Hodge Decomposition of Conformal Vector Fields on a Riemannian Manifold and Its Applications

Abstract

For a compact Riemannian m-manifold $(M^m,g),m>1,$ endowed with a nontrivial conformal vector field $\zeta$ with a conformal factor $\sigma$, there is an associated skew-symmetric tensor $\phi$ called the associated tensor, and also, $\zeta$ admits the Hodge decomposition $\zeta =\overline{\zeta }+\nabla \rho$, where $\overline{\zeta }$ satisfies $\mathrm{div}\overline{\zeta }=0$, which is called the Hodge vector, and $\rho$ is the Hodge potential of $\zeta$. The main purpose of this article is to initiate a study on the impact of the Hodge vector and its potential on $M^m$. The first result of this article states that a compact Riemannian m-manifold $M^m$ is an m-sphere $S^m\left(c\right)$ if and only if (1) for a nonzero constant c, the function $-\sigma /c$ is a solution of the Poisson equation $\Delta \rho =m\sigma$, and (2) the Ricci curvature satisfies $Ric\left(\overline{\zeta },\overline{\zeta }\right)\ge {∥\phi ∥}^2$. The second result states that if $M^m$ has constant scalar curvature $\tau =m(m-1)c>0$, then it is an $S^m\left(c\right)$ if and only if the Ricci curvature satisfies $Ric\left(\overline{\zeta },\overline{\zeta }\right)\ge {∥\phi ∥}^2$ and the Hodge potential $\rho$ satisfies a certain static perfect fluid equation. The third result provides another new characterization of $S^m\left(c\right)$ using the affinity tensor of the Hodge vector $\overline{\zeta }$ of a conformal vector field $\zeta$ on a compact Riemannian manifold $M^m$ with positive Ricci curvature. The last result states that a complete, connected Riemannian manifold $M^m$, $m>2$, is a Euclidean m-space if and only if it admits a nontrivial conformal vector field $\zeta$ whose affinity tensor vanishes identically and $\zeta$ annihilates its associated tensor $\phi$.

1. Introduction

A smooth vector field $\zeta$ on a Riemannian manifold $(M^m,g)$ is called a conformal vector field if its local flow consists of local conformal transformation; equivalently, it obeys

$${\mathcal{L}}_{\zeta }g=2\sigma g,$$

(1)

where ${\mathcal{L}}_{\zeta }$ stands for Lie differentiation and $\sigma$ is a smooth function on M called the conformal factor of $\zeta$ (cf. [1,2]). If the conformal factor $\sigma =0$, then $\zeta$ is called a Killing vector field. A conformal vector field $\zeta$ is called nontrivial if it is non-Killing, that is, if the conformal factor $\sigma \ne 0$. Note that if $M^m$ is compact, then the conformal vector field $\zeta$ admits a Hodge decomposition (cf. [3]):

$$\zeta =\overline{\zeta }+\nabla \rho ,$$

(2)

where $\overline{\zeta }$ is a divergence-free vector field called the Hodge vector of $\zeta$, and $\nabla \rho$ is the gradient of a smooth function $\rho$, called the Hodge potential, on $M^m$.

Consider a conformal vector field $\zeta$ on a Riemannian manifold $\left(M^m,g\right)$. Let $\eta$ be the 1-form dual to $\zeta$, that is, $\eta \left(U\right)=g\left(\zeta ,U\right)$, for any vector field U on $M^m$. Then there exists a skew-symmetric $\left(1,1\right)$-tensor $\phi$ associated with $\zeta$, given by

$$d\eta \left(U,V\right)=g\left(\phi U,V\right),U,V\in \mathsf{\Gamma }\left(T{M}^m\right),$$

(3)

This $\left(1,1\right)$-tensor $\phi$ is called the associated tensor of $\zeta$. Thus, the conformal vector field $\zeta$ on $M^m$ can be represented by the pentaplex $\left(\zeta ,\sigma ,\phi ,\overline{\zeta },\rho \right)$, where $\sigma$ is the conformal factor, $\phi$ is the associated tensor, $\overline{\zeta }$ is the Hodge vector and $\rho$ is the Hodge potential of $\zeta$.

Note that the geometry of Riemannian manifolds admitting conformal vector fields has been investigated by many mathematicians (cf. [4,5,6,7,8,9,10,11,12,13,14,15,16,17,18]). Furthermore, conformal vector fields play very natural roles in the Theory of General Relativity (cf. [8,19,20,21]). In contrast, as far as we know, there are no articles which study the impacts of Hodge vectors and the Hodge potentials of conformal vector fields on compact Riemannian manifolds. Therefore, the purpose of this article is to study the impact of the Hodge vector and Hodge potential of a nontrivial conformal vector field on a compact Riemannian manifold.

Next, we discuss some basic properties of conformal vector fields on an m-sphere $S^m\left(c\right)$. Let us consider $S^m\left(c\right)$ as a hypersphere of a Euclidean $(m+1)$-space ${\mathbb{E}}^{m+1}$ with a unit normal N and shape operator $A=-\sqrt{c}I$. We write the coordinate vector field ${\displaystyle \frac{\partial }{\partial x^1}}$ on the Euclidean $(m+1)$-space ${\mathbb{E}}^{m+1}$ as

$${\displaystyle \frac{\partial }{\partial x^1}}=\zeta +fN,f=⟨{\displaystyle \frac{\partial }{\partial x^1}},N⟩,$$

(4)

where $x^1,\dots ,x^{m+1}$ are Euclidean coordinates, and $⟨,⟩$ denotes the inner product associated with the Euclidean metric on ${\mathbb{E}}^{m+1}$.

Let g be the induced metric on $S^m\left(c\right)\subset {\mathbb{E}}^{m+1}$ and let ${\nabla }_X$ be the covariant derivative with respect to the vector field X on $S^m\left(c\right)$. Then, it follows from (4) that

$${\nabla }_X\zeta =-\sqrt{c}fX,\nabla f=\sqrt{c}\zeta ,$$

(5)

which implies that $\zeta$ is a conformal vector on $S^m\left(c\right)$ with a conformal factor $\sigma =-\sqrt{c}f$, and since it is a closed vector field, the associated tensor vanishes identically, that is $\phi =0$. Also, it follows from the second equation in (5) that the Hodge vector $\overline{\zeta }$ also vanishes, and the Hodge potential is given by $\rho =-\sigma /c$. Therefore, the pentaplex of the conformal vector field $\zeta$ on $S^m\left(c\right)$ is $\left(\zeta ,\sigma ,0,0,-{\displaystyle \frac{1}{c}}\sigma \right)$.

