Disaffinity Vectors on a Riemannian Manifold and Their Applications

Abstract

A disaffinity vector on a Riemannian manifold $(M,g)$ is a vector field whose affinity tensor vanishes. In this paper, we observe that nontrivial disaffinity functions offer obstruction to the topology of M and show that the existence of a nontrivial disaffinity function on M does not allow M to be compact. A characterization of the Euclidean space is also obtained by using nontrivial disaffinity functions. Further, we study properties of disaffinity vectors on M and prove that every Killing vector field is a disaffinity vector. Then, we prove that if the potential field $\zeta$ of a Ricci soliton $\left(M,g,\zeta ,\lambda \right)$ is a disaffinity vector, then the scalar curvature is constant. As an application, we obtain conditions under which a Ricci soliton $\left(M,g,\zeta ,\lambda \right)$ is trivial. Finally, we prove that a Yamabe soliton $\left(M,g,\xi ,\lambda \right)$ with a disaffinity potential field $\xi$ is trivial.

1. Introduction

Throughout this paper, we assume that $M^n$ is a connected Riemannian n-manifold with a Riemannian metric g. Let $\mathfrak{X}\left(M^n\right)$ denote the space of smooth vector fields on $M^n$.

The affinity tensor, ${\mathcal{L}}_{\xi }\nabla$, of a vector field $\xi$ on a Riemannian manifold M is defined by (cf. [1], p. 109):

$$\left({\mathcal{L}}_{\xi }\nabla \right)\left(X,Y\right)={\mathcal{L}}_{\xi }{\nabla }_{X}Y-{\nabla }_{{\mathcal{L}}_{\xi }X}Y-{\nabla }_X{\mathcal{L}}_{\xi }Y,X,Y\in \mathfrak{X}\left(M\right),$$

(1)

where ∇ is the Riemannian connection and $\mathcal{L}$ represents the Lie derivative. An affinity tensor was used in [2,3,4], where the authors used it to obtain a characterization of a trivial Ricci soliton and used differential equations to obtain different characterizations of a sphere, respectively. We were informed by one of the reviewers that W. A. Poor was not the first to use the affinity tensor of a vector field, but rather, slightly earlier, S. Kobayashi and K. Nomizu (cf. [5], Chapter IV, pp. 225–236) used these vector fields under the name “affine infinitesimal transformations” and extensively studied them on manifolds equipped with an affine connection, which is a more general context than that which we are considering here.

A vector field $\xi$ on M is called a disaffinity vector if its affinity tensor vanishes, that is ${\mathcal{L}}_{\xi }\nabla =0$. From (1), we obtain

$$\begin{array}{ccc}\hfill \left({\mathcal{L}}_{\xi }\nabla \right)\left(X,Y\right)& =& {\mathcal{L}}_{\xi }{\nabla }_{X}Y-{\nabla }_{{\mathcal{L}}_{\xi }X}Y-{\nabla }_X{\mathcal{L}}_{\xi }Y\hfill \\ & =& \left[\xi ,{\nabla }_{X}Y\right]-{\nabla }_{\left[\xi ,X\right]}Y-{\nabla }_X\left[\xi ,Y\right]\hfill \\ & =& {\nabla }_{\xi }{\nabla }_{X}Y-{\nabla }_{{\nabla }_{X}Y}\xi -{\nabla }_{\left[\xi ,X\right]}Y-{\nabla }_X{\nabla }_{\xi }Y+{\nabla }_X{\nabla }_Y\xi \hfill \\ & =& R\left(\xi ,X\right)Y+{\nabla }_X{\nabla }_Y\xi -{\nabla }_{{\nabla }_{X}Y}\xi .\hfill \end{array}$$

Hence, we obtain

$$R\left(X,\xi \right)Y={\nabla }_X{\nabla }_Y\xi -{\nabla }_{{\nabla }_{X}Y}\xi$$

(2)

for a disaffinity vector $\xi$ on M.

Now, we discuss a few examples of disaffinity vectors on some well-known Riemannian manifolds:

(i) Considering the Euclidean n-space ${\mathbb{E}}^n=\left(R^n,〈,〉\right)$ and the position vector field

$$\xi =\sum _{j=1}^n{u}^j{\displaystyle \frac{\partial }{\partial u^j}}$$

(3)

on ${\mathbb{E}}^n$, where $u^1,\dots ,u^n$ are Euclidean coordinates, then it follows that $\xi$ satisfies Equation (2) and, consequently, $\xi$ is a disaffinity vector on ${\mathbb{E}}^n$. Also, all parallel vector fields on the Euclidean n-space are disaffinity vectors.

