Mini review

A Place to Call Home: Bioengineering Pluripotential Stem Cell Cultures

  • Received: 01 December 2014 Accepted: 02 March 2015 Published: 11 March 2015
  • Pluripotent stem cells (PSCs) have the power to revolutionize the future of cell-based therapies and regenerative medicine. However, stem/progenitor cell use in the clinical arsenal has been hampered by discrepancies resulting from stem cell engineering and expansion, as well as in their (mass) differentiation in culture. Moreover, the manner in which external conditions affect PSC and induced-pluripotent stem cell lineage establishment as well as maturation remains controversial. In this review, we examine novel methods of cell engineering and the role of reprogramming transcription factors in PSC development. In addition, we explore the effect of external environmental signals on PSC cultivation and differentiation by elucidating key components of the primordial stem cell microenvironment, the blastocyst. Furthermore, we assess the effects of hypoxic conditions on DNA editing, gene expression, and protein function in PSC self-renewal and growth. Finally, we speculate on the principal use of gap junction subunit expression as relevant biomarkers of PSC fate. Improving bioreactor design and pertinent cell biomarker classification could vastly enhance manufactured stem cell yield and quality, thereby increasing the potency and safety of therapeutic cells to be used in regenerative medicine.

    Citation: Mark Weingarten, Nathan Akhavan, Joshua Hanau, Yakov Peter. A Place to Call Home: Bioengineering Pluripotential Stem Cell Cultures[J]. AIMS Bioengineering, 2015, 2(2): 15-28. doi: 10.3934/bioeng.2015.2.15

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  • Pluripotent stem cells (PSCs) have the power to revolutionize the future of cell-based therapies and regenerative medicine. However, stem/progenitor cell use in the clinical arsenal has been hampered by discrepancies resulting from stem cell engineering and expansion, as well as in their (mass) differentiation in culture. Moreover, the manner in which external conditions affect PSC and induced-pluripotent stem cell lineage establishment as well as maturation remains controversial. In this review, we examine novel methods of cell engineering and the role of reprogramming transcription factors in PSC development. In addition, we explore the effect of external environmental signals on PSC cultivation and differentiation by elucidating key components of the primordial stem cell microenvironment, the blastocyst. Furthermore, we assess the effects of hypoxic conditions on DNA editing, gene expression, and protein function in PSC self-renewal and growth. Finally, we speculate on the principal use of gap junction subunit expression as relevant biomarkers of PSC fate. Improving bioreactor design and pertinent cell biomarker classification could vastly enhance manufactured stem cell yield and quality, thereby increasing the potency and safety of therapeutic cells to be used in regenerative medicine.


    Let $ \mathbb{F}_{q} $ be the finite field of $ q $ elements, where $ q = p^{\alpha} $ with a prime $ p $ and a positive integer $ \alpha $. Suppose that

    $ q-1 = ef $

    with positive integers $ e $ and $ f $. Choose a generator $ \gamma $ of the multiplicative cyclic group

    $ \mathbb{F}_{q}^{*} = \mathbb{F}_{q}-\{0\}. $

    For $ v\in\mathbb{F}_{q}^{*} $, let $ \mathrm{ind}_{\gamma}(v) $ denote the unique non-negative integer $ m\leqslant q-2 $ such that $ v = \gamma^{m} $.

    For $ 0\leqslant i, j\leqslant e-1 $ (or rather for $ i, j $ modulo $ e $), the $ e^{2} $ cyclotomic numbers of order $ e $, denoted by $ A_{ij} $ (or $ A_{ij}^{(e)} $ to indicate the order $ e $), are defined as the cardinality of the set $ X_{ij} $, where

    $ X_{ij}: = \{v\in\mathbb{F}_{q}-\{0, -1\}\mid\mathrm{ind}_{\gamma}(v)\equiv i\, (\mathrm{mod}\, e), \ \mathrm{ind}_{\gamma}(v+1)\equiv j\, (\mathrm{mod}\, e)\}. $

    Introduced by Gauss [1,2] over two hundred years ago, cyclotomic numbers are a significant concept in number theory with deep connections to various mathematical areas. They have been extensively applied in combinatorial designs, coding theory, cryptography, and information theory (see [3,4,5,6]). For both theoretical and practical purposes, it is intriguing to determine all cyclotomic numbers of a given order $ e $ for all finite fields, a task usually called the cyclotomic problem.

    In the case $ q = p $, Gauss [1, §358] evaluated $ A_{ij}^{(3)} $ in terms of $ (L, M) $ satisfying the Diophantine system

    $ 4p = L^{2}+27M^{2}, \quad \text{with}\; \; \; L\equiv1\, (\mathrm{mod}\, 3). $

    This system determines $ L $ uniquely and $ M $ up to signs. Similarly, Gauss [2] evaluated $ A_{ij}^{(4)} $ in terms of $ (a, b) $ satisfying

    $ p = a^{2}+b^{2}, \quad \text{with}\; \; \; a\equiv1\, (\mathrm{mod}\, 4), $

    which fixes $ a $ uniquely and $ b $ up to signs. His results indicated that solving the cyclotomic problem over $ \mathbb{F}_{q} $ generally requires more than just the value of $ q $ and order $ e $; it also needs a quadratic partition of $ q $.

    Therefore, classical solutions to the cyclotomic problem are typically expressed using an appropriately chosen solution of a relevant Diophantine system (consisting of equations and congruences), often with the sign ambiguity. In this sense, many mathematicians have investigated the cyclotomic problem for various small orders $ \leqslant22 $ (see Dickson's early foundational work [7,8,9] and a good recent survey [10]), as well as for special orders such as [11,12,13] for $ l $, [13,14] for $ 2l $, [15] for $ l^{2} $, and [16,17] for $ 2l^{2} $ (with an odd prime $ l $).

