}{\alpha_3}, s = -\frac{\alpha_2\alpha_6^2}{\alpha_3^3}, t = 0, $ we have the representative $ \langle \nabla_3+\nabla_6 \rangle; $

    $ {\rm{(c)}}\ $if $ \alpha_1 = 0, \alpha_4\neq0, $ then by choosing $ x = \frac{\alpha_4}{\alpha_3}, s = -\frac{\alpha_2\alpha^2_4}{\alpha_3^3}, t = 0, $ we have the representative $ \langle \nabla_3+\nabla_4+\alpha\nabla_6 \rangle; $

    ${\rm{(d)}}\ $if $ \alpha_1\neq0, $ then by choosing $ x = \sqrt{\frac{\alpha_1}{\alpha_3}}, s = -\frac{\alpha_1\alpha_2}{\alpha_3^2}, t = 0, $ we have the representative $ \langle \nabla_1+\nabla_3+\alpha\nabla_4+\beta\nabla_6 \rangle. $

    2. $ \alpha_8 = 0, \alpha_5 = 0, \alpha_7\neq0, $ then we have the following subcases:

    $ {\rm{(a)}}\ $if $ \alpha_1 = 0, \alpha_3 = 0, \alpha_4 = 0, $ then by choosing $ x = 1, s = -\frac{\alpha_6}{2\alpha_7}, t = \frac{\alpha_3\alpha_6-2\alpha_2\alpha_7}{2\alpha_7^2}, $ we have the representative $ \langle \nabla_7 \rangle; $

    ${\rm{(b)}}\ $if $ \alpha_1 = 0, \alpha_3 = 0, \alpha_4\neq0, $ then by choosing $ x = \sqrt{\frac{\alpha_4}{\alpha_7}}, s = -\frac{\alpha_4\alpha_6}{2\alpha^2_7}, t = \frac{\sqrt{\alpha_4}(\alpha_3\alpha_6-2\alpha_2\alpha_4)}{2\alpha_7^2\sqrt{\alpha_7}}, $ we have the representative $ \langle \nabla_4+\nabla_7 \rangle; $

    ${\rm{(c)}}\ $if $ \alpha_1 = 0, \alpha_3\neq0, $ then by choosing $ x = \frac{\alpha_3}{\alpha_7}, s = -\frac{\alpha^2_3\alpha_6}{2\alpha^3_7}, t = \frac{\alpha_3^2\alpha_6-2\alpha_2\alpha_3\alpha_7}{2\alpha_7^3}, $ we have the representative $ \langle \nabla_3+\alpha\nabla_4+\nabla_7 \rangle; $

    $ {\rm{(d)}}\ $if $ \alpha_1\neq0, $ then by choosing $ x = \sqrt[3]{\frac{\alpha_1}{\alpha_7}}, s = -\sqrt[3]{\frac{\alpha_1^2\alpha^3_6}{8\alpha^5_7}}, t = \frac{\sqrt[3]{\alpha_1}(\alpha_3\alpha_6-2\alpha_2\alpha_7)}{2\alpha_7^2\sqrt[3]{\alpha_7}}, $ we have the representative $ \langle \nabla_1+\alpha\nabla_3+\beta\nabla_4+\nabla_7 \rangle. $

    3. $ \alpha_8 = 0, \alpha_5\neq0, $ then by choosing $ t = -\frac{\alpha_1}{\alpha_5}x, s = -\frac{\alpha_4}{\alpha_5}x^2, $ we have $ \alpha_1^* = \alpha_4^* = 0. $ Now we can suppose that $ \alpha_1 = 0, \alpha_4 = 0 $ and have the following subcases:

    $ {\rm{(a)}}\ $if $ \alpha_2 = 0, \alpha_3 = 0, \alpha_6 = 0, $ then we have the family of representatives $ \langle \nabla_5+\alpha\nabla_7 \rangle; $

    $ {\rm{(b)}}\ $if $ \alpha_2 = 0, \alpha_3 = 0, \alpha_6\neq0, $ then by choosing $ x = \sqrt{\frac{\alpha_6}{\alpha_5}}, s = 0, t = 0, $ we have the family of representatives $ \langle \nabla_5+\nabla_6+\alpha\nabla_7 \rangle; $

    ${\rm{(c)}}\ $if $ \alpha_2 = 0, \alpha_3\neq0, $ then by choosing $ x = \frac{\alpha_3}{\alpha_5}, s = 0, t = 0, $ we have the family of representatives $ \langle \nabla_3+\nabla_5+\alpha\nabla_6+\beta\nabla_7 \rangle; $

    $ {\rm{(d)}}\ $if $ \alpha_2\neq0, $ then by choosing $ x = \sqrt[3]{\frac{\alpha_2}{\alpha_5}}, s = 0, t = 0, $ we have the family of representatives $ \langle \nabla_2+\alpha\nabla_3+\nabla_5+\beta\nabla_6+\gamma\nabla_7 \rangle. $

    4. $ \alpha_8\neq0, $ then by choosing $ t = -\frac{\alpha_3}{\alpha_8}x, s = -\frac{\alpha_7}{\alpha_8}x^2, $ we have $ \alpha_3^* = \alpha_7^* = 0. $ Now we can suppose that $ \alpha_3 = 0, \alpha_7 = 0 $ and have the following subcases:

    ${\rm{(a)}}\ $if $ \alpha_1 = 0, \alpha_2 = 0, \alpha_4 = 0, \alpha_5 = 0, \alpha_6 = 0, $ then we have the representative $ \langle \nabla_8 \rangle; $

    $ {\rm{(b)}}\ $if $ \alpha_1 = 0, \alpha_2 = 0, \alpha_4 = 0, \alpha_5 = 0, \alpha_6\neq0, $ then by choosing $ x = \sqrt[4]{\frac{\alpha_6}{\alpha_8}}, s = 0, t = 0, $ we have the representative $ \langle \nabla_6+\nabla_8 \rangle; $

    $ {\rm{(c)}}\ $if $ \alpha_1 = 0, \alpha_2 = 0, \alpha_4 = 0, \alpha_5\neq0, $ then by choosing $ x = \sqrt{\frac{\alpha_5}{\alpha_8}}, s = 0, t = 0, $ we have the family of representatives $ \langle \nabla_5+\alpha\nabla_6+\nabla_8 \rangle; $

    $ {\rm{(d)}}\ $if $ \alpha_1 = 0, \alpha_2 = 0, \alpha_4\neq0, $ then by choosing $ x = \sqrt[4]{\frac{\alpha_4}{\alpha_8}}, s = 0, t = 0, $ we have the family of representatives $ \langle \nabla_4+\alpha\nabla_5+\beta\nabla_6+\nabla_8 \rangle; $

    $ {\rm{(e)}}\ $if $ \alpha_1 = 0, \alpha_2\neq0, $ then by choosing $ x = \sqrt[5]{\frac{\alpha_2}{\alpha_8}}, s = 0, t = 0, $ we have the family of representatives

    $ \langle \nabla_2+\alpha\nabla_4+\beta\nabla_5+\gamma\nabla_6+\nabla_8 \rangle; $

    $ {\rm{(f)}}\ $if $ \alpha_1\neq0, $ then by choosing $ x = \sqrt[5]{\frac{\alpha_1}{\alpha_8}}, s = 0, t = 0, $ we have the family of representatives

    $ \langle \nabla_1+\alpha\nabla_2+\beta\nabla_4+\gamma\nabla_5+\mu\nabla_6+\nabla_8 \rangle. $

    Summarizing all cases we have the following distinct orbits:

    $1+α2+β4+γ5+μ6+8O(α,β,γ,μ)=O(±α,±η45β,η25γ,η45μ)=O(±α,η35β,η45γ,η35μ)=O(±α,±η25β,η5γ,η25μ)=O(±α,η5β,η35γ,η5μ),1+3+α4+β6O(α,β)=O(α,β)=O(α,β)=O(α,β),1+α3+β4+7O(α,β)=O(η3α,η23β)=O(η23α,η3β),2+α3+5+β6+γ7O(α,β,γ)=O(α,β,γ)=O(η3α,η23β,γ)=O(η3α,η23β,γ)=O(η23α,η3β,γ)=O(η23α,η3β,γ),2+α4+β5+γ6+8O(α,β,γ)=O(α,β,γ)=O(η45α,η25β,η45γ)=O(η45α,η25β,η45γ)=O(η35α,η45β,η35γ)=O(η35α,η45β,η35γ)=O(η25α,η5β,η25γ)=O(η25α,η5β,η25γ)=O(η5α,η35β,η5γ)=O(η5α,η35β,η5γ),3,3+4+α6,3+α4+7O(α)=O(α),3+5+α6+β7O(α,β)=O(α,β),3+6,4+α5+β6+8O(α,β)=O(iα,β)=O(iα,β)=O(α,β),4+7,5+6+α7O(α,β)=O(α,β),5+α6+8,5+α7O(α)=O(α),6+8,7,8,$

    which gives the following new algebras:

    $ Nα,β,γ,μ150:e1e1=e2e1e2=e5e1e3=αe5e2e2=e4e2e3=βe5e2e4=γe5e3e3=e4+μe5e4e4=e5Nα,β151:e1e1=e2e1e2=e5e1e4=e5e2e2=e4e2e3=αe5e3e3=e4+βe5Nα,β152:e1e1=e2e1e2=e5e1e4=αe5e2e2=e4e2e3=βe5e3e3=e4e3e4=e5Nα,β,γ153:e1e1=e2e1e3=e5e1e4=αe5e2e2=e4e2e4=e5e3e3=e4+βe5e3e4=γe5Nα,β,γ154:e1e1=e2e1e3=e5e2e2=e4e2e3=αe5e2e4=βe5e3e3=e4+γe5e4e4=e5N155:e1e1=e2e1e4=e5e2e2=e4e3e3=e4Nα156:e1e1=e2e1e4=e5e2e2=e4e2e3=e5e3e3=e4+αe5Nα157:e1e1=e2e1e4=e5e2e2=e4e2e3=αe5e3e3=e4e3e4=e5Nα,β158:e1e1=e2e1e4=e5e2e2=e4e2e4=e5e3e3=e4+αe5e3e4=βe5N159:e1e1=e2e1e4=e5e2e2=e4e3e3=e4+e5Nα,β160:e1e1=e2e2e2=e4e2e3=e5e2e4=αe5e3e3=e4+βe5e4e4=e5N161:e1e1=e2e2e2=e4e2e3=e5e3e3=e4e3e4=e5Nα162:e1e1=e2e2e2=e4e2e4=e5e3e3=e4+e5e3e4=αe5Nα163:e1e1=e2e2e2=e4e2e4=e5e3e3=e4+αe5e4e4=e5Nα164:e1e1=e2e2e2=e4e2e4=e5e3e3=e4e3e4=αe5N165:e1e1=e2e2e2=e4e3e3=e4+e5e4e4=e5N166:e1e1=e2e2e2=e4e3e3=e4e3e4=e5N167:e1e1=e2e2e2=e4e3e3=e4e4e4=e5 $

    Here we will collect all information about $ {\mathbf N}_{13}^{4*}(\lambda): $

    $ N413(λ)e1e1=e2e1e2=e3e1e3=e4e2e2=λe4H2D(N413(2))=[Δ22],4[Δ23]+[Δ14],[Δ24],H2C(N413(2))=H2D(N413(2))[Δ23],[Δ33],[Δ34],[Δ44]H2D(N413(λ)λ2)=[Δ22],(3λ2)[Δ23]+[Δ14],H2C(N413(λ)λ2)=H2D(N413(λ)[Δ23],[Δ24],[Δ33],[Δ34],[Δ44]ϕ=(x000yx200z2xyx30tλy2+2xz(λ+2)x2yx4) $

    Let us use the following notations:

    $ 1=[Δ14]+(3λ2)[Δ23],2=[Δ22],3=[Δ23],4=[Δ24],5=[Δ33],6=[Δ34],7=[Δ44]. $

    Take $ \theta = \sum\limits_{i = 1}^{7}\alpha_i\nabla_i\in\mathrm{H}^2_{\mathfrak{C}}({\mathbf N}_{13}^{4*}(\lambda)) . $ Since

    $\phi^T(000α10α2(3λ2)α1+α3α40(3λ2)α1+α3α5α6α1α4α6α7)\phi = (αααα1αα2+λα(3λ2)α1+α3α4α(3λ2)α1+α3α5α6α1α4α6α7),$

    we have

    $ α1=(α1x+α4y+α6z+α7t)x4,α2=α2x4+4λ(α6y+α7z)xy2+λ2α7y4+4(α4z+(α3+(3λ2)α1)y)x3+2(4α6yz+2α7z2+(2α5+λα4)y2)x2λ((λ+2)(α4y+α6z+α7t)y+((α3+4λα1)y+α5z+α6t)x)x2,α3=[(λ+2)(α4x2+2α6xy+2α7xz+λα7y2)y+((α3+(3λ2)α1)x2+2α5xy+2α6xz+λα6y2)x]x2(3λ2)(α1x+α4y+α6z+α7t)x4,α4=(α4x2+2α6xy+2α7xz+λα7y2)x4,α5=(α5x2+2(λ+2)α6xy+(λ+2)2α7y2)x4,α6=(α6x+(λ+2)α7y)x6,α7=α7x8. $

    We are interested in

    $ (\alpha_3,\alpha_4,\alpha_5,\alpha_6,\alpha_7)\neq(0,0,0,0,0) {\text{ and }}(\alpha_1,\alpha_4,\alpha_6,\alpha_7)\neq(0,0,0,0).$

    Let us consider the following cases:

    $ 1.\ $$ \alpha_7 = 0, \alpha_6 = 0, \alpha_5 = 0, \alpha_4 = 0, $ then $ \alpha_1\neq0, $ $ \alpha_3\neq0 $ and

    $ {\rm{(a)}}\ $ if $ \lambda\notin \{ 1,2,4\}, $ $ (\lambda-4)\alpha_3\neq4(1-\lambda)(\lambda-2)\alpha_1, $ then by choosing $ y = \frac{\alpha_2x}{(\lambda-4)\alpha_3+4(\lambda-1)(\lambda-2)\alpha_1}, $ we have the family of representatives

    $ \langle \alpha\nabla_1+\nabla_3 \rangle_{\alpha\notin \Big\{ 0,\frac{(\lambda-4)}{4(1-\lambda)(\lambda-2) }\Big\};\, \lambda\neq 1,2,4}; $

    ${\rm{(b)}}\ $if $ \lambda\notin \{ 1,2,4\}, $ $ (\lambda-4)\alpha_3 = 4(1-\lambda)(\lambda-2)\alpha_1, $ $ \alpha_2 = 0, $ then we have the family of representatives $ \langle \frac{\lambda-4}{4(1-\lambda)(\lambda-2)}\nabla_1+\nabla_3 \rangle_{\lambda\neq 1,2,4}, $ which we will be jointed with the family from the case (1a);

    $ {\rm{(c)}}\ $if $ \lambda\notin \{ 1,2,4\}, $ $ (\lambda-4)\alpha_3 = 4(1-\lambda)(\lambda-2)\alpha_1, $ $ \alpha_2\neq0, $ then by choosing $ x = \frac{\alpha_2}{\alpha_3}, y = 0, z = 0, t = 0, $ we have the family of representatives $ \langle (\lambda-4)\nabla_1+4(1-\lambda)(\lambda-2) (\nabla_2+\nabla_3) \rangle_{\lambda\neq 1,2,4}; $

    ${\rm{(d)}}\ $if $ \lambda\in \{ 1,2,4\}, $ then by choosing some suitable $ x $ and $ y $ we have the family of representatives $ \langle \alpha\nabla_1+\nabla_3 \rangle_{\alpha\neq 0, \lambda\in \{1,2,4\}}, $ which will be jointed with the family from the case (1a).

    2. $ \alpha_7 = 0, \alpha_6 = 0, \alpha_5 = 0, \alpha_4\neq0, $ then we have

    ${\rm{(a)}}\ $if $ \alpha_3 = 2(2-\lambda)\alpha_1 , $ then by choosing

    $x = 4 \alpha_4^2, y = -4 \alpha_1 \alpha_4, z = \alpha_1 \alpha_3 (4-\lambda)-\alpha_2 \alpha_4-\alpha_1^2 (8-12 \lambda+3 \lambda^2), t = 0,$

    we have the representative $ \langle \nabla_4 \rangle; $

    ${\rm{(b)}}\ $if $ \alpha_3\neq2(2-\lambda)\alpha_1 , $ then by choosing

    $ x=α3+2(λ2)α1α4,y=α1(α3+2(λ2)α1)α24,z=(2(2λ)α1α3)(α2α4+(λ4)α1α3+(3λ212λ+8)α21)4α34,t=0,$

    we have the representative $ \langle \nabla_3+\nabla_4 \rangle . $

    3. $ \alpha_7 = 0, \alpha_6 = 0, \alpha_5\neq0, $ then

    ${\rm{(a)}}\ $if $ \alpha_4 = 0, $ then $ \alpha_1\neq0 $ and

    $ {\rm{(i)}}\ $ if $ \lambda\neq0, $ then by choosing

    $ x = \frac{\alpha_1}{\alpha_5}, y = -\frac{\alpha_1\alpha_3}{2\alpha_5^2}, z = \frac{\alpha_1(2\alpha_2\alpha_5+(\lambda-2)\alpha_3^2+4(\lambda^2-3\lambda+2)\alpha_1\alpha_3)}{2\lambda\alpha_5^3}, t = 0, $

    we have the family of representatives $ \langle \nabla_1+\nabla_5 \rangle_{\lambda\neq0}; $

    ${\rm{(ii)}}\ $if $ \lambda = 0, $ then by choosing $ x = \frac{\alpha_1}{\alpha_5}, y = -\frac{\alpha_1\alpha_3}{2\alpha_5^2}, z = 0, t = 0, $ we have the family representative $ \langle \nabla_1+\alpha\nabla_2+\nabla_5 \rangle_{\alpha\neq0,\lambda = 0} $ and the representative $ \langle \nabla_1+\nabla_5 \rangle_{\lambda = 0}, $ which will be jointed with the family from the case (3(a)i).

    (b) if $ \alpha_4\neq0 $ and $ \lambda = 0, $ then we have the followings:

    ${\rm{(i)}}\ $if $ \alpha_3\alpha_4 = 2\alpha_1(2\alpha_4+\alpha_5), $ then by choosing

    $x = 4 \alpha_4^3,$ $y = -4 \alpha_1 \alpha_4^2,$ $z = 4 \alpha_1 \alpha_3 \alpha_4-\alpha_2 \alpha_4^2-4 \alpha_1^2 (2 \alpha_4+\alpha_5),$ $t = 0,$

    we have the family of representatives $ \langle \alpha\nabla_4+\nabla_5 \rangle_{\alpha\neq0,\lambda = 0}; $

    ${\rm{(ii)}}\ $if $ \alpha_3\alpha_4\neq 2\alpha_1(2\alpha_4+\alpha_5), $ then by choosing

    $x=α3α42α1(2α4+α5)α4α5),y=α1(2α1(2α4+α5)α3α4)α24α5),z=(2α1(2α4+α5)α3α4)(α2α244α1α3α4+4α21(2α4+α5))4α44α5,t=0,$

    we have the family of representatives $ \langle \nabla_3+\alpha\nabla_4+\nabla_5 \rangle_{\alpha\neq0, \lambda = 0} ; $

    (c) if $ \alpha_4\neq0 $ and $ \lambda\neq 0, $ then we have the followings:

    $ {\rm{(i)}}\ $ if $ 4\alpha_4 = \lambda\alpha_5, $ $ 4\lambda(\lambda-4)\alpha_1\alpha_3+\lambda^2\alpha_2\alpha_5+4(3\lambda^3-12\lambda^2+8\lambda+16)\alpha_1^2 = 0, $ $ \lambda\alpha_3+2(\lambda^2-2\lambda-4)\alpha_1 = 0, $ then by choosing $ x = 1, y = -\frac{\alpha_1}{\alpha_4}, z = 0, t = 0, $ we have the family of representatives $ \langle \frac{\lambda}{4}\nabla_4+\nabla_5 \rangle_{\lambda\neq0} ; $

    ${\rm{(ii)}}\ $ if $ 4\alpha_4 = \lambda\alpha_5, $ $ 4\lambda(\lambda-4)\alpha_1\alpha_3+\lambda^2\alpha_2\alpha_5+4(3\lambda^3-12\lambda^2+8\lambda+16)\alpha_1^2 = 0, $ $ \lambda\alpha_3+2(\lambda^2-2\lambda-4)\alpha_1\neq0, $ then by choosing

    $x = \frac{\lambda\alpha_3+2(\lambda^2-2\lambda-4)\alpha_1}{\lambda\alpha_5}, y = -\frac{4\alpha_1(\lambda\alpha_3+2(\lambda^2-2\lambda-4)\alpha_1)}{\lambda^2\alpha^2_5}, z = 0, t = 0,$

    we have the family of representatives $ \langle \nabla_3+\frac{\lambda}{4}\nabla_4+\nabla_5 \rangle_{\lambda\neq0} ; $

    $ {\rm{(iii)}}\ $if $ 4\alpha_4 = \lambda\alpha_5, $ $ 4\lambda(\lambda-4)\alpha_1\alpha_3+\lambda^2\alpha_2\alpha_5+4(3\lambda^3-12\lambda^2+8\lambda+16)\alpha_1^2\neq0, $ then by choosing

    $x=4λ(λ4)α1α3+λ2α2α5+4(3λ312λ2+8λ+16)α21λα5,y=4α14λ(λ4)α1α3+λ2α2α5+4(3λ312λ2+8λ+16)α21λ2α25,z=0,t=0,$

    we have the family of representatives $ \langle \nabla_2+\alpha\nabla_3+\frac{\lambda}{4}\nabla_4+\nabla_5 \rangle_{\lambda\neq0} ; $

    $ {\rm{(iv)}}\ $if $ \lambda\neq0, 4\alpha_4\neq\lambda\alpha_5 , $ then by choosing

    $ y = -\frac{\alpha_1}{\alpha_4}x, z = -\frac{\alpha_2\alpha_4^2+(\lambda-4)\alpha_1\alpha_3\alpha_4+\alpha_1^2(4\alpha_5+(3\lambda^2-12\lambda+8)\alpha_4)}{\alpha^2_4(4\alpha_4-\lambda\alpha_5)}x, t = 0,$

    we have two families of representatives

    $ \langle \alpha\nabla_4+\nabla_5 \rangle_{\alpha\neq\frac{\lambda}{4}} {\text{ and }} \langle \nabla_3+\alpha\nabla_4+\nabla_5 \rangle_{\alpha\neq\frac{\lambda}{4}} $

    depending on $ \alpha_3\alpha_4-2\alpha_1\alpha_5+2(\lambda-2)\alpha_1\alpha_4 = 0 $ or not. These families will be jointed with representatives from cases (3(c)i) and (3(c)ii).

    4. $ \alpha_7 = 0, \alpha_6\neq0, $ then by choosing $ y = -\frac{\alpha_4}{2\alpha_6}x, z = -\frac{\alpha_4^2-2\alpha_1\alpha_6}{2\alpha^2_6}x, $ we have $ \alpha_1^* = 0, \alpha_4^* = 0. $ Since we can suppose that $ \alpha_1 = 0, \alpha_4 = 0 $ and

    $ {\rm{(a)}}\ $ if $ \lambda\neq0, \alpha_3 = 0, $ then by choosing $ t = {\frac{\alpha_2}{\lambda\alpha_6}}x, $ we have the representatives $ \langle \nabla_6 \rangle _{\lambda\neq0} $ and $ \langle \nabla_5+\nabla_6 \rangle _{\lambda\neq0} $ depending on $ \alpha_5 = 0 $ or not;

    ${\rm{(b)}}\ $ if $ \lambda\neq0, \alpha_3\neq0, $ then by choosing $ x = \sqrt{\frac{\alpha_3}{\alpha_6}}, t = {\frac{\alpha_2\sqrt{\alpha_3}}{\lambda\alpha_6\sqrt{\alpha_6}}}, $ we have the family of representatives $ \langle \nabla_3+\alpha\nabla_5+\nabla_6 \rangle _{\lambda\neq0}; $

    ${\rm{(c)}}\ $if $ \lambda = 0, \alpha_2 = 0, \alpha_3 = 0, $ then we have the representatives $ \langle \nabla_6 \rangle _{\lambda = 0} $ and $ \langle \nabla_5+\nabla_6 \rangle _{\lambda = 0} $ depending on $ \alpha_5 = 0 $ or not, which will be jointed with representatives from the case (4a);

    $ {\rm{(d)}}\ $if $ \lambda = 0, \alpha_2 = 0, \alpha_3\neq0, $ then by choosing $ x = \sqrt{\frac{\alpha_3}{\alpha_6}}, t = 0 , $ we have the family of representatives $ \langle \nabla_3+\alpha\nabla_5+\nabla_6 \rangle _{(\lambda = 0)}, $ which will be jointed with the family of representatives from the case (4b);

    $ {\rm{(e)}}\ $if $ \lambda = 0, \alpha_2\neq0, $ then by choosing $ x = \sqrt[3]{\frac{\alpha_2}{\alpha_6}}, t = 0, $ we have the family of representatives $ \langle \nabla_2+\alpha\nabla_3+\beta\nabla_5+\nabla_6 \rangle _{\lambda = 0}. $

    5. $ \alpha_7\neq0, \lambda\neq-2, $ then by choosing

    $y=α6α7(λ+2)x,z=2(λ+2)2α4α7(λ+4)α262(λ+2)2α27x,t=(λ2+6λ+8)α4α6α72(λ+2)2α1α27(λ+4)α362(λ+2)2α37x,$

    we have $ \alpha_1^* = 0, \alpha_4^* = 0, \alpha_6^* = 0. $ Now we can suppose that $ \alpha_1 = 0, \alpha_4 = 0, \alpha_6 = 0 $ then we have

    $ {\rm{(a)}}\ $ if $ \alpha_3 = 0, \alpha_5 = 0, \alpha_2 = 0, $ then we have the representative $ \langle \nabla_7 \rangle_{\lambda\neq-2} ; $

    $ {\rm{(b)}}\ $ if $ \alpha_3 = 0, \alpha_5 = 0, \alpha_2\neq0, $ then by choosing $ x = \sqrt[4]{\alpha_2\alpha_7^{-1}}, $ we have the representative $ \langle \nabla_2+\nabla_7 \rangle_{\lambda\neq-2} ; $

    $ {\rm{(c)}}\ $ if $ \alpha_5\neq0, $ then by choosing $ x = \sqrt{\alpha_5\alpha_7^{-1}}, $ $ y = -\frac{\alpha_3}{2 \sqrt{\alpha_5 \alpha_7}}, $ $ z = \frac{\alpha_3^2}{ 4 \sqrt{\alpha_5^3\alpha_7}}, $ $ t = 0, $ we have the family of representatives $ \langle \alpha\nabla_2+\nabla_5+\nabla_7 \rangle_{\lambda\neq-2} . $

    ${\rm{(d)}}\ $ if $ \alpha_3\neq0, \alpha_5 = 0 $ then by choosing $ x = \sqrt[3]{\alpha_3\alpha_7^{-1}}, $ we have the family of representatives $ \langle \alpha\nabla_2+\nabla_3+ \nabla_7 \rangle_{\lambda\neq-2} . $

    6. $ \alpha_7\neq0, \lambda = -2, $ then

    $ {\rm{(a)}}\ $if $ \alpha_6 = 0, \alpha_5 = 0, $ then by choosing $ z = \frac{y^2}{x}-\frac{\alpha_4x}{2\alpha_7}, t = -\frac{x \alpha_1+y \alpha_4}{\alpha_7}, $ we have $ \alpha_1^* = 0 $ and $ \alpha_4^* = 0. $ Now consider the followings:

    $ {\rm{(i)}}\ $if $ \alpha_3 = 8\alpha_1, \alpha_2\alpha_7-\alpha_4^2 = 0, $ then we have the representative $ \langle \nabla_7 \rangle_{\lambda = -2}, $ which will be jointed with the representative from the case (5a);

    $ {\rm{(ii)}}\ $if $ \alpha_3 = 8\alpha_1, \alpha_2\alpha_7-\alpha_4^2\neq0, $ then by choosing $ x = \sqrt[4]{\frac{\alpha_2\alpha_7-\alpha_4^2}{\alpha_7^2}}, y = 0, $ we have the representative $ \langle \nabla_2+\nabla_7 \rangle_{\lambda = -2}, $ which will be jointed with the representative from the case (5b);

    ${\rm{(iii)}}\ $if $ \alpha_3\neq8\alpha_1, $ then by choosing $ x = \sqrt[3]{\frac{\alpha_3-8\alpha_1}{\alpha_7}}, y = \frac{\alpha_2\alpha_7-\alpha_4^2}{48\alpha_1\alpha_7-6\alpha_3\alpha_7}x, $ we have the representative $ \langle \nabla_3+\nabla_7 \rangle_{\lambda = -2} . $

    (b) if $ \alpha_6 = 0, \alpha_5\neq0, $ then by choosing

    $x = \sqrt{\frac{\alpha_5}{\alpha_7}}, y = \frac{8\alpha_1-\alpha_3}{2\sqrt{\alpha_5\alpha_7}}, z = \frac{\alpha_7(\alpha_3-8\alpha_1)^2-2\alpha_4\alpha_5^2}{4\alpha_5\alpha_7\sqrt{\alpha_5\alpha_7}}, t = \frac{\alpha_3\alpha_4-2\alpha_1(4\alpha_4+\alpha_5)}{2\alpha_7\sqrt{\alpha_5\alpha_7}}, $

    we have the family of representatives $ \langle \alpha\nabla_2+\nabla_5+\nabla_7 \rangle_{\lambda = -2}, $ which will be jointed with the representative from the case (5c).

    (c) if $ \alpha_6\neq0, $ then we have the following cases:

    $ {\rm{(i)}}\ $ if $ \alpha_5\alpha_7 = \alpha_6^2, 8\alpha_1\alpha_7+\alpha_4\alpha_6 = \alpha_3\alpha_7, $ then by choosing

    $ x = \frac{\alpha_6}{\alpha_7}, y = 0, z = -\frac{\alpha_4\alpha_6}{2\alpha_7^2}, t = \frac{\alpha_6(\alpha_4\alpha_6-2\alpha_1\alpha_7)}{2\alpha_7^3}, $

    we have the family of representatives $ \langle \alpha\nabla_2+\nabla_5+\nabla_6+\nabla_7 \rangle_{\lambda = -2}; $

    $ {\rm{(ii)}}\ $if $ \alpha_5\alpha_7 = \alpha_6^2, 8\alpha_1\alpha_7+\alpha_4\alpha_6\neq \alpha_3\alpha_7, $ then by choosing

    $x=α6α7,y=α6(α2α7α242α1α6)6α7(α4α6+8α1α7α3α7)),z=y2xα4x2α7α6yα7,t=xα1+yα4+zα6α7,$

    we have the family of representatives

    $ \langle \alpha\nabla_3+\nabla_5+\nabla_6+\nabla_7 \rangle_{\alpha\neq0,\lambda = -2}; $

    ${\rm{(iii)}}\ $if $ \alpha_5\alpha_7-\alpha_6^2\neq0, $ then by choosing

    $x=α6α7,y=α6(α4α6+8α1α7α3α7)2α7(α26+α5α7),z=y2xα42α7xα6α7y,t=(α4α62α1α7)x22α6α7y2+2(α26α4α7)xy2α27x,$

    we have the family of representatives

    $ \langle \alpha\nabla_2+\beta\nabla_5+\nabla_6+\nabla_7 \rangle_{\beta\neq1,\lambda = -2},$

    which will be jointed with the family from the case (6(c)i).

    Summarizing all cases we have the following distinct orbits:

    $(λ4)1+4(1λ)(λ2)(2+3)λ{1;2;4},1+α2+5λ=0,α0,α1+3α0,1+5,2+α3+λ44+5O(α)=O(α)λ0,2+α3+β5+6O(α,β)=O(η23α,ηβ)=O(η3α,η23β)λ=0,α2+3+7O(α)=O(η3α)=O(η23α)λ2,α2+β5+6+7λ=2,α2+5+7,2+7,3+4,3+α4+5α0,3+α5+6,α3+5+6+7α0,λ=2,3+7λ=2,4λ2,α4+5α0,5+6,6,7.$

    Now we have the following new algebras

    $ Nλ1;2;4168:e1e1=e2e1e2=e3e1e3=e4e1e4=(λ4)e5e2e2=λe4+4(1λ)(λ2)e5e2e3=λ(λ+2)e5Nα0169:e1e1=e2e1e2=e3e1e3=e4e1e4=e5e2e2=αe5e2e3=2e5e3e3=e5Nλ,α0170:e1e1=e2e1e2=e3e1e3=e4e1e4=αe5e2e2=λe4e2e3=(1+α(3λ2))e5Nλ171:e1e1=e2e1e2=e3e1e3=e4e1e4=e5e2e2=λe4e2e3=(3λ2)e5e3e3=e5Nλ0,α172:e1e1=e2e1e2=e3e1e3=e4e2e2=λe4+e5e2e3=αe5e2e4=λ4e5e3e3=e5Nα,β173:e1e1=e2e1e2=e3e1e3=e4e2e2=e5e2e3=αe5e3e3=βe5e3e4=e5Nλ2,α174:e1e1=e2e1e2=e3e1e3=e4e2e2=λe4+αe5e2e3=e5e4e4=e5Nα,β175:e1e1=e2e1e2=e3e1e3=e4e2e2=2e4+αe5e3e3=βe5e3e4=e5e4e4=e5Nλ,α176:e1e1=e2e1e2=e3e1e3=e4e2e2=λe4+αe5e3e3=e5e4e4=e5Nλ177:e1e1=e2e1e2=e3e1e3=e4e2e2=λe4+e5e4e4=e5Nλ178:e1e1=e2e1e2=e3e1e3=e4e2e2=λe4e2e3=e5e2e4=e5Nλ,α0179:e1e1=e2e1e2=e3e1e3=e4e2e2=λe4e2e3=e5e2e4=αe5e3e3=e5Nλ,α180:e1e1=e2e1e2=e3e1e3=e4e2e2=λe4e2e3=e5e3e3=αe5e3e4=e5Nα0181:e1e1=e2e1e2=e3e1e3=e4e2e2=2e4e2e3=αe5e3e3=e5e3e4=e5e4e4=e5N182:e1e1=e2e1e2=e3e1e3=e4e2e2=2e4e2e3=e5e4e4=e5Nλ2183:e1e1=e2e1e2=e3e1e3=e4e2e2=λe4e2e4=e5Nλ,α0184:e1e1=e2e1e2=e3e1e3=e4e2e2=λe4e2e4=αe5e3e3=e5Nλ185:e1e1=e2e1e2=e3e1e3=e4e2e2=λe4e3e3=e5e3e4=e5Nλ186:e1e1=e2e1e2=e3e1e3=e4e2e2=λe4e3e4=e5Nλ187:e1e1=e2e1e2=e3e1e3=e4e2e2=λe4e4e4=e5 $

    Here we will collect all information about $ {\mathbf N}_{14}^{4*}: $

    $ \begin{array}{|l|l|l|l|}   \hline {\mathbf{N}}^{4*}_{14} & \begin{array}{l} e_1e_2 = e_3 \\ e_1e_3 = e_4 \end{array} & H2D(N414)=[Δ11],[Δ22],[Δ23],[Δ33]H2C(N414)=H2D(N414)[Δ14],[Δ24],[Δ34],[Δ44] & \phi = (x0000q000rxq0tsxrx2q)\\ \hline \end{array} $

    Let us use the following notations:

    $ 1=[Δ11],2=[Δ14],3=[Δ22],4=[Δ23],5=[Δ24],6=[Δ33],7=[Δ34],8=[Δ44]. $

    Take $ \theta = \sum\limits_{i = 1}^{8}\alpha_i\nabla_i\in\mathrm{H}^2_{\mathfrak{C}}(\mathbf{N}^{4*}_{14}) . $ Since

    $ \phi^T(α100α20α3α4α50α4α6α7α2α5α7α8)\phi = (α1ααα2αα3α4α5αα4α6α7α2α5α7α8), $

    we have

    $ α1=α1x2+2α2xt+α8t2,α2=(α2x+α8t)x2q,α3=(α3q+α4r+α5s)q+(α4q+α6r+α7s)r+(α5q+α7r+α8s)s,α4=(α4q+α6r+α7s)xq+(α5q+α7r+α8s)xr,α5=(α5q+α7r+α8s)x2q,α6=(α6q2+2α7qr+α8r2)x2,α7=(α7q+α8r)x3q,α8=α8x4q2. $

    We are interested in $ (\alpha_2,\alpha_5,\alpha_7,\alpha_8)\neq(0,0,0,0) . $ Let us consider the following cases:

    $1.\ $$ \alpha_8 = 0, \alpha_7 = 0, \alpha_5 = 0, $ then $ \alpha_2\neq0 $ and we have

    $ {\rm{(a)}}\ $if $ \alpha_6 = 0, \alpha_4 = 0, \alpha_3 = 0, $ then by choosing $ x = 2\alpha_2, q = 1, r = 0, s = 0, t = -\alpha_1, $ we have the representative $ \langle \nabla_2 \rangle; $

    ${\rm{(b)}}\ $if $ \alpha_6 = 0, \alpha_4 = 0, \alpha_3\neq0, $ then by choosing $ x = \alpha_3, q = \alpha_2\alpha_3^2, r = 0, s = 0, t = -\frac{\alpha_1\alpha_3}{2\alpha_2}, $ we have the representative $ \langle \nabla_2+\nabla_3 \rangle; $

    $ {\rm{(c)}}\ $if $ \alpha_6 = 0, \alpha_4\neq0, $ then by choosing $ x = \alpha_4, q = \alpha_2\alpha_4, r = -\frac{\alpha_2\alpha_3}{2}, s = 0, t = -\frac{\alpha_1\alpha_4}{2\alpha_2}, $ we have the representative $ \langle \nabla_2+\nabla_4 \rangle; $

    ${\rm{(d)}}\ $if $ \alpha_6\neq0, \alpha_3\alpha_6-\alpha_4^2 = 0, $ then by choosing $ x = \alpha_6, q = \alpha_2, r = -\frac{\alpha_2\alpha_4}{\alpha_6}, s = 0, t = -\frac{\alpha_1\alpha_6}{2\alpha_2}, $ we have the representative $ \langle \nabla_2+\nabla_6 \rangle; $

    $ {\rm{(e)}}\ $if $ \alpha_6\neq0, \alpha_3\alpha_6-\alpha_4^2\neq0, $ then by choosing

    $x=α3α6α24α6,q=α2α3α6α24α26,r=α2α4α3α6α24α36,s=0,t=α1α3α6α242α2α6,$

    we have the representative $ \langle \nabla_2+\nabla_3+\nabla_6 \rangle. $

    2. $ \alpha_8 = 0, \alpha_7 = 0, \alpha_5\neq0, $ then we have

    $ {\rm{(a)}}\ $ if $ \alpha_6 = 0, \alpha_2 = 0, $ then by choosing

    $x = 1, r = -\frac{\alpha_4}{\alpha_5}q, s = \frac{2\alpha_4^2-\alpha_3\alpha_5}{2\alpha_5^2}q, t = 0, $

    we have the representatives $ \langle \nabla_5 \rangle $ and $ \langle \nabla_1+\nabla_5 \rangle $ depending on whether $ \alpha_1 = 0 $ or not;

    ${\rm{(b)}}\ $ if $ \alpha_6 = 0, \alpha_2\neq0, $ then by choosing

    $ x = \alpha_5, q = \alpha_2, r = -\frac{\alpha_2\alpha_4}{\alpha_5}, s = \frac{\alpha_2(2\alpha_4^2-\alpha_3\alpha_5)}{2\alpha_5^2}, t = -\frac{\alpha_1\alpha_5}{2\alpha_2}, $

    we have the representatives $ \langle \nabla_2+\nabla_5 \rangle; $

    $ {\rm{(c)}}\ $if $ \alpha_6\neq0, \alpha_5 = -\alpha_6, $ then we have the following subcases:

    $ {\rm{(i)}}\ $ if $ \alpha_2 = 0, \alpha_4 = 0, \alpha_1 = 0, $ then we have the representative $ \langle \nabla_5-\nabla_6 \rangle; $

    ${\rm{(ii)}}\ $ if $ \alpha_2 = 0, \alpha_4 = 0, \alpha_1\neq0, $ then by choosing

    $ x = 1, q = \sqrt{\frac{\alpha_1}{\alpha_5}}, r = 0, s = -\frac{\alpha_3\sqrt{\alpha_1}}{2\alpha_5\sqrt{\alpha_5}}, t = 0,$

    we have the representative $ \langle \nabla_1+\nabla_5-\nabla_6 \rangle; $

    $ {\rm{(iii)}}\ $if $ \alpha_2 = 0, \alpha_4\neq0, \alpha_1 = 0, $ then by choosing

    $ x = {\frac{\alpha_4}{\alpha_5}}, q = 1, r = 0, s = -\frac{\alpha_3}{2\alpha_5}, t = 0, $

    we have the representative $ \langle \nabla_4+\nabla_5-\nabla_6 \rangle; $

    $ {\rm{(iv)}}\ $if $ \alpha_2 = 0, \alpha_4\neq0, \alpha_1\neq0, $ then by choosing

    $ x = \frac{\alpha_4}{\alpha_5}, q = \sqrt{\frac{\alpha_1}{\alpha_5}}, r = 0, s = -\frac{\alpha_3\sqrt{\alpha_1}}{2\alpha_5\sqrt{\alpha_5}}, t = 0, $

    we have the representative $ \langle \nabla_1+\nabla_4+\nabla_5-\nabla_6 \rangle; $

    $ {\rm{(v)}}\ $ if $ \alpha_2\neq0, \alpha_4 = 0, $ then by choosing

    $ x = \alpha_5, q = \alpha_2, r = 0, s = -\frac{\alpha_2\alpha_3}{2\alpha_5}, t = -\frac{\alpha_1\alpha_5}{2\alpha_2},$

    we have the representative $ \langle \nabla_2+\nabla_5-\nabla_6 \rangle; $

    ${\rm{(vi)}}\ $if $ \alpha_2\neq0, \alpha_4\neq0, $ then by choosing

    $ x = \frac{\alpha_4}{\alpha_5}, q = {\frac{\alpha_2\alpha_4}{\alpha^2_5}}, r = 0, s = -\frac{\alpha_2\alpha_3\alpha_4}{2\alpha_5^3}, t = -\frac{\alpha_1\alpha_4}{2\alpha_2\alpha_5}, $

    we have the representative $ \langle \nabla_2+\nabla_4+\nabla_5-\nabla_6 \rangle. $

    (d) if $ \alpha_6\neq0, \alpha_5\neq-\alpha_6, $ then we have the following subcases:

    $ {\rm{(i)}}\ $if $ \alpha_2 = 0, \alpha_1 = 0, $ then by choosing

    $x = 1, q = 1, s = \frac{ \alpha_4^2 (2 \alpha_5+\alpha_6)-\alpha_3 (\alpha_5+\alpha_6)^2}{2 \alpha_5 (\alpha_5+\alpha_6)^2}, r = -\frac{\alpha_4}{\alpha_5+\alpha_6}, t = 0,$

    we have the family of representatives $ \langle \nabla_5+\alpha\nabla_6 \rangle_{\alpha\neq0,-1}, $ which will be jointed with the cases (2a) and (2(c)i);

    ${\rm{(ii)}}\ $if $ \alpha_2 = 0, \alpha_1\neq0, $ then by choosing

    $x=1,q=α1α5,r=α4α1(α5+α6)α5,s=(α24(2α5+α6)α3(α5+α6)2)α12α5(α5+α6)2α5,t=0,$

    we have the family of representatives $ \langle \nabla_1+\nabla_5+\alpha\nabla_6 \rangle_{\alpha\neq0,-1}, $ which will be jointed with the cases (2a) and (2(c)ii);

    $ {\rm{(iii)}}\ $if $ \alpha_2\neq0, $ then by choosing

    $ x = \alpha_5, q = \alpha_2,$ $r = -\frac{\alpha_2\alpha_4}{\alpha_5+\alpha_6},$ $s = \frac{\alpha_2(\alpha_4^2(2\alpha_5+\alpha_6)-\alpha_3(\alpha_5+\alpha_6)^2)}{2\alpha_5(\alpha_5+\alpha_6)^2}, t = -\frac{\alpha_1\alpha_5}{2\alpha_2},$

    we have the family of representatives $ \langle \nabla_2+\nabla_5+\alpha\nabla_6 \rangle_{\alpha\neq0,-1}, $ which will be jointed with the cases (2b) and (2(c)v).

