1 Introduction

Because of their relevance in both geometric analysis and general relativity, the study of Riemannian (or semi-Riemannian) manifolds admitting an Einstein-like structure is an attractive subject in modern Mathematical Physics and Differential Geometry. Among the enormous literature on the topic, we indicate Deshmukh and Al-Sodais in [9], Andrade and de Melo in [1], Wylie in [15] and Araújo, Freitas and Santos in [2].

In this sense, an n-dimensional Riemannian manifold \((M^n,g)\), \(n\ge 2,\) is called an m-quasi-Einstein manifold, for a non-null constant \(m \in \mathbb {R}\), if there exist a smooth vector field X on M and a scalar \(\lambda \in \mathbb {R}\) satisfying the following equation

$$\begin{aligned} \text {Ric}+\frac{1}{2} \mathcal {L}_Xg-\frac{1}{m}X^\flat \otimes X^\flat =\lambda g, \end{aligned}$$
(1.1)

where \(\text {Ric}\), \(\mathcal {L}_Xg\), \(X^\flat \) and \(\otimes \) stand for the Ricci tensor, the Lie derivative of the metric g in the direction of X, the dual 1-form to X associated the metric g and the tensorial product, respectively. We notice that, when X vanishes identically \(M^n\) is an Einstein manifold.

These manifolds play an essential role in investigating the solutions of the Einstein field equations. Naturally, when m is a positive integer the m-quasi-Einstein manifolds correspond to some warped product Einstein metrics. See, for instance, Barros and Ribeiro in [5].

In this branch, a fundamental application of the called smooth metric measure spaces is as a description of quasi-Einstein manifolds (see Case in [7]). It is worth pointing out that, commonly, the study of m-quasi-Einstein manifolds is considered when X is a gradient of a smooth function on \(M^n\) (see Ribeiro and Tenenblat in [14]). Furthermore, non-gradient m-quasi-Einstein metrics has been playing an important role in the theory of near-horizon geometries and can be taken in the formulation of perfect fluid spacetimes. Moreover, when m goes to infinity, this structure reduces to the one associated to a Ricci soliton. For more details we refer to Bahuaud, Gunasekaran, Kunduri and Woolgar in [3] and references therein. See also Poddar, Sharma and Subramanian in [12].

In this setting, we quote to Bahuaud, Gunasekaran, Kunduri and Woolgar in [4], Güler and De in [11] and Ghosh in [10] for some examples of m-quasi-Einstein manifolds and related facts.

In what follows, S denotes the scalar curvature of \((M^n,g)\), defined as the trace of \(\text {Ric}.\) Recall that a smooth vector field X defined on a Riemannian manifold \((M^n,g)\) is said to be a Killing vector field if \(\mathcal {L}_Xg=0.\) Killing vector fields have a number of notable features. Besides, on the impact in the geometry as well as in the topology of a spacetime, there is a rich variety of questions where Killing vector fields play a central role. Equally, the assumption of admitting a Killing vector field on a compact manifold has been used in the literature from several points of view.

In this direction, very recently Cochran in [8] established, among other properties, the following interesting result.

Theorem 1.1

Let \((M^n, g)\) be a closed m-quasi-Einstein manifold. Then,

  1. a)

    S is constant if and only if X is a Killing vector field.

  2. b)

    If X is divergence-free, then X is a Killing vector field.

Our purpose in this note is to give an alternative and unified proof of the above fact, making use of a classical integral formula involving the Lie derivative and well-known identities referring to m-quasi-Einstein metrics. The new approach is valid for all \(m \ne 0\), in contrast to this previous proof which also assumed the constant \(m \ne -2\).

As a consequence of our discussion, we also reobtain a triviality result for m-quasi-Einstein manifolds due to Bahuaud et al. in [4].