    While Gauss initially approached cyclotomy via Gauss sums, Dickson's use of Jacobi sums [7] laid the groundwork for modern cyclotomy. The cyclotomic problem is, in fact, equivalent to the explicit evaluation of Jacobi sums of the same order. Let us recall the definiton of Jacobi sums. Let $ \zeta $ be a primitive complex $ e $-th root of unity fixed once for all. We define a multiplicative character $ \chi_{e} $ of order $ e $ on $ \mathbb{F}_{q}^{*} $ by

    $ \chi_{e}(\gamma^{m}) = \zeta^{m} \quad (\text{for any}\; m\in\mathbb{Z}), $

    and extend $ \chi_{e} $ to a character on $ \mathbb{F}_{q} $ by taking $ \chi_{e}(0) = 0 $. For convenience, we assume

    $ \chi_{e}^{m}(0) = 0 $

    for any integer $ m $. The Jacobi sums $ J(i, j) $ (or $ J_{e}(i, j) $ to indicate the order $ e $) of order $ e $, for $ 0\leqslant i, j\leqslant e-1 $ (or rather for $ i, j $ modulo $ e $), are defined by

    $ J(i, j) = \sum\limits_{v\in\mathbb{F}_{q}}\chi_{e}^{i}(v)\chi_{e}^{j}(v+1). $

    Jacobi sums and cyclotomic numbers are related by the following finite Fourier series expansions:

    $ J(a, b) = \sum\limits_{0\leqslant i, j\leqslant e-1}A_{ij}\zeta^{ai+bj}, \quad e^{2}A_{ab} = \sum\limits_{0\leqslant i, j\leqslant e-1}J(i, j)\zeta^{-(ai+bj)}. $

    To calculate all cyclotomic numbers of order $ e $, it suffices to calculate all Jacobi sums of order $ e $, and vice versa.

    Jacobi made significant contributions to mathematics, including the Jacobi symbol, the Jacobi triple product, the Jacobi elliptic functions, and the Jacobian in variable transformations. Among his notable discoveries are Jacobi sums, which he proposed in 1827 in a letter to Gauss and published ten years later. These sums were later extended by Cauchy, Gauss, and Eisenstein. While Gauss sums are pivotal in proving quadratic reciprocity, Jacobi sums are essential for proving cubic reciprocity and were generalized by Eisenstein for biquadratic reciprocity. Jacobi sums are also used to estimate the number of integer solutions to congruences like

    $ x^{3}+y^{3}\equiv1\, (\mathrm{mod}\, p), $

    which are crucial for developing the Weil conjectures [18]. In modern mathematics, Jacobi sums have found applications in primality testing [19].

    Wamelen [20] has provided an inductive arithmetic approach to characterize all Jacobi sums of any order, thereby solving the cyclotomic problem in theory. However, the Diophantine system he employed is notably large and intricate in general. In recent years, there has been growing interest in efficiently computing Jacobi sums [21], driven by their importance in applications such as primality testing, cryptosystems, combinatorial designs, and advanced number theory problems [19,22].

    For given $ \mathbb{F}_{q} $, $ \gamma $, and $ \zeta $, classical cyclotomic numbers and Jacobi sums are binary functions depending on two variables $ i, j\in\mathbb{Z}/e\mathbb{Z} $. The purpose of this paper is to investigate their trivariate analogs, so-called "ternary cyclotomic numbers" and "ternary Jacobi sums", defined in Section 2. As classical cyclotomic numbers and Jacobi sums are important both theoretically and practically, we want to study their ternary counterparts in order to explore theoretically interesting problems or potential applications. In Section 2, we obtain some ternary properties, which are analogous to those of classical cyclotomic numbers and Jacobi sums, with proofs similar to those of classical ones. Then we use these properties to solve the cyclotomic problem for ternary cyclotomic numbers. Section 3 provides explicit evaluations of all ternary cyclotomic numbers and ternary Jacobi sums for order $ e = 2 $, and Section 4 nearly completes the evaluation for order $ e = 3 $, except for an integer $ J_{3}(1, 1, 2) $, which remains unknown. Our calculations show that ternary Jacobi sums, which are some kind of character sums, cannot generally be transformed into classical Jacobi sums. So the cyclotomic problem for ternary cyclotomic numbers has its own interests in theory.

    Let $ \mathbb{F}_{q} $, $ \gamma $, $ \zeta $, $ \chi_{e} $ and

    $ q = p^{\alpha} = ef+1 $

    be as described in Section 1. We further assume

    $ q = ef+1 $

    is odd for convenience. (The upcoming definitions will be meaningless for $ \mathbb{F}_{2^{\alpha}} $, where $ v-1 = v+1 $.) Specifically, either $ e $ or $ f $ is even.

    For $ 0\leqslant i, j, k\leqslant e-1 $ (or rather for $ i, j, k\in\mathbb{Z}/e\mathbb{Z} $), we define the ternary cyclotomic numbers of order $ e $, denoted by $ A_{ijk} $ (or $ A_{ijk}^{(e)} $ to indicate the order $ e $), as the cardinality of the set

    $ Xijk:={vFq{0,±1}indγ(v1)i(mode), indγvj(mode), indγ(v+1)k(mode)}.
    $

    This definition relies on the choice of a multiplicative generator $ \gamma $ of $ \mathbb{F}_{q}^{*} $. We also define the ternary Jacobi sums of order $ e $, denoted by $ J(i, j, k) $ (or $ J_{e}(i, j, k) $ to indicate the order $ e $), as

    $ J(i, j, k): = \sum\limits_{v\in\mathbb{F}_{q}}\chi_{e}^{i}(v-1)\chi_{e}^{j}(v)\chi_{e}^{k}(v+1). $

    This definition depends on the character $ \chi_{e} $ on $ \mathbb{F}_{q} $, which is determined by the selections of $ \gamma $ and $ \zeta $ (as $ \chi_{e}(\gamma^{m}) = \zeta^{m} $ and $ \chi_{e}(0) = 0 $). Clearly, we have

    $ X_{ijk} = \{v\in\mathbb{F}_{q}\mid\chi_{e}(v-1) = \zeta^{i}, \ \chi_{e}(v) = \zeta^{j}, \ \chi_{e}(v+1) = \zeta^{k}\}. $

    Our definition of the ternary Jacobi sum can be viewed as a special case of the character sums studied in [23].

    To calculate all ternary cyclotomic numbers of order $ e $, it suffices to calculate all ternary Jacobi sums of order $ e $, and vice versa, by the following finite Fourier series expansions.

    Proposition 1. (Finite Fourier series expansions) The ternary cyclotomic numbers and ternary Jacobi sums of the same order $ e $ are related by: for any $ a, b, c\in\mathbb{Z}/e\mathbb{Z} $,

    $ J(a,b,c)=i,j,kZ/eZAijkζai+bj+ck,
    $
    (2.1)
    $ e3Aabc=i,j,kZ/eZJ(i,j,k)ζ(ai+bj+ck).
    $
    (2.2)

    Proof. Note that

    $ { \mathbb{F}_{q} = \{0, \pm1\}\cup\bigcup\limits_{i, j, k\in\mathbb{Z}/e\mathbb{Z}}X_{ijk}}. $

    For $ v\in\{0, \pm1\} $,

    $ \chi_{e}^{a}(v-1)\chi_{e}^{b}(v)\chi_{e}^{c}(v+1) = 0. $

    For the $ A_{ijk} $ elements $ v\in X_{ijk} $,

    $ \chi_{e}^{a}(v-1)\chi_{e}^{b}(v)\chi_{e}^{c}(v+1) = \zeta^{ai+bj+ck}. $

    Summing them all together, we obtain Eq (2.1). For the Eq (2.2), noting that $ \zeta = \chi_{e}(\gamma) $, we have

