Some Rigidity Results Related to Conformal Vector Fields

Abstract

This article explores properties of conformal vector fields on a Riemannian manifold, focusing on conditions that lead to the manifold being isometric to the Euclidean space. Given a conformal vector $\zeta$ with conformal factor $\sigma$ on a Riemannian manifold $\left(N,g\right)$, there is naturally associated a skew-symmetric tensor $\chi$ to $\zeta$ called the essential tensor of $\zeta$. It is shown that the essential tensor $\chi$ plays a vital role in our study. We intend to analyze when a conformal vector field becomes a Killing vector field. In a first result of this article, we obtain a necessary and sufficient geometric condition on a complete and connected Riemannian manifold $\left(N,g\right)$ admitting a conformal vector field $\zeta$ so that $\zeta$ is a Killing vector field. In the rest of the article, we obtain characterizations of a Euclidean space using conformal vector fields. In the first such result, it is shown that an n-dimensional complete and connected Riemannian manifold $\left(N,g\right)$, $n>2$ admits a conformal vector field $\zeta$ with conformal factor $\sigma \ne 0$ and essential tensor $\chi$ such that the affinity tensor of $\nabla \sigma$ is zero, the function $\zeta \left(\sigma \right)$ is a constant, and $\zeta$ annihilates $\chi$ if and only if $\left(N,g\right)$ is isometric to the Euclidean space $E^n$. Similarly, in a second characterization of the Euclidean space $E^n$ using a conformal vector field $\zeta$, we use the following conditions: $\zeta$ annihilates the Ricci operator S, $\nabla \sigma$ annihilates $\chi$, and the vector field $\chi \left(\zeta \right)$ is incompressible. Finally, we consider a conformal vector field $\zeta$ with conformal factor $\sigma \ne 0$ and essential tensor $\chi$ on a complete and connected Riemannian manifold $\left(N,g\right)$ such that the Hessian operator $H_{\sigma }$ is invariant under the local flow of $\zeta$ so that the function $\zeta \left(\sigma \right)-{\sigma }^2$ is a subharmonic function and $\zeta$ annihilates $\chi$, and show that $\left(N,g\right)$ is isometric to the Euclidean space $E^n$. The converse holds as well.

1. Introduction

A conformal vector field $\zeta$ on a Riemannian manifold $\left(N,g\right)$ is one with a local flow that consists of local conformal diffeomorphisms of $\left(N,g\right)$, or equivalently

$${\displaystyle \frac{1}{2}}{L}_{\zeta }g=\sigma g,$$

where $L_{\zeta }$ is the Lie derivative with respect to $\zeta$ and $\sigma$ is a smooth function on N, called the conformal factor. If the conformal factor $\sigma =0$, then $\zeta$ is a Killing vector field. It is known that by a conformal change of the metric, any conformal vector field on a Riemannian manifold $\left(N,g\right)$ that has no zeros can be made into a Killing vector field. Such a conformal vector field $\zeta$ having no zeros on $\left(N,g\right)$ is called an inessential conformal vector field. A conformal vector field that is not inessential is called an essential conformal vector field. Since inessential conformal vector fields on a Riemannian manifold $\left(N,g\right)$ can be made into Killing vector fields through conformal change of metric, our focus is on the study of essential conformal vector fields. If the Riemannian manifold $\left(N,g\right)$ is compact, the famous conjecture of Lichnerowicz (cf. [1]) about essential conformal vector fields states that a compact Riemannian manifold $\left(N,g\right)$$dimN\ge 3$ admitting an essential conformal vector field is conformally equivalent to a sphere. This conjecture, which is now a proved result in a more general form (cf. [2]), states that an n-dimensional complete Riemannian manifold $\left(N,g\right)$ that admits a complete and essential conformal vector field is conformally equivalent to either the sphere $S^n$ or to the Euclidean space $E^n$. It is worth noting that conformal vector fields not only reflect on the geometry of the manifold on which they are defined, but also play a significant role in physics and relativity (cf. [3,4,5,6,7,8,9]).

It is well known that conformal vector fields influence the geometry of the Riemannian manifold on which they exist. Among conformal vector fields, closed conformal vector fields on a Riemannian manifold have been extensively studied (cf. [2,5,6,7,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24]). Observe that the defining equation implies that the conformal factor $\sigma$ of the conformal vector $\zeta$ on an n-dimensional Riemannian manifold $\left(N,g\right)$ is given by $\sigma =n^{-1}div\zeta$; in the case that $\left(N,g\right)$ is Einstein and $\zeta$ is a closed conformal vector field, it turns out that $\nabla \sigma$ is again a conformal vector field and gives a differential equation, which is studied in detail in [15].

If a conformal vector field $\zeta$ with conformal factor $\sigma$ on a Riemannian manifold $\left(N,g\right)$ is not closed, then using the 1-form $\beta$ dual to $\zeta$, one comes across a skew-symmetric tensor $\chi$ on $\left(N,g\right)$ defined by

$$d\beta \left(F_1,F_2\right)=2g\left(\chi F_1,F_2\right)$$

for smooth vector fields $F_1,F_2$ on N. This skew-symmetric tensor $\chi$ is calle the essential tensor of the conformal vector field $\zeta$, and plays a vital role in the study of non-closed conformal vector fields (cf. [13,25,26]). In the next section, we discuss examples of non-closed conformal vector fields.

Note that spaces such as the sphere $S^n\left(c\right)$, the Euclidean space $E^n$, the hyperbolic space $H^n(-c)$, warped products $I\times {}_{f}S^n$, $S^1\times {}_{f}S^n$, and pseudo-ubmbilical submanifolds of complex space forms $M^{2n}\left(c\right)$ are equipped with conformal vector fields (cf. [11,13,14,25]). Taking a clue from the role of position vector fields of submanifolds of the Euclidean space, there has been interest in studying submanifolds of the Riemannian manifold $\left(N,g\right)$ that admits a conformal vector field (changing the role of position vector field of Euclidean space to the conformal vector field on $\left(N,g\right)$); cf. [11] and references therein.

An inessential conformal vector field $\zeta$ on a Riemannian manifold $\left(N,g\right)$ is made Killing on $\left(N,\overline{g}\right)$, where $\overline{g}$ is conformal to g. Instead of imposing the condition $\sigma =0$ resulting in making a conformal vector field $\zeta$ on a Riemannian manifold $\left(N,g\right)$ Killing, one would be interested in exploring geometric conditions on $\left(N,g\right)$ that will force $\sigma =0$. Such a question will require a specific geometric condition satisfied by a Killing vector field $\zeta$ on a Riemannian manifold $\left(N,g\right)$. In this article, we show that a Killing vector field $\zeta$ on a Riemannian manifold $\left(N,g\right)$ satisfies

$$Ric\left(\zeta ,\zeta \right)=div\left({\nabla }_{\zeta }\zeta \right)+{∥\chi ∥}^2,$$

(1)

where $Ric$ is the Ricci tensor and $\chi$ is the essential tensor of the Killing vector field $\zeta$. A natural question that arises is whether the property (1) is the characteristic property of a Killing vector field, or equivalently, whether condition (1) is satisfied by a conformal vector field $\zeta$ with essential tensor $\chi$ on a complete and connected Riemannian manifold $\left(N,g\right)$, and if this necessarily implies that $\zeta$ is a Killing vector field. We show that the answer to this question is in the affirmative (see Theorem 2).

We mentioned earlier that there is a focus on studying submanifolds of a Riemannian manifold $\left(N,g\right)$ that admits a conformal vector field (as a replacement of the role of position vector field for submanifolds of a Euclidean space). This initiates the question of finding conditions on an n-dimensional Riemannian manifold $\left(N,g\right)$ that admits a conformal vector field $\zeta$ with conformal factor $\sigma$ and essential tensor $\chi$ so that $\left(N,g\right)$ isometric to the Euclidean space $E^n$. In the rest of this article, we focus on this issue and find that the essential tensor $\chi$ plays an important role in this pursuit. We recall that for a smooth function f on a Riemannian manifold $\left(N,g\right)$, the affinity tensor of $\nabla f$ is the tensor (cf. [8])

$$\left(L_{\nabla f}\nabla \right),$$

and we call the function f a gairaffhinity function if the affinity tensor is zero. It is known that the Euclidean space $E^n$ admits a conformal vector field $\zeta$ where the conformal factor $\sigma$ is a ghairaffinity function (see next section). This motivates the question of finding conditions on a complete and connected Riemannian manifold $\left(N,g\right)$ that admits a conformal vector field $\zeta$ with a conformal factor $\sigma$ that is a ghairaffinity function and essential tensor $\chi$ so that $\left(N,g\right)$ is isometric to the Euclidean space $E^n$. We answer this question and show that an n-dimensional complete and connected Riemannian manifold $\left(N,g\right)$, $n>2$ admits a conformal vector field $\zeta$ with conformal factor $\sigma \ne 0$ a ghairaffinity function such that the function $\zeta \left(\sigma \right)$ is a constant and $\zeta$ annihilating the essential tensor $\chi$ is necessarily isometric to the Euclidean space $E^n$. The converse also holds (see Theorem 3).

