A Note on Killing Calculus on Riemannian Manifolds

Abstract

In this article, it has been observed that a unit Killing vector field $\xi$ on an n-dimensional Riemannian manifold $(M,g)$, influences its algebra of smooth functions $C^{\infty }\left(M\right)$. For instance, if h is an eigenfunction of the Laplace operator $\Delta$ with eigenvalue $\lambda$, then $\xi \left(h\right)$ is also eigenfunction with same eigenvalue. Additionally, it has been observed that the Hessian ${\mathcal{H}}^h(\xi ,\xi )$ of a smooth function $h\in C^{\infty }\left(M\right)$ defines a self adjoint operator ${⊡}_{\xi }$ and has properties similar to most of properties of the Laplace operator on a compact Riemannian manifold $(M,g)$. We study several properties of functions associated to the unit Killing vector field $\xi$. Finally, we find characterizations of the odd dimensional sphere using properties of the operator ${⊡}_{\xi }$ and the nontrivial solution of Fischer–Marsden differential equation, respectively.

1. Introduction

A smooth vector field $\xi$ on an n-dimensional Riemannian manifold $(M,g)$ is said to be a Killing vector field if its flow consists of isometries of $(M,g)$. We say that a Killing vector field $\xi$ is nontrivial if it is not parallel. It is known that a nontrivial Killing vector field on a compact Riemannian manifold restricts its topology and geometry, for example, it does not allow the Riemannian manifold $(M,g)$ to have negative Ricci curvature and that if $(M,g)$ has positive sectional curvatures, then its fundamental group contains a cyclic subgroup with constant index, depending only on the dimension of M (cf. [1,2,3]). Riemannian manifolds with Killing vector fields has been subject of interest for many mathematicians (cf. [2,4,5,6,7,8,9,10,11,12,13]). There are other important vector fields, such as Jacobi-type vector fields, geodesic vector fields and torqued vector fields, which play important roles in the geometry of a Riemannian manifold (cf. [10,11,14,15,16]). Moreover, incompressible vector fields have applications in Physics, and as Killing vector fields are incompressible, they have applications in Physics (cf. [17]).

Killing vector fields are found in abundance on Euclidean spaces ${\mathbf{E}}^n$; for instance all constant vector fields are Killing, though they are trivial Killing vector fields. If $u^1,\dots ,u^n$ are Euclidean coordinates on ${\mathbf{E}}^n$, then

$$\xi =u^j\frac{\partial }{\partial u^i}-u^i\frac{\partial }{\partial u^j},i\ne j$$

for fixed i and j, is a nontrivial Killing vector field on ${\mathbf{E}}^n$. Similarly, the vector field $\xi =J\Psi$ is a Killing vector field on the even dimensional Euclidean space ${\mathbf{E}}^{2n}$, J being the complex structure and $\Psi$ being the position (Euler) vector field on ${\mathbf{E}}^{2n}$. However, all these nontrivial Killing vector fields on the Euclidean spaces are of non-constant length. A natural question arises, whether there exists a nontrivial Killing vector field of constant length on a Euclidean space? The answer is negative.

In this paper, we exhibit several properties of a unit Killing vector field $\xi$ in relation to algebra $C^{\infty }\left(M\right)$ of smooth functions on $(M,g)$. In particular, we show that there is an operator ${⊡}_{\xi }:C^{\infty }\left(M\right)\to C^{\infty }\left(M\right)$ that has properties similar to that of the Laplace operator.

On the unit sphere ${\mathbf{S}}^{2n+1}$, there is a unit Killing vector field $\xi$ provided by the Sasakian structure (cf. [18]). This naturally raises a question of finding necessary and sufficient conditions on a compact $(M,g)$ that admits a unit Killing vector field to be isometric to ${\mathbf{S}}^{2n+1}$. In this paper we use the properties of the operator ${⊡}_{\xi }$ associated to the unit Killing vector field $\xi$ on a compact $(M,g)$ to find a characterization of the sphere ${\mathbf{S}}^{2n+1}\left(c\right)$. Additionally, we use properties of the nontrivial solution h of the Fischer–Marsden equation (cf. [19]) on a compact Riemannian manifold $(M,g)$ with Killing vector field $\xi$ and a suitable lower bound on the Ricci curvature $Ric\left(gradh,gradh\right)$ to find a characterization of the unit sphere ${\mathbf{S}}^{2n+1}\left(c\right)$. Note that even dimensional unit spheres ${\mathbf{S}}^{2n}$ do not admit unit Killing vector fields, owing to the fact that a Killing vector field on a positively curved even dimensional compact Riemannian manifold has a zero. However, other than unit sphere ${\mathbf{S}}^{2n+1}$, there are ellipsoids admitting unit Killing vector fields (cf. [5]).

2. Preliminaries

A smooth vector field $\xi$ on a Riemannian manifold $(M,g)$ is said to be a Killing vector field, if it satisfies

$${\pounds }_{\xi }g=0,$$

where ${\pounds }_{\xi }g$ is the Lie derivative with respect to $\xi$, or equivalently

$$g\left({\nabla }_X\xi ,Y\right)+g\left({\nabla }_Y\xi ,X\right)=0,X,Y\in \mathfrak{X}\left(M\right),$$

(1)

where ∇ is the Riemannian connection and $\mathfrak{X}\left(M\right)$ is the Lie algebra of smooth vector fields on M.

