A vector field $V$ on a Riemannian manifold $(M, g)$ is Killing if the Lie derivative of the metric with respect to $V$ vanishes as follows

$ \begin{equation} L_{V}g=0, \end{equation} $

(1.1)

which is equivalent to

$ \begin{equation} g(\nabla_X V, Y)+g(\nabla_Y V, X)=0, \end{equation} $

(1.2)

where $\nabla$ denotes the covariant differential operator of $(M, g)$ and $X, Y\in TM$. This is equivalent to the fact that the one-parameter group of diffeomorphisms associated to $V$ consists in isometries. Therefore, the space of the non-trivial Killing vector fields for $(M, g)$ in some sense measures the size of the isometry group of $(M, g)$.

The study of killing vector field has a long time. In 1946, Bochner [2] proved that when $(M, g)$ is compact and has negative Ricci curvature, every Killing vector field must vanish. Later, Bochner's result was extended by Yano to include conformal vector fields [10]. It is well known that the existence of non-trivial closed conformal vector fields also imposes many restrictions on a compact Riemannian manifold (see [9]). The Killing vector fields were generalized to the Killing $p$-form and conformal Killing $p$-form by some authors such as Bochner and Yano, Gallot and Meyer, Tachibana and Yamaguchi, Liu Jizhi and Cai Kairen (see [8]). In 1999, Gursky [5] proved a vanishing theorem for conformal vector fields on four-manifolds of negative scalar curvature, whose assumptions were conformally invariant, and in the case of locally conformally flat manifolds reduced to a sign condition on the Euler characteristic. The proof due to Gursky is actually a refinement of the Bochner method which had been used to prove classical vanishing theorems. Recently, inspired by Gursky's two papers [4, 5], Hu and Li [6] proved Theorem A as follows.

Theorem A Let $(M, g)$ be an $n$-dimensional compact oriented Riemannian manifold with scalar curvature $R<0$. If there exists a non-trivial Killing vector field on $(M, g)$, then we have

$ \begin{equation}\label{eq-theorem 1} \int_M\frac{1}{R}(|E|^2-\frac{R^2}{n(n-1)})\leq0, \end{equation} $

(1.3)

where $E$ denotes the trace-free Ricci tensor. Moreover, equality is attained in (1.3) if and only if $R$ is constant and the Riemannian universal cover of $(M, g)$ is isometric to a Riemannian product $\mathbb{R}\times N^{n-1}$ for some Einstein manifold $N^{n-1}$ with constant scalar curvature $R$.

We follow their methods [3, 5, 6] to improve Theorem A and obtain the following results.

Theorem 1.1 Let $(M, g)$ be an $n$-dimensional compact oriented Riemannian manifold with scalar curvature $R$. If there exists a non-trivial Killing vector field on $(M, g)$, then we have

$ \begin{equation} \int_M|E|\geq-\int_M\frac{R}{\sqrt{n(n-1)}}, \end{equation} $

(1.4)

where $E$ denotes the trace-free Ricci tensor. Moreover, equality holds in (1.4) if and only if $R$ is nonpositive constant and the Riemannian universal cover of $(M, g)$ is isometric to a Riemannian product $\mathbb{R}\times N^{n-1}$ for some Einstein manifold $N^{n-1}$ with constant scalar curvature $R$.

Corollary 1.2 Let $(M, g)$ be an $n$-dimensional compact oriented Riemannian manifold with scalar curvature $R<0$. If there exists a non-trivial Killing vector field on $(M, g)$, then we have

$ \begin{equation*} \int_M|E|\geq-\int_M\frac{R}{\sqrt{n(n-1)}}, \end{equation*} $

where $E$ denotes the trace-free Ricci tensor. Moreover, equality holds in the above if and only if $R$ is constant and the Riemannian universal cover of $(M, g)$ is isometric to a Riemannian product $\mathbb{R}\times N^{n-1}$ for some Einstein manifold $N^{n-1}$ with constant scalar curvature $R$.