Note that it follows from (5) that the conformal factor of $\zeta$ satisfies $\sigma \ne 0$, since if $\sigma =0$, then $f=0$, which implies that $\zeta =0$ in the second equation in (5). Hence, in view of the first equation in (4), we obtain ${\displaystyle \frac{\partial }{\partial x^1}}=0$, which is a contradiction. Consequently, we know that the conformal vector field $\zeta$ is nontrivial on $S^m\left(c\right)$.

Next, let us consider a conformal vector field $\left(\zeta ,\sigma ,\phi ,\overline{\zeta },\rho \right)$ on a Euclidean m-space ${\mathbb{E}}^m$ given by

$$\zeta =\mathsf{\Psi }+x^2{\displaystyle \frac{\partial }{\partial x^1}}-x^1{\displaystyle \frac{\partial }{\partial x^2}}$$

(6)

where $\mathsf{\Psi }={\displaystyle \sum _{j=1}^m}{x}^j{\displaystyle \frac{\partial }{\partial x^j}}$ is the position vector field and $x^1,\dots ,x^m$ are Euclidean coordinates on ${\mathbb{E}}^m$. From (6), we see that the covariant derivative of $\zeta$ is given by

$${\nabla }_X\zeta =X+\phi X,\mathrm{where}\phi X=\left(X{x}^2\right){\displaystyle \frac{\partial }{\partial x^1}}-\left(X{x}^1\right){\displaystyle \frac{\partial }{\partial x^2}}$$

is a skew-symmetric operator. It follows that $\zeta$ is a conformal vector field with conformal factor $\sigma =1$, and with $\phi$ as its associated operator. Moreover, with the function $\rho ={\displaystyle \frac{1}{2}}{∥\mathsf{\Psi }∥}^2$ on ${\mathbb{E}}^m$, we see that $\mathsf{\Psi }=\nabla \rho$ and $\zeta$ has decomposition

$$\zeta =\overline{\zeta }+\nabla \rho \mathrm{with}\overline{\zeta }=x^2{\displaystyle \frac{\partial }{\partial x^1}}-x^1{\displaystyle \frac{\partial }{\partial x^2}}.$$

Further, it is easy to verify that $\overline{\zeta }$ is divergence-free on ${\mathbb{E}}^m$. Hence, $\zeta$ is a conformal vector field on ${\mathbb{E}}^m$ with the pentaplex $\left(\zeta ,1,\phi ,\overline{\zeta },\rho \right)$.

Now, let us mention the following properties of the Hodge potential $\rho$ and the Hodge vector for the conformal vector field $\left(\zeta ,\sigma ,0,0,\rho \right)$ on $S^m\left(c\right)$.

(i)

The Hodge potential $\rho$ and conformal factor $\sigma$ are related by $\rho =-\sigma /c$ and $\Delta \rho =-\sigma /c$, which satisfy the Poisson equation $\Delta \rho =m\sigma$.

(ii)

Owing to Equation (5), $\rho =-\sigma /c$ and $\sigma =-\sqrt{c}f$, we know that the Hessian of the Hodge potential $\rho$ satisfies $Hess\left(\rho \right)=-c\rho g$ and the Poisson equation $\Delta \rho =-mc\rho$. Also, the Ricci tensor and the scalar curvature of $S^m\left(c\right)$ are given by $Ric=(m-1)cg$ and $\tau =m(m-1)c$, respectively. Moreover, the Hodge potential $\rho$ satisfies the static perfect fluid Equation (cf. [14,16]):

$$\rho Ric-Hess\left(\rho \right)={\displaystyle \frac{1}{m}}\left(\tau \rho -\Delta \rho \right)g.$$

(iii)

The affinity tensor ${\mathcal{L}}_X\nabla$ of a vector field X on a Riemannian manifold $(M^m,g)$ is defined by (cf. [13])

$$\left({\mathcal{L}}_X\nabla \right)\left(U,V\right)={\mathcal{L}}_{\zeta }{\nabla }_{U}V-{\mathcal{L}}_{{\nabla }_{\zeta }U}V-{\nabla }_U{\mathcal{L}}_{\zeta }V,$$

for vector fields $U,V$ on $M^m$, where $\mathcal{L}$ denotes the Lie derivative. Since the Hodge vector $\overline{\zeta }$ of the conformal vector field $\left(\zeta ,\sigma ,0,0,\rho \right)$ on $S^m\left(c\right)$ satisfies $\overline{\zeta }=0$, the affinity tensor of the Hodge vector $\overline{\zeta }$ vanishes identically, that is, ${\mathcal{L}}_{\overline{\zeta }}\nabla =0.$

From the three properties of the Hodge potential $\rho$ and the Hodge vector $\overline{\zeta }$ on $S^m\left(c\right)$ given above, we raise the following three questions for a compact, connected Riemannian manifold $M^m$ which admits a nontrivial conformal vector field $\left(\zeta ,\sigma ,\phi ,\overline{\zeta },\rho \right)$.

Question (i): If the Hodge potential $\rho =-\sigma /c$ satisfies the Poisson equation $\Delta \rho =m\sigma$ on $(M^m,g)$, under what condition is the Riemannian manifold $M^m$ an m-sphere $S^m\left(c\right)$?

Question (ii): If the Hodge potential ρ on $(M^m,g)$ satisfies the static perfect fluid equation

$$\rho Ric-Hess\left(\rho \right)={\displaystyle \frac{1}{m}}\left(\tau \rho -\Delta \rho \right)g,$$

under what condition is the Riemannian manifold $M^m$ an m-sphere $S^m\left(c\right)$?

Question (iii): If the Hodge vector $\overline{\zeta }$ has affinity tensor on $(M^m,g)$, under what condition is the Riemannian manifold $M^m$ an m-sphere $S^m\left(c\right)$?

In Section 3 of this article, we provide our answers to Questions (i) and (ii) as Theorems 1 and 2, respectively. In Section 4, we provide the answer to Question (iii) as Theorem 3. In Section 4, we give an additional result which states that a complete, connected Riemannian manifold $M^m$, $m>2$, is a Euclidean m-space if and only if it admits a nontrivial conformal vector field $\zeta$ whose affinity tensor vanishes identically and $\zeta$ annihilates its associated tensor $\phi$.