Note that on the Euclidean space $E^n$, the vector field

$$\zeta =\lambda \sum _i{u}^i{\displaystyle \frac{\partial }{\partial u^i}},$$

where $\lambda$ is a nonzero constant and $u^1,\dots ,u^n$ are Euclidean coordinates, is a conformal vector field on $E^n$ with conformal factor $\sigma =\lambda \ne 0$ and essential tensor $\chi =0$. We see that this example satisfies the following set of conditions:

(a) (i) $S\left(\zeta \right)=0$ (as $S=0$ for $E^n$), (ii) $\chi \left(\nabla \sigma \right)=0$, and (iii) $div\left(\chi \left(\zeta \right)\right)=0$

(b) (i) $div\left(\chi \left(\zeta \right)\right)=0$, (ii) $\zeta \left(\sigma \right)=0$, and (iii) $Ric\left(\zeta ,\zeta \right)\le 0$.

This naturally initiates the question of whether a complete and connected Riemannian manifold $\left(N,g\right)$ admitting a conformal vector field $\zeta$ with conformal factor $\sigma \ne 0$ and essential tensor $\chi$ (not necessarily zero), satisfying any one of the above set of conditions, is necessarily isometric to the Euclidean space $E^n$. We answer this question in the affirmative.

In the next result, we show that an n-dimensional complete and connected Riemannian manifold $\left(N,g\right)$, $n>2$ admits a conformal vector field $\zeta$ with conformal factor $\sigma \ne 0$ and essential tensor $\chi$ satisfying the following: (i) $\zeta$ annihilates the Ricci operator S, (ii) the gradient $\nabla \sigma$ annihilates the essential operator $\chi$, and (iii) the vector $\chi \left(\zeta \right)$ is incompressible and isometric to the Euclidean space $E^n$, with the converse also holding (see Theorem 4). In yet another characterization of the Euclidean space, we use a conformal vector field $\zeta$ on an n-dimensional complete and connected Riemannian manifold $\left(N,g\right)$, $n>2$ with conformal factor $\sigma$ and the essential tensor $\chi$, satisfying: (i) the vector $\chi \left(\zeta \right)$ is incompressible, (ii) the conformal factor $\sigma$ is a constant along the integral curves of $\zeta$, and (iii) the Ricci curvature $Ric\left(\zeta ,\zeta \right)\le 0$ (see Theorem 5).

Note that Tashiro [27] considered the closed conformal vector field $\zeta$ on an n-dimensional complete Riemannian manifold $\left(N,g\right)$, $n\ge 2$ and obtained conditions under which the manifold is isometric to the Euclidean space. Also, in [14] the authors considered a $\phi$-analytic conformal vector field $\zeta$, that is, a conformal vector field $\zeta$ on an n-dimensional complete and connected Riemannian manifold $\left(N,g\right)$ satisfying

$$L_{\zeta }\chi =0,$$

where $\chi$ is the essential tensor of the conformal vector field and, under additional assumptions that $\chi \left(\zeta \right)=0$ and $Ric\left(\zeta ,\zeta \right)=0$, showed that $\left(N,g\right)$ is isometric to the Euclidean space $E^n$ and that the converse also holds. In addition to this result in [14], the authors also showed that if $\zeta$ is a nontrivial $\phi$-analytic conformal vector field and satisfies $\Delta \zeta =0$, where $\Delta$ is the Laplace operator acting on vector fields, then $\left(N,g\right)$ is isometric to the Euclidean space $E^n$ and the converse holds as well.

Finally, we consider an n-dimensional complete and connected Riemannian manifold $\left(N,g\right)$ admitting a conformal vector field $\zeta$ with conformal factor $\sigma \ne 0$ and essential tensor $\chi$. We focus on the Hessian operator $H_{\sigma }$ of the conformal factor $\sigma$. We are interested in the operator $H_{\sigma }$ being invariant under the local flow of the conformal vector field $\zeta$, that is, equivalent to

$$L_{\zeta }{H}_{\sigma }=0,$$

and in this case we say that $H_{\sigma }$ is invariant under $\zeta$.

Tashiro in [27] considered an n-dimensional complete Riemannian manifold $\left(N,g\right)$, $n\ge 2$ admitting a closed conformal vector field $\zeta$ with conformal factor $\sigma$ satisfying

$$H_{\sigma }=bI,$$

where b is a nonzero constant. There, it was shown that $\left(N,g\right)$ is isometric to the Euclidean space $E^n$ (see Theorem 2). The above condition on the Hessian operator satisfies

$$L_{\zeta }{H}_{\sigma }=0,$$

which raises the natural question of what additional conditions on a conformal vector field $\zeta$ with conformal factor $\sigma$ and essential tensor $\chi$ (not necessarily zero) on an n-dimensional complete and connected Riemannian manifold $\left(N,g\right)$ will imply that $\left(N,g\right)$ is isometric to the Euclidean space $E^n$. We answer this question in the last result of this article, where we show that the n-dimensional complete and connected Riemannian manifold $\left(N,g\right)$ admitting a conformal vector field $\zeta$ with conformal factor $\sigma \ne 0$ and essential tensor $\chi$ with $H_{\sigma }$ is invariant under $\zeta$; the function $\left(\zeta \left(\sigma \right)-{\sigma }^2\right)$ is a subharmonic function, and $\chi \left(\zeta \right)=0$ is necessarily isometric to the Euclidean space $E^n$, with the converse also holding.

2. Preliminaries

On a Riemannian manifold $\left(N,g\right)$ with Riemannian metric g, we denote by ∇ the Riemannian connection on $\left(N,g\right)$. A vector field $\zeta$ on $\left(N,g\right)$ is called a conformal vector field if

$${\displaystyle \frac{1}{2}}{L}_{\zeta }g=\sigma g,$$

(2)

where $L_{\zeta }$ stands for Lie derivative with respect to $\zeta$ and $\sigma$ is a smooth function on N called the conformal factor of the conformal vector field $\zeta$. It follows that if $\sigma =0$, then $\zeta$ is a Killing vector field. We denote by $\beta$ the 1-form dual to the conformal vector field $\zeta$, that is, $\beta \left(F\right)=g\left(\zeta ,F\right)$ for $F\in \mathfrak{X}\left(N\right)$, where $\mathfrak{X}\left(N\right)$ is the set of smooth vector fields on N. It is natural to assign to a conformal vector field a skew-symmetric $(1,1)$-tensor field $\chi$ on N defined by

$$d\beta \left(F_1,F_2\right)=2g\left(\chi F_1,F_2\right),F_1,F_2\in \mathfrak{X}\left(N\right).$$

(3)

This tensor $\chi$ assigned to the conformal vector field $\zeta$ is called the essential tensor of $\zeta$, and has vital role to play in the geometry of the conformal vector field $\zeta$. Using the formulas

$$L_{\zeta }g\left(F_1,F_2\right)=g\left({\nabla }_{F_1}\zeta ,F_2\right)+g\left({\nabla }_{F_2}\zeta ,F_1\right),d\beta \left(F_1,F_2\right)=g\left({\nabla }_{F_1}\zeta ,F_2\right)-g\left({\nabla }_{F_2}\zeta ,F_1\right)$$

and adding them while using Equations (2) and (3), one immediately concludes

$${\nabla }_F\zeta =\sigma F+\chi F,F\in \mathfrak{X}\left(N\right).$$

(4)

The curvature tensor of the Riemannian manifold $\left(N,g\right)$

$$R\left(F_1,F_2\right)F_3={\nabla }_{F_1}{\nabla }_{F_2}{F}_3-{\nabla }_{F_2}{\nabla }_{F_1}{F}_3-{\nabla }_{\left[F_1,F_2\right]}{F}_3,$$

(5)

for $F_i\in \mathfrak{X}\left(N\right)$, $i=1,2,3$, and on taking a local frame $\left\{e_1,\dots ,e_n\right\}$ on $\left(N,g\right)$, the Ricci tensor $Ric$ of $\left(N,g\right)$ is defined by

$$Ric\left(F_1,F_2\right)=\sum _{i}g\left(R\left(e_i,F_1\right)F_2,e_i\right),F_1,F_2\in \mathfrak{X}\left(N\right).$$

(6)

Through the symmetries of the Riemannian curvature tensor, it follows that $Ric$ is a symmetric tensor. The Ricci operator S of $\left(N,g\right)$ is defined by

$$Ric\left(F_1,F_2\right)=g\left(S{F}_1,F_2\right),F_1,F_2\in \mathfrak{X}\left(N\right),$$

(7)

which is also a symmetric operator. The scalar curvature $\tau$ of $\left(N,g\right)$ is given by

$$\tau =\sum _{i}Ric\left(e_i,e_i\right).$$

(8)

On using the formula

$$\left({\nabla }_{F_1}S\right)\left(F_2\right)={\nabla }_{F_1}S{F}_2-S\left({\nabla }_{F_1}{F}_2\right),$$

we have the following expression for the gradient $\nabla \tau$ of the scalar curvature $\tau$ (cf. [9]):

$${\displaystyle \frac{1}{2}}\nabla \tau =\sum _i\left({\nabla }_{e_i}S\right)\left(e_i\right).$$

(9)

Given a conformal vector field $\zeta$ on an n-dimensional Riemannian manifold $\left(N,g\right)$ with essential tensor $\chi$, using Equation (4), we have

$$div\zeta =n\sigma ,$$

(10)

where we have used the skew-symmetry of the essential operator $\chi$. We find the following expression for the covariant derivative of the essential tensor $\chi$, though it is not new; yet, as it is so important to our work, we list it below as a Lemma.

