A Solitonic Study of Riemannian Manifolds Equipped with a Semi-Symmetric Metric ξ-Connection

Abstract

The aim of this paper is to characterize a Riemannian 3-manifold $M^3$ equipped with a semi-symmetric metric $\xi$-connection $\tilde{\nabla }$ with $\rho$-Einstein and gradient $\rho$-Einstein solitons. The existence of a gradient $\rho$-Einstein soliton in an $M^3$ admitting $\tilde{\nabla }$ is ensured by constructing a non-trivial example, and hence some of our results are verified. By using standard tensorial technique, we prove that the scalar curvature of $(M^3,\tilde{\nabla })$ satisfies the Poisson equation $\Delta R=\frac{4(2-\sigma -6\rho )}{\rho }$.

1. Introduction

The Ricci and other geometric flows are active topics of current research in mathematics, physics and engineering. The Ricci flow [1] is defined on a Riemannian n-manifold $(M^n,g)$ by an evolution equation for metric $g\left(t\right)$ of the form $\frac{\partial g}{\partial t}=-2S$, where S is the Ricci tensor of $M^n$ and t indicates the time. The metric g on $M^n$ satisfies the Ricci soliton (in short, RS) equation ${\pounds }_{E}g+2\sigma g+2S=0,$ where E is a vector field on $M^n$, $\sigma \in \mathcal{R}$ (the set of real numbers), and ${\pounds }_E$ represents the Lie derivative operator in the direction of E on $M^n$. A RS is called expanding (steady or shrinking) if $\sigma >0$ ($\sigma =0$ or $\sigma <0$). If $E=0$ or Killing, then the RS is called a trivial RS, and $M^n$ becomes an Einstein manifold. Thus the RS is a basic generalization of an Einstein manifold [2]. If $\mathcal{F}$ is a smooth function such that $E=\mathcal{D}\mathcal{F}$ for the gradient operator $\mathcal{D}$ of g, then the RS is described as a gradient Ricci soliton (GRS), E is referred to as the potential vector field, and $\mathcal{F}$ is called the potential function. Thus, the RS equation becomes $Hess\mathcal{F}+\sigma g+S=0$, where $Hess\mathcal{F}$ is the Hessian of $\mathcal{F}$ and $\left(Hess\mathcal{F}\right)({\zeta }_1,{\zeta }_2)=g({\nabla }_{{\zeta }_1}D\mathcal{F},{\zeta }_2)$ for all vector fields ${\zeta }_1$ and ${\zeta }_2$ on $M^n$. Here, ∇ stands for the Levi–Civita connection.

The notion of Ricci–Bourguignon flow, a natural generalization of Ricci flow, has been proposed in [3] and is described on an $M^n$ as:

$$\begin{array}{c}\hfill \frac{\partial g}{\partial t}=-2(S-\rho Rg),g\left(0\right)=g_0,\end{array}$$

(1)

where R is the scalar curvature and $\rho \in \mathcal{R}$. It is to be noticed that for the specific values of $\rho$, the following cases for the tensor $S-\rho Rg$ appeared in (1) [4] are obtained:

(i)

$\rho =\frac{1}{2}$, the Einstein tensor $S-\frac{R}{2}g$, (for Einstein soliton),

(ii)

$\rho =\frac{1}{n}$, the trace-less Ricci tensor $S-\frac{R}{n}g$,

(iii)

$\rho =\frac{1}{2(n-1)}$, the Schouten tensor $S-\frac{R}{2(n-1)}g,$ (for Schouten soliton),

(iv)

$\rho =0$, the Ricci tensor S (for RS).

An $(M^n,g)$, $n\ge 3$ is said to be a $\rho$-Einstein soliton (or $\rho$-ES) $(g,E,\sigma ,\rho )$ if

$$\begin{array}{c}\hfill {\pounds }_{E}g+2S+2(\sigma -\rho R)g=0.\end{array}$$

(2)

Similar to the RS, a $\rho$-ES is called expanding (steady or shrinking) if $\sigma >0$ ($\sigma =0$ or $\sigma <0$). If $E=\mathcal{D}\mathcal{F}$, then $(M^n,g)$ is called a gradient $\rho$-Einstein soliton (or gradient $\rho$-ES). Hence, (2) takes the form

$$\begin{array}{c}\hfill Hess\mathcal{F}+S+(\sigma -\rho R)g=0,\end{array}$$

(3)

where $Hess\mathcal{F}$ denotes the Hessian of $\mathcal{F}\in C^{\infty }\left(M^n\right)$ and defined by $Hess\mathcal{F}=\nabla \nabla \mathcal{F}$. Recently, $\rho$-Einstein solitons have been studied by several authors, such as [5,6,7,8,9,10,11,12]. On the other hand, we recommend the papers [13,14,15,16,17,18,19] for the studies of Ricci, Yamabe, Ricci-Yamabe, $\eta$-Ricci-Yamabe and quasi-Yamabe solitons on different geometric structures.

In this paper, we have made an effort to the solitonic study of a 3-dimensional Riemannian manifold $M^3$ equipped with a semi-symmetric metric $\xi$-connection $\tilde{\nabla }$. To achieve the goal, we present our work as follows: In Section 2, we gather the basic information of a Riemannian 3-manifold equipped with a semi-symmetric metric $\xi$-connection $(M^3,\tilde{\nabla },g)$, definitions and Lemmas. The properties of $\rho$-ES in $(M^3,\tilde{\nabla },g)$ are studied in Section 3. We address the properties of gradient $\rho$-ES in $(M^3,\tilde{\nabla },g)$ in Section 4. In the last section, we model a non-trivial example of $(M^3,\tilde{\nabla },g)$ admitting a gradient $\rho$-ES, and prove our results.