(ii) It is well known that on the standard unit sphere $S^{2n-1}$ in the complex n-space ${\mathbb{C}}^n$, the Reeb vector field $\xi =-JN$, where J is the complex structure on ${\mathbb{C}}^n$ and N is the unit normal to $S^{2n-1}$, defines a Sasakian structure on $S^{2n-1}$ (cf. [6]), and it satisfies Equation (2); therefore, $\xi$ is a disaffinity vector on $S^{2n-1}$.

Indeed, disaffinity vectors are in abundance and, as will be explored herein, we observe that they play an important role in shaping the geometry of the Riemannian manifold on which they live. Given a (smooth) function f on a Riemannian manifold $M^n$, we say f is a disaffinity function if its gradient $\nabla f$ is a disaffinity vector. Observe that if we consider the function $f={\displaystyle \frac{1}{2}}{∥\xi ∥}^2$ on ${\mathbb{E}}^n$, where $\xi$ is the position vector field appearing in Equation (3), then it follows that $\nabla f=\xi$. And, as we have seen, because $\xi$ is a disaffinity vector, it follows that f is a non-constant disaffinity function on ${\mathbb{E}}^n$. Since each constant function is a disaffinity function, we call a non-constant disaffinity function a nontrivial disaffinity function.

In this article, we are interested in the impact of the existence of a nontrivial disaffinity function on a Riemannian manifold. Indeed, we prove that if a Riemannian manifold $M^n$ admits a nontrivial disaffinity function, then it is non-compact (cf. Theorem 1), showing that the existence of a nontrivial disaffinity function provides an obstruction to the topology of $M^n$. This result may appear as a counter argument to example (ii); however, it is actually not so, as the Reeb vector field $\xi$ on compact $S^{2n-1}$ is not the gradient of a function. Next, we prove that each Killing vector is a disaffinity vector (cf. Proposition 1), supporting the statement that disaffinity vectors are in abundance. To show the importance of disaffinity functions on a complete Riemannian manifold $M^n$, we provide a characterization of ${\mathbb{E}}^n$ using a non-harmonic disaffinity function on $M^n$ (see Theorem 2). The next important question to consider involves finding conditionsunder which a Ricci soliton $\left(M^n,g,\zeta ,\lambda \right)$ is trivial (cf. Theorem 3). In addition, we prove that if the potential field $\xi$ of a Yamabe soliton $\left(M^n,g,\xi ,\lambda \right)$ (cf. [7]) is a disaffinity vector, then $\left(M^n,g,\xi ,\lambda \right)$ is trivial (see Theorem 4).

2. Preliminaries

The curvature tensor R of a Riemannian manifold $M^n$ is defined by

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

(4)

and the Ricci tensor $Ric$ is given by

$$Ric\left(X,Y\right)=\sum _{l}g\left(R\left(e_l,X\right)Y,e_l\right),$$

(5)

where $\left\{e_1,\dots ,e_n\right\}$ is a local orthonormal frame on $M^n$. Associated to the Ricci tensor is the Ricci operator Q, defined by

$$Ric\left(X,Y\right)=g\left(QX,Y\right).$$

(6)

The scalar curvature S of $M^n$ is

$$S=\sum _{l}Ric\left(e_l,e_l\right).$$

(7)

The following formula for the gradient $\nabla S$ of S is well known (cf. [7,8,9,10,11,12]):

$${\displaystyle \frac{1}{2}}\nabla S=\sum _l\left({\nabla }_{e_l}Q\right)\left(e_l\right),$$

(8)

where the covariant derivative ${\nabla }_{X}Q$ is defined by

$$\left({\nabla }_{X}Q\right)\left(Y\right)={\nabla }_{X}QY-Q\left({\nabla }_{X}Y\right).$$

Given a (smooth) function $f:M^n\to \mathbb{R}$, the Laplacian $\mathsf{\Delta }$ acting on f is given by

$$\mathsf{\Delta }f=\mathrm{div}\left(\nabla f\right),$$

(9)

where $\nabla f$ is the gradient of f and $\mathrm{div}X=\sum _{l}g\left({\nabla }_{e_l}X,e_l\right).$ If $M^n$ is compact, then the Stokes’s Theorem implies that

$${\int }_{M^n}\left(\mathrm{div}X\right)d{V}_g=0,$$

(10)

where $d{V}_g$ is the volume element of $M^n$.