Lemma 1. 

Let ζ be a conformal vector field on a Riemannian manifold $\left(N,g\right)$ with conformal factor σ and essential tensor χ. Then,

$$\left({\nabla }_{F_1}\chi \right)\left(F_2\right)=F_2\left(\sigma \right)F_1-g\left(F_1,F_2\right)\nabla \sigma +R\left(F_1,\zeta \right)F_2,F_1,F_2\in \mathfrak{X}\left(N\right).$$

Proof. 

On differentiating Equation (4), we have

$${\nabla }_{F_1}{\nabla }_{F_2}\zeta =F_1\left(\sigma \right)F_2+\left({\nabla }_{F_1}\chi \right)\left(F_2\right)+{\nabla }_{{\nabla }_{F_1}{F}_2}\zeta ,$$

which in view of Equation (5) reveals

$$R\left(F_1.F_2\right)\zeta =F_1\left(\sigma \right)F_2-F_2\left(\sigma \right)F_1+\left({\nabla }_{F_1}\chi \right)\left(F_2\right)-\left({\nabla }_{F_2}\chi \right)\left(F_1\right).$$

(11)

Note that the 2-form $d\beta$ in Equation (3) is closed; therefore, we have

$$g\left(\left({\nabla }_{F_1}\chi \right)\left(F_2\right),F_3\right)+g\left(\left({\nabla }_{F_2}\chi \right)\left(F_3\right),F_1\right)+g\left(\left({\nabla }_{F_3}\chi \right)\left(F_1\right),F_2\right)=0,$$

which, on using skew-symmetry of the essential tensor in second term and then using Equation (11), yields

$$g\left(R\left(F_1.F_2\right)\zeta -F_1\left(\sigma \right)F_2+F_2\left(\sigma \right)F_1,F_3\right)+g\left(\left({\nabla }_{F_3}\chi \right)\left(F_1\right),F_2\right)=0.$$

Transvecting the slot $F_2$ in the above equation, we get the result in the Lemma. □

Recall that the Riemannian curvature tensor of a Riemannian manifold $\left(N,g\right)$ is defined by

$$R\left(E_1,E_2;E_3,E_4\right)=h\left(R(E_1,E_2)E_3,E_4\right),E_i\in \mathfrak{X}\left(N\right),i=1,\dots ,4.$$

Lemma 2. 

Let ζ be a conformal vector field on a Riemannian manifold $\left(N,g\right)$ with conformal factor σ and essential tensor χ. Then,

$$-R\left(\zeta ,F_1;F_2,\zeta \right)+{\displaystyle \frac{1}{2}}\left(L_{{\nabla }_{\zeta }\zeta }g\right)\left(F_1,F_2\right)+g\left(\chi F_1,\chi F_2\right)=\zeta \left(\sigma \right)g\left(F_1,F_2\right)+{\sigma }^{2}g\left(F_1,F_2\right)$$

for smooth vector fields $F_1,F_2$ on N.

Proof. 

Following the techniques described in [28], on using

$$\left(L_{\zeta }g\right)\left(F_1,F_2\right)=g\left({\nabla }_{F_1}\zeta ,F_2\right)+g\left({\nabla }_{F_2}\zeta ,F_1\right)$$

we have that

$$\begin{array}{cc}\hfill \left(L_{\zeta }{L}_{\zeta }g\right)\left(F_1,F_2\right)& =\zeta \left(g\left({\nabla }_{F_1}\zeta ,F_2\right)+g\left({\nabla }_{F_2}\zeta ,F_1\right)\right)-\left(L_{\zeta }g\right)\left(\left[\zeta ,F_1\right],F_2\right)\hfill \\ \hfill & -\left(L_{\zeta }g\right)\left(F_1,\left[\zeta ,F_2\right]\right).\hfill \end{array}$$

(12)

Also, we use

$$\left(L_{{\nabla }_{\zeta }\zeta }g\right)\left(F_1,F_2\right)=g\left({\nabla }_{F_1}{\nabla }_{\zeta }\zeta ,F_2\right)+g\left({\nabla }_{F_2}{\nabla }_{\zeta }\zeta ,F_1\right)$$

in Equation (12), and upon simplification arrive at

$$\left(L_{\zeta }{L}_{\zeta }g\right)\left(F_1,F_2\right)=2R\left(\zeta ,F_1;\zeta ,F_2\right)+\left(L_{{\nabla }_{\zeta }\zeta }g\right)\left(F_1,F_2\right)+2g\left({\nabla }_{F_1}\zeta ,{\nabla }_{F_2}\zeta \right).$$

Using Equation (4) and the skew-symmetry of the tensor $\chi$ in the above equation, we obtain

$$\begin{array}{cc}\hfill \left(L_{\zeta }{L}_{\zeta }g\right)\left(F_1,F_2\right)& =2R\left(\zeta ,F_1;\zeta ,F_2\right)+\left(L_{{\nabla }_{\zeta }\zeta }g\right)\left(F_1,F_2\right)+2{\sigma }^{2}g\left(F_1,F_2\right)\hfill \\ \hfill & +2g\left(\chi F_1,\chi F_2\right).\hfill \end{array}$$

(13)

Now, taking the Lie derivative in Equation (2), we have

$$L_{\zeta }{L}_{\zeta }g=2\zeta \left(\sigma \right)g+4{\sigma }^{2}g,$$

and inserting this into Equation (13) gives the required expression. □

Let $\zeta$ be a conformal vector field on a Riemannian manifold $\left(N,g\right)$ with essential tensor $\chi$. Then, the squared length of $\chi$ is given by

$${∥\chi ∥}^2=\sum _{i}g\left(\chi e_i,\chi e_i\right)$$

(14)

for a local frame $\left\{e_1,\dots ,e_n\right\}$ on $\left(N,g\right)$.

Note that if the conformal factor $\sigma =0$, then $\zeta$ is called a Killing vector field. Thus, for a Killing vector field $\zeta$ on a Riemannian manifold $\left(N,g\right)$, Equation (4) takes the form

$${\nabla }_F\zeta =\chi F,F\in \mathfrak{X}\left(N\right).$$

(15)

Now, we derive the following important geometric characteristic property of a Killing vector field $\zeta$ on a Riemannian manifold $\left(N,g\right)$, which will be useful in deriving geometric conditions under which a conformal vector field is Killing.

Theorem 1. 

Let ζ be a Killing vector field on a Riemannian manifold $\left(N,g\right)$ with essential tensor χ. Then, the following holds:

$$Ric\left(\zeta ,\zeta \right)=div\left({\nabla }_{\zeta }\zeta \right)+{∥\chi ∥}^2.$$

Proof. 

As $\zeta$ is a Killing vector field, Lemma 1 gives

$$\left({\nabla }_{F_1}\chi \right)\left(F_2\right)=R\left(F_1,\zeta \right)F_2,F_1,F_2\in \mathfrak{X}\left(N\right).$$

(16)

Taking a local frame $\left\{e_1,\dots ,e_n\right\}$ on $\left(N,g\right)$ and using the expression

$$S\left(\zeta \right)=\sum _{i}R\left(\zeta ,e_i\right)e_i$$

in Equation (16), we get

$$S\left(\zeta \right)=-\sum _i\left({\nabla }_{e_i}\chi \right)\left(e_i\right),$$

that is,

$$Ric\left(\zeta ,\zeta \right)=-\sum _{i}g\left(\zeta ,\left({\nabla }_{e_i}\chi \right)\left(e_i\right)\right).$$

(17)

Now, using Equation (15) and the skew-symmetry of the essential tensor $\chi$, we have

$$\begin{array}{ccc}\hfill div\left({\nabla }_{\zeta }\zeta \right)& =& div\left(\chi \zeta \right)=\sum _{i}g\left({\nabla }_{e_i}\chi \zeta ,e_i\right)=\sum _{i}g\left(\left({\nabla }_{e_i}\chi \right)\left(\zeta \right)+\chi \left({\nabla }_{e_i}\zeta \right),e_i\right)\hfill \\ & =& \sum _{i}g\left(\left({\nabla }_{e_i}\chi \right)\left(\zeta \right),e_i\right)-\sum _{i}g\left(\chi e_i,\chi e_i\right).\hfill \end{array}$$

Thus, using Equations (14) and (17) in the above equation gives the desired result. □

