Vacuum static spaces with vanishing of complete divergence of Bach tensor and Weyl tensor

TL;DR

This paper investigates vacuum static spaces where the complete divergence of the Bach and Weyl tensors vanish, revealing implications for harmonicity, black hole non-existence, and the Besse conjecture.

Contribution

It establishes new links between divergence conditions of curvature tensors and geometric properties, including harmonicity and Bach-flatness, and proves the Besse conjecture under weaker assumptions.

Findings

Vanishing divergence implies harmonicity of the metric.

Non-existence of multiple black holes under these conditions.

Proof of the Besse conjecture with weaker assumptions.

Abstract

In this paper, we study vacuum static spaces with the complete divergence of the Bach tensor and Weyl tensor. First, we prove that the vanishing of complete divergence of the Bach tensor and Weyl tensor implies the harmonicity of the metric, and we present examples in which these conditions do not imply Bach flatness. As an application, we prove the non-existence of multiple black holes in vacuum static spaces with zero scalar curvature. On the other hand, we prove the Besse conjecture under these conditions, which are weaker than harmonicity or Bach flatness. Moreover, we show a rigidity result for vacuum static spaces and find a sufficient condition for the metric to be Bach-flat.

Equations

$$D_{g}df-\left(r_g-\frac{s_g}{n-1}g\right)f=\frac{1}{n}\left(\frac{s_g}{n-1}f+{\Delta }_{g}f\right)g,$$

$${\Delta }_{g}f=-\frac{s_g}{n-1}f,$$

$$D_{g}df-\left(r_g-\frac{s_g}{n-1}g\right)f=0.$$

$$s_g^{\prime }(h)=-{\Delta }_g{\mathrm{tr}}_{g}h+\delta \delta h-⟨r_g,h⟩$$

$$s_g^{\prime \ast }(f)=D_{g}df-({\Delta }_{g}f)g-f{r}_g$$

$$\text{mbox}{div}^4\mathcal{W}=0.$$

$$\frac{1}{2}|C{|}^2=-⟨\mathrm{div}C,z_g⟩,$$

$$C=d^D{r}_g-\frac{1}{2(n-1)}d{s}_g\wedge g.$$

$$(\varphi \wedge \eta )(X,Y,Z)=\varphi (X)\eta (Y,Z)-\varphi (Y)\eta (X,Z).$$

$$\frac{1}{2}|C{|}^2=-⟨\mathrm{div}C,z_g⟩$$

$$\displaystyle\begin{array}[]{ll}h\,r_{g}=D_{g}dh,\\ \Delta_{g}h=0.\end{array}$$

$$s_g^{\prime \ast }(f)=z_g$$

$$T=\frac{1}{n-2}df\wedge z+\frac{1}{(n-1)(n-2)}{i}_{\nabla f}z\wedge g,$$

$$B=\frac{1}{n-3}{\mathrm{div}}_1{\mathrm{div}}_4\mathcal{W}+\frac{1}{n-2}\mathring{\mathcal{W}}{r}_g,$$

$$\mathring{\mathcal{W}}{r}_g(X,Y)=\sum _{i=1}^n{r}_g(\mathcal{W}(X,E_i)Y,E_i)$$

$$\mathrm{div}\mathcal{W}=\frac{n-3}{n-2}C$$

$${\mathrm{div}}^4\mathcal{W}=\frac{n-3}{n-2}{\mathrm{div}}^{3}C,$$

$$(n-2)B=\mathrm{div}C+\mathring{\mathcal{W}}{r}_g=\mathrm{div}C+\mathring{\mathcal{W}}{z}_g.$$

$$\mathrm{div}B(X)=\frac{n-4}{(n-2{)}^2}⟨i_{X}C,z⟩.$$

$${\mathrm{div}}^{2}B=\frac{n-4}{(n-2{)}^2}\left(\frac{1}{2}|C{|}^2+⟨\mathrm{div}C,z⟩\right).$$

$$D^{\ast }Dz=-\sum _{i=1}^n{D}_{E_i,E_i}^{2}z\text{mbox}andz\circ z(X,Y)=\sum _{i=1}^{n}z(X,E_i)z(Y,E_i),$$

$$\mathrm{div}C=-D^{\ast }Dz-\frac{n}{n-2}z\circ z-\frac{s}{n-1}z+\mathring{\mathcal{W}}z+\frac{1}{n-2}|z{|}^{2}g.$$

$$\mathrm{div}C=\mathrm{div}(d^{D}r)=-D^{\ast }Dr-r\circ r+\mathring{R}r.$$

$$R=\frac{s}{2n(n-1)}g\text{owedge}g+\frac{1}{n-2}z\text{owedge}g+\mathcal{W},$$

$$\mathring{R}r=\mathring{\mathcal{W}}z+\frac{1}{n-2}|z{|}^{2}g+\frac{n-2}{n(n-1)}sz-\frac{2}{n-2}z\circ z+\frac{s^2}{n^2}g.$$

$${\mathrm{div}}^{2}C(X)=\frac{1}{2}⟨{\tilde{i}}_X\mathcal{W},C⟩-\frac{1}{n-2}⟨i_{X}C,z⟩$$

$${\mathrm{div}}^{2}C(X)$$

$$D_{X,Y,Z}^3\xi -D_{Y,X,Z}^3\xi =-R(X,Y)D_Z\xi +D_{R(X,Y)Z}\xi$$

$${\mathrm{div}}^{2}C(X)$$

$${\mathrm{div}}^{2}C(X)=\frac{1}{2}⟨{\tilde{i}}_{X}R,C⟩.$$

Full text

1 Introduction

An $n$-dimensional complete Riemannian manifold $(M,g)$ is said to be a static space with a perfect fluid if there exists a smooth non-trivial function $f$ on $M$ satisfying