For a function $f:M^n\to \mathbb{R}$, the Hessian operator $H_f$ is defined by

$$H_{f}X={\nabla }_X\nabla f,X\in \mathfrak{X}\left(M\right)$$

(11)

and the Hessian of f, $Hess\left(f\right)$, is given by

$$Hess\left(f\right)\left(X,Y\right)=g\left(H_{f}X,Y\right).$$

(12)

A function f on $M^n$ is called a disaffinity function if its gradient $\nabla f$ is a disaffinity vector. We see from the introduction that there is a non-constant disaffinity function on ${\mathbb{E}}^n$.

Recall that a vector field $\xi$ on a Riemannian manifold $M^n$ is said to be a Killing vector field if it satisfies (cf. [13]):

$${\mathcal{L}}_{\xi }g=0.$$

(13)

It is known that the Reeb vector field $\xi$ on the unit sphere $S^{2n-1}$ of example (ii) in the introduction is a Killing vector field, which is also a disaffinity vector. Indeed, we prove in the next result that every Killing vector field is a disaffinity vector, broadening the area of influence of disaffinity vectors.

Proposition 1. 

Every Killing vector field on a Riemannian manifold is a disaffinity vector.

Proof. 

Let $\xi$ be a Killing vector field on a Riemannian manifold $M^n$ and let $\eta$ be a 1-form dual to $\xi$, that is, $\eta \left(X\right)=g\left(X,\xi \right)$. Define $G:\mathfrak{X}\left(M\right)\to \mathfrak{X}\left(M\right)$ by

$${\displaystyle \frac{1}{2}}d\eta \left(X,Y\right)=g\left(GX,Y\right),X,Y\in \mathfrak{X}\left(M\right),$$

(14)

which shows that G is a skew symmetric operator. Now, using (13) and (14), we have

$$\begin{array}{cc}\hfill 2g\left({\nabla }_X\xi ,Y\right)& =g\left({\nabla }_X\xi ,Y\right)+g\left({\nabla }_Y\xi ,X\right)+g\left({\nabla }_X\xi ,Y\right)-g\left({\nabla }_Y\xi ,X\right)\hfill \\ \hfill & =2g\left(GX,Y\right).\hfill \end{array}$$

Thus, we have

$${\nabla }_X\xi =GX.$$

(15)

The above equation implies that

$${\nabla }_X{\nabla }_Y\xi ={\nabla }_{X}GY=\left({\nabla }_{X}G\right)\left(Y\right)+G\left({\nabla }_{X}Y\right),$$

which gives

$$R\left(X,Y\right)\xi =\left({\nabla }_{X}G\right)\left(Y\right)-\left({\nabla }_{Y}G\right)\left(X\right).$$

(16)

Note that as the 2-form $d\eta$ is closed, we can apply Equation (14) to confirm

$$g\left(\left({\nabla }_{X}G\right)\left(Y\right),Z\right)+g\left(\left({\nabla }_{Y}G\right)\left(Z\right),X\right)+g\left(\left({\nabla }_{Z}G\right)\left(X\right),Y\right)=0.$$

The above equation, on using skew symmetry of G, implies

$$g\left(\left({\nabla }_{X}G\right)\left(Y\right),Z\right)-g\left(\left({\nabla }_{Y}G\right)\left(X\right),Z\right)+g\left(\left({\nabla }_{Z}G\right)\left(X\right),Y\right)=0,$$

which, in view of Equation (16), confirms

$$g\left(R\left(X,Y\right)\xi ,Z\right)+g\left(\left({\nabla }_{Z}G\right)\left(X\right),Y\right)=0.$$

Thus, we have

$$\left({\nabla }_{Z}G\right)\left(X\right)=R\left(Z,\xi \right)X.$$

(17)

Now, Equation (2) for a Killing vector field $\xi$ and Equation (15) implies that

$$R\left(X,\xi \right)Y={\nabla }_{X}GY-G\left({\nabla }_{X}Y\right)=\left({\nabla }_{X}G\right)\left(Y\right),$$

which is Equation (17). Therefore, $\xi$ is a disaffinity vector on $M^n$. □

3. Disaffinity Functions on a Riemannian Manifolds

Recall that a function f on a Riemannian manifold $M^n$ is a disaffinity function if $\nabla f$ is a disaffinity vector. In this section, we study the influence of the existence of a nontrivial disaffinity function on $M^n$.