Recall that the affinity tensor of a vector field F on a Riemannian manifold $\left(N,g\right)$ is given by (cf. [9])

$$\left(L_F\nabla \right)\left(F_1,F_2\right)=L_F{\nabla }_{F_1}{F}_2-{\nabla }_{L_F{F}_1}{F}_2-{\nabla }_{F_1}{L}_F{F}_2,F_1,F_2\in \mathfrak{X}\left(N\right),$$

and we call a vector field F on $\left(N,g\right)$ a ghairaffinity vector if

$$\left(L_F\nabla \right)=0.$$

Also, we call a function $\rho$ on a Riemannian manifold $\left(N,g\right)$ a ghairaffinity function if

$$\left(L_{\nabla \rho }\nabla \right)=0,$$

which is equivalent to

$$R\left(F_1,\nabla \rho \right)F_2={\nabla }_{F_1}{\nabla }_{F_2}\nabla \rho -{\nabla }_{{\nabla }_{F_1}{F}_2}\nabla \rho ,F_1,F_2\in \mathfrak{X}\left(N\right).$$

(18)

Examples of ghairaffinity functions are in abundance on Euclidean spaces. For example, on the Euclidean space $E^n$ with Euclidean coordinates $u^1,\dots ,u^n$, Euclidean metric g, and Euclidean connection ∇, taking the smooth function $\rho ={\displaystyle \frac{1}{2}}{\left(u^i\right)}^2$ for a Euclidean coordinate $u^i$ gives

$$\nabla \rho =u^i{\displaystyle \frac{\partial }{\partial u^i}},$$

and we compute

$${\nabla }_{F_2}\nabla \rho =F_2\left(u^i\right){\displaystyle \frac{\partial }{\partial u^i}}\mathrm{and}{\nabla }_{F_1}{\nabla }_{F_2}\nabla \rho =F_1{F}_2\left(u^i\right){\displaystyle \frac{\partial }{\partial u^i}}.$$

Consequently, we arrive at

$${\nabla }_{F_1}{\nabla }_{F_2}\nabla \rho -{\nabla }_{{\nabla }_{E_1}{E}_2}\nabla \rho =Hess\left(u^i\right)\left(F_1,F_2\right){\displaystyle \frac{\partial }{\partial u^i}},$$

(19)

where $Hess\left(f\right)$ of a smooth function f is defined by

$$Hess\left(f\right)\left(F_1,F_2\right)=g\left({\nabla }_{E_1}\nabla f,F_2\right)$$

for smooth vector fields $F_1,F_2$. Since for the Euclidean coordinate $u^i$ we see that $Hess\left(u^i\right)\left(F_1,F_2\right)=0$ and the curvature tensor of the Euclidean space $E^n$ is zero, we conclude through Equations (18) and (19) that the function $\rho$ is a ghairaffinity function on $E^n$.

For a smooth function $\rho$ on a Riemannian manifold $\left(N,g\right)$, the symmetric operator $H_{\rho }$ defined by

$$H_{\rho }\left(F\right)={\nabla }_F\nabla \rho ,F\in \mathfrak{X}\left(N\right)$$

(20)

is called the Hessian operator, and is related to $Hess\left(\rho \right)$ by

$$Hess\left(\rho \right)\left(F_1,F_2\right)=g\left(H_{\rho }\left(F_1\right),F_2\right),F_1,F_2\in \mathfrak{X}\left(N\right).$$

(21)

Now, we wish to explore some examples of non-closed conformal vector fields, which will help in supporting the results in this article.

(1) Consider a $(2n-1)$-dimensional Riemannian manifold $\left(N,g\right)$ that has Sasakian structure $\left(\varphi ,\xi ,\eta \right)$, where $\varphi$ is a $\left(1,1\right)$-tensor field, $\xi$ is a unit vector field, $\eta$ is a 1-form dual to $\xi$ satisfying (cf. [10])

$${\varphi }^2=-I+\eta \otimes \xi ,\varphi \left(\xi \right)=0,\eta \circ \varphi =0,\eta \left(\xi \right)=1,$$

$$g\left(\varphi F_1,\varphi F_2\right)=g\left(F_1,F_2\right)-\eta \left(F_1\right)\eta \left(F_2\right),F_1,F_2\in \mathfrak{X}\left(N\right),$$

and

$$\left({\nabla }_{F_1}\varphi \right)\left(F_2\right)=g\left(F_1,F_2\right)\xi -\eta \left(F_2\right)F_1,F_1,F_2\in \mathfrak{X}\left(N\right).$$

Choose a closed conformal vector field $\mathbf{u}$ with conformal factor $\sigma$ on $\left(N,g\right)$, that is, satisfying

$${\nabla }_F\mathbf{u}=\sigma F,F\in \mathfrak{X}\left(N\right);$$

is this possible? The answer is yes: the unit sphere $S^{2n-1}$ has many closed conformal vector fields along with a Sasakian structure. Now, defining a vector field $\zeta$ on the Sasakian manifold $\left(N,g\right)$ by

$$\zeta =\mathbf{u}+\xi ,$$

it follows that

$${\nabla }_F\zeta =\sigma F-\varphi \left(F\right),F\in \mathfrak{X}\left(N\right),$$

that is, $\zeta$ is a conformal vector field with conformal factor $\sigma$ and essential tensor $\chi =-\varphi$. As $\chi \left(\zeta \right)=-\varphi \left(\mathbf{u}\right)$, it is straightforward to compute

$$div\chi \left(\zeta \right)=\left(2n-2\right)g\left(\mathbf{u},\xi \right).$$

Thus, if $\mathbf{u}$ is orthogonal to $\xi$, we get

$$div\chi \left(\zeta \right)=0.$$

(2) Consider the even-dimensional Euclidean space $\left(E^{2n},J,g\right)$, where J is the complex structure and g is the Euclidean metric, which is Hermitian on $E^{2n}$. Consider the position vector field

$$\xi =\sum _i{u}^i{\displaystyle \frac{\partial }{\partial u^i}},$$

where $u^1,\dots ,u^{2n}$ are the Euclidean coordinates on $E^{2n}$. Now, defining a vector field $\zeta$ on $E^{2n}$ by

$$\zeta =\xi +J\xi ,$$

it follows that

$${\nabla }_F\zeta =F+JF,F\in \mathfrak{X}\left(E^{2n}\right).$$

This proves that $\zeta$ is a conformal vector field on $\left(E^{2n},J,g\right)$ with conformal factor $\sigma =1$ and essential tensor $\chi =J$. We have $S\left(\zeta \right)=0$, $\chi \left(\nabla \sigma \right)=0$, and $\sigma$ is a ghairaffinity function; in addition, we have

$$div\chi \left(\zeta \right)=-2n.$$

3. Conformal Vector Fields on Complete Manifolds

We wish to explore geometric conditions on a complete Riemannian manifold $\left(N,g\right)$ so that a conformal vector field $\zeta$ with conformal factor $\sigma$ and essential tensor $\chi$ on $\left(N,g\right)$ becomes a Killing vector field. In particular, we shall establish that the geometric property described in Theorem 1 is an essential characteristic of a manifold admitting a Killing vector field, as seen in the following.

Theorem 2. 

Let ζ be a conformal vector field on a complete connected Riemannian manifold $\left(N,g\right)$ with essential tensor χ. If the condition

$$Ric\left(\zeta ,\zeta \right)=div\left({\nabla }_{\zeta }\zeta \right)+{∥\chi ∥}^2$$

holds, then ζ is a Killing vector field. The converse is also true.

Proof. 

Let $\zeta$ be a conformal vector field on a complete connected Riemannian manifold $\left(N,g\right)$ with conformal factor $\sigma$ and essential tensor $\chi$ with $dimN=n$. Suppose the condition

$$Ric\left(\zeta ,\zeta \right)=div\left({\nabla }_{\zeta }\zeta \right)+{∥\chi ∥}^2$$

(22)

holds. Then, by Lemma 2 we have

$$\begin{array}{cc}\hfill -R\left(\zeta ,F_1;F_2,\zeta \right)+{\displaystyle \frac{1}{2}}\left(L_{{\nabla }_{\zeta }\zeta }g\right)\left(F_1,F_2\right)+g\left(\chi F_1,\chi F_2\right)& =\zeta \left(\sigma \right)g\left(F_1,F_2\right)\hfill \\ \hfill & +{\sigma }^{2}g\left(F_1,F_2\right)\hfill \end{array}$$

(23)

for smooth vector fields $F_1,F_2$ on N. Taking the trace in the above equation and using Equation (6), we get

$$-Ric\left(\zeta ,\zeta \right)+div\left({\nabla }_{\zeta }\zeta \right)+{∥\chi ∥}^2=n\left(\zeta \left(\sigma \right)+{\sigma }^2\right).$$

Thus, in view of Equation (22), we get

$$\zeta \left(\sigma \right)+{\sigma }^2=0.$$

(24)

