Knowledge Organisers for learning: Examples, non-examples and concept maps in university mathematics

1.   Introduction

The Laplace operator $\Delta$ acting on smooth functions of a Riemannian manifold $\left(M^{n}, g\right)$, is defined by $\Delta f = \text{div}\left(\nabla f\right)$, where $\nabla f$ is the gradient of $f$. It is known that $\Delta$ is a self adjoint elliptic operator with respect to the inner product $\left(, \right)$ defined on the algebra of smooth functions $C^{\infty }\left(M^{n}\right)$ with compact support by

where $dV_{g}$ is the volume form on $M^{n}$ with respect to the metric $g$. If $\Delta f = -\lambda f$, for a constant $\lambda$, then $\lambda$ is said to be an eigenvalue of the Laplace operator $\Delta$, the negative sign in the definition chosen so that for non-zero eigenvalue $\lambda$, $\lambda > 0$. The set of eigenvalues $\lambda _{i}$ of the Laplace operator $\Delta$ on a Riemannian manifold $\left(M^{n}, g\right)$ is called the spectrum of $\left(M^{n}, g\right)$. Spectra of known Riemannian manifolds such as the sphere $S^{n}(c)$ and the real projective space $RP^{n}$ are known, and a nice description could be found in ^[2] (Chapter 2, Section 5). Two Riemannian manifolds of same dimension $\left(M^{n}, g\right)$ and $\left(\overline{M}^{n}, \overline{g}\right)$ having the same spectra are said to be isospectral manifolds, whereas they are said to be isometric if there exists a diffeomorphism $\phi : M^{n}\rightarrow \overline{M}^{n}$ that preserves the metric $\phi ^{\ast }\left(\overline{g}\right) = g\circ \phi$, that is, $\phi$ is an isometry. In the mid Nineteenth century, it was an open question whether isospectral Riemannian manifolds are isometric. We see that physicists followed this question under the topic "On hearing the shape of a drum" or "Can one hear the shape of a drum?" (cf. ^[5,11]), while Milnor constructed two flat tori in dimension $16$, which are isospectral but not isometric (cf. ^[12]). In ^[12], the author constructed two 2-dimensional compact manifolds of constant negative curvature which are isospectral and not isometric. Further, Ejiri constructed two non-flat compact Riemannian manifolds which are isospectral but not isometric (cf. ^[6]). This initiated an interest in comparing spectra of two Riemannian manifolds, for further results in this direction, we refer to (cf. ^{[1,7,8,9,10,15]}). One of the natural questions is under what conditions two isospectral Riemannian manifolds are isometric? We know that the $n$-sphere $S^{n}(c)$ has first non-zero eigenvalue $\lambda_{1} = nc$, and, supposing an $n$-dimensional Riemannian manifold $\left(M^{n}, g\right)$ also has non-zero eigenvalue $\lambda _{1} = nc$, we reduce the above general question to the following specific question: Under what condition is an $n$-dimensional Riemannian manifold $\left(M^{n}, g\right)$ that has non-zero eigenvalue $\lambda _{1} = -nc$ for a positive constant $c$ isometric to $S^{n}(c)$? In order to answer this question, naturally, we need a smooth function $\rho$ on an $n$-dimensional Riemannian manifold $\left(M^{n}, g\right)$ that satisfies $\Delta \rho = -nc\rho$. In order to address this issue of finding a smooth function, we consider two situations:

(ⅰ) Considering $\left(M^{n}, g\right)$ as an isometrically immersed hypersurface in the Euclidean space $\left(R^{n+1}, \langle, \rangle \right)$ with isometric immersion $f:\left(M^{n}, g\right) \rightarrow \left(R^{n+1}, \langle, \rangle \right)$, though there are several known functions on the hypersurface $\left(M^{n}, g\right)$, namely, the mean curvature function, the scalar curvature function, as well as the support function, but they cannot be eigenfunctions of the Laplace operator corresponding to a non-zero eigenvalue, while $\left(M^{n}, g\right)$ is isometric to the sphere $S^{n}(c)$. Then, we go for the alternative, namely, for a constant unit vector field $\overrightarrow{a}$ on the Euclidean space $R^{n+1}$, we get a smooth function $\rho = \langle f, \overrightarrow{a}\rangle$ on the hypersurface $\left(M^{n}, g\right)$, which we require to satisfy $\Delta \rho = -nc\rho$, $c > 0$.