The curvature tensor field R of $(M,g)$ is given by

$$R(X,Y)Z=\left[{\nabla }_X,{\nabla }_Y\right]Z-{\nabla }_{[X,Y]}Z,X,Y,Z\in \mathfrak{X}\left(M\right)$$

(2)

and Ricci tensor field is

$$Ric(X,Y)=\sum _{i=1}^{n}g\left(R(e_i,X)Y,e_i\right),$$

for a local orthonormal frame $\{e_1,\dots ,e_n\}$ on M.

The Ricci operator S is a symmetric operator associated to the Ricci tensor, defined by

$$g(SX,Y)=Ric(X,Y).$$

The trace $\mathbf{r}=TrS$ is the scalar curvature of M. Note that $grad\mathbf{r}$, the gradient of the scalar curvature $\mathbf{r}$, satisfies

$$\frac{1}{2}grad\mathbf{r}=\sum _{i=1}^n\left(\nabla S\right)(e_i,e_i),$$

(3)

where $\nabla S$ is given by

$$\left(\nabla S\right)(X,Y)={\nabla }_{X}SY-S{\nabla }_{X}Y.$$

We denote by $\eta ,$ 1-form dual to the Killing vector field $\xi$ on $(M,g)$ and define an operator $\mathbf{F}$ on $\left(M,g\right)$ by

$$2g\left(\mathbf{F}X,Y\right)=d\eta (X,Y),X,Y\in \mathfrak{X}\left(M\right).$$

(4)

We use bold faced letters for scalar curvature and specific vectors on a Euclidean space and some specific tensors. The operator $\mathbf{F}$ is skew-symmetric and using

$$d\eta (X,Y)=g\left({\nabla }_X\xi ,Y\right)-g\left({\nabla }_Y\xi ,X\right)$$

together with Equations (1) and (4), we conclude

$${\nabla }_X\xi =\mathbf{F}X,X\in \mathfrak{X}\left(M\right).$$

(5)

If the length of the Killing vector field $\xi$ is a constant, then on taking the inner product with $\xi$ in above Equation (5), we conclude $g\left({\nabla }_X\xi ,\xi \right)=g(\mathbf{F}X,\xi )=0$, and as $\mathbf{F}$ is skew-symmetric operator, we get

$$\mathbf{F}\left(\xi \right)=0.$$

(6)

Additionally, using Equation (5), we have

$$\left(\nabla \mathbf{F}\right)(X,Y)-\left(\nabla \mathbf{F}\right)(Y,X)=R(X,Y)\xi ,X,Y\in \mathfrak{X}\left(M\right),$$

(7)

where $\left(\nabla \mathbf{F}\right)(X,Y)={\nabla }_X\mathbf{F}Y-\mathbf{F}\left({\nabla }_{X}Y\right)$. Using the fact that the 2-form $d\eta$ is closed, $\mathbf{F}$ is skew-symmetric and, from Equation (7), we conclude

$$R(X,\xi )Y=\left(\nabla \mathbf{F}\right)(X,Y),X,Y\in \mathfrak{X}\left(M\right).$$

(8)

We denote by $C^{\infty }\left(M\right)$ the algebra of smooth functions on the Riemannian manifold $(M,g)$ and for a $h\in C^{\infty }\left(M\right)$, we denote its gradient by $gradh$. Then the Hessian operator $A^h$ of h is defined by

$$A^h\left(X\right)={\nabla }_{X}gradh,X\in \mathfrak{X}\left(M\right)$$

(9)

and it is a symmetric operator. Moreover the Hessian ${\mathcal{H}}^h$ of h is defined by

$${\mathcal{H}}^h(X,Y)=g\left(A^h\left(X\right),Y\right),X,Y\in \mathfrak{X}\left(M\right).$$

(10)

The Laplace operator $\Delta :$$C^{\infty }\left(M\right)\to C^{\infty }\left(M\right)$ on a Riemannian manifold $(M,g)$ is defined by $\Delta h=div\left(gradh\right)$ and we also have

$$\Delta h=Tr{A}^h.$$

(11)

If M is compact and $h\in C^{\infty }\left(M\right)$ is, such that

$${\int }_{M}h=0,$$

then by minimum principle, we have

$${\int }_M{∥gradh∥}^2\ge {\lambda }_1{\int }_M{h}^2$$

(12)

${\lambda }_1$ being first nonzero eigenvalue of $\Delta$.

3. Killing Calculus

Let $\xi$ be a unit Killing vector field on an n-dimensional Riemannian manifold $(M,g)$. For each $h\in C^{\infty }\left(M\right)$, we define $\overline{h}=\xi \left(h\right)$, $\overline{\overline{h}}$$=\xi \left(\overline{h}\right)$. We are interested in studying the properties of these functions $\overline{h}$, $\overline{\overline{h}}$. From Equation (5), it follows that if $\xi$ is a nontrivial Killing vector field, then the skew-symmetric operator $\mathbf{F}$ is non-vanishing. If the Euclidean space ${\mathbf{E}}^n$ admits a Killing vector field $\xi$ that has constant length, then Equation (8) implies

$$\left(\nabla \mathbf{F}\right)(X,Y)=0,X,Y\in \mathfrak{X}\left({\mathbf{E}}^n\right).$$

Choosing $Y=\xi$ in above equation and using Equation (6), we get $-{\mathbf{F}}^{2}X=0$, that is, ${∥\mathbf{F}∥}^2=0$, where

$${∥\mathbf{F}∥}^2=\underset{i=1}{\sum ^n}g\left(\mathbf{F}{e}_i,\mathbf{F}{e}_i\right),$$

$\left\{e_1,...,e_n\right\}$ being an orthonormal frame on the Euclidean space ${\mathbf{E}}^n$. Thus, we have $\mathbf{F}=0$ and the Killing vector field $\xi$ is trivial. Thus, we have the following:

Proposition 1.