    3. $ \alpha_8 = 0, \alpha_7\neq0 $ then by choosing $ r = -\frac{\alpha_5}{\alpha_7}q, s = \frac{\alpha_5\alpha_6-\alpha_4\alpha_7}{\alpha_7^2}q, $ we have $ \alpha_4^* = \alpha_5^* = 0. $ Therefore, we can suppose that $ \alpha_4 = 0, \alpha_5 = 0, $ thus we have

    $ {\rm{(a)}}\ $if $ \alpha_2 = 0, \alpha_1 = 0, \alpha_3 = 0, $ then we have the representatives $ \langle \nabla_7 \rangle $ and $ \langle \nabla_6+\nabla_7\rangle $ depending on whether $ \alpha_6 = 0 $ or not;

    $ {\rm{(b)}}\ $if $ \alpha_2 = 0, \alpha_1 = 0, \alpha_3\neq0, $ then by choosing $ x = \sqrt[3]{\frac{\alpha_3}{\alpha_7}}, q = 1, r = 0, s = 0, t = 0, $ we have the family of representatives $ \langle \nabla_3+\alpha\nabla_6+\nabla_7\rangle; $

    $ {\rm{(c)}}\ $if $ \alpha_2 = 0, \alpha_1\neq0, \alpha_3 = 0, $ then we have the representatives $ \langle \nabla_1+\nabla_7\rangle $ and $ \langle \nabla_1+\nabla_6+\nabla_7\rangle $ depending on whether $ \alpha_6 = 0 $ or not;

    ${\rm{(d)}}\ $if $ \alpha_2 = 0, \alpha_1\neq0, \alpha_3\neq0, $ then by choosing

    $x = \sqrt[3]{\frac{\alpha_3}{\alpha_7}}, q = \sqrt[6]{\frac{\alpha^3_1}{\alpha_3\alpha^2_7}}, r = 0, s = 0, t = 0, $

    we have the family of representatives $ \langle \nabla_1+\nabla_3+\alpha\nabla_6+\nabla_7\rangle; $

    $ {\rm{(e)}}\ $if $ \alpha_2\neq0, \alpha_3 = 0, $ then by choosing $ q = \frac{\alpha_2}{\alpha_7}, r = 0, s = 0, t = -\frac{\alpha_1}{2\alpha_2}x, $ we have the representatives $ \langle \nabla_2+\nabla_7 \rangle $ and $ \langle \nabla_2+\nabla_6+\nabla_7\rangle $ depending on whether $ \alpha_6 = 0 $ or not;

    $ {\rm{(f)}}\ $if $ \alpha_2\neq0, \alpha_3\neq0, $ then by choosing $ x = \sqrt[3]{\frac{\alpha_3}{\alpha_7}}, q = \frac{\alpha_2}{\alpha_7}, r = 0, s = 0, t = -\frac{\alpha_1\sqrt[3]{\alpha_3}}{2\alpha_2\sqrt[3]{\alpha_7}} $ we have the family of representatives $ \langle \nabla_2+\nabla_3+\alpha\nabla_6+\nabla_7\rangle. $

    4. $ \alpha_8\neq0 $ then by choosing $ r = -\frac{\alpha_7}{\alpha_8}q, s = \frac{\alpha_7^2-\alpha_5\alpha_8}{\alpha_8^2}q, t = -\frac{\alpha_2}{\alpha_8}x, $ we have $ \alpha_2^* = \alpha_5^* = \alpha_7^* = 0. $ Therefore, we can suppose that $ \alpha_2 = 0, \alpha_5 = 0, \alpha_7 = 0, $ then we have

    ${\rm{(a)}}\ $if $ \alpha_1 = 0, \alpha_3 = 0, \alpha_4 = 0, $ then we have the representatives $ \langle \nabla_8 \rangle $ and $ \langle \nabla_6+\nabla_8 \rangle $ depending on whether $ \alpha_6 = 0 $ or not;

    $ {\rm{(b)}}\ $if $ \alpha_1 = 0, \alpha_3 = 0, \alpha_4\neq0, $ then by choosing $ x = \sqrt[3]{\frac{\alpha_4}{\alpha_8}}, q = 1, r = 0, s = 0, t = 0, $ we have the family of representatives $ \langle \nabla_4+\alpha\nabla_6+\nabla_8 \rangle; $

    $ {\rm{(c)}}\ $if $ \alpha_1 = 0, \alpha_3\neq0, $ then by choosing $ x = \sqrt[4]{\frac{\alpha_3}{\alpha_8}}, q = 1, r = 0, s = 0, t = 0, $ we have the family of representatives $ \langle \nabla_3+\alpha\nabla_4+\beta\nabla_6+\nabla_8 \rangle; $

    ${\rm{(d)}}\ $if $ \alpha_1\neq0, \alpha_3 = 0, \alpha_4 = 0, $ then we have the representatives $ \langle \nabla_1+\nabla_8 \rangle $ and $ \langle \nabla_1+\nabla_6+\nabla_8 \rangle $ depending on whether $ \alpha_6 = 0 $ or not;

    $ {\rm{(e)}}\ $if $ \alpha_1\neq0, \alpha_3 = 0, \alpha_4\neq0, $ then by choosing $ x = \sqrt[3]{\frac{\alpha_4}{\alpha_8}}, q = \sqrt[6]{\frac{\alpha_1^3}{\alpha_4^2\alpha_8}}, r = 0, s = 0, t = 0, $ we have the family of representative $ \langle \nabla_1+\nabla_4+\alpha\nabla_6+\nabla_8 \rangle; $

    ${\rm{(f)}}\ $if $ \alpha_1\neq0, \alpha_3\neq0, $ then by choosing $ x = \sqrt[4]{\frac{\alpha_3}{\alpha_8}}, q = \frac{\sqrt{\alpha_1}}{\sqrt[4]{\alpha_3\alpha_8}}, r = 0, s = 0, t = 0, $ we have the family of representative $ \langle \nabla_1+\nabla_3+\alpha\nabla_4+\beta\nabla_6+\nabla_8 \rangle. $

    Summarizing, we have the following distinct orbits:

    $1+3+α4+β6+8O(α,β)=O(iα,β)=O(iα,β)=O(α,β),1+3+α6+7O(α)=O(η3α)=O(η23α),1+4+56,1+4+α6+8O(α)=O(η3α)=O(η23α),1+5+α6,1+6+7,1+6+8,1+7,1+8,2,2+3,2+3+6,2+3+α6+7O(α)=O(η3α)=O(η23α),2+4,2+4+56,2+5+α6,2+6,2+6+7,2+7,3+α4+β6+8O(α,β)=O(iα,β)=O(iα,β)=O(α,β),3+α6+7O(α)=O(η3α)=O(η23α),4+56,4+α6+8O(α)=O(η3α)=O(η23α),5+α6,6+7,6+8,7,8,$

    which gives the following new algebras:

    $ Nα,β188:e1e1=e5e1e2=e3e1e3=e4e2e2=e5e2e3=αe5e3e3=βe5e4e4=e5Nα189:e1e1=e5e1e2=e3e1e3=e4e2e2=e5e3e3=αe5e3e4=e5N190:e1e1=e5e1e2=e3e1e3=e4e2e3=e5e2e4=e5e3e3=e5Nα191:e1e1=e5e1e2=e3e1e3=e4e2e3=e5e3e3=αe5e4e4=e5Nα191:e1e1=e5e1e2=e3e1e3=e4e2e4=e5e3e3=αe5N192:e1e1=e5e1e2=e3e1e3=e4e3e3=e5e3e4=e5N193:e1e1=e5e1e2=e3e1e3=e4e3e3=e5e4e4=e5N194:e1e1=e5e1e2=e3e1e3=e4e3e4=e5N195:e1e1=e5e1e2=e3e1e3=e4e4e4=e5N196:e1e2=e3e1e3=e4e1e4=e5N197:e1e2=e3e1e3=e4e1e4=e5e2e2=e5N198:e1e2=e3e1e3=e4e1e4=e5e2e2=e5e3e3=e5Nα199:e1e2=e3e1e3=e4e1e4=e5e2e2=e5e3e3=αe5e3e4=e5N200:e1e2=e3e1e3=e4e1e4=e5e2e3=e5N201:e1e2=e3e1e3=e4e1e4=e5e2e3=e5e2e4=e5e3e3=e5Nα202:e1e2=e3e1e3=e4e1e4=e5e2e4=e5e3e3=αe5N203:e1e2=e3e1e3=e4e1e4=e5e3e3=e5N204:e1e2=e3e1e3=e4e1e4=e5e3e3=e5e3e4=e5N205:e1e2=e3e1e3=e4e1e4=e5e3e4=e5Nα,β206:e1e2=e3e1e3=e4e2e2=e5e2e3=αe5e3e3=βe5e4e4=e5Nα207:e1e2=e3e1e3=e4e2e2=e5e3e3=αe5e3e4=e5N208:e1e2=e3e1e3=e4e2e3=e5e2e4=e5e3e3=e5Nα209:e1e2=e3e1e3=e4e2e3=e5e3e3=αe5e4e4=e5Nα210:e1e2=e3e1e3=e4e2e4=e5e3e3=αe5N211:e1e2=e3e1e3=e4e3e3=e5e3e4=e5N212:e1e2=e3e1e3=e4e3e3=e5e4e4=e5N213:e1e2=e3e1e3=e4e3e4=e5N214:e1e2=e3e1e3=e4e4e4=e5 $

    Here we will collect all information about $ {\mathbf N}_{15}^{4*}: $

    $ \begin{array}{|l|l|l|l|}   \hline {\mathbf{N}}^{4*}_{15} & \begin{array}{l} e_1e_2 = e_3 \\ e_1e_3 = e_4 \\ e_2e_2 = e_4 \end{array} & H2D(N415)=[Δ11],[Δ22],[Δ23],[Δ33]H2C(N415)=H2D(N415)[Δ14],[Δ24],[Δ34],[Δ44] & \phi = (x0000x2000rx30tsxrx4)\\ \hline \end{array} $

    Let us use the following notations:

    $ 1=[Δ11],2=[Δ14],3=[Δ22],4=[Δ23],5=[Δ24],6=[Δ33],7=[Δ34],8=[Δ44]. $

    Take $ \theta = \sum\limits_{i = 1}^{8}\alpha_i\nabla_i\in\mathrm{H}^2_{\mathfrak{C}}(\mathbf{N}^{4*}_{15}) . $ Since

    $ \phi^T(α100α20α3α4α50α4α6α7α2α5α7α8)\phi = (α1ααα2αα3+αα4α5αα4α6α7α2α5α7α8), $

    we have

    $ α1=α1x2+2α2xt+α8t2,α2=(α2x+α8t)x4,α3=x4α3+2rx2α4+2sx2α5+r2α6+2rsα7+s2α8x(rxα2+tx2α7+rtα8),α4=(α4x2+α6r+α7s)x3+(α5x2+α7r+α8s)xr,α5=(α5x2+α7r+α8s)x4,α6=(α6x4+2α7x2r+α8r2)x2,α7=(α7x2+α8r)x5,α8=α8x8. $

    We are interested in $ (\alpha_2,\alpha_5,\alpha_7,\alpha_8)\neq(0,0,0,0) . $ Let us consider the following cases:

    $ 1.\ $$ \alpha_8 = 0, \alpha_7 = 0, \alpha_5 = 0, $ then $ \alpha_2\neq0 $ and we have

    ${\rm{(a)}}\ $ if $ \alpha_6 = 0, \alpha_4 = 0, $ then by choosing $ x = 2\alpha_2, r = 4\alpha_2\alpha_3, s = 0, t = -\alpha_1, $ we have the representative $ \langle \nabla_2 \rangle; $

    $ {\rm{(b)}}\ $ if $ \alpha_6 = 0, \alpha_4\neq0, \alpha_2 = 2\alpha_4, $ then we have the representatives $ \langle 2\nabla_2+\nabla_4 \rangle $ and $ \langle 2\nabla_2+\nabla_3+\nabla_4 \rangle $ depending on whether $ \alpha_3 = 0 $ or not;

    ${\rm{(c)}}\ $if $ \alpha_6 = 0, \alpha_4\neq0, \alpha_2\neq2\alpha_4, $ then by choosing $ x = \alpha_2-2\alpha_4, r = \alpha_3(\alpha_2-2\alpha_4), s = 0, t = \frac{\alpha_1(2\alpha_4-\alpha_2)}{2\alpha_2}, $ we have the family of representatives $ \langle \nabla_2+\alpha\nabla_4 \rangle_{\alpha\neq0,\frac{1}{2}}, $ which will be jointed with representatives from the cases (1a) and (1b);

    $ {\rm{(d)}}\ $if $ \alpha_6\neq0, $ then by choosing $ x = \frac{\alpha_2}{\alpha_6}, r = -\frac{\alpha_2^2\alpha_4}{\alpha_6^3}, s = 0, t = -\frac{\alpha_1}{2\alpha_6}, $ we have the representative $ \langle \nabla_2+\alpha\nabla_3+\nabla_6 \rangle. $

    2. $ \alpha_8 = 0, \alpha_7 = 0, \alpha_5\neq0 $ then we have

    ${\rm{(a)}}\ $if $ \alpha_5\neq-\alpha_6, $ then we have the following subcases:

    $ {\rm{(i)}}\ $ if $ \alpha_2 = 0, \alpha_1 = 0, $ then by choosing

    $x=2α5(α5+α6),s=2α5(α24(2α5+α6)α3(α5+α6)2),r=4α4α25(α5+α6),$

    we have the family of representatives $ \langle \nabla_5+\alpha\nabla_6 \rangle_{\alpha\neq-1} ; $

    ${\rm{(ii)}}\ $ if $ \alpha_2 = 0, \alpha_1\neq0, $ then by choosing

    $ x = \sqrt[4]{\frac{\alpha_1}{\alpha_5}},$ $r = -\frac{\alpha_4\sqrt{\alpha_1}}{(\alpha_5+\alpha_6)\sqrt{\alpha_5}},$ $s = \frac{((\alpha_5+\alpha_6)(2\alpha_4^2-\alpha_2\alpha_4)-\alpha_3(\alpha_5+\alpha_6)^2-\alpha_4^2\alpha_6)\sqrt{\alpha_1}}{2\alpha_5(\alpha_5+\alpha_6)^2\sqrt{\alpha_5}}, t = 0, $

    we have the family of representatives $ \langle \nabla_1+\nabla_5+\alpha\nabla_6 \rangle_{\alpha\neq-1} ; $

    $ {\rm{(iii)}}\ $ if $ \alpha_2\neq0, $ then by choosing

    $x=α2α5,r=α22α4α25(α5+α6),s=α22((α5+α6)(2α24α2α4)α3(α5+α6)2α24α6)2α35(α5+α6)2,t=α12α5,$

    we have the family of representatives $ \langle \nabla_2+\nabla_5+\alpha\nabla_6 \rangle_{\alpha\neq-1} . $

    (b) if $ \alpha_6 = -\alpha_5, $ then we have the following subcases:

    $ {\rm{(i)}}\ $if $ \alpha_4 = 0, \alpha_2 = 0, \alpha_1 = 0, $ then we have the representative $ \langle \nabla_5-\nabla_6 \rangle, $ which will be jointed with the family from the case (2(a)i);

    $ {\rm{(ii)}}\ $if $ \alpha_4 = 0, \alpha_2 = 0, \alpha_1\neq0, $ then by choosing $ x = \sqrt[4]{\frac{\alpha_1}{\alpha_5}}, r = 0, s = -\frac{\alpha_3\sqrt{\alpha_1}}{2\alpha_5\sqrt{\alpha_4}}, t = 0, $ we have the representative $ \langle \nabla_1+\nabla_5-\nabla_6 \rangle, $ which will be jointed with the family from the case (2(a)ii);

    $ {\rm{(iii)}}\ $if $ \alpha_4 = 0, \alpha_2\neq0, $ then by choosing $ x = \frac{\alpha_2}{\alpha_5}, r = 0, s = -\frac{\alpha^2_2\alpha_3}{2\alpha^3_5}, t = -\frac{\alpha_1}{2\alpha_5}, $ we have the representative $ \langle \nabla_2+\nabla_5-\nabla_6 \rangle, $ which will be jointed with the family from the case (2(a)iii);

    $ {\rm{(iv)}}\ $if $ \alpha_4\neq0, $ then by choosing $ x = \frac{\alpha_4}{\alpha_5}, $ $ s = -\frac{\alpha_3 \alpha_4^2}{ 2 \alpha_5^3}, $ $ r = 0, $ we have the families of representatives

    $ \langle \alpha\nabla_1+\nabla_4+\nabla_5-\nabla_6 \rangle {\text{ and }} \langle \alpha\nabla_2+\nabla_4+\nabla_5-\nabla_6 \rangle_{\alpha\neq0} $

    depending on $ \alpha_2 = 0 $ or not.

    3. $ \alpha_8 = 0, \alpha_7\neq0, $ then by choosing

    $ r = -\frac{\alpha_5}{\alpha_7}x^2, s = \frac{\alpha_5\alpha_6-\alpha_4\alpha_7}{\alpha_7^2}x^2, t = \frac{\alpha_3\alpha_7^2-2\alpha_4\alpha_5\alpha_7+\alpha_5^2\alpha_6+\alpha_2\alpha_5\alpha_7}{\alpha_7^3}x,$

    we have $ \alpha_3^* = \alpha_4^* = \alpha_5^* = 0. $ Therefore, we can suppose that $ \alpha_3 = 0, \alpha_4 = 0, \alpha_5 = 0, $ and we have

    $ {\rm{(a)}}\ $if $ \alpha_1 = 0, \alpha_2 = 0, $ then we have the representatives $ \langle \nabla_7 \rangle $ and $ \langle \nabla_6+\nabla_7 \rangle $ depending on whether $ \alpha_6 = 0 $ or not;

    $ {\rm{(b)}}\ $if $ \alpha_1 = 0, \alpha_2\neq0, $ then by choosing $ x = \sqrt{\alpha_2\alpha_7^{-1}}, r = 0, s = 0, t = 0, $ we have the family of representative $ \langle \nabla_2+\alpha\nabla_6+\nabla_7 \rangle; $

    $ {\rm{(c)}}\ $if $ \alpha_1\neq0, $ then by choosing $ x = \sqrt[5]{\frac{\alpha_1}{\alpha_7}}, r = 0, s = 0, t = 0, $ we have the family of representative $ \langle \nabla_1+\alpha\nabla_2+\beta\nabla_6+\nabla_7 \rangle. $

    4. $ \alpha_8\neq0, $ then by choosing $ r = -\frac{\alpha_7}{\alpha_8}x^2, t = -\frac{\alpha_2}{\alpha_8}x, s = -\frac{\alpha_5x^2+\alpha_7r}{\alpha_8} $ we have $ \alpha_2^* = \alpha_5^* = \alpha_7^* = 0. $ Therefore, we can suppose that $ \alpha_2 = 0, \alpha_5 = 0, \alpha_7 = 0, $ then we have

    $ {\rm{(a)}}\ $if $ \alpha_1 = 0, \alpha_3 = 0, \alpha_4 = 0, $ then we have the representatives $ \langle \nabla_8 \rangle $ and $ \langle \nabla_6+\nabla_8 \rangle $ depending on whether $ \alpha_6 = 0 $ or not;

    $ {\rm{(b)}}\ $if $ \alpha_1 = 0, \alpha_3 = 0, \alpha_4\neq0, $ then by choosing $ x = \sqrt[3]{{\alpha_4}{\alpha_8^{-1}}}, r = 0, s = 0, t = 0, $ we have the family of representative $ \langle \nabla_4+\alpha\nabla_6+\nabla_8 \rangle; $

    $ {\rm{(c)}}\ $if $ \alpha_1 = 0, \alpha_3\neq0, $ then by choosing $ x = \sqrt[4]{{\alpha_3}{\alpha_8^{-1}}}, r = 0, s = 0, t = 0, $ we have the family of representative $ \langle \nabla_3+\alpha\nabla_4+\beta\nabla_6+\nabla_8 \rangle; $

    $ {\rm{(d)}}\ $if $ \alpha_1\neq0, $ then by choosing $ x = \sqrt[6]{{\alpha_1}{\alpha_8^{-1}}}, r = 0, s = 0, t = 0, $ we have the family of representative $ \langle \nabla_1+\alpha\nabla_3+\beta\nabla_4+\gamma\nabla_6+\nabla_8 \rangle. $

    Summarizing all cases we have the following distinct orbits:

    $ 1+α2+β6+7O(α,β)=O(η25α,η5β)=O(η45α,η25β)=O(η5α,η35β)=O(η35α,η45β),1+α3+β4+γ6+8O(α,β,γ)=O(η3α,β,η23γ)=O(η3α,β,η23γ)=O(η23α,β,η3γ)=O(η23α,β,η3γ)=O(α,β,γ),α1+4+56,1+5+α6,22+3+4,2+α3+6,2+α4,α2+4+56α0,2+5+α6,2+α6+7O(α)=O(α),3+α4+β6+8O(α,β)=O(iα,β)=O(iα,β)=O(α,β),4+α6+8O(α)=O(η3α)=O(η23α),5+α6,6+7,6+8,7,8,$

    which gives the following new algebras:

    $ Nα,β215:e1e1=e5e1e2=e3e1e3=e4e1e4=αe5e2e2=e4e3e3=βe5e3e4=e5Nα,β,γ216:e1e1=e5e1e2=e3e1e3=e4e2e2=e4+αe5e2e3=βe5e3e3=γe5e4e4=e5Nα217:e1e1=αe5e1e2=e3e1e3=e4e2e2=e4e2e3=e5e2e4=e5e3e3=e5Nα218:e1e1=e5e1e2=e3e1e3=e4e2e2=e4e2e4=e5e3e3=αe5N219:e1e2=e3e1e3=e4e1e4=2e5e2e2=e4+e5e2e3=e5Nα220:e1e2=e3e1e3=e4e1e4=e5e2e2=e4+αe5e3e3=e5Nα221:e1e2=e3e1e3=e4e1e4=e5e2e2=e4e2e3=αe5Nα0222:e1e2=e3e1e3=e4e1e4=αe5e2e2=e4e2e3=e5e2e4=e5e3e3=e5Nα223:e1e2=e3e1e3=e4e1e4=e5e2e2=e4e2e4=e5e3e3=αe5Nα224:e1e2=e3e1e3=e4e1e4=e5e2e2=e4e3e3=αe5e3e4=e5Nα,β225:e1e2=e3e1e3=e4e2e2=e4+e5e2e3=αe5e3e3=βe5e4e4=e5Nα226:e1e2=e3e1e3=e4e2e2=e4e2e3=e5e3e3=αe5e4e4=e5Nα227:e1e2=e3e1e3=e4e2e2=e4e2e4=e5e3e3=αe5N228:e1e2=e3e1e3=e4e2e2=e4e3e3=e5e3e4=e5N229:e1e2=e3e1e3=e4e2e2=e4e3e3=e5e4e4=e5N230:e1e2=e3e1e3=e4e2e2=e4e3e4=e5N231:e1e2=e3e1e3=e4e2e2=e4e4e4=e5 $

    Here we will collect all information about $ {\mathbf N}_{16}^{4*}: $

    ${\mathbf{N}}^{4*}_{19} $ $\begin{array}{ll} e_1e_1 = e_4 & e_1e_2 = e_3 \\ e_2e_2 = e_4 & e_3e_3 = e_4 \end{array} $ $\begin{array}{l} \mathrm{H}^2_{\mathfrak{D}}(\mathbf{N}^{4*}_{19}) = \langle [\Delta_{11}],[\Delta_{13}],[\Delta_{22}],[\Delta_{23}]\rangle\\ \mathrm{H}^2_{\mathfrak{C}}(\mathbf{N}^{4*}_{19}) = \mathrm{H}^2_{\mathfrak{D}}(\mathbf{N}^{4*}_{19})\oplus \langle [\Delta_{14}], [\Delta_{24}], [\Delta_{34}], [\Delta_{44}] \rangle \end{array} $
    ${ \begin{array}{l} \begin{array}{l} \phi_1 = \begin{pmatrix} x&0&0&0\\ 0 & q&0&0\\ 0&0 & xq&0\\ t & s&0&1 \end{pmatrix}, \\ {x^2 = 1,q^2 = 1} \end{array} \begin{array}{l} \phi_2 = \begin{pmatrix} 0 & p&0&0\\ y&0&0&0\\ 0&0 & yp&0\\ t & s&0&1 \end{pmatrix}, \\ {y^2 = 1, p^2 = 1} \end{array} \end{array} } $

     | Show Table
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    Let us use the following notations:

    $ \begin{array}{llll} \nabla_1 = [\Delta_{11}], & \nabla_2 = [\Delta_{14}], & \nabla_3 = [\Delta_{22}], & \nabla_4 = [\Delta_{23}], \\ \nabla_5 = [\Delta_{24}], & \nabla_6 = [\Delta_{33}], & \nabla_7 = [\Delta_{34}], & \nabla_8 = [\Delta_{44}]. \end{array} $

    Take $ \theta = \sum\limits_{i = 1}^{8}\alpha_i\nabla_i\in\mathrm{H}^2_{\mathfrak{C}}(\mathbf{N}^{4*}_{16}) . $ Since

    $ \phi^T\begin{pmatrix} \alpha_1&0&0&\alpha_2\\ 0&\alpha_3&\alpha_4&\alpha_5\\ 0&\alpha_4&\alpha_6&\alpha_7\\ \alpha_2&\alpha_5&\alpha_7&\alpha_8 \end{pmatrix}\phi = \begin{pmatrix} \alpha_1^*&\alpha^{*}&\alpha^{**}&\alpha_2^*\\ \alpha^{*}&\alpha^*_3&\alpha^*_4+\alpha^{**}&\alpha_5^*\\ \alpha^{**}&\alpha^*_4+\alpha^{**}&\alpha_6^*&\alpha^*_7\\ \alpha^*_2&\alpha^*_5&\alpha^*_7&\alpha^*_8 \end{pmatrix}, $

    in the case $ \phi = \phi_1 $, we have

    $ \begin{array}{ll} \alpha_1^* = \alpha_1x^2+2\alpha_2xt+\alpha_8t^2, & \alpha_2^* = (\alpha_2x+\alpha_8t)x^3, \\ \alpha_3^* = \alpha_3x^2+2\alpha_5xs+\alpha_8s^2, & \alpha_4^* = (\alpha_4x+\alpha_7s)x^2-\alpha_7x^2t, \\ \alpha_5^* = (\alpha_5x+\alpha_8s)x^3, & \alpha_6^* = \alpha_6x^4, \\ \alpha_7^* = \alpha_7x^5, & \alpha_8^* = \alpha_8x^6; \end{array} $

    and on the opposite case, for $ \phi = \phi_2, $ we have

    $ \begin{array}{ll} \alpha_1^* = \alpha_3y^2+2 \alpha_5 t y+ \alpha_8 t^2, & \alpha_2^* = (\alpha_5 y + \alpha_8t )y^3, \\ \alpha_3^* = \alpha_1y^2+2 \alpha_2s y + \alpha_8s^2, & \alpha_4^* = ((s-t) \alpha_7-y \alpha_4)y^2, \\ \alpha_5^* = (y \alpha_2+s \alpha_8)y^3, & \alpha_6^* = \alpha_6 y^4, \\ \alpha_7^* = \alpha_7y^5, & \alpha_8^* = \alpha_8 y^6. \end{array} $

    We are interested in $ (\alpha_2,\alpha_5,\alpha_7,\alpha_8)\neq(0,0,0,0) . $ Let us consider the following cases:

    $ 1.\ $ $ \alpha_8 = 0, \alpha_7 = 0, \alpha_5 = 0, $ then $ \alpha_2\neq0 $ and

    $ {\rm{(a)}}\ $if $ \alpha_4\neq 0, $ then by choosing $ \phi = \phi_1, $ $ x = \alpha_4 \alpha_2^{-1}, $ $ t = -\frac{ \alpha_1 \alpha_4 }{2 \alpha_2^{2}}, $ we have the family of representatives $ \langle \nabla_2+\alpha\nabla_3+\nabla_4 +\beta \nabla_6 \rangle; $

    $ {\rm{(b)}}\ $if $ \alpha_4 = 0,\alpha_3\neq0, $ then by choosing $ \phi = \phi_1, $ $ x = \sqrt{ \alpha_3 \alpha_2^{-1}}, $ $ t = -\frac{ \alpha_1 \sqrt{ \alpha_3}}{2 \sqrt{\alpha_2^3}}, $ we have the family of representatives $ \langle \nabla_2+\nabla_3+\alpha \nabla_6 \rangle; $

    $ {\rm{(c)}}\ $if $ \alpha_4 = 0,\alpha_3 = 0, $ then by choosing $ \phi = \phi_1, $ $ x = 2 \alpha_2, $ $ t = - \alpha_1, $ $ s = 0, $ we have the family of representatives $ \langle \nabla_2+ \alpha \nabla_6 \rangle. $

    2. $ \alpha_8 = 0, \alpha_7 = 0, \alpha_5\neq0 $ and

    $ {\rm{(a)}}\ $if $ \alpha_2\neq 0,\alpha_4\neq 0, $ then by choosing

    $\phi = \phi_1,$ $x = \frac{\alpha_4}{\alpha_5},$ $t = -\frac{\alpha_1 \alpha_4}{ 2 \alpha_2 \alpha_5},$ $s = -\frac{ \alpha_3 \alpha_4}{ 2 \alpha_5^2},$

    we have the following family of representatives

    $ \langle \alpha \nabla_2+\nabla_4+\nabla_5+ \beta \nabla_6 \rangle_{\alpha\neq0};$

    $ {\rm{(b)}}\ $if $ \alpha_2\neq 0,\alpha_4 = 0, $ then by choosing

    $\phi = \phi_1,$ $x = 2 \alpha_2 \alpha_5,$ $t = - \alpha_1 \alpha_5,$ $s = - \alpha_2 \alpha_3,$

    we have the following family of representatives $ \langle \alpha \nabla_2+ \nabla_5+ \beta \nabla_6 \rangle_{\alpha\neq0}; $

    $ {\rm{(c)}}\ $if $ \alpha_2 = 0, $ then by choosing $ \phi = \phi_2, $ $ y = 1, $ $ t = 0, $ $ s = 0, $ we have the representative with $ \alpha_5^* = 0 $ and $ \alpha_2^*\neq0, $ which was considered above.

    3. $ \alpha_8 = 0, \alpha_7\neq0, $ then we have

    ${\rm{(a)}}\ $if $ \alpha_2 = 0, \alpha_5 = 0, \alpha_1 = 0, \alpha_3 = 0, $ then we have the representatives $ \langle \nabla_7 \rangle $ and $ \langle \nabla_6+\nabla_7\rangle $ depending on whether $ \alpha_6 = 0 $ or not;

    $ {\rm{(b)}}\ $if $ \alpha_2 = 0, \alpha_5 = 0, \alpha_1\neq0, $ then by choosing

    $\phi = \phi_1,$ $ x = \sqrt[3]{{\alpha_1}{\alpha_7^{-1}}}, s = 0, t = {\alpha_4\sqrt[3]{\alpha_1} \alpha_7^{-1}}, $

    we have the family of representatives $ \langle \nabla_1+\alpha\nabla_3+\beta\nabla_6+\nabla_7\rangle; $

    $ {\rm{(c)}}\ $if $ \alpha_2\neq0, $ then by choosing

    $\phi = \phi_1,$ $ x = {\alpha_2}{\alpha_7^{-1}}, s = -({\alpha_1\alpha_7+2\alpha_2\alpha_4}) /(2\alpha_7^2), t = -{\alpha_1}/ ({2\alpha_7}), $

    we have the family of representatives $ \langle \nabla_2+\alpha\nabla_3+\beta\nabla_5+\gamma\nabla_6+\nabla_7\rangle. $

    4. $ \alpha_8\neq0, $ then by choosing $ \phi = \phi_1, $ $ t = -\frac{\alpha_2}{\alpha_8}x, s = -\frac{\alpha_5}{\alpha_8}x, $ we have $ \alpha_2^* = \alpha_5^* = 0. $ Now we can suppose that $ \alpha_2 = 0, \alpha_5 = 0 $ and we have

    $ {\rm{(a)}}\ $if $ \alpha_1 = 0, \alpha_3 = 0, \alpha_4 = 0, \alpha_6 = 0, $ then we have the representatives $ \langle \nabla_8\rangle $ and $ \langle \nabla_7+\nabla_8\rangle $ depending on whether $ \alpha_7 = 0 $ or not;

    $ {\rm{(b)}}\ $if $ \alpha_1 = 0, \alpha_3 = 0, \alpha_4 = 0, \alpha_6\neq0, $ then by choosing $ \phi = \phi_1, $ $ x = \sqrt{{\alpha_6}{\alpha_8^{-1}}}, $ $ s = 0, $ $ t = 0, $ we have the family of representative $ \langle \nabla_6+\alpha\nabla_7+\nabla_8\rangle; $

    $ {\rm{(c)}}\ $if $ \alpha_1 = 0, \alpha_3 = 0, \alpha_4\neq0, $ then by choosing $ \phi = \phi_1, $ $ x = \sqrt[3]{{\alpha_4}{\alpha_8^{-1}}}, s = 0, t = 0, $ we have the family of representatives $ \langle \nabla_4+\alpha\nabla_6+\beta\nabla_7+\nabla_8\rangle; $

    $ {\rm{(d)}}\ $if $ \alpha_1\neq0, $ then by choosing $ \phi = \phi_1, $ $ x = \sqrt[4]{{\alpha_1}{\alpha_8^{-1}}}, s = 0, t = 0, $ we have the family of representatives $ \langle \nabla_1+\alpha\nabla_3+\beta\nabla_4+\gamma\nabla_6+\mu\nabla_7+\nabla_8\rangle. $

    Summarizing, we have the following distinct orbits:

    $\begin{array}{c} \langle \nabla_1+\alpha\nabla_3+\beta\nabla_4+\gamma\nabla_6+\mu\nabla_7+\nabla_8\rangle ^{ { \begin{array}{l} O(\alpha,\beta,\gamma,\mu) = O(\alpha,i\beta,-\gamma,-i\mu) = \\ O(\alpha,-i\beta,-\gamma,i\mu) = O(\alpha,-\beta,\gamma,-\mu) = \\ O(\frac{1}{\alpha},-\frac{\beta}{\sqrt[4]{\alpha^{3}}}, \frac{\gamma}{\sqrt{\alpha}},\frac{\mu}{\sqrt[4]{\alpha}}) = \\ O(\frac{1}{\alpha},-\frac{i\beta}{ \sqrt[4]{\alpha^{3}}},-\frac{\gamma}{\sqrt{\alpha}},-\frac{i\mu}{\sqrt[4]{\alpha}}) = \\ O(\frac{1}{\alpha},\frac{\beta}{\sqrt[4]{\alpha^{3}}}, \frac{\gamma}{\sqrt{\alpha}},-\frac{\mu}{\sqrt[4]{\alpha}}) = O(\frac{1}{\alpha},-\frac{\beta}{\sqrt[4]{\alpha^{3}}}, \frac{\gamma}{\sqrt{\alpha}},\frac{\mu}{\sqrt[4]{\alpha}})\end{array}}}, \\ \langle \nabla_1+\alpha\nabla_3+\beta\nabla_6+\nabla_7\rangle ^{{ \begin{array}{l} O(\alpha,\beta) = O(\alpha,-\eta_3\beta) = O(\alpha,\eta_3^2\beta) = \\ O(\alpha^{-1},-\eta_3\beta\sqrt[3]{\alpha^{-1}}) = O(\alpha^{-1},\eta_3^2\beta\sqrt[3]{\alpha^{-1}}) = O(\alpha^{-1},\beta\sqrt[3]{\alpha^{-1}})\end{array}}}, \\ \langle \nabla_2+\alpha\nabla_3+\nabla_4 +\beta \nabla_6 \rangle, \langle \nabla_2+\alpha\nabla_3+\beta\nabla_5+\gamma\nabla_6+\nabla_7\rangle^{O(\alpha,\beta,\gamma) = O(-\frac{\alpha}{\beta^4},\frac{1}{\beta},\frac{\gamma}{\beta})}, \\ \langle \nabla_2+\nabla_3+\alpha \nabla_6 \rangle, \langle \alpha \nabla_2+\nabla_4+\nabla_5+ \beta \nabla_6 \rangle_{\alpha\neq0}^{O(\alpha,\beta) = O(\alpha^{-1},\beta\alpha^{-1})}, \\ \langle \alpha \nabla_2+ \nabla_5+ \beta \nabla_6 \rangle_{\alpha\neq0}^{O(\alpha,\beta) = O(\alpha^{-1},\beta\alpha^{-1})}, \langle \nabla_2+ \alpha \nabla_6 \rangle, \\ \langle \nabla_4+\alpha\nabla_6+\beta\nabla_7+\nabla_8\rangle ^{{ \begin{array}{l} O(\alpha,\beta) = O(\eta_3^2\alpha,-\eta_3\beta) = O(-\eta_3\alpha,\eta_3^2\beta) = \\ O(\eta_3^2\alpha,\eta_3\beta) = O(-\eta_3\alpha,-\eta_3^2\beta) = O(\alpha,-\beta)\end{array} }}, \langle \nabla_6+\nabla_7\rangle, \\ \langle \nabla_6+\alpha\nabla_7+\nabla_8\rangle^{O(\alpha) = O(-\alpha)}, \langle \nabla_7 \rangle, \langle \nabla_7+\nabla_8\rangle, \langle \nabla_8\rangle, \end{array}$

    which gives the following new algebras:

    $ \begin{array}{llllllllllllllllll} {\mathbf{N}}_{232}^{\alpha, \beta, \gamma, \mu} & : & e_1e_1 = e_5 & e_1e_2 = e_3 & e_1e_3 = e_4 & e_2e_2 = \alpha e_5 \\ && e_2e_3 = e_4+\beta e_5 & e_3e_3 = \gamma e_5 & e_3e_4 = \mu e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{233}^{\alpha, \beta} & : & e_1e_1 = e_5 & e_1e_2 = e_3 & e_1e_3 = e_4 & e_2e_2 = \alpha e_5 \\ & & e_2e_3 = e_4 & e_3e_3 = \beta e_5 & e_3e_4 = e_5 \\ {\mathbf{N}}_{234}^{\alpha, \beta} & : & e_1e_2 = e_3 & e_1e_3 = e_4 & e_1e_4 = e_5 \\ && e_2e_2 = \alpha e_5 & e_2e_3 = e_4+e_5 & e_3e_3 = \beta e_5 \\ {\mathbf{N}}_{235}^{\alpha, \beta, \gamma} & : & e_1e_2 = e_3 & e_1e_3 = e_4 & e_1e_4 = e_5 & e_2e_2 = \alpha e_5 \\ & & e_2e_3 = e_4 & e_2e_4 = \beta e_5 & e_3e_3 = \gamma e_5 & e_3e_4 = e_5 \\ {\mathbf{N}}_{236}^{\alpha} & : & e_1e_2 = e_3 & e_1e_3 = e_4 & e_1e_4 = e_5 \\ && e_2e_2 = e_5 & e_2e_3 = e_4 & e_3e_3 = \alpha e_5 \\ {\mathbf{N}}_{237}^{\alpha\neq0, \beta} & : & e_1e_2 = e_3 & e_1e_3 = e_4 & e_1e_4 = \alpha e_5 \\ && e_2e_3 = e_4+e_5 & e_2e_4 = e_5 & e_3e_3 = \beta e_5 \\ {\mathbf{N}}_{238}^{\alpha\neq0, \beta} & : & e_1e_2 = e_3 & e_1e_3 = e_4 & e_1e_4 = \alpha e_5 \\ & & e_2e_3 = e_4 & e_2e_4 = e_5 & e_3e_3 = \beta e_5 \\ {\mathbf{N}}_{239}^{\alpha} & : & e_1e_2 = e_3 & e_1e_3 = e_4 & e_1e_4 = e_5 \\ && e_2e_3 = e_4 & e_3e_3 = \alpha e_5 \\ {\mathbf{N}}_{240}^{\alpha, \beta} & : & e_1e_2 = e_3 & e_1e_3 = e_4 & e_2e_3 = e_4+e_5 \\ && e_3e_3 = \alpha e_5 & e_3e_4 = \beta e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{241} & : & e_1e_2 = e_3 & e_1e_3 = e_4 & e_2e_3 = e_4 \\ && e_3e_3 = e_5 & e_3e_4 = e_5 \\ {\mathbf{N}}_{242}^{\alpha} & : & e_1e_2 = e_3 & e_1e_3 = e_4 & e_2e_3 = e_4 \\ && e_3e_3 = e_5 & e_3e_4 = \alpha e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{243} & : & e_1e_2 = e_3 & e_1e_3 = e_4 & e_2e_3 = e_4 & e_3e_4 = e_5 \\ {\mathbf{N}}_{244} & : & e_1e_2 = e_3 & e_1e_3 = e_4 & e_2e_3 = e_4 \\ & & e_3e_4 = e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{245} & : & e_1e_2 = e_3 & e_1e_3 = e_4 & e_2e_3 = e_4 & e_4e_4 = e_5 \\ \end{array} $

    Here we will collect all information about $ {\mathbf N}_{17}^{4*}: $

    ${\mathbf{N}}^{4*}_{19} $ $\begin{array}{ll} e_1e_1 = e_4 & e_1e_2 = e_3 \\ e_2e_2 = e_4 & e_3e_3 = e_4 \end{array} $ $\begin{array}{l} \mathrm{H}^2_{\mathfrak{D}}(\mathbf{N}^{4*}_{19}) = \langle [\Delta_{11}],[\Delta_{13}],[\Delta_{22}],[\Delta_{23}]\rangle\\ \mathrm{H}^2_{\mathfrak{C}}(\mathbf{N}^{4*}_{19}) = \mathrm{H}^2_{\mathfrak{D}}(\mathbf{N}^{4*}_{19})\oplus \langle [\Delta_{14}], [\Delta_{24}], [\Delta_{34}], [\Delta_{44}] \rangle \end{array} $
    ${ \begin{array}{l} \begin{array}{l} \phi_1 = \begin{pmatrix} x&0&0&0\\ 0 & q&0&0\\ 0&0 & xq&0\\ t & s&0&1 \end{pmatrix}, \\ {x^2 = 1,q^2 = 1} \end{array} \begin{array}{l} \phi_2 = \begin{pmatrix} 0 & p&0&0\\ y&0&0&0\\ 0&0 & yp&0\\ t & s&0&1 \end{pmatrix}, \\ {y^2 = 1, p^2 = 1} \end{array} \end{array} } $

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    Let us use the following notations:

    $ \begin{array}{llll} \nabla_1 = [\Delta_{11}], & \nabla_2 = [\Delta_{13}], & \nabla_3 = [\Delta_{14}], & \nabla_4 = [\Delta_{22}], \\ \nabla_5 = [\Delta_{23}], & \nabla_6 = [\Delta_{24}], & \nabla_7 = [\Delta_{34}], & \nabla_8 = [\Delta_{44}]. \end{array} $

    Take $ \theta = \sum\limits_{i = 1}^{8}\alpha_i\nabla_i\in\mathrm{H}^2_{\mathfrak{C}}(\mathbf{N}^{4*}_{17}) . $ Since

    $ \phi^T\begin{pmatrix} \alpha_1&0&\alpha_2&\alpha_3\\ 0&\alpha_4&\alpha_5&\alpha_6\\ \alpha_2&\alpha_5&0&\alpha_7\\ \alpha_3&\alpha_6&\alpha_7&\alpha_8 \end{pmatrix}\phi = \begin{pmatrix} \alpha_1^*&\alpha^{*}&\alpha^{*}_2&\alpha_3^*\\ \alpha^{*}&\alpha^*_4&\alpha^*_5&\alpha_6^*\\ \alpha^{*}_2&\alpha^*_5& 0 &\alpha^*_7\\ \alpha^*_3&\alpha^*_6&\alpha^*_7&\alpha^*_8 \end{pmatrix}, $

    then in the case $ \phi = \phi_1, $ we have

    $ \begin{array}{ll} \alpha_1^* = \alpha_1x^2+2\alpha_3xt+\alpha_8t^2, & \alpha_2^* = (\alpha_2x+\alpha_7t)xq, \\ \alpha_3^* = (\alpha_3x+\alpha_8t)x^2q^2, & \alpha_4^* = \alpha_4q^2+2\alpha_6qs+\alpha_8s^2, \\ \alpha_5^* = (\alpha_5q+\alpha_7s)xq, & \alpha_6^* = (\alpha_6q+\alpha_8s)x^2q^2, \\ \alpha_7^* = \alpha_7x^3q^3, & \alpha_8^* = \alpha_8x^4q^4; \end{array} $

    and in the opposite case $ \phi = \phi_2, $ we have

    $ \begin{array}{ll} \alpha_1^* = \alpha_4p^2+2 \alpha_6p t+ \alpha_8t^2, & \alpha_2^* = ( \alpha_5p + \alpha_7t)p y, \\ \alpha_3^* = ( \alpha_6p+ \alpha_8t)p^2 y^2 , & \alpha_4^* = \alpha_1y^2+2 \alpha_3s y+ \alpha_8s^2, \\ \alpha_5^* = ( \alpha_2y+ \alpha_7s)p y, & \alpha_6^* = ( \alpha_3y+ \alpha_8s)p^2 y^2 , \\ \alpha_7^* = \alpha_7p^3 y^3, & \alpha_8^* = \alpha_8p^4 y^4. \end{array} $

    We are interested in $ (\alpha_3,\alpha_6,\alpha_7,\alpha_8)\neq(0,0,0,0) . $ Let us consider the following cases:

    $ 1.\ $$ \alpha_8 = 0, \alpha_7 = 0, \alpha_6 = 0, $ then $ \alpha_3\neq0 $ and choosing $ \phi = \phi_1, $ $ t = -\frac{\alpha_1}{2\alpha_3}x, $ we get $ \alpha_1^* = 0. $ Now consider the following subcases:

    $ {\rm{(a)}}\ $if $ \alpha_2 = 0, \alpha_4 = 0, \alpha_5 = 0, $ then we have the representative $ \langle \nabla_3 \rangle; $

    $ {\rm{(b)}}\ $if $ \alpha_2 = 0, \alpha_4 = 0, \alpha_5\neq0, $ then by choosing $ \phi = \phi_1, $ $ x = \sqrt{\frac{\alpha_5}{\alpha_3}}, q = 1, s = 0, t = -\frac{\alpha_1\sqrt{\alpha_5}}{2\alpha_3\sqrt{\alpha_3}}, $ we have the representative $ \langle \nabla_3+\nabla_5 \rangle; $

    $ {\rm{(c)}}\ $if $ \alpha_2 = 0, \alpha_4\neq0, $ then by choosing $ \phi = \phi_1, $ $ x = \sqrt[3]{\frac{\alpha_4}{\alpha_3}}, q = 1, s = 0, t = -\frac{\alpha_1\sqrt[3]{\alpha_5}}{2\alpha_3\sqrt[3]{\alpha_3}}, $ we have the representative $ \langle \nabla_3+\nabla_4+\alpha\nabla_5 \rangle; $

    $ {\rm{(d)}}\ $if $ \alpha_2\neq0, \alpha_4 = 0, \alpha_5 = 0, $ then by choosing $ \phi = \phi_1, $ $ x = \alpha_2, q = \frac{1}{\alpha_3}, s = 0, t = -\frac{\alpha_1\alpha_2}{2\alpha_3}, $ we have the representative $ \langle \nabla_2+\nabla_3 \rangle; $

    ${\rm{(e)}}\ $if $ \alpha_2\neq0, \alpha_4 = 0, \alpha_5\neq0, $ then by choosing $ \phi = \phi_1, $ $ x = \sqrt{\frac{\alpha_5}{\alpha_3}}, q = \sqrt{\frac{\alpha_2^2}{\alpha_3\alpha_5}}, s = 0, t = -\frac{\alpha_1\sqrt{\alpha_5}}{2\alpha_3\sqrt{\alpha_3}}, $ we have the representative $ \langle \nabla_2+\nabla_3+\nabla_5 \rangle; $