    $ i,j,kZ/eZJ(i,j,k)ζ(ai+bj+ck)=i,j,kZ/eZvFqχie(v1)χje(v)χke(v+1)χe(γ)(ai+bj+ck)=vFq(e1i=0χie(v1γa))(e1j=0χje(vγb))(e1k=0χke(v+1γc)).
    $

    Note that

    $ e1j=0χje(vγb)={e,if v0 and indγ(v)b(mode),0,if v=0 or indγ(v)b(mode),
    $

    and similar for $ { \sum_{i = 0}^{e-1}\chi_{e}^{i}\Big(\frac{v-1}{\gamma^{a}}\Big)} $ and $ { \sum_{j = 0}^{e-1}\chi_{e}^{k}\Big(\frac{v+1}{\gamma^{c}}\Big)} $. So the terms of the above sum are non-zero only when $ v\in X_{abc} $, and thus

    $ \sum\limits_{i, j, k\in\mathbb{Z}/e\mathbb{Z}}J(i, j, k)\zeta^{-(ai+bj+ck)} = \sum\limits_{v\in X_{abc}}e^{3} = e^{3}A_{abc}. $

    Following arguments often involve the value of $ \chi_{e}(-1) $. Since

    $ \gamma^{\frac{q-1}{2}} = -1\in\mathbb{F}_{q}, $

    and when $ e $ is even,

    $ \zeta^{\frac{e}{2}} = -1\in\mathbb{C}, $

    we have

    $ \chi_{e}(-1)^{-1} = \chi_{e}(-1) = \chi_{e}(\gamma^{\frac{q-1}{2}}) = \zeta^{\frac{ef}{2}} = {1,if f is even,ζe2=1,if f is odd,
    = (-1)^{f}. $

    Classical cyclotomic numbers exhibit a symmetry property in their two variables:

    $ A_{ij} = A_{j+\frac{ef}{2}, i+\frac{ef}{2}} = {Aji,if f is even,Aj+e2,i+e2,if f is odd,
    $

    which implies that

    $ J(a, b) = (-1)^{f(a+b)}J(b, a). $

    It is similar for ternary cyclotomic numbers and ternary Jacobi sums.

    Proposition 2. For any $ i, j, k\in\mathbb{Z}/e\mathbb{Z} $,

    $ A_{ijk} = A_{k+\frac{ef}{2}, j+\frac{ef}{2}, i+\frac{ef}{2}} = {Akji,if f is even,Ak+e2,j+e2,i+e2,if f is odd.
    $

    Proof. For any $ v\in\mathbb{F}_{q} $, let $ w = -v $. Since

    $ \chi_{e}(-1) = \zeta^{\frac{ef}{2}}, $

    we have

    $ vXijk(χe(w1),χe(w),χe(w+1))=(ζi,ζj,ζk)(χe(w1),χe(w),χe(w+1))=(χe(1)ζk,χe(1)ζj,χe(1)ζi)wXk+ef2,j+ef2,i+ef2.
    $

    So $ v\mapsto-v $ induces a bijection from $ X_{ijk} $ to $ X_{k+\frac{ef}{2}, j+\frac{ef}{2}, i+\frac{ef}{2}} $.

    Corollary 3. For any $ a, b, c\in\mathbb{Z}/e\mathbb{Z} $,

    $ J(a, b, c) = (-1)^{f(a+b+c)}J(c, b, a). $

    As a consequence, when both $ f $ and $ b $ are odd, we have $ J(a, b, a) = 0 $.

    Proof. By Propositions 1 and 2, as

    $ \zeta^{-\frac{ef}{2}} = (-1)^{f}, $

    we have

    $ J(a,b,c)=i,j,kZ/eZAk+ef2,j+ef2,i+ef2ζai+bj+ck=l,m,nZ/eZAlmnζa(nef2)+b(mef2)+c(lef2)=ζef2(a+b+c)l,m,nZ/eZAlmnζcl+bm+an=(1)f(a+b+c)J(c,b,a).
    $

    Classical cyclotomic numbers exhibit another property

    $ A_{ij} = A_{-i, j-i}, $

    which implies that

    $ J(a, b) = J(-a-b, b). $

    It is similar for ternary cases as follows:

    Proposition 4. For any $ i, j, k\in\mathbb{Z}/e\mathbb{Z} $,

    $ A_{ijk} = A_{i-j+\frac{ef}{2}, -j, k-j} = {Aij,j,kj,if f is even,Aij+e2,j,kj,if f is odd.
    $

    Proof. For any $ v\in\mathbb{F}_{q} $, let $ w = v^{-1} $. Since

    $ \chi_{e}(-1) = \zeta^{\frac{ef}{2}}, $

    we have

    $ vXijk(χe(w11),χe(w1),χe(w1+1))=(ζi,ζj,ζk)(χe(w1)χe(1)χe(w),χe(w),χe(1+w)χe(w))=(ζi,ζj,ζk)(χe(w1),χe(w),χe(w+1))=(χe(1)ζij,ζj,ζkj)wXij+ef2,j,kj.
    $

    So $ v\mapsto v^{-1} $ induces a bijection from $ X_{ijk} $ to $ X_{i-j+\frac{ef}{2}, -j, k-j} $.

    Corollary 5. For any $ a, b, c\in\mathbb{Z}/e\mathbb{Z} $,

    $ J(a, b, c) = (-1)^{fa}J(a, -a-b-c, c). $

    Proof. By Propositions 1 and 4, as

    $ \zeta^{-\frac{ef}{2}} = (-1)^{f}, $

    we have

    $ J(a,b,c)=i,j,kZ/eZAij+ef2,j,kjζai+bj+ck=l,m,nZ/eZAl,m,nζa(lmef2)bm+c(nm)=(ζef2)al,m,nZ/eZAl,m,nζal+(abc)m+cn=(1)faJ(a,abc,c).
    $

    For classical cyclotomic numbers, we have

    $ iZ/eZAij={f1,if j0(mode),f,otherwise, and jZ/eZAij={f1,if ief2(mode),f,otherwise.
    $
    (2.3)

    We show similar equations for ternary cyclotomic numbers.