There does not exist a nontrivial Killing vector field of constant length on the Euclidean space ${\mathbf{E}}^n$.

Now, suppose $\xi$ is a Killing vector field on a Riemannian manifold $(M,g)$. Then as $\mathbf{F}$ is skew-symmetric, using Equation (5), we have $div\xi =0$, and for each $h\in C^{\infty }\left(M\right)$, we have $div\left(h\xi \right)=\overline{h}$. Thus, we get

Lemma 1.

Let ξ be a Killing vector field on a compact Riemannian manifold $(M,g)$. Then for each $h\in C^{\infty }\left(M\right)$

$${\int }_M\overline{h}=0.$$

As $\overline{h}=\xi \left(h\right)$, we find $X\left(\overline{h}\right)=Xg\left(\xi ,gradh\right)=g\left(\mathbf{F}X,gradh\right)+g\left(\xi ,A^h\left(X\right)\right)$, $X\in \mathfrak{X}\left(M\right)$ and get the following expression

$$grad\overline{h}=-\mathbf{F}\left(gradh\right)+A^h\left(\xi \right).$$

(13)

Lemma 2.

Let ξ be a Killing vector field on a Riemannian manifold $(M,g)$. Then for each $h\in C^{\infty }\left(M\right)$, $\Delta \overline{h}=\xi \left(\Delta h\right)$.

Proof. 

Using Equation (9), we have

$$R(X,Y)gradh=\left(\nabla A^h\right)(X,Y)-\left(\nabla A^h\right)(Y,X),X,Y\in \mathfrak{X}\left(M\right),$$

and using a local orthonormal frame $\left\{e_1,...,e_n\right\}$ on M, $n=dimM$ in above equation, we conclude

$$Ric\left(Y,gradh\right)=g\left(Y,\underset{i=1}{\sum ^n}\left(\nabla A^h\right)(e_i,e_i)\right)-Y\left(\Delta h\right),$$

where we have used the symmetry of the Hessian operator $A^h$ and $Tr{A}^h=\Delta h$. Thus, above equation implies

$$S\left(gradh\right)=-grad\Delta h+\underset{i=1}{\sum ^n}\left(\nabla A^h\right)(e_i,e_i).$$

(14)

Additionally, note that

$$Ric(\xi ,X)=\underset{i=1}{\sum ^n}R(e_i,\xi ;X,e_i)=-\underset{i=1}{\sum ^n}R(e_i,\xi ;e_i,X),X\in \mathfrak{X}\left(M\right),$$

which implies

$$S\left(\xi \right)=-\underset{i=1}{\sum ^n}R(e_i,\xi )e_i.$$

Using the above equation with Equation (8), we conclude

$$\underset{i=1}{\sum ^n}\left(\nabla \mathbf{F}\right)(e_i,e_i)=-S\left(\xi \right).$$

(15)

Now, using Equations (9) and (15), we get

$$div\left(\mathbf{F}\left(gradh\right)\right)=-\underset{i=1}{\sum ^n}g\left(A^h{e}_i,\mathbf{F}{e}_i\right)+Ric(gradh,\xi )=Ric(gradh,\xi ),$$

(16)

where the first term is zero, owing to the symmetry of $A^h$ and the skew-symmetry of $\mathbf{F}$. Similarly, using Equations (5) and (14), we compute

$$\begin{array}{cc}\hfill div{A}^h\left(\xi \right)& =\underset{i=1}{\sum ^n}g\left(A^h{e}_i,\mathbf{F}{e}_i\right)+g\left(\xi ,\underset{i=1}{\sum ^n}\left(\nabla A^h\right)(e_i,e_i)\right)\hfill \\ \hfill & =Ric(gradh,\xi )+\xi \left(\Delta h\right).\hfill \end{array}$$

(17)

Thus, using Equations (13), (16) and (17), we get $\Delta \overline{h}=\xi \left(\Delta h\right)$. □

Lemma 3.

Let ξ be a Killing vector field on a compact Riemannian manifold $(M,g)$. Then for each $h\in C^{\infty }\left(M\right)$,

$${\int }_M\overline{\overline{h}}\Delta h={\int }_M{∥grad\overline{h}∥}^2.$$

Proof. 

On using above Lemma, we have $\overline{\overline{h}}\Delta h=\xi \left(\overline{h}\right)\Delta h=\xi \left(\overline{h}\Delta h\right)-\overline{h}\xi (\Delta h)=\xi \left(\overline{h}\Delta h\right)-\overline{h}\Delta \overline{h}$. Now, using

$$\frac{1}{2}\Delta {\overline{h}}^2=\overline{h}\Delta \overline{h}+{∥grad\overline{h}∥}^2,$$

we get

$$\overline{\overline{h}}\Delta h=\xi \left(\overline{h}\Delta h\right)+{∥grad\overline{h}∥}^2-\frac{1}{2}\Delta {\overline{h}}^2.$$

Integrating the above equation while using Lemma 1, we get the desired result. □

It is interesting, as the following Lemma suggests, to note that for each $h\in C^{\infty }\left(M\right)$ on a compact $(M,g)$, functions h and $\overline{h}$ are orthogonal functions.

Lemma 4.