Theorem 1. 

If a Riemannian n-manifold $M^n$ admits a nontrivial disaffinity function, then it is non-compact.

Proof. 

Suppose that $M^n$ is a Riemannian manifold and f is a nontrivial disaffinity function on $M^n$. Then, $\nabla f$ satisfies Equation (2), which gives

$$R\left(X,\nabla f\right)Y=\left({\nabla }_X{H}_f\right)\left(Y\right),X,Y\in \mathfrak{X}\left(M\right).$$

(18)

Taking a local orthonormal frame $\left\{e_1,\dots ,e_n\right\}$ on $M^n$ and using the symmetry of the Hessian operator, Equation (18) implies

$$Ric\left(\nabla f,Y\right)=\sum _{j}g\left(Y,\left({\nabla }_{e_j}{H}_f\right)\left(e_j\right)\right).$$

(19)

Note that using Equations (4) and (11), we find

$$R\left(X,Y\right)\nabla f=\left({\nabla }_X{H}_f\right)\left(Y\right)-\left({\nabla }_Y{H}_f\right)\left(X\right).$$

(20)

Also, using $\mathsf{\Delta }f={\displaystyle \sum _j}g\left(H_f{e}_j,e_j\right)$, we obtain

$$\begin{array}{cc}\hfill X\left(\mathsf{\Delta }f\right)& =\sum _j\left(g\left(\left({\nabla }_X{H}_f\right)\left(e_j\right)+H_f\left({\nabla }_X{e}_j\right),e_j\right)+g\left(H_f{e}_j,{\nabla }_X{e}_j\right)\right)\hfill \\ \hfill & =\sum _{j}g\left(\left({\nabla }_X{H}_f\right)\left(e_j\right),e_j\right)+2\sum _{j}g\left(H_f{e}_j,{\nabla }_X{e}_j\right).\hfill \end{array}$$

(21)

Note that $H_f{e}_j={\displaystyle \sum _l}Hess\left(f\right)\left(e_j,e_l\right)e_l$, where $Hess\left(f\right)\left(e_j,e_l\right)$ is symmetric, and

$${\nabla }_X{e}_j=\sum _k{\omega }_j^k\left(X\right)e_k,$$

where the connection forms ${\omega }_j^k$ are skew symmetric. Consequently, we have

$$\sum _{j}g\left(H_f{e}_j,{\nabla }_X{e}_j\right)=0.$$

Thus, by using Equation (20) and the above equation in Equation (21), we find that

$$X\left(\mathsf{\Delta }f\right)=\sum _{j}g\left(R\left(X,e_j\right)\nabla f,e_j\right)+\sum _{j}g\left(\left({\nabla }_{e_j}{H}_f\right)X,e_j\right),$$

which, in view of Equation (5) and the symmetry of the Hessian operator $H_f$, yields

$$X\left(\mathsf{\Delta }f\right)=-Ric\left(X,\nabla f\right)+\sum _{j}g\left(X,\left({\nabla }_{e_j}{H}_f\right)\left(e_j\right)\right).$$

Using Equation (19) in the above equation, we obtain

$$X\left(\mathsf{\Delta }f\right)=0,X\in \mathfrak{X}\left(M\right),$$

(22)

and it confirms that $\mathsf{\Delta }f=c$, where c is a constant. Thus, after integrating $\mathsf{\Delta }f=c$, we obtain $c=0$ and $\mathsf{\Delta }f=0$ on a compact $M^n$. Therefore, f is a constant, which is a contradiction as f is a nontrivial disaffinity function. □

Now, we prove the following characterization of a Euclidean space.

Theorem 2. 

A complete Riemannian n-manifold $M^n$ admits a non-harmonic disaffinity function f such that $h={\displaystyle \frac{1}{2}}{∥\nabla f∥}^2$ satisfies the inequality

$$Ric\left(\nabla f,\nabla f\right)\ge \mathsf{\Delta }h-{\displaystyle \frac{1}{n}}{\left(\mathsf{\Delta }f\right)}^2$$

if and only if $M^n$ is isometric to the Euclidean n-space.

Proof. 