(ⅱ) We seek the Riemannian manifold $\left(M^{n}, g\right)$, and there exists a non-trivial concircular vector field $\xi$ with potential function $\rho$ and we require that the potential function $\rho$ satisfies $\Delta\rho = -nc\rho$, $c > 0$.

In this article, we explore the above two situations that an $n$-dimensional Riemannian manifold $\left(M^{n}, g\right)$ has an eigenvalue of the Laplace operator same as the first non-zero eigenvalue of the sphere $S^{n}(c)$, and find an additional condition so that these two are isometric (See Theorems 1, 2, 3). In the language of physics, "yes, we could hear a node through an apparatus to predict the shape of a drum".

2.   Preliminaries

On an $n$-dimensional Riemannian manifold $\left(M^{n}, g\right)$, denote by $\nabla$ the Riemannian connection. The curvature tensor $R$ of $\left(M^{n}, g\right)$ is given by

where $\Gamma \left(TM^{n}\right)$ is the space of smooth sections of the tangent bundle $TM^{n}$. The Ricci tensor $Ric$ of $\left(M^{n}, g\right)$ is a symmetric tensor given by

where $\left\{ e_{1}, .., e_{n}\right\}$ is a local orthonormal frame (or lof) on $\left(M^{n}, g\right)$. The Ricci operator $Q$ of $\left(M^{n}, g\right)$ is related to the Ricci tensor by

and therefore $Q$ is a symmetric $(1, 1)$ tensor field. The scalar curvature $\tau$ of the Riemannian manifold $\left(M^{n}, g\right)$ is given by

The following formula is well known (cf. ^[2]),

where $\nabla \tau$ is the gradient of $\tau$, and the covariant derivative is given by

Given a smooth function $f:M^{n}\rightarrow R$ on a Riemannian manifold $\left(M^{n}, g\right)$, the Laplace operator $\Delta$ acts on $f$, given by

where $\nabla f$ is the gradient of $f$ and

If $\left(M^{n}, g\right)$ is closed, then Stokes's theorem implies

where $dV_{g}$ is the volume element of $\left(M^{n}, g\right)$.

Given a symmetric $(1, 1)$ tensor field $T$ on an $n$-dimensional Riemannian manifold $\left(M^{n}, g\right)$ with trace $t$, that is,

then the Cauchy–Schwartz inequality is

where

Moreover, the equality in (2.3) holds if and only if

where $I$ is the identity $(1, 1)$ tensor.

Suppose on an $n$-dimensional closed Riemannian manifold $\left(M^{n}, g\right)$ there is a smooth function $f$ that satisfies $\Delta f = -ncf$ for a constant $c$. Then we have $f\Delta f = -ncf^{2}$. Then, integrating by parts, the last equation leads to

3.   Compact hypersurfaces in a Euclidean space with non-zero eigenvalue $nc$ of the Laplace operator

In this section, we consider an $n$-dimensional Riemannian manifold $\left(M^{n}, g\right)$ that admits an isometric immersion $f:\left(M^{n}, g\right)\rightarrow \left(R^{n+1}, \langle, \rangle \right)$ into the Euclidean space $\left(R^{n+1}, \langle, \rangle \right)$, where $\langle, \rangle$ is the Euclidean inner product. We denote by $\varsigma$ the unit normal to the hypersurface $\left(M^{n}, g\right)$, and by $B$ the shape operator with respect to the isometric immersion $f$. Then, denoting the Euclidean connection on $\left(R^{n+1}, \langle, \rangle \right)$ by $\overline{\nabla }$ and the Riemannian connection on the hypersurface $\left(M^{n}, g\right)$ by $\nabla$, we have the following fundamental equations for the hypersurface (cf. ^[2])

and

The mean curvature $\beta$ of the hypersurface $\left(M^{n}, g\right)$ is given by

where $\left\{ e_{1}, .., e_{n}\right\}$ is a local orthonormal frame on $\left(M^{n}, g\right)$.