Let ξ be a Killing vector field on a compact Riemannian manifold $(M,g)$. Then for each $h\in C^{\infty }\left(M\right)$

$$\left(i\right){\int }_{M}h\overline{h}=0,\left(\mathrm{ii}\right){\int }_{M}h\overline{\overline{h}}=-{\int }_M{\overline{h}}^2.$$

Proof. 

Note that $h\overline{h}=h\xi \left(h\right)=\frac{1}{2}\xi \left(h^2\right)$. Integrating this equation and using Lemma 1, we get (i). Additionally, we have

$$h\overline{\overline{h}}=h\xi \left(\overline{h}\right)=\xi \left(h\overline{h}\right)-\overline{h}\xi \left(h\right)=\xi \left(h\overline{h}\right)-{\overline{h}}^2.$$

Integrating the above equation and using Lemma 1, we get (ii). □

Proposition 2.

Let ξ be a unit Killing vector field on a compact Riemannian manifold $(M,g)$. Then for each $h\in C^{\infty }\left(M\right)$, the volume $V\left(M\right)$ of M satisfies

$$V\left(M\right)\le {\int }_M\left({∥grad\overline{h}-\xi ∥}^2-{\lambda }_1{\overline{h}}^2\right),$$

where ${\lambda }_1$ is the first nonzero eigenvalue of the Laplace operator Δ and the equality holds for h satisfying $\Delta h=-{\lambda }_{1}h.$

Proof. 

Using Lemma 1, and Equation (12), for any $h\in C^{\infty }\left(M\right)$, we get

$${\int }_M{∥grad\overline{h}∥}^2\ge {\lambda }_1{\int }_M{\overline{h}}^2.$$

(18)

We have

$${∥grad\overline{h}-\xi ∥}^2={∥grad\overline{h}∥}^2+1-2\xi \left(\overline{h}\right).$$

Integrating the above equation and using Lemma 1 and Equation (18), we get the result. Moreover, if $\Delta h=-{\lambda }_{1}h$, then by Lemma 2, we have $\Delta \overline{h}=-{\lambda }_1\overline{h}$ and that the equality in inequality (18) holds and consequently, the equality holds in the statement. □

Next, given a unit Killing vector field $\xi$ on a compact Riemannian manifold $(M,g)$, we define an operator ${⊡}_{\xi }:C^{\infty }\left(M\right)\to C^{\infty }\left(M\right)$ by ${⊡}_{\xi }\left(h\right)={\mathcal{H}}^h(\xi ,\xi )$, where ${\mathcal{H}}^h$ is the Hessian of the function h. We shall show that this operator ${⊡}_{\xi }$ is self adjoint operator with respect to the inner product

$$(f,h)={\int }_{M}fh,f,h\in C^{\infty }\left(M\right).$$

Proposition 3.

Let ξ be a unit Killing vector field on a compact Riemannian manifold $(M,g)$. Then the operator ${⊡}_{\xi }$ is a self adjoint operator on $C^{\infty }\left(M\right)$. Consequently,

$${\int }_M{⊡}_{\xi }h=0,h\in C^{\infty }\left(M\right).$$

Proof. 

For $f,h\in C^{\infty }\left(M\right)$, we have

$$\left({⊡}_{\xi }f,h\right)={\int }_M\left({⊡}_{\xi }f\right)h={\int }_{M}h{\mathcal{H}}^f(\xi ,\xi ).$$

(19)

In view of Equations (5) and (6), we have ${\nabla }_{\xi }\xi =0$, and, therefore,

$${\mathcal{H}}^f(\xi ,\xi )=\xi \xi \left(f\right).$$

Thus,

$$\begin{array}{ccc}\hfill h{\mathcal{H}}^f(\xi ,\xi )& =& h\xi \xi \left(f\right)=\xi \left(h\xi \left(f\right)\right)-\xi \left(h\right)\xi \left(f\right)\hfill \\ & =& \xi \left(h\xi \left(f\right)\right)-\xi \left(f\xi \left(h\right)\right)+f\xi \xi h\hfill \\ & =& \xi \left(h\xi \left(f\right)\right)-\xi \left(f\xi \left(h\right)\right)+f{\mathcal{H}}^h(\xi ,\xi ).\hfill \end{array}$$

Integrating the above equation and using Lemma 1 and Equation (19), we conclude

$$\left({⊡}_{\xi }f,h\right)={\int }_{M}h{\mathcal{H}}^f(\xi ,\xi )=(f,{⊡}_{\xi }h).$$

Hence, the operator ${⊡}_{\xi }$ is self adjoint operator on $C^{\infty }\left(M\right)$. Note for a constant c, we have ${⊡}_{\xi }c={\mathcal{H}}^c(\xi ,\xi )=0$ and, therefore,

$${\int }_M{⊡}_{\xi }h={\int }_{M}h{⊡}_{\xi }1=0.$$

□

Note that the Laplace operator satisfies $\frac{1}{2}\Delta h^2=h\Delta h+{∥gradh∥}^2$, $h\in C^{\infty }\left(M\right)$ and we will show that the operator ${⊡}_{\xi }$ has a similar property. Indeed, we have

$${\mathcal{H}}^{\frac{1}{2}{h}^2}(\xi ,\xi )=\xi \xi \left(\frac{1}{2}{h}^2\right)=\xi \left(h\xi \left(h\right)\right)=\xi {\left(h\right)}^2+h{\mathcal{H}}^h(\xi ,\xi ),$$

that is, the operator ${⊡}_{\xi }$ satisfies

$$\frac{1}{2}{⊡}_{\xi }{h}^2=h{⊡}_{\xi }h+{\overline{h}}^2.$$

(20)

Using Stokes’s Theorem, we know that, on a compact $(M,g)$, $\Delta h=0$ implies h is a constant. We have a similar result for the operator ${⊡}_{\xi }$ as a consequence of Proposition 3, as seen in the following result.