2. Preliminaries

For a nontrivial conformal vector $\zeta$ on a connected Riemannian m-manifold $(M^m,g)$ with the conformal factor $\sigma$ and associated tensor $\phi$, using Equations (1) and (3), we have

$${\nabla }_U\zeta =\sigma U+\phi U,U\in \mathsf{\Gamma }\left(T{M}^m\right),U,V\in \mathsf{\Gamma }\left(T{M}^m\right),$$

(7)

where $\mathsf{\Gamma }\left(T{M}^m\right)$ is the space of smooth sections of the tangent bundle $TM$. Taking the trace in Equation (7), we conclude that

$$\mathrm{div}\zeta =m\sigma .$$

(8)

If $(M^m,g)$ is compact, then the conformal vector $\zeta$ by virtue of the Hodge decomposition (2), is represented by the pentaplex $\left(\zeta ,\sigma ,\phi ,\overline{\zeta },\rho \right)$, where the Hodge vector $\overline{\zeta }$ is divergence-free, that is, $\mathrm{div}\overline{\zeta }=0$. Then using Equations (2), (7), and (8), for the conformal vector $\left(\zeta ,\sigma ,\phi ,\overline{\zeta },\rho \right)$, we have

$${\nabla }_U\overline{\zeta }=\sigma U+\phi U-H^{\rho }\left(U\right),\Delta \rho =m\sigma ,$$

(9)

where $H^{\rho }$ is the symmetric $(1,1)$-tensor associated with the Hodge potential $\rho$, defined by

$$H^{\rho }\left(U\right)={\nabla }_U\nabla \rho ,U\in \mathfrak{d}\left(T{M}^m\right),$$

and $\Delta \rho =Tr{H}^{\rho }$ is the Laplacian of the Hodge potential. Using Equation (7), we compute the curvature tensor $R\left(U,V\right)\zeta$ and obtain

$$R(U,V)\zeta =U\left(\sigma \right)V-V\left(\sigma \right)U+\left({\nabla }_U\phi \right)\left(V\right)-\left({\nabla }_v\phi \right)\left(U\right).$$

(10)

Taking the trace in the above equation, we obtain the following expression for the Ricci tensor $Ric\left(U,\zeta \right)$

$$Ric(U,\zeta )=-(m-1)U\left(\sigma \right)-\sum _{j=1}^{m}g\left(U,\left({\nabla }_{u_j}\phi \right)\left(u_j\right)\right),$$

(11)

where $dimM=m$, with $\phi$ being skew-symmetric $tr\phi =0$, and $\left\{u_1,\dots ,u_m\right\}$ is a local orthonormal frame. The Ricci operator S of a Riemannian manifold $\left(M^m,g\right)$ is a symmetric $\left(1,1\right)$-tensor field defined by $Ric\left(U,V\right)=g\left(SU,V\right)$, and thus, from Equation (11), we have

$$S\zeta =-(m-1)\nabla \sigma -\sum _{j=1}^m\left({\nabla }_{u_j}\phi \right)\left(u_j\right).$$

(12)

Note that the 2-form $g\left(\phi U,V\right)={\displaystyle \frac{1}{2}}d\eta \left(U,V\right)$ is a closed 2-form, and therefore, on using Equation (10) in

$$g\left(\left({\nabla }_U\phi \right)\left(V\right),W\right)+g\left(\left({\nabla }_V\phi \right)\left(W\right),U\right)+g\left(\left({\nabla }_W\phi \right)\left(U\right),V\right)=0,$$

we conclude that

$$\left({\nabla }_U\phi \right)\left(V\right)=R\left(U,\zeta \right)V+V\left(\sigma \right)U-g\left(U,V\right)\nabla \sigma .$$

(13)

Here, we would like to point out that equation $\Delta \rho =m\sigma$ in Equation (9) is a Poission equation satisfied by the Hodge potential of the conformal vector $\left(\zeta ,\sigma ,\phi ,\overline{\zeta },\rho \right)$ on a compact Riemannian manifold $\left(M^m,g\right)$, where according to Equation (8), the conformal factor $\sigma$ satisfies

$${\int }_M\sigma dV=0.$$

(14)

We recall Theorem-1 in ([4], p. 4), where the Poisson equation $\Delta \rho =m\sigma$ has a unique solution up to the addition of a constant, which will be useful to us in our study.

In this article, we are also interested in a very prudent differential equation defined on a Riemannian m-manifold $\left(M^m,g\right)$ called the static perfect fluid equation namely

$$fRic-Hess\left(f\right)={\displaystyle \frac{1}{m}}\left(\tau f-\Delta f\right)g,$$

(15)

where $Ric$ is the Ricci tensor, $Hess\left(f\right)\left(U,V\right)=g\left(H^{f}U,V\right)$ is the Hessian of f and $\tau$ is the scalar curvature of $\left(M^m,g\right)$ (cf. [14,16]), which we shall refer to as the SPF equation.

3. New Characterizations of m-Spheres via Conformal Vector Fields

Let $\left(\zeta ,\sigma ,\phi ,\overline{\zeta },\rho \right)$ be a conformal vector field on a compact Riemannian m-manifold $(M^m,g)$ with a conformal factor $\sigma$-associated operator $\phi$, the Hodge vector $\overline{\zeta }$ and the Hodge potential $\rho$. In this section, we shall concentrate on some specific properties of the Hodge potential $\rho$ to analyze its impact on the geometry of $(M^m,g)$. Observe that through Equation (9), we have

$$\Delta \rho =m\sigma ,$$

(16)

that is, the Hodge potential $\rho$ satisfies the Poisson Equation (16). First, we are interested in imposing the condition that for a nonzero constant c, the function ${\displaystyle \frac{-1}{c}}\sigma$ ($\sigma$ conformal factor) satisfies the Poisson Equation (16), and analyzing its impact on the geometry of $(M^m,g)$ possessing a nontrivial conformal vector field $\left(\zeta ,\sigma ,\phi ,\overline{\zeta },\rho \right)$. We prove the following.

Theorem 1. 

Let $(M^m,g),m>1,$ be a compact, connected Riemannian m-manifold which admits a nontrivial conformal vector field $\left(\zeta ,\sigma ,\phi ,\overline{\zeta },\rho \right)$. Then, $M^m$ is an m-sphere $S^m\left(c\right)$ if and only if the Ricci curvature $Ric\left(\overline{\zeta },\overline{\zeta }\right)$ in the direction of the Hodge vector $\overline{\zeta }$ satisfies $Ric\left(\overline{\zeta },\overline{\zeta }\right)\ge {∥\phi ∥}^2$ and there exists a constant $c\ne 0$ such that the function $-\sigma /c$ is a solution of the Poisson equation $\Delta \rho =m\sigma$.