Let $(M, g)$ be an $n$-dimensional Riemannian manifold. Let $\{e_{1}, \cdots, e_{n}\}$ with respect to the Riemannian metric $g$ be a local orthonormal basis of $TM$, and $\{\theta_{1}, \cdots, \theta_{n}\}$ be its dual basis. Let $\{\theta_{ij}\}$ be the connection forms of $(M, g)$, one has the structure equations

$ \begin{eqnarray}\label{eq-21} &&d\theta_{i}=\sum\limits_{j}\theta_{ij}\wedge\theta_{j}, ~~~\theta_{ij}+\theta_{ji}=0, \\\label{eq-22} \end{eqnarray} $

(2.1)

$ \begin{eqnarray} &&d\theta_{ij}-\sum\limits_{k}\theta_{ik}\wedge\theta_{kj}=-\frac{1}{2}\sum\limits_{k, l}R_{ijkl}\theta_{k}\wedge\theta_{l}, \end{eqnarray} $

(2.2)

where $R_{ijkl}$ are the components of the Riemannian curvature tensor of $(M, g)$. The Ricci curvature tensor $R_{ij}$ and the scalar curvature $R$ of $(M, g)$ are defined by

$ \begin{eqnarray}\label{eq-23} R_{ij}=\sum\limits_{k}R_{ikjk}~~{\rm and}~~R=\sum\limits_{i}R_{ii}, \end{eqnarray} $

(2.3)

respectively. For a vector field $V=\sum\limits_{i}V_{i}e_{i}$ on $(M, g)$, we define the covariant derivative $V_{i, j}$ and the second covariant derivative $V_{i, jk}$, respectively, by $\nabla V=\sum\limits_{i, j}V_{i, j}\theta_{j}\otimes e_{i}, $ i.e.,

$ \begin{eqnarray}\label{eq-24} \sum\limits_{j}V_{i, j}\theta_{_{j}}=dV_{i}+\sum\limits_{j}V_{j}\theta_{ji} \end{eqnarray} $

(2.4)

and

$ \begin{eqnarray}\label{eq-25} \sum\limits_{k}V_{i, jk}\theta_{k}=dV_{i, j}+\sum\limits_{k}V_{k, j}\theta_{ki}+\sum\limits_{k}V_{i, k}\theta_{kj}. \end{eqnarray} $

(2.5)

Using exterior derivation of (2.4), one gets the Ricci identity

$ \begin{eqnarray}\label{eq-26} V_{i, kl}=V_{i, lk}+\sum\limits_{j}V_{j}R_{jikl}. \end{eqnarray} $

(2.6)

Now we assume that $V=\sum\limits_{i}V_{i}e_{i}$ is a Killing vector. Note that from (1.2), $V=\sum\limits_{i}V_{i}e_{i}$ is a Killing vector field is equivalent to

$ \begin{eqnarray}\label{eq-27} V_{i, j}+V_{j, i}=0~~{\rm for ~all}~~ i, j. \end{eqnarray} $

(2.7)

From (2.5) and (2.7), we have

$ \begin{eqnarray} \sum\limits_{j, k}V_{j, jk}\theta_{k} &=&\sum\limits_{j}(dV_{j, j}+\sum\limits_{k}V_{k, j}\theta_{kj}+\sum\limits_{k}V_{j, k}\theta_{kj})\nonumber\\ &=&d(\sum\limits_{j}V_{j, j})+\sum\limits_{j, k}(V_{k, j}+V_{j, k})\theta_{kj}\nonumber\\ &=& 0, \nonumber \end{eqnarray} $

and thus for any $k$,

$ \begin{eqnarray}\label{eq-31} \sum\limits_{j}V_{j, jk}=0. \end{eqnarray} $

(2.8)

Combing (2.6), (2.7) with (2.8), we get the following Weitzenböck formula (see [1])