[TABLE]

where $D_{g}df$ is the Hessian of $f$, $r_{g}$ is the Ricci tensor of $g$ with its scalar curvature $s_{g}$, and $\Delta_{g}f$ is the (negative) Laplacian of $f$. In particular, if

[TABLE]

$(M,g)$ is said to be a vacuum static space. In this case, equation (1.1) reduces to

[TABLE]

The above equation was considered by Fischer and Marsden ([8]) in their study of the surjectivity of a linearized scalar curvature functional in the space of Riemannian metrics (cf. [10], [16]). More precisely, the linearized scalar curvature $s_{g}^{\prime}$ is given by

[TABLE]

for any symmetric bilinear form $h$ on $M$ (cf. [3]). Here, $\delta=-{\rm div}$ is the (negative) divergence, which is defined by $\displaystyle{\delta h(X)=-\sum_{i=1}^{n}D_{E_{i}}h(E_{i},X)}$ for any vector $X$ and a local frame $\{E_{i}\}$. Then, the $L^{2}$-adjoint operator $s_{g}^{\prime*}$ with respect to the metric $g$ is given by

[TABLE]

for any smooth function $f$ on $M$. Thus, if a smooth function $f$ on $M$ is a solution of the vacuum static equation (1.3), then $f$ is an element of the kernel space, $\ker s_{g}^{\prime*}$, of the operator $s_{g}^{\prime*}$. By taking the divergence of (1.3), we have $\frac{1}{2}f\,ds_{g}=0$, which implies that $s_{g}$ is constant on $M$ since there is no critical point of $f$ in $f^{-1}(0)$ (see, for example, [4]). When $M$ is compact, it is known ([4]) that a compact vacuum static space is either isometric to a Ricci-flat manifold with $\ker s_{g}^{\prime*}={\mathbb{R}}\cdot 1$ and $s_{g}=0$, or the scalar curvature $s_{g}$ is a strictly positive constant and $\frac{s_{g}}{n-1}$ is an eigenvalue of the Laplacian. It turns out that the warped product manifold $(M\times_{f}{\mathbb{R}},g\pm f^{2}dt^{2})$ is Einstein when $f\in\ker s_{g}^{\prime*}$ ([7]).

Some rigidity results related to vacuum static spaces have been found. For example, Qing and Yuan showed ([17]) that, if $(M,g,f)$ is a Bach-flat vacuum static space with compact level sets of $f$, then it is either Ricci-flat, isometric to $S^{n}$, $H^{n}$, or a warped product space. Recall that the Bach tensor (see Section 2 for definition) discussed first by Bach in [2] is deeply related to general relativity and conformal geometry (cf. [13]), and in dimension $n=4$, it is well known ([3]) that the Bach tensor is conformally invariant and arises as a gradient of the total Weyl curvature functional, which is given by the integral of the square norm of the Weyl tensor. On the other hand, Kim and Shin proved a local classification of four-dimensional vacuum static spaces with harmonic curvature, or ${\rm div}\,{R}=0$ ([11]). We say that a Riemannian manifold $(M,g)$ has harmonic curvature if ${\rm div}\,R=0$, or equivalently, that the Ricci tensor $r_{g}$ is a Codazzi tensor.

Therefore, it is interesting to investigate the rigidity of vacuum static spaces under weaker curvature conditions than $B=0$ or ${\rm div}\,{R}=0$. Towards this objective, Catino, Mastrolia, and Monticelli ([6]) considered the complete divergence of the Weyl tensor to classify gradient Ricci solitons. They proved a classification theorem for gradient shrinking Ricci solitons of dimensions $n\geq 4$ under the following curvature condition:

[TABLE]

The purpose of this paper is to find rigidity results for vacuum static spaces under weaker curvature conditions. First, we consider the vanishing of complete divergence of the Bach tensor with ${\rm div}^{4}{\mathcal{W}}=0$. It is natural to inquire whether ${\rm div}^{2}B=0$ with ${\rm div}^{4}{\mathcal{W}}=0$ on vacuum static spaces guarantees Bach flatness. However, because there exist $n$-dimensional vacuum static spaces that satisfy ${\rm div}^{2}B=0$ and ${\rm div}^{4}{\mathcal{W}}=0$ with $B\neq 0$ for $n\geq 4$ (Proposition 2.5), Bach flatness is not guaranteed. In this paper, we prove that ${\rm div}^{2}B=0$ with ${\rm div}^{4}{\mathcal{W}}=0$ on an $n$-dimensional complete (non-compact) vacuum static space does imply the harmonicity of the metric for $n\geq 5$.

Theorem 1.1**.**

Let $(M^{n},g,f),n\geq 5,$ be an $n$-dimensional complete vacuum static space with compact level sets of $f$. If ${\rm div}^{2}B=0$ and ${\rm div}^{4}{\mathcal{W}}=0$, then $(M,g)$ has harmonic curvature. In particular, ${\rm div}B=0$.