Let $\alpha \left(s\right)$ be an integral curve of $\zeta$. Then, on taking $\zeta \left(s\right)=\zeta \left(\alpha \left(s\right)\right)$ and $\sigma \left(s\right)=\left(\sigma \circ \alpha \right)\left(s\right)$, Equation (24) assumes the form

$${\displaystyle \frac{d\sigma }{ds}}=-{\sigma }^2\left(s\right).$$

(25)

Assume that $\sigma \left(0\right)={\sigma }_0$; then, the unique solution of the initial value problem

$${\displaystyle \frac{d\sigma }{ds}}=-{\sigma }^2\left(s\right),\sigma \left(0\right)={\sigma }_0$$

is

$$\sigma \left(s\right)={\displaystyle \frac{{\sigma }_0}{s{\sigma }_0+1}}.$$

(26)

Now, suppose that the constant ${\sigma }_0\ne 0$. Then, we compute

$${\displaystyle \frac{d}{ds}}∥\zeta \left(s\right)∥={\displaystyle \frac{1}{∥\zeta \left(s\right)∥}}g\left({\nabla }_{\zeta \left(s\right)}\zeta \left(s\right),\zeta \left(s\right)\right),$$

which in view of Equation (4) and the skew-symmetry of the essential tensor $\chi$ yields

$${\displaystyle \frac{d}{ds}}∥\zeta \left(s\right)∥=\sigma \left(s\right)∥\zeta \left(s\right)∥.$$

Using Equation (26) in the above equation, we arrive at

$${\displaystyle \frac{d}{ds}}∥\zeta \left(s\right)∥={\displaystyle \frac{{\sigma }_0}{s{\sigma }_0+1}}∥\zeta \left(s\right)∥,$$

that is,

$${\displaystyle \frac{d}{ds}}\left(\ln ∥\zeta \left(s\right)∥\right)={\displaystyle \frac{d}{ds}}\left(\ln \left(s{\sigma }_0+1\right)\right).$$

Integrating above equation yields

$$∥\zeta \left(s\right)∥=\left(s{\sigma }_0+1\right)∥\zeta \left(0\right)∥.$$

(27)

Note that the solution $\sigma \left(s\right)$ in Equation (26) blows away at

$$\overline{s}=-{\displaystyle \frac{1}{{\sigma }_0}},$$

and we let $L_0^{\overline{s}}$ be the length of the curve $\alpha \left(s\right)$ from 0 to $\overline{s}$. Then, using Equation (27), we have

$$L_0^{\overline{s}}={\displaystyle \underset{0}{\overset{\overline{s}}{\int }}}∥\zeta \left(s\right)∥ds={\displaystyle \frac{1}{2{\sigma }_0}}∥\zeta \left(0\right)∥\left(0-1\right)={\displaystyle \frac{1}{2}}∥\zeta \left(0\right)∥\overline{s}<\infty .$$

Hence, as $\left(N,g\right)$ is complete, there exists a point $x\in N$ such that $\alpha \left(\overline{s}\right)=x$. Note that as $s\to \overline{s}$, $\alpha \left(s\right)\to x$. Now, using Equation (26), we can see that

$$\sigma \left(x\right)=\underset{s\to \overline{s}}{lim}\sigma \left(\alpha \left(s\right)\right)=\underset{s\to \overline{s}}{lim}{\displaystyle \frac{{\sigma }_0}{s{\sigma }_0+1}}=\infty ,$$

which contradicts the fact that $\sigma$ is continuous (being a smooth function). Thus, the assumption ${\sigma }_0\ne 0$ leads to a contradiction. Hence, ${\sigma }_0=0$, which in view of Equation (26) implies that $\sigma \left(s\right)=0$ along each integral curve of $\zeta$. As N is connected, it leads to $\sigma =0$ on N, that is, $\zeta$ is a Killing vector field. The converse follows by Theorem 1. □

4. Conformal Vector Fields and Euclidean Spaces

The Euclidean space $E^n$ is known to admit many conformal vector fields. For instance, for a constant $\lambda$, the vector field given by

$$\zeta =\lambda \sum _i{u}^i{\displaystyle \frac{\partial }{\partial u^i}}$$

is a conformal vector field with conformal factor $\sigma =\lambda$ and essential tensor $\chi =0$. Similarly, on the even-dimensional Euclidean space $E^{2n}$ with complex structure J, for a constant $\mu$ the vector field

$$\mathbf{u}=\mu \sum _i\left(u^i{\displaystyle \frac{\partial }{\partial u^i}}+u^{i}J{\displaystyle \frac{\partial }{\partial u^i}}\right)$$

is a conformal vector field with conformal factor $\sigma =\mu$ and essential tensor $\chi =\mu J$. In both examples, it is easy to see that the conformal factor $\sigma$ is a ghairaffinity function. This naturally raises a question: Under what conditions is a complete and connected Riemannian manifold $\left(N,g\right)$ admitting a conformal vector field $\zeta$ and with conformal factor $\sigma$ being a ghairaffinity function isometric to a Euclidean space? We answer this question in the following.

Theorem 3. 

An n-dimensional complete connected Riemannian manifold $\left(N,g\right)$, $n>2$ admits a conformal vector field ζ with conformal factor $\sigma \ne 0$ a ghairaffinity function such that the function $\zeta \left(\sigma \right)$ is a constant and ζ annihilates the essential tensor χ if and only if $\left(N,g\right)$ is isometric to the Euclidean space $E^n$.

Proof. 

Since the conformal factor $\sigma$ is a ghairaffinity function, by Equation (18) we have

$$R\left(F_1,\nabla \sigma \right)F_2={\nabla }_{F_1}{\nabla }_{F_2}\nabla \sigma -{\nabla }_{{\nabla }_{F_1}{F}_2}\nabla \sigma ,F_1,F_2\in \mathfrak{X}\left(N\right).$$

Using Equation (20) in the above equation gives

$$R\left(F_1,\nabla \sigma \right)F_2=\left({\nabla }_{F_1}{H}_{\sigma }\right)\left(F_2\right).$$

(28)

Since $H_{\sigma }$ is symmetric, taking the trace in above equation leads to

$$Ric\left(\nabla \sigma ,F_2\right)=\sum _{i}g\left(F_2,\left({\nabla }_{e_i}{H}_{\sigma }\right)\left(e_i\right)\right).$$

(29)

Now, using a local frame $\left\{e_1,\dots ,e_n\right\}$ of normal coordinates and noting

$$\Delta \sigma =\sum _{i}g\left(H_{\sigma }\left(e_i\right),e_i\right),$$

we have

$$F\left(\Delta \sigma \right)=\sum _{i}Fg\left(H_{\sigma }\left(e_i\right),e_i\right)=\sum _{i}g\left(\left({\nabla }_F{H}_{\sigma }\right)\left(e_i\right),e_i\right).$$

(30)

Now, using the identity

$$R\left(F_1,F_2\right)\nabla \sigma =\left({\nabla }_{F_1}{H}_{\sigma }\right)\left(F_2\right)-\left({\nabla }_{F_2}{H}_{\sigma }\right)\left(F_1\right)$$

in Equation (30), we conclude

$$\begin{array}{ccc}\hfill F\left(\Delta \sigma \right)& =& \sum _{i}g\left(R\left(F,e_i\right)\nabla \sigma +\left({\nabla }_{e_i}{H}_{\sigma }\right)\left(F\right),e_i\right)\hfill \\ & =& -Ric\left(F,\nabla \sigma \right)+\sum _{i}g\left(F,\left({\nabla }_{e_i}{H}_{\sigma }\right)\left(e_i\right)\right),\hfill \end{array}$$

where we have used the symmetry of $H_{\sigma }$. Thus, we have

$$\sum _{i}g\left(F,\left({\nabla }_{e_i}{H}_{\sigma }\right)\left(e_i\right)\right)=F\left(\Delta \sigma \right)+Ric\left(F,\nabla \sigma \right),$$

(31)

which upon comparison with Equation (29) yields

$$F\left(\Delta \sigma \right)=0,F\in \mathfrak{X}\left(N\right).$$

(32)

Now, we can proceed to see the impact of the condition that the function $\zeta \left(\sigma \right)$ be a constant, which in particular implies $Fg\left(\zeta ,\nabla \sigma \right)=0$, $F\in \mathfrak{X}\left(N\right)$. Using Equation (4) in the last equation, we get

$$g\left(\sigma F+\chi F,\nabla \sigma \right)+g\left(\zeta ,H_{\sigma }\left(F\right)\right)=0,F\in \mathfrak{X}\left(N\right),$$

that is,

$$\sigma \nabla \sigma +H_{\sigma }\left(\zeta \right)=\chi \left(\nabla \sigma \right).$$

(33)

Note that

$$div\left(\sigma \nabla \sigma \right)={∥\nabla \sigma ∥}^2+\sigma \Delta \sigma$$