2. Riemannian Manifolds with a Semi-Symmetric Metric $\mathit{\xi }$-Connection

In 1970, Yano [20] investigated the properties of a semi-symmetric metric connection $\tilde{\nabla }$ on Riemannian n-manifolds $M^n$ and defined by ${\tilde{\nabla }}_{{\zeta }_1}{\zeta }_2={\nabla }_{{\zeta }_1}{\zeta }_2+\eta \left({\zeta }_2\right){\zeta }_1-g({\zeta }_1,{\zeta }_2)\xi$ for all ${\zeta }_1$ and ${\zeta }_2$ on $M^n$, where $\eta$ is a 1-form associated with the unit vector field $\xi$ such that $g(\xi ,\xi )=\eta \left(\xi \right)=1$ and $g({\zeta }_1,\xi )=\eta \left({\zeta }_1\right)$. Later, the properties of the semi-symmetric metric connection $\tilde{\nabla }$ have been explored by several researchers. One of these properties is the curvature invariant respecting to the semi-symmetric metric connection $\tilde{\nabla }$ and the Levi–Civita connection ∇. For example, the conformal curvature tensors corresponding to the semi-symmetric connection (Yano’s sense) and the Levi–Civita connection coincide. Similar results for different curvature tensors have been established by many geometers. A connection $\tilde{\nabla }$ is said to be semi-symmetric metric $\xi$-connection if and only if $\tilde{\nabla }\xi =0$. Afterwards, the properties of semi-symmetric metric $\xi$-connection have been studied in [21,22,23,24].

In an $(M^n,\tilde{\nabla },g)$, we have [21]

$$\begin{array}{c}\hfill {\nabla }_{{\zeta }_1}\xi =-{\zeta }_1+\eta \left({\zeta }_1\right)\xi ,g(\xi ,\xi )=1,\mathrm{and}\eta \left({\zeta }_1\right)=g({\zeta }_1,\xi )\end{array}$$

(4)

for any ${\zeta }_1$ on $M^n$. Next, we have [21]

$$\begin{array}{c}\hfill \left({\nabla }_{{\zeta }_1}\eta \right){\zeta }_2=-g({\zeta }_1,{\zeta }_2)+\eta \left({\zeta }_1\right)\eta \left({\zeta }_2\right),\end{array}$$

(5)

$$\begin{array}{c}\hfill K({\zeta }_1,{\zeta }_2)\xi =\eta \left({\zeta }_1\right){\zeta }_2-\eta \left({\zeta }_2\right){\zeta }_1,\end{array}$$

(6)

$$\begin{array}{c}\hfill K({\zeta }_1,\xi ){\zeta }_2=g({\zeta }_1,{\zeta }_2)\xi -\eta \left({\zeta }_2\right){\zeta }_1,\end{array}$$

(7)

$$\begin{array}{c}\hfill S({\zeta }_1,\xi )=-(n-1)\eta \left({\zeta }_1\right)\iff Q\xi =-(n-1)\xi ,\end{array}$$

(8)

$$\begin{array}{c}\hfill \left({\pounds }_{\xi }g\right)({\zeta }_1,{\zeta }_2)=2\{-g({\zeta }_1,{\zeta }_2)+\eta \left({\zeta }_1\right)\eta \left({\zeta }_2\right)\},\end{array}$$

(9)

for all ${\zeta }_1$, ${\zeta }_2$ on $M^n.$ Here, K and Q represent the curvature tensor and the Ricci operator of $M^n$, respectively.

Definition 1. 

An $M^n$ is said to be quasi-Einstein if its $S(\ne 0)$ satisfies

$$S({\zeta }_1,{\zeta }_2)=\mathfrak{l}g({\zeta }_1,{\zeta }_2)+\mathfrak{m}\eta \left({\zeta }_1\right)\eta \left({\zeta }_2\right),$$

where $\mathfrak{m}$ and $\mathfrak{l}$ are smooth functions on $M^n$. If $\mathfrak{m}=0$, then the manifold is called an Einstein manifold.

Definition 2. 

A partial differential equation $\Delta u=v$ on a complete $M^n$ is called a Poisson equation for some smooth functions u and v.

Remark 1 

([21,22]). An $(M^3,\tilde{\nabla },g)$ is a quasi-Einstein manifold of the form

$$S({\zeta }_1,{\zeta }_2)=\left(1+\frac{R}{2}\right)g({\zeta }_1,{\zeta }_2)-\left(3+\frac{R}{2}\right)\eta \left({\zeta }_1\right)\eta \left({\zeta }_2\right).$$

(10)

Remark 2 

([21,22]). In an $(M^3,\tilde{\nabla },g)$, we have

$$\begin{array}{c}\hfill \xi \left(R\right)=2(R+6),\end{array}$$

(11)

$$\begin{array}{c}\hfill \eta \left({\nabla }_{\xi }\mathcal{D}R\right)=4(R+6),\end{array}$$

(12)

where $\mathcal{D}$ is the gradient operator of g. From (11), it is noticed that R of $M^3$ is constant if and only if $R=-6$.

3. $\mathit{\rho }$-ES on $\mathbf{(}{\mathit{M}}^{\mathbf{3}},\tilde{\mathbf{\nabla }}\mathbf{,}\mathit{g}\mathbf{)}$

First, we prove the following theorem.

Theorem 1. 

If $(M^3,\tilde{\nabla },g)$ admits a ρ-ES $(g,E,\sigma ,\rho )$, then its scalar curvature R satisfies the Poisson equation $\Delta R=\frac{4(2-\sigma -6\rho )}{\rho },$ provided $\rho \ne 0$.

Proof. 

Let the metric of an $(M^3,\tilde{\nabla },g)$ be a ρ-ES $(g,E,\sigma ,\rho )$, then in view of (10), (2) leads to

$$\begin{array}{ccc}\hfill \left({\pounds }_{E}g\right)({\zeta }_1,{\zeta }_2)& =& -2\left\{1+\sigma +(\frac{1}{2}-\rho )R\right\}g({\zeta }_1,{\zeta }_2)\hfill \\ & & +(R+6)\eta \left({\zeta }_1\right)\eta \left({\zeta }_2\right),\hfill \end{array}$$

(13)

for any vector fields ${\zeta }_1$, ${\zeta }_2$ on $M^3$.