2 Proof of theorem

In this section we present the proof of theorem 1.1. To begin with, we recall some useful auxiliary results. All Riemannian manifolds \((M^n,g)\) are assumed to be connected and oriented. The Lie derivative of the metric g in the direction of Y reads as

$$\begin{aligned} (\mathcal {L}_Yg)(U,V)=\langle \nabla _UY,V\rangle +\langle U,\nabla _VY\rangle , \end{aligned}$$

for all smooth vector fields \(Y,\, U\) and V on \(M^n,\) where \(\nabla \) stands for the Levi-Civita connection of g. Taking trace in above equality, we see that \(tr (\mathcal {L}_Yg)=2\, \text {div}Y\), where \(\text {div}\) denotes the divergence of a vector field. In general, for an arbitrary smooth vector field Y on \(M^n\) one has

$$\begin{aligned} \text {div}(Y^\flat \otimes Y^\flat )=(\text {div}Y)Y^\flat +(\nabla _YY)^\flat . \end{aligned}$$
(2.1)

We also find that

$$\begin{aligned} \text {div}(|Y|^2Y)=|Y|^2\text {div}Y+2\langle \nabla _YY, Y \rangle . \end{aligned}$$
(2.2)

Similarly, we have

$$\begin{aligned} \text {div}((\text {div}Y) Y)= (\text {div}Y)^2+ \langle \nabla (\text {div}Y), Y \rangle . \end{aligned}$$
(2.3)

On the other hand, it can be proved (see Poor in [13], page 170) that for each vector field Y on a closed Riemannian manifold \((M^n, g)\) it holds that

$$\begin{aligned} \frac{1}{2}\int _M|\mathcal {L}_Yg|^2\,dM=\int _M(|\nabla Y|^2+(\text {div}Y)^2-\text {Ric}(Y,Y))\,dM. \end{aligned}$$
(2.4)

Next, notice that the trace of the Eq. (1.1) provide us

$$\begin{aligned} S+\text {div}X=\frac{1}{m}|X|^2+n\lambda . \end{aligned}$$
(2.5)

At this point, we remember the following key identity obtained by Barros and Ribeiro, which can be found in [5], page 215. Below, \(\Delta \) denotes the Laplacian operator on \(M^n\).

Lemma 2.1

Let \((M^n, g)\) be a m-quasi-Einstein manifold. Then,

$$ \frac{1}{2}\Delta |X|^2=|\nabla X|^2+\frac{2}{m}|X|^2\text {div}\, X-\text {Ric}(X,X).$$

As an application of Lemma 2.1 and for the sake of completeness, we would like to present a proof of Proposition 2.4 in [4], page 8, by following the argument used in that work. It is worth mentioning that essentially the authors in [4] also proved the above Lemma. We state the following result:

Proposition 2.2

Let \((M^n, g)\) be a closed m-quasi-Einstein manifold with divergence-free vector field X. If \(m<0\) and \(\lambda \le 0\), then X is identically zero.

Proof

According to (1.1),

$$\text {Ric}(X,X)+\langle \nabla _XX, X \rangle =\frac{1}{m}|X|^4+\lambda |X|^2.$$

Setting \(\text {div}X=0\) in (2.2), by the divergence theorem we have

$$\int _M\langle \nabla _XX, X \rangle \, dM=0.$$

Hence, Lemma 2.1 implies that

$$\int _M(|\nabla X|^2-\frac{1}{m}|X|^4-\lambda |X|^2)\, dM=0,$$

which gives, under the assumptions, \(X=0.\)

Alternatively, we may invoke the second item from Theorem 1.1 and apply the maximum principle for the Laplacian to deduce that \(|X|^2\) is constant. Again by Lemma 2.1, we verify that \(M^n\) is an Einstein manifold. \(\square \)

In order to prove the main result, we also need the following nice formula due to Barros and Gomes in [6], page 244. Namely,

Lemma 2.3

Let \((M^n, g)\) be a m-quasi-Einstein manifold. Then, for all vector field Y on \(M^n\), we have

$$\begin{aligned} \text {div}\, ((\mathcal {L}_Xg)Y)=2\text {div}\, \left( \lambda Y+\frac{1}{m}(X^\flat \otimes X^\flat )Y\right) -g(\nabla S, Y)-2\langle \nabla Y,\text {Ric} \rangle . \end{aligned}$$

Now, we are in position to procced with the proof of the theorem.