The curvature tensor, the Ricci tensor, and the scalar curvature of the hypersurface $\left(M^{n}, g\right)$ are given by (cf. ^[2])

and

Also, the shape operator $B$ has the following wonderful property (Codazzi equation of the hypersurfaces in flat spaces)

Using Eqs (3.3) and (3.4) as well as the symmetry of the shape operator, we have for any $X\in \Gamma \left(TM^{n}\right)$

Note that $Be_{l} = \sum\limits_{j}g\left(Be_{l}, e_{j}\right) e_{j}$ and $\nabla _{X}e_{l} = \sum\limits_{k}\omega _{l}^{k}(X)e_{k}$, where $g\left(Be_{l}, e_{j}\right)$ is symmetric while the connection forms $\omega_{l}^{k}$ are skew symmetric. Therefore,

Thus, above equation becomes

that is, we have

Treating the isometric immersion $f:\left(M^{n}, g\right) \rightarrow \left(R^{n+1}, \langle, \rangle \right)$ as a position vector of points of $M^{n}$ in $R^{n+1}$, and defining $\sigma = \langle f, \varsigma \rangle$, called the support function of the hypersurface $\left(M^{n}, g\right)$, we express $f$ as

where $\xi \in \Gamma \left(TM^{n}\right)$ is tangential to $\left(M^{n}, g\right)$. Differentiating equation (2.4), while using Eqs (3.1) and (3.2), we have upon equating the tangential and normal parts

and

Taking a constant unit vector field $\overrightarrow{a}$ on the Euclidean space $R^{n+1}$ (for instance a coordinate vector field), we define a smooth function $h$ on the hypersurface $\left(M^{n}, g\right)$, by $h = \langle \overrightarrow{a}, \varsigma \rangle$. Denoting the tangential part of $\overrightarrow{a}$ to the hypersurface $\left(M^{n}, g\right)$ by $\zeta$, we have

Differentiating the above equation with respect to $X\in \Gamma \left(TM^{n}\right)$, while using Eqs (3.1) and (3.2), we have upon equating the tangential and normal parts

and

Now, we prove the main result of this section.

Theorem 1. An $n$-dimensional compact and connected isometrically immersed hypersurface $f:(M^{n}, g)\rightarrow \left(R^{n+1}, \langle, \rangle \right)$ in the Euclidean space $\left(R^{n+1}, \langle, \rangle \right)$ with mean curvature $\beta$ and a constant unit vector $\overrightarrow{a} = \zeta +h\varsigma$ on $R^{n+1}$, where the function $\rho = \langle f, \overrightarrow{a}\rangle$ satisfies $\Delta \rho = -nc\rho$ for a positive constant $c$, is isometric to the sphere $S^{n}(c)$ if and only if the Ricci curvature $Ric\left(\zeta, \zeta \right)$ satisfies

Proof. Consider an $n$-dimensional compact and connected Riemannian manifold $(M^{n}, g)$ that admits an isometric immersion $f:(M^{n}, g)\rightarrow \left(R^{n+1}, \langle, \rangle \right)$ in the Euclidean space $\left(R^{n+1}, \langle, \rangle \right)$ with shape operator $B$, mean curvature $\beta$, and a constant unit vector $\overrightarrow{a} = \zeta +h\varsigma$ on $R^{n+1}$, where the function $\rho = \langle f, \overrightarrow{a}\rangle$ satisfies

for a positive constant $c$. Also, the Ricci curvature $Ric\left(\zeta, \zeta \right)$ satisfies

Now, differentiating $\rho = \langle f, \overrightarrow{a}\rangle$in the direction of $X\in \Gamma \left(TM^{n}\right)$, we get $X\left(\rho \right) = \langle X, \overrightarrow{a}\rangle = \langle f, \zeta \rangle$. This gives us the gradient of $\rho$ as