(34)

and

$$div{H}_{\sigma }\left(\zeta \right)=\sum _{i}g\left({\nabla }_{e_i}{H}_{\sigma }\left(\zeta \right),e_i\right)=\sum _{i}g\left(\left({\nabla }_{e_i}{H}_{\sigma }\right)\left(\zeta \right)+H_{\sigma }\left({\nabla }_{e_i}\zeta \right),e_i\right).$$

Using Equation (4) and the symmetry of the operator $H_{\sigma }$, we get

$$div{H}_{\sigma }\left(\zeta \right)=\sigma \Delta \sigma +\sum _{i}g\left(\zeta ,\left({\nabla }_{e_i}{H}_{\sigma }\right)\left(e_i\right)\right),$$

where we have used $Tr\left(H_{\sigma }\circ \chi \right)=0$. Using Equations (31) and (32) in the above equation yields

$$div{H}_{\sigma }\left(\zeta \right)=\sigma \Delta \sigma +Ric\left(\zeta ,\nabla \sigma \right).$$

(35)

Next, we compute

$$div\chi \left(\nabla \sigma \right)=\sum _{i}g\left({\nabla }_{e_i}\chi \left(\nabla \sigma \right),e_i\right)=\sum _{i}g\left(\left({\nabla }_{e_i}\chi \right)\left(\nabla \sigma \right)+\chi \left(H_{\sigma }{e}_i\right),e_i\right)$$

and, using the skew symmetry of the essential tensor, we get

$$div\chi \left(\nabla \sigma \right)=-\sum _{i}g\left(\nabla \sigma ,\left({\nabla }_{e_i}\chi \right)\left(e_i\right)\right).$$

(36)

Using Lemma 1, we have

$$\sum _i\left({\nabla }_{e_i}\chi \right)\left(e_i\right)=-(n-1)\nabla \sigma -S\zeta$$

which can be inserted into Equation (36) to yield

$$div\chi \left(\nabla \sigma \right)=(n-1){∥\nabla \sigma ∥}^2+Ric\left(\nabla \sigma ,\zeta \right).$$

(37)

Taking divergence in Equation (33) and using Equations (34), (35), and (37), we conclude

$${∥\nabla \sigma ∥}^2+\sigma \Delta \sigma +\sigma \Delta \sigma +Ric\left(\zeta ,\nabla \sigma \right)=(n-1){∥\nabla \sigma ∥}^2+Ric\left(\nabla \sigma ,\zeta \right),$$

that is,

$$(n-2){∥\nabla \sigma ∥}^2=2\sigma \Delta \sigma .$$

(38)

Also, taking the inner product in Equation (33) by $\nabla \sigma$ and noting that the essential tensor $\chi$ is skew-symmetric, we arrive at

$$\sigma {∥\nabla \sigma ∥}^2+g\left(H_{\sigma }\left(\zeta \right),\nabla \sigma \right)=0,$$

which is equivalent to

$$\sigma {∥\nabla \sigma ∥}^2+{\displaystyle \frac{1}{2}}\zeta \left({∥\nabla \sigma ∥}^2\right)=0.$$

(39)

Note that $\zeta \left(\sigma \right)=c$ is a constant (statement) and, by Equation (32), $\Delta \sigma =\overline{c}$ is another constant. As $n>2$, Equation (38) implies

$${∥\nabla \sigma ∥}^2={\displaystyle \frac{2}{n-2}}\overline{c}\sigma .$$

(40)

Also, the operation ${\displaystyle \frac{1}{2}}\zeta$ on Equation (40) provides

$${\displaystyle \frac{1}{2}}\zeta \left({∥\nabla \sigma ∥}^2\right)={\displaystyle \frac{1}{n-2}}\overline{c}c$$

and inserting the above equation in Equation (39) gives

$$\sigma {∥\nabla \sigma ∥}^2=-{\displaystyle \frac{1}{n-2}}\overline{c}c.$$

Multiplying Equation (40) by $\sigma$ and using it in the above equation, we conclude that

$${\displaystyle \frac{2}{n-2}}\overline{c}{\sigma }^2=-{\displaystyle \frac{1}{n-2}}\overline{c}c,$$

that is,

$$\overline{c}\left(2{\sigma }^2+c\right)=0.$$

We now have two choices: (i) if $\overline{c}=0$, then Equation (30) conveys that $\sigma$ is a constant; (ii) if $\overline{c}\ne 0$, then the above equation implies that $2{\sigma }^2+c=0$. The last choice gives $4\sigma F\left(\sigma \right)=0$, $F\in \mathfrak{X}\left(N\right)$; since $\sigma \ne 0$ and N is connected, we must have $F\left(\sigma \right)=0$, $F\in \mathfrak{X}\left(N\right)$, that is, in the second choice $\sigma$ is also a constant.

Finally, we define a smooth function f on N by

$$f={\displaystyle \frac{1}{2}}{∥\zeta ∥}^2,$$

which by virtue of Equation (4) and $\chi \zeta =0$ (statement) has the gradient

$$\nabla f=\sigma \zeta ,$$

(41)

where $\sigma$ is a constant (indeed, a nonzero constant by statement). Observe that if f were to be a constant, we get $\zeta =0$, which by Equation (4) puts us in the following situation:

$$\sigma F=-\chi F$$

with trace $n\sigma =0$, contrary to the assumption $\sigma \ne 0$. Hence, f is a non-constant function, which by Equation (41) satisfies

$$H_f\left(F\right)={\sigma }^{2}F+\sigma \chi F.$$

Taking the inner product by F in the above equation gives

$$Hess\left(f\right)\left(F,F\right)={\sigma }^{2}g\left(F,F\right),F\in \mathfrak{X}\left(N\right).$$

Polarization in the above equation yields

$$Hess\left(f\right)={\sigma }^{2}g$$

for a non-constant function f and nonzero constant ${\sigma }^2$. Hence, $\left(N,g\right)$ is isometric to the Euclidean space $E^n$ (cf. [29]).

The converse is trivial, as the Euclidean space $E^n$ admits a conformal vector field given by

$$\zeta =\lambda \sum _i{u}^i{\displaystyle \frac{\partial }{\partial u^i}}$$

for a constant $\lambda$ with conformal factor $\sigma =\lambda$, which satisfies Equation (18) and $\zeta \left(\sigma \right)=0$, which is a constant and the essential tensor $\chi =0$, guaranteeing that $\zeta$ annihilates $\chi$. □

It is interesting to note that if $\zeta$ is a conformal vector field with conformal factor $\sigma$ and essential tensor $\chi$ on an n-dimensional Riemannian manifold $\left(N,g\right)$ such that $\sigma$ is a ghairaffinity function, then one easily obtains that $\Delta \sigma$ is a constant. In particular, if $\left(N,g\right)$ is compact, it will imply $\Delta \sigma =0$, and consequently that $\sigma$ is a constant. However, on non-compact $\left(N,g\right)$, there are examples of non-constant function $\sigma$ with $\Delta \sigma$ being a constant, for instance the function

$$f={\displaystyle \frac{1}{2}}{∥\xi ∥}^2,$$

where $\xi$ is a position vector on the Euclidean space $E^n$. We see that $\Delta f=n$ is a constant and f a non-constant function. The combination where (i) $\sigma$ is a ghairaffinity function and (ii) $\zeta \left(\sigma \right)$ is a constant implies that $\sigma$ is a constant, as seen in the proof of Theorem 3. Thus, the role of the combination where (i) $\sigma$ is a ghairaffinity function and (ii) $\left(N,g\right)$ is compact is achieved in the non-compact case by the combination where (i) $\sigma$ is a ghairaffinity function and (ii) $\zeta \left(\sigma \right)$ is a constant.

On a Euclidean space $E^n$, there are conformal vector fields and the Ricci operator of $E^n$ is trivial; thus, it is natural to seek conditions on a complete and connected Riemannian manifold $\left(N,g\right)$ admitting a conformal vector field $\zeta$ that satisfies $S\left(\zeta \right)=0$ (a less restrictive condition than $S=0$ on Euclidean space $E^n$) so that $\left(N,g\right)$ is isometric to $E^n$. Recall that a vector field F on a Riemannian manifold $\left(N,g\right)$ is said to incompressible if

$$divF=0.$$

Our next result in this direction is provided below.

Theorem 4. 

Let $\left(N,g\right)$ be an n-dimensional, $n>1$, complete and connected Riemannian manifold that admits a conformal vector field ζ with conformal factor $\sigma \ne 0$ and essential tensor χ such that (i) ζ annihilates the Ricci operator S, (ii) $\nabla \sigma$ annihilates the essential tensor χ, and (iii) the vector field $\chi \left(\zeta \right)$ is incompressible. Then, $\left(N,g\right)$ is isometric to the Euclidean space $E^n$. The converse also holds.

Proof. 