Corollary 1.

Let ξ be a unit Killing vector field on a compact Riemannian manifold $(M,g)$. Then ${⊡}_{\xi }h=0$, if and only if, h is a constant on the integral curves of ξ.

Proof. 

Let $h\in C^{\infty }\left(M\right)$ be such that ${⊡}_{\xi }h=0$. Then, Equation (20) implies

$$\frac{1}{2}{⊡}_{\xi }{h}^2={\overline{h}}^2.$$

Integrating the above equation and using Proposition 3, we conclude

$${\int }_M{\overline{h}}^2=0.$$

Thus, $\overline{h}=0$, that is, h is a constant on the integral curves of $\xi$. The converse is obvious. □

Recall that the unit sphere ${\mathbf{S}}^{2n+1}$ possesses a unit Killing vector field $\xi$ provided by the Sasakian structure (cf. [18]). Additionally, there is a $h\in C^{\infty }\left({\mathbf{S}}^{2n+1}\right)$ satisfying

$${\nabla }_{X}gradh=-hX,X\in \mathfrak{X}\left({\mathbf{S}}^{2n+1}\right),$$

that is, $A^h\left(\xi \right)=-h\xi$. Note that h is the eigenfunction of the Laplace operator $\Delta$ corresponding to the first nonzero eigenvalue $2n+1$ and, also, we see that ${\mathcal{H}}^h(\xi ,\xi )=-h$. Thus, ${⊡}_{\xi }h=-h$, that is, h is an eigenfunction of the operator ${⊡}_{\xi }$ corresponding to eigenvalue 1.

Let $h\in C^{\infty }\left(M\right)$ be a non-constant function, satisfying ${⊡}_{\xi }h=\lambda h$ for a nonzero constant $\lambda$ and M be compact. Then in view of Proposition 3, Equation (20) implies

$${\int }_M\left(\lambda h^2+{\overline{h}}^2\right)=0.$$

As the constant $\lambda$ is nonzero and h is non-constant function, the above equation proves $\lambda <0$. Hence, if h is a non-constant eigenfunction, we have ${⊡}_{\xi }h=-\mu h$ for $\mu >0$ and we say $\mu$ is the eigenvalue of the operator ${⊡}_{\xi }$ and conclude that nonzero eigenvalues of the operator ${⊡}_{\xi }$ are positive.

Recall that, owing to Lemma 1, on a compact Riemannian manifold $(M,g)$ that admits a unit Killing vector field $\xi$, we have the Poisson equation $\Delta u=\overline{h}$ and this is known to have unique solution up to a constant. Additionally, we consider an analogue of the Poisson equation involving the operator ${⊡}_{\xi }$, the differential equation of the form

$${⊡}_{\xi }u=\overline{h},$$

for a $h\in C^{\infty }\left(M\right)$. We have the following result:

Proposition 4.

Let ξ be a unit Killing vector field on a compact Riemannian manifold $(M,g)$. Then $h\in C^{\infty }\left(M\right)$ is a solution of the differential equation ${⊡}_{\xi }u=\overline{h}$, if and only if, h is a constant on the integral curves of ξ.

Proof. 

Suppose h is a solution of ${⊡}_{\xi }u=\overline{h}$. Then, using Equations (5) and (6), we have ${\nabla }_{\xi }\xi =0$ and we get ${\mathcal{H}}^h(\xi ,\xi )=\xi \xi \left(h\right)=\overline{\overline{h}}$. Thus, using ${⊡}_{\xi }h=\overline{h}$, we get $\overline{\overline{h}}=\overline{h}$, that is,

$$h\overline{\overline{h}}=h\overline{h}.$$

Integrating the above equation and using (i) of Lemma 4, we get

$${\int }_{M}h\overline{\overline{h}}=0,$$

which in view of (ii) of Lemma 4, we conclude

$${\int }_M{\overline{h}}^2=0.$$

Hence, $\overline{h}=0$ and h is a constant on integral curves of $\xi$. The converse is trivial and it follows from Corollary 1. □

4. Characterizations of Odd Dimensional Spheres

In this section, we use the Killing calculus developed in the previous section to find a characterization of the odd dimensional sphere ${\mathbf{S}}^{2m+1}\left(c\right)$. We prove the following:

Theorem 1.

Let ξ be a unit Killing vector field on an n-dimensional compact Riemannian manifold $(M,g)$, $h\in C^{\infty }\left(M\right)$ be such that $\overline{h}$ is a non-constant function, and ${\lambda }_1$ be the first nonzero eigenvalue of the Laplace operator Δ. Then ${⊡}_{\xi }\Delta h=-nc\overline{\overline{h}}$, for a constant $c>0$ and the Ricci curvature in the direction of the vector field $grad\overline{h}$, is bounded below by $c\left(\frac{nc}{{\lambda }_1}+n-2\right)$, if and only if, n is odd ($n=2m+1)$ and $(M,g)$ is isometric to the sphere ${\mathbf{S}}^{2m+1}\left(c\right)$.

Proof. 