$ \begin{eqnarray}\label{eq-32} \frac{1}{2}\triangle |V|^{2} &=& \sum\limits_{i, j}V_{i, j}^{2}+\sum\limits_{i, j}V_{i}V_{i, jj}=|\nabla V|^{2}-\sum\limits_{i, j}V_{i}V_{j, ij} \nonumber\\ &=&|\nabla V|^{2}-\sum\limits_{i, j}V_{i}(V_{j, ji}+\sum\limits_{k}V_{k}R_{kjij})\nonumber\\ &=&|\nabla V|^{2}-\sum\limits_{i, j}V_{i}V_{j}R_{ij}. \end{eqnarray} $

(2.9)

Lemma 2.1 Let $V=\sum\limits_{i}V_{i}e_{i}$ be a Killing vector field on the $n$-dimensional Riemannian manifold $(M, g)$. Then we have

$ \begin{eqnarray}\label{eq-33} \sum\limits_{i}V_{i}^{2}\sum\limits_{i, j}V_{i, j}^{2}\geq 2\sum\limits_{j}(\sum\limits_{i}V_{i}V_{i, j})^{2}. \end{eqnarray} $

(2.10)

Remark 2.2 In [5], Gursky observed that (2.10) still holds for every conformal vector field $V$. In [6], Hu and Li proved Lemmas 2.1 and 2.3. For completeness, we write the proofs of Hu and Li.

Proof It suffices to prove (2.10) for any fixed point $p\in \Omega_{0}:=\{x\in M|V(x)\neq 0\}.$ Note that on $\Omega_{0}, $ (2.10) is equivalent to

$ \begin{eqnarray}\label{eq-34} |\nabla V|^{2}\geq 2|\nabla|V||^{2}. \end{eqnarray} $

(2.11)

Around $p$, we choose $\{e_{i}\}$ such that $V(p)=V_{1}(p)e_{1}(p)$; that is, $V_{2}=\cdots=V_{n}=0$ at $p$. From (2.7), we have

$ V_{1, 1}=\cdots=V_{n, n}=0;~~V_{1, j}=-V_{j, 1}, ~~2\leq j\leq n. $

Then at $p$, we have

$ \begin{eqnarray}2\sum\limits_{j}(\sum\limits_{i}V_{i}V_{i, j})^{2} =2V_{1}^{2}\sum\limits_{j}V_{1, j}^{2} =2V_{1}^{2}\sum\limits_{j=2}^{n}V_{1, j}^{2} \leq V_{1}^{2}\sum\limits_{i, j}V_{i, j}^{2} =\sum\limits_{i}V_{i}^{2}\sum\limits_{i, j}V_{i, j}^{2}.\nonumber \end{eqnarray} $

This proves (2.10), or equivalently proves (2.11) on $\Omega_{0}.$

Combing (2.6) and (2.7) with (2.8), we have

$ \begin{eqnarray}\label{eq-35} \triangle V_{i}=\sum\limits_{j}V_{i, jj}=-\sum\limits_{j}V_{j}R_{ij}, ~~~~~1\leq i\leq n. \end{eqnarray} $

(2.12)

Form (2.12) and the unique continuation result of Kazdan [7], we know that, for a non-trivial Killing vector field $V$ on the compact Riemannian manifold $(M, g)$, the set $M\backslash\Omega_{0}$ is of measure zero. Combining this fact with (2.9) and (2.11), one has the following lemma.

Lemma 2.3 Let $V$ be a non-trivial Killing vector field on a compact Riemannian manifold $(M, g)$. Then

$ \begin{eqnarray}\label{eq-36} \frac{1}{2}\triangle|V|^{2}\geq2|\nabla|V||^{2}-\sum\limits_{i, j}V_{i}V_{j}R_{ij} \end{eqnarray} $

(2.13)

holds on $M$ in the sense of distributions.

In order to prove Theorem 1.1, we need the following algebraic lemma, which can be proved by the standard method of Lagrange multipliers, and was observed by Hu and Li [6].