Note that ${\rm div}^{2}B=0$ is not an additional condition on a four-dimensional manifold, since ${\rm div}B=0$ always holds for $n=4$ (see Proposition 2.1). Thus, we consider a rather different condition for $n=4$. By Corollary 2.2, ${\rm div}^{2}B=0$ if and only if

[TABLE]

where $C$ is the Cotton tensor and $z_{g}$ is the traceless Ricci tensor. Thus, for $n=4$, instead of the condition ${\rm div}^{2}B=0$, (1.5) is an appropriate condition. Recall that the Cotton tensor $C\in\Gamma(\Lambda^{2}M\otimes T^{*}M)$ is defined as

[TABLE]

Here, $ds_{g}$ denotes the usual total differential of $s_{g}$, and for a $1$-form $\phi$ and a symmetric $2$-tensor $\eta\in C^{\infty}(S^{2}M)$, $\phi\wedge\eta$ is defined by

[TABLE]

Theorem 1.2**.**

Let $(M,g,f)$ be a four-dimensional complete vacuum static space with compact level sets of $f$. If

[TABLE]

and ${\rm div}^{4}{\mathcal{W}}=0$, then $M$ has harmonic curvature.

When the scalar curvature vanishes, by replacing $f$ with $h$, (1.3) reduces to the static vacuum Einstein equation:

[TABLE]

Let $(M,g,h)$ be a non-trivial solution of (1.9) with $h\geq 0$, which is connected and complete up to the boundary, and assume that the set $h^{-1}(0)=\partial M$ is compact and that $g$ extends smoothly to $\partial M$. The set $h^{-1}(0)$ is called the horizon and is the boundary of black holes in general relativity. It was proved in [9] that no multiple black holes exist in static vacuum spacetime if $(M,g)$ has harmonic curvature. As a consequence of Theorem 1.1 and 1.2, we show the nonexistence of multiple black holes under ${\rm div}^{4}{\mathcal{W}}=0$ and ${\rm div}^{2}B=0$ for $n\geq 5$ and $\frac{1}{2}|C|^{2}=-\langle{\rm div}C,z_{g}\rangle$ in the case of $n=4$.

Theorem 1.3**.**

Let $(g,h)$ be a non-trivial solution (1.9) on an $n$-dimensional manifold $M$ with the compact set $h^{-1}(0)=\partial M$. Assume that ${\rm div}^{2}B=0$ for $n\geq 5$, or $\frac{1}{2}|C|^{2}=-\langle{\rm div}\,C,z_{g}\rangle$ for $n=4$. If ${\rm div}^{4}{\mathcal{W}}=0$, then no multiple black holes exist in $M$.

As another application, we prove the Besse conjecture under these conditions. It has been conjectured in [3] that a non-trivial solution $(g,f)$ of the equation

[TABLE]

on a compact manifold $(M,g)$ is isometric to a standard sphere. We have the following theorem.

Theorem 1.4**.**

Let $(g,f)$ be a non-trivial solution of (1.10) on an $n$-dimensional compact manifold $M$. Assume that ${\rm div}^{2}B=0$ for $n\geq 5$, or $\frac{1}{2}|C|^{2}=-\langle{\rm div}\,C,z_{g}\rangle$ for $n=4$. If ${\rm div}^{4}{\mathcal{W}}=0$, then the Besse conjecture holds.

It was proved in [17] that, when $n=3$, the Besse conjecture holds if ${\rm div}^{2}B=0$.

On the other hand, for a vacuum static space $(M,g,f)$ satisfying ${\rm div}^{2}B=0$, we prove that $i_{\nabla f}B=0$ if and only if $B=0$ (Corollary 5.8). Moreover, we have the following rigidity result.

Theorem 1.5**.**

Let $(M^{n},g,f)$ be an $n$-dimensional compact vacuum static space with $i_{\nabla f}B=0$. If $|z|^{2}$ attains its local maximum on the set $M\setminus f^{-1}(0)$, then $(M,g)$ is isometric to a standard sphere.

To prove our main results, we introduce a $3$-tensor $T$ defined as

[TABLE]

which follows naturally from (1.3). The complete divergence of $T$ also implies Bach flatness.

Theorem 1.6**.**

Let $n\geq 5$ and $(M,g,f)$ be a non-trivial vacuum static space of compact level sets of $f$ with ${\rm div}^{2}B=0$ for $n\geq 5$, or $\frac{1}{2}|C|^{2}=-\langle{\rm div}\,C,z_{g}\rangle$ for $n=4$. If ${\rm div}^{3}T=0$ and $f$ has its minimum (or maximum) in $M$, then $(M,g)$ is Bach-flat.

2 Preliminaries

The Bach tensor $B$ of an $n$-dimensional Riemannian manifold $(M,g)$ is defined as

[TABLE]

where ${\mathcal{W}}$ is the Weyl tensor and

[TABLE]

for an orthonormal frame $\{E_{i}\}_{i=1}^{n}$. A Riemannian metric is said to be Bach-flat if the Bach tensor $B$ of the metric vanishes. It is easy to see that a locally and conformally flat metric has a vanishing Bach tensor. For $n=4$ dimensions, it is also well-known that metrics that are locally conformal to an Einstein metric as well as half-conformally flat metrics have vanishing Bach tensors.