Suppose $h\in C^{\infty }\left(M\right)$ is such that $\overline{h}$ is a non-constant function, satisfying

$${⊡}_{\xi }\Delta h=-nc\overline{\overline{h}}$$

(21)

for a positive constant c, and the Ricci curvature satisfies

$$Ric\left(grad\overline{h},grad\overline{h}\right)\ge c\left(\frac{nc}{{\lambda }_1}+n-2\right){∥grad\overline{h}∥}^2.$$

(22)

Note that, using Lemma 2, we have ${\left(\Delta \overline{h}\right)}^2={\left(\xi \left(\Delta h\right)\right)}^2={\left(\overline{\Delta h}\right)}^2$, and combining it with Equation (20), we have

$${\left(\overline{\Delta h}\right)}^2=\frac{1}{2}{⊡}_{\xi }{\left(\Delta h\right)}^2-\Delta h{⊡}_{\xi }\Delta h.$$

Integrating the above equation and using Equation (21), we get

$${\int }_M{\left(\overline{\Delta h}\right)}^2=nc{\int }_M\overline{\overline{h}}\Delta h$$

and the above equation in view of Lemma 3 implies

$${\int }_M{\left(\overline{\Delta h}\right)}^2=nc{\int }_M{∥grad\overline{h}∥}^2.$$

(23)

Using Bochner’s formula (cf. [20]), we have

$${\int }_M\left(Ric(\left(grad\overline{h},grad\overline{h}\right)+{∥A^{\overline{h}}∥}^2-{\left(\Delta \overline{h}\right)}^2\right)=0.$$

(24)

Additionally, we have

$${∥A^{\overline{h}}+c\overline{h}I∥}^2={∥A^{\overline{h}}∥}^2+n{c}^2{\overline{h}}^2+2c\overline{h}\Delta \overline{h}$$

and integrating the above equation while using Equation (24), we get

$${\int }_M{∥A^{\overline{h}}+c\overline{h}I∥}^2={\int }_M\left({\left(\Delta \overline{h}\right)}^2-Ric(\left(grad\overline{h},grad\overline{h}\right)+n{c}^2{\overline{h}}^2+2c\overline{h}\Delta \overline{h}\right).$$

Using Lemma 1 and inequality (12), in the above equation, we get

$${\int }_M{∥A^{\overline{h}}+c\overline{h}I∥}^2\le {\int }_M\left({\left(\Delta \overline{h}\right)}^2-Ric(\left(grad\overline{h},grad\overline{h}\right)+\frac{n{c}^2}{{\lambda }_1}{∥grad\overline{h}∥}^2+2c\overline{h}\Delta \overline{h}\right)$$

and using $\frac{1}{2}\Delta {\overline{h}}^2=\overline{h}\Delta \overline{h}+{∥grad\overline{h}∥}^2$ in last term of above inequality, we conclude

$${\int }_M{∥A^{\overline{h}}+c\overline{h}I∥}^2\le {\int }_M\left({\left(\Delta \overline{h}\right)}^2-Ric(\left(grad\overline{h},grad\overline{h}\right)+\frac{n{c}^2}{{\lambda }_1}{∥grad\overline{h}∥}^2-2c{∥grad\overline{h}∥}^2\right).$$

Now, using Equation (23) in the above inequality, we arrive at

$${\int }_M{∥A^{\overline{h}}+c\overline{h}I∥}^2\le {\int }_M\left(c\left(\frac{nc}{{\lambda }_1}+n-2\right){∥grad\overline{h}∥}^2-Ric(\left(grad\overline{h},grad\overline{h}\right)\right).$$

Finally, inequality (22) and the above inequality implies $A^{\overline{h}}+c\overline{h}I=0$ and we have

$${\nabla }_{X}grad\overline{h}=-c\overline{h}X,X\in \mathfrak{X}\left(M\right).$$

Thus, the function $\overline{h}$ satisfies the Obata’s differential equation (cf. [21,22]) and, therefore, $(M,g)$ is isometric to ${\mathbf{S}}^n\left(c\right)$. However, if n is even, it is known that on an even dimensional Riemannian manifold of positive sectional curvature, a Killing vector field has a zero (cf. [12]) and we get a contradiction to the fact that $\xi$ is a unit Killing vector field. Hence, n must be odd, and $2m+1$ and $(M,g)$ are isometric to the sphere ${\mathbf{S}}^{2m+1}\left(c\right)$.

Conversely, suppose $(M,g)$ is isometric to ${\mathbf{S}}^{2m+1}\left(c\right)$. Treating ${\mathbf{S}}^{2m+1}\left(c\right)$ as a hypersurface of the complex space $C^{(m+1)}$ with unit normal vector field $\varsigma$ and shape operator $A=-\sqrt{c}I$. Using complex structure J on $C^{(m+1)}$, we get unit vector field $\xi =-J\varsigma$ on ${\mathbf{S}}^{2m+1}\left(c\right)$. Denote the Euclidean connection on $C^{(m+1)}$ by D and the Hermitian Euclidean metric by $⟨,⟩$, we have

$$D_X\xi =\sqrt{c}JX,X\in \mathfrak{X}\left({\mathbf{S}}^{2m+1}\left(c\right)\right).$$