    $ {\rm{(f)}}\ $if $ \alpha_2\neq0, \alpha_4\neq0, $ then by choosing $ \phi = \phi_1, $ $ x = \sqrt[3]{{\alpha_4}{\alpha_3^{-3}}}, q = \sqrt[3]{{\alpha_2^3}{\alpha^{-2}_3\alpha_4^{-1}}}, $ we have the family of representative $ \langle \nabla_2+\nabla_3+\nabla_4+\alpha\nabla_5 \rangle. $

    2. $ \alpha_8 = 0, \alpha_7 = 0, \alpha_6\neq0, $ and $ \alpha_3 = 0 , $ then by choosing some suitable automorphism $ \phi_2 $ we have $ \alpha_3^*\neq0 $ which is the case considered above. Now we can suppose that $ \alpha_3\neq0, $ and choosing $ t = -\frac{\alpha_1}{2\alpha_3}x, s = -\frac{\alpha_4}{2\alpha_6}x, $ we have $ \alpha_1^* = 0, \alpha_4^* = 0. $ Therefore, we can suppose that $ \alpha_1 = 0, \alpha_4 = 0. $ Consider the following subcases:

    $ {\rm{(a)}}\ $$ \alpha_2 = 0, \alpha_5 = 0, $ then by choosing $ \phi = \phi_1, $ $ x = \alpha_6, q = \alpha_3, s = 0, t = 0, $ we have the representative $ \langle \nabla_3+\nabla_6 \rangle; $

    $ {\rm{(b)}}\ $$ \alpha_2\neq0, $ then by choosing $ \phi = \phi_1, $ $ x = {\alpha_3^{-1}}{\sqrt{\alpha_2\alpha_6}}, q = \sqrt{\alpha_2\alpha_6^{-1}}, s = 0, t = 0, $ we have the family of representatives $ \langle \nabla_2+\nabla_3+\alpha\nabla_5+\nabla_6 \rangle. $

    3. $ \alpha_8 = 0, \alpha_7\neq0, $ then by choosing $ \phi = \phi_1, $ $ t = -{\alpha_2}{\alpha_7^{-1}}x, s = -{\alpha_5}{\alpha_7^{-1}}q, $ we have $ \alpha_2^* = 0, \alpha_5^* = 0. $ Therefore, we can suppose that $ \alpha_2 = 0, \alpha_5 = 0. $ Consider the following subcases:

    $ {\rm{(a)}}\ $if $ \alpha_1 = 0, \alpha_4 = 0, \alpha_3 = 0, \alpha_6 = 0, $ then we have the representative $ \langle \nabla_7 \rangle ; $

    $ {\rm{(b)}}\ $if $ \alpha_1 = 0, \alpha_4 = 0, \alpha_3 = 0, \alpha_6\neq0, $ then by choosing $ \phi = \phi_1, $ $ x = {\alpha_6}{\alpha_7^{-1}}, q = 1, s = 0, t = 0, $ we have the representative $ \langle \nabla_6+\nabla_7 \rangle ; $

    $ {\rm{(c)}}\ $if $ \alpha_1 = 0, \alpha_4 = 0, \alpha_3\neq0, $ and $ \alpha_6 = 0, $ then by choosing some suitable automorphism $ \phi_2, $ we have $ \alpha_6^*\neq0 . $ Thus we can consider the case $ \alpha_6\neq0 $ and choosing $ \phi = \phi_1, $ $ x = {\alpha_6}{\alpha_7^{-1}}, q = {\alpha_3}{\alpha_7^{-1}}, s = 0, t = 0, $ we have the representative $ \langle \nabla_3+\nabla_6+\nabla_7 \rangle ; $

    $ {\rm{(d)}}\ $if $ \alpha_1 = 0, \alpha_4\neq0, \alpha_3 = 0, \alpha_6 = 0, $ then by choosing $ \phi = \phi_1, $ $ x = 1, q = {\alpha_4}{\alpha_7^{-1}}, s = 0, t = 0, $ we have the representative $ \langle \nabla_4+\nabla_7 \rangle; $

    $ {\rm{(e)}}\ $if $ \alpha_1 = 0, \alpha_4\neq0, \alpha_3 = 0, \alpha_6\neq0, $ then by choosing $ \phi = \phi_1, $ $ x = {\alpha_6}{\alpha_7}^{-1}, q = {\alpha_4\alpha_7^2}{\alpha_6^{-3}}, $ we have the representative $ \langle \nabla_4+\nabla_6+\nabla_7 \rangle; $

    $ {\rm{(f)}}\ $if $ \alpha_1 = 0, \alpha_4\neq0, \alpha_3\neq0, $ then by choosing

    $\phi = \phi_1,$ $ x = \sqrt[3]{{\alpha_4}{\alpha_3^{-1}}}, q = {\alpha_3}{\alpha_7^{-1}}, s = 0, t = 0, $

    we have the family of representatives $ \langle \nabla_3+\nabla_4+\alpha\nabla_6+\nabla_7 \rangle ; $

    $ {\rm{(g)}}\ $if $ \alpha_1\neq0. $ In case of $ \alpha_4 = 0, $ choosing some suitable automorphism $ \phi_2, $ we have $ \alpha_4^*\neq0. $ Thus, we can suppose $ \alpha_4\neq0 , $ and choosing

    $\phi = \phi_1,$ $ x = \sqrt[8]{{\alpha^3_4}{\alpha_1^{-1}\alpha_7^{-2}}}, q = \sqrt[8]{{\alpha_1^3}{\alpha_4^{-1}\alpha^{-2}_7}}, s = 0, t = 0, $

    we have the family of representatives $ \langle \nabla_1+\alpha\nabla_3+\nabla_4+\beta\nabla_6+\nabla_7 \rangle . $

    4. $ \alpha_8\neq0, $ then by choosing $ \phi = \phi_1, $ $ t = -\frac{\alpha_3}{\alpha_8}x, s = -\frac{\alpha_6}{\alpha_8}q, $ we have $ \alpha_3^* = 0, \alpha_6^* = 0. $ Consider the following cases:

    $ {\rm{(a)}}\ $if $ \alpha_1 = 0, \alpha_4 = 0, \alpha_2 = 0, \alpha_5 = 0, $ then we have the representatives $ \langle \nabla_8 \rangle $ and $ \langle \nabla_7+\nabla_8 \rangle $ depending on whether $ \alpha_7 = 0 $ or not;

    $ {\rm{(b)}}\ $if $ \alpha_1 = 0, \alpha_4 = 0, \alpha_2 = 0, \alpha_5\neq0, $ then we have the representatives $ \langle \nabla_5+\nabla_8 \rangle $ and $ \langle \nabla_5+\nabla_7+\nabla_8 \rangle $ depending on whether $ \alpha_7 = 0 $ or not;

    $ {\rm{(c)}}\ $if $ \alpha_1 = 0, \alpha_4 = 0, \alpha_2\neq0. $ In case of $ \alpha_5 = 0, $ choosing some suitable automorphism $ \phi_2, $ we have $ \alpha_5^*\neq0 . $ Thus, we can suppose $ \alpha_5\neq0 , $ and choosing

    $\phi = \phi_1,$ $ x = \sqrt[5]{{\alpha^3_5}{\alpha_2^{-2}\alpha_8^{-1}}}, q = \sqrt[5]{{\alpha^3_2}{\alpha_5^{-2}\alpha_8^{-1}}}, s = 0, t = 0,$

    we have the family of representatives $ \langle \nabla_2+\nabla_5+\alpha\nabla_7+\nabla_8 \rangle ; $

    ${\rm{(d)}}\ $if $ \alpha_1 = 0, \alpha_4\neq0, \alpha_2 = 0, \alpha_5 = 0, $ then we have the representatives $ \langle \nabla_4+\nabla_8 \rangle $ and $ \langle \nabla_4+\nabla_7+\nabla_8 \rangle $ depending on whether $ \alpha_7 = 0 $ or not;

    $ {\rm{(e)}}\ $if $ \alpha_1 = 0, \alpha_4\neq0, \alpha_2 = 0, \alpha_5\neq0 , $ then by choosing

    $\phi = \phi_1,$ $x = {\alpha_4}{\alpha_5^{-1}}, q = {\alpha^2_5}{\alpha_4^{-1}\sqrt{\alpha_4^{-1}\alpha_8^{-1}}}, s = 0, t = 0, $

    we have the family of representatives $ \langle \nabla_4+\nabla_5+\alpha\nabla_7+\nabla_8 \rangle ; $

    ${\rm{(f)}}\ $if $ \alpha_1 = 0, \alpha_4\neq0, \alpha_2\neq0 , $ then by choosing

    $\phi = \phi_1,$ $ x = \sqrt[8]{{\alpha_4^3}{\alpha_2^{-2}\alpha_8^{-1}}}, q = \sqrt[4]{{\alpha^2_2}{\alpha_4^{-1}\alpha_8^{-1}}}, s = 0, t = 0, $

    we have the family of representatives $ \langle \nabla_2+\nabla_4+\alpha\nabla_5+\beta\nabla_7+\nabla_8 \rangle ; $

    $ {\rm{(g)}}\ $if $ \alpha_1\neq0 $ then by choosing some suitable automorphism $ \phi_2, $ we have $ \alpha_4^*\neq0 $. Thus, we can suppose $ \alpha_4\neq0 , $ and choosing

    $\phi = \phi_1,$ $x = \sqrt[6]{{\alpha^2_4}{\alpha_1^{-1}\alpha_8^{-1}}}, q = \sqrt[6]{{\alpha_1^2}{\alpha_4^{-1}\alpha_8^{-1}}}, s = 0, t = 0, $

    we have the family of representatives

    $ \langle \nabla_1+\alpha\nabla_2+\nabla_4+\beta\nabla_5+\gamma\nabla_7+\nabla_8 \rangle.$

    Summarizing, we have the following distinct orbits:

    $\begin{array}{c} \langle \nabla_1+\alpha\nabla_2+\nabla_4+\beta\nabla_5+\gamma\nabla_7+\nabla_8 \rangle^{ { \begin{array}{l} O(\alpha,\beta,\gamma) = O(\eta_3^2\alpha, \eta_3^2\beta, \eta_3^2\gamma) = O(-\eta_3^2\alpha, \eta_3^2\beta, -\eta_3^2\gamma) = \\ O(\eta_3^2\alpha, -\eta_3^2\beta, -\eta_3^2\gamma) = O(-\eta_3^2\alpha, \eta_3^2\beta, \eta_3^2\gamma) = \\ O(\eta_3\alpha, \eta_3\beta, -\eta_3\gamma) = O(-\eta_3\alpha, \eta_3\beta, \eta_3\gamma) = \\ O(\eta_3\alpha, -\eta_3\beta, \eta_3\gamma) = O(-\eta_3\alpha, -\eta_3\beta, -\eta_3\gamma) = \\ O(-\alpha, \beta, - \gamma) = O(\alpha, -\beta, - \gamma) = \\ O(-\alpha, -\beta, \gamma) = O(\beta,\alpha,\gamma) = \\ O(\eta_3^2\beta, \eta_3^2\alpha, \eta_3^2\gamma) = O(-\eta_3^2\beta, \eta_3^2\alpha, -\eta_3^2\gamma) = \\ O(\eta_3^2\beta, -\eta_3^2\alpha, -\eta_3^2\gamma) = O(-\eta_3^2\beta, \eta_3^2\alpha, \eta_3^2\gamma) = \\ O(\eta_3\beta, \eta_3\alpha, -\eta_3\gamma) = O(-\eta_3\beta, \eta_3\alpha, \eta_3\gamma) = \\ O(\eta_3\beta, -\eta_3\alpha, \eta_3\gamma) = O(-\eta_3\beta, -\eta_3\alpha, -\eta_3\gamma) = \\ O(-\beta, \alpha, - \gamma) = O(\beta, -\alpha, - \gamma) = O(-\beta, -\alpha, \gamma) \end{array}}}, \\ \langle \nabla_1+\alpha\nabla_3+\nabla_4+\beta\nabla_6 +\nabla_7 \rangle^{ {\begin{array}{l} O(\alpha,\beta) = O(\eta_4\alpha,-\eta_4\beta) = O(-\eta_4\alpha,\eta_4\beta) = O(\eta_4^3\alpha,-\eta_4^3\beta) = \\ O(-\eta_4^3\alpha,\eta_4^3\beta) = O(-i\alpha,-i\beta) = O(i\alpha,i\beta) = O(-\alpha,-\beta) = \\ O(\beta,\alpha) = O(\eta_4\beta,-\eta_4\alpha) = O(-\eta_4\beta,\eta_4\alpha) = O(\eta_4^3\beta,-\eta_4^3\alpha) = \\ O(-\eta_4^3\beta,\eta_4^3\alpha) = O(-i\beta,-i\alpha) = O(i\beta,i\alpha) = O(-\beta,-\alpha) \end{array}}}, \\ \langle \nabla_2+\nabla_3 \rangle, \langle \nabla_2+\nabla_3+\nabla_4+\alpha\nabla_5 \rangle^{O(\alpha) = O(-\eta_3 \alpha) = O(\eta_3^2 \alpha)}, \\ \langle \nabla_2+\nabla_3+\nabla_5 \rangle, \langle \nabla_2+\nabla_3 +\alpha\nabla_5+\nabla_6 \rangle^{O(\alpha) = O(\alpha^{-1})}, \\ \langle \nabla_2+\nabla_4+\alpha\nabla_5+\beta\nabla_7+\nabla_8 \rangle ^{{ \begin{array}{l} O(\alpha,\beta) = O(\eta_4^3\alpha,-\eta_4^3\beta) = O(-\eta_4^3\alpha,\eta_4^3\beta) = O(\eta_4\alpha,-\eta_4\beta) = \\ O(-\eta_4\alpha,\eta_4\beta) = O(i\alpha,i\beta) = O(-i\alpha,-i\beta) = O(-\alpha,-\beta) \end{array}}}, \\ \langle \nabla_2+\nabla_5+\alpha\nabla_7+\nabla_8 \rangle ^{ { \begin{array}{l} O(\alpha) = O(\eta_5^2\alpha) = O(\eta_5^4\alpha) = \\ O(-\eta_5\alpha) = O(-\eta_5^3\alpha) \end{array}}}, \langle \nabla_3 \rangle, \\ \langle \nabla_3+\nabla_4+\alpha\nabla_5 \rangle ^{O(\alpha) = O(-\eta_3\alpha) = O(\eta_3^2\alpha)}, \\ \langle \nabla_3+\nabla_4+\alpha\nabla_6+\nabla_7 \rangle^{O(\alpha) = O(-\eta_3\alpha) = O(\eta_3^2\alpha)}, \langle \nabla_3+\nabla_5 \rangle, \langle \nabla_3+\nabla_6 \rangle, \\ \langle \nabla_3+\nabla_6+\nabla_7 \rangle, \langle \nabla_4+\nabla_5+\alpha\nabla_7+\nabla_8 \rangle^{O(\alpha) = O(-\alpha)}, \langle \nabla_4+\nabla_6+\nabla_7 \rangle, \\ \langle \nabla_4+\nabla_7 \rangle, \langle \nabla_4+\nabla_7+\nabla_8 \rangle, \langle \nabla_4+\nabla_8 \rangle, \langle \nabla_5+\nabla_7+\nabla_8 \rangle, \langle \nabla_5+\nabla_8 \rangle, \\ \langle \nabla_6+\nabla_7 \rangle, \langle \nabla_7 \rangle, \langle \nabla_7+\nabla_8 \rangle, \langle \nabla_8 \rangle, \end{array}$

    which gives the following new algebras:

    $ \begin{array}{llllllllllllllllll} {\mathbf{N}}_{246}^{\alpha, \beta ,\gamma} & : & e_1e_1 = e_5 & e_1e_2 = e_3 & e_1e_3 = \alpha e_5 & e_2e_2 = e_5 \\ && e_2e_3 = \beta e_5 & e_3e_3 = e_4 & e_3e_4 = \gamma e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{247}^{\alpha, \beta} & : & e_1e_1 = e_5 & e_1e_2 = e_3 & e_1e_4 = \alpha e_5 & e_2e_2 = e_5 \\ & & e_2e_4 = \beta e_5 & e_3e_3 = e_4 & e_3e_4 = e_5 \\ {\mathbf{N}}_{248} & : & e_1e_2 = e_3 & e_1e_3 = e_5 & e_1e_4 = e_5 & e_3e_3 = e_4 \\ {\mathbf{N}}_{249}^{\alpha} & : & e_1e_2 = e_3 & e_1e_3 = e_5 & e_1e_4 = e_5 \\ && e_2e_2 = e_5 & e_2e_3 = \alpha e_5 & e_3e_3 = e_4 \\ {\mathbf{N}}_{250} & : & e_1e_2 = e_3 & e_1e_3 = e_5 & e_1e_4 = e_5 & e_2e_3 = e_5 & e_3e_3 = e_4 \\ {\mathbf{N}}_{251}^{\alpha} & : & e_1e_2 = e_3 & e_1e_3 = e_5 & e_1e_4 = e_5 \\ && e_2e_3 = \alpha e_5 & e_2e_4 = e_5 & e_3e_3 = e_4 \\ {\mathbf{N}}_{252}^{\alpha, \beta} & : & e_1e_2 = e_3 & e_1e_3 = e_5 & e_2e_2 = e_5 & e_2e_3 = \alpha e_5 \\ & & e_3e_3 = e_4 & e_3e_4 = \beta e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{253}^{\alpha} & : & e_1e_2 = e_3 & e_1e_3 = e_5 & e_2e_3 = e_5 \\ && e_3e_3 = e_4 & e_3e_4 = \alpha e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{254} & : & e_1e_2 = e_3 & e_1e_4 = e_5 & e_3e_3 = e_4 \\ {\mathbf{N}}_{255}^{\alpha} & : & e_1e_2 = e_3 & e_1e_4 = e_5 & e_2e_2 = e_5 & e_2e_3 = \alpha e_5 & e_3e_3 = e_4 \\ {\mathbf{N}}_{256}^{\alpha} & : & e_1e_2 = e_3 & e_1e_4 = e_5 & e_2e_2 = e_5 \\ && e_2e_4 = \alpha e_5 & e_3e_3 = e_4 & e_3e_4 = e_5 \\ {\mathbf{N}}_{257} & : & e_1e_2 = e_3 & e_1e_4 = e_5 & e_2e_3 = e_5 & e_3e_3 = e_4 \\ {\mathbf{N}}_{258} & : & e_1e_2 = e_3 & e_1e_4 = e_5 & e_2e_4 = e_5 & e_3e_3 = e_4 \\ {\mathbf{N}}_{259} & : & e_1e_2 = e_3 & e_1e_4 = e_5 & e_2e_4 = e_5 & e_3e_3 = e_4 & e_3e_4 = e_5 \\ {\mathbf{N}}_{260}^{\alpha} & : & e_1e_2 = e_3 & e_2e_2 = e_5 & e_2e_3 = e_5 \\ && e_3e_3 = e_4 & e_3e_4 = \alpha e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{261} & : & e_1e_2 = e_3 & e_2e_2 = e_5 & e_2e_4 = e_5 & e_3e_3 = e_4 & e_3e_4 = e_5 \\ {\mathbf{N}}_{262} & : & e_1e_2 = e_3 & e_2e_2 = e_5 & e_3e_3 = e_4 & e_3e_4 = e_5 \\ {\mathbf{N}}_{263} & : & e_1e_2 = e_3 & e_2e_2 = e_5 & e_3e_3 = e_4 & e_3e_4 = e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{264} & : & e_1e_2 = e_3 & e_2e_2 = e_5 & e_3e_3 = e_4 & e_4e_4 = e_5 \\ {\mathbf{N}}_{265} & : & e_1e_2 = e_3 & e_2e_3 = e_5 & e_3e_3 = e_4 & e_3e_4 = e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{266} & : & e_1e_2 = e_3 & e_2e_3 = e_5 & e_3e_3 = e_4 & e_4e_4 = e_5 \\ {\mathbf{N}}_{267} & : & e_1e_2 = e_3 & e_2e_4 = e_5 & e_3e_3 = e_4 & e_3e_4 = e_5 \\ {\mathbf{N}}_{268} & : & e_1e_2 = e_3 & e_3e_3 = e_4 & e_3e_4 = e_5 \\ {\mathbf{N}}_{269} & : & e_1e_2 = e_3 & e_3e_3 = e_4 & e_3e_4 = e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{270} & : & e_1e_2 = e_3 & e_3e_3 = e_4 & e_4e_4 = e_5 \\ \end{array} $

    Here we will collect all information about $ {\mathbf N}_{18}^{4*}: $

    $ \begin{array}{|l|l|l|l|} \hline {\mathbf{N}}^{4*}_{18} & \begin{array}{l} e_1e_1 = e_4 \\ e_1e_2 = e_3 \\ e_3e_3 = e_4 \end{array} & \begin{array}{lcl} \mathrm{H}^2_{\mathfrak{D}}(\mathbf{N}^{4*}_{18})& = &\\ {\langle [\Delta_{11}],[\Delta_{13}],[\Delta_{22}],[\Delta_{23}]\rangle}\\ \mathrm{H}^2_{\mathfrak{C}}(\mathbf{N}^{4*}_{18})& = &\mathrm{H}^2_{\mathfrak{D}}(\mathbf{N}^{4*}_{18})\oplus\\ {\langle [\Delta_{14}], [\Delta_{24}], [\Delta_{34}], [\Delta_{44}] \rangle} \end{array} & \phi_{\pm} = \begin{pmatrix} x&0&0&0\\ 0&\pm 1&0&0\\ 0&0&\pm x &0\\ t&s&0&x^2 \end{pmatrix}\\ \hline \end{array} $

    Let us use the following notations:

    $ \begin{array}{llll} \nabla_1 = [\Delta_{11}], & \nabla_2 = [\Delta_{13}], & \nabla_3 = [\Delta_{14}], & \nabla_4 = [\Delta_{22}], \\ \nabla_5 = [\Delta_{23}], & \nabla_6 = [\Delta_{24}], & \nabla_7 = [\Delta_{34}], & \nabla_8 = [\Delta_{44}]. \end{array} $

    Take $ \theta = \sum\limits_{i = 1}^{8}\alpha_i\nabla_i\in\mathrm{H}^2_{\mathfrak{C}}(\mathbf{N}^{4*}_{18}) . $ Since

    $ \phi_{\pm}^T\begin{pmatrix} \alpha_1&0&\alpha_2&\alpha_3\\ 0&\alpha_4&\alpha_5&\alpha_6\\ \alpha_2&\alpha_5&0&\alpha_7\\ \alpha_3&\alpha_6&\alpha_7&\alpha_8 \end{pmatrix}\phi_{\pm} = \begin{pmatrix} \alpha_1^*&\alpha{*}&\alpha^{*}_2&\alpha_3^*\\ \alpha{*}&\alpha^*_4&\alpha^*_5&\alpha_6^*\\ \alpha^{*}_2&\alpha^*_5&0&\alpha^*_7\\ \alpha^*_3&\alpha^*_6&\alpha^*_7&\alpha^*_8 \end{pmatrix}, $

    we have

    $ \begin{array}{ll} \alpha_1^* = \alpha_1x^2+2\alpha_3xt+\alpha_8t^2, & \alpha_2^* = \pm (\alpha_2x+\alpha_7t) x, \\ \alpha_3^* = (\alpha_3x+\alpha_8t)x^2, & \alpha_4^* = \alpha_4\pm2\alpha_6s+\alpha_8s^2, \\ \alpha_5^* = (\alpha_5\pm\alpha_7s)x, & \alpha_6^* = (\pm\alpha_6+\alpha_8s)x^2, \\ \alpha_7^* = \pm\alpha_7x^3, & \alpha_8^* = \alpha_8x^4. \end{array} $

    We are interested in $ (\alpha_3,\alpha_6,\alpha_7,\alpha_8)\neq(0,0,0,0) . $ Let us consider $ \phi = \phi_+ $ and the following cases:

    $ 1.\ $$ \alpha_8 = 0, \alpha_7 = 0, \alpha_6 = 0, $ then $ \alpha_3\neq0 $ and choosing $ t = -\frac{\alpha_1}{2\alpha_3}x, $ we get $ \alpha_1^* = 0. $ Now consider the following subcases:

    ${\rm{(a)}}\ $if $ \alpha_2 = 0, \alpha_4 = 0, \alpha_5 = 0, $ then we have the representative $ \langle \nabla_3 \rangle ; $

    ${\rm{(b)}}\ $if $ \alpha_2 = 0, \alpha_4 = 0, \alpha_5\neq0, $ then by choosing $ x = \sqrt{\frac{\alpha_5}{\alpha_3}}, s = 0, t = -\frac{\alpha_1\sqrt{\alpha_5}}{2\alpha_3\sqrt{\alpha_3}}, $ we have the representative $ \langle \nabla_3+\nabla_5 \rangle; $

    $ {\rm{(c)}}\ $if $ \alpha_2 = 0, \alpha_4\neq0, $ then by choosing $ x = \sqrt[3]{\frac{\alpha_4}{\alpha_3}}, s = 0, t = -\frac{\alpha_1\sqrt[3]{\alpha_4}}{2\alpha_3\sqrt[3]{\alpha_3}}, $ we have the family of representatives $ \langle \nabla_3+\nabla_4+\alpha\nabla_5 \rangle; $

    $ {\rm{(d)}}\ $if $ \alpha_2\neq0, $ then by choosing $ x = {\alpha_2}{\alpha_3^{-1}}, s = 0, t = -\frac{\alpha_1\alpha_2}{2\alpha^2_3}, $ we have the family of representatives $ \langle \nabla_2+\nabla_3+\alpha\nabla_4+\beta\nabla_5 \rangle. $

    2. $ \alpha_8 = 0, \alpha_7 = 0, \alpha_6\neq0, $ then by choosing $ s = -\frac{\alpha_4}{2\alpha_6}x, $ we have $ \alpha_4^* = 0. $ Consider the following cases:

    $ {\rm{(a)}}\ $$ \alpha_3 = 0, $ then we have two families of representatives $ \langle \alpha\nabla_1+\beta\nabla_2+\nabla_6 \rangle $ and $ \langle \alpha\nabla_1+\beta\nabla_2+\nabla_5+\nabla_6 \rangle $ depending on whether $ \alpha_5 = 0 $ or not;

    $ {\rm{(b)}}\ $$ \alpha_3\neq0 $ then by choosing $ x = \frac{\alpha_6}{\alpha_3}, s = -\frac{\alpha_4}{2\alpha_6}, t = -\frac{\alpha_1\alpha_6}{2\alpha_3^2}, $ we have the family of representatives $ \langle \alpha\nabla_2+\nabla_3+\beta\nabla_5+\nabla_6 \rangle. $

    3. $ \alpha_8 = 0, \alpha_7\neq0, $ then by choosing $ t = -{\alpha_2}{\alpha_7^{-1}}x, s = -{\alpha_5}{\alpha_7^{-1}}, $ we have $ \alpha_2^* = 0, \alpha_5^* = 0. $ Thus, we can suppose that $ \alpha_2 = 0, \alpha_5 = 0 $ and now consider the following cases:

    $ {\rm{(a)}}\ $if $ \alpha_1 = 0, \alpha_4 = 0, \alpha_6 = 0, $ then we have the family of representatives $ \langle \alpha\nabla_3+\nabla_7 \rangle; $

    $ {\rm{(b)}}\ $if $ \alpha_1 = 0, \alpha_4 = 0, \alpha_6\neq0, $ then by choosing $ x = {\alpha_6}{\alpha_7^{-1}}, s = 0, t = 0, $ we have the family of representatives $ \langle \alpha\nabla_3+\nabla_6+\nabla_7 \rangle; $

    $ {\rm{(c)}}\ $if $ \alpha_1 = 0, \alpha_4\neq0, $ then by choosing $ x = \sqrt[3]{{\alpha_4}{\alpha_7^{-1}}}, s = 0, t = 0, $ we have the family of representatives $ \langle \alpha\nabla_3+\nabla_4+\beta\nabla_6+\nabla_7 \rangle; $

    $ {\rm{(d)}}\ $if $ \alpha_1\neq0, $ then by choosing $ x = {\alpha_1}{\alpha_7^{-1}}, s = 0, t = 0, $ we have the family of representatives $ \langle \nabla_1+\alpha\nabla_3+\beta\nabla_4+\gamma\nabla_6+\nabla_7 \rangle. $

    4. $ \alpha_8\neq0, $ then by choosing $ t = -\frac{\alpha_3}{\alpha_8}x, s = -\frac{\alpha_6}{\alpha_8}, $ we have $ \alpha_3^* = 0, \alpha_6^* = 0. $ Thus, we can suppose that $ \alpha_3 = 0, \alpha_6 = 0. $ Consider the following cases:

    $ {\rm{(a)}}\ $if $ \alpha_1 = 0, \alpha_2 = 0, \alpha_4 = 0, \alpha_5 = 0, $ then we have the representatives $ \langle \nabla_8 \rangle $ and $ \langle \nabla_7+\nabla_8 \rangle $ depending on whether $ \alpha_7 = 0 $ or not;

    $ {\rm{(b)}}\ $if $ \alpha_1 = 0, \alpha_2 = 0, \alpha_4 = 0, \alpha_5\neq0, $ then by choosing $ x = \sqrt[3]{{\alpha_5}{\alpha_8^{-1}}}, s = 0, t = 0, $ we have the family of representatives $ \langle \nabla_5+\alpha\nabla_7+\nabla_8 \rangle; $

    $ {\rm{(c)}}\ $if $ \alpha_1 = 0, \alpha_2 = 0, \alpha_4\neq0 $ then by choosing $ x = \sqrt[4]{{\alpha_4}{\alpha_8^{-1}}}, s = 0, t = 0, $ we have the family of representatives $ \langle \nabla_4+\alpha\nabla_5+\beta\nabla_7+\nabla_8 \rangle; $

    $ {\rm{(d)}}\ $if $ \alpha_1 = 0, \alpha_2\neq0, $ then by choosing $ x = \sqrt{{\alpha_2}{\alpha_8^{-1}}}, s = 0, t = 0, $ we have the family of representatives $ \langle \nabla_2+\alpha\nabla_4+\beta\nabla_5+\gamma\nabla_7+\nabla_8 \rangle; $

    $ {\rm{(e)}}\ $if $ \alpha_1\neq0, $ then by choosing $ x = \sqrt{{\alpha_1}{\alpha_8}^{-1}}, s = 0, t = 0, $ we have the family of representatives $ \langle \nabla_1+\alpha\nabla_2+\beta\nabla_4+\gamma\nabla_5+\mu\nabla_7+\nabla_8 \rangle . $

    Summarizing all cases, we have the following distinct orbits:

    $\begin{array}{c} \langle \nabla_1+\alpha\nabla_2+\beta\nabla_4+\gamma\nabla_5+\mu\nabla_7+\nabla_8 \rangle ^{ { \begin{array}{l}O(\alpha,\beta,\gamma,\mu) = O(-\alpha,-\beta,-\gamma,\mu) = \\ O(-\alpha,\beta,\gamma,-\mu) = O(\alpha,-\beta,-\gamma,-\mu) \end{array}}}, \\ \langle \alpha\nabla_1+\beta\nabla_2+\nabla_5+\nabla_6 \rangle^{O(\alpha,\beta) = O(-\alpha,\beta)}, \langle \alpha\nabla_1+\beta\nabla_2+\nabla_6 \rangle^{O(\alpha,\beta) = O(-\alpha,\beta)}, \\ \langle \nabla_1+\alpha\nabla_3+\beta\nabla_4+\gamma\nabla_6+\nabla_7 \rangle^{O(\alpha,\beta,\gamma) = O(-\alpha,\beta,-\gamma)}, \langle \nabla_2+\nabla_3+\alpha\nabla_4+\beta\nabla_5 \rangle^{O(\alpha,\beta) = O(-\alpha,\beta)}, \langle \alpha\nabla_2+\nabla_3 +\beta\nabla_5+\nabla_6 \rangle, \\ \langle \nabla_2+\alpha\nabla_4+\beta\nabla_5+\gamma\nabla_7+\nabla_8 \rangle^{O(\alpha,\beta,\gamma) = O(\alpha,i\beta, i\gamma) = O(\alpha,-i\beta, -i\gamma)}, \langle \nabla_3 \rangle, \\ \langle \nabla_3+\nabla_4+\alpha\nabla_5 \rangle ^{O(\alpha) = O(-\eta_3\alpha) = O(\eta_3^2\alpha)}, \\ \langle \alpha\nabla_3+\nabla_4+\beta\nabla_6+\nabla_7 \rangle^{{ \begin{array}{l} O(\alpha,\beta) = O(-\alpha,-\beta) = O(-\alpha,\eta_3\beta) = \\O(-\alpha,-\eta_3^2\beta) = O(\alpha,-\eta_3\beta) = O(\alpha,\eta_3^2\beta)\end{array}}}, \langle \nabla_3+\nabla_5 \rangle, \\ \langle \alpha\nabla_3+\nabla_6+\nabla_7 \rangle ^{O(\alpha,\beta) = O(-\alpha,\beta)}, \langle \alpha\nabla_3+\nabla_7 \rangle^{O(\alpha) = O(-\alpha)}, \\ \langle \nabla_4+\alpha\nabla_5+\beta\nabla_7+\nabla_8 \rangle ^{ { \begin{array}{l}O(\alpha,\beta) = O(i\alpha,-i\beta) = O(-i\alpha,i\beta) = O(-\alpha,-\beta) = \\ O(\alpha,-\beta) = O(i\alpha,i\beta) = O(-i\alpha,-i\beta) = O(-\alpha,\beta) \end{array}}}, \\ \langle \nabla_5+\alpha\nabla_7+\nabla_8 \rangle^{ O(\alpha) = O(\eta_3\alpha) = O(-\eta_3^2\alpha) = O(-\alpha) = O(-\eta_3\alpha) = O(\eta_3^2\alpha)}, \langle \nabla_7+\nabla_8 \rangle, \langle \nabla_8\rangle, \end{array}$

    which gives the following new algebras:

    $ \begin{array}{llllllllllllllllll} {\mathbf{N}}_{271}^{\alpha, \beta, \gamma,\mu} & : & e_1e_1 = e_4+e_5 & e_1e_2 = e_3 & e_1e_3 = \alpha e_5 & e_2e_2 = \beta e_5 \\ && e_2e_3 = \gamma e_5 & e_3e_3 = e_4 & e_3e_4 = \mu e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{272}^{\alpha, \beta} & : & e_1e_1 = e_4+\alpha e_5 & e_1e_2 = e_3 & e_1e_3 = \beta e_5 \\ & & e_2e_3 = e_5 & e_2e_4 = e_5 & e_3e_3 = e_4 \\ {\mathbf{N}}_{273}^{\alpha, \beta} & : & e_1e_1 = e_4+\alpha e_5 & e_1e_2 = e_3 & e_1e_3 = \beta e_5 & e_2e_4 = e_5 & e_3e_3 = e_4 \\ {\mathbf{N}}_{274}^{\alpha, \beta ,\gamma} & : & e_1e_1 = e_4+e_5 & e_1e_2 = e_3 & e_1e_4 = \alpha e_5 & e_2e_2 = \beta e_5 \\ & & e_2e_4 = \gamma e_5 & e_3e_3 = e_4 & e_3e_4 = e_5 \\ {\mathbf{N}}_{275}^{\alpha, \beta} & : & e_1e_1 = e_4 & e_1e_2 = e_3 & e_1e_3 = e_5 & e_1e_4 = e_5 \\ && e_2e_2 = \alpha e_5 & e_2e_3 = \beta e_5 & e_3e_3 = e_4 \\ {\mathbf{N}}_{276}^{\alpha, \beta} & : & e_1e_1 = e_4 & e_1e_2 = e_3 & e_1e_3 = \alpha e_5 & e_1e_4 = e_5 \\ && e_2e_3 = \beta e_5 & e_2e_4 = e_5 & e_3e_3 = e_4 \\ {\mathbf{N}}_{277}^{\alpha, \beta ,\gamma} & : & e_1e_1 = e_4 & e_1e_2 = e_3 & e_1e_3 = e_5 & e_2e_2 = \alpha e_5 \\ && e_2e_3 = \beta e_5 & e_3e_3 = e_4 & e_3e_4 = \gamma e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{278} & : & e_1e_1 = e_4 & e_1e_2 = e_3 & e_1e_4 = e_5 & e_3e_3 = e_4 \\ {\mathbf{N}}_{279}^{\alpha} & : & e_1e_1 = e_4 & e_1e_2 = e_3 & e_1e_4 = e_5 \\ && e_2e_2 = e_5 & e_2e_3 = \alpha e_5 & e_3e_3 = e_4 \\ {\mathbf{N}}_{280}^{\alpha, \beta} & : & e_1e_1 = e_4 & e_1e_2 = e_3 & e_1e_4 = \alpha e_5 & e_2e_2 = e_5 \\ && e_2e_4 = \beta e_5 & e_3e_3 = e_4 & e_3e_4 = e_5 \\ {\mathbf{N}}_{281} & : & e_1e_1 = e_4 & e_1e_2 = e_3 & e_1e_4 = e_5 & e_2e_3 = e_5 & e_3e_3 = e_4 \\ {\mathbf{N}}_{282}^{\alpha} & : & e_1e_1 = e_4 & e_1e_2 = e_3 & e_1e_4 = \alpha e_5 \\ && e_2e_4 = e_5 & e_3e_3 = e_4 & e_3e_4 = e_5 \\ {\mathbf{N}}_{283}^{\alpha} & : & e_1e_1 = e_4 & e_1e_2 = e_3 & e_1e_4 = \alpha e_5 & e_3e_3 = e_4 & e_3e_4 = e_5 \\ {\mathbf{N}}_{284}^{\alpha, \beta} & : & e_1e_1 = e_4 & e_1e_2 = e_3 & e_2e_2 = e_5 & e_2e_3 = \alpha e_5 \\ & & e_3e_3 = e_4 & e_3e_4 = \beta e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{285}^{\alpha} & : & e_1e_1 = e_4 & e_1e_2 = e_3 & e_2e_3 = e_5 \\ && e_3e_3 = e_4 & e_3e_4 = \alpha e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{286} & : & e_1e_1 = e_4 & e_1e_2 = e_3 & e_3e_3 = e_4 & e_3e_4 = e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{287} & : & e_1e_1 = e_4 & e_1e_2 = e_3 & e_3e_3 = e_4 & e_4e_4 = e_5 \\ \end{array} $

    Here we will collect all information about $ {\mathbf N}_{19}^{4*}: $

    ${\mathbf{N}}^{4*}_{19} $ $\begin{array}{ll} e_1e_1 = e_4 & e_1e_2 = e_3 \\ e_2e_2 = e_4 & e_3e_3 = e_4 \end{array} $ $\begin{array}{l} \mathrm{H}^2_{\mathfrak{D}}(\mathbf{N}^{4*}_{19}) = \langle [\Delta_{11}],[\Delta_{13}],[\Delta_{22}],[\Delta_{23}]\rangle\\ \mathrm{H}^2_{\mathfrak{C}}(\mathbf{N}^{4*}_{19}) = \mathrm{H}^2_{\mathfrak{D}}(\mathbf{N}^{4*}_{19})\oplus \langle [\Delta_{14}], [\Delta_{24}], [\Delta_{34}], [\Delta_{44}] \rangle \end{array} $
    ${ \begin{array}{l} \begin{array}{l} \phi_1 = \begin{pmatrix} x&0&0&0\\ 0 & q&0&0\\ 0&0 & xq&0\\ t & s&0&1 \end{pmatrix}, \\ {x^2 = 1,q^2 = 1} \end{array} \begin{array}{l} \phi_2 = \begin{pmatrix} 0 & p&0&0\\ y&0&0&0\\ 0&0 & yp&0\\ t & s&0&1 \end{pmatrix}, \\ {y^2 = 1, p^2 = 1} \end{array} \end{array} } $

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    Let us use the following notations:

    $ \begin{array}{llll} \nabla_1 = [\Delta_{11}], & \nabla_2 = [\Delta_{13}], & \nabla_3 = [\Delta_{14}], & \nabla_4 = [\Delta_{22}], \\ \nabla_5 = [\Delta_{23}], & \nabla_6 = [\Delta_{24}], & \nabla_7 = [\Delta_{34}], & \nabla_8 = [\Delta_{44}]. \end{array} $

    Take $ \theta = \sum\limits_{i = 1}^{8}\alpha_i\nabla_i\in\mathrm{H}^2_{\mathfrak{C}}(\mathbf{N}^{4*}_{19}) . $ Since

    $ \phi^T\begin{pmatrix} \alpha_1&0&\alpha_2&\alpha_3\\ 0&\alpha_4&\alpha_5&\alpha_6\\ \alpha_2&\alpha_5&0&\alpha_7\\ \alpha_3&\alpha_6&\alpha_7&\alpha_8 \end{pmatrix}\phi = \begin{pmatrix} \alpha_1^*&\alpha^{*}&\alpha^{*}_2&\alpha_3^*\\ \alpha^{*}&\alpha^*_4&\alpha^*_5&\alpha_6^*\\ \alpha^{*}_2&\alpha^*_5&0&\alpha^*_7\\ \alpha^*_3&\alpha^*_6&\alpha^*_7&\alpha^*_8 \end{pmatrix}, $

    then, in the case $ \phi = \phi_1^{x = 1,q = 1}, $ we have

    $ \begin{array}{llll} \alpha_1^* = \alpha_1+2\alpha_3t+\alpha_8t^2, & \alpha_2^* = \alpha_2+\alpha_7t, & \alpha_3^* = \alpha_3+\alpha_8t, & \alpha_4^* = \alpha_4+2\alpha_6s+\alpha_8s^2, \\ \alpha_5^* = \alpha_5+\alpha_7s, & \alpha_6^* = \alpha_6+\alpha_8s, & \alpha_7^* = \alpha_7, & \alpha_8^* = \alpha_8. \end{array} $

    For define the main families of representatives, we will use $ \phi = \phi_1^{x = 1,q = 1} $ and for find equal orbits we will use other automorphisms. We are interested in

    $ (\alpha_3,\alpha_6,\alpha_7,\alpha_8)\neq(0,0,0,0) .$

    Let us consider the following cases:

    $1.\ $if $ \alpha_8 = 0, \alpha_7 = 0, \alpha_6 = 0, $ then $ \alpha_3\neq0 $ and choosing $ t = -\frac{\alpha_1}{2\alpha_3}, $ we have the family of representatives $ \langle \alpha\nabla_2+\nabla_3+\beta\nabla_4+\gamma\nabla_5 \rangle; $

    $ 2.\ $if $ \alpha_8 = 0, \alpha_7 = 0, \alpha_6\neq0 $ and $ \alpha_3 = 0, $ then by choosing some suitable automorphism $ \phi_2 $ we have $ \alpha_3^*\neq0 , $ $ \alpha_6^* = 0, $ which is the case considered above;

    $ 3.\ $if $ \alpha_8 = 0, \alpha_7 = 0, \alpha_6\neq0, \alpha_3\neq0, $ then by choosing $ t = -\frac{\alpha_1}{2\alpha_3}, s = \frac{\alpha_4}{2\alpha_6}, $ we have the family of representatives $ \langle \alpha\nabla_2+\beta\nabla_3+\gamma\nabla_5+\nabla_6 \rangle_{\beta\neq0} ; $

    $ 4.\ $if $ \alpha_8 = 0, \alpha_7\neq0, $ then by choosing $ t = -{\alpha_2}{\alpha_7}^{-1}, s = -{\alpha_5}{\alpha_7}^{-1}, $ we have the family of representatives $ \langle \alpha\nabla_1+\beta\nabla_3+\gamma\nabla_4+\mu\nabla_6+\nabla_7 \rangle; $

    $5.\ $if $ \alpha_8\neq0, $ then by choosing $ t = -{\alpha_3}{\alpha_8}^{-1}, s = -{\alpha_6}{\alpha_8}^{-1}, $ we have the family of representatives $ \langle \alpha\nabla_1+\beta\nabla_2+\gamma\nabla_4+\mu\nabla_5+\nu\nabla_7+\nabla_8 \rangle. $

    Summarizing, we have the following distinct orbits:

    $ \begin{array}{c} \langle \alpha\nabla_1+\beta\nabla_2+\gamma\nabla_4+\mu\nabla_5+\nu\nabla_7+\nabla_8 \rangle ^{{ \begin{array}{l} O(\alpha,\beta,\gamma,\mu,\nu) = O(\alpha,-\beta,\gamma,\mu,-\nu) = \\ O(\alpha,\beta,\gamma,-\mu,-\nu) = O(\alpha,-\beta,\gamma,-\mu,\nu) = \\ O(\gamma,\mu,\alpha,\beta,\nu) = O(\gamma,-\mu,\alpha,\beta,-\nu) = \\ O(\gamma,\mu,\alpha,-\beta,-\nu) = O(\gamma,-\mu,\alpha,-\beta,\nu) \end{array}}}, \\ \langle \alpha\nabla_1+\beta\nabla_3+\gamma\nabla_4+\mu\nabla_6+\nabla_7 \rangle ^{{ \begin{array}{l} O(\alpha,\beta,\gamma,\mu) = O(-\alpha,-\beta,-\gamma,\mu) = \\ O(-\alpha,\beta,-\gamma,-\mu) = O(\alpha,-\beta,\gamma,-\mu) = \\ O(\gamma,\mu,\alpha,\beta) = O(-\gamma,-\mu,-\alpha,\beta) = \\ O(-\gamma,\mu,-\alpha,-\beta) = O(\gamma,-\mu,\alpha,-\beta) \end{array}}}, \\ \langle \alpha\nabla_2+\nabla_3+\beta\nabla_4+\gamma\nabla_5 \rangle^{{ \begin{array}{l} O(\alpha,\beta,\gamma) = O(-\alpha,\beta,\gamma) = O(-\alpha,-\beta,\gamma) = O(\alpha,-\beta,\gamma) = \end{array}}}, \\ \langle \alpha\nabla_2+\beta\nabla_3+\gamma\nabla_5+\nabla_6 \rangle_{\beta \neq0}^{{ \begin{array}{l} O(\alpha,\beta,\gamma) = O(\alpha,-\beta,-\gamma) = O(\frac{\gamma}{\beta},\frac{1}{\beta},\frac{\alpha}{\beta}) = O(\frac{\gamma}{\beta},-\frac{1}{\beta},-\frac{\alpha}{\beta}) \end{array}}},\end{array}$

    which gives the following new algebras:

    $ \begin{array}{llllllllllllllllll} {\mathbf{N}}_{288}^{\alpha, \beta, \gamma,\mu,\nu} & : & e_1e_1 = e_4+\alpha e_5 & e_1e_2 = e_3 & e_1e_3 = \beta e_5 & e_2e_2 = e_4+\gamma e_5 \\ & & e_2e_3 = \mu e_5 & e_3e_3 = e_4 & e_3e_4 = \nu e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{289}^{\alpha, \beta, \gamma,\mu} & : & e_1e_1 = e_4+\alpha e_5 & e_1e_2 = e_3 & e_1e_4 = \beta e_5 & e_2e_2 = e_4+\gamma e_5 \\ & & e_2e_4 = \mu e_5 & e_3e_3 = e_4 & e_3e_4 = e_5 \\ {\mathbf{N}}_{290}^{\alpha, \beta} & : & e_1e_1 = e_4 & e_1e_2 = e_3 & e_1e_3 = \alpha e_5 & e_1e_4 = e_5 \\ & & e_2e_2 = e_4+\beta e_5 & e_2e_3 = \gamma e_5 & e_3e_3 = e_4 \\ {\mathbf{N}}_{291}^{\alpha, \beta\neq0 ,\gamma} & : & e_1e_1 = e_4 & e_1e_2 = e_3 & e_1e_3 = \alpha e_5 & e_1e_4 = \beta e_5 \\ & & e_2e_3 = \gamma e_5 & e_2e_4 = e_5 & e_3e_3 = e_4 \\ \end{array} $

    Here we will collect all information about $ {\mathbf N}_{01}^{4}: $

    $ \begin{array}{|l|l|l|l|} \hline \rm{ } & \rm{ } & \rm{Cohomology} & \rm{Automorphisms} \\ \hline {\mathbf{N}}^{4}_{01} & \begin{array}{l} e_1e_1 = e_2 \\ e_1e_2 = e_3 \\ e_2e_3 = e_4 \end{array} &\begin{array}{l}\mathrm{H}^2_{\mathfrak{C}}(\mathbf{N}^{4}_{01}) = \Big \langle [\Delta_{ij}] \Big\rangle\\ (i,j) \notin \{ (1,1),(1,2),(2,3)\} \end{array} & \phi = \begin{pmatrix} x&0&0&0\\ 0 & x^2&0&0\\ z&0 & x^3&0\\ t&0 & x^2z & x^5 \end{pmatrix}\\ \hline \end{array} $

    Let us use the following notations:

    $ \begin{array}{llll} \nabla_1 = [\Delta_{13}], & \nabla_2 = [\Delta_{14}], & \nabla_3 = [\Delta_{22}], & \nabla_4 = [\Delta_{24}], \\ \nabla_5 = [\Delta_{33}], & \nabla_6 = [\Delta_{34}], & \nabla_7 = [\Delta_{44}]. \end{array} $

    Take $ \theta = \sum\limits_{i = 1}^{7}\alpha_i\nabla_i\in\mathrm{H}^2_{\mathfrak{C}}(\mathbf{N}^{4}_{01}) . $ Since

    $ \phi^T\begin{pmatrix} 0&0&\alpha_1&\alpha_2\\ 0&\alpha_3&0&\alpha_4\\ \alpha_1&0&\alpha_5&\alpha_6\\ \alpha_2&\alpha_4&\alpha_6&\alpha_7 \end{pmatrix}\phi = \begin{pmatrix} \alpha^*&\alpha^{**}&\alpha^{*}_1&\alpha^*_2\\ \alpha^{**}&\alpha^*_3&\alpha^{***}&\alpha^*_4\\ \alpha^{*}_1&\alpha^{***}&\alpha^*_5&\alpha^*_6\\ \alpha^*_2&\alpha^*_4&\alpha^*_6&\alpha^*_7 \end{pmatrix} $

    we have

    $ \begin{array}{lll} { \alpha_1^* = \big((\alpha_1x+\alpha_5z+\alpha_6t)x+(\alpha_2x+\alpha_6z+\alpha_7t)z\big)x^2, }\\ \alpha_2^* = (\alpha_2x+\alpha_6z+\alpha_7t)x^5, & \alpha_3^* = \alpha_3x^4, & \alpha_4^* = \alpha_4x^7, \\ \alpha_5^* = (\alpha_5x^{2}+2\alpha_6xz+\alpha_7z^2)x^4, & \alpha_6^* = (\alpha_6x+\alpha_7z)x^{7}, & \alpha_7^* = \alpha_7x^{10}. \end{array} $

    We are interested in $ (\alpha_2,\alpha_4,\alpha_6,\alpha_7)\neq(0,0,0,0) $ and consider following cases:

    $ 1.\ $$ \alpha_7 = \alpha_6 = \alpha_4 = 0, $ then $ \alpha_2\neq0, $ and we have the following subcases:

    ${\rm{(a)}}\ $if $ \alpha_5 = -\alpha_2, $ then we have

    $ {\rm{(i)}}\ $ if $ \alpha_1 = 0, \alpha_3 = 0, $ then we have the representative $ \langle \nabla_2-\nabla_5 \rangle; $

    $ {\rm{(ii)}}\ $ if $ \alpha_1 = 0, \alpha_3\neq0, $ then by choosing $ x = \sqrt{{\alpha_3}{\alpha_2}^{-1}}, z = 0, t = 0, $ we have the representative $ \langle \nabla_2+\nabla_3-\nabla_5 \rangle; $

    $ {\rm{(iii)}}\ $if $ \alpha_1\neq0, $ then by choosing $ x = \sqrt{{\alpha_1}{\alpha_2}^{-1}}, z = 0, t = 0, $ we have the family of representatives $ \langle \nabla_1+\nabla_2+\alpha\nabla_3-\nabla_5 \rangle; $

    (b) if $ \alpha_5\neq-\alpha_2, $ then by choosing $ z = -\frac{\alpha_1}{\alpha_5+\alpha_2}x, t = 0, $ we have the family of representatives $ \langle \nabla_2+\alpha\nabla_5 \rangle_{\alpha\neq-1} $ and $ \langle \nabla_2+\nabla_3+\alpha\nabla_5 \rangle_{\alpha\neq-1} $ depending on whether $ \alpha_3 = 0 $ or not, which will be jointed with the cases (1(a)i) and (1(a)ii).