The Hessian operator $\mathcal{H}_{\rho }$ of the function $\rho$ is given by $\mathcal{H}_{\rho }X = \nabla _{X}\nabla \rho$, $X\in \Gamma \left(TM^{n}\right)$, and using Eqs (3.8) and (3.12), we arrive at

Taking the trace in the above equation and taking account of Eq (3.10), we get

and therefore, through Eq (3.13), we conclude

From the above equation, we reach

Next, using Eq (3.8), we have

which yields

Inserting $\text{div}\zeta = nh\beta$ (an outcome of Eq (3.8)), in the above equation, we arrive at

Recalling the following well known integral formula (cf. ^[16])

and integrating Eq (3.16) while using the above integral formula, we conclude

Using Eq (3.8), we compute

and consequently, we have

Thus, inserting above equation and $\text{div}\zeta = nh\beta$ in Eq (3.17) confirms

that is,

For a local orthonormal frame $\left\{ e_{1}, .., e_{n}\right\}$, we have

Utilizing the above equation in (3.18), we arrive at

Inserting from Eq (3.15) in the above equation, we have

and treating it with inequality (3.11) allows us to reach the conclusion

Note that $\rho$ satisfies Eq (3.10), that is, $\Delta \rho = -nc\rho$ for a non-zero constant $c$. We claim that $\rho$ can not be a constant, for if it were, Equation (3.10) will imply $\rho = 0$, and then Eq (3.14) will imply $h\beta = 0$. Note that by Eq (3.6) we have $\text{div}\xi = n\left(1+\sigma \beta \right)$, and therefore on the compact hypersurface $\left(M^{n}, g\right)$, we have (cf. ^[4])

which does not allow $\beta = 0$. Hence, in the situation where $\rho$ is a constant, we have $h = 0$, and, also, by Eq (3.12), $\zeta = 0$, and by virtue of Eq (3.7), we will reach the conclusion $\overrightarrow{a} = 0$, contrary to the assumption that $\overrightarrow{a}$ is a unit vector. Thus, $\rho$ is a non-constant function which satisfies Obata's equation (3.8) proving that $\left(M^{n}, g\right)$ is isometric to the sphere $S^{n}(c)$ (cf. ^[13,14]).

Conversely, consider the isometric immersion $f:S^{n}(c)\rightarrow \left(R^{n+1}, \langle, \rangle \right)$ of the sphere $S^{n}(c)$ in the Euclidean space $\left(R^{n+1}, \langle, \rangle \right)$ given by $f(x) = x$. Then, the unit normal $\varsigma = \sqrt{c}f$, the shape operator $B = -\sqrt{c}I$, and the mean curvature $\beta = -\sqrt{c}$. Consider the unit constant vector $\overrightarrow{a}$ given by the first Euclidean coordinate vector field, that is,

where $\zeta$ is tangent to the sphere $S^{n}(c)$ and $h = \langle \overrightarrow{a}, \varsigma \rangle = \langle \overrightarrow{a}, \sqrt{c}f\rangle = \sqrt{c}\langle f, \overrightarrow{a} \rangle$. Thus, defining $\rho = \langle f, \overrightarrow{a}\rangle$, we have

Now, differentiating (3.19) in the direction of $X\in \Gamma \left(TS^{n}(c)\right)$ and equating the tangential and normal parts, we confirm

Using Eqs (3.19) and (3.20), we see $\sqrt{c}\zeta = \sqrt{c}\nabla \rho$, that is,

which, in view of the first equation in (3.20), provides

Finally, using Eq (3.21), the Ricci curvature $Ric\left(\zeta, \zeta \right)$ of the sphere $S^{n}(c)$ is given by

However, Equations (3.19) and (3.22) confirm

Now, integrating Eq (3.23) while using above equation yields

Hence, the converse also holds. □

Next, we prove the following result for the complete hypersurface $(M^{n}, g)$ of the Euclidean space $\left(R^{n+1}, \langle, \rangle \right)$.