Let $M^n$ be a complete Riemannian n-manifold and $f:M^n\to R$ be a non-harmonic disaffinity function. Suppose that the function $h={\displaystyle \frac{1}{2}}{∥\nabla f∥}^2$ satisfies the inequality

$$Ric\left(\nabla f,\nabla f\right)\ge \mathsf{\Delta }h-{\displaystyle \frac{1}{n}}{\left(\mathsf{\Delta }f\right)}^2.$$

(23)

Note that for $X\in \mathfrak{X}\left(M\right)$, we have $X\left(h\right)=g\left({\mathcal{H}}_{f}X,\nabla f\right),$ which gives

$$\nabla h={\mathcal{H}}_f\left(\nabla f\right).$$

Differentiating the above equation with respect to $X\in \mathfrak{X}\left(M\right)$ gives

$${\mathcal{H}}_h\left(X\right)=\left({\nabla }_X{\mathcal{H}}_f\right)\left(\nabla f\right)+{\mathcal{H}}_f^2\left(\nabla f\right).$$

(24)

Now, since f is a disaffinity function, Equation (18) implies

$$R\left(X,\nabla f\right)\nabla f=\left({\nabla }_X{H}_f\right)(\nabla f)$$

and inserting this equation in Equation (24) gives

$${\mathcal{H}}_h\left(X\right)=R\left(X,\nabla f\right)\nabla f+{\mathcal{H}}_f^2\left(\nabla f\right).$$

Contracting the above equation yields

$$\mathsf{\Delta }h=Ric\left(\nabla f,\nabla f\right)+{\parallel {\mathcal{H}}_f\parallel }^2.$$

We then rearrange the above equation into the form

$$\parallel {\mathcal{H}}_f{\parallel }^2-{\displaystyle \frac{1}{n}}{\left(\mathsf{\Delta }f\right)}^2=\mathsf{\Delta }h-{\displaystyle \frac{1}{n}}{\left(\mathsf{\Delta }f\right)}^2-Ric\left(\nabla f,\nabla f\right).$$

Now, using the inequality (23) in the above equation, we find

$$\parallel {\mathcal{H}}_f{\parallel }^2-{\displaystyle \frac{1}{n}}{\left(\mathsf{\Delta }f\right)}^2\le 0,$$

which, by virtue of Cauchy–Schwartz’s inequality $\parallel {\mathcal{H}}_f{\parallel }^2\ge {\displaystyle \frac{1}{n}}{\left(\mathsf{\Delta }f\right)}^2$, gives

$$\parallel {\mathcal{H}}_f{\parallel }^2={\displaystyle \frac{1}{n}}{\left(\mathsf{\Delta }f\right)}^2.$$

The above equality holds if and only if

$${\mathcal{H}}_f={\displaystyle \frac{\mathsf{\Delta }f}{n}}I.$$

(25)

Since f is a disaffinity function by (23), we see that $\mathsf{\Delta }f=$ is a constant, say, $nc$. Moreover, since f is non-harmonic function, we obtain $c\ne 0$. Thus, Equation (25) becomes

$$Hess\left(f\right)=cg$$

and it guarantees that $M^n$ is isometric to the Euclidean n-space (cf. [12]).

Conversely, suppose that $M^n$ is isometric to the Euclidean space ${\mathbb{E}}^n$. Then, we know that the position vector field $\xi$ given by Equation (3) is a disaffinity vector. If we consider

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

then we obtain $\nabla f=\xi$. This shows that f is a disaffinity function on ${\mathbb{E}}^n$. Also, $\mathsf{\Delta }f=\mathrm{div}\xi =n$ implies that f is a non-harmonic disaffinity function. Further, here,

$$h={\displaystyle \frac{1}{2}}{∥\nabla f∥}^2={\displaystyle \frac{1}{2}}{∥\xi ∥}^2=f.$$

Thus, $\mathsf{\Delta }h=\mathsf{\Delta }f=n$ and $Ric\left(\nabla f,\nabla f\right)=0$ for ${\mathbb{E}}^n$ as well as

$$\mathsf{\Delta }h-{\displaystyle \frac{1}{n}}{\left(\mathsf{\Delta }f\right)}^2=0.$$

Consequently, the converse holds. □

4. Disaffinity Vectors and Triviality of Ricci Solitons

For a Ricci soliton $\left(M^n,g,\zeta ,\lambda \right)$, the potential field $\zeta$ and constant $\lambda$ satisfy (cf. [7]),

$${\displaystyle \frac{1}{2}}{\mathcal{L}}_{\zeta }g+Ric=\lambda g.$$

(26)

A Ricci soliton $\left(M^n,g,\zeta ,\lambda \right)$ is called trivial if either $\zeta =0$ or ${\mathcal{L}}_{\zeta }g=0$.