    2. $ \alpha_7 = 0, \alpha_6 = 0, \alpha_4\neq0, $ then we have the following subcases:

    ${\rm{(a)}}\ $if $ \alpha_5 = -\alpha_2, \alpha_1 = 0, $

    $ {\rm{(i)}}\ $ if $ \alpha_3 = 0, $ then we have the representatives $ \langle \nabla_4 \rangle $ and $ \langle \nabla_2+\nabla_4-\nabla_5 \rangle $ depending on whether $ \alpha_2 = 0 $ or not;

    $ {\rm{(ii)}}\ $ if $ \alpha_3\neq0, $ then by choosing $ x = \sqrt[3]{{\alpha_3}{\alpha_4}^{-1} }, $ we have the family of representatives $ \langle \alpha\nabla_2+\nabla_3+ \nabla_4-\alpha\nabla_5 \rangle; $

    (b) if $ \alpha_5 = -\alpha_2, \alpha_1\neq0, $ then by choosing $ x = \sqrt[3]{{\alpha_1}{\alpha_4}^{-1} }, $ we have the family of representatives $ \langle \nabla_1+\alpha\nabla_2+\beta\nabla_3+ \nabla_4-\alpha\nabla_5 \rangle; $

    (c) if $ \alpha_5\neq-\alpha_2, $ then we have

    ${\rm{(i)}}\ $if $ \alpha_3 = 0, \alpha_2 = 0, $ then by choosing $ x = \frac{\alpha_5}{\alpha_4}, z = -\frac{\alpha_1\alpha_5}{\alpha_4(\alpha_2+\alpha_5)}, t = 0, $ we have the representative $ \langle \nabla_4+\nabla_5 \rangle; $

    $ {\rm{(ii)}}\ $if $ \alpha_3 = 0, \alpha_2\neq0, $ then by choosing $ x = \frac{\alpha_2}{\alpha_4}, z = -\frac{\alpha_1\alpha_2}{\alpha_4(\alpha_2+\alpha_5)}, t = 0, $ we have the family of representatives $ \langle \nabla_2+\nabla_4+\alpha\nabla_5 \rangle_{\alpha\neq-1}, $ which will be jointed with a representative from the case (2(a)i);

    $ {\rm{(iii)}}\ $if $ \alpha_3\neq0, $ then by choosing $ x = \sqrt[3]{{\alpha_3}{\alpha_4}^{-1}}, z = -\frac{\alpha_1\sqrt[3]{\alpha_3}}{(\alpha_2+\alpha_5)\sqrt[3]{\alpha_4}}, t = 0, $ we have the family of representatives $ \langle \alpha\nabla_2+\nabla_3+\nabla_4+\beta\nabla_5 \rangle_{\beta\neq-\alpha}, $ which will be jointed with the family from the case (2(a)i).

    3. $ \alpha_7 = 0, \alpha_6\neq0, $ then we consider the following subcases:

    $ {\rm{(a)}}\ $if $ \alpha_3 = 0, \alpha_4 = 0, $ then choosing $ z = -\frac{\alpha_2}{\alpha_6}x, t = -\frac{\alpha_1x+\alpha_5z}{\alpha_6}, $ we have representatives $ \langle \nabla_6 \rangle $ and $ \langle \nabla_5+\nabla_6 \rangle $ depending on whether $ \alpha_5 = 2\alpha_2 $ or not;

    ${\rm{(b)}}\ $if $ \alpha_3 = 0, \alpha_4\neq0, $ then by choosing $ x = \frac{\alpha_4}{\alpha_6}, z = -\frac{\alpha_2\alpha_4}{\alpha^2_6}, t = \frac{\alpha_4(\alpha_2\alpha_5-\alpha_1\alpha_6)}{\alpha^3_6}, $ we have the family of representatives $ \langle \nabla_4+\alpha\nabla_5+\nabla_6 \rangle; $

    $ {\rm{(c)}}\ $if $ \alpha_3\neq0, $ then by choosing $ x = \sqrt[4]{\frac{\alpha_3}{\alpha_6}}, z = -\frac{\alpha_2\sqrt[4]{\alpha_3}}{\alpha_6\sqrt[4]{\alpha_6}}, t = \frac{(\alpha_2\alpha_5-\alpha_1\alpha_6)\sqrt[4]{\alpha_3}}{\alpha_6^2\sqrt[4]{\alpha_6}}, $ we have the family of representatives $ \langle \nabla_3+\alpha\nabla_4+\beta\nabla_5+\nabla_6 \rangle. $

    4. $ \alpha_7\neq0, $ then by choosing $ z = -\frac{\alpha_6}{\alpha_7}x, t = \frac{\alpha_6^2+\alpha_2\alpha_7}{\alpha^2_7}x, $ we have $ \alpha_2^* = 0, \alpha_6^* = 0. $ Thus, we can suppose that $ \alpha_2 = 0, \alpha_6 = 0 $ and now consider following subcases:

    $ {\rm{(a)}}\ $$ \alpha_1 = 0, \alpha_3 = 0, \alpha_4 = 0, $ then we have representatives $ \langle \nabla_7 \rangle $ and $ \langle \nabla_5+\nabla_7 \rangle $ depending on whether $ \alpha_5\alpha_7-\alpha^2_6 = 0 $ or not;

    $ {\rm{(b)}}\ $$ \alpha_1 = 0, \alpha_3 = 0, \alpha_4\neq0, $ then by choosing $ x = \sqrt[3]{{\alpha_4}{\alpha_7}^{-1}}, $ we have the family of representatives $ \langle \nabla_4+\alpha\nabla_5+\nabla_7 \rangle; $

    ${\rm{(c)}}\ $$ \alpha_1 = 0, \alpha_3\neq0, $ then by choosing $ x = \sqrt[6]{{\alpha_3}{\alpha_7}^{-1}}, $ we have the family of representatives $ \langle \nabla_3+\alpha\nabla_4+\beta\nabla_5+\nabla_7 \rangle; $

    $ {\rm{(d)}}\ $$ \alpha_1\neq0, $ then by choosing $ x = \sqrt[6]{{\alpha_1}{\alpha_7^{-3}}}, $ we have the family of representatives $ \langle \nabla_1+\alpha\nabla_3+\beta\nabla_4+\gamma\nabla_5+\nabla_7 \rangle. $

    Summarizing all cases, we have the following distinct orbits:

    $\begin{array}{c} \langle \nabla_1+ \alpha \nabla_2 + \beta \nabla_3 + \nabla_4 -\alpha \nabla_5 \rangle^ {O(\alpha, \beta) = O(-\eta_3\alpha, \eta_3 \beta) = O(\eta_3^2\alpha,-\eta_3^2\beta)}, \\ \langle \nabla_1+ \nabla_2 + \alpha \nabla_3 - \nabla_5 \rangle, \\ \langle \nabla_1+ \alpha \nabla_3 + \beta \nabla_4 + \gamma \nabla_5 + \nabla_7 \rangle^ {{ \begin{array}{l} O(\alpha, \beta, \gamma) = O(\alpha, \beta, -\eta_3\gamma) = O(\alpha, -\beta, -\eta_3\gamma) = \\ O(\alpha, -\beta, \eta_3^2 \gamma) = O(\alpha, \beta, \eta_3^2\gamma) = O(\alpha, -\beta, \gamma) \end{array} }}, \\ \langle \alpha \nabla_2 + \nabla_3 + \nabla_4 +\beta \nabla_5 \rangle ^{O(\alpha, \beta) = O(-\eta_3\alpha, -\eta_3 \beta) = O(\eta_3^2\alpha, \eta_3^2\beta)}, \langle \nabla_2+ \nabla_3 + \alpha \nabla_5 \rangle, \\ \langle \nabla_2+ \nabla_4 + \alpha \nabla_5 \rangle, \langle \nabla_2 + \alpha \nabla_5 \rangle, \langle \nabla_3+\alpha\nabla_4+\beta\nabla_5+\nabla_6 \rangle ^{{ \begin{array}{l} O(\alpha, \beta) = O(-i\alpha, -\beta) = \\ O(i\alpha, -\beta) = O(-\alpha, \beta) \end{array}}}, \\ \langle \nabla_3 + \alpha\nabla_4 + \beta \nabla_5 + \nabla_7 \rangle^ {{ \begin{array}{l} O(\alpha, \beta) = O(\alpha, -\eta_3 \beta) = O(-\alpha, -\eta_3\beta) = \\ O(-\alpha, \eta_3^2 \beta) = O(\alpha, \eta_3^2 \beta) = O(-\alpha, \beta) \end{array}}}, \langle \nabla_4 \rangle, \langle \nabla_4 + \nabla_5 \rangle, \\ \langle \nabla_4+\alpha\nabla_5+\nabla_6 \rangle, \langle \nabla_4 + \alpha \nabla_5 + \nabla_7 \rangle^{O(\alpha) = O(-\eta_3\alpha) = O(\eta_3^2\alpha)}, \langle \nabla_5+\nabla_6 \rangle, \langle \nabla_5 + \nabla_7 \rangle, \\ \langle \nabla_6 \rangle, \langle \nabla_7 \rangle. \end{array}$

    Hence, we have the following new algebras:

    $ \begin{array}{llllllllllllllllll} {\mathbf{N}}_{292}^{\alpha, \beta} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_1e_3 = e_5 & e_1e_4 = \alpha e_5 \\ && e_2e_2 = \beta e_5 & e_2e_3 = e_4 & e_2e_4 = e_5 & e_3e_3 = -\alpha e_5 \\ {\mathbf{N}}_{293}^{\alpha} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_1e_3 = e_5 & e_1e_4 = e_5 \\ && e_2e_2 = \alpha e_5 & e_2e_3 = e_4 & e_3e_3 = -e_5 \\ {\mathbf{N}}_{294}^{\alpha, \beta ,\gamma} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_1e_3 = e_5 & e_2e_2 = \alpha e_5 \\ && e_2e_3 = e_4 & e_2e_4 = \beta e_5 & e_3e_3 = \gamma e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{295}^{\alpha, \beta} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_1e_4 = \alpha e_5 & e_2e_2 = e_5 \\ && e_2e_3 = e_4 & e_2e_4 = e_5 & e_3e_3 = \beta e_5 \\ {\mathbf{N}}_{296}^{\alpha} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_1e_4 = e_5 \\ && e_2e_2 = e_5 & e_2e_3 = e_4 & e_3e_3 = \alpha e_5 \\ {\mathbf{N}}_{297}^{\alpha} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_1e_4 = e_5 \\ && e_2e_3 = e_4 & e_2e_4 = e_5 & e_3e_3 = \alpha e_5 \\ {\mathbf{N}}_{298}^{\alpha} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_1e_4 = e_5 & e_2e_3 = e_4 & e_3e_3 = \alpha e_5 \\ {\mathbf{N}}_{299}^{\alpha, \beta} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_2e_2 = e_5 & e_2e_3 = e_4 \\ && e_2e_4 = \alpha e_4 & e_3e_3 = \beta e_5 & e_3e_4 = e_5 \\ {\mathbf{N}}_{300}^{\alpha, \beta} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_2e_2 = e_5 & e_2e_3 = e_4 \\ && e_2e_4 = \alpha e_4 & e_3e_3 = \beta e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{301} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_2e_3 = e_4 & e_2e_4 = e_5 \\ {\mathbf{N}}_{302} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_2e_3 = e_4 & e_2e_4 = e_5 & e_3e_3 = e_5 \\ {\mathbf{N}}_{303}^{\alpha} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_2e_3 = e_4 \\ & & e_2e_4 = e_5 & e_3e_3 = \alpha e_5 & e_3e_4 = e_5 \\ {\mathbf{N}}_{304}^{\alpha} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_2e_3 = e_4 \\ & & e_2e_4 = e_5 & e_3e_3 = \alpha e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{305} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_2e_3 = e_4 & e_3e_3 = e_5 & e_3e_4 = e_5 \\ {\mathbf{N}}_{306} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_2e_3 = e_4 & e_3e_3 = e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{307} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_2e_3 = e_4 & e_4e_4 = e_5 \\ {\mathbf{N}}_{308} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_2e_3 = e_4 & e_4e_4 = e_5 \\ \end{array} $

    Here we will collect all information about $ {\mathbf N}_{02}^{4}: $

    $ \begin{array}{|l|l|l|l|} \hline \rm{ } & \rm{ } & \rm{Cohomology} & \rm{Automorphisms} \\ \hline {\mathbf{N}}^{4}_{02} & \begin{array}{l} e_1e_1 = e_2 \\ e_1e_2 = e_3 \\ e_1e_3 = e_4 \\ e_2e_3 = e_4 \end{array} & \begin{array}{l} \mathrm{H}^2_{\mathfrak{C}}(\mathbf{N}^{4}_{02}) = \Big \langle [\Delta_{ij}] \Big\rangle\\ (i,j) \notin \{ (1,1),(1,2),(1,3)\} \end{array} & \phi = \begin{pmatrix} 1&0&0&0\\ 0&1&0&0\\ z&0&1&0\\ t&2z&z&1 \end{pmatrix}\\ \hline \end{array} $

    Let us use the following notations:

    $ \begin{array}{llll} \nabla_1 = [\Delta_{14}], & \nabla_2 = [\Delta_{22}], & \nabla_3 = [\Delta_{23}], & \nabla_4 = [\Delta_{24}], \\ \nabla_5 = [\Delta_{33}], & \nabla_6 = [\Delta_{34}], & \nabla_7 = [\Delta_{44}]. \end{array} $

    Take $ \theta = \sum\limits_{i = 1}^{7}\alpha_i\nabla_i\in\mathrm{H}^2_{\mathfrak{C}}(\mathbf{N}^{4}_{02}) . $ Since

    $ \phi^T\begin{pmatrix} 0&0&0&\alpha_1\\ 0&\alpha_2&\alpha_3&\alpha_4\\ 0&\alpha_3&\alpha_5&\alpha_6\\ \alpha_1&\alpha_4&\alpha_6&\alpha_7 \end{pmatrix}\phi = \begin{pmatrix} \alpha^*&\alpha^{**}&\alpha^{***}&\alpha^*_1\\ \alpha^{**}&\alpha^*_2&\alpha^*_3+\alpha^{***}&\alpha^*_4\\ \alpha^{***}&\alpha^*_3+\alpha^{***}&\alpha^*_5&\alpha^*_6\\ \alpha^*_1&\alpha^*_4&\alpha^*_6&\alpha^*_7 \end{pmatrix} $

    we have

    $ \begin{array}{lcl} \alpha_1^* & = & \alpha_1+\alpha_6z+\alpha_7t, \\ \alpha_2^* & = & \alpha_2+4\alpha_4z+4\alpha_7z^2, \\ \alpha_3^* & = & \alpha_3+2\alpha_6z+(\alpha_4+2\alpha_7z)z-(\alpha_5z+\alpha_6t)-(\alpha_1+\alpha_6z+\alpha_7t)z, \\ \alpha_4^* & = & \alpha_4+2\alpha_7z, \\ \alpha_5^* & = & \alpha_5+2\alpha_6z+\alpha_7z^2, \\ \alpha_6^* & = & \alpha_6+\alpha_7z, \\ \alpha_7^* & = & \alpha_7. \end{array} $

    We are interested in $ (\alpha_1,\alpha_4,\alpha_6,\alpha_7)\neq(0,0,0,0) $ and consider following cases:

    $1.\ $if $ \alpha_7 = \alpha_6 = \alpha_4 = 0, $ then $ \alpha_1\neq0, $ and we have

    $ {\rm{(a)}}\ $if $ \alpha_5 = -\alpha_1 , $ then we have the family of representatives

    $ \langle \nabla_1+\alpha\nabla_2+\beta\nabla_3-\nabla_5 \rangle;$

    $ {\rm{(b)}}\ $if $ \alpha_5\neq-\alpha_1 , $ then by choosing $ z = -\frac{\alpha_3}{\alpha_1+\alpha_5}, t = 0, $ we have the family of representatives $ \langle \nabla_1+\alpha\nabla_2+\beta\nabla_5 \rangle_{\beta\neq-1}; $

    2. if $ \alpha_7 = 0, \alpha_6 = 0, \alpha_4\neq0, $ then by choosing $ z = -\frac{\alpha_2}{4\alpha_4}, t = 0, $ we have the family of representatives $ \langle \alpha\nabla_1+\beta\nabla_3+\nabla_4+\gamma\nabla_5 \rangle; $

    3. if $ \alpha_7 = 0, \alpha_6\neq0, $ then by choosing

    $ z = -{\alpha_1}{\alpha_6}^{-1}, t = ({\alpha_3\alpha_6-\alpha_1(2\alpha_6+\alpha_4-\alpha_5)}){\alpha_6^{-1}},$

    we have the family of representatives $ \langle \alpha\nabla_2+\beta\nabla_4+\gamma\nabla_5+\nabla_6 \rangle; $

    4. if $ \alpha_7\neq0, $ then by choosing $ z = -{\alpha_6}{\alpha_7}^{-1}, t = ({\alpha^2_6-\alpha_1\alpha_7}){\alpha^{-2}_7}, $ we have the family of representatives $ \langle \alpha\nabla_2+\beta\nabla_3+\gamma\nabla_4+\mu\nabla_5+\nabla_7 \rangle. $

    Summarizing, we have the following distinct orbits:

    $ \begin{array}{c} \langle \nabla_1+ \alpha \nabla_2 + \beta \nabla_3 - \nabla_5 \rangle, \langle \nabla_1+ \alpha \nabla_2 + \beta \nabla_5 \rangle_{\beta\neq -1}, \langle \alpha \nabla_1+ \beta \nabla_3 + \nabla_4 + \gamma\nabla_5 \rangle, \\ \langle \alpha \nabla_2 + \beta \nabla_3 + \gamma\nabla_4 + \mu \nabla_5 + \nabla_7\rangle, \langle \alpha \nabla_2 + \beta \nabla_4 +\gamma\nabla_5+ \nabla_6\rangle, \end{array} $

    which gives the following new algebras:

    $ \begin{array}{llllllllllllllllll} {\mathbf{N}}_{309}^{\alpha, \beta} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_1e_3 = e_4 & e_1e_4 = e_5 \\ & & e_2e_2 = \alpha e_5 & e_2e_3 = e_4+\beta e_5 & e_3e_3 = -e_5 \\ {\mathbf{N}}_{310}^{\alpha, \beta\neq-1} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_1e_3 = e_4 & e_1e_4 = e_5 \\ && e_2e_2 = \alpha e_5 & e_2e_3 = e_4 & e_3e_3 = \beta e_5 \\ {\mathbf{N}}_{311}^{\alpha, \beta,\gamma} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_1e_3 = e_4 & e_1e_4 = \alpha e_5 \\ & & e_2e_3 = e_4+\beta e_5 & e_2e_4 = e_5 & e_3e_3 = \gamma e_5 \\ {\mathbf{N}}_{312}^{\alpha, \beta,\gamma, \mu} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_1e_3 = e_4 & e_2e_2 = \alpha e_5 \\ && e_2e_3 = e_4+\beta e_5 & e_2e_4 = \gamma e_5 & e_3e_3 = \mu e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{313}^{\alpha, \beta,\gamma} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_1e_3 = e_4 & e_2e_2 = \alpha e_5 \\ & & e_2e_3 = e_4 & e_2e_4 = \beta e_5 & e_3e_3 = \gamma e_5 & e_3e_4 = e_5 \\ \end{array} $

    Here we will collect all information about $ {\mathbf N}_{03}^{4}: $

    $ \begin{array}{|l|l|l|l|} \hline \rm{ } & \rm{ } & \rm{Cohomology} & \rm{Automorphisms} \\ \hline {\mathbf{N}}^{4}_{03} & \begin{array}{l} e_1e_1 = e_2 \\ e_1e_2 = e_3 \\ e_3e_3 = e_4 \end{array} &\begin{array}{l} \mathrm{H}^2_{\mathfrak{C}}(\mathbf{N}^{4}_{03}) = \Big \langle [\Delta_{ij}] \Big\rangle\\ {(i,j) \notin \{ (1,1),(1,2),(3,3)}\} \end{array} & \phi = \begin{pmatrix} x&0&0&0\\ 0&x^2&0&0\\ 0&0&x^3&0\\ t&0&0&x^6 \end{pmatrix}\\ \hline \end{array} $

    Let us use the following notations:

    $ \begin{array}{llll} \nabla_1 = [\Delta_{13}], & \nabla_2 = [\Delta_{14}], & \nabla_3 = [\Delta_{22}], & \nabla_4 = [\Delta_{23}], \\ \nabla_5 = [\Delta_{24}], & \nabla_6 = [\Delta_{34}], & \nabla_7 = [\Delta_{44}]. \end{array} $

    Take $ \theta = \sum\limits_{i = 1}^{7}\alpha_i\nabla_i\in\mathrm{H}^2_{\mathfrak{C}}(\mathbf{N}^{4}_{03}) . $ Since

    $ \phi^T\begin{pmatrix} 0&0&\alpha_1&\alpha_2\\ 0&\alpha_3&\alpha_4&\alpha_5\\ \alpha_1&\alpha_4&0&\alpha_6\\ \alpha_2&\alpha_5&\alpha_6&\alpha_7 \end{pmatrix}\phi = \begin{pmatrix} \alpha^*&\alpha^{**}&\alpha^{*}_1&\alpha^*_2\\ \alpha^{**}&\alpha^*_3&\alpha^{*}_4&\alpha^*_5\\ \alpha^{*}_1&\alpha^{*}_4&0&\alpha^*_6\\ \alpha^*_2&\alpha^*_5&\alpha^*_6&\alpha^*_7 \end{pmatrix} $

    we have

    $ \begin{array}{llll} \alpha_1^* = (\alpha_1x+\alpha_6t)x^3, & \alpha_2^* = (\alpha_2x+\alpha_7t)x^6, & \alpha_3^* = \alpha_3x^4, & \alpha_4^* = \alpha_4x^5, \\ \alpha_5^* = \alpha_5x^8, & \alpha_6^* = \alpha_6x^{9}, & \alpha_7^* = \alpha_7x^{12}. \end{array} $

    We are interested in $ (\alpha_2,\alpha_5,\alpha_6,\alpha_7)\neq(0,0,0,0) $ and consider following cases:

    $ 1.\ $$ \alpha_7 = \alpha_6 = \alpha_5 = 0, $ then $ \alpha_2\neq0, $ and we have the following subcases:

    $ {\rm{(a)}}\ $if $ \alpha_1 = 0, \alpha_3 = 0, $ then we have the representatives $ \langle \nabla_2 \rangle $ and $ \langle \nabla_2+\nabla_4 \rangle $ depending on whether $ \alpha_4 = 0 $ or not;

    $ {\rm{(b)}}\ $if $ \alpha_1 = 0, \alpha_3\neq0, $ then by choosing $ x = \sqrt[3]{{\alpha_3}{\alpha_2}^{-1}}, t = 0, $ we have the family of representatives $ \langle \nabla_2+\nabla_3+\alpha\nabla_4 \rangle; $

    $ {\rm{(c)}}\ $if $ \alpha_1\neq0, $ then by choosing $ x = \sqrt[3]{{\alpha_1}{\alpha_2^{-1}}}, t = 0, $ we have the family of representatives $ \langle \nabla_1+\nabla_2+\alpha\nabla_3+\beta\nabla_4 \rangle. $

    2. $ \alpha_7 = 0, \alpha_6 = 0, \alpha_5\neq0, $ then we have the following subcases:

    $ {\rm{(a)}}\ $if $ \alpha_1 = 0, \alpha_2 = 0, \alpha_3 = 0, $ then we have the representatives $ \langle \nabla_5 \rangle $ and $ \langle \nabla_4+\nabla_5 \rangle $ depending on whether $ \alpha_5 = 0 $ or not;

    $ {\rm{(b)}}\ $if $ \alpha_1 = 0, \alpha_2 = 0, \alpha_3\neq0, $ then by choosing $ x = \sqrt[4]{{\alpha_3}{\alpha_5^{-1}}}, t = 0, $ we have the family of representatives $ \langle \nabla_3+\alpha\nabla_4+\nabla_5 \rangle; $

    $ {\rm{(c)}}\ $if $ \alpha_1 = 0, \alpha_2\neq0, $ then by choosing $ x = {\alpha_2}{\alpha_5^{-1}}, t = 0, $ we have the family of representatives $ \langle \nabla_2+\alpha\nabla_3+\beta\nabla_4+\nabla_5 \rangle; $

    $ {\rm{(d)}}\ $if $ \alpha_1\neq0, $ then by choosing $ x = \sqrt[4]{{\alpha_1}{\alpha_5^{-1}}}, t = 0, $ we have the family of representatives $ \langle \nabla_1+\alpha\nabla_2+\beta\nabla_3+\gamma\nabla_4+\nabla_5 \rangle. $

    3. $ \alpha_7 = 0, \alpha_6\neq0, $ then we have the following subcases:

    $ {\rm{(a)}}\ $if $ \alpha_2 = 0, \alpha_3 = 0, \alpha_4 = 0, $ then we have representatives $ \langle \nabla_6 \rangle $ and $ \langle \nabla_5+\nabla_6 \rangle $ depending on whether $ \alpha_5 = 0 $ or not;

    $ {\rm{(b)}}\ $if $ \alpha_2 = 0, \alpha_3 = 0, \alpha_4\neq0, $ then by choosing $ x = \sqrt[4]{{\alpha_4}{\alpha_6}^{-1}}, t = -\alpha_1 \sqrt[4]{\alpha_4 \alpha_6^{-5}} , $ we have the family of representatives $ \langle \nabla_4+\alpha\nabla_5+\nabla_6 \rangle; $

    $ {\rm{(c)}}\ $if $ \alpha_2 = 0, \alpha_3\neq0, $ then by choosing $ x = \sqrt[5]{{\alpha_3}{\alpha_6^{-1}}}, t = - \alpha_1\sqrt[5]{\alpha_3 \alpha_6^{-6}}, $ we have the family of representatives $ \langle \nabla_3+\alpha\nabla_4+\beta\nabla_5+\nabla_6 \rangle; $

    ${\rm{(d)}}\ $if $ \alpha_2\neq0, $ then by choosing $ x = {{\alpha_2}{\alpha_6^{-1}}}, t = -{\alpha_1\sqrt{\alpha_2 \alpha_6^{-3}}}, $ we have the family of representatives $ \langle \nabla_2+\alpha\nabla_3+\beta\nabla_4+\gamma\nabla_5+\nabla_6 \rangle. $

    4. $ \alpha_7\neq0, $ then we have the following subcases:

    $ {\rm{(a)}}\ $$ \alpha_1\alpha_7-\alpha_2\alpha_6 = 0, \alpha_3 = 0, \alpha_4 = 0, \alpha_5 = 0, $ then we have representatives $ \langle \nabla_7 \rangle $ and $ \langle \nabla_6+\nabla_7 \rangle $ depending on whether $ \alpha_6 = 0 $ or not;

    $ {\rm{(b)}}\ $$ \alpha_1\alpha_7-\alpha_2\alpha_6 = 0, \alpha_3 = 0, \alpha_4 = 0, \alpha_5\neq0, $ then by choosing $ x = \sqrt[4]{{\alpha_5}{\alpha_7^{-1}}}, $ $ t = -{\alpha_2\sqrt[4]{\alpha_5 \alpha_7^{-5}}}, $ we have the family of representatives $ \langle \nabla_5+\alpha\nabla_6+\nabla_7 \rangle; $

    ${\rm{(c)}}\ $$ \alpha_1\alpha_7-\alpha_2\alpha_6 = 0, \alpha_3 = 0, \alpha_4\neq0, $ then by choosing $ x = \sqrt[7]{{\alpha_4\alpha_7^{-1}}}, $ $ t = -{\alpha_2\sqrt[7]{\alpha_4 \alpha_7^{-8}}}, $ we have the family of representatives $ \langle \nabla_4+\alpha\nabla_5+\beta\nabla_6+\nabla_7 \rangle; $

    $ {\rm{(d)}}\ $$ \alpha_1\alpha_7-\alpha_2\alpha_6 = 0, \alpha_3\neq0, $ then by choosing $ x = \sqrt[8]{{\alpha_3}{\alpha_7^{-1}}}, $ $ t = -{\alpha_2\sqrt[8]{\alpha_3 \alpha_7^{-9}}}, $ we have the family of representatives $ \langle \nabla_3+\alpha\nabla_4+\beta\nabla_5+\gamma\nabla_6+\nabla_7 \rangle; $

    $ {\rm{(e)}}\ $$ \alpha_1\alpha_7-\alpha_2\alpha_6\neq0, $ then by choosing

    $ x = \sqrt[8]{{(\alpha_1\alpha_7-\alpha_2\alpha_6)}{\alpha^{-2}_7}}, t = -{\alpha_2\sqrt[8]{(\alpha_1\alpha_7-\alpha_2\alpha_6)\alpha_7^{-10}}}, $

    we have the family of representatives $ \langle \nabla_1+\alpha\nabla_3+\beta\nabla_4+\gamma\nabla_5+\mu\nabla_6+\nabla_7 \rangle. $

    Summarizing, we have the following distinct orbits:

    $\begin{array}{c} \langle \nabla_1+\nabla_2 + \alpha \nabla_3 + \beta \nabla_4 \rangle ^{O(\alpha, \beta) = O(\alpha, -\eta_3 \beta) = O(\alpha, \eta_3^2\beta)}, \\ \langle \nabla_1+ \alpha \nabla_2 + \beta \nabla_3 + \gamma \nabla_4 + \nabla_5\rangle ^{O(\alpha, \beta, \gamma) = O(-i\alpha, \beta, i\gamma) = O(i\alpha, \beta,-i \gamma) = O(-\alpha, \beta, -\gamma)}, \\ \langle \nabla_1+\alpha \nabla_3+ \beta \nabla_4 + \gamma\nabla_5 + \mu\nabla_6 + \nabla_7\rangle^{ {\begin{array}{l} O(\alpha,\beta,\gamma, \mu) = O(\alpha,\eta_4^3\beta,-\gamma, -\eta_4^3\mu) = \\ O(\alpha,-\eta_4^3\beta,-\gamma, \eta_4^3\mu) = O(\alpha,\eta_4 \beta,-\gamma, -\eta_4\mu) = \\ O(\alpha,-\eta_4\beta,-\gamma, \eta_4\mu) = O(\alpha,i\beta,\gamma, i\mu) = \\ O(\alpha,-i\beta,\gamma, -i\mu) = O(\alpha,-\beta, \gamma, -\mu) \end{array}} }, \\ \langle \nabla_2\rangle, \langle \nabla_2+\nabla_3 +\alpha\nabla_4\rangle^{O(\alpha) = O(-\eta_3\alpha) = O(\eta_3^2\alpha)}, \langle \nabla_2+\alpha\nabla_3 +\beta\nabla_4 + \nabla_5\rangle, \\ \langle \nabla_2+\alpha \nabla_3+ \beta \nabla_4 + \gamma\nabla_5 + \nabla_6\rangle^{O(\alpha,\beta,\gamma) = O(-\alpha,\beta,-\gamma)}, \langle \nabla_2+ \nabla_4\rangle, \\ \langle \nabla_3+\alpha \nabla_4+\nabla_5\rangle^{ { \begin{array}{l} O(\alpha) = O(-\alpha) = \\ O(i\alpha) = O(-i\alpha) \end{array}}}, \\ \langle \nabla_3+\alpha \nabla_4+ \beta \nabla_5+ \nabla_6\rangle ^{{ \begin{array}{l} O(\alpha, \beta) = O(\eta_5^4\alpha, -\eta_5\beta) = O(-\eta_5^3\alpha, \eta_5^2\beta) = \\ O(\eta_5^2\alpha, -\eta_5^3\beta) = O(-\eta_5\alpha, \eta_5^4\beta) \end{array}}}, \\ \langle \nabla_3+\alpha \nabla_4+ \beta \nabla_5 +\gamma\nabla_6 + \nabla_7\rangle^{ {\begin{array}{l} O(\alpha,\beta,\gamma) = O(\eta_4^3\alpha,-\beta,-\eta_4^3\gamma) = O(-\eta_4^3\alpha,-\beta,\eta_4^3\gamma) = \\ O(\eta_4\alpha,-\beta,-\eta_4\gamma) = O(-\eta_4\alpha,-\beta,\eta_4\gamma) = \\ O(i\alpha,\beta,i\gamma) = O(-i\alpha,\beta,-i\gamma) = O(-\alpha,\beta,-\gamma) \end{array}} }, \\ \langle \nabla_4+ \nabla_5\rangle, \langle \nabla_4+\alpha \nabla_5+ \nabla_6\rangle^{O(\alpha) = O(i\alpha) = O(-\alpha) = O(-i\alpha)}, \\ \langle \nabla_4+\alpha \nabla_5+ \beta\nabla_6 + \nabla_7\rangle^{ {\begin{array}{l} O(\alpha,\beta) = O(\eta^4_7\alpha,-\eta^3_7\beta) = O(-\eta_7\alpha,\eta_7^6\beta) = O(-\eta_7^5\alpha,\eta^2_7\beta) = \\ O(\eta^2_7\alpha,-\eta^5_7\beta) = O(\eta_7^6\alpha,-\eta_7\beta) = O(-\eta^3_7\alpha,\eta^4_7\beta) \end{array}}}, \langle \nabla_5\rangle, \\ \langle \nabla_5 + \nabla_6\rangle, \langle \nabla_5+\alpha \nabla_6+ \nabla_7\rangle^{O(\alpha) = O(i\alpha) = O(-\alpha) = O(-i\alpha)}, \langle \nabla_6\rangle, \langle \nabla_6 +\nabla_7\rangle, \langle \nabla_7\rangle, \end{array}$

    which gives the following new algebras:

    $ \begin{array}{llllllllllllllllll} {\mathbf{N}}_{314}^{\alpha, \beta} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_1e_3 = e_5 & e_1e_4 = e_5 \\ & & e_2e_2 = \alpha e_5 & e_2e_3 = \beta e_5 & e_3e_3 = e_4 \\ {\mathbf{N}}_{315}^{\alpha, \beta ,\gamma} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_1e_3 = e_5 & e_1e_4 = \alpha e_5 \\ && e_2e_2 = \beta e_5 & e_2e_3 = \gamma e_5 & e_2e_4 = e_5 & e_3e_3 = e_4 \\ {\mathbf{N}}_{316}^{\alpha, \beta, \gamma, \mu} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_1e_3 = e_5 & e_2e_2 = \alpha e_5 & e_2e_3 = \beta e_5 \\ & & e_2e_4 = \gamma e_5 & e_3e_3 = e_4 & e_3e_4 = \mu e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{317} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_1e_4 = e_5 & e_3e_3 = e_4 \\ {\mathbf{N}}_{318}^{\alpha} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_1e_4 = e_5 \\ && e_2e_2 = e_5 & e_2e_3 = \alpha e_5 & e_3e_3 = e_4 \\ {\mathbf{N}}_{319}^{\alpha, \beta} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_1e_4 = e_5 & e_2e_2 = \alpha e_5 \\ && e_2e_3 = \beta e_5 & e_2e_4 = e_5 & e_3e_3 = e_4 \\ {\mathbf{N}}_{320}^{\alpha, \beta ,\gamma} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_1e_4 = e_5 & e_2e_2 = \alpha e_5 \\ && e_2e_3 = \beta e_5 & e_2e_4 = \gamma e_5 & e_3e_3 = e_4 & e_3e_4 = e_5 \\ {\mathbf{N}}_{321} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_1e_4 = e_5 & e_2e_3 = e_5 & e_3e_3 = e_4 \\ {\mathbf{N}}_{322}^{\alpha} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_2e_2 = e_5 \\ && e_2e_3 = \alpha e_5 & e_2e_4 = e_5 & e_3e_3 = e_4 \\ {\mathbf{N}}_{323}^{\alpha, \beta} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_2e_2 = e_5 & e_2e_3 = \alpha e_5 \\ && e_2e_4 = \beta e_5 & e_3e_3 = e_4 & e_3e_4 = e_5 \\ {\mathbf{N}}_{324}^{\alpha, \beta ,\gamma} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_2e_2 = e_5 & e_2e_3 = \alpha e_5 \\ && e_2e_4 = \beta e_5 & e_3e_3 = e_4 & e_3e_4 = \gamma e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{325} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_2e_3 = e_5 \\ && e_2e_4 = e_5 & e_3e_3 = e_4 \\ {\mathbf{N}}_{326}^{\alpha} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_2e_3 = e_5 \\ && e_2e_4 = \alpha e_5 & e_3e_3 = e_4 & e_3e_4 = e_5 \\ {\mathbf{N}}_{327}^{\alpha, \beta} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_2e_3 = e_5 & e_2e_4 = \alpha e_5 \\ & & e_3e_3 = e_4 & e_3e_4 = \beta e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{328} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_2e_4 = e_5 & e_3e_3 = e_4 \\ {\mathbf{N}}_{329} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_2e_4 = e_5 & e_3e_3 = e_4 & e_3e_4 = e_5 \\ {\mathbf{N}}^\alpha_{330} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_2e_4 = e_5 \\ & & e_3e_3 = e_4 & e_3e_4 = \alpha e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{331} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_3e_3 = e_4 & e_3e_4 = e_5 \\ {\mathbf{N}}_{332} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_3e_3 = e_4 & e_3e_4 = e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{333} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_3e_3 = e_4 & e_4e_4 = e_5 \\ \end{array} $

    Here we will collect all information about $ {\mathbf N}_{04}^{4}: $

    $ \begin{array}{|l|l|l|l|} \hline \rm{ } & \rm{ } & \rm{Cohomology} & \rm{Automorphisms} \\ \hline {\mathbf{N}}^{4}_{04} & \begin{array}{l} e_1e_1 = e_2 \\ e_1e_2 = e_3 \\ e_2e_2 = e_4 \\ e_3e_3 = e_4 \end{array} & \begin{array}{l} \mathrm{H}^2_{\mathfrak{C}}(\mathbf{N}^{4}_{04}) = \Big \langle [\Delta_{ij}] \Big\rangle\\ (i,j) \notin \{(1,1),(1,2),(3,3)\} \end{array} & \phi_{\pm} = \begin{pmatrix} \pm1&0&0&0\\ 0&1&0&0\\ 0&0&\pm1&0\\ t&0&0&1 \end{pmatrix}\\ \hline \end{array} $

    Let us use the following notations:

    $ \begin{array}{llll} \nabla_1 = [\Delta_{13}], & \nabla_2 = [\Delta_{14}], & \nabla_3 = [\Delta_{22}], & \nabla_4 = [\Delta_{23}], \\ \nabla_5 = [\Delta_{24}], & \nabla_6 = [\Delta_{34}], & \nabla_7 = [\Delta_{44}]. \end{array} $

    Take $ \theta = \sum\limits_{i = 1}^{7}\alpha_i\nabla_i\in\mathrm{H}^2_{\mathfrak{C}}(\mathbf{N}^{4}_{04}) . $ Since

    $ \phi^T\begin{pmatrix} 0&0&\alpha_1&\alpha_2\\ 0&\alpha_3&\alpha_4&\alpha_5\\ \alpha_1&\alpha_4&0&\alpha_6\\ \alpha_2&\alpha_5&\alpha_6&\alpha_7 \end{pmatrix}\phi = \begin{pmatrix} \alpha^*&\alpha^{**}&\alpha^{*}_1&\alpha^*_2\\ \alpha^{**}&\alpha^*_3&\alpha^{*}_4&\alpha^*_5\\ \alpha^{*}_1&\alpha^{*}_4&0&\alpha^*_6\\ \alpha^*_2&\alpha^*_5&\alpha^*_6&\alpha^*_7 \end{pmatrix} $

    we have

    $ \begin{array}{llll} \alpha_1^* = \alpha_1\pm\alpha_6t, & \alpha_2^* = \pm\alpha_2+\alpha_7t, & \alpha_3^* = \alpha_3, & \alpha_4^* = \pm\alpha_4, \\ \alpha_5^* = \alpha_5, & \alpha_6^* = \pm \alpha_6, & \alpha_7^* = \alpha_7. \end{array} $

    We are interested in $ (\alpha_2,\alpha_5,\alpha_6,\alpha_7)\neq(0,0,0,0) $ and consider following cases:

    $ 1.\ $if $ \alpha_7 = \alpha_6 = \alpha_5 = 0, $ then $ \alpha_2\neq0, $ and we have the family of representatives