    Proposition 6. Let $ g = \mathrm{ind}_{\gamma}(2) $. For any $ i, j, k\in\mathbb{Z}/e\mathbb{Z} $,

    $ tZ/eZAtjk={Ajk1,if j0(mode) and kg(mode),Ajk,otherwise,tZ/eZAijt={Aij1,if ig+ef2(mode) and jef2(mode),Aij,otherwise,tZ/eZAitk={Aig,kg1,if i ef2(mode) and k0(mode),Aig,kg,otherwise.
    $

    Proof. Note that

    $ { \bigcup\limits_{t\in\mathbb{Z}/e\mathbb{Z}}X_{tjk}} = X_{jk}-\{1\}. $

    Also note that $ 1\in X_{jk} $ if and only if

    $ j\equiv0\, (\mathrm{mod}\, e)\; \; \; \text{and}\; \; \; k\equiv g\, (\mathrm{mod}\, e). $

    These $ X_{tjk} $ are pairwise disjoint, which gives the first equation.

    For the second equation,

    $ { \bigcup\limits_{t\in\mathbb{Z}/e\mathbb{Z}}X_{ijt}} = \{v\in\mathbb{F}_{q}-\{0, \pm1\}\mid\chi_{e}(v-1) = \zeta^{i}, \ \chi_{e}(v) = \zeta^{j}\}. $

    So

    $ v\in{ \bigcup\limits_{t\in\mathbb{Z}/e\mathbb{Z}}X_{ijt}} $

    if and only if

    $ v-1\in X_{ij}-\{-2\}. $

    Also note that $ -2\in X_{ij} $ if and only if

    $ i\equiv g+\frac{ef}{2}\, (\mathrm{mod}\, e)\; \; \text{and}\; \; j\equiv\frac{ef}{2}\, (\mathrm{mod}\, e). $

    For the third equation,

    $ { \bigcup\limits_{t\in\mathbb{Z}/e\mathbb{Z}}X_{itk}} = \{v\in\mathbb{F}_{q}-\{0, \pm1\}\mid\chi_{e}(v-1) = \zeta^{i}, \ \chi_{e}(v+1) = \zeta^{k}\}. $

    Then

    $ v\in{ \bigcup\limits_{t\in\mathbb{Z}/e\mathbb{Z}}X_{itk}} $

    if and only if

    $ \frac{v-1}{2}\in X_{i-g, k-g}-\{-2^{-1}\}. $

    Here

    $ v\mapsto\frac{v-1}{2} $

    induces a bijection on $ \mathbb{F}_{q} $. Also note that

    $ -2^{-1}\in X_{i-g, k-g} $

    if and only if

    $ i\equiv\frac{ef}{2}\, (\mathrm{mod}\, e) \; \; \text{and}\; \; k\equiv0\, (\mathrm{mod}\, e). $

    Note that

    $ \sum\limits_{v\in\mathbb{F}_{q}}\chi_{e}^{i}(v) = \sum\limits_{m = 0}^{q-2}\chi_{e}^{i}(\gamma^{m}) = \sum\limits_{m = 0}^{q-2}(\zeta^{i})^{m} = {q1,if i0(mode),0,otherwise.
    $

    So classical Jacobi sums $ J(i, 0) $ and $ J(0, i) $ can be easily evaluated by

    $ (1)fiJ(i,0)=J(0,i)=vFqχ0e(v)χie(v+1)=vFqχie(v+1)χie(1)={q2,if i0(mode),1,otherwise.
    $
    (2.4)

    Similarly, ternary Jacobi sums $ J(i, j, k) $, with either $ i $, $ j $ or $ k $ divided by $ e $, can be evaluated in terms of classical Jacobi sums of the same order $ e $.

    Proposition 7. Let $ g = \mathrm{ind}_{\gamma}(2) $. For $ i, j, k\in\mathbb{Z}/e\mathbb{Z} $, we have

    $ J(0,j,k)=J(j,k)ζgk,J(i,j,0)=J(i,j)(1)f(i+j)ζgi,J(i,0,k)=ζg(i+k)J(i,k)(1)fi.
    $

    Proof. Recall that

    $ \chi_{e}^{m}(0) = 0 $

    for any $ m\in\mathbb{Z} $, we have

    $ \chi_{e}(2) = \zeta^{g} \; \; \text{and}\; \; \chi_{e}(-1) = (-1)^{f}. $

    So

    $ J(0,j,k)=vFqχ0e(v1)χje(v)χke(v+1)=vFq{1}χje(v)χke(v+1)=vFqχje(v)χke(v+1)χje(1)χke(2)=J(j,k)ζgk,J(i,j,0)=vFq{1}χie(v1)χje(v)=vFqχie(v1)χje(v)χie(2)χje(1)=J(i,j)(1)f(i+j)ζgi.
    $

    In the following we let

    $ w = \frac{v-1}{2}. $

    Note that

    $ v\mapsto\frac{v-1}{2} $

    induces a bijection on $ \mathbb{F}_{q} $. Then

    $ J(i,0,k)=vFqχie(v1)χke(v+1)χie(1)χke(1)=wFqχie(2w)χke(2w+2)(1)fi=ζg(i+k)J(i,k)(1)fi.
    $

    Corollary 8.

    $ J(0, 0, 0) = q-3. $

    For $ i\not\equiv0\, (\mathrm{mod}\, e) $, we have

    $ J(i,0,0)=(1)fi+1(1+ζgi),J(0,i,0)=1(1)fi,J(0,0,i)=1ζgi.
    $

    To conclude this section, it is important to highlight a key result that will facilitate our subsequent calculations, as it is self-evident from the definition.

    Lemma 9. Let $ k $ be an integer coprime to $ e $, and $ \sigma_{k} $ denote the $ \mathbb{Q} $-automorphism of the field $ \mathbb{Q}(\zeta) $ with

    $ \sigma_{k}(\zeta) = \zeta^{k}. $

    Then for any $ a, b, c\in\mathbb{Z}/e\mathbb{Z} $, we have

    $ J(ka,kb)=σk(J(a,b)),J(ka,kb,kc)=σk(J(a,b,c)).
    $

    In this section, assuming

    $ e = 2 \quad \text{and} \quad q = p^{\alpha} = 2f+1, $

    we calculate all ternary cyclotomic numbers and ternary Jacobi sums of order $ 2 $ for a generator $ \gamma $ of $ \mathbb{F}_{q}^{*} $. We fix $ \zeta = -1 $, and let

    $ g = \mathrm{ind}_{\gamma}(2). $

    By Eq (2.4) and

    $ J_{2}(a, b) = J_{2}(-a-b, b), $

    all classical Jacobi sums of order 2 are:

    $ J2(0,0)=q2,J2(1,0)=(1)f+1,J2(1,1)=J2(0,1)=1.
    $
    (3.1)

    First, let us calculate $ J_{2}(1, 1, 1) $ when $ f $ is even (which is this section's most challenging part). We will see that it is related to classical Jacobi sums of order $ 4 $. Let us take the imaginary unit