Taking covariant derivative of (13) respecting to ${\zeta }_3$, we find

$$\begin{array}{ccc}\hfill \left({\nabla }_{{\zeta }_3}{\pounds }_{E}g\right)({\zeta }_1,{\zeta }_2)& =& \left({\zeta }_{3}R\right)\left\{(2\rho -1)g({\zeta }_1,{\zeta }_2)+\eta \left({\zeta }_1\right)\eta \left({\zeta }_2\right)\right\}\hfill \\ & & -(R+6)\left\{g({\zeta }_1,{\zeta }_3)\eta \left({\zeta }_2\right)+g({\zeta }_2,{\zeta }_3)\eta \left({\zeta }_1\right)-2\eta \left({\zeta }_1\right)\eta \left({\zeta }_2\right)\eta \left({\zeta }_3\right)\right\}.\hfill \end{array}$$

(14)

As g is parallel with respect to ∇, then the formula [25]

$$({\pounds }_E{\nabla }_{{\zeta }_1}g-{\nabla }_{{\zeta }_1}{\pounds }_{E}g-{\nabla }_{[E,{\zeta }_1]}g)({\zeta }_2,{\zeta }_3)=-g(({\pounds }_E\nabla )({\zeta }_1,{\zeta }_2),{\zeta }_3)-g(({\pounds }_E\nabla )({\zeta }_1,{\zeta }_3),{\zeta }_2)$$

turns to

$$\left({\nabla }_{{\zeta }_1}{\pounds }_{E}g\right)({\zeta }_2,{\zeta }_3)=g(({\pounds }_E\nabla )({\zeta }_1,{\zeta }_2),{\zeta }_3)+g(({\pounds }_E\nabla )({\zeta }_1,{\zeta }_3),{\zeta }_2).$$

Since ${\pounds }_E\nabla$ is symmetric, therefore we have

$$2g(({\pounds }_E\nabla )({\zeta }_1,{\zeta }_2),{\zeta }_3)=\left({\nabla }_{{\zeta }_1}{\pounds }_{E}g\right)({\zeta }_2,{\zeta }_3)+\left({\nabla }_{{\zeta }_2}{\pounds }_{E}g\right)({\zeta }_1,{\zeta }_3)-\left({\nabla }_{{\zeta }_3}{\pounds }_{E}g\right)({\zeta }_1,{\zeta }_2),$$

which in view of (14) gives

$$\begin{array}{ccc}\hfill 2g(({\pounds }_E\nabla )({\zeta }_1,{\zeta }_2),{\zeta }_3)& =& \left({\zeta }_{1}R\right)\left\{(2\rho -1)g({\zeta }_2,{\zeta }_3)+\eta \left({\zeta }_2\right)\eta \left({\zeta }_3\right)\right\}\hfill \\ & & +\left({\zeta }_{2}R\right)\left\{(2\rho -1)g({\zeta }_1,{\zeta }_3)+\eta \left({\zeta }_1\right)\eta \left({\zeta }_3\right)\right\}\hfill \\ & & -\left({\zeta }_{3}R\right)\left\{(2\rho -1)g({\zeta }_1,{\zeta }_2)+\eta \left({\zeta }_1\right)\eta \left({\zeta }_2\right)\right\}\hfill \\ & & -2(R+6)\left\{g({\zeta }_1,{\zeta }_2)\eta \left({\zeta }_3\right)-\eta \left({\zeta }_1\right)\eta \left({\zeta }_2\right)\eta \left({\zeta }_3\right)\right\},\hfill \end{array}$$

from which it follows that

$$\begin{array}{ccc}\hfill 2({\pounds }_E\nabla )({\zeta }_1,{\zeta }_2)& =& \left({\zeta }_{1}R\right)\left\{(2\rho -1){\zeta }_2+\eta \left({\zeta }_2\right)\xi \right\}\hfill \\ & & +\left({\zeta }_{2}R\right)\left\{(2\rho -1){\zeta }_1+\eta \left({\zeta }_1\right)\xi \right\}\hfill \\ & & -\left(\mathcal{D}R\right)\left\{(2\rho -1)g({\zeta }_1,{\zeta }_2)+\eta \left({\zeta }_1\right)\eta \left({\zeta }_2\right)\right\}\hfill \\ & & -2(R+6)\left\{g({\zeta }_1,{\zeta }_2)\xi -\eta \left({\zeta }_1\right)\eta \left({\zeta }_2\right)\xi \right\}.\hfill \end{array}$$

(15)

Replacing ${\zeta }_2$ by ξ and ${\zeta }_1$ by ${\zeta }_2$ in (15), we have

$$\begin{array}{ccc}\hfill ({\pounds }_E\nabla )({\zeta }_2,\xi )& =& \rho g(\mathcal{D}R,{\zeta }_2)\xi -\rho \left(\mathcal{D}R\right)\eta \left({\zeta }_2\right)\hfill \\ & & +(R+6)\left\{(2\rho -1){\zeta }_2+\eta \left({\zeta }_2\right)\xi \right\}.\hfill \end{array}$$

(16)

The covariant differentiation of (16) respecting to ${\zeta }_1$ yields

$$\begin{array}{ccc}\hfill ({\nabla }_{{\zeta }_1}{\pounds }_E\nabla )({\zeta }_2,\xi )& =& 2\left({\zeta }_{1}R\right)\left\{(2\rho -1){\zeta }_2+\eta \left({\zeta }_2\right)\xi \right\}\hfill \\ & & +\left({\zeta }_{2}R\right)\left\{(\rho -1){\zeta }_1+\eta \left({\zeta }_1\right)\xi \right\}\hfill \\ & & -\left(\mathcal{D}R\right)\left\{(\rho -1)g({\zeta }_1,{\zeta }_2)+\eta \left({\zeta }_1\right)\eta \left({\zeta }_2\right)\right\}\hfill \\ & & -3(R+6)\left\{g({\zeta }_1,{\zeta }_2)\xi -\eta \left({\zeta }_1\right)\eta \left({\zeta }_2\right)\xi \right\}\hfill \\ & & -(R+6)\left\{(2\rho -1)\eta \left({\zeta }_1\right){\zeta }_2+\eta \left({\zeta }_2\right){\zeta }_1\right\}\hfill \\ & & +\rho g({\nabla }_{{\zeta }_1}\mathcal{D}R,{\zeta }_2)\xi -\rho \left({\nabla }_{{\zeta }_1}\mathcal{D}R\right)\eta \left({\zeta }_2\right),\hfill \end{array}$$

(17)

where (4), (5) and (16) being used.