Proof. 

Let $\left(\zeta ,\sigma ,\phi ,\overline{\zeta },\rho \right)$ be a nontrivial conformal vector field on compact $(M^m,g)$, $m>1$, such that the Ricci curvature $Ric\left(\overline{\zeta },\overline{\zeta }\right)$ in the direction of the Hodge vector $\overline{\zeta }$ satisfies

$$Ric\left(\overline{\zeta },\overline{\zeta }\right)\ge {∥\phi ∥}^2$$

(17)

and for a nonzero constant c, the function $-\sigma /c$ satisfies the Poisson equation $\Delta \rho =m\sigma$, that is,

$$\Delta \sigma =-cm\sigma .$$

(18)

Note that the Hodge vector $\overline{\zeta }$ satisfies (cf. Equation (9))

$${\nabla }_U\overline{\zeta }=\sigma U+\phi U-H^{\rho }\left(U\right)\mathrm{and}\mathrm{div}\overline{\zeta }=0,$$

(19)

which gives

$$\left({\mathcal{L}}_{\overline{\zeta }}g\right)\left(U.V\right)=2\sigma g\left(U,V\right)-2g\left(H^{\rho }\left(U\right),V\right),U,V\in \mathsf{\Gamma }\left(T{M}^m\right).$$

(20)

Now, using a local orthonormal frame $\left\{u_1,\dots ,u_m\right\}$ and the expression

$$|{\mathcal{L}}_{\overline{\zeta }}g{|}^2=\sum _{j,k=1}^m\left({\mathcal{L}}_{\overline{\zeta }}g\right){\left(u_j,u_k\right)}^2$$

in Equation (20), we reach

$$|{\mathcal{L}}_{\overline{\zeta }}g{|}^2=4\left\{m{\sigma }^2-2\sigma \Delta \rho +{∥H^{\rho }∥}^2\right\},$$

which, in view of Equation (9), yields

$${\displaystyle \frac{1}{2}}|{\mathcal{L}}_{\overline{\zeta }}g{|}^2=2\left\{{∥H^{\rho }∥}^2-m{\sigma }^2\right\}.$$

(21)

Also, using the expression

$${∥\nabla \overline{\zeta }∥}^2=\sum _{j=1}^{m}g\left({\nabla }_{u_j}\overline{\zeta },{\nabla }_{u_j}\overline{\zeta }\right)$$

with Equation (19), we conclude that

$${∥\nabla \overline{\zeta }∥}^2={∥\phi ∥}^2+{∥H^{\rho }∥}^2-m{\sigma }^2,$$

(22)

where we used

$$\sum _{j}g\left(H^{\rho }\left(u_j\right),\phi u_j\right)=0,$$

due to the symmetric and antisymmetric nature of the operators and the Poisson equationn $\Delta \rho =m\sigma$. We recall the following integral formula

$${\int }_{M^m}\left(Ric\left(\overline{\zeta },\overline{\zeta }\right)+{\displaystyle \frac{1}{2}}|{\mathcal{L}}_{\overline{\zeta }}g{|}^2-{∥\nabla \overline{\zeta }∥}^2-{\left(\mathrm{div}\overline{\zeta }\right)}^2\right)=0,$$

and owing to $\mathrm{div}\overline{\zeta }=0$ and Equations (21) and (22), the above equation reduces to

$${\int }_{M^m}\left(Ric\left(\overline{\zeta },\overline{\zeta }\right)-{∥\phi ∥}^2\right)+{\int }_{M^m}\left({∥H^{\rho }∥}^2-m{\sigma }^2\right)=0.$$

(23)

Using the inequality (17) and the Schwartz’s inequality ${∥H^{\rho }∥}^2\ge {\displaystyle \frac{1}{m}}{\left(\Delta \rho \right)}^2=m{\sigma }^2$ in the above equation allows to conclude that

$$Ric\left(\overline{\zeta },\overline{\zeta }\right)={∥\phi ∥}^2\mathrm{and}\left({∥H^{\rho }∥}^2-m{\sigma }^2\right)=0.$$

However, the equality in the Schwartz’s inequality ${∥H^{\rho }∥}^2\ge {\displaystyle \frac{1}{m}}{\left(\Delta \rho \right)}^2$ holds if and only if

$$H^{\rho }={\displaystyle \frac{\Delta \rho }{m}}I=\sigma I.$$

(24)

Moreover, as $-\sigma /c$ is a solution of the Poisson equation $\Delta \rho =m\sigma$, due to the uniqueness of the solution of the Poisson equation (cf. [22]), we have

$$-{\displaystyle \frac{1}{c}}\sigma =\rho +\alpha ,$$

where $\alpha$ is a constant. We have

$$\nabla \sigma =-c\nabla \rho .$$

Using Equation (24) with the above equation yields

$${\nabla }_U\nabla \sigma =-c\sigma U.$$

(25)

Moreover, Equation (18) gives $\sigma \Delta \sigma =-mc{\sigma }^2$, which, on integration by parts, yields

$${\int }_{M^m}{∥\nabla \sigma ∥}^2=mc{\int }_{M^m}{\sigma }^2.$$

(26)

Note that with constant $c\ne 0$, the above equation confirms that the conformal factor $\sigma$ cannot be a constant. Hence, $\sigma$ is a nonconstant function, and Equation (25) implies that $c>0$, and with these data, Equation (25) is Obata’s differential equation (cf. [23,24]), which guarantees that $(M^m,g)$ is a sphere $S^m\left(c\right)$.

Conversely, on $S^m\left(c\right)$, by virtue of Equation (5), we have the conformal vector field $\left(\zeta ,\sigma ,0,0,-{\displaystyle \frac{1}{c}}\sigma \right)$ on $S^m\left(c\right)$, with the conformal factor $\sigma$, associated tensor $\phi =0$, Hodge vector $\overline{\zeta }=0$ and Hodge potential $\rho =-\sigma /c$. We have

$$\Delta \rho =-{\displaystyle \frac{1}{c}}\Delta \sigma =m\sigma .$$

Consequently, we see that both conditions are met and the converse holds. □

In the next step, we consider a conformal vector field $\left(\zeta ,\sigma ,\phi ,\overline{\zeta },\rho \right)$ on a compact Riemannian m-manifold $(M^m,g)$ with the conformal factor $\sigma$-associated operator $\phi$, Hodge vector $\overline{\zeta }$ and Hodge potential $\rho$, such that the Hodge potential satisfies SPF Equation (15). Indeed we prove the following:

Theorem 2. 