Denoting the induced Riemannian connection on ${\mathbf{S}}^{2m+1}\left(c\right)$ by ∇ and defining $\mathbf{F}X={\left(JX\right)}^T$, the tangential component of $JX$, in the above equation gives

$${\nabla }_X\xi -\sqrt{c}g\left(X,\xi \right)\varsigma =\sqrt{c}\mathbf{F}X+\sqrt{c}{\left(JX\right)}^{\perp },$$

where ${\left(JX\right)}^{\perp }$ is the normal component of $JX$ and g is the induced metric on ${\mathbf{S}}^{2m+1}\left(c\right)$. Equating tangential components, we have

$${\nabla }_X\xi =\sqrt{c}\mathbf{F}X,X\in \mathfrak{X}\left(M\right),$$

and since, by definition of $\mathbf{F}$, it is skew-symmetric and we conclude that $\xi$ is a unit Killing vector field on ${\mathbf{S}}^{2m+1}\left(c\right)$ and that

$$\mathbf{F}\left(\xi \right)=0.$$

(25)

Now, choose a nonzero constant vector field $\mathbf{u}$ on the complex space $C^{(m+1)}$ and define smooth function h on ${\mathbf{S}}^{2m+1}\left(c\right)$ by $h=⟨\mathbf{u},\varsigma ⟩$ and define a vector field $\mathbf{v}$ on ${\mathbf{S}}^{2m+1}\left(c\right)$ by $\mathbf{v}={\mathbf{u}}^T$, the tangential component of $\mathbf{u}$ to ${\mathbf{S}}^{2m+1}\left(c\right)$. Then, we have $\mathbf{u}=\mathbf{v}+h\varsigma$. Differentiating this equation with respect to $X\in \mathfrak{X}\left({\mathbf{S}}^{2m+1}\left(c\right)\right)$, we get

$$0={\nabla }_X\mathbf{v}-\sqrt{c}g(X,\mathbf{v})\varsigma +X\left(h\right)\varsigma +\sqrt{c}hX$$

and we conclude

$${\nabla }_X\mathbf{v}=-\sqrt{c}hX,X\left(h\right)=\sqrt{c}g(X,\mathbf{v}),X\in \mathfrak{X}\left({\mathbf{S}}^{2m+1}\left(c\right)\right),$$

that is, $\mathbf{v}=\frac{1}{\sqrt{c}}gradh$ and

$${\nabla }_{X}gradh=-chX,X\in \mathfrak{X}\left({\mathbf{S}}^{2m+1}\left(c\right)\right).$$

(26)

We claim that h is not a constant, if h is a constant, Equation (26) implies $h=0$ and then $\mathbf{v}=0$, which means constant vector field $\mathbf{u}=0$ on ${\mathbf{S}}^{2m+1}\left(c\right)$. However, $\mathbf{u}$ being a constant vector field, it will be zero on $C^{(m+1)}$, a contradiction to our assumption that $\mathbf{u}\ne 0$. Hence, h is a non-constant function. Additionally, we have $\overline{h}=\xi \left(h\right)=g(\xi ,gradh)=\sqrt{c}g(\xi ,\mathbf{v})$, and it implies

$$X\left(\overline{h}\right)=cg\left(\mathbf{F}X,\mathbf{v}\right)-chg(\xi ,X),$$

that is,

$$grad\overline{h}=-c\mathbf{Fv}-ch\xi .$$

If $\overline{h}$ is a constant, it will imply $\mathbf{Fv}=-h\xi$, that is, $h=g\left(\mathbf{v},\mathbf{F}\xi \right)=0$ (in view of Equation (25)) and is a contradiction as h is non-constant. Hence, $\overline{h}$ is non-constant. Additionally, Equation (26) implies $\Delta h=-(2m+1)ch$ and, therefore, we have

$${⊡}_{\xi }\Delta h=-(2m+1)c{⊡}_{\xi }h=-(2m+1)c{\mathcal{H}}^h(\xi ,\xi )=-(2m+1)c\overline{\overline{h}}.$$

Using the expression for the Ricci curvature of the sphere ${\mathbf{S}}^{2m+1}\left(c\right)$, we have

$$Ric\left(grad\overline{h},grad\overline{h}\right)=2mc{∥grad\overline{h}∥}^2.$$

Moreover, the first nonzero eigenvalue ${\lambda }_1$ of the sphere ${\mathbf{S}}^{2m+1}\left(c\right)$ is ${\lambda }_1=(2m+1)c$ and, therefore, with $n=2m+1$, we have

$$\left(\frac{nc}{{\lambda }_1}+n-2\right)=2m$$

and, consequently,

$$Ric\left(grad\overline{h},grad\overline{h}\right)=c\left(\frac{nc}{{\lambda }_1}+n-2\right){∥grad\overline{h}∥}^2$$

and all the requirements of the statement are met. □

Recall that, Fischer and Marsden considered a differential equation

$$\left(\Delta h\right)g+hRic={\mathcal{H}}^h$$

(27)

on a Riemannian manifold $(M,g)$ (cf. [19]), and have shown that, if a Riemannian manifold admits a nontrivial solution of this differential equation, then its scalar curvature $\mathbf{r}$ is a constant.

Definition 1.

We call a Riemannian manifold $(M,g)$ admitting a nontrivial solution of the differential Equation $\left(27\right)$ a Fischer–Marsden manifold.

Observe that, on an n-dimensional Fischer–Marsden manifold $(M,g)$, the nontrivial solution h satisfies

$$\Delta h=-\frac{\mathbf{r}}{n-1}h.$$

(28)

Suppose a Fischer–Marsden manifold $(M,g)$ admits a unit Killing vector field $\xi$, then using Equations (27) and (28), we observe that the nontrivial solution h of differential Equation (27), satisfies

$${⊡}_{\xi }h=\left(Ric(\xi ,\xi )-\frac{\mathbf{r}}{n-1}\right)h.$$

Using Equation (20), we conclude

$$\frac{1}{2}{⊡}_{\xi }{h}^2=\left(Ric(\xi ,\xi )-\frac{\mathbf{r}}{n-1}\right)h^2+{\overline{h}}^2.$$

Thus, we have the following.