On the other hand, the Bach tensor $B$ is related to the Cotton tensor $C$. Note that, since

[TABLE]

under the identification of $\Gamma(T^{*}M\otimes\Lambda^{2}M)$ with $\Gamma(\Lambda^{2}M\otimes T^{*}M)$, we have

[TABLE]

and from (2.1),

[TABLE]

Here, $\mathring{\mathcal{W}}z_{g}$ is defined similarly to $\mathring{\mathcal{W}}r_{g}$. In particular, ${\rm div}^{4}{\mathcal{W}}=0$ if and only if ${\rm div}^{3}\,C=0$.

Hereafter, we denote curvatures $r_{g},z_{g},s_{g}$, and $D_{g}df$ by $r,z,s$, and $Ddf$, respectively, for convenience and simplicity, if there is no ambiguity.

The following shows that the divergence of the Bach tensor is related to the Cotton tensor and the traceless Ricci tensor $z$. The second equation (2.2) follows by taking the divergence of ${\rm div}\,B$.

Proposition 2.1** (cf. [5]).**

The divergence of the Bach tensor is given by

[TABLE]

Here, ${i}_{X}$ is the usual interior product to the first factor defined by ${i}_{X}{C}(Y,Z)={C}(X,Y,Z)$ for any vector fields $X,Y$, and $Z$.

By taking the divergence again, we have

Corollary 2.2**.**

[TABLE]

For a Riemannian $n$-manifold $(M^{n},g)$ and its traceless Ricci tensor $z$, $D^{*}Dz$ and $z\circ z$ are defined as

[TABLE]

respectively, for a local frame $\{E_{i}\}_{i=1}^{n}$ of $M$. With these notations, the divergence of the Cotton tensor, ${\rm div}\,C$, can be expressed as follows when the scalar curvature of a Riemmanian manifold is constant.

Lemma 2.3**.**

Let $(M^{n},g)$ be a Riemannian manifold with a constant scalar curvature. Then,

[TABLE]

Proof.

Since $s$ is constant, it follows from Remark 4.71 in [3] that

[TABLE]

Here, $r\circ r$ and $\mathring{R}r$ are defined similarly to $z\circ z$ and $\mathring{\mathcal{W}}r$, respectively. From the curvature decomposition

[TABLE]

we have

[TABLE]

Therefore, we can obtain (2.4) by substituting (2.6) and $r=z+\frac{s}{n}g$ into (2.5). ∎

Proposition 2.4**.**

Let $(M^{n},g)$ be a Riemannian manifold with a constant scalar curvature. Then, we have

[TABLE]

for any vector field $X$.

Proof.

Note that $C=d^{D}z$ since $s$ is constant. Thus, for a geodesic orthonormal frame $\{E_{i}\}_{i=1}^{n}$,

[TABLE]

Since

[TABLE]

for any tensor $\xi$ (cf. [3]), we have

[TABLE]

Here, $z_{ij}=z(E_{i},E_{j})$. This can be written as

[TABLE]

Next, multiplying the curvature decomposition

[TABLE]

by $C(E_{i},E_{j},E_{k})=C_{ijk}$, we have

[TABLE]

Here, we have used the fact that $\sum_{i}C(E_{i},Y,E_{i})=0$ for any $Y$. ∎

Before closing this section, we present examples of $n$-dimensional vacuum static spaces satisfying ${\rm div}^{2}B=0$ and ${\rm div}^{4}{\mathcal{W}}=0$, but are not Bach-flat for $n\geq 4$.

Proposition 2.5**.**

There exists an $n$-dimensional vacuum static space satisfying ${\rm div}^{2}B=0$ and ${\rm div}^{4}{\mathcal{W}}=0$ with $B\neq 0$ for $n\geq 4$.

Proof.

Note that, on $(M_{1}^{n},g_{1})\times(M_{2}^{m},g_{2})$ with the product metric $g=g_{1}+g_{2}$, a function $f\in\ker s_{g_{1}}^{\prime*}$ generates a function $f\in\ker s_{g}^{\prime*}$ if $(M_{2},g_{2})$ is Einstein with $\frac{s_{1}}{n-1}=\frac{s_{2}}{m}$ (see B.1 in [12]).

Thus, for dimension $2n$, where $n\geq 2$, let $M^{2n}=S^{n}(\frac{a}{kl})\times S^{n}(\frac{a}{l})$ be a product manifold with a standard product metric $g_{0}$, where $l=\frac{3n(n-1)}{2}$, $k=\frac{n}{n-1}$, $a>0$. Therefore, $(M,g_{0},f)$ is a vacuum static space with $f\in\ker s_{g}^{\prime*}$. Here, $s_{g}=\frac{2(2n-1)a}{3n}$, $r_{ii}=\frac{2(n-1)a}{3n^{2}}$ for $1\leq i\leq n$, and $r_{jj}=\frac{2a}{3n}$ for $n+1\leq j\leq 2n$. It is also easy to see that $C=0$ since $g=g_{1}+g_{2}$ is a standard product metric with Einstein metrics $g_{i}$, $i=1,2$. Thus, choosing a local frame $\{E_{1},\cdots,E_{n},E_{n+1},\cdots,E_{2n}\}$ on $M^{2n}$ such that $\{E_{1},\cdots,E_{n}\}$ and $\{E_{n+1},\cdots,E_{2n}\}$ are local frames on $S^{n}(\frac{a}{kl})$ and $S^{n}(\frac{a}{l})$, respectively, we have