Let $(M^m,g),m>2$, be a compact, connected Riemannian m-manifold with positive scalar curvature τ which admits a nontrivial conformal vector field $\left(\zeta ,\sigma ,\phi ,\overline{\zeta },\rho \right)$. Then, $M^m$ is an m-sphere $S^m\left(c\right)$ for a constant $c>0$ if and only if the Ricci curvature satisfies $Ric\left(\overline{\zeta },\overline{\zeta }\right)\ge {∥\phi ∥}^2$ and the Hodge potential ρ satisfies the static perfect fluid equation:

$$\rho Ric-Hess\left(\rho \right)={\displaystyle \frac{1}{m}}\left(\tau \rho -\Delta \rho \right)g.$$

Proof. 

Let $\left(\zeta ,\sigma ,\phi ,\overline{\zeta },\rho \right)$ be a nontrivial conformal vector field on compact $(M^m,g)$, $m>2$, such that the Ricci curvature $Ric\left(\overline{\zeta },\overline{\zeta }\right)$ in the direction of the Hodge vector $\overline{\zeta }$ satisfies inequality (17) and the Hodge function $\rho$ satisfies the SPF equation

$$\rho Ric-Hess\left(\rho \right)={\displaystyle \frac{1}{m}}\left(\tau \rho -\Delta \rho \right)g.$$

(27)

Using Equations (16) and (17), we have

$$H^{\rho }\left(U\right)=\rho SU-{\displaystyle \frac{1}{m}}\left(\tau \rho -m\sigma \right)U,U\in \mathsf{\Gamma }\left(T{M}^m\right)$$

(28)

and on taking a local orthonormal frame $\left\{u_1,\dots ,u_m\right\}$ with the above equation, we compute

$$\begin{array}{ccc}{∥H^{\rho }∥}^2\hfill & =& \sum _{j=1}^{m}g\left(H^{\rho }\left(u_j\right),H^{\rho }\left(u_j\right)\right)\hfill \\ & =& \sum _{j=1}^{m}g\left(\rho S{u}_j-{\displaystyle \frac{1}{m}}\left(\tau \rho -m\sigma \right)u_j,\rho S{u}_j-{\displaystyle \frac{1}{m}}\left(\tau \rho -m\sigma \right)u_j\right)\hfill \\ & =& {\rho }^2{∥S∥}^2+{\displaystyle \frac{1}{m}}{\left(\tau \rho -m\sigma \right)}^2-2{\displaystyle \frac{\rho \tau }{m}}\left(\tau \rho -m\sigma \right).\hfill \end{array}$$

Rearranging the above equation, we arrive at

$${∥H^{\rho }∥}^2={\rho }^2\left({∥S∥}^2-{\displaystyle \frac{1}{m}}{\tau }^2\right)+m{\sigma }^2.$$

Inserting the above equation in Equation (23), it gives

$${\int }_{M^m}\left(Ric\left(\overline{\zeta },\overline{\zeta }\right)-{∥\phi ∥}^2\right)+{\int }_{M^m}{\rho }^2\left({∥S∥}^2-{\displaystyle \frac{1}{m}}{\tau }^2\right)=0.$$

(29)

Note that the Schwartz’s inequality implies that

$${\rho }^2\left({∥S∥}^2-{\displaystyle \frac{1}{m}}{\tau }^2\right)\ge 0$$

Together with inequality (17), this implies that both integrands in Equation (29) are non-negative, and consequently, we conclude that

$$Ric\left(\overline{\zeta },\overline{\zeta }\right)-{∥\phi ∥}^2=0\mathrm{and}{\rho }^2\left({∥S∥}^2-{\displaystyle \frac{1}{m}}{\tau }^2\right)=0.$$

(30)

Here, we notice that if the Hodge potential $\rho =0$, we have $\zeta =\overline{\zeta }$, which would imply that $\mathrm{div}\zeta =m\sigma =0$, which contradicts the fact that $\left(\zeta ,\sigma ,\phi ,\overline{\zeta },\rho \right)$ is a nontrivial conformal vector field. Hence, the Hodge potential $\rho \ne 0$, and when combined with Equation (30) it leads to

$${∥S∥}^2={\displaystyle \frac{1}{m}}{\tau }^2.$$

(31)

Notice that the above equation is the equality in the Schwartz’s inequality

$${∥S∥}^2\ge {\displaystyle \frac{1}{m}}{\tau }^2$$

and equality (31) holds if and only if

$$S={\displaystyle \frac{\tau }{m}}I.$$

(32)

An interesting implication of Equation (32) and the restriction on the dimension $m>2$ is that $\tau$ is a constant; we denote it as $\tau =m(m-1)c$ for a constant c. Using Equation (32) in Equation (28) yields

$$H^{\rho }=\sigma I,$$

(33)

that is,

$$\left({\nabla }_U{H}^{\rho }\right)\left(V\right)=U\left(\sigma \right)V,U,V\in \mathsf{\Gamma }\left(T{M}^m\right).$$

Utilizing the above equation in the identity

$$R\left(U,V\right)\nabla \rho =\left({\nabla }_U{H}^{\rho }\right)\left(V\right)-\left({\nabla }_V{H}^{\rho }\right)\left(U\right),$$

we arrive at

$$R\left(U,V\right)\nabla \rho =U\left(\sigma \right)V-V\left(\sigma \right)U,U,V\in \mathsf{\Gamma }\left(T{M}^m\right).$$

Taking the trace in the above equation, we have

$$Ric\left(V,\nabla \rho \right)=-(m-1)V\left(\sigma \right),$$

that is,

$$S\nabla \rho =-(m-1)\nabla \sigma$$

and combining it with Equation (32), we conclude that

$$c\nabla \rho =-\nabla \sigma ,$$

where we used $m>2$. The above equation, in view of Equation (33), confirms that

$${\nabla }_U\nabla \sigma =-c\sigma U,U\in \mathsf{\Gamma }\left(T{M}^m\right).$$

(34)

If the conformal factor $\sigma$ were a constant, Equation (8) would imply that $\sigma =0$, a contradiction to the fact that $\left(\zeta ,\sigma ,\phi ,\overline{\zeta },\rho \right)$ is a nontrivial conformal vector field. Then, taking the trace in Equation (34), we obtain $\Delta \sigma =-mc\sigma$, which guarantees that $\sigma$ is a nonconstant eigenfunction of the Laplace operator $\Delta$ on compact $M^m$ with the eigenvalue $mc$. Hence, $c>0$, and as such, Equation (34) assures that $(M^m,g)$ is a sphere $S^m\left(c\right)$.