    $ \mathfrak{i} = \sqrt{-1} $

    as the primitive $ 4 $-th root $ \zeta_{4} $ of unity. We write $ J_{4}(i, j) $ (with $ i, j\in\mathbb{Z}/4\mathbb{Z} $) for the {classical} Jacobi sums of order $ 4 $ with respect to $ \gamma $ and $ \zeta_{4} $. Let $ \chi_{4} $ be the character of $ \mathbb{F}_{q} $ defined by

    $ \chi_{4}(0) = 0, \quad \chi_{4}(\gamma^{m}) = \zeta_{4}^{m} = \mathfrak{i}^{m} \; \; (\text{for any} \; m\in\mathbb{Z}). $

    By definition,

    $ J_{4}(i, j) = \sum\limits_{v\in\mathbb{F}_{q}}\chi_{4}^{i}(v)\chi_{4}^{j}(v+1). $

    A critical insight is that

    $ \chi_{2} = \chi_{4}^{2} $

    on $ \mathbb{F}_{q} $. Also note that

    $ \mathbb{F}_{q}^{*} = \{\gamma^{m}\mid0\leqslant m\leqslant2f-1\}. $

    So

    $ J2(1,1,1)=vFqχ2(v1)χ2(v)χ2(v+1)=vFqχ2(v)χ2(v21)=vFqχ4(v2)χ24(v21)=2f1m=0χ4(γ2m)χ24(γ2m1)=2f1m=0χ4(γ2m)χ24(γ2m1).
    $

    The final expression is a part of the following formula:

    $ J4(2,1)=vFqχ24(v)χ4(v+1)=wFqχ24(w1)χ4(w)=wFqχ4(w)χ24(w1)=2f1n=0χ4(γn)χ24(γn1)=f1m=0χ4(γ2m)χ24(γ2m1)+f1m=0χ4(γ2m+1)χ24(γ2m+11).
    $

    Since

    $ \chi_{4}(\gamma^{2m}) = (-1)^{m}, \quad \chi_{4}(\gamma^{2m+1}) = \mathfrak{i}^{2m+1}\in\{\pm\mathfrak{i}\} $

    and

    $ \chi_{4}^{2}(v)\in\{0, \pm1\} $

    for any $ v\in\mathbb{F}_{q} $,

    $ {\bf Re}(J_{4}(2, 1)) = \sum\limits_{m = 0}^{f-1}\chi_{4}(\gamma^{2m})\chi_{4}^{2}(\gamma^{2m}-1) = \frac{J_{2}(1, 1, 1)}{2}, $

    where $ {\bf Re}(z) $ represents the real component of a complex number $ z $.

    From [24], we extract the evaluation required. Take special note that the Jacobi sums as defined in [24, §2] are distinct from our own definitions. In fact, their Jacobi sums $ R(m, n) $ equal our

    $ J(n, -m-n) = (-1)^{fm}J(m, n). $

    So their finding for $ R(1, 1) $ [24, Propositions 1, 2] can be reinterpreted as a result for our $ J_{4}(1, 2) $ as follows.

    Lemma 10. [24, Propositions 1, 2] Let $ q = p^{\alpha}\equiv1\, (\mathrm{mod}\, 4) $ with $ p $ prime and $ \alpha\geqslant1 $, and let

    $ s = -{\bf Re}(J_{4}(1, 2)). $

    If $ p\equiv3\, (\mathrm{mod}\, 4) $, then $ \alpha $ is even,

    $ s = (-p)^{\alpha/2}\equiv1\, (\mathrm{mod}\, 4) $

    and

    $ J_{4}(1, 2) = -s = -(-p)^{\alpha/2}. $

    If $ p\equiv1\, (\mathrm{mod}\, 4) $, then $ s $ is the unique integer coprime to $ q $ such that $ s\equiv1\, (\mathrm{mod}\, 4) $ and $ q-s^{2} $ is a perfect square.

    Remark 11. The earlier literature [25, p.298] also presented the results of Lemma 10, but erred in the sign when $ p\equiv3\, (\mathrm{mod}\, 4) $.

    When

    $ f = \dfrac{q-1}{2} $

    is even, by Lemma 10,

    $ J_{2}(1, 1, 1) = 2{\bf Re}(J_{4}(2, 1)) = 2{\bf Re}\big((-1)^{\frac{q-1}{4}}J_{4}(1, 2)\big) = (-1)^{\frac{q-1}{4}}(-2s). $

    When $ f $ is odd, by Corollary 3, $ J_{2}(1, 1, 1) = 0 $.

    Evaluating the ternary Jacobi sums of order $ 2 $, excluding $ J_{2}(1, 1, 1) $, is now straightforward by Proposition 7, Corollary 8, and Eq (3.1). Combining all these, we formulate the results into the following theorem.

    Theorem 12. All ternary Jacobi sums of order $ 2 $ over $ \mathbb{F}_{q} $ with respect to a generator $ \gamma $ of $ \mathbb{F}_{q}^{*} $ are explicitly given as follows, with $ g = \mathrm{ind}_{\gamma}(2) $ and $ s $ defined as in Lemma 10.

    $ J2(0,0,0)=q3,J2(1,0,0)=(1)f+1(1+(1)g),J2(0,0,1)=J2(0,1,1)=J2(1,1,0)=1(1)g,J2(0,1,0)=J2(1,0,1)=1(1)f,J2(1,1,1)={0,if q3(mod4),2s,if q1(mod8),2s,if q5(mod8).
    $

    Using finite Fourier series expansions from Proposition 1, along with Theorem 12, we derive a complete and explicit evaluation of all ternary cyclotomic numbers of order $ 2 $ as follows.

    Theorem 13. All ternary cyclotomic numbers of order $ 2 $ over $ \mathbb{F}_{q} $, corresponding to a generator $ \gamma $ of $ \mathbb{F}_{q}^{*} $, are explicitly evaluated as follows, with $ g = \mathrm{ind}_{\gamma}(2) $ and $ s $ defined as in Lemma 10.