Since $\zeta$ annihilates the Ricci operator S, we have

$$Ric\left(\zeta ,F\right)=0,F\in \mathfrak{X}\left(N\right).$$

(42)

Now, $\nabla \sigma$ annihilates the essential tensor $\chi$, that is, $\chi \left(\nabla \sigma \right)=0$, which on differentiation gives

$$\left({\nabla }_E\chi \right)\left(\nabla \sigma \right)+\chi \left(H_{\sigma }E\right)=0.$$

Utilizing Lemma 1 in the above equation yields

$${∥\nabla \sigma ∥}^{2}E-E\left(\sigma \right)\nabla \sigma +R\left(E,\zeta \right)\nabla \sigma +\chi \left(H_{\sigma }E\right)=0,F\in \mathfrak{X}\left(N\right),$$

which on taking the trace yields

$$(n-1){∥\nabla \sigma ∥}^2+Ric\left(\zeta ,\nabla \sigma \right)=0.$$

Inserting Equation (42) in the above equation and noting $n>1$, we conclude that $\sigma$ is a constant. Now, we can use Equation (4) in computing

$$\begin{array}{cc}\hfill div\chi \left(\zeta \right)& =\sum _{i}g\left({\nabla }_{e_i}\chi \left(\zeta \right),e_i\right)=\sum _{i}g\left(\left({\nabla }_{e_i}\chi \right)\left(\zeta \right)+\chi \left(\sigma e_i+\chi \left(e_i\right)\right),e_i\right)\hfill \\ \hfill & =-{∥\chi ∥}^2-\sum _{i}g\left(\zeta ,\left({\nabla }_{e_i}\chi \right)\left(e_i\right)\right),\hfill \end{array}$$

(43)

where we have used the skew-symmetry of the essential tensor $\chi$. Using Lemma 1, we have the following on account of $\sigma$ being a constant:

$$\sum _i\left({\nabla }_{e_i}\chi \right)\left(e_i\right)=\sum _{i}R\left(e,\zeta \right)e_i$$

and taking the inner product with $\zeta$ in the above equation yields

$$\sum _{i}g\left(\zeta ,\left({\nabla }_{e_i}\chi \right)\left(e_i\right)\right)=-Ric\left(\zeta ,\zeta \right).$$

Using Equation (42), we conclude

$$\sum _{i}g\left(\zeta ,\left({\nabla }_{e_i}\chi \right)\left(e_i\right)\right)=0.$$

Inserting this equation in Equation (43) and noting that the vector field $\chi \left(\zeta \right)$ is incompressible, we immediately confirm $\chi =0$. Finally, on taking the function f defined by

$$f={\displaystyle \frac{1}{2}}{∥\zeta ∥}^2$$

together with $\sigma \ne 0$ a constant and $\chi =0$, as in the proof of Theorem 3, we confirm that $\left(N,g\right)$ is isometric to the Euclidean space $E^n$. The converse is trivial. □

We observe that one of the strategies for showing an n-dimensional complete and connected Riemannian manifold $\left(N,g\right)$ admitting a conformal vector field $\zeta$ with conformal factor $\sigma$ and essential tensor $\chi$ isometric to the Euclidean space $E^n$ requires (i) $\sigma$ a nonzero constant and (ii) $\chi =0$. Thus, in Theorem 4 we see that the first two conditions and $n>1$ imply that $\sigma$ is a constant and that the remaining requirement is fulfilled by the restriction

$$div\chi \left(\zeta \right)=0.$$

It is possible to achieve the remaining requirement by some other suitable restriction; therefore, it cannot be said that the condition that the vector field $\chi \left(\zeta \right)$ be incompressible is an optimal condition.

Finally, in the following result we observe that the incompressibility of the vector field $\chi \left(\zeta \right)$ appears to be an interesting tool in obtaining a characterization of the Euclidean space.

Theorem 5. 

Let $\left(N,g\right)$ be an n-dimensional complete and connected Riemannian manifold that admits a conformal vector field ζ with conformal factor $\sigma \ne 0$ and essential tensor χ such that (i) the vector field $\chi \left(\zeta \right)$ is incompressible, (ii) the conformal factor σ is constant along the integral curves of ζ, and (iii) the Ricci curvature $Ric\left(\zeta ,\zeta \right)$ is nonpositive. Then $\left(N,g\right)$ is isometric to the Euclidean space $E^n$. The converse also holds.

Proof. 

Using Equations (4) and (10), we have

$${\nabla }_{\zeta }\zeta =\sigma \zeta +\chi \left(\zeta \right).$$

(44)

Since, $\chi \left(\zeta \right)$ is incompressible, using Equation (10) in (44) yields

$$div\left({\nabla }_{\zeta }\zeta \right)=\zeta \left(\sigma \right)+n{\sigma }^2.$$

(45)

Now, Lemma 2 gives

$$-R\left(\zeta ,F_1;F_2,\zeta \right)+{\displaystyle \frac{1}{2}}\left(L_{{\nabla }_{\zeta }\zeta }g\right)\left(F_1,F_2\right)+g\left(\chi F_1,\chi F_2\right)=\zeta \left(\sigma \right)g\left(F_1,F_2\right)+{\sigma }^{2}g\left(F_1,F_2\right)$$

for smooth vector fields $F_1,F_2$ on N. Taking the trace in the above equation, we get

$$-Ric\left(\zeta ,\zeta \right)+div\left({\nabla }_{\zeta }\zeta \right)+{∥\chi ∥}^2=n\zeta \left(\sigma \right)+n{\sigma }^2,$$

and inserting Equation (45) into the above gives

$$-Ric\left(\zeta ,\zeta \right)+{∥\chi ∥}^2=\left(n-1\right)\zeta \left(\sigma \right).$$

As $\sigma$ is a constant along the integral curves of $\zeta$, the above equation takes the form

$$-Ric\left(\zeta ,\zeta \right)+{∥\chi ∥}^2=0.$$

(46)

Since, $Ric\left(\zeta ,\zeta \right)\le 0$, we get $\chi =0$. Thus, by Lemma 1, we have

$$R\left(F_1,\zeta \right)F_2=-F_2\left(\sigma \right)F_1+g\left(F_1,F_2\right)\nabla \sigma ,$$

and we can use $F_2=\zeta$ and $\zeta \left(\sigma \right)=0$ in the above equation and conclude

$$R\left(F_1,\zeta \right)\zeta =g\left(F_1,\zeta \right)\nabla \sigma .$$

Note that $R\left(F_1,\zeta \right)\zeta$ is symmetric; in view of this fact, the above equation implies

$$g\left(F_1,\zeta \right)g\left(\nabla \sigma ,F_2\right)=g\left(F_2,\zeta \right)g\left(\nabla \sigma ,F_1\right),F_1,F_2\in \mathfrak{X}\left(N\right).$$

Consequently, we have

$$g\left(F_1,\zeta \right)\nabla \sigma =g\left(\nabla \sigma ,F_1\right)\zeta ,$$

and taking $F_1=\zeta$ in above equation yields

$${∥\zeta ∥}^2\nabla \sigma =0.$$

(47)

If $\zeta =0$, then the first equation in (44) will imply $\sigma =0$, a contradiction to the hypothesis that $\sigma \ne 0$. Hence, on connected N, Equation (47) confirms that $\sigma$ is a nonzero constant, and we have already established that the essential tensor $\chi =0$. Thus, as in the proof of Theorem 3, $\left(N,g\right)$ is isometric to the Euclidean space $E^n$. The converse is trivial. □

5. An Invariance Under a Conformal Vector Field

Given a conformal vector field $\zeta$ with conformal factor $\sigma$ and essential tensor $\chi$ on a Riemannian manifold $\left(N,g\right)$, there is a natural operator, namely, the Hessian operator $H_{\sigma }$ associated with the conformal vector field. Note that if the operator $H_{\sigma }$ is invariant under the local flow of $\zeta$, we say that $H_{\sigma }$ is invariant under the conformal vector field $\zeta$, which is equivalent to

$$L_{\zeta }{H}_{\sigma }=0.$$

(48)

Note that the Euclidean space $E^n$ has a conformal vector field $\zeta$ with conformal factor $\sigma =\lambda$ a constant and $H_{\sigma }=0$, which satisfies Equation (48). This naturally raises a question: Under what conditions is a complete and connected Riemannian manifold $\left(N,g\right)$ possessing a conformal vector field with a conformal factor satisfying condition (48) isometric to a Euclidean space? Recall that a smooth function f on a Riemannian manifold $\left(N,g\right)$ is said to be a subharmonic function if it satisfies

$$\Delta f\ge 0.$$

(49)

In this section, we use the notion of a subharmonic function to answer the above question through the following result.

Theorem 6. 

Let $\left(N,g\right)$ be an n-dimensional complete and connected Riemannian manifold that admits a conformal vector field ζ with conformal factor $\sigma \ne 0$ and essential tensor χ such that (i) $H_{\sigma }$ is invariant under the conformal vector field ζ, (ii) the function $\zeta \left(\sigma \right)-{\sigma }^2$ is subharmonic, and (iii) ζ annihilates the essential tensor χ. Then, $\left(N,g\right)$ is isometric to the Euclidean space $E^n$. The converse also holds.

Proof. 