Again from [25], we have

$$\begin{array}{c}\hfill \left({\pounds }_{E}K\right)({\zeta }_1,{\zeta }_2){\zeta }_3=({\nabla }_{{\zeta }_1}{\pounds }_E\nabla )({\zeta }_2,{\zeta }_3)-({\nabla }_{{\zeta }_2}{\pounds }_E\nabla )({\zeta }_1,{\zeta }_3),\end{array}$$

(18)

which by putting ${\zeta }_3=\xi$ and using (17) becomes

$$\begin{array}{ccc}\hfill \left({\pounds }_{E}K\right)({\zeta }_1,{\zeta }_2)\xi & =& g(\mathcal{D}R,{\zeta }_1)\left\{(3\rho -1){\zeta }_2+\eta \left({\zeta }_2\right)\xi \right\}\hfill \\ & & -g(\mathcal{D}R,{\zeta }_2)\left\{(3\rho -1){\zeta }_1+\eta \left({\zeta }_1\right)\xi \right\}\hfill \\ & & +2(R+6)(\rho -1)\left\{\eta \left({\zeta }_2\right){\zeta }_1-\eta \left({\zeta }_1\right){\zeta }_2\right\}\hfill \\ & & +\rho g({\nabla }_{{\zeta }_1}\mathcal{D}R,{\zeta }_2)\xi -\rho g({\nabla }_{{\zeta }_2}\mathcal{D}R,{\zeta }_1)\xi \hfill \\ & & -\rho \left({\nabla }_{{\zeta }_1}\mathcal{D}R\right)\eta \left({\zeta }_2\right)+\rho \left({\nabla }_{{\zeta }_2}\mathcal{D}R\right)\eta \left({\zeta }_1\right).\hfill \end{array}$$

(19)

Contracting (19) respecting to ${\zeta }_1$ then using (4) and (11) we lead to

$$\begin{array}{ccc}\hfill \left({\pounds }_{E}S\right)({\zeta }_2,\xi )& =& (1-6\rho ){\zeta }_2\left(R\right)+2(R+6)(2\rho -1)\eta \left({\zeta }_2\right)\hfill \\ & & +\rho g({\nabla }_{\xi }\mathcal{D}R,{\zeta }_2)\xi -\rho (\Delta R)\eta \left({\zeta }_2\right).\hfill \end{array}$$

(20)

By putting ${\zeta }_2=\xi$ in (20) then using (4), (11) and (12), we find

$$\begin{array}{ccc}\hfill \left({\pounds }_{E}S\right)(\xi ,\xi )& =& -4\rho (R+6)-\rho (\Delta R).\hfill \end{array}$$

(21)

The Lie derivative of (8) respecting to E leads to

$$\begin{array}{c}\hfill \left({\pounds }_{E}S\right)(\xi ,\xi )=4\eta \left({\pounds }_E\xi \right).\end{array}$$

(22)

Putting ${\zeta }_1={\zeta }_2=\xi$ in (13) infers

$$\begin{array}{c}\hfill \left({\pounds }_{E}g\right)(\xi ,\xi )=-2\sigma +2\rho R+4.\end{array}$$

(23)

The Lie derivative of $g(\xi ,\xi )=1$ gives

$$\begin{array}{c}\hfill \left({\pounds }_{E}g\right)(\xi ,\xi )=-2\eta \left({\pounds }_E\xi \right).\end{array}$$

(24)

Now combining (21)–(25) we deduce

$$\begin{array}{c}\hfill \Delta R=\frac{4(2-\sigma -6\rho )}{\rho },\mathrm{provided}\rho \ne 0.\end{array}$$

(25)

This completes the proof. □

It is well-known that the $\rho$-ES Equation (2) on $M^n$ with the soliton constant $\rho =\frac{1}{2},\frac{1}{n},\frac{1}{2(n-1)}$ reduces to the Einstein soliton, traceless Ricci soliton, Schouten soliton, respectively. It is also known that a smooth function $\mathfrak{f}$ on an $M^n$ is called harmonic, subharmonic or superharmonic if $\Delta \mathfrak{f}=0,\ge 0\mathrm{or}\le 0$, respectively. These facts together with Theorem 1 state the following:

Corollary 1. 

Let $(M^3,\tilde{\nabla },g)$ admit a ρ-ES, then we have

Value of $\rho$	Solitons	Poisson equation	Condition for R to be subharmonic and superharmonic
$\rho =\frac{1}{2}$	Einstein soliton	$\Delta R=-8(\sigma +1)$	$\left(i\right)$  R is subharmonic if $\sigma \le -1$, $\left(ii\right)$  R is superharmonic if $\sigma \ge -1$,
$\rho =\frac{1}{3}$	traceless Ricci soliton	$\Delta R=-12\sigma$	$\left(i\right)$  R is subharmonic if $\sigma \le 0$, $\left(ii\right)$  R is superharmonic if $\sigma \ge 0$,
$\rho =\frac{1}{4}$	Schouten soliton	$\Delta R=16(\frac{1}{2}-\sigma )$	$\left(i\right)$  R is subharmonic if $\sigma \le \frac{1}{2}$, $\left(ii\right)$  R is superharmonic if $\sigma \ge \frac{1}{2}$.

Remark 3. 