[TABLE]

for $i=2,...,n$, and

[TABLE]

for $j=n+1,...,2n$. Thus, since $C=0$, by (2.3) we have

[TABLE]

Now, for dimension $2n+1$, $n\geq 2$, it is easy to see that $(S^{n}\times S^{n+1},g,f)$ with a standard product metric $g$ satisfies $f\in\ker s_{g}^{\prime*}$. Similarly, $C=0$. Thus, we have $s_{g}=2n^{2}$. Therefore,

[TABLE]

for $2\leq i\leq n$, and

[TABLE]

for $n+1\leq j\leq 2n$. Since $C=0$, we have

[TABLE]

∎

3 Harmonicity

In this section, we will give a proof of Theorem 1.1. It suffices to prove that the Cotton tensor $C$ vanishes in view of (1.6). First, by applying $d^{D}$ to both sides of (1.3), or

[TABLE]

we obtain the following result.

Lemma 3.1**.**

Let $(M,g,f)$ be a non-trivial vacuum static space. Then,

[TABLE]

Proof.

From (1.3),

[TABLE]

Note that

[TABLE]

where ${\tilde{i}}_{\nabla f}R$ is defined similarly to ${\tilde{i}}_{\nabla f}{\mathcal{W}}$. Next, from the curvature decomposition

[TABLE]

we have

[TABLE]

Substituting these into (3.3) and using the definition of $T$, our Lemma follows. ∎

Corollary 3.2**.**

On a vacuum static space, we have

[TABLE]

Proof.

Note that

[TABLE]

Thus, from Proposition 2.4 and Lemma 3.1, we have

[TABLE]

∎

For any real numbers $t$ and $t^{\prime}$, let $M_{t,t^{{}^{\prime}}}=\{x\in M\,|\,t\leq f(x)\leq t^{{}^{\prime}}\}$ and $\Gamma_{t}=\{x\in M\,|\ f(x)=t\}$.

Lemma 3.3**.**

Assume that ${\rm div}^{2}B=0$ for $n\geq 5$ or $\frac{1}{2}|C|^{2}=-\langle{\rm div}\,C,z\rangle$ for $n=4$ on a vacuum static space $M$ with compact level sets of $f$. Then, we have

[TABLE]

for regular values $t$ and $t^{{}^{\prime}}$ of $f$ with $t<t^{\prime}$.

Proof.

By Corollary 2.2 and the assumption that ${\rm div}^{2}B=0$ for $n\geq 5$, or $\frac{1}{2}|C|^{2}=-\langle{\rm div}\,C,z\rangle$ for $n=4$, we have

[TABLE]

For an orthonormal frame $\{E_{i}\}_{i=1}^{n}$, we compute

[TABLE]

Thus, from (3.5), we have

[TABLE]

on $M$. This implies that

[TABLE]

∎

Now, we are ready to prove Theorem 1.1 and 1.2, which state that the vanishing of complete divergences of the Bach tensor and Cotton tensor on a vacuum static space implies that $(M,g)$ has harmonic curvature for $n\geq 5$, if the level sets of $f$ are compact.

For regular values $t$ and $t^{\prime}$ of $f$ with $t<t^{\prime}$, from Corollary 3.2, we have

[TABLE]

Here, $N=\frac{\nabla f}{|\nabla f|}$, and we used the result of Lemma 3.3 in the last equality. Therefore, we have

[TABLE]

By taking $t^{\prime}=0$, we may conclude that $C=0$ on $\Gamma_{t}$ for all regular values $t$ of $f$ with $t<0$. Similarly, $C=0$ on $\Gamma_{t}$ for regular values of $f$ with $t>0$. Hence, we may conclude that $C=0$ on all of $M$ by continuity. In other words, $M$ has harmonic curvature. This proves our theorems.

4 Uniqueness of black holes and Besse conjecture

In this section, as applications of Theorem 1.1 and 1.2, we will prove Theorem 1.3 and 1.4. First, we prove that multiple black holes do not exist in an $n$-dimensional static vacuum spacetime under the vanishing of complete divergence of the Bach tensor and Weyl tensor (Theorem 1.3). The $n$-dimensional static vacuum Einstein equation is given by (1.9), or

[TABLE]

In particular, the scalar curvature ${s}_{g}$ of ${g}$ vanishes. In fact, a non-trivial solution $(g,h)$ of (1.9) is a vacuum static space with zero scalar curvature. Note that solutions to these equations constitute a Ricci-flat $(n+1)$-dimensional manifold $\overline{M}$ of the form $\overline{M}=M\times_{h}S^{1}$ or $\overline{M}=M\times_{h}{\mathbb{R}}$, with a Riemannian or Lorentzian metric of the form

[TABLE]

If the manifold $(M,g)$ satisfying (1.9) is geodesically complete, then $h$ is known to be a constant function ([1]). A vacuum static space $(M,g,h)$ is said to be $h$-weakly harmonic with a function $h$ if the Ricci curvature $r$ satisfies $d^{D}r(\nabla h,\cdot,\nabla h)=0$.

Theorem 4.1** ([9]).**

Assume that $(M,g,h)$ is a vacuum static space with vanishing scalar curvature and $h^{-1}(0)=\partial M$ is compact. If $(M,g,h)$ has $h$-weakly harmonic curvature, then multiple black holes do not exist in $M$.