    If $ q\equiv3\, (\mathrm{mod}\, 4) $, then

    $ A(2)000=A(2)100=A(2)110=A(2)111=q52(1)g8,A(2)001=A(2)010=A(2)011=A(2)101=q1+2(1)g8.
    $

    If $ q\equiv1\, (\mathrm{mod}\, 8) $, then

    $ A(2)000=q112s4(1)g8,A(2)011=A(2)110=q+12s8,A(2)101=q32s+4(1)g8,A(2)001=A(2)100=A(2)010=A(2)111=q3+2s8.
    $

    If $ q\equiv5\, (\mathrm{mod}\, 8) $, then

    $ A(2)000=A(2)101=q7+2s8,A(2)011=A(2)110=q+1+2s8,A(2)001=A(2)100=A(2)010=A(2)111=q32s8.
    $

    This section aims to compute as many ternary Jacobi sums of order $ 3 $ as feasible. Let

    $ e = 3\; \; \text{and}\; \; q = p^{\alpha} = 3f+1 $

    be an odd prime power. Clearly, $ f $ is even. Let

    $ g = \mathrm{ind}_{\gamma}(2). $

    We choose a cube root of unity $ \zeta_{3}\in\{e^{\pm\frac{2\pi}{3}\mathfrak{i}}\} $. Then

    $ \zeta_{3}^{2} = -1-\zeta_{3}. $

    Also note that

    $ \sigma_{2}(\zeta_{3}) = \zeta_{3}^{2} = \zeta_{3}^{-1} = \overline{\zeta_{3}}, $

    for the $ \mathbb{Q} $-automorphism $ \sigma_{2} $: $ \mathbb{Q}(\zeta)\to\mathbb{Q}(\zeta) $. So $ \sigma_{2} $ is {just} the complex conjugation.

    First, recall the evaluation of classical Jacobi sums of order $ 3 $. Since $ f $ is even,

    $ J_{3}(a, b) = J_{3}(b, a) = J_{3}(-a-b, b) = J_{3}(b, -a-b) = J_{3}(-a-b, a) = J_{3}(a, -a-b). $

    By Eq (2.4),

    $ J_{3}(0, 0) = q-2, $

    and

    $ J_{3}(0, 1) = J_{3}(1, 0) = J_{3}(2, 1) = J_{3}(1, 2) = J_{3}(2, 0) = J_{3}(0, 2) = -1. $

    The value of

    $ J_{3}(1, 1) = \overline{J_{3}(2, 2)} $

    is also known as the following lemma.

    Lemma 14. [12,25] For

    $ q = p^{\alpha} = 3f+1 $

    with $ p $ odd prime and $ \alpha\geqslant1 $, we can write

    $ J_{3}(1, 1) = \frac{L+3M}{2}+3M\zeta_{3}, $

    for some $ L, M\in\mathbb{Z} $ with $ L\equiv1\, (\mathrm{mod}\, 3) $.

    (1) If $ p\equiv2\, (\mathrm{mod}\, 3) $, then $ \alpha $ is even,

    $ M = 0, \quad L = -2(-p)^{\alpha/2}, \; \; \mathit{\text{and}}\; \; J_{3}(1, 1) = -(-p)^{\alpha/2}. $

    (2) If $ p\equiv1\, (\mathrm{mod}\, 3) $, then $ (L, M) $ is the unique solution of the Diophantine system:

    $ {4q=L2+27M2,L1(mod3),pL,γq13(L+9M)/(L9M)(modp).
    $
    (4.1)

    Proof. For $ p\equiv1\, (\mathrm{mod}\, 3) $, this result is a part of [12, Proposition 1]. For $ p\equiv2\, (\mathrm{mod}\, 3) $, [25, p.297] provided the value of $ J_{3}(1, 1) $ but erred in the sign. To rectify this mistake, we present an elementary proof below.

    By definition,

    $ J_{3}(1, 1) = a+b\zeta_{3} $

    with some $ a, b\in\mathbb{Z} $. Write $ A_{ij} $ (with $ 0\leqslant i, j\leqslant2 $) for the classical cyclotomic numbers of order $ 3 $. Note that $ f $ is even. So

    $ A_{01} = A_{10} = A_{22}\; \; \text{and}\; \; A_{02} = A_{20} = A_{11}. $

    By definition,

    $ b = A_{01}+A_{10}+A_{22}-(A_{02}+A_{11}+A_{20}) = 3(A_{01}-A_{02})\equiv0\, (\mathrm{mod}\, 3). $

    By Eq (2.3),

    $ f1=A00+A01+A02,f=A22+A02+A12=A01+A02+A12.
    $

    So

    $ A_{12} = A_{00}+1. $

    By definition,

    $ a = A_{00}+A_{12}+A_{21}-(A_{02}+A_{11}+A_{20}) = 2+3(A_{00}-A_{02})\equiv2\, (\mathrm{mod}\ 3). $

    By [26, Theorem 2.1.3], if none of $ i $, $ j $, and $ i+j $ are multiples of $ e $, then

    $ |J_{e}(i, j)| = \sqrt{q}. $

    Therefore,

    $ p^{\alpha} = q = (a+b\zeta_{3})(\overline{a+b\zeta_{3}}) = (a+b\zeta_{3})(a+b\zeta_{3}^{2}). $

    The Eisenstein integer ring $ \mathbb{Z}[\zeta_{3}] $ is a unique factorization domain (UFD), where the rational prime $ p\equiv2\, (\mathrm{mod}\, 3) $ is an Eisenstein prime, which is irreducible in $ \mathbb{Z}[\zeta_{3}] $. So

    $ a+b\zeta_{3} = p^{n}u $

    for an integer $ n\geqslant0 $ and an Eisenstein unit

    $ u\in\{\pm1, \pm\zeta_{3}, \pm(1+\zeta_{3})\}. $

    Since

    $ p^{n} = |a+b\zeta_{3}| = p^{\alpha/2}, $

    we have

    $ a+b\zeta_{3} = p^{\alpha/2}u \in\{\pm p^{\alpha/2}, \pm p^{\alpha/2}\zeta_{3}, \pm p^{\alpha/2} (1+\zeta_{3})\}. $

    The condition

    $ b\equiv0\not\equiv\pm p^{\alpha/2}\, (\mathrm{mod}\, 3) $

    requires that $ u = \pm1 $ and thus $ b = 0 $. Moreover,

    $ J_{3}(1, 1) = a = \pm p^{\alpha/2}\equiv2\equiv p\, (\mathrm{mod}\ 3), $

    and hence

    $ J_{3}(1, 1) = -(-p)^{\alpha/2}. $

    As $ f $ is even, by Corollaries 3 and 5, for any $ a, b, c\in\mathbb{Z}/3\mathbb{Z} $ we have

    $ J_{3}(a, b, c) = J_{3}(c, b, a) = J_{3}(a, -a-b-c, c). $

    Along with Proposition 7, Corollary 8, and Lemma 9, we obtain:

    $ J3(0,0,0)=q3,J3(2,0,1)=J3(1,0,2)=J3(0,2,0)=J3(0,1,0)=2,J3(1,2,0)=J3(0,2,1)=J3(1,0,0)=J3(0,0,1)=1ζg3,J3(2,1,0)=J3(0,1,2)=J3(2,0,0)=J3(0,0,2)=1ζ2g3,J3(1,1,0)=J3(0,1,1)=L+3M2+3Mζ3ζg3,J3(2,2,0)=J3(0,2,2)=L+3M2+3Mζ23ζ2g3,J3(1,1,1)=J3(1,0,1)=ζ2g3(L+3M2+3Mζ3)1,J3(2,2,2)=J3(2,0,2)=ζg3(L+3M2+3Mζ23)1.
    $

    Next, we calculate

    $ J_{3}(1, 2, 1) = \overline{J_{3}(2, 1, 2)}, $

    following a similar approach to that for $ J_{2}(1, 1, 1) $ in Section 3. Note that $ f $ is even and

    $ \mathbb{F}_{q}^{*} = \{\gamma^{m}\mid0\leqslant m\leqslant3f-1\}. $

    So

    $ J3(1,2,1)=vFqχ23(v)χ3(v21)=3f1m=0χ3(γ2m)χ3(γ2m1)=232f1m=0χ3(γ2m1)χ3(γ2m).
    $
    (4.2)

    Also note that

    $ J3(1,1)=3f1n=0χ3(γn1)χ3(γn)=32f1m=0χ3(γ2m1)χ3(γ2m)+32f1m=0χ3(γ2m+11)χ3(γ2m+1).
    $
    (4.3)

    Let us choose the primitive complex $ 6 $-th root of unity $ \zeta_{6} $ such that $ \zeta_{6}^{2} = \zeta_{3} $. As $ \zeta_{6}^{3} = -1 $,

    $ \zeta_{6} = -\zeta_{6}^{-2} = -\zeta_{3}^{-1} = -\zeta_{3}^{2} = 1+\zeta_{3}. $

    Write $ J_{6}(i, j) $ ($ i, j\in\mathbb{Z}/6\mathbb{Z} $) for the {classical} Jacobi sums of order $ 6 $ with respect to $ \gamma $ and $ \zeta_{6} $. Let $ \chi_{6} $ be the character of $ \mathbb{F}_{q} $ defined by $ \chi_{6}(0) = 0 $ and

    $ \chi_{6}(\gamma^{m}) = \zeta_{6}^{m} \quad (\text{for any}\; m\in\mathbb{Z}). $

    Note that $ \chi_{6}^{2} = \chi_{3} $ on $ \mathbb{F}_{q} $. By definition,

    $ J6(2,5)=3f1n=0χ26(γn1)χ56(γn)=3f1n=0χ3(γn1)χ56(γn)=32f1m=0χ3(γ2m1)χ56(γ2m)+32f1m=0χ3(γ2m+11)χ56(γ2m+1).
    $

    Since

    $ χ56(γ2m)=χ3(γ2m),χ56(γ2m+1)=ζ56χ3(γ2m)=ζ3χ3(γ2m)=χ3(γ2m+1),
    $

    we obtain

    $ J6(2,5)=32f1m=0χ3(γ2m1)χ3(γ2m)32f1m=0χ3(γ2m+11)χ3(γ2m+1).
    $
    (4.4)

    Theorem 15. Let $ g = \mathrm{ind}_{\gamma}(2) $, and $ (L, M) $ be defined as in Lemma 14 for

    $ q = p^{\alpha} = 3f+1 $

    with $ p $ odd prime and $ \alpha\geqslant1 $. Then

    $ J_{3}(1, 2, 1) = J_{3}(1, 1)+\overline{J_{6}(1, 1)} = {L,if g0(mod3),L+9M2ζ3,if g1(mod3),L+9M2(1+ζ3),if g2(mod3).
    $

    Moreover, if $ p\equiv2\, (\mathrm{mod}\, 3) $, then $ g\equiv0\, (\mathrm{mod}\, 3) $ and

    $ J_{3}(1, 2, 1) = L = -2(-p)^{\alpha/2}. $

    Proof. By Eqs (4.2)–(4.4), we have

    $ J_{3}(1, 2, 1) = J_{3}(1, 1)+J_{6}(2, 5). $

    We know

    $ J_{3}(1, 1) = \frac{L+3M}{2}+3M\zeta_{3}. $

    Also note that

    $ J_{6}(2, 5) = J_{6}(5, 5) = \sigma_{5}(J_{6}(1, 1)) = \overline{J_{6}(1, 1)}. $

    To determine $ J_{6}(1, 1) $, we note that

    $ J_{6}(1, 1)\in\mathbb{Z}[\zeta_{6}] = \mathbb{Z}[\zeta_{3}] $

    by definition. Let

    $ J_{6}(1, 1) = \frac{E+F}{2}+F\zeta_{3} $

    for some $ E, F\in\mathbb{Z} $. By finite Fourier series expansions, $ E $ and $ F $ are indeed $ \mathbb{Z} $-linear combinations of classical cyclotomic numbers $ A_{ij}^{(6)} $ (with $ 0\leqslant i, j\leqslant5 $).

    For $ p\equiv1\, (\mathrm{mod}\, 3) $, [14, Theorem 2] provides

    $ J_{6}(1, 1) = \frac{(-E+F)\zeta_{3}-(E+F)\zeta_{3}^{2}}{2} = \frac{E+F}{2}+F\zeta_{3}, $

    where

    $ (E,F)={(L,3M),if g0(mod3),(L9M2,L3M2),if g1(mod3),(L+9M2,L3M2),if g2(mod3).
    $
    (4.5)

    So we only need to show that Eq (4.5) also holds for $ p\equiv2\, (\mathrm{mod}\, 3) $.

    In an earlier work, Dickson [7, §17–19] established Eq (4.5) in the setting of $ q = p $, utilizing certain linear relations among cyclotomic numbers $ A_{ij}^{(6)} $ (with $ 0\leqslant i, j\leqslant5 $). These relations, in fact, are valid for $ A_{ij}^{(6)} $ over any finite field $ \mathbb{F}_{q} $ with $ q\equiv1\, (\mathrm{mod}\, 6) $. Thus, Dickson's proof of Eq (4.5) is universally applicable to all fields $ \mathbb{F}_{q} $ with $ q\equiv1\, (\mathrm{mod}\, 6) $.

    Note that the Jacobi sum $ R(11) $ in [7, §17–19] corresponds to our $ (-1)^{f}J_{6}(1, 1) $. Consequently, the pair $ (E, F) $ in [7, §18] (with $ f $ even) matches our pair, while the pair in §19 (with $ f $ odd) is our $ (-E, -F) $.