    $ \langle \alpha\nabla_1+\nabla_2+\beta\nabla_3+\gamma\nabla_4 \rangle;$

    $ 2.\ $if $ \alpha_7 = 0, \alpha_6 = 0, \alpha_5\neq0, $ then we have the family of representatives

    $ \langle \alpha\nabla_1+\beta\nabla_2+\gamma\nabla_3+\mu\nabla_4+\nabla_5 \rangle;$

    $ 3.\ $if $ \alpha_7 = 0, \alpha_6\neq0, $ then by choosing $ \phi = \phi_+, t = -{\alpha_1}{\alpha_6^{-1}}, $ we have the family of representatives

    $ \langle \alpha\nabla_2+\beta\nabla_3+\gamma\nabla_4+\mu\nabla_5+\nabla_6 \rangle;$

    $ 4.\ $if $ \alpha_7\neq0, $ then by choosing $ \phi = \phi_+, t = -{\alpha_2}{\alpha_7^{-1}} $ we have the family of representatives

    $\langle \alpha\nabla_1+\beta\nabla_3+\gamma\nabla_4+\mu\nabla_5+\nu\nabla_6+\nabla_7 \rangle.$

    Summarizing, we have the following distinct orbits:

    $\begin{array}{c} \langle \alpha\nabla_1+\nabla_2+\beta\nabla_3+\gamma\nabla_4 \rangle^{O(\alpha, \beta, \gamma) = O(-\alpha, -\beta, \gamma)}, \\\langle \alpha\nabla_1+\beta\nabla_2+\gamma\nabla_3+\mu\nabla_4+\nabla_5 \rangle ^{O(\alpha, \beta, \gamma,\mu) = O(\alpha, -\beta, \gamma, -\mu)} \\ \langle \alpha\nabla_1+\beta\nabla_3+\gamma\nabla_4+\mu\nabla_5+\nu\nabla_6+\nabla_7 \rangle^{O(\alpha, \beta, \gamma,\mu,\nu) = O(\alpha, \beta, -\gamma,\mu,-\nu)}, \\ \langle \alpha\nabla_2+\beta\nabla_3+\gamma\nabla_4+\mu\nabla_5+\nabla_6 \rangle ^{O(\alpha, \beta, \gamma, \mu) = O(\alpha, -\beta, \gamma, -\mu)}, \end{array} $

    which gives the following new algebras:

    $ \begin{array}{llllllllllllllllll} {\mathbf{N}}_{334}^{\alpha, \beta ,\gamma} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_1e_3 = \alpha e_5 & e_1e_4 = e_5 \\ & & e_2e_2 = e_4+\beta e_5 & e_2e_3 = \gamma e_5 & e_3e_3 = e_4 \\ {\mathbf{N}}_{335}^{\alpha, \beta,\gamma, \mu} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_1e_3 = \alpha e_5 & e_1e_4 = \beta e_5 \\ && e_2e_2 = e_4+\gamma e_5 & e_2e_3 = \mu e_5 & e_2e_4 = e_5 & e_3e_3 = e_4 \\ {\mathbf{N}}_{336}^{\alpha, \beta,\gamma, \mu, \nu} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_1e_3 = \alpha e_5 \\& & e_2e_2 = e_4+\beta e_5 & e_2e_3 = \gamma e_5 & e_2e_4 = \mu e_5 \\ && e_3e_3 = e_4 & e_3e_4 = \nu e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{337}^{\alpha, \beta,\gamma, \mu} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_1e_4 = \alpha e_5 & e_2e_2 = e_4+\beta e_5 \\ && e_2e_3 = \gamma e_5 & e_2e_4 = \mu e_5 & e_3e_3 = e_4 & e_3e_4 = e_5 \\ \end{array} $

    Here we will collect all information about $ {\mathbf N}_{05}^{4}: $

    $ \begin{array}{|l|l|l|l|} \hline \rm{ } & \rm{ } & \rm{Cohomology} & \rm{Automorphisms}\\ \hline {\mathbf{N}}^{4}_{05} & \begin{array}{l} e_1e_1 = e_2 \\ e_1e_3 = e_4 \\ e_2e_2 = e_3 \end{array} & \begin{array}{l} \mathrm{H}^2_{\mathfrak{C}}(\mathbf{N}^{4}_{05}) = \Big \langle [\Delta_{ij}] \Big\rangle\\ {(i,j) \notin \{ (1,1),(1,3),(2,2)} \} \end{array} & \phi = \begin{pmatrix} x&0&0&0\\ 0&x^2&0&0\\ z&0&x^4&0\\ t&2xz&0&x^5 \end{pmatrix}\\ \hline \end{array} $

    Let us use the following notations:

    $ \begin{array}{llll} \nabla_1 = [\Delta_{12}] & \nabla_2 = [\Delta_{14}] & \nabla_3 = [\Delta_{23}] & \nabla_4 = [\Delta_{24}] \\ \nabla_5 = [\Delta_{33}] & \nabla_6 = [\Delta_{34}] & \nabla_7 = [\Delta_{44}]. \end{array} $

    Take $ \theta = \sum\limits_{i = 1}^{7}\alpha_i\nabla_i\in\mathrm{H}^2_{\mathfrak{C}}(\mathbf{N}^{4}_{05}) . $ Since

    $ \phi^T\begin{pmatrix} 0&\alpha_1&0&\alpha_2\\ \alpha_1&0&\alpha_3&\alpha_4\\ 0&\alpha_3&\alpha_5&\alpha_6\\ \alpha_2&\alpha_4&\alpha_6&\alpha_7 \end{pmatrix}\phi = \begin{pmatrix} \alpha^*&\alpha^*_1&\alpha^{***}&\alpha^*_2\\ \alpha^*_1&\alpha^{**}&\alpha^*_3&\alpha^*_4\\ \alpha^{***}&\alpha^*_3&\alpha^*_5&\alpha^*_6\\ \alpha^*_2&\alpha^*_4&\alpha^*_6&\alpha^*_7 \end{pmatrix} $

    we have

    $ \begin{array}{lll} { \alpha_1^* = (\alpha_1x+\alpha_3z+\alpha_4t)x^2+2(\alpha_2x+\alpha_6z+\alpha_7t)xz, }\\ \alpha_2^* = (\alpha_2x+\alpha_6z+\alpha_7t)x^5, & \alpha_3^* = (\alpha_3x+2\alpha_6z)x^5, & \alpha_4^* = (\alpha_4x+2\alpha_7z)x^6, \\ \alpha_5^* = \alpha_5x^8, & \alpha_6^* = \alpha_6x^9, & \alpha_7^* = \alpha_7x^{10}. \end{array} $

    We are interested in $ (\alpha_2,\alpha_4,\alpha_6,\alpha_7)\neq(0,0,0,0) $ and consider following cases:

    $ 1.\ $$ \alpha_7 = \alpha_6 = \alpha_4 = 0, $ then $ \alpha_2\neq0 $ and we have the following subcases:

    $ {\rm{(a)}}\ $$ \alpha_3 = -2\alpha_2, $

    $ {\rm{(i)}}\ $ if $ \alpha_1 = 0, \alpha_5 = 0, $ then we have the representative $ \langle \nabla_2-2\nabla_3 \rangle; $

    $ {\rm{(ii)}}\ $ if $ \alpha_1 = 0, \alpha_5\neq0, $ then by choosing $ x = \sqrt{{\alpha_2}{\alpha_5^{-1}}}, z = 0, t = 0, $ we have the representative $ \langle \nabla_2-2\nabla_3+\nabla_5 \rangle; $

    $ {\rm{(iii)}}\ $if $ \alpha_1\neq0, $ then by choosing $ x = \sqrt[3]{{\alpha_1}{\alpha_2^{-1}}}, z = 0, t = 0, $ we have the family of representatives $ \langle \nabla_1+\nabla_2-2\nabla_3+\alpha\nabla_5 \rangle. $

    (b) $ \alpha_3\neq -2\alpha_2, $

    $ {\rm{(i)}}\ $if $ \alpha_5 = 0, $ then choosing $ x = 1, z = -\frac{\alpha_1}{\alpha_3+2\alpha_2}, $ we have the family of representatives $ \langle \nabla_2+\alpha\nabla_3 \rangle_{\alpha\neq-2}, $ which will be jointed with the case (1(a)i);

    $ {\rm{(ii)}}\ $if $ \alpha_5\neq0, $ then by choosing $ x = \sqrt{\frac{\alpha_2}{\alpha_5}}, z = -\frac{\alpha_1\sqrt{\alpha_2}}{(\alpha_3+2\alpha_2)\sqrt{\alpha_5}}, $ we have the family of representatives $ \langle\nabla_2+\alpha\nabla_3+\nabla_5 \rangle_{\alpha\neq-2}, $ which will be jointed with the case (1(a)ii).

    2. $ \alpha_7 = \alpha_6 = 0, \alpha_4\neq0, $ then we have the following subcases:

    $ {\rm{(a)}}\ $if $ \alpha_2 = 0, \alpha_3 = 0, $ then we have representatives $ \langle \nabla_4 \rangle $ and $ \langle \nabla_4+\nabla_5 \rangle $ depending on whether $ \alpha_5 = 0 $ or not;

    $ {\rm{(b)}}\ $if $ \alpha_2 = 0, \alpha_3\neq0 , $ then by choosing $ x = {\alpha_3}{\alpha_4^{-1}}, z = 0, t = -{\alpha_1\alpha_3}{\alpha_4^{-2}}, $ we have the family of representatives $ \langle \nabla_3+ \nabla_4+\alpha\nabla_5 \rangle; $

    $ {\rm{(c)}}\ $if $ \alpha_2\neq0 , $ then by choosing $ x = {\alpha_2}{\alpha_4^{-1}}, z = 0, t = -{\alpha_1\alpha_2}{\alpha_4^{-2}}, $ we have the family of representatives $ \langle \nabla_2+\alpha\nabla_3+ \nabla_4+\beta\nabla_5 \rangle. $

    3. $ \alpha_7 = 0, \alpha_6\neq0, $ then by choosing $ z = -{\alpha_2}x{\alpha_6^{-1}} , $ we have $ \alpha_2^* = 0. $ Thus, we can suppose that $ \alpha_2 = 0 $ and consider following subcases:

    $ {\rm{(a)}}\ $$ \alpha_4 = 0, $

    $ {\rm{(i)}}\ $if $ \alpha_1 = 0, \alpha_3 = 0, $ then we have representatives $ \langle \nabla_6 \rangle $ and $ \langle \nabla_5+\nabla_6 \rangle $ depending on whether $ \alpha_5 = 0 $ or not;

    ${\rm{(ii)}}\ $if $ \alpha_1 = 0, \alpha_3\neq0 $ then by choosing $ x = \sqrt[3]{{\alpha_3}{\alpha_6^{-1}}}, z = 0, t = 0, $ we have the family of representatives $ \langle \nabla_3+\alpha\nabla_5+\nabla_6 \rangle; $

    $ {\rm{(iii)}}\ $if $ \alpha_1\neq0, $ then by choosing $ x = \sqrt[6]{{\alpha_1}{\alpha_6^{-1}}}, z = 0, t = 0, $ we have the family of representatives $ \langle \nabla_1+\alpha\nabla_3+\beta\nabla_5+\nabla_6 \rangle. $

    (b) $ \alpha_4\neq0 , $ then by choosing $ x = \sqrt{{\alpha_4}{\alpha_6^{-1}}}, t = -{\alpha_1\sqrt{\alpha_4 \alpha_6^{-3}}}, $ we have the family of representatives $ \langle \alpha\nabla_3+\nabla_4+\beta\nabla_5+\nabla_6 \rangle. $

    4. $ \alpha_7\neq0, $

    $ {\rm{(a)}}\ $if $ \alpha_1\alpha_7-\alpha_2\alpha_4 = 0, \alpha_3 = 0, \alpha_6 = 0, $ then then by choosing $ z = -\frac{\alpha_4}{2\alpha_7}x, t = \frac{\alpha_4\alpha_6-2\alpha_2\alpha_7}{2\alpha^2_7}x, $ we have representatives $ \langle \nabla_7 \rangle $ and $ \langle \nabla_5+\nabla_7 \rangle $ depending on whether $ \alpha_5 = 0 $ or not;

    $ {\rm{(b)}}\ $if $ \alpha_1\alpha_7-\alpha_2\alpha_4 = 0, \alpha_3\alpha_7-\alpha_4\alpha_6 = 0, \alpha_6\neq0, $ then by choosing

    $ x = \frac{\alpha_6}{\alpha_7}, z = -\frac{\alpha_4\alpha_6}{2\alpha_7^2}, t = \frac{\alpha_6(\alpha_4\alpha_6-2\alpha_2\alpha_7)}{2\alpha^3_7},$

    we have the family of representatives $ \langle \alpha\nabla_5+\nabla_6+\nabla_7 \rangle; $

    ${\rm{(c)}}\ $if $ 2\alpha_1\alpha_7^2-\alpha_3\alpha_4\alpha_7+\alpha_4^2\alpha_6-2\alpha_2\alpha_4\alpha_7 = 0, \alpha_3\alpha_7-\alpha_4\alpha_6\neq0, $ then by choosing

    $ x = \sqrt[4]{\frac{\alpha_3\alpha_7-\alpha_4\alpha_7}{\alpha^2_7}}, z = -\frac{\alpha_4\sqrt[4]{\alpha_3\alpha_7-\alpha_4\alpha_7}}{2\alpha_7\sqrt[4]{2\alpha_7^2}}, t = \frac{(\alpha_4\alpha_6-2\alpha_2\alpha_7)\sqrt[4]{\alpha_3\alpha_7-\alpha_4\alpha_6}}{2\alpha^2_7\sqrt[4]{2\alpha_7^2}}, $

    we have the family of representatives $ \langle \nabla_3+\alpha\nabla_5+\beta\nabla_6+\nabla_7 \rangle; $

    $ {\rm{(d)}}\ $if $ 2\alpha_1\alpha_7^2-\alpha_3\alpha_4\alpha_7+\alpha_4^2\alpha_6-2\alpha_2\alpha_4\alpha_7\neq0, $ then by choosing

    $\begin{array}{c} x = \sqrt[7]{\frac{2\alpha_1\alpha_7^2-\alpha_3\alpha_4\alpha_7+\alpha_4^2\alpha_6-2\alpha_2\alpha_4\alpha_7}{2\alpha^3_7}}, \\ z = -\frac{\alpha_4\sqrt[7]{2\alpha_1\alpha_7^2-\alpha_3\alpha_4\alpha_7+\alpha_4^2\alpha_6-2\alpha_2\alpha_4\alpha_7}}{2\alpha_7\sqrt[7]{2\alpha_7^3}}, \\ t = \frac{(\alpha_4\alpha_6-2\alpha_2\alpha_7)\sqrt[7]{2\alpha_1\alpha_7^2-\alpha_3\alpha_4\alpha_7+\alpha_4^2\alpha_6-2\alpha_2\alpha_4\alpha_7}}{2\alpha^3_7\sqrt[7]{2\alpha_7^3}}, \end{array}$

    we have the family of representatives $ \langle \nabla_1+\alpha\nabla_3+\beta\nabla_5+\gamma\nabla_6+\nabla_7 \rangle. $

    Summarizing, we have the following distinct orbits:

    $\begin{array}{c} \langle \nabla_1+\nabla_2-2\nabla_3+\alpha\nabla_5 \rangle^{O(\alpha) = O(-\eta_3\alpha) = O(\eta^2_3\alpha)}, \\ \langle \nabla_1+\alpha\nabla_3+\beta\nabla_5+\nabla_6 \rangle^{{\begin{array}{l} O(\alpha,\beta) = O(\alpha,-\eta_3\beta) = O(-\alpha,\eta_3\beta) = \\ O(-\alpha,-\eta_3^2\beta) = O(\alpha,\eta_3^2\beta) = O(-\alpha,-\beta) \end{array}}}, \\ \langle \nabla_1+\alpha\nabla_3+\beta\nabla_5+\gamma\nabla_6+\nabla_7 \rangle^{{\begin{array}{l} O(\alpha,\beta,\gamma) = O(\eta_7^4\alpha,\eta^2_7\beta,-\eta_7\gamma) = O(-\eta_7\alpha,\eta^4_7\beta,\eta^2_7\gamma) = \\ O(-\eta_7^5\alpha,\eta^6_{7}\beta,-\eta^3_7\gamma) = O(\eta^2_7\alpha,-\eta_7\beta,\eta^4_7\gamma) = \\ O(\eta_7^6\alpha,-\eta^3_7\beta,-\eta^5_7\gamma) = O(-\eta^3_7\alpha,-\eta^5_7\beta,\eta^6_7\gamma) \end{array}}}, \\ \langle \nabla_2+\alpha\nabla_3 \rangle, \langle \nabla_2+\alpha\nabla_3+ \nabla_4+\beta\nabla_5 \rangle, \langle\nabla_2+\alpha\nabla_3+\nabla_5 \rangle, \langle \nabla_3+ \nabla_4+\alpha\nabla_5 \rangle, \\ \langle \alpha\nabla_3+\nabla_4+\beta\nabla_5+\nabla_6 \rangle^{O(\alpha,\beta) = O(-\alpha,-\beta)}, \langle \nabla_3+\alpha\nabla_5+\nabla_6 \rangle^{O(\alpha) = O(-\eta_3\alpha) = O(\eta^2_3\alpha)}, \\ \langle \nabla_3+\alpha\nabla_5+\beta\nabla_6+\nabla_7 \rangle^{O(\alpha,\beta) = O(-\alpha,-i\beta) = O(-\alpha,i \beta) = O(\alpha,-\beta)}, \langle \nabla_4 \rangle, \langle \nabla_4+\nabla_5 \rangle, \\ \langle \nabla_5+\nabla_6 \rangle, \langle \alpha\nabla_5+\nabla_6+\nabla_7 \rangle, \langle \nabla_5+\nabla_7 \rangle, \langle\nabla_6 \rangle, \langle\nabla_7 \rangle,\end{array}$

    which gives the following new algebras:

    $ \begin{array}{llllllllllllllllll} {\mathbf{N}}_{338}^{\alpha} & : & e_1e_1 = e_2 & e_1e_2 = e_5 & e_1e_3 = e_4 & e_1e_4 = e_5 \\ & & e_2e_2 = e_3 & e_2e_3 = -2e_5 & e_3e_3 = \alpha e_5 \\ {\mathbf{N}}_{339}^{\alpha, \beta} & : & e_1e_1 = e_2 & e_1e_2 = e_5 & e_1e_3 = e_4 & e_2e_2 = e_3 \\ && e_2e_3 = \alpha e_5 & e_3e_3 = \beta e_5 & e_3e_4 = e_5 \\ {\mathbf{N}}_{340}^{\alpha, \beta ,\gamma} & : & e_1e_1 = e_2 & e_1e_2 = e_5 & e_1e_3 = e_4 & e_2e_2 = e_3 \\ & & e_2e_3 = \alpha e_5 & e_3e_3 = \beta e_5 & e_3e_4 = \gamma e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{341}^{\alpha} & : & e_1e_1 = e_2 & e_1e_3 = e_4 & e_1e_4 = e_5 & e_2e_2 = e_3 & e_2e_3 = \alpha e_5 \\ {\mathbf{N}}_{342}^{\alpha, \beta} & : & e_1e_1 = e_2 & e_1e_3 = e_4 & e_1e_4 = e_5 & e_2e_2 = e_3 \\ && e_2e_3 = \alpha e_5 & e_2e_4 = e_5 & e_3e_3 = \beta e_5 \\ {\mathbf{N}}_{343}^{\alpha} & : & e_1e_1 = e_2 & e_1e_3 = e_4 & e_1e_4 = e_5 \\ && e_2e_2 = e_3 & e_2e_3 = \alpha e_5 & e_3e_3 = e_5 \\ {\mathbf{N}}_{344}^{\alpha} & : & e_1e_1 = e_2 & e_1e_3 = e_4 & e_2e_2 = e_3 \\ && e_2e_3 = e_5 & e_2e_4 = e_5 & e_3e_3 = \alpha e_5 \\ {\mathbf{N}}_{345}^{\alpha, \beta} & : & e_1e_1 = e_2 & e_1e_3 = e_4 & e_2e_2 = e_3 & e_2e_3 = \alpha e_5 \\ && e_2e_4 = e_5 & e_3e_3 = \beta e_5 & e_3e_4 = e_5 \\ {\mathbf{N}}_{346}^{\alpha} & : & e_1e_1 = e_2 & e_1e_3 = e_4 & e_2e_2 = e_3 \\ && e_2e_3 = e_5 & e_3e_3 = \alpha e_5 & e_3e_4 = e_5 \\ {\mathbf{N}}_{347}^{\alpha, \beta} & : & e_1e_1 = e_2 & e_1e_3 = e_4 & e_2e_2 = e_3 & e_2e_3 = e_5 \\ && e_3e_3 = \alpha e_5 & e_3e_4 = \beta e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{348} & : & e_1e_1 = e_2 & e_1e_3 = e_4 & e_2e_2 = e_3 & e_2e_4 = e_5 \\ {\mathbf{N}}_{349} & : & e_1e_1 = e_2 & e_1e_3 = e_4 & e_2e_2 = e_3 & e_2e_4 = e_5 & e_3e_3 = e_5 \\ {\mathbf{N}}_{350} & : & e_1e_1 = e_2 & e_1e_3 = e_4 & e_2e_2 = e_3 & e_3e_3 = e_5 & e_3e_4 = e_5 \\ {\mathbf{N}}_{351}^{\alpha} & : & e_1e_1 = e_2 & e_1e_3 = e_4 & e_2e_2 = e_3 \\ && e_3e_3 = \alpha e_5 & e_3e_4 = e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{352} & : & e_1e_1 = e_2 & e_1e_3 = e_4 & e_2e_2 = e_3 & e_3e_3 = e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{353} & : & e_1e_1 = e_2 & e_1e_3 = e_4 & e_2e_2 = e_3 & e_3e_4 = e_5 \\ {\mathbf{N}}_{354} & : & e_1e_1 = e_2 & e_1e_3 = e_4 & e_2e_2 = e_3 & e_4e_4 = e_5 \\ \end{array} $

    Here we will collect all information about $ {\mathbf N}_{06}^{4}: $

    $ \begin{array}{|l|l|l|l|} \hline \rm{ } & \rm{ } & \rm{Cohomology} & \rm{Automorphisms} \\ \hline {\mathbf{N}}^{4}_{06} & \begin{array}{l} e_1e_1 = e_2 \\ e_1e_2 = e_4 \\ e_1e_3 = e_4 \\ e_2e_2 = e_3 \end{array} &\begin{array}{l} \mathrm{H}^2_{\mathfrak{C}}(\mathbf{N}^{4}_{06}) = \Big \langle [\Delta_{ij}] \Big\rangle\\ (i,j) \notin \{(1,1),(1,2),(2,2)\} \end{array} & \phi_{\pm} = \begin{pmatrix} \pm1&0&0&0\\ 0&1&0&0\\ z&0&1&0\\ t&\pm2z&0&\pm1 \end{pmatrix}\\ \hline \end{array} $

    Let us use the following notations:

    $ \begin{array}{llll} \nabla_1 = [\Delta_{13}], & \nabla_2 = [\Delta_{14}], & \nabla_3 = [\Delta_{23}], & \nabla_4 = [\Delta_{24}], \\ \nabla_5 = [\Delta_{33}], & \nabla_6 = [\Delta_{34}], & \nabla_7 = [\Delta_{44}]. \end{array} $

    Take $ \theta = \sum\limits_{i = 1}^{7}\alpha_i\nabla_i\in\mathrm{H}^2_{\mathfrak{C}}(\mathbf{N}^{4}_{06}). $ Since

    $ \phi_{\pm}^T\begin{pmatrix} 0&0&\alpha_1&\alpha_2\\ 0&0&\alpha_3&\alpha_4\\ \alpha_1&\alpha_3&\alpha_5&\alpha_6\\ \alpha_2&\alpha_4&\alpha_6&\alpha_7 \end{pmatrix}\phi_{\pm} = \begin{pmatrix} \alpha^*&\alpha^{**}&\alpha^{*}_1+\alpha^{**}&\alpha^*_2\\ \alpha^{**}&\alpha^{***}&\alpha^*_3&\alpha^*_4\\ \alpha^{*}_1+\alpha^{**}&\alpha^*_3&\alpha^*_5&\alpha^*_6\\ \alpha^*_2&\alpha^*_4&\alpha^*_6&\alpha^*_7 \end{pmatrix} $

    we have

    $ \begin{array}{lll} { \alpha_1^* = \pm \alpha_1- \alpha_3z- \alpha_4t+ \alpha_5z+ \alpha_6t-2 ( \alpha_2\pm \alpha_6z\pm \alpha_7t)z, }\\ \alpha_2^* = \alpha_2\pm\alpha_6z \pm\alpha_7t, & \alpha_3^* = \alpha_3 \pm 2\alpha_6z, & \alpha_4^* = 2\alpha_7z \pm \alpha_4, \\ \alpha_5^* = \alpha_5, & \alpha_6^* = \pm \alpha_6, & \alpha_7^* = \alpha_7. \end{array} $

    Since $ (\alpha_2,\alpha_4,\alpha_6,\alpha_7)\neq(0,0,0,0) $ and for $ \phi = \phi_+, $ we have the following cases:

    $ 1.\ $if $ \alpha_7 = \alpha_6 = \alpha_4 = 0, $ then $ \alpha_2\neq0, $ and we have the following subcase:

    $ {\rm{(a)}}\ $if $ \alpha_5 = \alpha_3+2\alpha_2, $ then we have the family of representatives

    $ \langle \alpha\nabla_1+\nabla_2+\beta\nabla_3+(\beta+2)\nabla_5 \rangle;$

    $ {\rm{(b)}}\ $if $ \alpha_5\neq\alpha_3+2\alpha_2, $ then by choosing $ z = 0, t = \frac{\alpha_1}{\alpha_3+2\alpha_2-\alpha_5} , $ we have the family of representative $ \langle \nabla_2+\alpha\nabla_3+\beta\nabla_5 \rangle_{\beta\neq\alpha+2}; $

    2. if $ \alpha_7 = 0, \alpha_6 = 0, \alpha_4\neq0, $ then by choosing $ z = 0, t = \frac{\alpha_1}{\alpha_4}, $ we have the family of representatives $ \langle \alpha\nabla_2+\beta\nabla_3+\nabla_4+\gamma\nabla_5 \rangle; $

    3. if $ \alpha_7 = 0, \alpha_6\neq0, $ then we have the following subcases:

    $ {\rm{(a)}}\ $if $ \alpha_6 = \alpha_4, $ then by choosing $ z = -{\alpha_2}{\alpha_6^{-1}}, t = 0, $ we have the family of representatives $ \langle \alpha\nabla_1+\beta\nabla_3+\nabla_4+\gamma\nabla_5+\nabla_6 \rangle; $

    $ {\rm{(b)}}\ $if $ \alpha_6\neq\alpha_4, $ then by choosing $ z = -\frac{\alpha_2}{\alpha_6}, t = \frac{\alpha_1\alpha_6-\alpha_2\alpha_5+\alpha_2\alpha_3}{\alpha_6(\alpha_4-\alpha_6)}, $ we have the family of representatives $ \langle \alpha\nabla_3+\beta\nabla_4+\gamma\nabla_5+\nabla_6 \rangle_{\beta\neq1}; $

    4. if $ \alpha_7\neq0, $ then by choosing $ z = -\frac{\alpha_4}{2\alpha_7}, t = \frac{\alpha_4\alpha_6-2\alpha_2\alpha_7}{2\alpha^2_7}, $ we have the family of representatives

    $ \langle \alpha\nabla_1+\beta\nabla_3+\gamma\nabla_5+\mu\nabla_6+\nabla_7 \rangle.$

    Summarizing, we have the following distinct orbits:

    $\begin{array}{c} \langle \alpha\nabla_1+\nabla_2+\beta\nabla_3+(\beta+2)\nabla_5 \rangle_{\alpha \neq 0}^{O(\alpha, \beta) = O(-\alpha, \beta)}, \\ \langle \alpha\nabla_1+\beta\nabla_3+\nabla_4+\gamma\nabla_5+\nabla_6 \rangle^{O(\alpha, \beta, \gamma) = O(\alpha, -\beta,-\gamma)}_{\alpha\neq 0}, \\ \langle \alpha\nabla_1+\beta\nabla_3+\gamma\nabla_5+\mu\nabla_6+\nabla_7 \rangle^{O(\alpha, \beta, \gamma, \mu) = O(-\alpha, \beta, \gamma, -\mu)}, \langle \nabla_2+\alpha\nabla_3+\beta\nabla_5 \rangle, \\ \langle \alpha\nabla_2+\beta\nabla_3+\nabla_4+\gamma\nabla_5 \rangle^{O(\alpha, \beta, \gamma) = O(-\alpha, -\beta,- \gamma)}, \\ \langle \alpha \nabla_3+\beta \nabla_4+\gamma\nabla_5+\nabla_6 \rangle^{O(\alpha, \beta, \gamma) = O(-\alpha, \beta, -\gamma)}, \end{array}$

    which gives the following new algebras:

    $ \begin{array}{llllllllllllllllll} {\mathbf{N}}_{355}^{\alpha\neq0, \beta} & : & e_1e_1 = e_2 & e_1e_3 = e_4+\alpha e_5 & e_1e_4 = e_5 \\ && e_2e_2 = e_3 & e_2e_3 = \beta e_5 & { e_3e_3 = (\beta+2)e_5 } \\ {\mathbf{N}}_{356}^{\alpha\neq0, \beta,\gamma} & : & e_1e_1 = e_2 & e_1e_3 = e_4+\alpha e_5 & e_2e_2 = e_3 & e_2e_3 = \beta e_5 \\ & & e_2e_4 = e_5 & e_3e_3 = \gamma e_5 & e_3e_4 = e_5 & \\ {\mathbf{N}}_{357}^{\alpha, \beta,\gamma,\mu} & : & e_1e_1 = e_2 & e_1e_3 = e_4+\alpha e_5 & e_2e_2 = e_3 & e_2e_3 = \beta e_5 \\ & & e_3e_3 = \gamma e_5 & e_3e_4 = \mu e_5 & e_4e_4 = e_5 & \\ {\mathbf{N}}_{358}^{\alpha,\beta} & : & e_1e_1 = e_2 & e_1e_3 = e_4 & e_1e_4 = e_5 \\ && e_2e_2 = e_3 & e_2e_3 = \alpha e_5 & e_3e_3 = \beta e_5 \\ {\mathbf{N}}_{359}^{\alpha, \beta,\gamma} & : & e_1e_1 = e_2 & e_1e_3 = e_4 & e_1e_4 = \alpha e_5 & e_2e_2 = e_3 \\ & & e_2e_3 = \beta e_5 & e_2e_4 = e_5 & e_3e_3 = \gamma e_5 & \\ {\mathbf{N}}_{360}^{\alpha,\beta,\gamma} & : & e_1e_1 = e_2 & e_1e_3 = e_4 & e_2e_2 = e_3 & e_2e_3 = \alpha e_5 \\ & & e_2e_4 = \beta e_5 & e_3e_3 = \gamma e_5 & e_3e_4 = e_5 & \\ \end{array} $

    Here we will collect all information about $ {\mathbf N}_{07}^{4}: $

    $ \begin{array}{|l|l|l|l|} \hline \rm{ } & \rm{ } & \rm{Cohomology} & \rm{Automorphisms} \\ \hline {\mathbf{N}}^{4}_{07} & \begin{array}{l} e_1e_1 = e_2 \\ e_2e_2 = e_3 \\ e_2e_3 = e_4 \end{array} &\begin{array}{l} \mathrm{H}^2_{\mathfrak{C}}(\mathbf{N}^{4}_{07}) = \Big \langle [\Delta_{ij}] \Big\rangle\\ (i,j) \notin \{ (1,1),(2,2),(2,3) \} \end{array} & \phi = \begin{pmatrix} x&0&0&0\\ 0&x^2&0&0\\ 0&0&x^4&0\\ t&0&0&x^6 \end{pmatrix}\\ \hline \end{array} $

    Let us use the following notations:

    $ \begin{array}{llll} \nabla_1 = [\Delta_{12}], & \nabla_2 = [\Delta_{13}], & \nabla_3 = [\Delta_{14}], & \nabla_4 = [\Delta_{24}], \\ \nabla_5 = [\Delta_{33}], & \nabla_6 = [\Delta_{34}], & \nabla_7 = [\Delta_{44}]. \end{array} $

    Take $ \theta = \sum\limits_{i = 1}^{7}\alpha_i\nabla_i\in\mathrm{H}^2_{\mathfrak{C}}(\mathbf{N}^{4}_{07}) . $ Since

    $ \phi^T\begin{pmatrix} 0&\alpha_1&\alpha_2&\alpha_3\\ \alpha_1&0&0&\alpha_4\\ \alpha_2&0&\alpha_5&\alpha_6\\ \alpha_3&\alpha_4&\alpha_6&\alpha_7 \end{pmatrix}\phi = \begin{pmatrix} \alpha^*&\alpha^*_1&\alpha^{*}_2&\alpha^*_3\\ \alpha^*_1&0&0&\alpha^*_4\\ \alpha^{*}_2&0&\alpha^*_5&\alpha^*_6\\ \alpha^*_3&\alpha^*_4&\alpha^*_6&\alpha^*_7 \end{pmatrix} $

    we have

    $ \begin{array}{llll} \alpha_1^* = (\alpha_1x+\alpha_4t)x^2, & \alpha_2^* = (\alpha_2x+\alpha_6t)x^4, & \alpha_3^* = (\alpha_3x+\alpha_7t)x^6, & \alpha_4^* = \alpha_4x^8, \\ \alpha_5^* = \alpha_5x^8, & \alpha_6^* = \alpha_6x^{10}, & \alpha_7^* = \alpha_7x^{12}. \end{array} $

    We are interested in $ (\alpha_3,\alpha_4,\alpha_6,\alpha_7)\neq(0,0,0,0) $ and consider following cases:

    $ 1.\ $$ \alpha_7 = \alpha_6 = \alpha_4 = 0, $ then $ \alpha_3\neq0, $ and we have the following subcases:

    ${\rm{(a)}}\ $if $ \alpha_1 = 0, \alpha_2 = 0, $ then we have representatives $ \langle \nabla_3 \rangle $ and $ \langle \nabla_3+\nabla_5 \rangle $ depending on whether $ \alpha_5 = 0 $ or not;

    $ {\rm{(b)}}\ $if $ \alpha_1 = 0, \alpha_2\neq0, $ then by choosing $ x = \sqrt{{\alpha_2}{\alpha_3^{-1}}}, t = 0, $ we have the family of representatives $ \langle \nabla_2+\nabla_3+\alpha\nabla_5 \rangle; $

    $ {\rm{(c)}}\ $if $ \alpha_1\neq0, $ then by choosing $ x = \sqrt[4]{{\alpha_1}{\alpha_3^{-1}}}, t = 0, $ we have the family of representatives $ \langle \nabla_1+\alpha\nabla_2+\nabla_3+\beta\nabla_5 \rangle. $

    2. $ \alpha_7 = 0, \alpha_6 = 0, \alpha_4\neq0, $ then we have the following subcases:

    $ {\rm{(a)}}\ $if $ \alpha_2 = 0, \alpha_3 = 0 , $ then we have the familty of representative $ \langle \nabla_4+\alpha\nabla_5 \rangle; $

    ${\rm{(b)}}\ $if $ \alpha_2 = 0, \alpha_3\neq0 , $ then by choosing $ x = {\alpha_3}{\alpha_4^{-1}}, t = -{\alpha_1\alpha_3}{\alpha_4^{-2}}, $ we have the family of representatives $ \langle \nabla_3+ \nabla_4+\alpha\nabla_5 \rangle; $

    $ {\rm{(c)}}\ $if $ \alpha_2\neq0, $ then by choosing $ x = \sqrt[3]{{\alpha_2}{\alpha_4^{-1}}}, t = -{\alpha_1\sqrt[3]{\alpha_2 \alpha_4^{-4}}}, $ we have the family of representatives $ \langle \nabla_2+\alpha\nabla_3+ \nabla_4+\beta\nabla_5 \rangle. $

    3. $ \alpha_7 = 0, \alpha_6\neq0, $ then we have the following subcases:

    $ {\rm{(a)}}\ $if $ \alpha_1 = 0, \alpha_3 = 0, \alpha_4 = 0, $ then we have representatives $ \langle \nabla_6 \rangle $ and $ \langle \nabla_5+\nabla_6 \rangle $ depending on whether $ \alpha_5 = 0 $ or not;

    $ {\rm{(b)}}\ $if $ \alpha_1\alpha_6 = \alpha_2\alpha_4, \alpha_3 = 0, \alpha_4\neq0, $ then choosing $ x = \sqrt{{\alpha_4}{\alpha_6^{-1}}}, t = -{\alpha_2\sqrt{\alpha_4} \alpha_6^{-3}}, $ we have the family of representatives $ \langle \nabla_4+\alpha\nabla_5+\nabla_6 \rangle; $

    $ {\rm{(c)}}\ $if $ \alpha_1\alpha_6 = \alpha_2\alpha_4, \alpha_3\neq0, $ then by choosing $ x = \sqrt[3]{\frac{\alpha_3}{\alpha_6}}, t = -\frac{\alpha_2\sqrt[3]{\alpha_3}}{\alpha_6\sqrt[3]{\alpha_6}}, $ we have the family of representatives $ \langle \nabla_3+\alpha\nabla_4+\beta\nabla_5+\nabla_6 \rangle; $

    ${\rm{(d)}}\ $if $ \alpha_1\alpha_6\neq\alpha_2\alpha_4 , $ then choosing $ x = \sqrt[7]{\frac{\alpha_1\alpha_6-\alpha_2\alpha_4}{\alpha^2_6}}, t = -\frac{\alpha_2\sqrt[7]{\alpha_1\alpha_6-\alpha_2\alpha_4}}{\alpha_6\sqrt[7]{\alpha^2_6}}, $ we have the family of representatives $ \langle \nabla_1+\alpha\nabla_3+\beta\nabla_4+\gamma\nabla_5+\nabla_6 \rangle. $

    4. $ \alpha_7\neq0, $ then we have the following subcases:

    $ {\rm{(a)}}\ $$ \alpha_1 = 0, \alpha_2\alpha_7 = \alpha_3\alpha_6, \alpha_4 = 0, \alpha_5 = 0, $ then we have representatives $ \langle \nabla_7 \rangle $ and $ \langle \nabla_6+\nabla_7 \rangle $ depending on whether $ \alpha_6 = 0 $ or not;

    ${\rm{(b)}}\ $$ \alpha_1 = 0, \alpha_2\alpha_7 = \alpha_3\alpha_6, \alpha_4 = 0, \alpha_5\neq0, $ then by choosing $ x = \sqrt[4]{{\alpha_5}{\alpha_7^{-1}}}, t = -{\alpha_3\sqrt[4]{\alpha_5 \alpha_7^{-5}}}, $ we have family of representatives $ \langle \nabla_5+\alpha\nabla_6+\nabla_7 \rangle; $

    $ {\rm{(c)}}\ $$ \alpha_1\alpha_7 = \alpha_3\alpha_4, \alpha_2\alpha_7 = \alpha_3\alpha_6, \alpha_4\neq0, $ then by choosing $ x = \sqrt[4]{{\alpha_4}{\alpha_7^{-1}}}, t = -{\alpha_3\sqrt[4]{\alpha_4 \alpha_7^{-5}}}, $ we have family of representatives $ \langle \nabla_4+\alpha\nabla_5+\beta\nabla_6+\nabla_7 \rangle; $

    ${\rm{(d)}}\ $$ \alpha_1\alpha_7 = \alpha_3\alpha_4, \alpha_2\alpha_7\neq\alpha_3\alpha_6, $ then by choosing $ x = \sqrt[7]{(\alpha_2\alpha_7-\alpha_3\alpha_6)\alpha^{-2}_7}, $ $ t = -{\alpha_3\sqrt[7]{(\alpha_2\alpha_7-\alpha_3\alpha_6)\alpha_7^{-9}}}, $ we have the family of representatives

    $ \langle \nabla_2+\alpha\nabla_4+\beta\nabla_5+\gamma\nabla_6+\nabla_7 \rangle;$

    $ {\rm{(e)}}\ $$ \alpha_1\alpha_7\neq \alpha_3\alpha_4, $ then by choosing

    $ x = \sqrt[9]{(\alpha_1\alpha_7-\alpha_3\alpha_4)\alpha^{-2}_7}, t = -\alpha_3\sqrt[9]{(\alpha_1\alpha_7-\alpha_3\alpha_4)\alpha_7^{-11} }, $

    we have family of representatives $ \langle \nabla_1+\alpha\nabla_2+\beta\nabla_4+\gamma\nabla_5+\mu\nabla_6+\nabla_7 \rangle. $

    Summarizing, we have the following distinct orbits:

    $\begin{array}{c} \langle \nabla_1+\alpha\nabla_2+\nabla_3+\beta\nabla_5 \rangle ^{O(\alpha,\beta) = O(-\alpha,i\beta) = O(-\alpha,-i\beta) = O(\alpha,-\beta)}, \\ \langle \nabla_1+\alpha\nabla_2+\beta\nabla_4+\gamma\nabla_5+\mu\nabla_6+\nabla_7 \rangle ^{{\begin{array}{l} O(\alpha,\beta,\gamma,\mu) = O(-\eta_9^7\alpha,\eta^4_9\beta,\eta^4_9\gamma,\eta^2_9\mu) = \\ O(-\eta^5_9\alpha,\eta_9^8\beta,\eta_9^8\gamma,\eta^4_9\mu) = O(-\eta_3\alpha,-\eta_3\beta,-\eta_3\gamma,\eta^2_3\mu) = \\ O(-\eta_9\alpha,-\eta^7_9\beta,-\eta^7_9\gamma,\eta^8_9\mu) = O(\eta_9^8\alpha,\eta^2_9\beta,\eta^2_9\gamma,-\eta_9\mu) = \\ O(\eta^2_3\alpha,\eta^2_3\beta,\eta^2_3\gamma,-\eta_3\mu) = O(\eta^4_9\alpha,-\eta_9\beta,-\eta_9\gamma,-\eta^5_9\mu) = \\ O(\eta^2_9\alpha,-\eta^5_9\beta,-\eta^5_9\gamma,-\eta^7_9\mu) \end{array}}}, \\ \langle \nabla_1+\alpha\nabla_3+\beta\nabla_4+\gamma\nabla_5+\nabla_6 \rangle ^{{\begin{array}{l} O(\alpha,\beta,\gamma) = O(-\eta_7^3\alpha,\eta^2_7\beta,\eta^2_7\gamma) = \\ O(\eta_7^6\alpha,\eta^4_7\beta,\eta^4_7\gamma) = O(\eta^2_7\alpha,\eta^6_{7}\beta,\eta^6_7\gamma) = \\ O(-\eta^5_7\alpha,-\eta_7\beta,-\eta_7\gamma) = O(-\eta_7\alpha,-\eta^3_7\beta,-\eta^3_7\gamma) = \\ O(\eta^4_7\alpha,-\eta^5_7\beta,-\eta^5_7\gamma) \end{array}}}, \\ \langle \nabla_2+\alpha\nabla_3+ \nabla_4+\beta\nabla_5 \rangle^{O(\alpha,\beta) = O(-\eta_3\alpha,\beta) = O(\eta^2_3\alpha,\beta)}, \langle \nabla_2+\nabla_3+\alpha\nabla_5 \rangle^{O(\alpha) = O(-\alpha)}, \\ \langle \nabla_2+\alpha\nabla_4+\beta\nabla_5+\gamma\nabla_6+\nabla_7 \rangle^{{\begin{array}{l} O(\alpha,\beta,\gamma) = O(\eta_7^4\alpha,\eta^4_7\beta,\eta^2_7\gamma) = O(-\eta_7\alpha,-\eta_7\beta,\eta^4_7\gamma) = \\ O(-\eta^5_7\alpha,-\eta^5_{7}\beta,\eta^6_7\gamma) = O(\eta^2_7\alpha,\eta^2_7\beta,-\eta_7\gamma) = \\ O(\eta_7^6\alpha,\eta_7^6\beta,-\eta^3_7\gamma) = O(-\eta^3_7\alpha,-\eta^3_7\beta,-\eta^5_7\gamma) \end{array}}}, \langle \nabla_3 \rangle, \\ \langle \nabla_3+ \nabla_4+\alpha\nabla_5 \rangle, \langle \nabla_3+\alpha\nabla_4+\beta\nabla_5+\nabla_6 \rangle^{O(\alpha,\beta) = O(-\eta_3\alpha,-\eta_3\beta) = O(\eta^2_3\alpha,\eta^2_3\beta)}, \\ \langle \nabla_3+\nabla_5 \rangle, \langle \nabla_4+\alpha\nabla_5 \rangle, \langle \nabla_4+\alpha\nabla_5+\nabla_6 \rangle, \langle \nabla_4+\alpha\nabla_5+\beta\nabla_6+\nabla_7 \rangle^{O(\alpha,\beta) = O(\alpha,-\beta)}, \\ \langle \nabla_5+\nabla_6 \rangle, \langle \nabla_5+\alpha\nabla_6+\nabla_7 \rangle^{O(\alpha) = O(-\alpha) }, \langle \nabla_6 \rangle, \langle \nabla_6 +\nabla_7 \rangle, \langle \nabla_7 \rangle,\end{array} $

    which gives the following new algebras:

    $ \begin{array}{llllllllllllllllll} {\mathbf{N}}_{361}^{\alpha, \beta} & : & e_1e_1 = e_2 & e_1e_2 = e_5 & e_1e_3 = \alpha e_5 & e_1e_4 = e_5 \\ & & e_2e_2 = e_3 & e_2e_3 = e_4 & e_3e_3 = \beta e_5 \\ {\mathbf{N}}_{362}^{\alpha, \beta,\gamma,\mu} & : & e_1e_1 = e_2 & e_1e_2 = e_5 & e_1e_3 = \alpha e_5 & e_2e_2 = e_3 & e_2e_3 = e_4 \\ & & e_2e_4 = \beta e_5 & e_3e_3 = \gamma e_5 & e_3e_4 = \mu e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{363}^{\alpha, \beta,\gamma} & : & e_1e_1 = e_2 & e_1e_2 = e_5 & e_1e_4 = \alpha e_5 & e_2e_2 = e_3 \\ & & e_2e_3 = e_4 & e_2e_4 = \beta e_5 & e_3e_3 = \gamma e_5 & e_3e_4 = e_5 \\ {\mathbf{N}}_{364}^{\alpha, \beta} & : & e_1e_1 = e_2 & e_1e_3 = e_5 & e_1e_4 = \alpha e_5 & e_2e_2 = e_3 \\ & & e_2e_3 = e_4 & e_2e_4 = e_5 & e_3e_3 = \beta e_5 \\ {\mathbf{N}}_{365}^{\alpha} & : & e_1e_1 = e_2 & e_1e_3 = e_5 & e_1e_4 = e_5 \\ && e_2e_2 = e_3 & e_2e_3 = e_4 & e_3e_3 = \alpha e_5 \\ {\mathbf{N}}_{366}^{\alpha, \beta,\gamma} & : & e_1e_1 = e_2 & e_1e_3 = e_5 & e_2e_2 = e_3 & e_2e_3 = e_4 \\ & & e_1e_4 = \alpha e_5 & e_3e_3 = \beta e_5 & e_3e_4 = \gamma e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{367} & : & e_1e_1 = e_2 & e_1e_4 = e_5 & e_2e_2 = e_3 & e_2e_3 = e_4 \\ {\mathbf{N}}_{368}^{\alpha} & : & e_1e_1 = e_2 & e_1e_4 = e_5 & e_2e_2 = e_3 \\ && e_2e_3 = e_4 & e_2e_4 = e_5 & e_3e_3 = \alpha e_5 \\ {\mathbf{N}}_{369}^{\alpha,\beta} & : & e_1e_1 = e_2 & e_1e_4 = e_5 & e_2e_2 = e_3 & e_2e_3 = e_4 \\ & & e_2e_4 = \alpha e_5 & e_3e_3 = \beta e_5 & e_3e_4 = e_5 & \\ {\mathbf{N}}_{370} & : & e_1e_1 = e_2 & e_1e_4 = e_5 & e_2e_2 = e_3 & e_2e_3 = e_4 & e_3e_3 = e_5 \\ {\mathbf{N}}_{371}^{\alpha} & : & e_1e_1 = e_2 & e_2e_2 = e_3 & e_2e_3 = e_4 & e_2e_4 = e_5 & e_3e_3 = \alpha e_5 \\ {\mathbf{N}}_{372}^{\alpha} & : & e_1e_1 = e_2 & e_2e_2 = e_3 & e_2e_3 = e_4 \\ && e_2e_4 = e_5 & e_3e_3 = \alpha e_5 & e_3e_4 = e_5 \\ {\mathbf{N}}_{373}^{\alpha, \beta} & : & e_1e_1 = e_2 & e_2e_2 = e_3 & e_2e_3 = e_4 & e_2e_4 = e_5 \\ & & e_3e_3 = \alpha e_5 & e_3e_4 = \beta e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{374} & : & e_1e_1 = e_2 & e_2e_2 = e_3 & e_2e_3 = e_4 & e_3e_3 = e_5 & e_3e_4 = e_5 \\ {\mathbf{N}}_{375}^{\alpha} & : & e_1e_1 = e_2 & e_2e_2 = e_3 & e_2e_3 = e_4 \\ & & e_3e_3 = e_5 & e_3e_4 = \alpha e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{376} & : & e_1e_1 = e_2 & e_2e_2 = e_3 & e_2e_3 = e_4 & e_3e_4 = e_5 \\ {\mathbf{N}}_{377} & : & e_1e_1 = e_2 & e_2e_2 = e_3 & e_2e_3 = e_4 & e_3e_4 = e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{378} & : & e_1e_1 = e_2 & e_2e_2 = e_3 & e_2e_3 = e_4 & e_4e_4 = e_5 \\ \end{array} $

    Here we will collect all information about $ {\mathbf N}_{08}^{4}: $

    $ \begin{array}{|l|l|l|l|} \hline \rm{ } & \rm{ } & \rm{Cohomology} & \rm{Automorphisms}\\ \hline {\mathbf{N}}^{4}_{08} & \begin{array}{l} e_1e_1 = e_2 \\ e_1e_3 = e_4 \\ e_2e_2 = e_3 \\ e_2e_3 = e_4 \end{array} & \begin{array}{l} \mathrm{H}^2_{\mathfrak{C}}(\mathbf{N}^{4}_{08}) = \Big \langle [\Delta_{ij}] \Big\rangle\\ {(i,j) \notin \{(1,1),(1,3),(2,2)}\} \end{array} & \phi = \begin{pmatrix} 1&0&0&0\\ 0&1&0&0\\ 0&0&1&0\\ t&0&0&1 \end{pmatrix}\\ \hline \end{array} $

    Let us use the following notations:

    $ \begin{array}{llll} \nabla_1 = [\Delta_{12}], & \nabla_2 = [\Delta_{14}], & \nabla_3 = [\Delta_{23}], & \nabla_4 = [\Delta_{24}], \\ \nabla_5 = [\Delta_{33}], & \nabla_6 = [\Delta_{34}], & \nabla_7 = [\Delta_{44}]. \end{array} $

    Take $ \theta = \sum\limits_{i = 1}^{7}\alpha_i\nabla_i\in\mathrm{H}^2_{\mathfrak{C}}(\mathbf{N}^{4}_{08}) . $ Since

    $ \phi^T\begin{pmatrix} 0&\alpha_1&0&\alpha_2\\ \alpha_1&0&\alpha_3&\alpha_4\\ 0&\alpha_3&\alpha_5&\alpha_6\\ \alpha_2&\alpha_4&\alpha_6&\alpha_7 \end{pmatrix}\phi = \begin{pmatrix} \alpha^*&\alpha^{*}_1&\alpha^{**}&\alpha^*_2\\ \alpha^{*}_1&0&\alpha^*_3+\alpha^{**}&\alpha^*_4\\ \alpha^{**}&\alpha^*_3+\alpha^{**}&\alpha^*_5&\alpha^*_6\\ \alpha^*_2&\alpha^*_4&\alpha^*_6&\alpha^*_7 \end{pmatrix} $

    we have

    $ \begin{array}{llll} \alpha_1^* = \alpha_1+\alpha_4t, & \alpha_2^* = \alpha_2+\alpha_7t, & \alpha_3^* = \alpha_3-\alpha_6t, & \alpha_4^* = \alpha_4, \\ \alpha_5^* = \alpha_5, & \alpha_6^* = \alpha_6, & \alpha_7^* = \alpha_7. \end{array} $

    Since $ (\alpha_2,\alpha_4,\alpha_6,\alpha_7)\neq(0,0,0,0), $ we have the following cases:

    $1.\ $if $ \alpha_7 = \alpha_6 = \alpha_4 = 0, $ then $ \alpha_2\neq0, $ and we have the family of representatives

    $ \langle \alpha\nabla_1+\nabla_2+\beta\nabla_3+\gamma\nabla_5 \rangle;$

    $2.\ $if $ \alpha_7 = 0, \alpha_6 = 0, \alpha_4\neq0, $ then by choosing $ t = -{\alpha_1}{\alpha_4^{-1}}, $ we have the family of representatives

    $ \langle \alpha\nabla_2+\beta\nabla_3+\nabla_4+\gamma\nabla_5 \rangle;$

    $ 3.\ $if $ \alpha_7 = 0, \alpha_6\neq0, $ then by choosing $ t = {\alpha_3}{\alpha_6^{-1}}, $ we have the family of representatives

    $ \langle \alpha\nabla_1+\beta\nabla_2+\gamma\nabla_4+\mu\nabla_5+\nabla_6 \rangle;$

    $ 4.\ $if $ \alpha_7\neq0, $ then by choosing $ t = -{\alpha_2}{\alpha_7^{-1}}, $ we have the family of representatives

    $ \langle \alpha\nabla_1+\beta\nabla_3+\gamma\nabla_4+\mu\nabla_5+\nu\nabla_6+\nabla_7 \rangle.$

    Summarizing, we have the following distinct orbits:

    $\begin{array}{c} \langle \alpha\nabla_1+\nabla_2+\beta\nabla_3+\gamma\nabla_5 \rangle,\langle \alpha\nabla_1+\beta\nabla_2+\gamma\nabla_4+\mu\nabla_5+\nabla_6 \rangle, \\ \langle \alpha\nabla_1+\beta\nabla_3+\gamma\nabla_4+\mu\nabla_5+\nu\nabla_6+\nabla_7 \rangle,\langle \alpha\nabla_2+\beta\nabla_3+\nabla_4+\gamma\nabla_5 \rangle,\end{array}$

    which gives the following new algebras:

    $ \begin{array}{llllllllllllllllll} {\mathbf{N}}_{379}^{\alpha, \beta,\gamma} & : & e_1e_1 = e_2 & e_1e_2 = \alpha e_5 & e_1e_3 = e_4 & e_1e_4 = e_5 \\ & & e_2e_2 = e_3 & e_2e_3 = e_4+\beta e_5 & e_3e_3 = \gamma e_5 \\ {\mathbf{N}}_{380}^{\alpha, \beta,\gamma,\mu} & : & e_1e_1 = e_2 & e_1e_2 = \alpha e_5 & e_1e_3 = e_4 \\ & & e_1e_4 = \beta e_5 & e_2e_2 = e_3 & e_2e_3 = e_4 \\ & & e_2e_4 = \gamma e_5 & e_3e_3 = \mu e_5 & e_3e_4 = e_5 \\ {\mathbf{N}}_{381}^{\alpha, \beta,\gamma,\mu, \nu} & : & e_1e_1 = e_2 & e_1e_2 = \alpha e_5 & e_1e_3 = e_4 \\ && e_2e_2 = e_3 & e_2e_3 = e_4+\beta e_5 & e_2e_4 = \gamma e_5 \\ && e_3e_3 = \mu e_5 & e_3e_4 = \nu e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{382}^{\alpha, \beta,\gamma} & : & e_1e_1 = e_2 & e_1e_3 = e_4 & e_1e_4 = \alpha e_5 & e_2e_2 = e_3 \\ & & e_2e_3 = e_4+\beta e_5 & e_2e_4 = e_5 & e_3e_3 = \gamma e_5 \\ \end{array} $

    Here we will collect all information about $ {\mathbf N}_{09}^{4}: $

    $ \begin{array}{|l|l|l|l|} \hline \rm{ } & \rm{ } & \rm{Cohomology} & \rm{Automorphisms} \\ \hline {\mathbf{N}}^{4}_{09} & \begin{array}{l} e_1e_1 = e_2 \\ e_2e_2 = e_3 \\ e_3e_3 = e_4 \end{array} &\begin{array}{l} \mathrm{H}^2_{\mathfrak{C}}(\mathbf{N}^{4}_{09}) = \Big \langle [\Delta_{ij}] \Big\rangle\\ (i,j) \notin\{ (1,1),(2,2),(3,3)\} \end{array} & \phi = \begin{pmatrix} x&0&0&0\\ 0&x^2&0&0\\ 0&0&x^4&0\\ t&0&0&x^8 \end{pmatrix}\\ \hline \end{array} $

    Let us use the following notations:

    $ \begin{array}{llll} \nabla_1 = [\Delta_{12}], & \nabla_2 = [\Delta_{13}], & \nabla_3 = [\Delta_{14}], & \nabla_4 = [\Delta_{23}], \\ \nabla_5 = [\Delta_{24}], & \nabla_6 = [\Delta_{34}], & \nabla_7 = [\Delta_{44}]. \end{array} $

    Take $ \theta = \sum\limits_{i = 1}^{7}\alpha_i\nabla_i\in\mathrm{H}^2_{\mathfrak{C}}(\mathbf{N}^{4}_{09}) . $ Since

    $ \phi^T\begin{pmatrix} 0&\alpha_1&\alpha_2&\alpha_3\\ \alpha_1&0&\alpha_4&\alpha_5\\ \alpha_2&\alpha_4&0&\alpha_6\\ \alpha_3&\alpha_5&\alpha_6&\alpha_7 \end{pmatrix}\phi = \begin{pmatrix} \alpha^*&\alpha^*_1&\alpha^{*}_2&\alpha^*_3\\ \alpha^*_1&0&\alpha^*_4&\alpha^*_5\\ \alpha^{*}_2&\alpha^*_4&0&\alpha^*_6\\ \alpha^*_3&\alpha^*_5&\alpha^*_6&\alpha^*_7 \end{pmatrix} $

    we have

    $ \begin{array}{llll} \alpha_1^* = (\alpha_1x+\alpha_5t)x^2, & \alpha_2^* = (\alpha_2x+\alpha_6t)x^4, & \alpha_3^* = (\alpha_3x+\alpha_7t)x^8, & \alpha_4^* = \alpha_4x^6, \\ \alpha_5^* = \alpha_5x^{10}, & \alpha_6^* = \alpha_6x^{12}, & \alpha_7^* = \alpha_7x^{16}. \end{array} $

    Since $ (\alpha_3,\alpha_5,\alpha_6,\alpha_7)\neq(0,0,0,0), $ we have the following cases:

    $ 1.\ $$ \alpha_7 = \alpha_6 = \alpha_5 = 0, $ then $ \alpha_3\neq0, $ and we have the following subcases:

    $ {\rm{(a)}}\ $if $ \alpha_1 = 0, \alpha_2 = 0, $ then we have representatives $ \langle \nabla_3 \rangle $ and $ \langle \nabla_3+\nabla_4 \rangle $ depending on whether $ \alpha_4 = 0 $ or not;

    $ {\rm{(b)}}\ $if $ \alpha_1 = 0, \alpha_2\neq0, $ then by choosing $ x = \sqrt[4]{{\alpha_2}{\alpha_3^{-1}}}, t = 0, $ we have the family of representatives $ \langle \nabla_2+\nabla_3+\alpha\nabla_4 \rangle; $

    $ {\rm{(c)}}\ $if $ \alpha_1\neq0, $ then by choosing $ x = \sqrt[6]{{\alpha_1}{\alpha_3^{-1}}}, t = 0, $ we have the family of representatives $ \langle \nabla_1+\alpha\nabla_2+\nabla_3+\beta\nabla_4 \rangle. $

    2. $ \alpha_7 = 0, \alpha_6 = 0, \alpha_5\neq0, $ then we have the following cases:

    ${\rm{(a)}}\ $if $ \alpha_2 = 0, \alpha_3 = 0, $ then we have representatives $ \langle \nabla_5 \rangle $ and $ \langle \nabla_4+\nabla_5 \rangle $ depending on whether $ \alpha_4 = 0 $ or not;

    $ {\rm{(b)}}\ $if $ \alpha_2 = 0, \alpha_3\neq0, $ then choosing $ x = {\alpha_3}{\alpha_5^{-1}}, t = -{\alpha_1\alpha_3}{\alpha_5^{-2}}, $ we have the family of representatives $ \langle \nabla_3+ \alpha\nabla_4+\nabla_5 \rangle; $

    $ {\rm{(c)}}\ $if $ \alpha_2\neq0, $ then by choosing $ x = \sqrt[5]{{\alpha_2}{\alpha_5^{-1}}}, t = -{\alpha_1\sqrt[5]{\alpha_2 \alpha_5^{-6} }}, $ we have the family of representatives $ \langle \nabla_2+\alpha\nabla_3+ \beta\nabla_4+\nabla_5 \rangle. $

    3. $ \alpha_7 = 0, \alpha_6\neq0, $ then we have the following cases:

    $ {\rm{(a)}}\ $if $ \alpha_1\alpha_6 = \alpha_2\alpha_5, \alpha_3 = 0, \alpha_4 = 0, $ then we have representatives $ \langle \nabla_6 \rangle $ and $ \langle \nabla_5+\nabla_6 \rangle $ depending on whether $ \alpha_5 = 0 $ or not;

    $ {\rm{(b)}}\ $if $ \alpha_1\alpha_6 = \alpha_2\alpha_5, \alpha_3 = 0, \alpha_4\neq0, $ then by choosing $ x = \sqrt[6]{{\alpha_4}{\alpha_6^{-1}}}, t = -{\alpha_2\sqrt[6]{\alpha_4 \alpha_6^{-7} }}, $ we have the family of representatives $ \langle \nabla_4+\alpha\nabla_5+\nabla_6 \rangle; $

    $ {\rm{(c)}}\ $if $ \alpha_1\alpha_6 = \alpha_2\alpha_5, \alpha_3\neq0, $ then by choosing $ x = \sqrt[3]{{\alpha_3}{\alpha_6^{-1}}}, t = -{\alpha_2\sqrt[3]{\alpha_3 \alpha_6^{-4}}}, $ we have the family of representatives $ \langle \nabla_3+\alpha\nabla_4+\beta\nabla_5+\nabla_6 \rangle; $

    $ {\rm{(d)}}\ $if $ \alpha_1\alpha_6\neq \alpha_2\alpha_5 , $ then by choosing

    $ x = \sqrt[9]{(\alpha_1\alpha_6-\alpha_2\alpha_5)\alpha^{-2}_6},$ $t = -{\alpha_2\sqrt[9]{(\alpha_1\alpha_6-\alpha_2\alpha_5)\alpha_6^{-11}}},$

    we have the family of representatives

    $ \langle \nabla_1+\alpha\nabla_3+\beta\nabla_4+\gamma\nabla_5+\nabla_6 \rangle.$

    4. $ \alpha_7\neq0, $ then we have the following cases:

    $ {\rm{(a)}}\ $if $ \alpha_1 = 0, \alpha_2\alpha_7 = \alpha_3\alpha_6, \alpha_4 = 0, \alpha_5 = 0, $ then choosing $ x = 1, t = -\frac{\alpha_3}{\alpha_7}, $ we have representatives $ \langle \nabla_7 \rangle $ and $ \langle \nabla_6+\nabla_7 \rangle $ depending on whether $ \alpha_6 = 0 $ or not;

    $ {\rm{(b)}}\ $if $ \alpha_1\alpha_7 = \alpha_3\alpha_5, \alpha_2\alpha_7 = \alpha_3\alpha_6, \alpha_4 = 0, \alpha_5\neq0, $ then by choosing $ x = \sqrt[6]{{\alpha_5}{\alpha_7^{-1}}}, t = -{\alpha_3\sqrt[6]{\alpha_5 \alpha_7^{-7} }}, $ we have the family of representatives $ \langle \nabla_5+\alpha\nabla_6+\nabla_7 \rangle; $

    $ {\rm{(c)}}\ $if $ \alpha_1\alpha_7 = \alpha_3\alpha_5, \alpha_2\alpha_7 = \alpha_3\alpha_6, \alpha_4\neq0, $ then by choosing

    $ x = \sqrt[10]{{\alpha_4}{\alpha_7^{-1}}},$ $t = -{\alpha_3\sqrt[10]{\alpha_4 \alpha_7^{-11}}},$

    we have the family of representatives $ \langle \nabla_4+\alpha\nabla_5+\beta\nabla_6+\nabla_7 \rangle; $

    $ {\rm{(d)}}\ $$ \alpha_1\alpha_7 = \alpha_3\alpha_5, \alpha_2\alpha_7\neq\alpha_3\alpha_6, $ then by choosing

    $ x = \sqrt[11]{(\alpha_2\alpha_7-\alpha_3\alpha_6)\alpha^{-2}_7 },$ $t = -{\alpha_3\sqrt[11]{(\alpha_2\alpha_7-\alpha_3\alpha_6) \alpha_7^{-13} }}, $

    we have the family of representatives

    $\langle \nabla_2+\alpha\nabla_4+\beta\nabla_5+\gamma\nabla_6+\nabla_7 \rangle;$

    $ {\rm{(e)}}\ $$ \alpha_1\alpha_7\neq\alpha_3\alpha_5, $ then by choosing

    $ x = \sqrt[13]{({\alpha_1\alpha_7-\alpha_3\alpha_5}){\alpha^{-2}_7}},$ $t = -{\alpha_3\sqrt[13]{(\alpha_1\alpha_7-\alpha_3\alpha_5)\alpha_7^{-15} }}, $

    we have the family of representatives

    $ \langle \nabla_1+\alpha\nabla_2+\beta\nabla_4+\gamma\nabla_5+\mu\nabla_6+\nabla_7 \rangle.$

    Summarizing, we have the following distinct orbits:

    $\begin{array}{c} \langle \nabla_1+\alpha\nabla_2+\nabla_3+\beta\nabla_4 \rangle ^{O(\alpha,\beta) = O(-\eta_3\alpha,\beta) = O(-\eta_3\alpha,-\beta) = O(\eta_3^2\alpha,-\beta) = O(\eta^2_3\alpha,\beta) = O(\alpha,-\beta)}, \\ \langle \nabla_1+\alpha\nabla_2+\beta\nabla_4+\gamma\nabla_5+\mu\nabla_6+ \\ \nabla_7 \rangle^{{\begin{array}{l} O(\alpha,\beta,\gamma,\mu) = O(-\eta_{13}^{11}\alpha,\eta^{10}_{13}\beta,\eta^6_{13}\gamma,\eta^4_{13}\mu) = \\ O(-\eta^9_{13}\alpha,-\eta^7_{13}\beta,\eta^{12}_{13}\gamma,\eta^8_{13}\mu) = O(-\eta^7_{13}\alpha,\eta^4_{13}\beta,-\eta^5_{13}\gamma,\eta^{12}_{13}\mu) = \\ O(-\eta^5_{13}\alpha,-\eta^{1}_{13}\beta,-\eta^{11}_{13}\gamma,-\eta^{3}_{13}\mu) = O(-\eta^{3}_{13}\alpha,-\eta^{11}_{13}\beta,\eta^4_{13}\gamma,-\eta^7_{13}\mu) = \\ O(-\eta^{1}_{13}\alpha,\eta^8_{13}\beta,\eta^{10}_{13}\gamma,-\eta^{11}_{13}\mu) = O(\eta^{12}_{13}\alpha,-\eta^5_{13}\beta,-\eta^{3}_{13}\gamma,\eta^{2}_{13}\mu) = \\ O(\eta^{10}_{13}\alpha,\eta^{2}_{13}\beta,-\eta^{9}_{13}\gamma,\eta^6_{13}\mu) = O(\eta^8_{13}\alpha,\eta^{12}_{13}\beta,\eta^{2}_{13}\gamma,\eta^{10}_{13}\mu) = \\ O(\eta^6_{13}\alpha,-\eta^{9}_{13}\beta,\eta^{8}_{13}\gamma,-\eta^{1}_{13}\mu) = O(\eta^4_{13}\alpha,\eta^{6}_{13}\beta,-\eta^{1}_{13}\gamma,-\eta^5_{13}\mu) = \\ O(\eta^{2}_{13}\alpha,-\eta^{3}_{13}\beta,-\eta^{7}_{13}\gamma,-\eta^9_{13}\mu) \end{array}}}, \\ \langle \nabla_1+\alpha\nabla_3+\beta\nabla_4+\gamma\nabla_5+\nabla_6 \rangle^{{\begin{array}{l} O(\alpha,\beta,\gamma,\mu) = O(-\eta_3\alpha,\eta^2_3\beta,\eta^2_9\gamma) = O(\eta^2_3\alpha,-\eta_3\beta,\eta^4_9\gamma) = \\ O(\alpha,\beta,\eta^2_3\gamma) = O(-\eta_3\alpha,\eta^2_3\beta,\eta^8_9\gamma) = O(\eta^2_3\alpha,-\eta_3\beta,-\eta_9\gamma) = \\ O(\alpha,\beta,-\eta_3\gamma) = O(-\eta_3\alpha,\eta^2_3\beta,-\eta^5_9\gamma) = O(\eta^2_3\alpha,-\eta_3\beta,-\eta^7_9\gamma) \end{array}}}, \\ \langle \nabla_2+\nabla_3+\alpha\nabla_4 \rangle^{O(\alpha) = O(i\alpha) = O(-\alpha) = (-i\alpha)}, \\ \langle \nabla_2+\alpha\nabla_3+ \beta\nabla_4+\nabla_5 \rangle^{ O(\alpha,\beta) = O(-\eta_5\alpha,\eta^4_5\beta) = O(\eta^2_5\alpha,-\eta^3_5\beta) = O(-\eta^3_5\alpha,\eta^2_5\beta) = O(\eta_5^4\alpha,-\eta_5\beta)},\\ \langle \nabla_2+\alpha\nabla_4+\beta\nabla_5+\gamma\nabla_6+\nabla_7 \rangle^{{\begin{array}{l} O(\alpha,\beta,\gamma) = O(\eta_{11}^{10}\alpha,\eta^6_{11}\beta,\eta^4_{11}\gamma = \\ O(-\eta^9_{11}\alpha,-\eta_{11}\beta,\eta^38_{11}\gamma) = O(\eta^8_{11}\alpha,-\eta^7_{11}\beta,-\eta_{11}\gamma) = \\ O(-\eta^7_{11}\alpha,\eta^2_{11}\beta,-\eta^5_{11}\gamma) = O(\eta^{6}_{11}\alpha,\eta^8_{11}\beta,-\eta^9_{11}\gamma) = \\ O(-\eta^{5}_{11}\alpha,-\eta^3_{11}\beta,\eta^2_{11}\gamma) = O(\eta^4_{11}\alpha,-\eta^9_{13}\beta,\eta^{6}_{11}\gamma) = \\ O(-\eta^3_{11}\alpha,\eta^{4}_{11}\beta,\eta^{10}_{11}\gamma) = O(\eta^2_{11}\alpha,\eta^{10}_{11}\beta,-\eta^{3}_{11}\gamma) = \\ O(-\eta_{11}\alpha,-\eta^{5}_{11}\beta,-\eta^{7}_{11}\gamma) \end{array}}}, \langle \nabla_3 \rangle, \\ \langle \nabla_3+\nabla_4 \rangle, \langle \nabla_3+ \alpha\nabla_4+\nabla_5 \rangle, \langle \nabla_3+\alpha\nabla_4+\beta\nabla_5+\nabla_6 \rangle^{O(\alpha,\beta) = O(\alpha,-\eta_3\beta) = O(\alpha,\eta^2_3\beta)}, \\ \langle \nabla_4+\nabla_5 \rangle \langle \nabla_4+\alpha\nabla_5+\nabla_6 \rangle^{{ \begin{array}{l}O(\alpha) = O(-\eta_3\alpha) = \\O(\eta^2_3\alpha) \end{array}}}, \\ \langle \nabla_4+\alpha\nabla_5+\beta\nabla_6+\nabla_7 \rangle^{ { \begin{array}{l} O(\alpha,\beta) = O(-\eta_5\alpha,\eta^4_5\beta) = O(\eta^2_5\alpha,-\eta^3_5\beta) = \\ O(-\eta^3_5\alpha,\eta^2_5\beta) = O(\eta^4_5\alpha,-\eta_5 \beta) \end{array}}}, \langle \nabla_5 \rangle, \langle \nabla_5+\nabla_6 \rangle, \\ \langle \nabla_5+\alpha\nabla_6+\nabla_7 \rangle^{O(\alpha) = O(-\eta_3\alpha) = O(\eta^2_3\alpha)}, \langle \nabla_6 \rangle, \langle \nabla_6+\nabla_7 \rangle, \langle \nabla_7 \rangle,\end{array}$

    which gives the following new algebras:

    $ \begin{array}{llllllllllllllllll} {\mathbf{N}}_{383}^{\alpha, \beta} & : & e_1e_1 = e_2 & e_1e_2 = e_5 & e_1e_3 = \alpha e_5 & e_1e_4 = e_5 \\ && e_2e_2 = e_3 & e_2e_3 = \beta e_5 & e_3e_3 = e_4 \\ {\mathbf{N}}_{384}^{\alpha, \beta,\gamma,\mu} & : & e_1e_1 = e_2 & e_1e_2 = e_5 & e_1e_3 = \alpha e_5 & e_2e_2 = e_3 & e_2e_3 = \beta e_5 \\ & & e_2e_4 = \gamma e_5 & e_3e_3 = e_4 & e_3e_4 = \mu e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{385}^{\alpha, \beta,\gamma} & : & e_1e_1 = e_2 & e_1e_2 = e_5 & e_1e_4 = \alpha e_5 & e_2e_2 = e_3 \\ && e_2e_3 = \beta e_5 & e_2e_4 = \gamma e_5 & e_3e_3 = e_4 & e_3e_4 = e_5 \\ {\mathbf{N}}_{386}^{\alpha} & : & e_1e_1 = e_2 & e_1e_3 = e_5 & e_1e_4 = e_5 \\ && e_2e_2 = e_3 & e_2e_3 = \alpha e_5 & e_3e_3 = e_4 \\ {\mathbf{N}}_{387}^{\alpha, \beta} & : & e_1e_1 = e_2 & e_1e_3 = e_5 & e_1e_4 = \alpha e_5 & e_2e_2 = e_3 \\ && e_2e_3 = \beta e_5 & e_2e_4 = e_5 & e_3e_3 = e_4 \\ {\mathbf{N}}_{388}^{\alpha, \beta,\gamma} & : & e_1e_1 = e_2 & e_1e_3 = e_5 & e_2e_2 = e_3 & e_2e_3 = \alpha e_5 \\ && e_2e_4 = \beta e_5 & e_3e_3 = e_4 & e_3e_4 = \gamma e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{389} & : & e_1e_1 = e_2 & e_1e_4 = e_5 & e_2e_2 = e_3 & e_3e_3 = e_4 \\ {\mathbf{N}}_{390} & : & e_1e_1 = e_2 & e_1e_4 = e_5 & e_2e_2 = e_3 & e_2e_3 = e_5 & e_3e_3 = e_4 \\ {\mathbf{N}}_{391}^{\alpha} & : & e_1e_1 = e_2 & e_1e_4 = e_5 & e_2e_2 = e_3 \\ & & e_2e_3 = \alpha e_5 & e_2e_4 = e_5 & e_3e_3 = e_4 \\ {\mathbf{N}}_{392}^{\alpha, \beta} & : & e_1e_1 = e_2 & e_1e_4 = e_5 & e_2e_2 = e_3 & e_2e_3 = \alpha e_5 \\ && e_2e_4 = \beta e_5 & e_3e_3 = e_4 & e_3e_4 = e_5 \\ {\mathbf{N}}_{393} & : & e_1e_1 = e_2 & e_2e_2 = e_3 & e_2e_3 = e_5 & e_2e_4 = e_5 & e_3e_3 = e_4 \\ {\mathbf{N}}_{394}^{\alpha} & : & e_1e_1 = e_2 & e_2e_2 = e_3 & e_2e_3 = e_5 \\ && e_2e_4 = \alpha e_5 & e_3e_3 = e_4 & e_3e_4 = e_5 \\ {\mathbf{N}}_{395}^{\alpha, \beta} & : & e_1e_1 = e_2 & e_2e_2 = e_3 & e_2e_3 = e_5 & e_2e_4 = \alpha e_5 \\ & & e_3e_3 = e_4 & e_3e_4 = \beta e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{396} & : & e_1e_1 = e_2 & e_2e_2 = e_3 & e_2e_4 = e_5 & e_3e_3 = e_4 \\ {\mathbf{N}}_{397} & : & e_1e_1 = e_2 & e_2e_2 = e_3 & e_2e_4 = e_5 & e_3e_3 = e_4 & e_3e_4 = e_5 \\ {\mathbf{N}}_{398}^{\alpha} & : & e_1e_1 = e_2 & e_2e_2 = e_3 & e_2e_4 = e_5 \\ && e_3e_3 = e_4 & e_3e_4 = \alpha e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{399} & : & e_1e_1 = e_2 & e_2e_2 = e_3 & e_3e_3 = e_4 & e_3e_4 = e_5 \\ {\mathbf{N}}_{400} & : & e_1e_1 = e_2 & e_2e_2 = e_3 & e_3e_3 = e_4 & e_3e_4 = e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{401} & : & e_1e_1 = e_2 & e_2e_2 = e_3 & e_3e_3 = e_4 & e_4e_4 = e_5 \\ \end{array} $

    Here we will collect all information about $ {\mathbf N}_{10}^{4}: $

    $ \begin{array}{|l|l|l|l|} \hline \rm{ } & \rm{ } & \rm{Cohomology} & \rm{Automorphisms} \\ \hline {\mathbf{N}}^{4}_{10} & \begin{array}{l} e_1e_1 = e_2 \\ e_1e_2 = e_4 \\ e_2e_2 = e_3 \\ e_3e_3 = e_4 \end{array} & \begin{array}{l} \mathrm{H}^2_{\mathfrak{C}}(\mathbf{N}^{4}_{10}) = \Big \langle [\Delta_{ij}] \Big\rangle\\ (i,j) \notin \{ (1,1),(1,2),(2,2)\} \end{array} & \begin{array}{l} \phi_k = \begin{pmatrix} \eta^k&0&0&0\\ 0& \eta^{2k}&0&0\\ 0&0& \eta^{4k}&0\\ t&0&0& \eta^{8k} \end{pmatrix} \\ {\eta = -\eta_5, \ k = 0,1,2,3,4} \end{array}\\ \hline \end{array} $

    Let us use the following notations:

    $ \begin{array}{llll} \nabla_1 = [\Delta_{13}], & \nabla_2 = [\Delta_{14}], & \nabla_3 = [\Delta_{23}], & \nabla_4 = [\Delta_{24}], \\ \nabla_5 = [\Delta_{33}], & \nabla_6 = [\Delta_{34}], & \nabla_7 = [\Delta_{44}]. \end{array} $

    Take $ \theta = \sum\limits_{i = 1}^{7}\alpha_i\nabla_i\in\mathrm{H}^2_{\mathfrak{C}}(\mathbf{N}^{4}_{10}) . $ Since

    $ \phi_k^T\begin{pmatrix} 0&0&\alpha_1&\alpha_2\\ 0&0&\alpha_3&\alpha_4\\ \alpha_1&\alpha_3&\alpha_5&\alpha_6\\ \alpha_2&\alpha_4&\alpha_6&\alpha_7 \end{pmatrix}\phi_k = \begin{pmatrix} \alpha^*&\alpha^{**}&\alpha^{*}_1&\alpha^*_2\\ \alpha^{**}&0&\alpha^*_3&\alpha^*_4\\ \alpha^{*}_1&\alpha^*_3&\alpha^*_5+\alpha^{**}&\alpha^*_6\\ \alpha^*_2&\alpha^*_4&\alpha^*_6&\alpha^*_7 \end{pmatrix}, $

    we have

    $ \begin{array}{llll} \alpha_1^* = \eta^{4 k} (\eta^k \alpha_1+t \alpha_6), & \alpha_2^* = \eta^{8 k} (\eta^k \alpha_2+t \alpha_7), & \alpha_3^* = \eta^{6 k} \alpha_3, & \alpha_4^* = \eta^{10 k} \alpha_4, \\ \alpha_5^* = -t \eta^{2 k} \alpha_4+\eta^{8 k} \alpha_5, & \alpha_6^* = \eta^{12 k} \alpha_6, & \alpha_7^* = \eta^{16 k} \alpha_7. \end{array} $

    Since $ (\alpha_2,\alpha_4,\alpha_6,\alpha_7)\neq(0,0,0,0), $ we have the following cases:

    $ 1.\ $if $ \alpha_7 = 0, \alpha_6 = 0, \alpha_4 = 0, $ then $ \alpha_2\neq0, $ and we have the family of representatives

    $ \langle \alpha\nabla_1+\nabla_2+\beta\nabla_3+\gamma\nabla_5\rangle;$

    $ 2.\ $if $ \alpha_7 = 0, \alpha_6 = 0, \alpha_4\neq0, $ then we have the family of representatives

    $ \langle \alpha\nabla_1+\beta\nabla_2+\gamma\nabla_3+\nabla_4 \rangle;$

    $ 3.\ $if $ \alpha_7 = 0, \alpha_6\neq0, $ then we have the family of representatives

    $ \langle \alpha\nabla_2+\beta\nabla_3+\gamma\nabla_4+\mu\nabla_5+\nabla_6 \rangle;$

    $ 4.\ $if $ \alpha_7\neq0, $ then we have the family of representatives

    $ \langle \alpha\nabla_1+\beta\nabla_3+\gamma\nabla_4+\mu\nabla_5+\nu\nabla_6+\nabla_7 \rangle.$

    Summarizing, we have the following distinct orbits:

    $\begin{array}{c} \langle \alpha\nabla_1+\beta\nabla_2+\gamma\nabla_3+\nabla_4 \rangle^{ {\begin{array}{l} O(\alpha, \beta, \gamma) = O(\alpha, \eta^4_5\beta, -\eta_5 \gamma) = O(\alpha, -\eta^3_5\beta, \eta^2_5 \gamma) = \\ O(\alpha, \eta^2_5\beta, -\eta^3_5 \gamma) = O(\alpha, -\eta_5\beta, \eta^4_5 \gamma)\end{array}}}, \\ \langle \alpha\nabla_1+\nabla_2+\beta\nabla_3+\gamma\nabla_5 \rangle^{ {\begin{array}{l} O(\alpha, \beta, \gamma) = O(-\eta_5\alpha, \eta^2_5\beta, \eta^4_5 \gamma) = O(\eta^2_5\alpha, \eta^4_5\beta, -\eta^3_5 \gamma) = \\ O(-\eta^3_5\alpha, -\eta_5\beta, \eta^2_5 \gamma) = O(\eta^4_5\alpha, -\eta^3_5\beta, -\eta_5 \gamma) \end{array}}}, \\ \langle \alpha\nabla_1+\beta\nabla_3+\gamma\nabla_4+\mu\nabla_5+\nu\nabla_6+ \\ \nabla_7 \rangle^{{\begin{array}{l} O(\alpha, \beta, \gamma, \mu, \nu) = O(\eta^4_5\alpha, \beta, \eta^4_5 \gamma, \eta^2_5\mu, -\eta_5\nu) = \\ O(-\eta^3_5\alpha, \beta, -\eta^3_5 \gamma, \eta^4_5\mu, \eta^2_5\nu) = O(\eta^2_5\alpha, \beta, \eta^2_5 \gamma,- \eta_5\mu, -\eta^3_5\nu) = \\ O(-\eta_5\alpha, \beta, -\eta_5 \gamma, -\eta^3_5\mu, \eta^4_5\nu) \end{array}}}, \\ \langle \alpha\nabla_2+\beta\nabla_3+\gamma\nabla_4+\mu\nabla_5+ \\ \nabla_6 \rangle^{{\begin{array}{l} O(\alpha, \beta, \gamma, \mu) = O(\eta_5^2\alpha, \eta_5^4\beta, -\eta_5^3 \gamma, -\eta_5\mu) = O(\eta_5^4\alpha, -\eta_5^3\beta, -\eta_5\gamma, \eta^2_5\mu) = \\ O(-\eta_5\alpha, \eta_5^2\beta, \eta_5^4 \gamma, -\eta^3_5\mu) = O(-\eta_5^3\alpha, -\eta_5\beta, \eta_5^2 \gamma, \eta^4_5\mu) \end{array}}},\end{array}$

    which gives the following new algebras:

    $ \begin{array}{llllllllllllllllll} {\mathbf{N}}_{402}^{\alpha, \beta,\gamma} & : & e_1e_1 = e_2 & e_1e_2 = e_4 & e_1e_3 = \alpha e_5 & e_1e_4 = \beta e_5 \\ & & e_2e_2 = e_3 & e_2e_3 = \gamma e_5 & e_2e_4 = e_5 & e_3e_3 = e_4 \\ {\mathbf{N}}_{403}^{\alpha, \beta,\gamma} & : & e_1e_1 = e_2 & e_1e_2 = e_4 & e_1e_3 = \alpha e_5 & e_1e_4 = e_5 \\ & & e_2e_2 = e_3 & e_2e_3 = \beta e_5 & e_3e_3 = e_4+\gamma e_5 \\ {\mathbf{N}}_{404}^{\alpha, \beta,\gamma,\mu,\nu} & : & e_1e_1 = e_2 & e_1e_2 = e_4 & e_1e_3 = \alpha e_5 \\ && e_2e_2 = e_3 & e_2e_3 = \beta e_5 & e_2e_4 = \gamma e_5 \\ & & e_3e_3 = e_4+\mu e_5 & e_3e_4 = \nu e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{405}^{\alpha, \beta,\gamma,\mu} & : & e_1e_1 = e_2 & e_1e_2 = e_4 & e_1e_4 = \alpha e_5 & e_2e_2 = e_3 \\ & & e_2e_3 = \beta e_5 & e_2e_4 = \gamma e_5 & e_3e_3 = e_4+\mu e_5 & e_3e_4 = e_5 \\ \end{array} $

    Here we will collect all information about $ {\mathbf N}_{11}^{4}(\lambda): $

    $ \begin{array}{|l|l|l|l|} \hline \rm{ } & \rm{ } & \rm{Cohomology} & \rm{Automorphisms} \\ \hline {\mathbf{N}}^{4}_{11}(\lambda) & \begin{array}{l} e_1e_1 = e_2 \\ e_1e_2 = \lambda e_4 \\ e_2e_2 = e_3\\ e_2e_3 = e_4 \\ e_3e_3 = e_4 \end{array} & \begin{array}{l} \mathrm{H}^2_{\mathfrak{C}}(\mathbf{N}^{4}_{11}(\lambda)) = \Big \langle [\Delta_{ij}] \Big\rangle\\ (i,j) \notin \{ (1,1),(2,2),(3,3) \} \end{array} & \phi = \begin{pmatrix} 1&0&0&0\\ 0&1&0&0\\ 0&0&1&0\\ t&0&0&1 \end{pmatrix}\\ \hline \end{array} $

    Let us use the following notations:

    $ \begin{array}{lll l} \nabla_1 = [\Delta_{12}], & \nabla_2 = [\Delta_{13}], & \nabla_3 = [\Delta_{14}], & \nabla_4 = [\Delta_{23}], \\ \nabla_5 = [\Delta_{24}], & \nabla_6 = [\Delta_{34}], & \nabla_7 = [\Delta_{44}]. \end{array} $

    Take $ \theta = \sum\limits_{i = 1}^{7}\alpha_i\nabla_i\in\mathrm{H}^2_{\mathfrak{C}}(\mathbf{N}^{4}_{11}(\lambda)) . $ Since

    $ \phi^T\begin{pmatrix} 0&\alpha_1&\alpha_2&\alpha_3\\ \alpha_1&0&\alpha_4&\alpha_5\\ \alpha_2&\alpha_4&0&\alpha_6\\ \alpha_3&\alpha_5&\alpha_6&\alpha_7 \end{pmatrix}\phi = \begin{pmatrix} \alpha^*&\alpha_1^{*}&\alpha^{*}_2&\alpha^*_3\\ \alpha_1^{*}&0&\alpha^*_4&\alpha^*_5\\ \alpha^{*}_2&\alpha^*_4&0&\alpha^*_6\\ \alpha^*_3&\alpha^*_5&\alpha^*_6&\alpha^*_7 \end{pmatrix} $

    we have

    $ \begin{array}{llll} \alpha_1^* = \alpha_1+\alpha_5t, & \alpha_2^* = \alpha_2+\alpha_6t, & \alpha_3^* = \alpha_3+\alpha_7t, & \alpha_4^* = \alpha_4, \\ \alpha_5^* = \alpha_5, & \alpha_6^* = \alpha_6, & \alpha_7^* = \alpha_7. \end{array} $

    Since $ (\alpha_3,\alpha_5,\alpha_6,\alpha_7)\neq(0,0,0,0), $ we have the following cases:

    $ 1.\ $if $ \alpha_7 = 0, \alpha_6 = 0, \alpha_5 = 0 , $ then $ \alpha_3\neq0 $ and we have the family of representatives

    $ \langle \alpha\nabla_1+\beta\nabla_2+\nabla_3+\gamma\nabla_4 \rangle; $

    $ 2.\ $if $ \alpha_7 = 0, \alpha_6 = 0, \alpha_5\neq0 $ then by choosing $ t = -{\alpha_1}{\alpha_5^{-1}}, $ we have the family of representatives

    $ \langle \alpha\nabla_2+\beta\nabla_3+\gamma\nabla_4+\nabla_5 \rangle;$

    $ 3.\ $if $ \alpha_7 = 0, \alpha_6\neq0 $ then by choosing $ t = -{\alpha_2}{\alpha_6^{-1}}, $ we have the family of representatives

    $ \langle \alpha\nabla_1+\beta\nabla_3+\gamma\nabla_4+\mu\nabla_5+\nabla_6 \rangle;$

    $ 4.\ $if $ \alpha_7\neq0 $ then by choosing $ t = -{\alpha_3}{\alpha_7^{-1}}, $ we have the family of representatives

    $ \langle \alpha\nabla_1+\beta\nabla_2+\gamma\nabla_4+\mu\nabla_5+\nu\nabla_6+\nabla_7 \rangle. $

    Summarizing, we have the following distinct orbits:

    $\begin{array}{c} \langle \alpha\nabla_1+\beta\nabla_2+\nabla_3+\gamma\nabla_4 \rangle,\langle \alpha\nabla_1+\beta\nabla_2+\gamma\nabla_4+\mu\nabla_5+\nu\nabla_6+\nabla_7 \rangle, \\\langle \alpha\nabla_1+\beta\nabla_3+\gamma\nabla_4+\mu\nabla_5+\nabla_6 \rangle,\langle \alpha\nabla_2+\beta\nabla_3+\gamma\nabla_4+\nabla_5 \rangle,\end{array}$

    which gives the following new algebras:

    $ \begin{array}{llllllllllllllllll} {\mathbf{N}}_{406}^{\lambda,\alpha, \beta,\gamma} & : & e_1e_1 = e_2 & e_1e_2 = \lambda e_4+\alpha e_5 & e_1e_3 = \beta e_5 & e_1e_4 = e_5 \\ && e_2e_2 = e_3 & e_2e_3 = e_4+\gamma e_5 & e_3e_3 = e_4 \\ {\mathbf{N}}_{407}^{\lambda,\alpha, \beta,\gamma,\mu,\nu} & : & e_1e_1 = e_2 & e_1e_2 = \lambda e_4+\alpha e_5 & e_1e_3 = \beta e_5 \\ && e_2e_2 = e_3 & e_2e_3 = e_4+\gamma e_5 & e_2e_4 = \mu e_5 \\ && e_3e_3 = e_4 & e_3e_4 = \nu e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{408}^{\lambda,\alpha, \beta,\gamma} & : & e_1e_1 = e_2 & e_1e_2 = \lambda e_4+\alpha e_5 & e_1e_4 = \beta e_5 & e_2e_2 = e_3 \\ & & e_2e_3 = e_4+\gamma e_5 & e_2e_4 = \mu e_5 & e_3e_3 = e_4 & e_3e_4 = e_5 \\ {\mathbf{N}}_{409}^{\lambda,\alpha, \beta,\gamma} & : & e_1e_1 = e_2 & e_1e_2 = \lambda e_4 & e_1e_3 = \alpha e_5 & e_1e_4 = \beta e_5 \\ && e_2e_2 = e_3 & e_2e_3 = e_4+\gamma e_5 & e_2e_4 = e_5 & e_3e_3 = e_4 \end{array} $

    Remark 2. Note that the algebras $ \mathbf{N}^{4}_{11}(\lambda) $ and $ \mathbf{N}^{4}_{11}(-\lambda) $ are isomorphic. Hence, there are some additional isomorphism relations for algebras from the present subsection $ \mathbf{N}^{\lambda, \Xi} \cong \mathbf{N}^{-\lambda, \Xi} . $

    Theorem 5.1. Let $ \mathbf N $ be a complex $ 5 $-dimensional nilpotent commutative algebra. Then we have one of the following situations.