Conversely, suppose that $(M^m,g)$ is a sphere $S^m\left(c\right)$. Then, as seen earlier, $S^m\left(c\right)$ admits a nontrivial conformal vector field $\left(\zeta ,\sigma ,0,0,-{\displaystyle \frac{1}{c}}\sigma \right)$ with inequality (17), which holds vacuously, and the Hodge potential $\rho =$$-{\displaystyle \frac{1}{c}}\sigma$ satisfies the SPF equation (see Equation (5)). □

4. Affinity Tensor of a Conformal Vector Field

In this section, we are interested in the affinity tensor of a vector field $\zeta$ on a Riemannian manifold $\left(M^m,g\right)$ defined by

$$\left({\mathcal{L}}_{\zeta }\nabla \right)\left(U,V\right)={\mathcal{L}}_{\zeta }{\nabla }_{U}V-{\mathcal{L}}_{{\nabla }_{\zeta }U}V-{\nabla }_U{\mathcal{L}}_{\zeta }V,U,V\in \mathsf{\Gamma }\left(T{M}^m\right)$$

(cf. [13], p. 109), and the properties of this tensor are known to have significance in shaping the geometry of $\left(M^m,g\right)$ (cf. [15]). It is easy to see that the affinity tensor is given by

$$\left({\mathcal{L}}_{\zeta }\nabla \right)\left(U,V\right)=R\left(\zeta ,U\right)V+{\nabla }_U{\nabla }_V\zeta -{\nabla }_{{\nabla }_{U}V}\zeta .$$

(35)

Consider a conformal vector field $\left(\zeta ,\sigma ,\phi ,\overline{\zeta },\rho \right)$ with a conformal factor $\sigma$, an associated tensor $\phi$, the Hodge vector $\overline{\zeta }$ and the Hodge potential $\rho$ on a compact Riemannian manifold $\left(M^m,g\right)$. In our first result, we intend to study the impact of the vanishing of the affinity tensor of the Hodge vector $\overline{\zeta }$ on the geometry of $\left(M^m,g\right)$. Indeed, we prove the following:

Theorem 3. 

Let $(M^m,g)$, $m>1$, be a compact, connected Riemannian m-manifold of positive Ricci curvature which admits a nontrivial conformal vector field $\left(\zeta ,\sigma ,\phi ,\overline{\zeta },\rho \right)$. Then $M^m$ is an m-sphere $S^m\left(c\right)$ if and only if the affinity tensor of the Hodge vector $\overline{\zeta }$ vanished identically, that is, ${\mathcal{L}}_{\overline{\zeta }}\nabla =0$, and the Ricci curvature satisfies

$${\int }_{M^m}Ric\left(\nabla \sigma ,\nabla \sigma \right)\le (m-1)c{\int }_{M^m}\left(2{∥\nabla \sigma ∥}^2-mc{\sigma }^2\right)$$

for a positive constant c.

Proof. 

Suppose that the compact and connected Riemannian manifold $(M^m,g)$, $m>1$, admits a nontrivial conformal vector field $\left(\zeta ,\sigma ,\phi ,\overline{\zeta },\rho \right)$ such that the affinity tensor of the Hodge vector $\overline{\zeta }$ satisfies

$${\mathcal{L}}_{\overline{\zeta }}\nabla =0$$

(36)

and for a nonzero constant c, the Ricci curvature $Ric\left(\nabla \sigma ,\nabla \sigma \right)$ satisfies

$${\int }_{M^m}Ric\left(\nabla \sigma ,\nabla \sigma \right)\le (m-1)c{\int }_{M^m}\left(2{∥\nabla \sigma ∥}^2-mc{\sigma }^2\right).$$

(37)

Then, using Equations (35) and (36), we obtain

$$R\left(\overline{\zeta },U\right)V+{\nabla }_U{\nabla }_V\overline{\zeta }-{\nabla }_{{\nabla }_{U}V}\overline{\zeta }=0,U,V\in \mathfrak{d}\left(T{M}^m\right),$$

which, in view of Equation (9), yields

$$R\left(\overline{\zeta },U\right)V+U\left(\sigma \right)V+\left({\nabla }_U\phi \right)\left(V\right)-\left({\nabla }_U{H}^{\rho }\right)\left(V\right)=0.$$

Substituting Equation (13) in the above equation gives

$$R\left(\overline{\zeta },U\right)V+U\left(\sigma \right)V+R\left(U,\zeta \right)V+V\left(\sigma \right)U-g\left(U,V\right)\nabla \sigma -\left({\nabla }_U{H}^{\rho }\right)\left(V\right)=0,$$

which, in view of $\zeta =\overline{\zeta }+\nabla \rho$, implies that

$$R\left(U,\nabla \rho \right)V+U\left(\sigma \right)V+V\left(\sigma \right)U-g\left(U,V\right)\nabla \sigma -\left({\nabla }_U{H}^{\rho }\right)\left(V\right)=0,$$

that is,

$$R\left(\nabla \rho ,U\right)V=U\left(\sigma \right)V+V\left(\sigma \right)U-g\left(U,V\right)\nabla \sigma -\left({\nabla }_U{H}^{\rho }\right)\left(V\right).$$

(38)

We know that for a local frame $\left\{u_1,\dots ,u_m\right\}$, we have the following expression for the Ricci operator

$$SW=\sum _{j=1}^{m}R\left(W,u_j\right)u_j,W\in \mathsf{\Gamma }\left(T{M}^m\right).$$

Consequently, Equation (38) confirms that

$$S\left(\nabla \rho \right)=-(m-2)\nabla \sigma -\sum _{j=1}^m\left({\nabla }_{u_j}{H}^{\rho }\right)\left(u_j\right).$$

(39)

Now, we use the formula

$$R\left(U,V\right)\nabla \rho =\left({\nabla }_U{H}^{\rho }\right)\left(V\right)-\left({\nabla }_V{H}^{\rho }\right)\left(U\right),U,V\in \mathsf{\Gamma }\left(T{M}^m\right),$$

which, on tracing and using $\Delta \rho =m\sigma$ as well as the symmetry of the operator $H^{\rho }$, yields

$$Ric\left(V,\nabla \rho \right)=\sum _{j=1}^{m}g\left(V,\left({\nabla }_{u_j}{H}^{\rho }\right)\left(u_j\right)\right)-V\left(m\sigma \right),$$

that is,

$$\sum _{j=1}^m\left({\nabla }_{u_j}{H}^{\rho }\right)\left(u_j\right)=S\left(\nabla \rho \right)+m\nabla \sigma .$$