In particular, if $M$ has harmonic curvature, then the same result as Theorem 4.1 holds. For the proof of Theorem 1.3, it suffices to prove that $C=0$ on $M$. We simply follow the proof of Theorem 1.1 and 1.2 to show that $(M,g)$ has harmonic curvature. Then, from Theorem 4.1, we may conclude that $M$ has no multiple black holes, given that $|\nabla h|^{2}$ is constant on each level set of $h$.

Secondly, we prove the Besse conjecture under the vanishing of complete divergence of the Bach tensor and Weyl tensor (Theorem 1.4). The proof is similar to the case of vacuum static spaces. Let $(g,f)$ be a non-trivial solution of (1.10), or

[TABLE]

Note that, if $C=0$, we have the following result.

Theorem 4.2** ([18]).**

Let $(g,f)$ be a non-trivial solution of (1.10) on an $n$-dimensional compact manifold $M$, $n\geq 4$. If $M$ has harmonic curvature, then the Besse conjecture holds.

Therefore, it suffices to prove that $C=0$ on all of $M$. For the proof, we apply $d^{D}$ to (1.10) to obtain

[TABLE]

where $T$ is defined as (1.11) (compare with (3.3)). Moreover, we have

[TABLE]

(compare with Corollary 3.2). If ${\rm div}^{2}B=0$, then for regular values $t$ and $t^{\prime}$ of $f$ with $t<t^{\prime}$,

[TABLE]

(compare with Lemma 3.3). Therefore, for regular values $t$ and $t^{\prime}$ of $f$ with $t<t^{\prime}$, we have

[TABLE]

By taking $t^{\prime}=-1$ with the assumption that ${\rm div}^{4}{\mathcal{w}}=0$ and ${\rm div}^{2}B=0$ for $n\geq 5$, or $\frac{1}{2}|C|^{2}=-\langle{\rm div}\,C,z\rangle$ for $n=4$, we may conclude that, for $t\neq-1$,

[TABLE]

This implies that $C=0$ on each regular level set of $f$, or on all of $M$ by continuity, since critical points of $f$ on $M$ have measure zero (see Proposition 2.2 of [18]). Consequently, we may conclude that $(M,g,f)$ is Einstein or isometric to a standard sphere.

5 Radially Bach-flat vacuum static spaces

In this section, we consider vacuum static spaces $(M,g,f)$ satisfying (1.3) with radial Bach flatness. The notion of radial Bach flatness originated from [15], in which Petersen and Wiley defined the notion of a radially flat curvature, a radially and conformally flat curvature, or radial Ricci flatness for gradient Ricci solitons. As mentioned in the introduction, Qing and Yuan ([17]) classified Bach-flat vacuum static spaces $(M,g,f)$ with compact level sets of $f$. For the proof, they introduced a $3$-tensor that is identical to $fT$ in our case, which is defined in Lemma 3.1. By showing an integral identity (Proposition 2.3), they proved that the tensor $T$ must be vanishing for Bach-flat vacuum static spaces. We can also show the same identity (Lemma 5.3), and we will give a proof for self-containedness. In view of those identities, we can see that, for a vacuum static space, the vanishing of $T$ is implied only by the vanishing of $i_{\nabla f}B$. We also prove the converse under slightly weaker conditions. That is, for a vacuum static space satisfying ${\rm div}^{2}B=0$, the condition ${\rm div}^{3}T=0$ implies $(M,g)$ is Bach-flat. In view of Proposition 2.4, we can see that the vanishing of the Cotton tensor does not imply Bach flatness nor the vanishing of the tensor $T$ for a vacuum static space, even though it satisfies ${\rm div}^{2}B=0$.

First, we present various properties of the Bach tensor $B$, tensor $T$, and their divergences for vacuum static spaces $(M,g,f)$. As an application, we will prove a rigidity for compact vacuum static spaces with radial Bach flatness by using the maximum principle.

Lemma 5.1**.**

Let $(g,f)$ be a solution of (1.3). Then,

[TABLE]

where ${\widehat{C}}$ is a $2$-tensor defined as

[TABLE]

for any vectors $X,Y$.

Proof.

Let $\{E_{i}\}$ be a local geodesic frame on $M$. From (2.2) and (3.1), together with the fact that ${\rm tr}_{14}{\mathcal{W}}=0$, we have

[TABLE]

∎

By a straightforward computation, we obtain the following.

Lemma 5.2**.**

Let $(M,g,f)$ be a non-trivial vacuum static space. Then,

[TABLE]

and

[TABLE]

Lemma 5.3**.**

Let $(M,g,f)$ be a complete vacuum static space with compact level sets of $f$. Then,

[TABLE]

for any regular values $t$ and $t^{\prime}$ with $t<t^{\prime}$.

Proof.