Theorem 2. An $n$-dimensional complete and simply connected isometrically immersed hypersurface $f:(M^{n}, g)\rightarrow \left(R^{n+1}, \langle, \rangle \right)$, $n > 1$, in the Euclidean space $\left(R^{n+1}, \langle, \rangle \right)$ with mean curvature $\beta$ and a constant unit vector $\overrightarrow{a} = \zeta +h\varsigma$ on $R^{n+1}$, where the function $h = \langle \overrightarrow{a}, \varsigma \rangle \neq 0$ satisfies $\Delta h = -nch$ for a positive constant $c$, is isometric to the sphere $S^{n}(c)$ if and only if the mean curvature $\beta$ is a constant along the integral curves of $\zeta$ and $\beta^{2}\geq c$ holds.

Proof. Consider an $n$-dimensional complete and simply connected Riemannian manifold $(M^{n}, g)$ that admits an isometric immersion $f:(M^{n}, g)\rightarrow \left(R^{n+1}, \langle, \rangle \right)$ in the Euclidean space $\left(R^{n+1}, \langle, \rangle \right)$ such that the function $h = \langle \overrightarrow{a}, \varsigma \rangle \neq 0$ satisfies

with mean curvature $\beta$ satisfying

and

We use Eqs (3.5) and (3.8), the symmetry of the shape operator $B$, and a local orthonormal frame $\left\{e_{1}, .., e_{n}\right\}$ in order to find $\text{div}\left(B\zeta \right)$,

Using Eq (3.15), we get

Now, taking the divergence in Eq (3.9) and using the above equation with Eq (3.24) yields

that is,

Combining above equation with inequality (3.25), while keeping in view Cauchy–Schwartz's inequality $\left\Vert B\right\Vert ^{2}\geq n\beta ^{2}$, we get

Since $h\neq 0$, we get

which, being an inequality in Cauchy–Schwartz's inequality $\left\Vert B\right\Vert ^{2}\geq n\beta ^{2}$, we must have

The above equation implies

which gives

Combining the above equation with Eq (3.5) yields

and, as $n > 1$, we get that $\beta$ is a constant, and by virtue of Eqs (3.26) and (3.27), we have

Now, using Eq (3.27) in the expression of the curvature tensor of the hypersurface with the above equation gives

that is, $\left(M^{n}, g\right)$ is a complete and simply connected space of constant positive curvature $c$. Hence, $\left(M^{n}, g\right)$ is isometric to $S^{n}(c)$. The converse is trivial. □

4.   Concircular vector field with potential function $\sigma$ an eigenfunction of Laplace operator with eigenvalue $nc$

Consider an $n$-dimensional Riemannian manifold $\left(M^{n}, g\right)$ that possesses a concircular vector field $\xi$ (cf. ^[3]), that is, the vector field satisfies

where $\sigma$ is a smooth function, called the potential function of the concircular vector field. A concircular vector field is said to be non-trivial if the potential function $\sigma \neq 0$. Using Eq (4.1), we immediately have

In this section, we are interested in an $n$-dimensional compact Riemannian manifold $\left(M^{n}, g\right)$ that possesses a non-trivial concircular vector field $\xi$ with potential function $\sigma$ satisfying

where $c > 0$ is a constant, that is, $\sigma$ is an eigenfunction of the Laplace operator with eigenvalue the same as the first non-zero eigenvalue of the sphere $S^{n}(c)$, and we find a condition under which $\left(M^{n}, g\right)$ is isometric to the sphere $S^{n}(c)$.

Before we approach this issue, we first prepare an auxiliary result to prove the main result. First, for Eq (4.1), using Eq (2.1) immediately gives the following expression of the curvature tensor, namely

Taking the trace in the above equation and using Eq (2.2), we reach

and this equation gives the following expression for the Ricci operator $Q$ operating on $\xi$, namely

where $\nabla \sigma$ is the gradient of the potential function $\sigma$.