Lemma 2.4 Let $A=({a_{ij}})_{n\times n}$ be a real symmetric matrix with $\sum\limits_{i}a_{ii}=0$ and $x_{1}, \cdots, x_{n}\in\mathbb{R}$. Then

$ \begin{eqnarray}\label{eq-212} -\sqrt{\frac{n-1}{n}\sum\limits_{i, j}a_{ij}^{2}}(\sum\limits_{i}x_{i}^{2})\leq \sum\limits_{i, j}a_{ij}x_{i}x_{j}\leq \sqrt{\frac{n-1}{n}\sum\limits_{i, j}a_{ij}^{2}}(\sum\limits_{i}x_{i}^{2}). \end{eqnarray} $

(2.14)

Moreover, when $\sum\limits_{i}x^{2}_{i}\neq 0$, one of the equalities in (2.14) is attained if and only if there exists an orthogonal $n\times n$ matrix $T$ such that

$ \begin{eqnarray} TAT^{-1}=\left( \begin{array}{cccc} (n-1)\lambda&&\\ &-\lambda&&\\ &&\ddots&\\ &&-\lambda \end{array} \right) \end{eqnarray} $

(2.15)

and $(x_{1}, \cdots, x_{n})$ correspondingly takes the value $((n-1)\lambda, 0, \cdots, 0)$, where $ \lambda >0 $ holds if there is equality in the right-hand side of (2.14), and $\lambda<0$ holds if there is equality in the left-hand side of (2.14).

Proof of Theorem 1.1 Let $V=\sum\limits_iV_ie_i$ be a non-trivial Killing vector field on $(M, g)$. Denote by $E$ the trace-free part of the Ricci tensor Ric, i.e., $E_{ij}=R_{ij}-(R/n)\delta_{ij}$. Then, applying Lemma 2.4, we get that

$ \begin{eqnarray}\label{eq-41} -\sum\limits_{i, j}V_iV_jR_{ij} = -\sum\limits_{i, j}V_iV_jE_{ij}-\frac{R}{n}|V|^2 \geq -\sqrt{\frac{n-1}{n}}|E||V|^2-\frac{R}{n}|V|^2 \end{eqnarray} $

(2.16)

and the second equality holds at a point $p\in M$ with $V(p)\neq0$ if and only if by choosing suitable $\{ e_i \}$, $E$ can be diagonalized as

$ \begin{eqnarray}\label{eq-42} E=\left( \begin{array}{cccc} (n-1)\lambda&&\\ &-\lambda&&\\ &&\ddots&\\ &&-\lambda \end{array} \right), \end{eqnarray} $

(2.17)

correspondingly, $V_1=(n-1)\lambda$ and $V_2=\ldots=V_n=0$ for some $\lambda>0$.

Combining (2.13) with (2.16), we get

$ \begin{eqnarray} \frac{1}{2}\triangle |V|^{2}\geq 2|\nabla |V||^{2}-\sqrt{\frac{n-1}{n}}|E||V|^{2}-\frac{R}{n}|V|^{2} . \end{eqnarray} $

(2.18)

For $\varepsilon>0$, we define a function $f_{\varepsilon}=\sqrt{|V|^{2}+\varepsilon^{2}}$. Thus we

$ \begin{eqnarray} |\nabla f_{\varepsilon}|^{2}=\frac{|V|^2|\nabla |V||^2}{|V|^{2}+\varepsilon^{2}}\leq|\nabla |V||^2. \end{eqnarray} $

(2.19)