Proof

Initially, suppose that S is constant. We have from (2.5) that

$$\nabla (\text {div}X)=\frac{1}{m}\nabla |X|^2.$$

Thus, taking into account the expression (2.3) we get that

$$\text {div}((\text {div}X) X)= (\text {div}X)^2+ \frac{2}{m}\langle \nabla _XX, X \rangle .$$

From (2.2) and the above equality, applying the divergence theorem we obtain

$$\begin{aligned} \int _M|X|^2\text {div}X\, dM=m\int _M(\text {div}X)^2\, dM. \end{aligned}$$
(2.6)

Note that, from identity (2.4), Lemma 2.1 and by the divergence theorem, we have

$$\frac{1}{2}\int _M|\mathcal {L}_Xg|^2\,dM=-\frac{2}{m}\int _M|X|^2\text {div}X\, dM+\int _M(\text {div}X)^2\, dM.$$

Plugging the relation (2.6) in the above equality we derive that

$$\frac{1}{2}\int _M|\mathcal {L}_Xg|^2\,dM=-\int _M(\text {div}X)^2\, dM \le 0.$$

Therefore, we deduce that \(\mathcal {L}_Xg=0\).

Reciprocally, let us suppose that X is a Killing vector field on \(M^n.\) In this case, the fundamental Eq. (1.1) becomes

$$\begin{aligned} \text {Ric}=\lambda g-J_m, \end{aligned}$$
(2.7)

where \(J_m=-\frac{1}{m}X^\flat \otimes X^\flat .\) Next, choosing \(Y=\nabla S\) and using again the divergence theorem, we are able to use Lemma 2.3 and identity (2.7) to obtain that

$$\begin{aligned} \int _M|\nabla S|^2\, dM=-2\int _M\langle \nabla ^2S,\text {Ric} \rangle \, dM=2\int _M\langle \nabla ^2S, J_m \rangle \, dM. \end{aligned}$$
(2.8)

Here, \(\nabla ^2\) stands for the Hessian operator on \(M^n.\) Now, on the one hand, we have that

$$\begin{aligned} \text {div}(J_m(\nabla S))=(\text {div}J_m)(\nabla S)+\langle \nabla ^2S, J_m \rangle . \end{aligned}$$
(2.9)

On the other hand, according to formula (2.1) and \(\text {div}X=0\), from (2.9) we infer that

$$\begin{aligned} \text {div}(J_m(\nabla S))=-\frac{1}{m}\langle \nabla _XX, \nabla S \rangle +\langle \nabla ^2S, J_m \rangle . \end{aligned}$$
(2.10)

Thus, by integrating in (2.10) and using that X is a Killing field we have

$$\int _M\langle \nabla ^2S, J_m \rangle \, dM=\frac{1}{m}\int _M \langle \nabla _XX, \nabla S \rangle \, dM=\frac{1}{2m}\int _MS\Delta |X|^2\, dM. $$

Therefore, from (2.8) and (2.5) we get that

$$\int _M|\nabla S|^2\, dM=\frac{1}{m^2}\int _M|X|^2\Delta |X|^2\, dM=-\frac{1}{m^2}\int _M|\nabla |X|^2|^2\, dM \le 0.$$

Hence, \((M^n,g)\) has constant scalar curvature (implying that |X| also is constant), and we finish the proof of the first item.

The second item is immediate. Indeed, supposing \(\text {div}X=0\), we make use of Lemma 2.1 to conclude

$$ \frac{1}{2}\Delta |X|^2=|\nabla X|^2-\text {Ric}(X,X).$$

Finally, on integrating this identity and comparing with the formula (2.4) we can write

$$\int _M|\mathcal {L}_Xg|^2\,dM=0.$$

So, X is a Killing vector field, completing the proof of the theorem. \(\square \)