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eISSN 0454-8124
pISSN 1225-6951

### Articles

Kyungpook Mathematical Journal 2017; 57(4): 677-682

Published online December 23, 2017

### Generalised Ricci Solitons on Sasakian Manifolds

Mohammed El Amine Mekki and Ahmed Mohammed Cherif

Department of Mathematics, University Mustapha Stambouli, Mascara, 29000, Algeria

Received: January 12, 2017; Accepted: July 18, 2017

### Abstract

In this paper, we show that a Sasakian manifold which also satisfies the generalised gradient Ricci soliton equation, satisfying some conditions, is necessarily Einstein.

Keywords: Sasakian manifolds, Ricci soliton, Einstein manifolds

### 1. Introduction and Main Results

By R and Ric we denote respectively the Riemannian curvature tensor and the Ricci tensor of a Riemannian manifold (M,g). Then R and Ric are defined by:

R(X,Y)Z=XYZ-YXZ-[X,Y]Z,Ric(X,Y)=g(R(X,ei)ei,Y),

where ∇ is the Levi-Civita connection with respect to g, {ei} is an orthonormal frame, and X, Y,Z ∈ Γ(TM). Given a smooth function f on M, the gradient of f is defined by:

the Hessian of f is defined by:

where X, Y ∈ Γ(TM). For X ∈ Γ(TM), we define X ∈ Γ(T*M) by:

X(Y)=g(X,Y).

(For more details, see for example [11]).

- The generalised Ricci soliton equation in Riemannian manifold (M,g) is defined by (see [10]):

Xg=-2c1XX+2c2Ric+2λg,

where X ∈ Γ(TM), ℒXg is the Lie-derivative of g along X given by:

(Xg)(Y,Z)=g(YX,Z)+g(ZX,Y),

for all Y,Z ∈ Γ(TM), and c1, c2, λ ∈ ℝ. Equation (1.6), is a generalization of:

• Killing’s equation (c1 = c2 = λ = 0);

• Equation for homotheties (c1 = c2 = 0);

• Ricci soliton (c1 = 0, c2 = −1);

• Cases of Einstein-Weyl (c1 = 1, c2=-1n-2);

• Metric projective structures with skew-symmetric Ricci tensor in projective class (c1 = 1, c2=-1n-1, λ = 0);

• Vacuum near-horzion geometry equation (c1 = 1, c2=12).

(For more details, see [5], [8], [9], [10]). Equation (1.6), is also a generalization of Einstein manifolds (see [1]). Note that, if X = grad f, where fC(M), the generalised Ricci soliton equation is given by:

Hess f=-c1dfdf+c2Ric+λg.

- An (2n + 1)-dimensional Riemannian manifold (M,g) is said to be an almost contact metric manifold if there exist a (1, 1)-tensor field ϕ on M, a vector field ξ ∈ Γ(TM) and a 1-form η ∈ Γ(T*M), such that:

η(ξ)=1,         ϕ2X=-X+η(X)ξg(ϕX,ϕY)=g(X,Y)-η(X)η(Y),

for all X, Y ∈ Γ(TM). In particular, in an almost contact metric manifold we also have ϕξ = 0, ηϕ = 0 and η(X) = g(ξ,X). It can be proved that an almost contact metric manifold is Sasakian if and only if (see [2], [3], [12]):

(Xϕ)Y=g(X,Y)ξ-η(Y)X,

for any X, Y ∈ Γ(TM). For a Sasakian manifold the following equations hold:

R(X,Y)ξ=η(Y)X-η(X)Y.Xξ=-ϕX,         (Xη)Y=-g(ϕX,Y).

The main result of this paper is the following:

### Theorem 1.1

Suppose (M,ϕ, ξ, η, g) is a Sasakian manifold of dimension (2n+1), and satisfies the generalised Ricci soliton equation (1.8) with c1(λ + 2c2n) = −1, then f is a constant function. Furthermore, if c2 = 0, then (M,g) is an Einstein manifold.

From Theorem 1.1, we get the following:

• Suppose (M,g) is a Sasakian manifold and satisfies the gradient Ricci soliton equation (i.e. Hess f = −Ric+λg), then f is a constant function and (M,g) is an Einstein manifold (this result is obtained by P. Nurowski and M. Randall [10]).

• In a Sasakian manifold (M,ϕ, ξ, η, g), there is no non-constant smooth function f, such that Hess f = λg, for some constant λ.

### 2. Proof of the Result

For the proof of Theorem 1.1, we need the following lemmas.

### Lemma 2.1

Let (M,ϕ, ξ, η, g) be a Sasakian manifold. Then:

(ξ(Xg))(Y,ξ)=g(X,Y)+g(ξξX,Y)+Yg(ξX,ξ),

where X, Y ∈ Γ(TM), with Y is orthogonal to ξ.