The ρ-ES on an $M^n$ with $\rho =0$ reduces to the RS. The properties of RS on $(M^3,\tilde{\nabla },g)$ have been explored by Chaubey and De [22]. Thus, we can say that the Theorem 1 generalizes the study of Einstein soliton, traceless RS and the Schouten soliton on $(M^3,\tilde{\nabla },g)$.

It is well-known that the Poisson equation $\Delta u=v$ with $v=0$ becomes a Laplace equation. Suppose that an $(M^3,\tilde{\nabla },g)$ does not admit RS. Then, Theorem 1 and above discussion state:

Corollary 2. 

If $(M^3,\tilde{\nabla },g)$ admits a ρ-ES, which is not a RS $(\rho \ne 0)$, then R of $M^3$ satisfies Laplace equation if and only if $\sigma =2(1-3\rho )$.

Let $(M^3,\tilde{\nabla },g)$ admit a $\rho$-ES. If R of $M^3$ satisfies the Laplace equation, then $\sigma =2(1-3\rho )$. The $\rho$-ES under consideration to be steady, shrinking or expanding if $\rho$ is equal to, less than or greater than $\frac{1}{3}$. Thus, we write our corollary as

Corollary 3. 

Let the metric of an $(M^3,\tilde{\nabla },g)$ be ρ-ES, which is not a RS $(\rho \ne 0)$. If R of $M^3$ satisfies the Laplace equation, then the ρ-ES is steady, shrinking or expanding if $\rho =\frac{1}{3}$, $\rho <\frac{1}{3}$ or $\rho >\frac{1}{3}$, respectively.

4. Gradient $\mathit{\rho }$-ES on $\mathbf{(}{\mathit{M}}^{\mathbf{3}},\tilde{\mathbf{\nabla }}\mathbf{,}\mathit{g}\mathbf{)}$

Theorem 2. 

Let $(M^3,\tilde{\nabla },g)$ admit a gradient ρ-ES. Then, either $M^3$ is Einstein or the gradient ρ-ES is steady type gradient traceless RS.

Proof. 

Let the metric of an $(M^3,\tilde{\nabla },g)$ be a gradient ρ-ES. Then, (3) can be written as

$$\begin{array}{c}\hfill {\nabla }_{{\zeta }_1}\mathcal{D}\mathcal{F}+Q{\zeta }_1+(\sigma -\rho R){\zeta }_1=0,\end{array}$$

(26)

for all ${\zeta }_1$ on $M^3$.

The covariant differentiation of (26) with respect to ${\zeta }_2$ leads to

$$\begin{array}{c}\hfill {\nabla }_{{\zeta }_2}{\nabla }_{{\zeta }_1}\mathcal{D}\mathcal{F}=-\left({\nabla }_{{\zeta }_2}Q\right){\zeta }_1-Q\left({\nabla }_{{\zeta }_2}{\zeta }_1\right)-(\sigma -\rho R){\nabla }_{{\zeta }_2}{\zeta }_1+\rho {\zeta }_2\left(R\right){\zeta }_1.\end{array}$$

(27)

Interchanging ${\zeta }_1$ and ${\zeta }_2$ in (27) leads to

$$\begin{array}{c}\hfill {\nabla }_{{\zeta }_1}{\nabla }_{{\zeta }_2}\mathcal{D}\mathcal{F}=-\left({\nabla }_{{\zeta }_1}Q\right){\zeta }_2-Q\left({\nabla }_{{\zeta }_1}{\zeta }_2\right)-(\sigma -\rho R){\nabla }_{{\zeta }_1}{\zeta }_2+\rho {\zeta }_1\left(R\right){\zeta }_2.\end{array}$$

(28)

By plugging of (26)–(28), we find

$$\begin{array}{c}\hfill K({\zeta }_1,{\zeta }_2)\mathcal{D}\mathcal{F}=-\left({\nabla }_{{\zeta }_1}Q\right){\zeta }_2+\left({\nabla }_{{\zeta }_2}Q\right){\zeta }_1+\rho \left\{{\zeta }_1\left(R\right){\zeta }_2-{\zeta }_2\left(R\right){\zeta }_1\right\}.\end{array}$$

Contracting the forgoing equation along ${\zeta }_1$, we obtain

$$\begin{array}{c}\hfill S({\zeta }_2,\mathcal{D}\mathcal{F})=\frac{(1-4\rho )}{2}{\zeta }_2\left(R\right).\end{array}$$

(29)

In account of (10), we have

$$\begin{array}{c}\hfill S({\zeta }_2,\mathcal{D}\mathcal{F})=(1+\frac{R}{2}){\zeta }_2\left(\mathcal{F}\right)-(3+\frac{R}{2})\eta \left({\zeta }_2\right)\xi \left(\mathcal{F}\right).\end{array}$$

(30)

Thus, from (29) and (30), it follows that

$$\begin{array}{c}\hfill (1-4\rho ){\zeta }_2\left(R\right)=(R+2){\zeta }_2\left(\mathcal{F}\right)-(R+6)\eta \left({\zeta }_2\right)\xi \left(\mathcal{F}\right).\end{array}$$

(31)

By putting ${\zeta }_2=\xi$ in (31), then using (4) and (11), we find

$$\begin{array}{c}\hfill \xi \left(\mathcal{F}\right)=-\frac{1}{2}(1-4\rho )(R+6).\end{array}$$

(32)

By using (32) and (31) turns to

$$\begin{array}{c}\hfill (1-4\rho ){\zeta }_2\left(R\right)=(R+2){\zeta }_2\left(\mathcal{F}\right)+\frac{1}{2}{(R+6)}^2(1-4\rho )\eta \left({\zeta }_2\right).\end{array}$$

(33)