From Lemma 3.1 and Lemma 5.1,

[TABLE]

Therefore, from (2.3),

[TABLE]

In particular,

[TABLE]

Let $\{E_{1},\cdots,E_{n-1},N=\frac{\nabla f}{|\nabla f|}\}$ be a local frame around a regular point of $f$. Then,

[TABLE]

In the second equality, we used the fact that $T(E_{i},E_{j},\nabla f)=0$, together with ${\rm tr}_{12}T={\rm tr}_{13}T=0$ and (3.1). In the last equality, we used (5.1). Therefore, by the divergence theorem we have

[TABLE]

implying our conclusion. ∎

Next, we will show that, for a non-trivial vacuum static space $(M,g,f)$ satisfying ${\rm div}^{2}B=0$, the vanishing of $B$ is equivalent to $B(\nabla f,\nabla f)=0$. First, from the definition of $T$, we have, on each level hypersurface $f^{-1}(t)$ for a regular value $t$ of $f$,

[TABLE]

where $\alpha:=z(N,N)=z\left(\frac{\nabla f}{|\nabla f|},\frac{\nabla f}{|\nabla f|}\right)$. Note that the function $\alpha$ is defined only on the set $M\setminus{\rm Crit}(f)$, where ${\rm Crit}(f)$ is the set of all critical points of $f$. However, since $|\alpha|\leq|z|$, $\alpha$ can be extended to a $C^{0}$ function on the whole manifold $M$ (for more details, see [18]). Therefore, when $T=0$, it follows from (5.3) that

[TABLE]

for a local frame $\{E_{1},\cdots,E_{n-1},N=\frac{\nabla f}{|\nabla f|}\}$. Moreover, by substituting the triple $(\nabla f,X,\nabla f)$ into $T$ for a vector field $X$ with $\langle X,\nabla f\rangle=0$, we obtain

[TABLE]

which implies

[TABLE]

for any vector $X$ with $X\perp\nabla f$. Thus, the traceless Ricci tensor $z$ can be expressed as a diagonal matrix and

[TABLE]

We can also derive this identity from (5.2) in Lemma 5.2.

Lemma 5.4**.**

Let $(M,g,f)$ be an $n$-dimensional vacuum static space. Then,

[TABLE]

Proof.

By taking the divergence of both sides in (3.2) and using Lemma 5.1, we obtain

[TABLE]

Therefore, (5.5) follows from (2.3). ∎

Lemma 5.5**.**

We have

[TABLE]

for any vector $X$.

Proof.

First, note that

[TABLE]

Thus, by taking the divergence of both sides of (5.5) and using the fact that $\langle z,{\tilde{i}}_{X}C\rangle=0$ and ${\rm div}\,i_{\nabla f}C=-i_{\nabla f}{\rm div}C$, together with (3.1), we obtain

[TABLE]

Thus,

[TABLE]

Since from (2.3), ${\rm div}\,C$ is a symmetric $2$-tensor, by using (3.1), one can also compute

[TABLE]

By substituting these in Proposition LABEL:em2018-9-25-1 and replacing $B$ with (2.3), we obtain

[TABLE]

Finally, from Lemma 3.1 together with (5.7), we have

[TABLE]

Hence,

[TABLE]

∎

Lemma 5.6**.**

Let $(M,g,f)$ be an $n$-dimensional vacuum static space. If $T=0$, then

• (1)

$\langle i_{X}C,z\rangle=0$* for any vector $X$; therefore, ${\rm div}\,B=0$.*

• (2)

$\langle{\mathcal{W}}_{N},z\rangle=0$, where ${\mathcal{W}}_{N}$ is defined by ${\mathcal{W}}_{N}(X,Y)={\mathcal{W}}(N,X,N,Y)$ with $N=\frac{\nabla f}{|\nabla f|}$.

Proof.

From Lemma 5.5, we have $\langle i_{X}C,z\rangle=0$ for any vector $X$; therefore, from Propostion 2.1 and (3.4),

[TABLE]

In particular, since $\langle i_{\nabla f}C,z\rangle=0$, from (3.2),

[TABLE]

∎

Proposition 5.7**.**

Let $(M,g,f)$ be an $n$-dimensional vacuum static space satisfying ${\rm div}^{2}B=0$. If $T=0$, then $B=0$ and $C=0$.

Proof.

From Lemma 3.1, $T=0$ implies that

[TABLE]

Therefore,

[TABLE]

for any vectors $X$ and $Y$. Since the cyclic summation of $C$ vanishes, we have

[TABLE]

In other words,

[TABLE]

By taking the divergence of (5.8) and using Lemma LABEL:lem201_-2-10-1, we have

[TABLE]

Since ${\mathring{\mathcal{W}}}z(N,N)=\langle{\mathcal{W}}_{N},z\rangle=0$, from Lemma 5.6, we have

[TABLE]

Finally, from Corollary 2.2 and (5.4), together with $\displaystyle{\sum_{j=1}^{n-1}C_{ijj;i}=-C_{inn;i}}$, we have

[TABLE]

Thus, $C=d^{D}z=0$; therefore, $B=0$ from Lemma 5.4. ∎

Corollary 5.8**.**

Let $(M,g,f)$ be an $n$-dimensional vacuum static space satisfying ${\rm div}^{2}B=0$. Then, $B(\nabla f,\nabla f)=0$ if and only if $B=0$.

Proof.

The proof follows from Lemma 5.3 and Proposition 5.7. ∎

Next, we will show that, for a vacuum static space $(M,g,f)$ with $i_{\nabla f}B=0$, the square norm of the traceless Ricci tensor, $|z|^{2}$, attains its maximum on the set $f^{-1}(0)$. As a corollary, if $|z|^{2}$ attains its local maximum on the set $M\setminus f^{-1}(0)$, $(M,g)$ is Einstein and isometric to a standard sphere when $M$ is compact.

To show this property, we first compute the divergence of the tensor $T$.