Substituting this equation in Equation (39) gives

$$S\left(\nabla \rho \right)=-\left(m-1\right)\nabla \sigma ,$$

that is,

$$Ric\left(\nabla \rho ,\nabla \sigma \right)=-(m-1){∥\nabla \sigma ∥}^2.$$

(40)

Next, for a nonzero constant c, we have

$$Ric\left(\nabla \sigma +c\nabla \rho ,\nabla \sigma +c\nabla \rho \right)=Ric\left(\nabla \sigma ,\nabla \sigma \right)+2cRic\left(\nabla \rho ,\nabla \sigma \right)+c^{2}Ric\left(\nabla \rho ,\nabla \rho \right).$$

Integrating the above equation, while using Bochner’s formula

$${\int }_{M^m}\left(Ric\left(\nabla \rho ,\nabla \rho \right)+{∥H^{\rho }∥}^2-{\left(\Delta \rho \right)}^2\right)=0$$

and Equation (40) and $\Delta \rho =m\sigma$, we reach

$$\begin{array}{ccc}{\int }_{M^m}Ric\left(\nabla \sigma +c\nabla \rho ,\nabla \sigma +c\nabla \rho \right)\hfill & =& {\int }_{M^m}\left(Ric\left(\nabla \sigma ,\nabla \sigma \right)-2(m-1)c{∥\nabla \sigma ∥}^2\right.\hfill \\ & & \left.+c^2\left(m^2{\sigma }^2-{∥H^{\rho }∥}^2\right)\right).\hfill \end{array}$$

We, use $\Delta \rho =m\sigma$ and rearrange the above equation as

$$\begin{array}{ccc}& & {\int }_{M^m}Ric\left(\nabla \sigma +c\nabla \rho ,\nabla \sigma +c\nabla \rho \right)+c^2{\int }_{M^m}\left({∥H^{\rho }∥}^2-{\displaystyle \frac{1}{m}}{\left(\Delta \rho \right)}^2\right)\hfill \\ & =& {\int }_{M^m}Ric\left(\nabla \sigma ,\nabla \sigma \right)-(m-1)c{\int }_{M^m}\left(2{∥\nabla \sigma ∥}^2-mc{\sigma }^2\right).\hfill \end{array}$$

Using the fact that the Ricci curvature of $\left(M^m,g\right)$ is positive and the Schwartz’s inequality ${∥H^{\rho }∥}^2\ge {\displaystyle \frac{1}{m}}{\left(\Delta \rho \right)}^2$, as well as the inequality (37) in the above equation, allows us to conclude that

$$\nabla \sigma +c\nabla \rho =0\mathrm{and}{∥H^{\rho }∥}^2={\displaystyle \frac{1}{m}}{\left(\Delta \rho \right)}^2.$$

(41)

Note that the second equation in (41) is the equality in the Schwartz’s inequality, and it holds if and only if $H^{\rho }={\displaystyle \frac{1}{m}}\Delta \rho I$, which, in view of $\Delta \rho =m\sigma$, leads to

$$H^{\rho }=\sigma I.$$

(42)

Also, using the first equation in (41), we have $\Delta \sigma =-c\Delta \rho =-mc\sigma$ with a nonzero constant c, which implies either that $\sigma$ is nonconstant or $\sigma =0$. However, $\sigma =0$ is forbidden owing to the fact that $\left(\zeta ,\sigma ,\phi ,\overline{\zeta },\rho \right)$ is nontrivial. Hence, $\sigma$ is a nonconstant eigenfunction of the Laplace operator $\Delta$ on compact $\left(M^m,g\right)$ with the eigenvalue $mc$, and therefore, we have $c>0$. Also, using the first equation in (41) together with Equation (42), we have

$${\nabla }_U\nabla \sigma =-c{H}^{\rho }\left(U\right)=-c\sigma U,U\in \mathsf{\Gamma }\left(T{M}^m\right).$$

Therefore, $\sigma$ is a nonconstant function, and the constant $c>0$ in the above equation implies that $\left(M^m,g\right)$ is a sphere $S^m\left(c\right)$.

Conversely, on an m-sphere $S^m\left(c\right)$, we have the following nontrivial conformal vector field $\left(\zeta ,\sigma ,0,0,-{\displaystyle \frac{1}{c}}\sigma \right)$ coming from Equation (5), with the conformal factor satisfying $\Delta \sigma =-mc\sigma$, that is,

$${\int }_{S^m\left(c\right)}{∥\nabla \sigma ∥}^2=mc{\int }_{S^m\left(c\right)}{\sigma }^2.$$

(43)

Since the Hodge vector $\overline{\zeta }=0$, its affinity tensor vanishes. Also, for the sphere $S^m\left(c\right)$, we have

$$Ric\left(\nabla \sigma ,\nabla \sigma \right)=(m-1)c{∥\nabla \sigma ∥}^2,$$

and therefore, using Equation (43), we see that the following holds:

$${\int }_{S^m\left(c\right)}Ric\left(\nabla \sigma ,\nabla \sigma \right)=(m-1)c{\int }_{S^m\left(c\right)}\left(2{∥\nabla \sigma ∥}^2-mc{\sigma }^2\right),$$

which finishes the proof. □

Finally, in this section, we consider a noncompact complete and connected Riemannian manifold $\left(M^m,g\right)$. However, on a Euclidean m-space ${\mathbb{E}}^m$, through Equation (6), we see that there is a conformal vector field $\left(\zeta ,\sigma ,\phi ,\overline{\zeta },\rho \right)$, which has a Hodge vector $\overline{\zeta }$ and Hodge potential $\sigma$. However, in general, given a conformal vector field $\zeta$ on a complete and connected Riemannian manifold $\left(M^m,g\right)$, there is no guarantee of the Hodge decomposition of $\zeta$. In this section, we consider a nontrivial conformal vector field $\zeta$ on a complete and connected Riemannian manifold $\left(M^m,g\right)$, whose affinity tensor vanishes, and seek the following characterization of the Euclidean space ${\mathbb{E}}^m$.

Theorem 4. 

A complete, connected Riemannian manifold $M^m$, $m>2$, is a Euclidean m-space if and only if it admits a nontrivial conformal vector field ζ whose affinity tensor vanishes identically and ζ annihilates its associated tensor φ.