    ${\mathit{1.}} \ $If $ \mathbf N $ is associative, then $ \mathbf N $ is isomorphic to one algebra listed in [15].

    ${\mathit{2.}} \ $If $ \mathbf N $ is a non-associative Jordan algebra, then $ \mathbf N $ is isomorphic to one algebra listed in [9].

    $ {\mathit{3.}} \ $If $ \mathbf N $ is a non-Jordan $ \mathfrak{CD} $-algebra, then $ \mathbf N $ is isomorphic to one algebra listed in [11].

    ${\mathit{4.}}\ $If $ \mathbf N $ is a non-$ \mathfrak{CD} $-algebra, then $ \mathbf N $ is isomorphic to one algebra listed in the following list.

    $ \begin{array}{llllllll} {\mathbf N}_{01} & : & e_1 e_1 = e_2 & e_1 e_2 = e_3 & e_2e_3 = e_4 \\ {\mathbf N}_{02} & : & e_1 e_1 = e_2 & e_1 e_2 = e_3 & e_1e_3 = e_4 & e_2 e_3 = e_4 && \\ {\mathbf N}_{03} & : & e_1 e_1 = e_2 & e_1 e_2 = e_3 & e_3 e_3 = e_4 &&\\ {\mathbf N}_{04} & : & e_1 e_1 = e_2 & e_1 e_2 = e_3 & e_2e_2 = e_4 & e_3 e_3 = e_4 &&\\ {\mathbf N}_{05} & : & e_1 e_1 = e_2 & e_1 e_3 = e_4 & e_2 e_2 = e_3 && \\ {\mathbf N}_{06} & : & e_1 e_1 = e_2 & e_1e_2 = e_4 & e_1 e_3 = e_4 & e_2 e_2 = e_3 && \\ {\mathbf N}_{07} & : & e_1 e_1 = e_2 & e_2 e_2 = e_3 & e_2 e_3 = e_4 && \\ {\mathbf N}_{08} & : & e_1 e_1 = e_2 & e_1e_3 = e_4 & e_2 e_2 = e_3 & e_2 e_3 = e_4 && \\ {\mathbf N}_{09} & : & e_1 e_1 = e_2 & e_2 e_2 = e_3 & e_3 e_3 = e_4 && \\ {\mathbf N}_{10} & : & e_1 e_1 = e_2 & e_2e_2 = e_3 & e_1e_2 = e_4 & e_3 e_3 = e_4 &&\\ {\mathbf N}_{11}^{\lambda} & : & e_1 e_1 = e_2 & e_1e_2 = \lambda e_4 & e_2 e_2 = e_3 \\ && e_2e_3 = e_4 & e_3 e_3 = e_4 &\\ \end{array} $
    $ \begin{array}{llllllll} {\mathbf{N}}_{12} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_1e_3 = e_4 \\ && e_2e_2 = e_5 & e_3e_3 = e_5 \\ {\mathbf{N}}_{13}^{\alpha} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_1e_3 = \alpha e_4 \\ && e_2e_2 = e_5 & e_2e_3 = e_4 & e_3e_3 = e_5 \\ {\mathbf{N}}_{14} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_1e_3 = e_4 \\ && e_2e_3 = e_4 & e_3e_3 = e_5 \\ {\mathbf{N}}_{15} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_2e_3 = e_4 & e_3e_3 = e_5 \\ {\mathbf{N}}_{16} & : & e_1e_1 = e_2 & e_1e_2 = e_4 & e_1e_3 = e_5 \\ && e_2e_2 = e_3 & e_2e_3 = e_5 \\ {\mathbf{N}}_{17} & : & e_1e_1 = e_2 & e_1e_2 = e_4 & e_2e_2 = e_3 & e_2e_3 = e_5 \\ {\mathbf{N}}_{18} & : & e_1e_1 = e_2 & e_1e_2 = e_4 & e_2e_2 = e_3 \\ & & e_2e_3 = e_5 & e_3e_3 = e_5 \\ {\mathbf{N}}_{19} & : & e_1e_1 = e_2 & e_1e_2 = e_4 & e_2e_2 = e_3 & e_3e_3 = e_5 \\ {\mathbf{N}}_{20}^{\alpha} & : & e_1e_1 = e_2 & { e_1e_2 = e_4+\alpha e_5 }& e_1e_3 = e_4 \\ && e_2e_2 = e_3 & e_2e_3 = e_5 & e_3e_3 = e_5 \\ {\mathbf{N}}_{21} & : & e_1e_1 = e_2 & e_1e_2 = e_4 & e_1e_3 = e_4 \\ && e_2e_2 = e_3 & e_2e_3 = e_5 \\ {\mathbf{N}}_{22}^{\alpha\neq 1} & : & e_1e_1 = e_2 & e_1e_2 = e_4 & e_1e_3 = e_4 \\ && e_2e_2 = e_3 & e_2e_3 = \alpha e_5 & e_3e_3 = e_5 \\ {\mathbf{N}}_{23}^{\alpha, \beta} & : & e_1e_1 = e_2 & { e_1e_2 = \beta e_4 +\alpha e_5 }& e_1e_3 = e_4 \\ && e_2e_2 = e_3 & e_2e_3 = e_4 & e_3e_3 = e_5 \\ {\mathbf{N}}_{24}^{\alpha} & : & e_1e_1 = e_2 & { e_1e_2 = \alpha e_4+e_5 }& e_2e_2 = e_3 \\ && e_2e_3 = e_4 & e_3e_3 = e_5 \\ {\mathbf{N}}_{25} & : & e_1e_1 = e_2 & e_1e_3 = e_4 & e_2e_2 = e_3 \\ && e_2e_3 = e_4 & e_3e_3 = e_5 \\ {\mathbf{N}}_{26} & : & e_1e_1 = e_2 & e_1e_2 = e_4 & e_1e_3 = e_5 \\ && e_2e_2 = e_3 & e_3e_3 = e_4 \\ {\mathbf{N}}_{27} & : & e_1e_1 = e_2 & e_1e_3 = e_4 & e_2e_2 = e_3 & e_2e_3 = e_5 \\ {\mathbf{N}}_{28} & : & e_1e_1 = e_2 & e_1e_3 = e_4 & e_2e_2 = e_3 \\ & & e_2e_3 = e_5 & e_3e_3 = e_5 \\ {\mathbf{N}}_{29} & : & e_1e_1 = e_2 & e_1e_3 = e_4 & e_2e_2 = e_3 & e_3e_3 = e_5 \\ {\mathbf{N}}_{30} & : & e_1e_1 = e_2 & e_2e_2 = e_3 & e_2e_3 = e_4 & e_3e_3 = e_5 \\ {\mathbf{N}}_{31} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_1e_4 = e_5 \\ && e_2e_2 = e_5 & e_2e_3 = e_5 \\ {\mathbf{N}}_{32}^{\alpha} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_1e_4 = e_5 \\ && e_2e_2 = \alpha e_5 & e_2e_4 = e_5 & e_3e_3 = e_5 \\ \end{array} $
    $ \begin{array}{llllllll} {\mathbf{N}}_{33} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_1e_4 = e_5 \\ && e_2e_2 = e_5 & e_3e_3 = e_5 \\ {\mathbf{N}}_{34} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_1e_4 = e_5 & e_2e_3 = e_5 \\ {\mathbf{N}}_{35} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_1e_4 = e_5 & e_3e_3 = e_5 \\ {\mathbf{N}}_{36} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_2e_2 = e_5 \\ && e_2e_3 = e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{37} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_2e_2 = e_5 \\ && e_2e_4 = e_5 & e_3e_3 = e_5 \\ {\mathbf{N}}_{38}^{\alpha} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_2e_2 = \alpha e_5 \\ && e_2e_4 = e_5 & e_3e_3 = e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{39} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_2e_2 = e_5 \\ && e_3e_3 = e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{40} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_2e_2 = e_5 & e_3e_4 = e_5 \\ {\mathbf{N}}_{41} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_2e_3 = e_5 \\ && e_2e_4 = e_5 & e_3e_4 = e_5 \\ {\mathbf{N}}_{42} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_2e_3 = e_5 & e_3e_4 = e_5 \\ {\mathbf{N}}_{43} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_2e_3 = e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{44} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_2e_4 = e_5 & e_3e_3 = e_5 \\ {\mathbf{N}}_{45} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_3e_3 = e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{46} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_3e_4 = e_5 \\ {\mathbf{N}}_{47} & : & e_1e_1 = e_2 & e_1e_2 = e_5 & e_1e_3 = e_5 \\ && e_2e_2 = e_3 & e_4e_4 = e_5 \\ {\mathbf{N}}_{48} & : & e_1e_1 = e_2 & e_1e_2 = e_5 & e_1e_4 = e_5 \\ & & e_2e_2 = e_3 & e_3e_3 = e_5 \\ {\mathbf{N}}_{49}^{\alpha} & : & e_1e_1 = e_2 & e_1e_2 = \alpha e_5 & e_1e_4 = e_5 \\ && e_2e_2 = e_3 & e_2e_3 = e_5 & e_3e_3 = e_5 \\ {\mathbf{N}}_{50}^{\alpha, \beta} & : & e_1e_1 = e_2 & e_1e_2 = \alpha e_5 & e_2e_2 = e_3 & e_2e_3 = \beta e_5 \\ & & e_2e_4 = e_5 & e_3e_3 = e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{51}^{\alpha} & : & e_1e_1 = e_2 & e_1e_2 = \alpha e_5 & e_2e_2 = e_3 \\ && e_2e_3 = e_5 & e_3e_3 = e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{52} & : & e_1e_1 = e_2 & e_1e_2 = e_5 & e_2e_2 = e_3 \\ && e_2e_3 = e_5 & e_3e_4 = e_5 \\ {\mathbf{N}}_{53}^{\alpha} & : & e_1e_1 = e_2 & e_1e_2 = \alpha e_5 & e_2e_2 = e_3 \\ && e_2e_3 = e_5 & e_2e_4 = e_5 & e_3e_4 = e_5 \\ {\mathbf{N}}_{54} & : & e_1e_1 = e_2 & e_1e_2 = e_5 & e_2e_2 = e_3 \\ && e_2e_4 = e_5 & e_3e_4 = e_5 \\ {\mathbf{N}}_{55} & : & e_1e_1 = e_2 & e_1e_2 = e_5 & e_2e_2 = e_3 \\ && e_3e_3 = e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{56} & : & e_1e_1 = e_2 & e_1e_2 = e_5 & e_2e_2 = e_3 & e_3e_4 = e_5 \\ {\mathbf{N}}_{57} & : & e_1e_1 = e_2 & e_1e_3 = e_5 & e_1e_4 = e_5 \\ && e_2e_2 = e_3 & e_2e_3 = e_5 \\ {\mathbf{N}}_{58} & : & e_1e_1 = e_2 & e_1e_3 = e_5 & e_2e_2 = e_3 \\ && e_2e_3 = e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{59} & : & e_1e_1 = e_2 & e_1e_3 = e_5 & e_2e_2 = e_3 & e_2e_4 = e_5 \\ {\mathbf{N}}_{60} & : & e_1e_1 = e_2 & e_1e_3 = e_5 & e_2e_2 = e_3 \\ & & e_2e_4 = e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{61} & : & e_1e_1 = e_2 & e_1e_3 = e_5 & e_2e_2 = e_3 & e_4e_4 = e_5 \\ {\mathbf{N}}_{62} & : & e_1e_1 = e_2 & e_1e_4 = e_5 & e_2e_2 = e_3 & e_2e_3 = e_5 \\ {\mathbf{N}}_{63}^{\alpha} & : & e_1e_1 = e_2 & e_1e_4 = \alpha e_5 & e_2e_2 = e_3 \\ && e_2e_3 = e_5 & e_2e_4 = e_5 & e_3e_3 = e_5 \\ \end{array} $
    $ \begin{array}{llllllll} {\mathbf{N}}_{64} & : & e_1e_1 = e_2 & e_1e_4 = e_5 & e_2e_2 = e_3 \\ & & e_2e_4 = e_5 & e_3e_3 = e_5 \\ {\mathbf{N}}_{65} & : & e_1e_1 = e_2 & e_1e_4 = e_5 & e_2e_2 = e_3 & e_3e_3 = e_5 \\ {\mathbf{N}}_{66} & : & e_1e_1 = e_2 & e_2e_2 = e_3 & e_2e_3 = e_5 & e_3e_4 = e_5 \\ {\mathbf{N}}_{67} & : & e_1e_1 = e_2 & e_2e_2 = e_3 & e_2e_3 = e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{68} & : & e_1e_1 = e_2 & e_2e_2 = e_3 & e_2e_4 = e_5 & e_3e_3 = e_5 \\ {\mathbf{N}}_{69} & : & e_1e_1 = e_2 & e_2e_2 = e_3 & e_2e_4 = e_5 & e_3e_4 = e_5 \\ {\mathbf{N}}_{70} & : & e_1e_1 = e_2 & e_2e_2 = e_3 & e_3e_3 = e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{71} & : & e_1e_1 = e_2 & e_2e_2 = e_3 & e_3e_4 = e_5 & \\ {\mathbf{N}}_{72} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_1e_3 = e_5 \\ && e_1e_4 = \frac{3}{4}e_5 & e_2e_2 = e_4 & e_2e_3 = -\frac{3}{4}e_5 \\ {\mathbf{N}}_{73}^{\alpha, \beta} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_1e_3 = \alpha e_5 & e_1e_4 = e_5 \\ && e_2e_2 = e_4 & e_2e_3 = 3e_5 & e_2e_4 = \beta e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{74}^{\alpha} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_1e_3 = e_5 \\ && e_1e_4 = \alpha e_5 & e_2e_2 = e_4 & e_2e_4 = e_5 \\ {\mathbf{N}}_{75}^{\alpha} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_1e_3 = e_5 \\ && e_2e_2 = e_4 & e_2e_4 = \alpha e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{76}^{\alpha} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & { e_1e_4 = (1+\alpha) e_5 } \\ && e_2e_2 = e_4 & e_2e_3 = 3\alpha e_5 \\ {\mathbf{N}}_{77}^{\alpha} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_1e_4 = e_5 & e_2e_2 = e_4 \\ && e_2e_3 = 3e_5 & e_2e_4 = \alpha e_5 & e_3e_3 = e_5 \\ {\mathbf{N}}_{78}^{\alpha, \beta} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_1e_4 = \alpha e_5 & e_2e_2 = e_4 \\ && e_2e_3 = 3\alpha e_5 & e_2e_4 = \beta e_5 & e_3e_3 = e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{79} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_1e_4 = e_5 \\ && e_2e_2 = e_4 & e_2e_4 = e_5 & e_3e_3 = e_5 \\ {\mathbf{N}}_{80}^{\alpha \neq 1} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_2e_2 = e_4 \\ && e_2e_4 = \alpha e_5 & e_3e_3 = e_5 \\ {\mathbf{N}}_{81}^{\alpha} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_2e_2 = e_4 \\ && e_2e_4 = e_5 & e_3e_3 = \alpha e_5 & e_3e_4 = e_5 \\ {\mathbf{N}}_{82} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_2e_2 = e_4 & e_3e_4 = e_5 \\ {\mathbf{N}}_{83}^{\alpha, \beta ,\gamma} & : & e_1e_1 = e_2 & e_1e_2 = e_5 & e_1e_3 = e_5 & e_2e_2 = \alpha e_5 \\ & & e_2e_3 = e_4 & e_2e_4 = \beta e_5 & e_3e_3 = \gamma e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{84} & : & e_1e_1 = e_2 & e_1e_2 = e_5 & e_1e_4 = e_5 & e_2e_3 = e_4 \\ {\mathbf{N}}_{85}^{\alpha, \beta} & : & e_1e_1 = e_2 & e_1e_2 = e_5 & e_1e_4 = e_5 & e_2e_2 = \alpha e_5 \\ & & e_2e_3 = e_4 & e_2e_4 = \beta e_5 & e_3e_4 = e_5 \\ {\mathbf{N}}_{86}^{\alpha} & : & e_1e_1 = e_2 & e_1e_2 = e_5 & e_1e_4 = e_5 \\ && e_2e_2 = e_5 & e_2e_3 = e_4 & e_3e_3 = \alpha e_5 \\ {\mathbf{N}}_{87} & : & e_1e_1 = e_2 & e_1e_2 = e_5 & e_1e_4 = e_5 \\ && e_2e_3 = e_4 & e_3e_3 = e_5 \\ {\mathbf{N}}_{88}^{\alpha, \beta} & : & e_1e_1 = e_2 & e_1e_2 = e_5 & e_2e_2 = e_5 & e_2e_3 = e_4 \\ & & e_2e_4 = \alpha e_5 & e_3e_3 = \beta e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{89}^{\alpha} & : & e_1e_1 = e_2 & e_1e_2 = e_5 & e_2e_2 = e_5 \\ && e_2e_3 = e_4 & e_2e_4 = \alpha e_5 & e_3e_4 = e_5 \\ {\mathbf{N}}_{90}^{\alpha} & : & e_1e_1 = e_2 & e_1e_2 = e_5 & e_2e_3 = e_4 \\ && e_2e_4 = e_5 & e_3e_3 = \alpha e_5 & e_4e_4 = e_5 \\ \end{array} $
    $\begin{array}{llllllll} {\mathbf{N}}_{91} & : & e_1e_1 = e_2 & e_1e_2 = e_5 & e_2e_3 = e_4 \\ && e_2e_4 = e_5 & e_3e_4 = e_5 \\ {\mathbf{N}}_{92} & : & e_1e_1 = e_2 & e_1e_2 = e_5 & e_2e_3 = e_4 \\ && e_3e_3 = e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{93} & : & e_1e_1 = e_2 & e_1e_2 = e_5 & e_2e_3 = e_4 & e_3e_4 = e_5 \\ {\mathbf{N}}_{94} & : & e_1e_1 = e_2 & e_1e_2 = e_5 & e_2e_3 = e_4 & e_4e_4 = e_5 \\ {\mathbf{N}}_{95}^{\alpha} & : & e_1e_1 = e_2 & e_1e_3 = e_5 & e_2e_2 = e_5 \\ && e_2e_3 = e_4 & e_2e_4 = e_5 & e_3e_3 = \alpha e_5 \\ {\mathbf{N}}_{96}^{\alpha, \beta} & : & e_1e_1 = e_2 & e_1e_3 = e_5 & e_2e_2 = e_5 & e_2e_3 = e_4 \\ & & e_2e_4 = \alpha e_5 & e_3e_3 = \beta e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{97} & : & e_1e_1 = e_2 & e_1e_3 = e_5 & e_2e_3 = e_4 & e_2e_4 = e_5 \\ {\mathbf{N}}_{98} & : & e_1e_1 = e_2 & e_1e_3 = e_5 & e_2e_3 = e_4 \\ & & e_2e_4 = e_5 & e_3e_3 = e_5 \\ {\mathbf{N}}_{99}^{\alpha} & : & e_1e_1 = e_2 & e_1e_3 = e_5 & e_2e_3 = e_4 \\ && e_2e_4 = e_5 & e_3e_3 = \alpha e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{100} & : & e_1e_1 = e_2 & e_1e_3 = e_5 & e_2e_3 = e_4 \\ && e_3e_3 = e_5 & e_4e_4 = e_5 \\ \end{array} $
    $ \begin{array}{llllllll} {\mathbf{N}}_{101} & : & e_1e_1 = e_2 & e_1e_3 = e_5 & e_2e_3 = e_4 & e_4e_4 = e_5 \\ {\mathbf{N}}_{102} & : & e_1e_1 = e_2 & e_1e_4 = e_5 & e_2e_3 = e_4 \\ {\mathbf{N}}_{103} & : & e_1e_1 = e_2 & e_1e_4 = e_5 & e_2e_2 = e_5 & e_2e_3 = e_4 \\ {\mathbf{N}}_{104}^{\alpha} & : & e_1e_1 = e_2 & e_1e_4 = e_5 & e_2e_2 = e_5 \\ && e_2e_3 = e_4 & e_2e_4 = e_5 & e_3e_3 = \alpha e_5 \\ {\mathbf{N}}_{105}^{\alpha} & : & e_1e_1 = e_2 & e_1e_4 = e_5 & e_2e_2 = \alpha e_5 \\ && e_2e_3 = e_4 & e_2e_4 = e_5 & e_3e_4 = e_5 \\ {\mathbf{N}}_{106} & : & e_1e_1 = e_2 & e_1e_4 = e_5 & e_2e_2 = e_5 \\ && e_2e_3 = e_4 & e_3e_3 = e_5 \\ {\mathbf{N}}_{107} & : & e_1e_1 = e_2 & e_1e_4 = e_5 & e_2e_3 = e_4 & e_2e_4 = e_5 \\ {\mathbf{N}}_{108} & : & e_1e_1 = e_2 & e_1e_4 = e_5 & e_2e_3 = e_4 \\ && e_2e_4 = e_5 & e_3e_3 = e_5 \\ {\mathbf{N}}_{109} & : & e_1e_1 = e_2 & e_1e_4 = e_5 & e_2e_3 = e_4 & e_3e_3 = e_5 \\ {\mathbf{N}}_{110} & : & e_1e_1 = e_2 & e_2e_2 = e_5 & e_2e_3 = e_4 & e_2e_4 = e_5 \\ \end{array} $
    $ \begin{array}{llllllll} {\mathbf{N}}_{111} & : & e_1e_1 = e_2 & e_2e_2 = e_5 & e_2e_3 = e_4 \\ & & e_2e_4 = e_5 & e_3e_3 = e_5 \\ {\mathbf{N}}_{112}^{\alpha} & : & e_1e_1 = e_2 & e_2e_2 = e_5 & e_2e_3 = e_4 \\ && e_2e_4 = \alpha e_5 & e_3e_3 = e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{113} & : & e_1e_1 = e_2 & e_2e_2 = e_5 & e_2e_3 = e_4 \\ && e_2e_4 = e_5 & e_3e_4 = e_5 \\ {\mathbf{N}}_{114} & : & e_1e_1 = e_2 & e_2e_2 = e_5 & e_2e_3 = e_4 \\ && e_2e_4 = e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{115} & : & e_1e_1 = e_2 & e_2e_2 = e_5 & e_2e_3 = e_4 & e_3e_4 = e_5 \\ {\mathbf{N}}_{116} & : & e_1e_1 = e_2 & e_2e_2 = e_5 & e_2e_3 = e_4 & e_4e_4 = e_5 \\ {\mathbf{N}}_{117} & : & e_1e_1 = e_2 & e_2e_3 = e_4 & e_2e_4 = e_5 \\ {\mathbf{N}}_{118} & : & e_1e_1 = e_2 & e_2e_3 = e_4 & e_2e_4 = e_5 & e_3e_3 = e_5 \\ {\mathbf{N}}_{119} & : & e_1e_1 = e_2 & e_2e_3 = e_4 & e_2e_4 = e_5 \\ && e_3e_3 = e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{120} & : & e_1e_1 = e_2 & e_2e_3 = e_4 & e_2e_4 = e_5 & e_3e_4 = e_5 \\ {\mathbf{N}}_{121} & : & e_1e_1 = e_2 & e_2e_3 = e_4 & e_2e_4 = e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{122} & : & e_1e_1 = e_2 & e_2e_3 = e_4 & e_3e_3 = e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{123} & : & e_1e_1 = e_2 & e_2e_3 = e_4 & e_3e_4 = e_5 \\ {\mathbf{N}}_{124} & : & e_1e_1 = e_2 & e_2e_3 = e_4 & e_4e_4 = e_5 \\ {\mathbf{N}}^{\alpha}_{125} & : & e_1e_1 = e_2 & e_1e_2 = e_4 & e_1e_3 = e_5 \\ && e_2e_2 = \alpha e_5 & e_2e_4 = e_5 & e_3e_3 = e_4 \\ {\mathbf{N}}^{\alpha,\beta}_{126} & : & e_1e_1 = e_2 & e_1e_2 = e_4 & e_1e_3 = \alpha e_5 & e_2e_2 = e_5 \\ & & e_3e_3 = e_4 + \beta e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}^{\alpha}_{127} & : & e_1e_1 = e_2 & e_1e_2 = e_4 & e_1e_3 = \alpha e_5 \\ && e_2e_3 = e_5 & e_3e_3 = e_4 & e_4e_4 = e_5 \\ {\mathbf{N}}_{128} & : & e_1e_1 = e_2 & e_1e_2 = e_4 & e_1e_3 = e_5 \\ && e_3e_3 = e_4 & e_4e_4 = e_5 \\ {\mathbf{N}}_{129} & : & e_1e_1 = e_2 & e_1e_2 = e_4 & e_2e_2 = e_5 \\ && e_2e_4 = e_5 & e_3e_3 = e_4 \\ {\mathbf{N}}_{130} & : & e_1e_1 = e_2 & e_1e_2 = e_4 & e_2e_2 = e_5 \\ & & e_3e_3 = e_4 & e_3e_4 = e_5 \\ {\mathbf{N}}_{131} & : & e_1e_1 = e_2 & e_1e_2 = e_4 & e_2e_4 = e_5 & e_3e_3 = e_4 \\ {\mathbf{N}}_{132} & : & e_1e_1 = e_2 & e_1e_2 = e_4 & e_3e_3 = e_4+e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{133} & : & e_1e_1 = e_2 & e_1e_2 = e_4 & e_3e_3 = e_4 & e_3e_4 = e_5 \\ {\mathbf{N}}_{134} & : & e_1e_1 = e_2 & e_1e_2 = e_4 & e_3e_3 = e_4 & e_4e_4 = e_5 \\ {\mathbf{N}}_{135}^{\alpha} & : & e_1e_1 = e_2 & e_1e_2 = e_5 & e_1e_3 = e_4 & e_1e_4 = e_5 \\ & & e_2e_2 = e_4 & e_2e_3 = -2e_5 & e_3e_3 = \alpha e_5 \\ {\mathbf{N}}_{136}^{\alpha, \beta ,\gamma} & : & e_1e_1 = e_2 & e_1e_2 = e_5 & e_1e_3 = e_4 \\ && e_2e_2 = e_4+\alpha e_5 & e_2e_3 = \beta e_5 \\ && e_3e_3 = \gamma e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{137}^{\alpha, \beta} & : & e_1e_1 = e_2 & e_1e_2 = e_5 & e_1e_3 = e_4 & e_2e_2 = e_4 \\ & & e_2e_3 = \alpha e_5 & e_2e_4 = \beta e_5 & e_3e_4 = e_5 \\ {\mathbf{N}}_{138}^{\alpha} & : & e_1e_1 = e_2 & e_1e_3 = e_4 & e_1e_4 = \alpha e_5 \\ && e_2e_2 = e_4+e_5 & e_2e_4 = e_5 & e_3e_3 = 4e_5 \\ {\mathbf{N}}_{139}^{\alpha} & : & e_1e_1 = e_2 & e_1e_3 = e_4 & e_1e_4 = e_5 \\ && e_2e_2 = e_4 & e_2e_3 = \alpha e_5 \\ {\mathbf{N}}_{140}^{\alpha} & : & e_1e_1 = e_2 & e_1e_3 = e_4 & e_1e_4 = e_5 \\ && e_2e_2 = e_4 & e_2e_3 = \alpha e_5 & e_3e_3 = e_5 \\ \end{array} $
    $ \begin{array}{llllllll} {\mathbf{N}}_{141}^{\alpha} & : & e_1e_1 = e_2 & e_1e_3 = e_4 & e_1e_4 = e_5 \\ && e_2e_2 = e_4 & e_2e_4 = e_5 & e_3e_3 = \alpha e_5 \\ {\mathbf{N}}_{142}^{\alpha, \beta} & : & e_1e_1 = e_2 & e_1e_3 = e_4 & e_2e_2 = e_4+e_5 \\ && e_2e_3 = \alpha e_5 & e_3e_3 = \beta e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{143}^{\alpha} & : & e_1e_1 = e_2 & e_1e_3 = e_4 & e_2e_2 = e_4 \\ && e_2e_3 = e_5 & e_2e_4 = \alpha e_5 & e_3e_4 = e_5 \\ {\mathbf{N}}_{144}^{\alpha} & : & e_1e_1 = e_2 & e_1e_3 = e_4 & e_2e_2 = e_4 \\ && e_2e_3 = e_5 & e_3e_3 = \alpha e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{145}^{\alpha} & : & e_1e_1 = e_2 & e_1e_3 = e_4 & e_2e_2 = e_4 \\ && e_2e_4 = e_5 & e_3e_3 = \alpha e_5 \\ {\mathbf{N}}_{146} & : & e_1e_1 = e_2 & e_1e_3 = e_4 & e_2e_2 = e_4 \\ && e_2e_4 = e_5 & e_3e_4 = e_5 \\ {\mathbf{N}}_{147} & : & e_1e_1 = e_2 & e_1e_3 = e_4 & e_2e_2 = e_4 \\ && e_3e_3 = e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{148} & : & e_1e_1 = e_2 & e_1e_3 = e_4 & e_2e_2 = e_4 & e_3e_4 = e_5 \\ {\mathbf{N}}_{149} & : & e_1e_1 = e_2 & e_1e_3 = e_4 & e_2e_2 = e_4 & e_4e_4 = e_5 \\ {\mathbf{N}}_{150}^{\alpha, \beta, \gamma, \mu } & : & e_1e_1 = e_2 & e_1e_2 = e_5 & e_1e_3 = \alpha e_5 & e_2e_2 = e_4 \\ && e_2e_3 = \beta e_5 & e_2e_4 = \gamma e_5 & e_3e_3 = e_4+\mu e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{151}^{\alpha, \beta} & : & e_1e_1 = e_2 & e_1e_2 = e_5 & e_1e_4 = e_5 \\ && e_2e_2 = e_4 & e_2e_3 = \alpha e_5 & e_3e_3 = e_4+\beta e_5 \\ {\mathbf{N}}_{152}^{\alpha, \beta} & : & e_1e_1 = e_2 & e_1e_2 = e_5 & e_1e_4 = \alpha e_5 & e_2e_2 = e_4 \\ && e_2e_3 = \beta e_5 & e_3e_3 = e_4 & e_3e_4 = e_5 \\ {\mathbf{N}}_{153}^{\alpha, \beta, \gamma} & : & e_1e_1 = e_2 & e_1e_3 = e_5 & e_1e_4 = \alpha e_5 & e_2e_2 = e_4 \\ && e_2e_4 = e_5 & { e_3e_3 = e_4+\beta e_5 } & e_3e_4 = \gamma e_5 \\ {\mathbf{N}}_{154}^{\alpha, \beta, \gamma} & : & e_1e_1 = e_2 & e_1e_3 = e_5 & e_2e_2 = e_4 & e_2e_3 = \alpha e_5 \\ && e_2e_4 = \beta e_5 & { e_3e_3 = e_4+\gamma e_5 } & e_4e_4 = e_5 \\ {\mathbf{N}}_{155} & : & e_1e_1 = e_2 & e_1e_4 = e_5 & e_2e_2 = e_4 & e_3e_3 = e_4 \\ {\mathbf{N}}_{156}^{\alpha} & : & e_1e_1 = e_2 & e_1e_4 = e_5 & e_2e_2 = e_4 & e_2e_3 = e_5 \\ & & { e_3e_3 = e_4+\alpha e_5 } \\ {\mathbf{N}}_{157}^{\alpha} & : & e_1e_1 = e_2 & e_1e_4 = e_5 & e_2e_2 = e_4 \\ && e_2e_3 = \alpha e_5 & e_3e_3 = e_4 & e_3e_4 = e_5 \\ {\mathbf{N}}_{158}^{\alpha, \beta} & : & e_1e_1 = e_2 & e_1e_4 = e_5 & e_2e_2 = e_4 & e_2e_4 = e_5 \\ & & e_3e_3 = e_4+\alpha e_5 & e_3e_4 = \beta e_5 \\ {\mathbf{N}}_{159} & : & e_1e_1 = e_2 & e_1e_4 = e_5 & e_2e_2 = e_4 \\ & & e_3e_3 = e_4+e_5 \\ {\mathbf{N}}_{160}^{\alpha, \beta} & : & e_1e_1 = e_2 & e_2e_2 = e_4 & e_2e_3 = e_5 & e_2e_4 = \alpha e_5 \\ & & e_3e_3 = e_4+\beta e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{161} & : & e_1e_1 = e_2 & e_2e_2 = e_4 & e_2e_3 = e_5 \\ & & e_3e_3 = e_4 & e_3e_4 = e_5 \\ {\mathbf{N}}_{162}^{\alpha} & : & e_1e_1 = e_2 & e_2e_2 = e_4 & e_2e_4 = e_5 \\ && e_3e_3 = e_4+e_5 & e_3e_4 = \alpha e_5 \\ {\mathbf{N}}_{163}^{\alpha} & : & e_1e_1 = e_2 & e_2e_2 = e_4 & e_2e_4 = e_5 \\ & & e_3e_3 = e_4+\alpha e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{164}^{\alpha} & : & e_1e_1 = e_2 & e_2e_2 = e_4 & e_2e_4 = e_5 \\ && e_3e_3 = e_4 & e_3e_4 = \alpha e_5 \\ {\mathbf{N}}_{165} & : & e_1e_1 = e_2 & e_2e_2 = e_4 & e_3e_3 = e_4+e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{166} & : & e_1e_1 = e_2 & e_2e_2 = e_4 & e_3e_3 = e_4 & e_3e_4 = e_5 \\ {\mathbf{N}}_{167} & : & e_1e_1 = e_2 & e_2e_2 = e_4 & e_3e_3 = e_4 & e_4e_4 = e_5 \\ {\mathbf{N}}_{168}^{\lambda \neq 1; 2} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_1e_3 = e_4 \\ && e_1e_4 = (\lambda-4)e_5 & { e_2e_2 = \lambda e_4 + 4(1-\lambda)(\lambda-2)e_5 } \\ && { e_2e_3 = - \lambda(\lambda+2)e_5 }\\ {\mathbf{N}}_{169}^{\alpha\neq0} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_1e_3 = e_4 & e_1e_4 = e_5 \\ & & e_2e_2 = \alpha e_5 & e_2e_3 = -2 e_5 & e_3e_3 = e_5 \\ {\mathbf{N}}_{170}^{\lambda, \alpha} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_1e_3 = e_4 \\ & & e_1e_4 = \alpha e_5 & e_2e_2 = \lambda e_4 & { e_2e_3 = (1+\alpha(3\lambda-2)) e_5 } \\ \end{array} $
    $\begin{array}{llllllll} {\mathbf{N}}_{171}^{\lambda} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_1e_3 = e_4 \\ && e_1e_4 = e_5 & e_2e_2 = \lambda e_4 \\ && e_2e_3 = (3\lambda-2) e_5 & e_3e_3 = e_5 \\ {\mathbf{N}}_{172}^{\lambda \neq 0,\alpha} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_1e_3 = e_4 \\ && e_2e_2 = \lambda e_4 + e_5 & e_2e_3 = \alpha e_5 & e_2e_4 = \frac{\lambda} {4} e_5 & e_3e_3 = e_5 \\ {\mathbf{N}}_{173}^{\alpha, \beta} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_1e_3 = e_4 & e_2e_2 = e_5 \\ & & e_2e_3 = \alpha e_5 & e_3e_3 = \beta e_5 & e_3e_4 = e_5 \\ {\mathbf{N}}_{174}^{\lambda\neq-2, \alpha} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_1e_3 = e_4 \\ && e_2e_2 = \lambda e_4 + \alpha e_5 & e_2e_3 = e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{175}^{\alpha, \beta} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_1e_3 = e_4 \\ & & e_2e_2 = -2 e_4+\alpha e_5 & e_3e_3 = \beta e_5 & e_3e_4 = e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{176}^{\lambda,\alpha} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_1e_3 = e_4 \\ && e_2e_2 = \lambda e_4+\alpha e_5 & e_3e_3 = e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{177}^{\lambda} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_1e_3 = e_4 \\ && e_2e_2 = \lambda e_4+ e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{178}^{\lambda} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_1e_3 = e_4 \\ & & e_2e_2 = \lambda e_4 & e_2e_3 = e_5 & e_2e_4 = e_5 \\ {\mathbf{N}}_{179}^{\lambda,\alpha\neq0} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_1e_3 = e_4 & e_2e_2 = \lambda e_4 \\ && e_2e_3 = e_5 & e_2e_4 = \alpha e_5 & e_3e_3 = e_5 \\ {\mathbf{N}}_{180}^{\lambda,\alpha} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_1e_3 = e_4 & e_2e_2 = \lambda e_4 \\ && e_2e_3 = e_5 & e_3e_3 = \alpha e_5 & e_3e_4 = e_5 \\ {\mathbf{N}}_{181}^{\alpha \neq 0 } & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_1e_3 = e_4 & e_2e_2 = -2 e_4 \\ && e_2e_3 = \alpha e_5 & e_3e_3 = e_5 & e_3e_4 = e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{182} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_1e_3 = e_4 \\ && e_2e_2 = -2 e_4 & e_2e_3 = e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{183}^{\lambda\neq 2} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_1e_3 = e_4 \\ && e_2e_2 = \lambda e_4 & e_2e_4 = e_5 \\ {\mathbf{N}}_{184}^{\lambda, \alpha} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_1e_3 = e_4 \\ && e_2e_2 = \lambda e_4 & e_2e_4 = \alpha e_5 & e_3e_3 = e_5 \\ {\mathbf{N}}_{185}^{\lambda} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_1e_3 = e_4 \\ && e_2e_2 = \lambda e_4 & e_3e_3 = e_5 & e_3e_4 = e_5 \\ {\mathbf{N}}_{186}^{\lambda} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_1e_3 = e_4 \\ && e_2e_2 = \lambda e_4 & e_3e_4 = e_5 \\ {\mathbf{N}}_{187}^{\lambda} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_1e_3 = e_4 \\ && e_2e_2 = \lambda e_4 & e_4e_4 = e_5 \\ {\mathbf{N}}_{188}^{\alpha, \beta} & : & e_1e_1 = e_5 & e_1e_2 = e_3 & e_1e_3 = e_4 & e_2e_2 = e_5 \\ && e_2e_3 = \alpha e_5 & e_3e_3 = \beta e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{189}^{\alpha} & : & e_1e_1 = e_5 & e_1e_2 = e_3 & e_1e_3 = e_4 \\ & & e_2e_2 = e_5 & e_3e_3 = \alpha e_5 & e_3e_4 = e_5 \\ {\mathbf{N}}_{190} & : & e_1e_1 = e_5 & e_1e_2 = e_3 & e_1e_3 = e_4 \\ && e_2e_3 = e_5 & e_2e_4 = e_5 & e_3e_3 = -e_5 \\\end{array} $
    $ \begin{array}{llllllll} {\mathbf{N}}_{191}^{\alpha} & : & e_1e_1 = e_5 & e_1e_2 = e_3 & e_1e_3 = e_4 \\ && e_2e_3 = e_5 & e_3e_3 = \alpha e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{191}^{\alpha} & : & e_1e_1 = e_5 & e_1e_2 = e_3 & e_1e_3 = e_4 \\ && e_2e_4 = e_5 & e_3e_3 = \alpha e_5 \\ {\mathbf{N}}_{192} & : & e_1e_1 = e_5 & e_1e_2 = e_3 & e_1e_3 = e_4 \\ && e_3e_3 = e_5 & e_3e_4 = e_5 \\ {\mathbf{N}}_{193} & : & e_1e_1 = e_5 & e_1e_2 = e_3 & e_1e_3 = e_4 \\ && e_3e_3 = e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{194} & : & e_1e_1 = e_5 & e_1e_2 = e_3 & e_1e_3 = e_4 & e_3e_4 = e_5 \\ {\mathbf{N}}_{195} & : & e_1e_1 = e_5 & e_1e_2 = e_3 & e_1e_3 = e_4 & e_4e_4 = e_5 \\ {\mathbf{N}}_{196} & : & e_1e_2 = e_3 & e_1e_3 = e_4 & e_1e_4 = e_5 \\ {\mathbf{N}}_{197} & : & e_1e_2 = e_3 & e_1e_3 = e_4 & e_1e_4 = e_5 & e_2e_2 = e_5 \\ {\mathbf{N}}_{198} & : & e_1e_2 = e_3 & e_1e_3 = e_4 & e_1e_4 = e_5 \\ && e_2e_2 = e_5 & e_3e_3 = e_5 \\ {\mathbf{N}}_{199}^{\alpha} & : & e_1e_2 = e_3 & e_1e_3 = e_4 & e_1e_4 = e_5 \\ && e_2e_2 = e_5 & e_3e_3 = \alpha e_5 & e_3e_4 = e_5 \\ {\mathbf{N}}_{200} & : & e_1e_2 = e_3 & e_1e_3 = e_4 & e_1e_4 = e_5 & e_2e_3 = e_5 \\ {\mathbf{N}}_{201} & : & e_1e_2 = e_3 & e_1e_3 = e_4 & e_1e_4 = e_5 \\ && e_2e_3 = e_5 & e_2e_4 = e_5 & e_3e_3 = -e_5 \\ {\mathbf{N}}_{202}^{\alpha} & : & e_1e_2 = e_3 & e_1e_3 = e_4 & e_1e_4 = e_5 \\ & & e_2e_4 = e_5 & e_3e_3 = \alpha e_5 \\ {\mathbf{N}}_{203} & : & e_1e_2 = e_3 & e_1e_3 = e_4 & e_1e_4 = e_5 & e_3e_3 = e_5 \\ {\mathbf{N}}_{204} & : & e_1e_2 = e_3 & e_1e_3 = e_4 & e_1e_4 = e_5 \\ && e_3e_3 = e_5 & e_3e_4 = e_5 \\ {\mathbf{N}}_{205} & : & e_1e_2 = e_3 & e_1e_3 = e_4 & e_1e_4 = e_5 & e_3e_4 = e_5 \\ {\mathbf{N}}_{206}^{\alpha, \beta} & : & e_1e_2 = e_3 & e_1e_3 = e_4 & e_2e_2 = e_5 \\ && e_2e_3 = \alpha e_5 & e_3e_3 = \beta e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{207}^{\alpha} & : & e_1e_2 = e_3 & e_1e_3 = e_4 & e_2e_2 = e_5 \\ && e_3e_3 = \alpha e_5 & e_3e_4 = e_5 \\ {\mathbf{N}}_{208} & : & e_1e_2 = e_3 & e_1e_3 = e_4 & e_2e_3 = e_5 \\ && e_2e_4 = e_5 & e_3e_3 = -e_5 \\ {\mathbf{N}}_{209}^{\alpha} & : & e_1e_2 = e_3 & e_1e_3 = e_4 & e_2e_3 = e_5 \\ && e_3e_3 = \alpha e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{210}^{\alpha} & : & e_1e_2 = e_3 & e_1e_3 = e_4 & e_2e_4 = e_5 & e_3e_3 = \alpha e_5 \\ {\mathbf{N}}_{211} & : & e_1e_2 = e_3 & e_1e_3 = e_4 & e_3e_3 = e_5 & e_3e_4 = e_5 \\ {\mathbf{N}}_{212} & : & e_1e_2 = e_3 & e_1e_3 = e_4 & e_3e_3 = e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{213} & : & e_1e_2 = e_3 & e_1e_3 = e_4 & e_3e_4 = e_5 \\ {\mathbf{N}}_{214} & : & e_1e_2 = e_3 & e_1e_3 = e_4 & e_4e_4 = e_5 \\ {\mathbf{N}}_{215}^{\alpha, \beta} & : & e_1e_1 = e_5 & e_1e_2 = e_3 & e_1e_3 = e_4 & e_1e_4 = \alpha e_5 \\ && e_2e_2 = e_4 & e_3e_3 = \beta e_5 & e_3e_4 = e_5 \\ {\mathbf{N}}_{216}^{\alpha, \beta ,\gamma} & : & e_1e_1 = e_5 & e_1e_2 = e_3 & e_1e_3 = e_4 \\ && e_2e_2 = e_4+\alpha e_5 & e_2e_3 = \beta e_5 \\ & & e_3e_3 = \gamma e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{217}^{\alpha} & : & e_1e_1 = \alpha e_5 & e_1e_2 = e_3 & e_1e_3 = e_4 & e_2e_2 = e_4 \\ & & e_2e_3 = e_5 & e_2e_4 = e_5 & e_3e_3 = -e_5 \\ {\mathbf{N}}_{218}^\alpha & : & e_1e_1 = e_5 & e_1e_2 = e_3 & e_1e_3 = e_4 \\ && e_2e_2 = e_4 & e_2e_4 = e_5 & e_3e_3 = \alpha e_5 \\ {\mathbf{N}}_{219} & : & e_1e_2 = e_3 & e_1e_3 = e_4 & e_1e_4 = 2e_5 \\ && e_2e_2 = e_4+e_5 & e_2e_3 = e_5 \\ {\mathbf{N}}_{220}^{\alpha} & : & e_1e_2 = e_3 & e_1e_3 = e_4 & e_1e_4 = e_5 \\ && e_2e_2 = e_4+\alpha e_5 & e_3e_3 = e_5 \\ \end{array} $