The covariant differentiation of (33) along ${\zeta }_1$ leads to

$$\begin{array}{ccc}\hfill (1-4\rho )g({\nabla }_{{\zeta }_1}\mathcal{D}R,{\zeta }_2)& =& {\zeta }_1\left(R\right){\zeta }_2\left(\mathcal{F}\right)+(R+2)g({\nabla }_{{\zeta }_1}\mathcal{D}\mathcal{F},{\zeta }_2)\hfill \\ & & +(R+6)(1-4\rho ){\zeta }_1\left(R\right)\eta \left({\zeta }_2\right)\hfill \\ & & +\frac{1}{2}{(R+6)}^2(1-4\rho )\left\{\eta \left({\zeta }_1\right)\eta \left({\zeta }_2\right)-g({\zeta }_1,{\zeta }_2)\right\}.\hfill \end{array}$$

(34)

Interchanging ${\zeta }_1$ and ${\zeta }_2$ in (34), we have

$$\begin{array}{ccc}\hfill (1-4\rho )g({\nabla }_{{\zeta }_2}\mathcal{D}R,{\zeta }_1)& =& {\zeta }_2\left(R\right){\zeta }_1\left(\mathcal{F}\right)+(R+2)g({\nabla }_{{\zeta }_2}\mathcal{D}\mathcal{F},{\zeta }_1)\hfill \\ & & +(R+6)(1-4\rho ){\zeta }_2\left(R\right)\eta \left({\zeta }_1\right)\hfill \\ & & +\frac{1}{2}{(R+6)}^2(1-4\rho )\left\{\eta \left({\zeta }_1\right)\eta \left({\zeta }_2\right)-g({\zeta }_1,{\zeta }_2)\right\}.\hfill \end{array}$$

(35)

Equating the left hand sides of last two equations gives

$$\begin{array}{c}\hfill {\zeta }_1\left(R\right){\zeta }_2\left(\mathcal{F}\right)+(R+6)(1-4\rho ){\zeta }_1\left(R\right)\eta \left({\zeta }_2\right)\\ \hfill -{\zeta }_2\left(R\right){\zeta }_1\left(\mathcal{F}\right)-(R+6)(1-4\rho ){\zeta }_2\left(R\right)\eta \left({\zeta }_1\right)=0,\end{array}$$

which by replacing ${\zeta }_2=\xi$ then using (4), (11) and (32) takes the form

$$\begin{array}{c}\hfill (R+6)\{(1-4\rho ){\zeta }_1\left(R\right)-4{\zeta }_1\left(\mathcal{F}\right)-4(R+6)(1-4\rho )\eta \left({\zeta }_1\right)\}=0.\end{array}$$

Thus, we have either $R=-6$, or $(1-4\rho ){\zeta }_1\left(R\right)=4{\zeta }_1\left(\mathcal{F}\right)+4(R+6)(1-4\rho )\eta \left({\zeta }_1\right)$. If we firstly suppose that $R\ne -6$ and $(1-4\rho ){\zeta }_1\left(R\right)=4{\zeta }_1\left(\mathcal{F}\right)+4(R+6)(1-4\rho )\eta \left({\zeta }_1\right)$, which by virtue of (33) turns to

$$\begin{array}{c}\hfill (R-2)\left\{2{\zeta }_1\left(\mathcal{F}\right)+(R+6)(1-4\rho )\eta \left({\zeta }_1\right)\right\}=0,\end{array}$$

(36)

which refers that either $R=2$ or ${\zeta }_1\left(\mathcal{F}\right)=-\frac{1}{2}(R+6)(1-4\rho )\eta \left({\zeta }_1\right)$. From (11), it is obvious that if R is constant, then its value must be $-6$, which shows that $R=2$ is inadmissible. Thus, we have ${\zeta }_1\left(\mathcal{F}\right)=-\frac{1}{2}(R+6)(1-4\rho )\eta \left({\zeta }_1\right)$, which is equivalent to

$$\begin{array}{c}\hfill \mathcal{D}\mathcal{F}=-\frac{1}{2}(R+6)(1-4\rho )\xi =\xi \left(\mathcal{F}\right)\xi .\end{array}$$

(37)

Thus, the gradient of $\mathcal{F}$ is pointwise collinear with $\xi .$ Now, taking the covariant derivative of (37) with respect to ${\zeta }_1$ and using (4), we have

$$\begin{array}{c}\hfill {\nabla }_{{\zeta }_1}\mathcal{D}\mathcal{F}={\zeta }_1\left(\xi \left(\mathcal{F}\right)\right)\xi -\xi \left(\mathcal{F}\right)({\zeta }_1-\eta \left({\zeta }_1\right)\xi ).\end{array}$$

(38)

Therefore, from (26) and (38), we obtain

$$\begin{array}{c}\hfill Q{\zeta }_1+(\sigma -\rho R){\zeta }_1=-{\zeta }_1\left(\xi \left(\mathcal{F}\right)\right)\xi +\xi \left(\mathcal{F}\right)({\zeta }_1-\eta \left({\zeta }_1\right)\xi ).\end{array}$$

(39)

Now, by replacing ${\zeta }_1$ by ξ in (39) then using (8), (11) and (32) we lead to

$$\begin{array}{c}\hfill \sigma =(1-3\rho )(R+8).\end{array}$$

(40)

Let us suppose that $\rho =\frac{1}{3}$, that is, the gradient ρ-ES on an $M^3$ is gradient traceless RS. This fact together with Equation (40) leads to $\sigma =0$. Thus, the gradient traceless RS is steady. This completes the proof. □

Theorem 3. 

Let an $(M^3,\tilde{\nabla },g)$ be a non-gradient traceless RS. Then, the gradient ρ-ES is trivial soliton with constant $\sigma =2(1-3\rho )$. Also, the ρ-ES is shrinking and expanding according to $\rho >\frac{1}{3}$ and $\rho <\frac{1}{3}$.

Proof. 

Now, we suppose that $\rho \ne \frac{1}{3}$. Thus, (40) leads to

$$\begin{array}{c}\hfill R=\frac{\sigma }{1-3\rho }-8,\end{array}$$

(41)

which informs that R is constant and hence (11) infers that $R=-6$. This contradicts our hypothesis $R\ne -6$.