Proof. 

Suppose that $\zeta$ is a nontrivial conformal vector field on an m-dimensional complete and connected Riemannian manifold $(M^m,g)$, $m>2$, such that it satisfies

$${\mathcal{L}}_{\zeta }\nabla =0$$

(44)

and

$$\phi \zeta =0.$$

(45)

On using Equation (35), we have

$${\nabla }_U{\nabla }_V\zeta -{\nabla }_{{\nabla }_{U}V}\zeta =R\left(U,\zeta \right)V.$$

Using Equation (7) in the above equation, we arrive at

$${\nabla }_U\left(\sigma V+\phi V\right)-\sigma {\nabla }_{U}V-\phi {\nabla }_{U}V=R\left(U,\zeta \right)V,$$

that is,

$$U\left(\sigma \right)V+\left({\nabla }_U\phi \right)\left(V\right)=R\left(U,\zeta \right)V.$$

Now, inserting Equation (13), we conclude that

$$U\left(\sigma \right)V+R\left(U,\zeta \right)V+V\left(\sigma \right)U-g\left(U,V\right)\nabla \sigma =R\left(U,\zeta \right)V,$$

that is,

$$U\left(\sigma \right)V+V\left(\sigma \right)U=g\left(U,V\right)\nabla \sigma ,U,V\in \mathsf{\Gamma }\left(T{M}^m\right).$$

Taking the trace in the above equation, it gives

$$2\nabla \sigma =m\nabla \sigma$$

and as $m>2$, we conclude that the conformal factor $\sigma$ is a constant. Moreover, with $\zeta$ being a nontrivial conformal vector field, we see the constant $\sigma \ne 0$. Next, we define the function $\varphi ={\displaystyle \frac{1}{2}}{∥\zeta ∥}^2$, and the gradient of this function $\varphi$ can be found using Equation (7) as follows:

$$\nabla \varphi =\sigma \zeta -\phi \zeta ,$$

which, in view of Equation (45), gives

$$\nabla \varphi =\sigma \zeta .$$

(46)

Differentiating the above equation with respect to a vector field U and using the fact that $\sigma$ is a constant and Equation (7), we obtain

$${\nabla }_U\nabla \varphi =\sigma \left(\sigma U+\phi U\right),$$

that is,

$$Hess\left(\varphi \right)\left(U,V\right)={\sigma }^{2}g\left(U,V\right)+\sigma g\left(\phi U,V\right),U,V\in \mathsf{\Gamma }\left(T{M}^m\right).$$

Since in the equation

$$Hess\left(\varphi \right)\left(U,V\right)-{\sigma }^{2}g\left(U,V\right)=\sigma g\left(\phi U,V\right)$$

the left hand side is symmetric while the right hand side is skew-symmetric, we conclude that

$$Hess\left(\varphi \right)\left(U,V\right)-{\sigma }^{2}g\left(U,V\right)=0\mathrm{and}\sigma g\left(\phi U,V\right)=0.$$

Moreover, as the constant $\sigma \ne 0$, we conclude that $\phi =0$ and

$$Hess\left(\varphi \right)\left(U,V\right)={\sigma }^{2}g\left(U,V\right),U,V\in \mathsf{\Gamma }\left(T{M}^m\right).$$

(47)

Now, if the function $\varphi$ is a constant, and the constant $\sigma \ne 0$, using Equation (46), we arrive at $\zeta =0$, which is contrary to the assumption that $\zeta$ is a nontrivial conformal vector field. Thus, the nonconstant function $\varphi$ on complete and connected $\left(M^m,g\right)$ satisfies Equation (47) for the nonzero constant $c={\sigma }^2\ne 0$. Hence, $\left(M^m,g\right)$ is a Euclidean m-space ${\mathbb{E}}^m$ (cf. [15]).

Conversely, on a Euclidean m-space ${\mathbb{E}}^m$, we have the conformal vector field

$$\zeta =\sum _{j=1}^m{x}^j{\displaystyle \frac{\partial }{\partial x^j}},$$

where $x^1,\dots ,x^m$ are Euclidean coordinates, which satisfies

$${\nabla }_U\zeta =U,U\in \mathsf{\Gamma }\left(T{\mathbb{E}}^m\right).$$

It is a closed conformal vector field with a conformal factor 1 and an associated tensor $\phi =0$. It is easy to see that the affinity tensor of $\zeta$ vanishes identically, that is, ${\mathcal{L}}_{\zeta }\nabla =0.$ □

5. Conclusions

Given a nontrivial conformal vector field $\left(\zeta ,\sigma ,\phi ,\overline{\zeta },\rho \right)$ on a compact Riemannian manifold $\left(M^m,g\right)$, in Theorems 1 and 2, we have asked the function ${\displaystyle \frac{-1}{c}}\sigma$ to satisfy the Poisson equation

$$\Delta \rho =m\sigma$$

and the Hodge potential $\rho$ to satisfy the static perfect fluid equation

$$\rho Ric-Hess\left(\rho \right)={\displaystyle \frac{1}{m}}\left(\tau \rho -\Delta \rho \right)g,$$

(48)

to obtain characterizations of the sphere $S^m\left(c\right)$. There is yet another important differential equation on a Riemannian manifold $\left(M^m,g\right)$, $m>1$, namely the Fischer–Marsden equation (cf. [25])

$$(\Delta \rho )g+\rho Ric=Hess\left(\rho \right).$$

(49)

However, a compact $\left(M^m,g\right)$ admitting the solution of the Fischer–Marsden equation forces $\left(M^m,g\right)$ to have constant scalar curvature $\tau$. Moreover, the trace of Equation (49), gives

$$\Delta \rho =-{\displaystyle \frac{\tau }{m-1}}\rho .$$

(50)

The importance of the above Poisson equation lies in the fact that when Equations (48) and (50) are considered together, it reduces the static perfect fluid equation to the Fischer–Marsden equation. Observe that the Poisson Equation (50) is not studied, and therefore, it is not known what impact it has on the geometry of a Riemannian manifold $\left(M^m,g\right)$. It will be very interesting to study the following question:

Under what conditions is a compact Riemannian manifold $\left(M^m,g\right)$ equipped with a nontrivial conformal vector field $\left(\zeta ,\sigma ,\phi ,\overline{\zeta },\rho \right)$ with Hodge potential $\rho$ satisfying the Poisson Equation (50) necessarily isometric to the sphere $S^m\left(c\right)$?