Lemma 5.9**.**

Let $(M,g,f)$ be an $n$-dimensional vacuum static space. Then,

[TABLE]

Proof.

By using (1.2) and (3.1), we can compute

[TABLE]

From the fact that $\delta z=0$ and from (3.1), we can obtain

[TABLE]

From the definition of the tensor $T$, we have

[TABLE]

∎

We can also show the following by using (3.2).

Lemma 5.10**.**

Let $(M^{n},g,f)$ be a non-trivial vacuum static space. Then,

[TABLE]

Proof.

From (3.2) together with Lemma 2.3 and Lemma 5.1, we have

[TABLE]

∎

Comparing Lemma 5.9 with Lemma 5.10, we obtain the following.

Corollary 5.11**.**

Let $(M,g,f)$ be an $n$-dimensional vacuum static space. Then,

[TABLE]

Lemma 5.12**.**

Let $(M,g,f)$ be an $n$-dimensional vacuum static space satisfying ${\rm div}^{2}B=0$. If $i_{\nabla f}B=0$, then the function $|z|^{2}$ attains its maximum on the set $f^{-1}(0)$.

Proof.

From Lemma 5.3 and Proposition 5.7, $T=0=C$. Therefore,

[TABLE]

from Corollary 5.11. Since $\displaystyle{\frac{1}{2}\Delta|z|^{2}=-\langle D^{*}Dz,z\rangle+|Dz|^{2}}$, we have

[TABLE]

This shows that, on the set $f^{-1}(0)$,

[TABLE]

Now, on the set $\{f\geq\epsilon\}$, for a sufficiently small positive real number $\epsilon>0$, (5.10) can be expressed as

[TABLE]

By applying the maximum principle, we obtain

[TABLE]

By letting $\epsilon\to 0$, we finally obtain

[TABLE]

Similarly, by applying the maximum principle on the set $\{f\leq-\epsilon\}$, we have

[TABLE]

Hence we prove our Lemma. ∎

Now, we are ready to prove Theorem 1.5.

From Lemma 5.12, $|z|^{2}$ should be constant on $M$; therefore, $\langle D^{*}Dz,z\rangle=|Dz|^{2}$. From Lemma 5.10, we have

[TABLE]

Fix a point $p\in M$ and choose a local frame that diagonalizes $z$ at $p$. Then, (5.11) becomes

[TABLE]

at the point $p$, which implies that $|z|(p)=0$. Since the point $p$ is arbitrary, we can conclude that $z=0$ on $M$. Finally, from (3.1),

[TABLE]

which implies that $(M,g)$ is isometric to a standard sphere, as shown by Obata ([14]).

Finally, we will show some rigidity results for vacuum static spaces with complete divergence of the tensor $T$. That is, we prove that, if $(M^{n},g,f)$ is a non-trivial vacuum static space with compact level sets of $f$ and if ${\rm div}^{2}B=0$ and ${\rm div}^{3}T=0$, then $T=0$. Therefore, $(M,g)$ must be Bach-flat.

To show this, we first need the following property on the complete divergence of $T$.

Lemma 5.13**.**

[TABLE]

Proof.

By substituting the equation in Proposition 2.1 into (5.6), we obtain

[TABLE]

By taking the divergence of this equation again, we have

[TABLE]

Our Lemma follows from Proposition 2.1 and (3.4). ∎

Let $M_{t}=\{x\in M\,|\,f(x)<t\}$.

Theorem 5.14**.**

Let $n\geq 5$ and $(M,g,f)$ be an $n$-dimensional vacuum static space of compact level sets of $f$ with ${\rm div}^{2}B=0$. If ${\rm div}^{3}T=0$ and $f$ has its minimum (or maximum) in $M$, then $(M,g)$ is Bach-flat.

Proof.

From Lemma 5.13,

[TABLE]

Since

[TABLE]

we have

[TABLE]

Therefore, from (5.12) and (5.13), we have

[TABLE]

On the other hand, it is easy to see that

[TABLE]

Note that, from Proposition 2.1 and (3.4), we have

[TABLE]

Since, from Lemma 5.2,

[TABLE]

we have

[TABLE]

As a result, we may conclude that $T=0$ on $\Gamma_{t}$ for all $t$, which implies that $T=0$ on all of $M$. Therefore, the proof follows from Proposition 5.7. ∎

Theorem 5.15**.**

Let $(M,g,f)$ be a four-dimensional non-trivial complete vacuum static space with compact level sets of $f$. If

[TABLE]

${\rm div}^{3}\,T=0$, and $f$ has its minimum (or maximum) in $M$, then $(M,g)$ is Bach-flat.

Proof.

Recall that, for $n=4$, we always have ${\rm div}B=0$. Thus, similarly to Lemma 5.5, we have

[TABLE]

and

[TABLE]

Thus, if ${\rm div}^{3}T=0$,

[TABLE]

Next, as in the proof of Lemma 3.3, we have

[TABLE]

when $n=4$ under the condition $\frac{1}{2}|C|^{2}=-\langle{\rm div}C,z\rangle$. Therefore, from the co-area formula, we have

[TABLE]

where $a=\min_{M}f$. Finally, as in the proof of Theorem 5.14, we have

[TABLE]

Since

[TABLE]

we may conclude that $T$ vanishes identically on $M$.

∎

By combining Theorem 5.14 and 5.15, Theorem 1.6 follows.