In the following paragraph, we show that for each concircular vector field $\xi$ on a connected Riemannian manifold $\left(M^{n}, g\right)$ there corresponds a smooth function $f$, which we call a concircular function of the concircular vector field $\xi$. Note that Eq (4.2) implies

and the operator $R\left(X, \xi \right) \xi$, $X\in \Gamma \left(TM^{n}\right)$ is symmetric in $X$, and, therefore, the above equation implies

The above equation implies

and taking the inner product in the above equation with $\nabla \sigma$ and replacing $X$ by $\xi$, we conclude

that is

This proves that vector fields $\nabla \sigma$ and $\xi$ are parallel, and, consequently, there exists a smooth function $f$ such that

We call this function $f$ the concircular function of the concircular vector field $\xi$.

First, we prove the following proposition.

Proposition 1. Let $\xi$ be a non-trivial concircular vector field on an $n$-dimensional compact Riemannian manifold $\left(M^{n}, g\right)$ with potential function $\sigma$ and concircular function $f$. If the potential function $\sigma$ satisfies

for a positive constant $c$, then

Proof. Let $\xi$ be a non-trivial concircular vector field on an $n$-dimensional compact Riemannian manifold $\left(M^{n}, g\right)$ with potential function $\sigma$ and concircular function $f$, and the potential function $\sigma$ satisfies

for a positive constant $c$. Using Eqs (4.2) and (4.3), we have

and integrating the above equation, we confirm

Using the integral formula in ^[16], we have for a vector field $\zeta$ on $\left(M^{n}, g\right)$

Replacing $\zeta$ in the above equation by $\nabla \sigma$ and noting that

we conclude

Thus, we have

and, inserting Eq (4.4), we reach

Now, inserting from Eq (4.4) in the above equation takes us to

Using Eqs (4.3) and (4.4), we have

and using this in Eq (4.5) leads to

Note that taking the divergence on both sides of Eq (4.4) and using Eq (4.2) gives

and combining this with Eq (4.4) allows us to conclude

Using Eqs (4.2), (4.4), and (4.7), we compute

and, integrating the above equation, gives

Inserting the above equation into Eq (4.6) leads to

and combining it with Eq (4.7) yields

□

As a straightforward application of the above result, we have the following theorem.

Theorem 3. An $n$-dimensional compact and connected Riemannian manifold $\left(M^{n}, g\right)$ admits a non-trivial concircular vector field $\xi$ with potential function $\sigma$ and concircular function $f$ such that the potential function $\sigma$ satisfies

for a positive constant $c$, and the concircular function $f$ is a constant along the integral curves of $\xi$ if and only if $\left(M^{n}, g\right)$ is isometric to $S^{n}(c)$.

Proof. Suppose an $n$-dimensional compact and connected Riemannian manifold $\left(M^{n}, g\right)$ admits a non-trivial concircular vector field $\xi$ with potential function $\sigma$ and concircular function $f$ such that the potential function $\sigma$ satisfies

for a positive constant $c$, and the concircular function $f$ is a constant along the integral curves of $\xi$. Then, by Proposition 1 we have

The Cauchy–Schwartz inequality implies

and equality holds if and only if

In view of inequality (4.10) and Eq (4.9), we are ready to conclude the equality

and, therefore, Eq (4.11) holds. Combining Eqs (4.8) and (4.11), we arrive at

Note that the potential function $\sigma$ can not be a constant, for if it were a constant, the above equation would imply $\sigma = 0$, which is contrary to the assumption that $\xi$ is a non-trivial concircular vector field. Thus, Equation (4.12) is Obata's equation, and therefore $\left(M^{n}, g\right)$ is isometric to the sphere $S^{n}(c)$ (cf. ^[13,14]).