In this section, we are interested in studying the impart of the condition that the potential field $\zeta$ of the Ricci soliton $\left(M^n,g,\zeta ,\lambda \right)$ is a disaffinity vector. It is worth noting that if $\left(M^n,g,\zeta ,\lambda \right)$ is nontrivial and the potential field $\zeta$ is a disaffinity vector, then the Ricci soliton is non-compact. For a compact Ricci soliton $\left(M^n,g,\zeta ,\lambda \right)$ to be a gradient Ricci soliton, that is $\zeta =\nabla \phi$ for a function $\phi$ (cf. [7]), this implies that $\phi$ is a nontrivial disaffinity function. Thus, Theorem 1 implies that $M^n$ is non-compact.

On a Ricci soliton $\left(M^n,g,\zeta ,\lambda \right)$, let $\beta$ denote the 1-form dual to the potential field $\zeta$, that is, $\beta \left(X\right)=g\left(X,\zeta \right)$. Define a skew symmetric operator $G:\mathfrak{X}\left(M\right)\to \mathfrak{X}\left(M\right)$ by

$${\displaystyle \frac{1}{2}}d\beta \left(X,Y\right)=g\left(GX,Y\right),X,Y\in \mathfrak{X}\left(M\right).$$

(27)

Then, using

$$2g\left({\nabla }_X\zeta ,Y\right)=\left({\mathcal{L}}_{\zeta }g\right)\left(X,Y\right)+d\beta \left(X,Y\right)$$

and Equations (6), (26), and (27), we arrive at

$${\nabla }_X\zeta =\lambda X-QX+GX.$$

(28)

Using the above equation, we obtain

$$R\left(X,Y\right)\zeta =-\left({\nabla }_{X}Q\right)\left(Y\right)+\left({\nabla }_{Y}Q\right)\left(X\right)+\left({\nabla }_{X}G\right)\left(Y\right)-\left({\nabla }_{Y}G\right)\left(X\right)$$

and, using a local orthonormal frame $\left\{e_1,\dots ,e_n\right\}$, the symmetry of Q, skew symmetry of G, and Equation (8) in the above equation, we obtain

$$Ric\left(Y,\zeta \right)=-{\displaystyle \frac{1}{2}}Y\left(S\right)+Y\left(S\right)-\sum _{j}g\left(Y,\left({\nabla }_{e_j}G\right)\left(e_j\right)\right).$$

Thus, we have

$$Q\left(\zeta \right)={\displaystyle \frac{1}{2}}\nabla S-\sum _j\left({\nabla }_{e_j}G\right)\left(e_j\right).$$

(29)

Proposition 2. 

If $\left(M^n,g,\zeta ,\lambda \right)$ is a Ricci soliton with the potential field ζ as a disaffinity vector, then the scalar curvature S of $M^n$ is constant.

Proof. 

Suppose that the potential field $\zeta$ of the Ricci soliton $\left(M^n,g,\zeta ,\lambda \right)$ is a disaffinity vector. Then, using Equations (2) and (28), we obtain

$$\begin{array}{cc}\hfill R\left(X,\zeta \right)Y& ={\nabla }_X\left(\lambda Y-QY+GY\right)-\left(\lambda {\nabla }_{X}Y-Q\left({\nabla }_{X}Y\right)+G\left({\nabla }_{X}Y\right)\right)\hfill \\ \hfill & =-\left({\nabla }_{X}Q\right)\left(Y\right)+\left({\nabla }_{X}G\right)\left(Y\right).\hfill \end{array}$$

Using a local orthonormal frame $\left\{e_1,\dots ,e_n\right\}$, the symmetry of Q, skew symmetry of G, and Equation (8) in the above equation, we obtain

$$\begin{array}{c}\hfill Ric\left(Y,\zeta \right)=-{\displaystyle \frac{1}{2}}Y\left(S\right)-\sum _{j}g\left(Y,\left({\nabla }_{e_j}G\right)\left(e_j\right)\right),\end{array}$$

that is,

$$Q\left(\zeta \right)=-{\displaystyle \frac{1}{2}}\nabla S-\sum _j\left({\nabla }_{e_j}G\right)\left(e_j\right).$$

(30)

From Equations (29) and (30), we arrive at $\nabla S=0$, and hence, S is a constant. □

Note that one of the interesting questions in the geometry of Ricci solitons is to find conditions under which a Ricci soliton $\left(M^n,g,\zeta ,\lambda \right)$ is trivial. This question is relatively easier if $M^n$ is compact, and hence, it is important to find conditions under which a non-compact Ricci soliton $\left(M^n,g,\zeta ,\lambda \right)$ is trivial.