Secondly, we consider that $R=-6$ and $(1-4\rho ){\zeta }_1\left(R\right)\ne 4{\zeta }_1\left(\mathcal{F}\right)+4(R+6)(1-4\rho )\eta \left({\zeta }_1\right)$. For $R=-6$, (33) informs that $\mathcal{F}\in \mathcal{R}$ and hence the GRBS on the manifold is trivial. Moreover, the Riemannian 3-manifold under assumption is an Einstein manifold with $\sigma =2(1-3\rho )$. This completes the proof. □

Let us suppose that an $(M^3,\tilde{\nabla },g)$ admits a proper gradient $\rho$-ES. Then, the $\rho$-ES reduces to the gradient traceless RS and $\rho =\frac{1}{3},\sigma =0$. Using these facts in (26) and then contracting the foregoing equation over ${\zeta }_1$ gives $\Delta \mathcal{F}=0$.

A smooth function $\mathfrak{h}$ on an $M^n$ is called harmonic if $\Delta \mathfrak{h}=0.$

The above discussions state the following:

Corollary 4. 

Let a complete $(M^3,\tilde{\nabla },g)$ admit a proper gradient ρ-ES. Then the gradient function of the gradient ρ-ES is harmonic.

Contracting (38) over ${\zeta }_1$, we find

$$\Delta \mathcal{F}=\xi \left(\xi \right(\mathcal{F}\left)\right)-2\xi \left(\mathcal{F}\right).$$

Again, considering $\sigma =0,$ $\rho =\frac{1}{3}$ and then contracting (26) over ${\zeta }_1$, we conclude that

$$\Delta \mathcal{F}=0.$$

The last two equations show that $\xi \left(\xi \right(\mathcal{F}\left)\right)-2\xi \left(\mathcal{F}\right)=0$. Let $\xi =\frac{\partial }{\partial t}$. Thus, we notice that the potential function $\mathcal{F}$ satisfies the PDE

$$\frac{{\partial }^2\mathcal{F}}{\partial t^2}-2\frac{\partial \mathcal{F}}{\partial t}=0.$$

It is obvious that $\mathcal{F}=A{e}^{2t}+B$ for smooth functions A and B, which are independent of t, is the solution of the above PDE. Now, we list our results in the following:

Corollary 5. 

Let the metric of a complete $(M^3,\tilde{\nabla },g)$ admit a proper gradient ρ-ES. Then, the potential function $\mathcal{F}$ of such soliton satisfies the PDE $\frac{{\partial }^2\mathcal{F}}{\partial t^2}-2\frac{\partial \mathcal{F}}{\partial t}=0$, and it can be evaluated by $\mathcal{F}=A{e}^{2t}+B$.

5. Example

We consider the manifold ${\mathit{M}}^3=\{(w_1,w_2,w_3)\in {\mathcal{R}}^3\}$, where $(w_1,w_2,w_3)$ are the usual coordinates in ${\mathbb{R}}^3$. Let $u_1,$$u_2$ and $u_3$ be the vector fields on $M^3$ given by

$$u_1=e^{\mathfrak{b}{w}_3+w_1}\frac{\partial }{\partial w_1},u_2=e^{\mathfrak{b}{w}_3+w_2}\frac{\partial }{\partial w_2},u_3=\frac{1}{\mathfrak{b}}\frac{\partial }{\partial w_3}=\xi ,$$

where $\mathfrak{b}(\ne 0)\in \mathcal{R}$. Then, $\{u_1,u_2,u_3\}$ forms a basis in the module of the vector fields of $M^3$.

Let the Riemannian metric g be defined by

$$\begin{array}{c}\hfill g(u_p,u_q)=\left\{\begin{array}{c}1,1\le p=q\le 3,\hfill \\ 0,otherwise.\hfill \end{array}\right.\end{array}$$

Hence, $M^3$ is a Riemannian manifold of dimension 3. Let the 1-form $\eta$ on $M^3$ be defined by $\eta \left({\zeta }_1\right)=g({\zeta }_1,u_3)=g({\zeta }_1,\xi )$ for all ${\zeta }_1$ on $M^3$. Now, by direct computations, we obtain

$$[u_1,u_2]=0,[u_1,u_3]=-u_1,[u_2,u_3]=-u_2.$$

By using Koszul’s formula, we obtain

$$\begin{array}{c}\hfill {\nabla }_{u_p}{u}_q=\left\{\begin{array}{c}-u_p,p=1,2,q=3,\hfill \\ u_3,1\le p=q\le 2,\hfill \\ 0,otherwise.\hfill \end{array}\right.\end{array}$$

Now we suppose that ${\zeta }_1={\zeta }_1^1{u}_1+{\zeta }_1^2{u}_2+{\zeta }_1^3{u}_3$, then for $\xi =u_3$ it follows that ${\nabla }_{{\zeta }_1}\xi =-{\zeta }_1+\eta \left({\zeta }_1\right)\xi .$ It can be easily seen that $\tilde{\nabla }$ defined on $M^3$ satisfies the conditions

$$\tilde{T}({\zeta }_1,{\zeta }_2)=-\eta \left({\zeta }_1\right){\zeta }_2+\eta \left({\zeta }_2\right){\zeta }_1,\tilde{\nabla }g=0,\mathrm{and}\tilde{\nabla }\xi =0,$$

for arbitrary vector fields ${\zeta }_1$ and ${\zeta }_2$ on $M^3$, where $\tilde{T}$ indicates the torsion tensor of $\tilde{\nabla }$. Thus, we can say that $\tilde{\nabla }$ is a semi-symmetric metric $\xi$-connection on $M^3$.

The non-zero constituents of K are obtained as follows:

$$K(u_1,u_3)u_1=u_3,K(u_1,u_2)u_1=u_2,K(u_2,u_3)u_2=u_3,$$

$$K(u_1,u_2)u_2=K(u_1,u_3)u_3=-u_1,K(u_2,u_3)u_3=-u_2.$$

By using above components of the curvature tensor K we obtain

$$S(u_p,u_q)=-2,1\le p=q\le 3,$$

from which we obtain $R=-6$.