Conversely, take a constant unit vector $\overrightarrow{a}$ on the Euclidean space $\left(R^{n+1}, \langle, \rangle \right)$ while treating $S^{n}(c)$ as a hypersurface of $\left(R^{n+1}, \langle, \rangle \right)$ with unit normal $\varsigma$, shape operator $B = -\sqrt{c}I$, and expressing $\overrightarrow{a}$ as

where $\xi$ is tangent to $S^{n}(c)$ and $h = \langle \overrightarrow{a}, \varsigma \rangle$. Differentiating equation (4.13) with respect to $X\in \Gamma \left(TS^{n}(c)\right)$ and equating the tangential and normal parts, we arrive at

This confirms that $\xi$ is a concircular vector field on $S^{n}(c)$ with potential function $\sigma = -\sqrt{c}h$, and the second equation gives $\nabla \sigma = -c\xi$. This proves that the concircular function $f = -c$. Moreover, if the potential function $\sigma = 0$, we get $h = 0$, and by the second equation in (4.14), we get $\xi = 0$. In this case, Eq (4.13) confirms $\overrightarrow{a} = 0$, a contradiction to the fact that $\overrightarrow{a}$ is a unit vector. Thus, the potential function $\sigma \neq 0$, that is, the concircular vector field $\xi$ on $S^{n}(c)$ is non-trivial. Note that $\text{div}\xi = -n\sqrt{c}h = n\sigma$, and, combining it with the equation $\nabla \sigma = -c\xi$, we get $\Delta\sigma = -nc\sigma$ with $c$ a positive constant. Hence, the converse holds. □

5.   Conclusions

We have initiated the study of an $n$-dimensional compact Riemannian manifold $\left(M^{n}, g\right)$ that has an eigenvalue $nc$ for a positive constant $c$ of the Laplace operator the same as the first non-zero eigenvalue of the $n$-sphere $S^{n}(c)$ of constant curvature $c$, and searched for an additional condition under which $\left(M^{n}, g\right)$ is isometric to the sphere $S^{n}(c)$. The main aim was to find an appropriate smooth function on $\left(M^{n}, g\right)$ that will become the eigenfunction of the Laplace operator with eigenvalue $nc$ as seen in Theorems 1 and 3. Naturally, the scope of this study is quite modest, for instance one can consider an $n$-dimensional compact Riemannian manifold $\left(M^{n}, g\right)$ that admits a torse forming vector field $\xi$ (cf. ^[17]). Recall that a torse forming vector field $\xi$ on $\left(M^{n}, g\right)$ satisfies

where $\sigma$ is a smooth function defined on $M^{n}$ called the conformal scalar and $\omega$ is a smooth $1$-form on $M^{n}$ called the generating form of the a torse forming vector field $\xi$. It will be an interesting question to consider torse forming vector field $\xi$ on an $n$-dimensional compact Riemannian manifold $\left(M^{n}, g\right)$ such that its conformal scalar $\sigma$ satisfies $\Delta \sigma = -nc\sigma$ for a positive constant $c$, and find conditions under which $\left(M^{n}, g\right)$ is isometric to $S^{n}(c)$.

We know that the second non-zero eigenvalue of the sphere $S^{n}(c)$ is given by $\lambda _{2} = 2(n+1)c$, and another aspect of our work could be, if there is a smooth function $f$ on an $n$-dimensional compact Riemannian manifold $\left(M^{n}, g\right)$ such that $\Delta f = -2(n+1)cf$, that is, $\left(M^{n}, g\right)$ has an eigenvalue same as second non-zero eigenvalue of the sphere $S^{n}(c)$, to find additional condition on $(M^{n}, g)$ so that $(M^{n}, g)$ is isometric to $S^{n}(c)$.

Author contributions

Sharief Deshmukh: Conceptualization, Methodology, Writing-original draft, Writing-review and editing, Supervision; Amira Ishan: Conceptualization, Methodology, Writing-review and editing; Olga Belova: Formal analysis, Writing-original draft, Writing-review and editing. All authors have read and agreed to the published version of the manuscript.

Acknowledgments

The authors would like to acknowledge the Deanship of Graduate Studies and Scientific Research, Taif University for funding this work.

Conflict of interest

The authors declare no conflicts of interest in this paper.