Theorem 3. 

A Ricci soliton $\left(M^n,g,\zeta ,\lambda \right)$ with ζ as a disaffinity vector and the Ricci operator Q satisfying $Q\left(\zeta \right)=\lambda \zeta$ is trivial.

Proof. 

Suppose that $\left(M^n,g,\zeta ,\lambda \right)$ is a Ricci soliton with potential field $\zeta$ as a disaffinity vector satisfying

$$Q\left(\zeta \right)=\lambda \zeta .$$

(31)

Differentiating the above equation and using Equation (28), we arrive at

$$\left({\nabla }_{X}Q\right)\left(\zeta \right)+Q\left(\lambda X-QX+GX\right)=\lambda \left(\lambda X-QX+GX\right).$$

Let us choose a local orthonormal frame $\{e_1,\dots ,e_n\}$ with $X=e_j$. Then, after taking the inner product with $e_j$ and summing up the resulting equations, we find

$$\sum _{j}g\left(\zeta ,\left({\nabla }_{e_j}Q\right)\left(e_j\right)\right)+\lambda S-{∥Q∥}^2=n{\lambda }^2-\lambda S,$$

(32)

where we have applied the skew symmetry of G and the symmetry of Q to confirm

$$\sum _{j}g\left(QG{e}_j,e_j\right)=0\mathrm{and}\sum _{j}g\left(G{e}_j,e_j\right)=0.$$

Using Proposition 2, we see that S is constant. Thus, on using Equation (8) in Equation (32), we find that

$$-{∥Q∥}^2=n{\lambda }^2-2\lambda S,$$

which could be rearranged as

$${∥Q∥}^2-{\displaystyle \frac{1}{n}}{S}^2=-{\displaystyle \frac{1}{n}}{\left(n\lambda -S\right)}^2.$$

(33)

Now, applying Cauchy–Schwartz’s inequality

$${∥Q∥}^2\ge {\displaystyle \frac{1}{n}}{S}^2$$

in (33), we obtain

$${∥Q∥}^2={\displaystyle \frac{1}{n}}{S}^2\mathrm{and}S=n\lambda .$$

(34)

The first equation in (34) is the equality case in Cauchy–Schwartz’s inequality, which holds if and only if $Q={\displaystyle \frac{S}{n}}I$. Thus, by virtue of the second equation in (34) gives $Q=\lambda I$. This gives ${\mathcal{L}}_{\zeta }g=0.$ Consequently, the Ricci soliton $\left(M^n,g,\zeta ,\lambda \right)$ is trivial. □

5. Disaffinity Vectors and Yamabe Solitons

In this section, we study Yamabe solitons $\left(M^n,g,\xi ,\lambda \right)$, $n\ge 2$, whose potential field $\zeta$ satisfies (cf. [7,10])

$${\displaystyle \frac{1}{2}}{\mathcal{L}}_{\xi }g=\left(S-\lambda \right)g,$$

(35)

for a constant $\lambda$, where S is the scalar curvature of $M^n$.

A Yamabe soliton $\left(M^n,g,\xi ,\lambda \right)$ is called trivial if its scalar curvature S is a constant. If the potential field $\xi$ of the Yamabe soliton $\left(M^n,g,\xi ,\lambda \right)$ is the gradient of a function $\phi$, that is, $\xi =\nabla \phi$, then $\left(M^n,g,\nabla \phi ,\lambda \right)$ is called a gradient Yamabe soliton. It is known that a compact gradient Yamabe soliton is always trivial (cf. [7]). In this section, we are interested in the impact of the potential field $\xi$ being a disaffinity vector on the geometry of Yamabe soliton $\left(M^n,g,\xi ,\lambda \right)$.