As $H_{\sigma }$ is invariant under the conformal vector field $\zeta$, through Equation (48) we get

$$\left[\zeta ,H_{\sigma }F\right]=H_{\sigma }\left[\zeta ,F\right],F\in \mathfrak{X}\left(N\right),$$

and using Equation (4) in the above equation, we arrive at

$$\left({\nabla }_{\zeta }{H}_{\sigma }\right)\left(F\right)=\chi H_{\sigma }F-H_{\sigma }\chi F,F\in \mathfrak{X}\left(N\right).$$

(50)

Since, the trace $Tr{H}_{\sigma }=\Delta \sigma$, on taking the trace in Equation (50) and noting that $Tr\chi H_{\sigma }=Tr{H}_{\sigma }\chi =0$, we conclude

$$\zeta \left(\nabla \sigma \right)=0.$$

(51)

Now, consider the function $\phi =\zeta \left(\sigma \right)$; on using Equation (4), we can compute the gradient $\nabla \phi$ to be

$$\nabla \phi =\sigma \nabla \sigma -\chi \left(\nabla \sigma \right)+H_{\sigma }\zeta .$$

(52)

As we are interested computing $\Delta \phi$, we shall compute the divergence of each term in Equation (52). We have

$$div\left(\sigma \nabla \sigma \right)={∥\nabla \sigma ∥}^2+\sigma \Delta \sigma$$

(53)

and

$$\begin{array}{ccc}\hfill div\chi \left(\nabla \sigma \right)& =& \sum _{i}g\left({\nabla }_{e_i}\chi \left(\nabla \sigma \right),e_i\right)=\sum _{i}g\left(\left({\nabla }_{e_i}\chi \right)\left(\nabla \sigma \right)+\chi H_{\sigma }{e}_i,e_i\right)\hfill \\ & =& \sum _{i}g\left(\left({\nabla }_{e_i}\chi \right)\left(\nabla \sigma \right),e_i\right)=-\sum _{i}g\left(\nabla \sigma ,\left({\nabla }_{e_i}\chi \right)\left(e_i\right)\right),\hfill \end{array}$$

where we have used the skew-symmetry of $\chi$ and $Tr\chi H_{\sigma }=0$. Now, using Lemma 1 in the above equation, we reach

$$\begin{array}{cc}\hfill div\chi \left(\nabla \sigma \right)& =g\left(\nabla \sigma ,(n-1)\nabla \sigma \right)-\sum _{i}g\left(\nabla \sigma ,R\left(e_i,\zeta \right)e_i\right)\hfill \\ \hfill & =(n-1){∥\nabla \sigma ∥}^2+Ric\left(\nabla \sigma ,\zeta \right).\hfill \end{array}$$

(54)

Also, using Equation (4), we have

$$\begin{array}{ccc}\hfill div{H}_{\sigma }\zeta & =& \sum _{i}g\left({\nabla }_{e_i}{H}_{\sigma }\zeta ,e_i\right)=\sum _{i}g\left(\left({\nabla }_{e_i}{H}_{\sigma }\right)\left(\zeta \right)+H_{\sigma }\left(\sigma e_i+\chi e_i\right),e_i\right)\hfill \\ & =& \sigma \Delta \sigma +\sum _{i}g\left(\zeta ,\left({\nabla }_{e_i}{H}_{\sigma }\right)\left(e_i\right)\right),\hfill \end{array}$$

where we have used $Tr{H}_{\sigma }=\Delta \sigma$, $Tr{H}_{\sigma }\chi =0$, and the symmetry of $H_{\sigma }$. Next, we take $F=\zeta$ in Equation (31) and use it in above equation to confirm

$$div{H}_{\sigma }\zeta =\sigma \Delta \sigma +\zeta \left(\Delta \sigma \right)+Ric\left(\zeta ,\nabla \sigma \right),$$

which in view of Equation (51) changes to

$$div{H}_{\sigma }\zeta =\sigma \Delta \sigma +Ric\left(\zeta ,\nabla \sigma \right).$$

(55)

Thus, on combining Equations (53)–(55) with Equation (52), we have the conclusion

$$\Delta \phi =2\sigma \nabla \sigma -(n-2){∥\nabla \sigma ∥}^2,$$

which on using

$$\Delta {\sigma }^2=2\sigma \Delta \sigma +2{∥\nabla \sigma ∥}^2$$

gives

$$\Delta \phi =\Delta {\sigma }^2-n{∥\nabla \sigma ∥}^2.$$

Thus, we conclude

$$\Delta \left(\zeta \left(\sigma \right)-{\sigma }^2\right)=-n{∥\nabla \sigma ∥}^2.$$

Now, invoking the fact that the function $\zeta \left(\sigma \right)-{\sigma }^2$ is subharmonic, we get that $\sigma$ is a constant. Though we could follow the line taken in the proof of Theorem 3 that using $\chi \left(\zeta \right)=0$ implies $\chi =0$, we instead exhibit another proof of this proposition in the following arguments.

Using Equation (4) and $\chi \left(\zeta \right)=0$, we have

$${\nabla }_{\zeta }\zeta =\sigma \zeta$$

with $\sigma$ being a constant. Differentiating the above equation, we get

$${\nabla }_F{\nabla }_{\zeta }\zeta ={\sigma }^{2}F+\sigma \chi F,F\in \mathfrak{X}\left(N\right),$$

which implies

$$R\left(F,\zeta \right)\zeta +{\nabla }_{\zeta }{\nabla }_F\zeta +{\nabla }_{\left[F,\zeta \right]}\zeta ={\sigma }^{2}F+\sigma \chi F,$$

and on using Equation (4) with constant $\sigma$, after simple steps we get

$$R\left(F,\zeta \right)\zeta +\left({\nabla }_{\zeta }\chi \right)\left(F\right)+\sigma \chi F+{\chi }^{2}F=0,F\in \mathfrak{X}\left(N\right).$$

(56)

Note that reflecting Lemma 1 onto $\left({\nabla }_{\zeta }\chi \right)\left(F\right),$ with $\sigma$ being a constant, yields

$$\left({\nabla }_{\zeta }\chi \right)\left(F\right)=0.$$

Hence, Equation (56) becomes

$$R\left(F,\zeta \right)\zeta +{\chi }^{2}F=-\sigma \chi F,F\in \mathfrak{X}\left(N\right).$$

Since, $R\left(F,\zeta \right)\zeta$ is a symmetric operator and ${\chi }^{2}F$ is as well, it is apparent that $R\left(F,\zeta \right)\zeta +{\chi }^{2}F$ is symmetric, while the operator $\sigma \chi F$ is skew-symmetric; this gives

$$\sigma \chi F=0,F\in \mathfrak{X}\left(N\right).$$

Next, invoking the hypothesis that $\sigma \ne 0$, the above equation implies $\chi =0$. Hence, as in the proof of Theorem 3, we get that $\left(N,g\right)$ is isometric to the Euclidean space $E^n$. The converse is trivial. □

6. Concluding Remarks

In this article, the result in Theorem 2 characterizes a Killing vector field using a conformal vector field $\zeta$ with essential tensor $\chi$ on a complete and connected Riemannian manifold $\left(N,g\right)$. This may be a humble beginning of the fact that the property

$$Ric\left(\zeta ,\zeta \right)=div\left({\nabla }_{\zeta }\zeta \right)+{∥\chi ∥}^2$$

(57)

is an exclusive property of a Killing vector field. Note that for any nonzero vector field $\zeta$ on a Riemannian manifold $\left(N,g\right)$, the essential tensor $\chi$ of $\zeta$ is related to the 1-form $\beta$ dual to $\zeta$ by

$$d\beta \left(F_1,F_2\right)=2g\left(\chi F_1,F_2\right),$$

(58)

meaning that on using Equation (14) we have

$$\begin{array}{ccc}\hfill {∥d\beta ∥}^2& =& \sum _{i,j}d\beta {\left(e_i,e_j\right)}^2=4\sum _{i,j}g{\left(\chi e_i,e_j\right)}^2=4\sum _i\sum _{j}g\left(\chi e_i,g\left(\chi e_i,e_j\right)e_j\right)\hfill \\ & =& 4\sum _{i}g\left(\chi e_i,\chi e_i\right)=4{∥\chi ∥}^2.\hfill \end{array}$$

Thus, the condition in (57) for any arbitrary vector field $\zeta$ on $\left(N,g\right)$ takes the form

$$Ric\left(\zeta ,\zeta \right)=div\left({\nabla }_{\zeta }\zeta \right)+{\displaystyle \frac{1}{4}}{∥d\beta ∥}^2.$$

Note that a closed Killing vector field is parallel; therefore, for a nontrivial Killing vector field, we require that it should not be closed. We propose the following interesting problem.

Problem 1. 

A nonzero non-closed vector field ζ on an n-dimensional complete and connected Riemannian manifold $\left(N,g\right)$ with dual 1-form β satisfying

$$Ric\left(\zeta ,\zeta \right)=div\left({\nabla }_{\zeta }\zeta \right)+{\displaystyle \frac{1}{4}}{∥d\beta ∥}^2$$

is a Killing vector field.