From (2.18) and (2.19), we directly compute

$ \begin{eqnarray} f_{\varepsilon}\triangle f_{\varepsilon}+|\nabla f_{\varepsilon}|^{2} &=&\frac{1}{2}\triangle f_{\varepsilon}^{2} =\frac{1}{2}\triangle |V|^{2} \nonumber \\ &\geq&2|\nabla |V||^{2}-\sqrt{\frac{n-1}{n}}|E||V|^{2}-\frac{R}{n}|V|^{2} \nonumber \\ &\geq& 2|\nabla f_{\varepsilon}|^{2}-\sqrt{\frac{n-1}{n}}|E|f_{\varepsilon}^{2}-\frac{R}{n}f_{\varepsilon}^{2} \end{eqnarray} $

(2.20)

and thus

$ \begin{eqnarray} f_{\varepsilon}\triangle f_{\varepsilon}\geq |\nabla f_{\varepsilon}|^{2}-\sqrt{\frac{n-1}{n}}|E|f_{\varepsilon}^{2}-\frac{R}{n}f_{\varepsilon}^{2} \end{eqnarray} $

(2.21)

and thus

$ \begin{eqnarray} \triangle \log f_{\varepsilon}=f_{\varepsilon}^{-2}(f_{\varepsilon}\triangle f_{\varepsilon}-|\nabla f_{\varepsilon}|^{2}) \geq -\sqrt{\frac{n-1}{n}}|E|-\frac{R}{n}. \end{eqnarray} $

(2.22)

Integrating (2.22) on $ M$, we obtain

$ \begin{eqnarray*} 0=\int_{M} \Delta \log f_{\varepsilon}\geq -\int_{M}(\sqrt{\frac{n-1}{n}}|E|+\frac{R}{n}), \end{eqnarray*} $

i.e.,

$ \begin{eqnarray} \int_{M}|E|\geq -\frac{1}{\sqrt{(n-1){n}}}\int_{M}R. \end{eqnarray} $

(2.23)

If the equality holds in (2.23), then equality (2.16) must hold at each point of $M$. Thus at each point of $M$, $E$ can be diagonalized as in (2.17). So it satisfies

$ \begin{eqnarray}\label{eq-46} |E|\equiv-\frac{R}{\sqrt{n(n-1)}} \ \ \rm{on}\ \ M. \end{eqnarray} $

(2.24)

Furthermore, combining (2.17) with (2.24) gives $\lambda=-R/n(n-1)$. Then we must have $V_1=-R/n$, $V_2=\ldots=V_n=0$ and

$ \begin{eqnarray}\label{eq-47} {\rm Ric}=(R_{ij})=\left( \begin{array}{cccc} 0&&\\ &\frac{R}{n-1}&&\\ &&\ddots&\\ &&\frac{R}{n-1} \end{array} \right). \end{eqnarray} $

(2.25)

This implies that $\sum\limits_{ij}V_iV_jR_{ij}=0$, and thus from (2.9), by the maximum principle we get that $|V|=-R/n$ is constant, then $\nabla V=0$, i.e., $V$ is a parallel vector field on $(M, g)$. By the de Rham decomposition theorem and (2.25), we see that the Riemannian universal cover $\widetilde{M}$ of $M$ is the product of $\mathbb{R}$ and an Einstein manifold $\widetilde{N}$ with nonpositive constant scalar curvature $R$.

Remark 2.5 We see that if $R<0$, one has

$ \begin{eqnarray*}\label{eq-43} -(\sqrt{\frac{n-1}{n}}|E|+\frac{R}{n})|V|^2&=&-\frac{n-1}{2R}[(|E|+\frac{R}{\sqrt{n(n-1)}})^2-(|E|^2-\frac{R^2}{n(n-1)})]|V|^2\nonumber\\ &\geq&\frac{n-1}{2R}(|E|^2-\frac{R^2}{n(n-1)})|V|^2. \end{eqnarray*} $

Then

$ -(\sqrt{\frac{n-1}{n}}|E|+\frac{R}{n})\geq\frac{n-1}{R}(|E|^2-\frac{R^2}{n(n-1)}) $

holds on $M$ in the sense of distributions. Hence Corollary 1.2 can be considered as generalization of Theorem 1 in [6], i.e., Theorem A.