    Using Eq (4.5), we obtain

    $ J3(1,2,1)=J3(1,1)+J6(2,5)=J3(1,1)+¯J6(1,1)=L+3M2+3Mζ3+E+F2+Fζ23=L+3M+EF2+(3MF)ζ3={L,if g0(mod3),L+9M2ζ3,if g1(mod3),L+9M2(1+ζ3),if g2(mod3).
    $

    For the special case of $ p\equiv2\, (\mathrm{mod}\, 3) $, we have

    $ 2\equiv2^{1+2(p-1)}\, (\mathrm{mod}\, p) $

    by Fermat's little theorem. As

    $ 1+2(p-1)\equiv0\, (\mathrm{mod}\, 3), $

    let

    $ 1+2(p-1) = 3t $

    with some $ t\in\mathbb{Z} $. Then

    $ g = \mathrm{ind}_{\gamma}(2) = \mathrm{ind}_{\gamma}(2^{3t})\equiv3tg\, (\mathrm{mod}\, (q-1)). $

    Since $ 3\mid(q-1) $, we have $ g\equiv0\, (\mathrm{mod}\, 3) $, and hence

    $ J_{3}(1, 2, 1) = L = -2(-p)^{\alpha/2}. $

    This completes the proof.

    Remark 16. For $ p\equiv1\, (\mathrm{mod}\, 3) $ and $ q = p^{\alpha} $, Acharya and Katre [14, Theorem 2] proved that $ (E, F) $ is the unique solution of the Diophantine system

    $ \left\{ 4q=E2+3F2,E1(mod3),pE,Fg(mod3),γq13(E+F)/(E+F)(modp).
    \right. $

    The remaining issue now is to evaluate

    $ J_{3}(1, 1, 2) = J_{3}(2, 1, 1) = J_{3}(2, 2, 1) = J_{3}(1, 2, 2). $

    Here, the second equality stems from Corollary 5, while the other two follow from Corollary 3. First, proving it to be an integer is straightforward.

    Proposition 17. $ J_{3}(1, 1, 2) $ is an integer.

    Proof. By Corollaries 3, 5 and Lemma 9,

    $ J_{3}(1, 1, 2) = J_{3}(2, 1, 1) = J_{3}(2, 2, 1) = \sigma_{2}(J_{3}(1, 1, 2)) = \overline{J_{3}(1, 1, 2)}. $

    So $ J_{3}(1, 1, 2)\in\mathbb{R} $. By definition,

    $ J_{6}(1, 1)\in\mathbb{Z}[\zeta_{6}] = \mathbb{Z}[\zeta_{3}]. $

    Let

    $ J_{3}(1, 1, 2) = a+b\zeta_{3} $

    for some $ a, b\in\mathbb{Z} $, whose imaginary part is $ 0 $ only if $ b = 0 $. So $ J_{3}(1, 1, 2) = a\in\mathbb{Z} $.

    We have explored various approaches, but the exact value of $ J_{3}(1, 1, 2) $ still eludes us. Finally, let us elaborate on two unsuccessful ideas regarding its calculation:

    (1) Drawing from the prior computation of $ J_{3}(1, 2, 1) $, we might guess: Can $ J_{3}(1, 1, 2) $ be expressed as a linear combination of $ J_{6}(i, j) $, with coefficients independent of $ \mathbb{F}_{q} $? More precisely, let us consider 36 absolute constants $ c_{ij}\in\mathbb{C} $ (for $ 0\leqslant i, j\leqslant5 $) such that the equality

    $ J_{3}(1, 1, 2) = \sum\limits_{0\leqslant i, j\leqslant5}c_{ij}J_{6}(i, j) $

    holds for any finite field $ \mathbb{F}_{q} $ with $ q\equiv1\:(\mathrm{mod}\, 6) $. For each $ q $, this equality yields a linear relation among the coefficients $ c_{ij} $. Unfortunately, computational solutions (by a computer program) to these linear equations (for a sufficient number of $ q $) reveal that such constants $ c_{ij} $ do not exist.

    (2) For $ v = \gamma^{n}\in\mathbb{F}_{q}^{*} $, we note that

    $ \sum\limits_{i = 0}^{2}\chi_{3}^{i}(v) = \sum\limits_{i = 0}^{2}\chi_{3}^{i}(\gamma^{n}) = \sum\limits_{i = 0}^{2}(\zeta_{3}^{n})^{i} = {3,if 3n,0,otherwise.
    $

    Also note that

    $ J(1,1,2)=J(2,1,1)=J(2,2,1),J(2,0,1)=2.
    $

    So

    $ 2J(1,1,2)2=J(2,1,1)+J(2,2,1)+J(2,0,1)=vFqχ23(v1)(2i=0χi3(v))χ3(v+1)=3f1m=0χ23(γ3m1)χ3(γ3m+1)=3f1m=1χ3(γ3m+1γ3m1)=3f1m=1χ3(2γ3m1+1).
    $

    Similar computations for $ J(1, 1, 2)+J(0, 1, 2)+J(2, 1, 2) $ and $ J(1, 1, 2)+J(1, 1, 0)+J(1, 1, 1) $ yields similar character sums involving cubic elements of $ \mathbb{F}_{q}^{*} $. To evaluate $ J_{3}(1, 1, 2) $, it suffices to evaluate any one of them. Currently, we have not found a good way to compute them in general.

    In this paper, we introduce the trivariate counterparts of classical cyclotomic numbers and Jacobi sums, named "ternary cyclotomic numbers" and "ternary Jacobi sums". We present their basic properties that mirror those of the classical cyclotomic numbers and Jacobi sums. In Section 3 we provide explicit evaluations for all ternary Jacobi sums (Theorem 12) and ternary cyclotomic numbers (Theorem 13) of order $ e = 2 $. Section 4 delivers near-complete results for order $ e = 3 $, with the exception of the elusive integer $ J_{3}(1, 1, 2) $ for us. To solve the cyclotomic problem for ternary cyclotomic numbers of order $ 3 $, one only needs to calculate $ J_{3}(1, 1, 2) $. {Determining the precise value of the integer $ J_{3}(1, 1, 2) $ stands as our initial objective for upcoming endeavors. In the future, we will investigate more general methods for the ternary cyclotomic problem, as well as its potential applications in other fields.}

    Zhichao Tang: writing–original draft, conceptualization, investigation, software, validation; Xiang Fan: writing–review & editing, methodology, supervision, funding acquisition. All authors have read and agreed to the published version of the manuscript.

    The authors gratefully acknowledge the editor and the anonymous reviewers for their insightful feedback, greatly enhancing the manuscript. This work is funded by Guangzhou Science and Technology Programme (No. 202102021218). The second author was also sponsored by the National Natural Science Foundation of China (No. 11801579) and Guangdong Basic and Applied Basic Research Foundation (No. 2020B1515310017).

    All authors declare no conflicts of interest in this paper.

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