Now, by taking $\mathcal{D}\mathcal{F}=\left(u_1\mathcal{F}\right)u_1+\left(u_2\mathcal{F}\right)u_2+\left(u_3\mathcal{F}\right)u_3$, we have

$$\begin{array}{c}\hfill {\nabla }_{u_1}\mathcal{D}\mathcal{F}=(u_1\left(u_1\mathcal{F}\right)-u_3\mathcal{F})u_1+\left(u_1\left(u_2\mathcal{F}\right)\right)u_2+(u_1\left(u_3\mathcal{F}\right)+u_1\mathcal{F})u_3,\end{array}$$

$$\begin{array}{c}\hfill {\nabla }_{{\mathcal{E}}_2}\mathcal{D}\mathcal{F}=\left(u_2\left(u_1\mathcal{F}\right)\right)u_1+(u_2\left(u_2\mathcal{F}\right)-u_3\mathcal{F})u_2+(u_2\left(u_3\mathcal{F}\right)+{\mathcal{F}}_2\mathcal{F}){\mathcal{F}}_3,\end{array}$$

$$\begin{array}{c}\hfill {\nabla }_{{\mathcal{E}}_3}\mathcal{D}\mathcal{F}=\left(u_3\left(u_1\mathcal{F}\right)\right)u_1+\left(u_3\left(u_2\mathcal{F}\right)\right)u_2+\left(u_3\left(u_3\mathcal{F}\right)\right)u_3.\end{array}$$

Thus, by virtue of (26), we obtain

$$\begin{array}{c}\hfill \left\{\begin{array}{c}{u}_1\left(u_1\mathcal{F}\right)-u_3\mathcal{F}=2-6\rho -\sigma ,\hfill \\ u_2\left(u_2\mathcal{F}\right)-u_3\mathcal{F}=2-6\rho -\sigma ,\hfill \\ u_3\left(u_3\mathcal{F}\right)=2-6\rho -\sigma ,\hfill \\ u_1\left(u_2\mathcal{F}\right)=0,\hfill \\ u_2\left(u_1\mathcal{F}\right)=0,\hfill \\ u_2\left(u_3\mathcal{F}\right)+u_2\mathcal{F}=0.\hfill \end{array}\right.\end{array}$$

(42)

Thus, the relations in (42) are, respectively, amounting to

$$\begin{array}{c}\hfill e^{2(\mathfrak{b}{w}_3+w_1)}\left[\frac{{\partial }^2\mathcal{F}}{\partial w_1^2}+\frac{\partial \mathcal{F}}{\partial w_1}\right]-\frac{1}{\mathfrak{b}}\frac{\partial \mathcal{F}}{\partial w_3}=2-6\rho -\sigma ,\end{array}$$

$$\begin{array}{c}\hfill e^{2(\mathfrak{b}{w}_3+w_1)}\left[\frac{{\partial }^2\mathcal{F}}{\partial w_2^2}+\frac{\partial \mathcal{F}}{\partial w_2}\right]-\frac{1}{\mathfrak{b}}\frac{\partial \mathcal{F}}{\partial w_3}=2-6\rho -\sigma ,\end{array}$$

$$\begin{array}{c}\hfill \frac{1}{{\mathfrak{b}}^{\mathfrak{2}}}\frac{{\partial }^2\mathcal{F}}{\partial w_3^2}=2-6\rho -\sigma ,\end{array}$$

$$\begin{array}{c}\hfill \frac{{\partial }^2\mathcal{F}}{\partial w_1\partial w_2}=0,\end{array}$$

$$\begin{array}{c}\hfill \frac{{\partial }^2\mathcal{F}}{\partial w_2\partial w_1}=0,\end{array}$$

$$\begin{array}{c}\hfill \frac{1}{\mathfrak{b}}[\frac{{\partial }^2\mathcal{F}}{\partial w_2\partial w_3}+\frac{\partial \mathcal{F}}{\partial w_2}]=0.\end{array}$$

From the above relations, it is noticed that $\mathcal{F}\in \mathcal{R}$ for $\sigma =2-6\rho$. Hence, the Equation (26) is satisfied. Thus, g is a gradient $\rho$-ES with the soliton vector field $E=\mathcal{D}\mathcal{F}$, where $\mathcal{F}\in \mathcal{R}$ and $\sigma =2-6\rho$. For $\rho =\frac{1}{3},$ we obtain $\sigma =0$, i.e., the gradient $\rho$-ES is trivial with constant $\sigma =2-6\rho$. Thus, Theorem 2 is verified.

6. Results and Discussion

It is well known that the $\rho$-Einstein soliton Equation (2) with $\rho =0$ becomes the Ricci soliton equation, which has been studied in [22]. Thus, we can say that the $\rho$-Einstein soliton is a natural generalization of Ricci soliton. In this manuscript, we have explored the properties of $\rho$-Einstein solitons in Riemannian geometry, which generalizes the results of [22].

7. Conclusions

To prove the curvatures invariant, Chauey et al. [23] defined the notion of semi-symmetric metric P-connection in Riemannian setting, which is a particular case of Riemannian concircular structure manifold [26]. This topic has great applications in differential equations. We proved that the scalar curvature of Riemannian 3-manifolds endowed with a semi-symmetric metric $\xi$-connection and Ricci–Bourguignon soliton satisfies the Poisson and Laplace equations. It is well known that the Poisson and Laplace equations play a crucial role in the development of engineering, physics, mathematics, etc. We have also established the conditions for which the scalar curvature is harmonic, sub-harmonic and super-harmonic. We also established the existence condition of a gradient $\rho$-Einstein soliton in the Riemannian 3-manifolds, and consequently we proved some results. To verify our results, we constructed a non-trivial example of a three-dimensional Riemannian manifold equipped with a semi-symmetric metric $\xi$-connection. These topics are modern and have a lot of scope for researchers.