Corollary 2.

Let ξ be a unit Killing vector field on an n-dimensional compact Fischer–Marsden manifold $(M,g)$ with constant Ricci curvature $Ric(\xi ,\xi )$. Then

$$Ric(\xi ,\xi )\le \frac{\mathbf{r}}{n-1}$$

and the equality holds if, and only if, the nontrivial solution h of the Fischer–Marsden equation is a constant on the integral curves of ξ.

In [19], Fischer and Marsden conjectured that a compact Fischer–Marsden manifold is an Einstein manifold. Recall that a Riemannian manifold $(M,g)$ is said to be an Einstein manifold if $Ric=\lambda g$, where $\lambda$ is a constant. In the rest of this section, we show that some additional conditions of Fischer–Marsden manifold gives additional outcomes to the Einstein manifold—namely, with additional conditions, we show that a compact Fischer–Marsden manifold is not only Einstein but also a sphere. Note that scalar curvature $\mathbf{r}$ is a constant and on a compact Fischer–Marsden manifold $(M,g)$, Equation (28) implies

$${\int }_M{∥gradh∥}^2=\frac{\mathbf{r}}{n-1}{\int }_M{h}^2,$$

(29)

that is, $\mathbf{r}>0$ (as h is a nontrivial solution of differential Equation (27)). On an n-dimensional compact Fischer–Marsden manifold $(M,g)$, we put $\mathbf{r}=n(n-1)c$, where constant $c>0$.

Theorem 2.

Let ξ be a unit Killing vector field on an n-dimensional compact Fischer–Marsden manifold $(M,g)$ with scalar curvature $\mathbf{r}=n(n-1)c$. Then, the Ricci curvature in the direction of the vector field $gradh$ is bounded below by $(n-1)c$, if and only if, n is odd $(n=2m+1)$ and $(M,g)$ is isometric to the sphere ${\mathbf{S}}^{2m+1}\left(c\right)$.

Proof. 

Let $(M,g)$ be a compact Fischer–Marsden manifold with scalar curvature $\mathbf{r}=n(n-1)c$ and h be a nontrivial solution of the Equation (27). Now,

$${∥A^h+chI∥}^2={∥A^h∥}^2+n{c}^2{h}^2+2ch\Delta h={∥A^h∥}^2+n{c}^2{h}^2+2c\left(\frac{1}{2}\Delta h^2-{∥gradh∥}^2\right).$$

Integrating the above equation and using Equation (29), we conclude

$${\int }_M{∥A^h+hI∥}^2={\int }_M\left({∥A^h∥}^2-c{∥gradh∥}^2\right).$$

(30)

Additionally, the Bochner’s formula gives

$${\int }_M{∥A^h∥}^2={\int }_M\left({\left(\Delta h\right)}^2-Ric\left(gradh,gradh\right)\right),$$

and in view of Equations (28) and (29), the above equation takes the form

$${\int }_M{∥A^h∥}^2={\int }_M\left(nc{∥gradh∥}^2-Ric\left(gradh,gradh\right)\right).$$

Using above equation in Equation (30), we conclude

$${\int }_M{∥A^h+chI∥}^2={\int }_M\left((n-1)c{∥gradh∥}^2-Ric\left(gradh,gradh\right)\right)$$

and using the bound on the Ricci curvature, $Ric(gradh,gradh)\ge (n-1)c{∥gradh∥}^2$, in the above equation, we get $A^h=-chI$. Thus,

$${\nabla }_{X}gradh=-chX,X\in \mathfrak{X}\left(M\right),$$

which is Obata’s differential equation (cf. [21,22]). This proves that $(M,g)$ is isometric to the sphere ${\mathbf{S}}^n\left(c\right)$. As seen in the proof of Theorem 1, we see that n is odd, $n=2m+1$ and $(M,g)$ is isometric to ${\mathbf{S}}^{2m+1}\left(c\right)$.

Conversely, we have shown in the proof of Theorem 1, that there exists a unit Killing vector field $\xi$ on the sphere ${\mathbf{S}}^{2m+1}\left(c\right)$ and the eigenfunction h of $\Delta$ corresponding to first nonzero eigenvalue $(2m+1)c$. Moreover, using Equation (26), we have

$${\mathcal{H}}^h=-chg,$$

that is,

$$\left(\Delta h\right)g+hRic=-(2m+1)chg+2mchg=-chg.$$

Hence, the Fischer–Marsden differential Equation (27) holds and consequently, ${\mathbf{S}}^{2m+1}\left(c\right)$ is a Fischer–Marsden manifold with Ricci curvature equal to $2mc$. Thus, all the conditions in the statement are met. □

5. Conclusions

We have seen that, given a unit Killing vector field $\xi$ on a compact Riemannian manifold $(M,g)$, there is associated a self adjoint operator ${⊡}_{\xi }:C^{\infty }\left(M\right)\to C^{\infty }\left(M\right)$ that has similar properties to that of the Laplace operator. As an application of this operator ${⊡}_{\xi }$ we get a characterization of an odd dimensional sphere (Theorem 1). There are questions related to this operator ${⊡}_{\xi }$ those could be subject of future research, such as showing that eigenspaces of this operator ${⊡}_{\xi }$ are finite dimensional and are mutually orthogonal with respect to different eigenvalues, as well as the relation between volume and the first nonzero eigenvalue.