As before, we denote by $\beta$ the dual of the potential field $\xi$, and define the skew symmetric operator $F:\mathfrak{X}\left(M\right)\to \mathfrak{X}\left(M\right)$ by

$${\displaystyle \frac{1}{2}}d\beta \left(X,Y\right)=g\left(FX,Y\right).$$

(36)

Then, using (35) and (36), we have the following expression:

$${\nabla }_X\xi =\left(S-\lambda \right)X+FX.$$

(37)

Theorem 4. 

A Yamabe soliton $\left(M^n,g,\xi ,\lambda \right)$, $n\ge 2$, with potential field ξ as a disaffinity vector is trivial.

Proof. 

Suppose that $\left(M^n,g,\xi ,\lambda \right)$ is a Yamabe soliton, $n\ge 2$ whose potential field $\xi$ satisfies Equation (2). Then, using Equations (2) and (37), we have

$$\begin{array}{cc}\hfill R\left(X,\xi \right)Y& ={\nabla }_X\left((S-\lambda )Y+FY\right)-\left(S-\lambda \right){\nabla }_{X}Y-F\left({\nabla }_{X}Y\right)\hfill \\ \hfill & =X\left(S\right)Y+\left({\nabla }_{X}F\right)\left(Y\right).\hfill \end{array}$$

Using the skew symmetry of F and a local orthonormal frame $\{e_1,\dots ,e_n\}$ of $M^n$, we may take the trace from the above equation to obtain

$$Ric\left(\xi ,Y\right)=Y\left(S\right)-\sum _{j}g\left(Y,\left({\nabla }_{e_j}F\right)\left(e_j\right)\right),$$

which is equivalent to

$$Q\left(\xi \right)=\nabla S-\sum _j\left({\nabla }_{e_j}F\right)\left(e_j\right).$$

(38)

Next, we use Equations (4) and (37) to obtain

$$R\left(X,Y\right)\xi =X\left(S\right)Y-Y\left(S\right)X+\left({\nabla }_{X}F\right)\left(Y\right)-\left({\nabla }_{Y}F\right)\left(X\right),$$

which, on using a local orthonormal frame $\left\{e_1,\dots ,e_n\right\}$ and skew symmetry of the operator F while taking the trace from the above equation, we obtain

$$Ric\left(Y,\xi \right)=-(n-1)Y\left(S\right)-\sum _{j}g\left(Y,\left({\nabla }_{e_j}F\right)\left(e_j\right)\right).$$

(39)

Equation (39) implies

$$Q\left(\xi \right)=-(n-1)\nabla S-\sum _j\left({\nabla }_{e_j}F\right)\left(e_j\right).$$

(40)

Combining Equations (38) and (40) gives $\nabla S=0$. Therefore, the Yamabe soliton $\left(M^n,g,\xi ,\lambda \right)$ is trivial. □

6. Epilogue

In this article, we have seen the impact of a nontrivial disaffinity function as well as disaffinity vector on the geometry of the Riemannian manifold $M^n$. In particular, we use the non-harmonic disaffinity function to derive a characterization of the Euclidean space. We also use the constraint on the potential field of a Ricci or Yamabe soliton to be a disaffinity vector to obtain the triviality of the solitons. We shall elaborate two situations for future study, one for nontrivial disaffinity functions and the other for the disaffinity vectors on a Riemannian manifold.

Recall that the notion of Eikonal equation $∥\nabla f∥=1$ on a Riemannian manifold $M^n$ comes from physics, specially medical imaging (cf. [9,11]). Note that the existence of the Eikonal equation $∥\nabla f∥=1$ on $M^n$ forces $M^n$ to be non-compact. A similar role is played by the nontrivial disaffinity function as well. Thus, it would be interesting to analyze the geometry of the Riemannian manifold that admits a nontrivial disaffinity function f, which satisfies the Eikonal equation $∥\nabla f∥=1$.

Recall that a vector field $\xi$ on $M^n$ is called torse-forming (cf. [14]) if it satisfies

$${\nabla }_X\xi =\sigma X+\omega \left(X\right)\xi ,X\in \mathfrak{X}\left(M^n\right),$$

where $\sigma$ is a function called the conformal scalar and $\omega$ is a 1-form called the generating form. For another situation, we may consider a compact Riemannian manifold $M^n$ that admits a torse-forming vector field $\xi$. It would be interesting to study a torse-forming vector field $\xi$ on a Riemannian manifold $M^n$ that is also the disaffinity vector, and to investigate its impact on the geometry of $M^n$.