Proof

First note that:

(ξ(Xg))(Y,ξ)=ξ((Xg)(Y,ξ))-(Xg)(ξY,ξ)-(Xg)(Y,ξξ),

since ℒξY = [ξ, Y ], ℒξξ = [ξ, ξ] = 0, by equations (1.7) and (2.1), we have:

(ξ(Xg))(Y,ξ)=ξg(YX,ξ)+ξg(ξX,Y)-g([ξ,Y]X,ξ)-g(ξX,[ξ,Y])=g(ξYX,ξ)+g(YX,ξξ)+g(ξξX,Y)+g(ξX,ξY)-g([ξ,Y]X,ξ)-g(ξX,ξY)+g(ξX,Yξ),

from equation (1.13), we get ∇ξξ = ϕξ = 0, so that:

(ξ(Xg))(Y,ξ)=g(ξYX,ξ)+g(ξξX,Y)-g([ξ,Y]X,ξ)+Yg(ξX,ξ)-g(YξX,ξ),

by the definition of the Riemannian curvature tensor (1.1), and (2.2), we obtain:

(ξ(Xg))(Y,ξ)=g(R(ξ,Y)X,ξ)+g(ξξX,Y)+Yg(ξX,ξ),

from equation (1.12), with g(Y, ξ) = 0, we have:

g(R(ξ,Y)X,ξ)=g(R(Y,ξ)ξ,X)=g(X,Y).

the Lemma follows from equations (2.3) and (2.4).

### Lemma 2.2

Let (M,g) be a Riemannian manifold, and let fC(M). Then:

(ξ(dfdf))(Y,ξ)=Y(ξ(f))ξ(f)+Y(f)ξ(ξ(f)),

where ξ, Y ∈ Γ(TM).

Proof

We compute:

(ξ(dfdf))(Y,ξ)=ξ(Y(f)ξ(f))-[ξ,Y](f)ξ(f)-Y(f)[ξ,ξ](f)=ξ(Y(f))ξ(f)+Y(f)ξ(ξ(f))-[ξ,Y](f)ξ(f),

since [ξ, Y ](f) = ξ(Y (f)) − Y (ξ(f)), we get:

(ξ(dfdf))(Y,ξ)=[ξ,Y](f)ξ(f)+Y(ξ(f))ξ(f)+Y(f)ξ(ξ(f))-[ξ,Y](f)ξ(f)=Y(ξ(f))ξ(f)+Y(f)ξ(ξ(f)).

The proof is completed.

### Lemma 2.3

Let (M,ϕ, ξ, η, g) be a Sasakian manifold of dimension (2n+1), and satisfies the generalised Ricci soliton equation (1.8). Then:

Proof

Let Y ∈ Γ(TM), from the definition of Ricci curvature (1.2), and the curvature condition (1.12), we have:

Ric(ξ,Y)=g(R(ξ,ei)ei,Y)=g(R(ei,Y)ξ,ei)=η(Y)g(ei,ei)-η(ei)g(X,ei)=(2n+1)η(Y)-η(Y)=2nη(Y)=2ng(ξ,Y),

where {ei} is an orthonormal frame on M, implies that:

λg(ξ,Y)+c2Ric(ξ,Y)=λg(ξ,Y)+2c2ng(ξ,Y)=(λ+2c2n)g(ξ,Y),

from equations (1.8) and (2.5), we obtain:

the Lemma follows from equation (2.6) and the definition of the Hessian (1.3).

Proof of Theorem 1.1

Let Y ∈ Γ(TM), such that g(ξ, Y ) = 0, from Lemma 2.1, with X = grad f, we have:

by Lemma 2.3, and equation (2.7), we get:

since ∇ξξ = 0 and g(ξ, ξ) = 1, from equation (2.8), we obtain:

from Lemma 2.3, equation (2.9), and since g(ξ, Y ) = 0, we have:

2(ξ(Hess f))(Y,ξ)=Y(f)-c1ξ(ξ(f))Y(f)+c12ξ(f)2Y(f)-2c1ξ(f)Y(ξ(f)).

Note that, from equation (1.13), we have ℒξg = 0 (i.e. ξ is a Killing vector field), implies that ℒξ Ric = 0, taking the Lie derivative to the generalised Ricci soliton equation (1.8) yields:

2(ξ(Hess f))(Y,ξ)=-2c1(ξ(dfdf))(Y,ξ),

thus, from equations (2.10), (2.11) and Lemma 2.2, we have:

Y(f)-c1ξ(ξ(f))Y(f)+c12ξ(f)2Y(f)-2c1ξ(f)Y(ξ(f))=-2c1Y(ξ(f))ξ(f)-2c1Y(f)ξ(ξ(f)),

is equivalent to:

Y(f)[1+c1ξ(ξ(f))+c12ξ(f)2]=0,

according to Lemma 2.3, we have:

by equations (2.13) and (2.14), we obtain:

Y(f)[1+c1(λ+2c2n)]=0,

since c1(λ+2c2n) ≠ −1, we fined that Y (f) = 0, i.e., grad f is parallel to ξ. Hence grad f = 0 as D = kerη is nowhere integrable, i.e., f is a constant function.

### Acknowledgements

The authors would like to thank the reviewers for their useful remarks and suggestions. Partially supported by National Agency Scientific Research of Algeria.

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