    $ \begin{array}{llllllll} {\mathbf{N}}_{221}^{\alpha} & : & e_1e_2 = e_3 & e_1e_3 = e_4 & e_1e_4 = e_5 \\ & & e_2e_2 = e_4 & e_2e_3 = \alpha e_5 \\ {\mathbf{N}}_{222}^{\alpha\neq 0} & : & e_1e_2 = e_3 & e_1e_3 = e_4 & e_1e_4 = \alpha e_5 & e_2e_2 = e_4 \\ & & e_2e_3 = e_5 & e_2e_4 = e_5 & e_3e_3 = -e_5 \\ {\mathbf{N}}_{223}^{\alpha} & : & e_1e_2 = e_3 & e_1e_3 = e_4 & e_1e_4 = e_5 \\ & & e_2e_2 = e_4 & e_2e_4 = e_5 & e_3e_3 = \alpha e_5 \\ {\mathbf{N}}_{224}^{\alpha} & : & e_1e_2 = e_3 & e_1e_3 = e_4 & e_1e_4 = e_5 \\ && e_2e_2 = e_4 & e_3e_3 = \alpha e_5 & e_3e_4 = e_5 \\ {\mathbf{N}}_{225}^{\alpha, \beta} & : & e_1e_2 = e_3 & e_1e_3 = e_4 & e_2e_2 = e_4+e_5 \\ && e_2e_3 = \alpha e_5 & e_3e_3 = \beta e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{226}^{\alpha} & : & e_1e_2 = e_3 & e_1e_3 = e_4 & e_2e_2 = e_4 \\ & & e_2e_3 = e_5 & e_3e_3 = \alpha e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{227}^{\alpha} & : & e_1e_2 = e_3 & e_1e_3 = e_4 & e_2e_2 = e_4 \\ && e_2e_4 = e_5 & e_3e_3 = \alpha e_5 \\ {\mathbf{N}}_{228} & : & e_1e_2 = e_3 & e_1e_3 = e_4 & e_2e_2 = e_4 \\ && e_3e_3 = e_5 & e_3e_4 = e_5 \\ {\mathbf{N}}_{229} & : & e_1e_2 = e_3 & e_1e_3 = e_4 & e_2e_2 = e_4 \\ && e_3e_3 = e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{230} & : & e_1e_2 = e_3 & e_1e_3 = e_4 & e_2e_2 = e_4 & e_3e_4 = e_5 \\ {\mathbf{N}}_{231} & : & e_1e_2 = e_3 & e_1e_3 = e_4 & e_2e_2 = e_4 & e_4e_4 = e_5 \\ {\mathbf{N}}_{232}^{\alpha, \beta, \gamma, \mu} & : & e_1e_1 = e_5 & e_1e_2 = e_3 & e_1e_3 = e_4 & e_2e_2 = \alpha e_5 \\ && e_2e_3 = e_4+\beta e_5 & e_3e_3 = \gamma e_5 & e_3e_4 = \mu e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{233}^{\alpha, \beta} & : & e_1e_1 = e_5 & e_1e_2 = e_3 & e_1e_3 = e_4 & e_2e_2 = \alpha e_5 \\ & & e_2e_3 = e_4 & e_3e_3 = \beta e_5 & e_3e_4 = e_5 \\ {\mathbf{N}}_{234}^{\alpha, \beta} & : & e_1e_2 = e_3 & e_1e_3 = e_4 & e_1e_4 = e_5 & e_2e_2 = \alpha e_5 \\ & & e_2e_3 = e_4+e_5 & e_3e_3 = \beta e_5 \\ {\mathbf{N}}_{235}^{\alpha, \beta, \gamma} & : & e_1e_2 = e_3 & e_1e_3 = e_4 & e_1e_4 = e_5 & e_2e_2 = \alpha e_5 \\ & & e_2e_3 = e_4 & e_2e_4 = \beta e_5 & e_3e_3 = \gamma e_5 & e_3e_4 = e_5 \\ {\mathbf{N}}_{236}^{\alpha} & : & e_1e_2 = e_3 & e_1e_3 = e_4 & e_1e_4 = e_5 \\ && e_2e_2 = e_5 & e_2e_3 = e_4 & e_3e_3 = \alpha e_5 \\ {\mathbf{N}}_{237}^{\alpha\neq0, \beta} & : & e_1e_2 = e_3 & e_1e_3 = e_4 & e_1e_4 = \alpha e_5 \\ & & e_2e_3 = e_4+e_5 & e_2e_4 = e_5 & e_3e_3 = \beta e_5 \\ {\mathbf{N}}_{238}^{\alpha\neq0, \beta} & : & e_1e_2 = e_3 & e_1e_3 = e_4 & e_1e_4 = \alpha e_5 \\ & & e_2e_3 = e_4 & e_2e_4 = e_5 & e_3e_3 = \beta e_5 \\ {\mathbf{N}}_{239}^{\alpha} & : & e_1e_2 = e_3 & e_1e_3 = e_4 & e_1e_4 = e_5 \\ & & e_2e_3 = e_4 & e_3e_3 = \alpha e_5 \\ {\mathbf{N}}_{240}^{\alpha, \beta} & : & e_1e_2 = e_3 & e_1e_3 = e_4 & e_2e_3 = e_4+e_5 \\ & & e_3e_3 = \alpha e_5 & e_3e_4 = \beta e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{241} & : & e_1e_2 = e_3 & e_1e_3 = e_4 & e_2e_3 = e_4 \\ && e_3e_3 = e_5 & e_3e_4 = e_5 \\ {\mathbf{N}}_{242}^{\alpha} & : & e_1e_2 = e_3 & e_1e_3 = e_4 & e_2e_3 = e_4 \\ & & e_3e_3 = e_5 & e_3e_4 = \alpha e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{243} & : & e_1e_2 = e_3 & e_1e_3 = e_4 & e_2e_3 = e_4 & e_3e_4 = e_5 \\ {\mathbf{N}}_{244} & : & e_1e_2 = e_3 & e_1e_3 = e_4 & e_2e_3 = e_4 \\ & & e_3e_4 = e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{245} & : & e_1e_2 = e_3 & e_1e_3 = e_4 & e_2e_3 = e_4 & e_4e_4 = e_5 \\ {\mathbf{N}}_{246}^{\alpha, \beta ,\gamma} & : & e_1e_1 = e_5 & e_1e_2 = e_3 & e_1e_3 = \alpha e_5 & e_2e_2 = e_5 \\ && e_2e_3 = \beta e_5 & e_3e_3 = e_4 & e_3e_4 = \gamma e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{247}^{\alpha, \beta} & : & e_1e_1 = e_5 & e_1e_2 = e_3 & e_1e_4 = \alpha e_5 & e_2e_2 = e_5 \\ & & e_2e_4 = \beta e_5 & e_3e_3 = e_4 & e_3e_4 = e_5 \\ {\mathbf{N}}_{248} & : & e_1e_2 = e_3 & e_1e_3 = e_5 & e_1e_4 = e_5 & e_3e_3 = e_4 \\ {\mathbf{N}}_{249}^{\alpha} & : & e_1e_2 = e_3 & e_1e_3 = e_5 & e_1e_4 = e_5 \\ && e_2e_2 = e_5 & e_2e_3 = \alpha e_5 & e_3e_3 = e_4 \\ {\mathbf{N}}_{250} & : & e_1e_2 = e_3 & e_1e_3 = e_5 & e_1e_4 = e_5 \\ && e_2e_3 = e_5 & e_3e_3 = e_4 \\ \end{array} $
    $ \begin{array}{llllllll} {\mathbf{N}}_{251}^{\alpha} & : & e_1e_2 = e_3 & e_1e_3 = e_5 & e_1e_4 = e_5 \\ & & e_2e_3 = \alpha e_5 & e_2e_4 = e_5 & e_3e_3 = e_4 \\ {\mathbf{N}}_{252}^{\alpha, \beta} & : & e_1e_2 = e_3 & e_1e_3 = e_5 & e_2e_2 = e_5 & e_2e_3 = \alpha e_5 \\ & & e_3e_3 = e_4 & e_3e_4 = \beta e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{253}^{\alpha} & : & e_1e_2 = e_3 & e_1e_3 = e_5 & e_2e_3 = e_5 \\ && e_3e_3 = e_4 & e_3e_4 = \alpha e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{254} & : & e_1e_2 = e_3 & e_1e_4 = e_5 & e_3e_3 = e_4 \\ {\mathbf{N}}_{255}^{\alpha} & : & e_1e_2 = e_3 & e_1e_4 = e_5 & e_2e_2 = e_5 \\ && e_2e_3 = \alpha e_5 & e_3e_3 = e_4 \\ {\mathbf{N}}_{256}^{\alpha} & : & e_1e_2 = e_3 & e_1e_4 = e_5 & e_2e_2 = e_5 \\ && e_2e_4 = \alpha e_5 & e_3e_3 = e_4 & e_3e_4 = e_5 \\ {\mathbf{N}}_{257} & : & e_1e_2 = e_3 & e_1e_4 = e_5 & e_2e_3 = e_5 & e_3e_3 = e_4 \\ {\mathbf{N}}_{258} & : & e_1e_2 = e_3 & e_1e_4 = e_5 & e_2e_4 = e_5 & e_3e_3 = e_4 \\ {\mathbf{N}}_{259} & : & e_1e_2 = e_3 & e_1e_4 = e_5 & e_2e_4 = e_5 \\ & & e_3e_3 = e_4 & e_3e_4 = e_5 \\ {\mathbf{N}}_{260}^{\alpha} & : & e_1e_2 = e_3 & e_2e_2 = e_5 & e_2e_3 = e_5 \\ && e_3e_3 = e_4 & e_3e_4 = \alpha e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{261} & : & e_1e_2 = e_3 & e_2e_2 = e_5 & e_2e_4 = e_5 \\ && e_3e_3 = e_4 & e_3e_4 = e_5 \\ {\mathbf{N}}_{262} & : & e_1e_2 = e_3 & e_2e_2 = e_5 & e_3e_3 = e_4 & e_3e_4 = e_5 \\ {\mathbf{N}}_{263} & : & e_1e_2 = e_3 & e_2e_2 = e_5 & e_3e_3 = e_4 \\ & & e_3e_4 = e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{264} & : & e_1e_2 = e_3 & e_2e_2 = e_5 & e_3e_3 = e_4 & e_4e_4 = e_5 \\ {\mathbf{N}}_{265} & : & e_1e_2 = e_3 & e_2e_3 = e_5 & e_3e_3 = e_4 \\ & & e_3e_4 = e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{266} & : & e_1e_2 = e_3 & e_2e_3 = e_5 & e_3e_3 = e_4 & e_4e_4 = e_5 \\ {\mathbf{N}}_{267} & : & e_1e_2 = e_3 & e_2e_4 = e_5 & e_3e_3 = e_4 & e_3e_4 = e_5 \\ {\mathbf{N}}_{268} & : & e_1e_2 = e_3 & e_3e_3 = e_4 & e_3e_4 = e_5 \\ {\mathbf{N}}_{269} & : & e_1e_2 = e_3 & e_3e_3 = e_4 & e_3e_4 = e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{270} & : & e_1e_2 = e_3 & e_3e_3 = e_4 & e_4e_4 = e_5 \\ {\mathbf{N}}_{271}^{\alpha, \beta, \gamma,\mu} & : & e_1e_1 = e_4+e_5 & e_1e_2 = e_3 & e_1e_3 = \alpha e_5 & e_2e_2 = \beta e_5 \\ && e_2e_3 = \gamma e_5 & e_3e_3 = e_4 & e_3e_4 = \mu e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{272}^{\alpha, \beta} & : & e_1e_1 = e_4+\alpha e_5 & e_1e_2 = e_3 & e_1e_3 = \beta e_5 \\ && e_2e_3 = e_5 & e_2e_4 = e_5 & e_3e_3 = e_4 \\ {\mathbf{N}}_{273}^{\alpha, \beta} & : & e_1e_1 = e_4+\alpha e_5 & e_1e_2 = e_3 & e_1e_3 = \beta e_5 \\ & & e_2e_4 = e_5 & e_3e_3 = e_4 \\ {\mathbf{N}}_{274}^{\alpha, \beta ,\gamma} & : & e_1e_1 = e_4+e_5 & e_1e_2 = e_3 & e_1e_4 = \alpha e_5 & e_2e_2 = \beta e_5 \\ & & e_2e_4 = \gamma e_5 & e_3e_3 = e_4 & e_3e_4 = e_5 \\ {\mathbf{N}}_{275}^{\alpha, \beta} & : & e_1e_1 = e_4 & e_1e_2 = e_3 & e_1e_3 = e_5 & e_1e_4 = e_5 \\ && e_2e_2 = \alpha e_5 & e_2e_3 = \beta e_5 & e_3e_3 = e_4 \\ {\mathbf{N}}_{276}^{\alpha, \beta} & : & e_1e_1 = e_4 & e_1e_2 = e_3 & e_1e_3 = \alpha e_5 & e_1e_4 = e_5 \\ && e_2e_3 = \beta e_5 & e_2e_4 = e_5 & e_3e_3 = e_4 \\ {\mathbf{N}}_{277}^{\alpha, \beta ,\gamma} & : & e_1e_1 = e_4 & e_1e_2 = e_3 & e_1e_3 = e_5 & e_2e_2 = \alpha e_5 \\ && e_2e_3 = \beta e_5 & e_3e_3 = e_4 & e_3e_4 = \gamma e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{278} & : & e_1e_1 = e_4 & e_1e_2 = e_3 & e_1e_4 = e_5 & e_3e_3 = e_4 \\ {\mathbf{N}}_{279}^{\alpha} & : & e_1e_1 = e_4 & e_1e_2 = e_3 & e_1e_4 = e_5 \\ && e_2e_2 = e_5 & e_2e_3 = \alpha e_5 & e_3e_3 = e_4 \\ {\mathbf{N}}_{280}^{\alpha, \beta} & : & e_1e_1 = e_4 & e_1e_2 = e_3 & e_1e_4 = \alpha e_5 & e_2e_2 = e_5 \\ && e_2e_4 = \beta e_5 & e_3e_3 = e_4 & e_3e_4 = e_5 \\ \end{array} $
    $ \begin{array}{llllllll} {\mathbf{N}}_{281} & : & e_1e_1 = e_4 & e_1e_2 = e_3 & e_1e_4 = e_5 \\ && e_2e_3 = e_5 & e_3e_3 = e_4 \\ {\mathbf{N}}_{282}^{\alpha} & : & e_1e_1 = e_4 & e_1e_2 = e_3 & e_1e_4 = \alpha e_5 \\ & & e_2e_4 = e_5 & e_3e_3 = e_4 & e_3e_4 = e_5 \\ {\mathbf{N}}_{283}^{\alpha} & : & e_1e_1 = e_4 & e_1e_2 = e_3 & e_1e_4 = \alpha e_5 \\ & & e_3e_3 = e_4 & e_3e_4 = e_5 \\ {\mathbf{N}}_{284}^{\alpha, \beta} & : & e_1e_1 = e_4 & e_1e_2 = e_3 & e_2e_2 = e_5 & e_2e_3 = \alpha e_5 \\ & & e_3e_3 = e_4 & e_3e_4 = \beta e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{285}^{\alpha} & : & e_1e_1 = e_4 & e_1e_2 = e_3 & e_2e_3 = e_5 \\ & & e_3e_3 = e_4 & e_3e_4 = \alpha e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{286} & : & e_1e_1 = e_4 & e_1e_2 = e_3 & e_3e_3 = e_4 \\ & & e_3e_4 = e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{287} & : & e_1e_1 = e_4 & e_1e_2 = e_3 & e_3e_3 = e_4 & e_4e_4 = e_5 \\ {\mathbf{N}}_{288}^{\alpha, \beta, \gamma,\mu,\nu} & : & e_1e_1 = e_4+\alpha e_5 & e_1e_2 = e_3 & e_1e_3 = \beta e_5 \\ && e_2e_2 = e_4+\gamma e_5 & e_2e_3 = \mu e_5 & e_3e_3 = e_4 \\ & & e_3e_4 = \nu e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{289}^{\alpha, \beta, \gamma,\mu} & : & e_1e_1 = e_4+\alpha e_5 & e_1e_2 = e_3 & e_1e_4 = \beta e_5 \\ && e_2e_2 = e_4+\gamma e_5 & e_2e_4 = \mu e_5 \\ & & e_3e_3 = e_4 & e_3e_4 = e_5 \\ {\mathbf{N}}_{290}^{\alpha, \beta} & : & e_1e_1 = e_4 & e_1e_2 = e_3 & e_1e_3 = \alpha e_5 & e_1e_4 = e_5 \\ & & e_2e_2 = e_4+\beta e_5 & e_2e_3 = \gamma e_5 & e_3e_3 = e_4 \\ {\mathbf{N}}_{291}^{\alpha, \beta\neq0 ,\gamma} & : & e_1e_1 = e_4 & e_1e_2 = e_3 & e_1e_3 = \alpha e_5 & e_1e_4 = \beta e_5 \\ & & e_2e_3 = \gamma e_5 & e_2e_4 = e_5 & e_3e_3 = e_4 \\ {\mathbf{N}}_{292}^{\alpha, \beta} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_1e_3 = e_5 \\ & & e_1e_4 = \alpha e_5 & e_2e_2 = \beta e_5 & e_2e_3 = e_4 \\ & & e_2e_4 = e_5 & e_3e_3 = -\alpha e_5 \\ {\mathbf{N}}_{293}^{\alpha} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_1e_3 = e_5 & e_1e_4 = e_5 \\ & & e_2e_2 = \alpha e_5 & e_2e_3 = e_4 & e_3e_3 = -e_5 \\ {\mathbf{N}}_{294}^{\alpha, \beta ,\gamma} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_1e_3 = e_5 & e_2e_2 = \alpha e_5 \\ && e_2e_3 = e_4 & e_2e_4 = \beta e_5 & e_3e_3 = \gamma e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{295}^{\alpha, \beta} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_1e_4 = \alpha e_5 & e_2e_2 = e_5 \\ && e_2e_3 = e_4 & e_2e_4 = e_5 & e_3e_3 = \beta e_5 \\ {\mathbf{N}}_{296}^{\alpha} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_1e_4 = e_5 \\ & & e_2e_2 = e_5 & e_2e_3 = e_4 & e_3e_3 = \alpha e_5 \\ {\mathbf{N}}_{297}^{\alpha} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_1e_4 = e_5 \\ & & e_2e_3 = e_4 & e_2e_4 = e_5 & e_3e_3 = \alpha e_5 \\ {\mathbf{N}}_{298}^{\alpha} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_1e_4 = e_5 \\ & & e_2e_3 = e_4 & e_3e_3 = \alpha e_5 \\ {\mathbf{N}}_{299}^{\alpha, \beta} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_2e_2 = e_5 & e_2e_3 = e_4 \\ && e_2e_4 = \alpha e_4 & e_3e_3 = \beta e_5 & e_3e_4 = e_5 \\ {\mathbf{N}}_{300}^{\alpha, \beta} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_2e_2 = e_5 & e_2e_3 = e_4 \\ && e_2e_4 = \alpha e_4 & e_3e_3 = \beta e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{301} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_2e_3 = e_4 & e_2e_4 = e_5 \\ {\mathbf{N}}_{302} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_2e_3 = e_4 \\ & & e_2e_4 = e_5 & e_3e_3 = e_5 \\ {\mathbf{N}}_{303}^{\alpha} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_2e_3 = e_4 \\ & & e_2e_4 = e_5 & e_3e_3 = \alpha e_5 & e_3e_4 = e_5 \\ {\mathbf{N}}_{304}^{\alpha} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_2e_3 = e_4 \\ & & e_2e_4 = e_5 & e_3e_3 = \alpha e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{305} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_2e_3 = e_4 \\ && e_3e_3 = e_5 & e_3e_4 = e_5 \\ {\mathbf{N}}_{306} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_2e_3 = e_4 \\ && e_3e_3 = e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{307} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_2e_3 = e_4 & e_4e_4 = e_5 \\ {\mathbf{N}}_{308} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_2e_3 = e_4 & e_4e_4 = e_5 \\ {\mathbf{N}}_{309}^{\alpha, \beta} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_1e_3 = e_4 \\ & & e_1e_4 = e_5 & e_2e_2 = \alpha e_5 \\ & & e_2e_3 = e_4+\beta e_5 & e_3e_3 = -e_5 \\ {\mathbf{N}}_{310}^{\alpha, \beta\neq-1} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_1e_3 = e_4 & e_1e_4 = e_5 \\ && e_2e_2 = \alpha e_5 & e_2e_3 = e_4 & e_3e_3 = \beta e_5 \\ \end{array} $
    $ \begin{array}{llllllll} {\mathbf{N}}_{311}^{\alpha, \beta,\gamma} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_1e_3 = e_4 & e_1e_4 = \alpha e_5 \\ & & e_2e_3 = e_4+\beta e_5 & e_2e_4 = e_5 & e_3e_3 = \gamma e_5 \\ {\mathbf{N}}_{312}^{\alpha, \beta,\gamma, \mu} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_1e_3 = e_4 & e_2e_2 = \alpha e_5 \\ && e_2e_3 = e_4+\beta e_5 & e_2e_4 = \gamma e_5 & e_3e_3 = \mu e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{313}^{\alpha, \beta,\gamma} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_1e_3 = e_4 & e_2e_2 = \alpha e_5 \\ & & e_2e_3 = e_4 & e_2e_4 = \beta e_5 & e_3e_3 = \gamma e_5 & e_3e_4 = e_5 \\ {\mathbf{N}}_{314}^{\alpha, \beta} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_1e_3 = e_5 & e_1e_4 = e_5 \\ & & e_2e_2 = \alpha e_5 & e_2e_3 = \beta e_5 & e_3e_3 = e_4 \\ {\mathbf{N}}_{315}^{\alpha, \beta ,\gamma} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_1e_3 = e_5 & e_1e_4 = \alpha e_5 \\ && e_2e_2 = \beta e_5 & e_2e_3 = \gamma e_5 & e_2e_4 = e_5 & e_3e_3 = e_4 \\ {\mathbf{N}}_{316}^{\alpha, \beta, \gamma, \mu} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_1e_3 = e_5 \\ && e_2e_2 = \alpha e_5 & e_2e_3 = \beta e_5 & e_2e_4 = \gamma e_5 \\ & & e_3e_3 = e_4 & e_3e_4 = \mu e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{317} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_1e_4 = e_5 & e_3e_3 = e_4 \\ {\mathbf{N}}_{318}^{\alpha} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_1e_4 = e_5 \\ && e_2e_2 = e_5 & e_2e_3 = \alpha e_5 & e_3e_3 = e_4 \\ {\mathbf{N}}_{319}^{\alpha, \beta} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_1e_4 = e_5 & e_2e_2 = \alpha e_5 \\ && e_2e_3 = \beta e_5 & e_2e_4 = e_5 & e_3e_3 = e_4 \\ {\mathbf{N}}_{320}^{\alpha, \beta ,\gamma} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_1e_4 = e_5 & e_2e_2 = \alpha e_5 \\ && e_2e_3 = \beta e_5 & e_2e_4 = \gamma e_5 & e_3e_3 = e_4 & e_3e_4 = e_5 \\ {\mathbf{N}}_{321} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_1e_4 = e_5 \\ & & e_2e_3 = e_5 & e_3e_3 = e_4 \\ {\mathbf{N}}_{322}^{\alpha} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_2e_2 = e_5 \\ && e_2e_3 = \alpha e_5 & e_2e_4 = e_5 & e_3e_3 = e_4 \\ {\mathbf{N}}_{323}^{\alpha, \beta} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_2e_2 = e_5 & e_2e_3 = \alpha e_5 \\ && e_2e_4 = \beta e_5 & e_3e_3 = e_4 & e_3e_4 = e_5 \\ {\mathbf{N}}_{324}^{\alpha, \beta ,\gamma} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_2e_2 = e_5 & e_2e_3 = \alpha e_5 \\ && e_2e_4 = \beta e_5 & e_3e_3 = e_4 & e_3e_4 = \gamma e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{325} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_2e_3 = e_5 \\ & & e_2e_4 = e_5 & e_3e_3 = e_4 \\ {\mathbf{N}}_{326}^{\alpha} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_2e_3 = e_5 \\ && e_2e_4 = \alpha e_5 & e_3e_3 = e_4 & e_3e_4 = e_5 \\ {\mathbf{N}}_{327}^{\alpha, \beta} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_2e_3 = e_5 & e_2e_4 = \alpha e_5 \\ & & e_3e_3 = e_4 & e_3e_4 = \beta e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{328} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_2e_4 = e_5 & e_3e_3 = e_4 \\ {\mathbf{N}}_{329} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_2e_4 = e_5 \\ & & e_3e_3 = e_4 & e_3e_4 = e_5 \\ \end{array} $
    $ \begin{array}{llllllll} {\mathbf{N}}^\alpha_{330} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_2e_4 = e_5 \\ & & e_3e_3 = e_4 & e_3e_4 = \alpha e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{331} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_3e_3 = e_4 & e_3e_4 = e_5 \\ {\mathbf{N}}_{332} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_3e_3 = e_4 \\ & & e_3e_4 = e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{333} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_3e_3 = e_4 & e_4e_4 = e_5 \\ {\mathbf{N}}_{334}^{\alpha, \beta ,\gamma} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_1e_3 = \alpha e_5 & e_1e_4 = e_5 \\ && e_2e_2 = e_4+\beta e_5 & e_2e_3 = \gamma e_5 & e_3e_3 = e_4 \\ {\mathbf{N}}_{335}^{\alpha, \beta,\gamma, \mu} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_1e_3 = \alpha e_5 & e_1e_4 = \beta e_5 \\ && e_2e_2 = e_4+\gamma e_5 & e_2e_3 = \mu e_5 & e_2e_4 = e_5 & e_3e_3 = e_4 \\ {\mathbf{N}}_{336}^{\alpha, \beta,\gamma, \mu, \nu} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_1e_3 = \alpha e_5 \\ & & e_2e_2 = e_4+\beta e_5 & e_2e_3 = \gamma e_5 & e_2e_4 = \mu e_5 \\ & & e_3e_3 = e_4 & e_3e_4 = \nu e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{337}^{\alpha, \beta,\gamma, \mu} & : & e_1e_1 = e_2 & e_1e_2 = e_3 & e_1e_4 = \alpha e_5 \\ && e_2e_2 = e_4+\beta e_5 & e_2e_3 = \gamma e_5 & e_2e_4 = \mu e_5 \\ && e_3e_3 = e_4 & e_3e_4 = e_5 \\ {\mathbf{N}}_{338}^{\alpha} & : & e_1e_1 = e_2 & e_1e_2 = e_5 & e_1e_3 = e_4 & e_1e_4 = e_5 \\ & & e_2e_2 = e_3 & e_2e_3 = -2e_5 & e_3e_3 = \alpha e_5 \\ {\mathbf{N}}_{339}^{\alpha, \beta} & : & e_1e_1 = e_2 & e_1e_2 = e_5 & e_1e_3 = e_4 & e_2e_2 = e_3 \\ && e_2e_3 = \alpha e_5 & e_3e_3 = \beta e_5 & e_3e_4 = e_5 \\ {\mathbf{N}}_{340}^{\alpha, \beta ,\gamma} & : & e_1e_1 = e_2 & e_1e_2 = e_5 & e_1e_3 = e_4 & e_2e_2 = e_3 \\ & & e_2e_3 = \alpha e_5 & e_3e_3 = \beta e_5 & e_3e_4 = \gamma e_5 & e_4e_4 = e_5 \\ \end{array} $
    $ \begin{array}{llllllll} {\mathbf{N}}_{341}^{\alpha} & : & e_1e_1 = e_2 & e_1e_3 = e_4 & e_1e_4 = e_5 \\ & & e_2e_2 = e_3 & e_2e_3 = \alpha e_5 \\ {\mathbf{N}}_{342}^{\alpha, \beta} & : & e_1e_1 = e_2 & e_1e_3 = e_4 & e_1e_4 = e_5 & e_2e_2 = e_3 \\ && e_2e_3 = \alpha e_5 & e_2e_4 = e_5 & e_3e_3 = \beta e_5 \\ {\mathbf{N}}_{343}^{\alpha} & : & e_1e_1 = e_2 & e_1e_3 = e_4 & e_1e_4 = e_5 \\ & & e_2e_2 = e_3 & e_2e_3 = \alpha e_5 & e_3e_3 = e_5 \\ {\mathbf{N}}_{344}^{\alpha} & : & e_1e_1 = e_2 & e_1e_3 = e_4 & e_2e_2 = e_3 \\ && e_2e_3 = e_5 & e_2e_4 = e_5 & e_3e_3 = \alpha e_5 \\ {\mathbf{N}}_{345}^{\alpha, \beta} & : & e_1e_1 = e_2 & e_1e_3 = e_4 & e_2e_2 = e_3 & e_2e_3 = \alpha e_5 \\ && e_2e_4 = e_5 & e_3e_3 = \beta e_5 & e_3e_4 = e_5 \\ {\mathbf{N}}_{346}^{\alpha} & : & e_1e_1 = e_2 & e_1e_3 = e_4 & e_2e_2 = e_3 \\ && e_2e_3 = e_5 & e_3e_3 = \alpha e_5 & e_3e_4 = e_5 \\ {\mathbf{N}}_{347}^{\alpha, \beta} & : & e_1e_1 = e_2 & e_1e_3 = e_4 & e_2e_2 = e_3 & e_2e_3 = e_5 \\ && e_3e_3 = \alpha e_5 & e_3e_4 = \beta e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{348} & : & e_1e_1 = e_2 & e_1e_3 = e_4 & e_2e_2 = e_3 & e_2e_4 = e_5 \\ {\mathbf{N}}_{349} & : & e_1e_1 = e_2 & e_1e_3 = e_4 & e_2e_2 = e_3 \\ && e_2e_4 = e_5 & e_3e_3 = e_5 \\ {\mathbf{N}}_{350} & : & e_1e_1 = e_2 & e_1e_3 = e_4 & e_2e_2 = e_3 \\ && e_3e_3 = e_5 & e_3e_4 = e_5 \\ {\mathbf{N}}_{351}^{\alpha} & : & e_1e_1 = e_2 & e_1e_3 = e_4 & e_2e_2 = e_3 \\ & & e_3e_3 = \alpha e_5 & e_3e_4 = e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{352} & : & e_1e_1 = e_2 & e_1e_3 = e_4 & e_2e_2 = e_3 \\ && e_3e_3 = e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{353} & : & e_1e_1 = e_2 & e_1e_3 = e_4 & e_2e_2 = e_3 & e_3e_4 = e_5 \\ {\mathbf{N}}_{354} & : & e_1e_1 = e_2 & e_1e_3 = e_4 & e_2e_2 = e_3 & e_4e_4 = e_5 \\ {\mathbf{N}}_{355}^{\alpha\neq0, \beta} & : & e_1e_1 = e_2 & { e_1e_3 = e_4+\alpha e_5 } \\ && e_1e_4 = e_5 & e_2e_2 = e_3 \\ & & e_2e_3 = \beta e_5 & { e_3e_3 = (\beta+2)e_5 } \\ {\mathbf{N}}_{356}^{\alpha\neq0, \beta,\gamma} & : & e_1e_1 = e_2 & { e_1e_3 = e_4+\alpha e_5 } \\ && e_2e_2 = e_3 & e_2e_3 = \beta e_5 & e_2e_4 = e_5 \\ & & e_3e_3 = \gamma e_5 & e_3e_4 = e_5 & \\ {\mathbf{N}}_{357}^{\alpha, \beta,\gamma,\mu} & : & e_1e_1 = e_2 & { e_1e_3 = e_4+\alpha e_5 } \\ & & e_2e_2 = e_3 & e_2e_3 = \beta e_5 & e_3e_3 = \gamma e_5 \\ & & e_3e_4 = \mu e_5 & e_4e_4 = e_5 & \\ {\mathbf{N}}_{358}^{\alpha,\beta} & : & e_1e_1 = e_2 & e_1e_3 = e_4 & e_1e_4 = e_5 \\ & & e_2e_2 = e_3 & e_2e_3 = \alpha e_5 & e_3e_3 = \beta e_5 \\ {\mathbf{N}}_{359}^{\alpha, \beta,\gamma} & : & e_1e_1 = e_2 & e_1e_3 = e_4 & e_1e_4 = \alpha e_5 & e_2e_2 = e_3 \\ & & e_2e_3 = \beta e_5 & e_2e_4 = e_5 & e_3e_3 = \gamma e_5 & \\ {\mathbf{N}}_{360}^{\alpha,\beta,\gamma} & : & e_1e_1 = e_2 & e_1e_3 = e_4 & e_2e_2 = e_3 & e_2e_3 = \alpha e_5 \\ & & e_2e_4 = \beta e_5 & e_3e_3 = \gamma e_5 & e_3e_4 = e_5 & \\ {\mathbf{N}}_{361}^{\alpha, \beta} & : & e_1e_1 = e_2 & e_1e_2 = e_5 & e_1e_3 = \alpha e_5 & e_1e_4 = e_5 \\ & & e_2e_2 = e_3 & e_2e_3 = e_4 & e_3e_3 = \beta e_5 \\ \end{array} $
    $ \begin{array}{llllllll} {\mathbf{N}}_{362}^{\alpha, \beta,\gamma,\mu} & : & e_1e_1 = e_2 & e_1e_2 = e_5 & e_1e_3 = \alpha e_5 \\ & & e_2e_2 = e_3 & e_2e_3 = e_4 & e_2e_4 = \beta e_5 \\ & & e_3e_3 = \gamma e_5 & e_3e_4 = \mu e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{363}^{\alpha, \beta,\gamma} & : & e_1e_1 = e_2 & e_1e_2 = e_5 & e_1e_4 = \alpha e_5 & e_2e_2 = e_3 \\ & & e_2e_3 = e_4 & e_2e_4 = \beta e_5 & e_3e_3 = \gamma e_5 & e_3e_4 = e_5 \\ {\mathbf{N}}_{364}^{\alpha, \beta} & : & e_1e_1 = e_2 & e_1e_3 = e_5 & e_1e_4 = \alpha e_5 & e_2e_2 = e_3 \\ & & e_2e_3 = e_4 & e_2e_4 = e_5 \ & e_3e_3 = \beta e_5 \\ {\mathbf{N}}_{365}^{\alpha} & : & e_1e_1 = e_2 & e_1e_3 = e_5 & e_1e_4 = e_5 \\ & & e_2e_2 = e_3 & e_2e_3 = e_4 & e_3e_3 = \alpha e_5 \\ {\mathbf{N}}_{366}^{\alpha, \beta,\gamma} & : & e_1e_1 = e_2 & e_1e_3 = e_5 & e_2e_2 = e_3 & e_2e_3 = e_4 \\ & & e_1e_4 = \alpha e_5 & e_3e_3 = \beta e_5 & e_3e_4 = \gamma e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{367} & : & e_1e_1 = e_2 & e_1e_4 = e_5 & e_2e_2 = e_3 & e_2e_3 = e_4 \\ {\mathbf{N}}_{368}^{\alpha} & : & e_1e_1 = e_2 & e_1e_4 = e_5 & e_2e_2 = e_3 \\ & & e_2e_3 = e_4 & e_2e_4 = e_5 & e_3e_3 = \alpha e_5 \\ {\mathbf{N}}_{369}^{\alpha,\beta} & : & e_1e_1 = e_2 & e_1e_4 = e_5 & e_2e_2 = e_3 & e_2e_3 = e_4 \\ && e_2e_4 = \alpha e_5 & e_3e_3 = \beta e_5 & e_3e_4 = e_5 & \\ {\mathbf{N}}_{370} & : & e_1e_1 = e_2 & e_1e_4 = e_5 & e_2e_2 = e_3 \\ && e_2e_3 = e_4 & e_3e_3 = e_5 \\ {\mathbf{N}}_{371}^{\alpha} & : & e_1e_1 = e_2 & e_2e_2 = e_3 & e_2e_3 = e_4 \\ && e_2e_4 = e_5 & e_3e_3 = \alpha e_5 \\ {\mathbf{N}}_{372}^{\alpha} & : & e_1e_1 = e_2 & e_2e_2 = e_3 & e_2e_3 = e_4 \\ & & e_2e_4 = e_5 & e_3e_3 = \alpha e_5 & e_3e_4 = e_5 \\ {\mathbf{N}}_{373}^{\alpha, \beta} & : & e_1e_1 = e_2 & e_2e_2 = e_3 & e_2e_3 = e_4 & e_2e_4 = e_5 \\ && e_3e_3 = \alpha e_5 & e_3e_4 = \beta e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{374} & : & e_1e_1 = e_2 & e_2e_2 = e_3 & e_2e_3 = e_4 \\ && e_3e_3 = e_5 & e_3e_4 = e_5 \\ {\mathbf{N}}_{375}^{\alpha} & : & e_1e_1 = e_2 & e_2e_2 = e_3 & e_2e_3 = e_4 \\ && e_3e_3 = e_5 & e_3e_4 = \alpha e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{376} & : & e_1e_1 = e_2 & e_2e_2 = e_3 & e_2e_3 = e_4 & e_3e_4 = e_5 \\ {\mathbf{N}}_{377} & : & e_1e_1 = e_2 & e_2e_2 = e_3 & e_2e_3 = e_4 \\ & & e_3e_4 = e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{378} & : & e_1e_1 = e_2 & e_2e_2 = e_3 & e_2e_3 = e_4 & e_4e_4 = e_5 \\ {\mathbf{N}}_{379}^{\alpha, \beta,\gamma} & : & e_1e_1 = e_2 & e_1e_2 = \alpha e_5 & e_1e_3 = e_4 \\ && e_1e_4 = e_5 & e_2e_2 = e_3 \\ && e_2e_3 = e_4+\beta e_5 & e_3e_3 = \gamma e_5 \\ {\mathbf{N}}_{380}^{\alpha, \beta,\gamma,\mu} & : & e_1e_1 = e_2 & e_1e_2 = \alpha e_5 & e_1e_3 = e_4 \\ & & e_1e_4 = \beta e_5 & e_2e_2 = e_3 & e_2e_3 = e_4 \\ & & e_2e_4 = \gamma e_5 & e_3e_3 = \mu e_5 & e_3e_4 = e_5 \\ {\mathbf{N}}_{381}^{\alpha, \beta,\gamma,\mu, \nu} & : & e_1e_1 = e_2 & e_1e_2 = \alpha e_5 & e_1e_3 = e_4 & e_2e_2 = e_3 \\ & & e_2e_3 = e_4+\beta e_5 & e_2e_4 = \gamma e_5 & e_3e_3 = \mu e_5 \\ & & e_3e_4 = \nu e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{382}^{\alpha, \beta,\gamma} & : & e_1e_1 = e_2 & e_1e_3 = e_4 & e_1e_4 = \alpha e_5 & e_2e_2 = e_3 \\ & & e_2e_3 = e_4+\beta e_5 & e_2e_4 = e_5 & e_3e_3 = \gamma e_5 \\ {\mathbf{N}}_{383}^{\alpha, \beta} & : & e_1e_1 = e_2 & e_1e_2 = e_5 & e_1e_3 = \alpha e_5 & e_1e_4 = e_5 \\ && e_2e_2 = e_3 & e_2e_3 = \beta e_5 & e_3e_3 = e_4 \\ {\mathbf{N}}_{384}^{\alpha, \beta,\gamma,\mu} & : & e_1e_1 = e_2 & e_1e_2 = e_5 & e_1e_3 = \alpha e_5 \\ & & e_2e_2 = e_3 & e_2e_3 = \beta e_5 & e_2e_4 = \gamma e_5 \\ && e_3e_3 = e_4 & e_3e_4 = \mu e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{385}^{\alpha, \beta,\gamma} & : & e_1e_1 = e_2 & e_1e_2 = e_5 & e_1e_4 = \alpha e_5 & e_2e_2 = e_3 \\ && e_2e_3 = \beta e_5 & e_2e_4 = \gamma e_5 & e_3e_3 = e_4 & e_3e_4 = e_5 \\ {\mathbf{N}}_{386}^{\alpha} & : & e_1e_1 = e_2 & e_1e_3 = e_5 & e_1e_4 = e_5 \\ && e_2e_2 = e_3 & e_2e_3 = \alpha e_5 & e_3e_3 = e_4 \\ {\mathbf{N}}_{387}^{\alpha, \beta} & : & e_1e_1 = e_2 & e_1e_3 = e_5 & e_1e_4 = \alpha e_5 & e_2e_2 = e_3 \\ && e_2e_3 = \beta e_5 & e_2e_4 = e_5 & e_3e_3 = e_4 \\ {\mathbf{N}}_{388}^{\alpha, \beta,\gamma} & : & e_1e_1 = e_2 & e_1e_3 = e_5 & e_2e_2 = e_3 & e_2e_3 = \alpha e_5 \\ && e_2e_4 = \beta e_5 & e_3e_3 = e_4 & e_3e_4 = \gamma e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{389} & : & e_1e_1 = e_2 & e_1e_4 = e_5 & e_2e_2 = e_3 & e_3e_3 = e_4 \\ {\mathbf{N}}_{390} & : & e_1e_1 = e_2 & e_1e_4 = e_5 & e_2e_2 = e_3 \\ && e_2e_3 = e_5 & e_3e_3 = e_4 \\ \end{array} $
    $ \begin{array}{llllllll} {\mathbf{N}}_{391}^{\alpha} & : & e_1e_1 = e_2 & e_1e_4 = e_5 & e_2e_2 = e_3 \\ && e_2e_3 = \alpha e_5 & e_2e_4 = e_5 & e_3e_3 = e_4 \\ {\mathbf{N}}_{392}^{\alpha, \beta} & : & e_1e_1 = e_2 & e_1e_4 = e_5 & e_2e_2 = e_3 & e_2e_3 = \alpha e_5 \\ && e_2e_4 = \beta e_5 & e_3e_3 = e_4 & e_3e_4 = e_5 \\ {\mathbf{N}}_{393} & : & e_1e_1 = e_2 & e_2e_2 = e_3 & e_2e_3 = e_5 \\ && e_2e_4 = e_5 & e_3e_3 = e_4 \\ {\mathbf{N}}_{394}^{\alpha} & : & e_1e_1 = e_2 & e_2e_2 = e_3 & e_2e_3 = e_5 \\ & & e_2e_4 = \alpha e_5 & e_3e_3 = e_4 & e_3e_4 = e_5 \\ {\mathbf{N}}_{395}^{\alpha, \beta} & : & e_1e_1 = e_2 & e_2e_2 = e_3 & e_2e_3 = e_5 & e_2e_4 = \alpha e_5 \\ & & e_3e_3 = e_4 & e_3e_4 = \beta e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{396} & : & e_1e_1 = e_2 & e_2e_2 = e_3 & e_2e_4 = e_5 & e_3e_3 = e_4 \\ {\mathbf{N}}_{397} & : & e_1e_1 = e_2 & e_2e_2 = e_3 & e_2e_4 = e_5 \\ & & e_3e_3 = e_4 & e_3e_4 = e_5 \\ {\mathbf{N}}_{398}^{\alpha} & : & e_1e_1 = e_2 & e_2e_2 = e_3 & e_2e_4 = e_5 \\ && e_3e_3 = e_4 & e_3e_4 = \alpha e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{399} & : & e_1e_1 = e_2 & e_2e_2 = e_3 & e_3e_3 = e_4 & e_3e_4 = e_5 \\ {\mathbf{N}}_{400} & : & e_1e_1 = e_2 & e_2e_2 = e_3 & e_3e_3 = e_4 \\ && e_3e_4 = e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{401} & : & e_1e_1 = e_2 & e_2e_2 = e_3 & e_3e_3 = e_4 & e_4e_4 = e_5 \\ {\mathbf{N}}_{402}^{\alpha, \beta,\gamma} & : & e_1e_1 = e_2 & e_1e_2 = e_4 & e_1e_3 = \alpha e_5 & e_1e_4 = \beta e_5 \\ & & e_2e_2 = e_3 & e_2e_3 = \gamma e_5 & e_2e_4 = e_5 & e_3e_3 = e_4 \\ {\mathbf{N}}_{403}^{\alpha, \beta,\gamma} & : & e_1e_1 = e_2 & e_1e_2 = e_4 & e_1e_3 = \alpha e_5 & e_1e_4 = e_5 \\ & & e_2e_2 = e_3 & e_2e_3 = \beta e_5 & e_3e_3 = e_4+\gamma e_5 \\ {\mathbf{N}}_{404}^{\alpha, \beta,\gamma,\mu,\nu} & : & e_1e_1 = e_2 & e_1e_2 = e_4 & e_1e_3 = \alpha e_5 \\ & & e_2e_2 = e_3 & e_2e_3 = \beta e_5 & e_2e_4 = \gamma e_5 \\ & & e_3e_3 = e_4+\mu e_5 & e_3e_4 = \nu e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{405}^{\alpha, \beta,\gamma,\mu} & : & e_1e_1 = e_2 & e_1e_2 = e_4 & e_1e_4 = \alpha e_5 & e_2e_2 = e_3 \\ & & e_2e_3 = \beta e_5 & e_2e_4 = \gamma e_5 & e_3e_3 = e_4+\mu e_5 & e_3e_4 = e_5 \\ {\mathbf{N}}_{406}^{\lambda,\alpha, \beta,\gamma} & : & e_1e_1 = e_2 & { e_1e_2 = \lambda e_4+\alpha e_5 } \\ && e_1e_3 = \beta e_5 & e_1e_4 = e_5 & e_2e_2 = e_3 \\ && e_2e_3 = e_4+\gamma e_5 & e_3e_3 = e_4 \\ {\mathbf{N}}_{407}^{\lambda,\alpha, \beta,\gamma,\mu,\nu} & : & e_1e_1 = e_2 & { e_1e_2 = \lambda e_4+\alpha e_5 } \\ && e_1e_3 = \beta e_5 & e_2e_2 = e_3 & { e_2e_3 = e_4+\gamma e_5 } \\ & & e_2e_4 = \mu e_5 & e_3e_3 = e_4 & e_3e_4 = \nu e_5 & e_4e_4 = e_5 \\ {\mathbf{N}}_{408}^{\lambda,\alpha, \beta,\gamma} & : & e_1e_1 = e_2 & { e_1e_2 = \lambda e_4+\alpha e_5 } \\ & & e_1e_4 = \beta e_5 & e_2e_2 = e_3 & e_2e_3 = e_4+\gamma e_5 \\ && e_2e_4 = \mu e_5 & e_3e_3 = e_4 & e_3e_4 = e_5 \\ {\mathbf{N}}_{409}^{\lambda,\alpha, \beta,\gamma} & : & e_1e_1 = e_2 & e_1e_2 = \lambda e_4 & e_1e_3 = \alpha e_5 \\ && e_1e_4 = \beta e_5 & e_2e_2 = e_3 & { e_2e_3 = e_4+\gamma e_5 } \\ & & e_2e_4 = e_5 & e_3e_3 = e_4 \end{array} $
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