## Pohozaev-Schoen identity

In this short note I record a proof of the Pohozaev-Schoen identity (see this paper of Schoen, Prop 1.4). This proof is actually standard and I put it here for my own benefit. If I have time I will give some applications of this identity later.

 Theorem 1 (Pohozaev-Schoen identity) Let ${(M^n,g)}$ ${(n> 2)}$ be a compact Riemannian manifold (possibly with boundary) and ${X}$ is a vector field on ${M}$, then $\displaystyle \int_M X(R)= -\frac{n}{n-2}\int_M \langle \stackrel\circ {\mathrm{Ric}},\mathcal {L}_X g\rangle+\frac{2n}{n-2} \int_{\partial M} \stackrel \circ {\mathrm{Ric}}(X, \nu).$ where ${R}$ is the scalar curvature, ${\stackrel\circ {\mathrm{Ric}}}$ is the traceless Ricci tensor and $\mathcal L_X g$ is the Lie derivative of $g$.

Proof:

We will calculate using a local orthnormal frame. Let ${\stackrel\circ{\mathrm {Ric}}= \mathrm{Ric}-\frac{R}{n}g}$ be the traceless Ricci tensor. By the (twice contracted) second Bianchi identity $R_{ij;j}=\frac 12 R_{;i}$, we have

$\displaystyle \begin{array}{rcl} \stackrel\circ R_{ij;j}= R_{ij;j}-\frac{1}{n}R_{;i}= \frac{1}{2}R_{;i}-\frac{1}{n}R_{;i}=\frac{n-2}{2n}R_{;i}. \end{array}$

Therefore by integration by parts,

$\displaystyle \begin{array}{rcl} \int_M X(R ) =\int_M X_i R_{;i}= \frac{2n}{n-2}\int_M \stackrel\circ R_{ij;j} X_i &=\frac{2n}{n-2}\left(-\int_M \stackrel\circ R_{ij}X_{i;j} + \int_{\partial M} \stackrel \circ R_{ij}X_i \nu_j\right). \end{array}$

As ${ \stackrel \circ R_{ij}}$ is a symmetric tensor, we have

$\displaystyle \stackrel \circ R_{ij}X_{i;j}= \frac{1}{2} \stackrel\circ R_{ij}(X_{i;j}+X_{j;i})= \frac{1}{2}\stackrel\circ R_{ij}(\mathcal L_X g)_{ij}.$

(Here we have used the fact that $(\mathcal L_X g)(Y,Z)=\langle \nabla_Y X, Z\rangle + \langle \nabla _Z X, Y\rangle$ (Exercise). ) Putting this into the above equation, we have

$\displaystyle \int_M X(R)= -\frac{n}{n-2}\int_M \langle \stackrel\circ {\mathrm{Ric}},\mathcal {L}_X g\rangle+\frac{2n}{n-2} \int_{\partial M} \stackrel \circ {\mathrm{Ric}}(X, \nu).$

$\Box$

 Corollary 2 If ${X}$ is a conformal vector field (see also here), i.e. ${\mathcal L_X g= f g}$ for some function ${f}$ on ${M}$, then $\displaystyle \int_M X(R)= \frac{2n}{n-2} \int_{\partial M} \stackrel \circ {\mathrm{Ric}}(X, \nu).$ In particular, ${\int_M X(R)=0}$ if ${M}$ is closed.

 Remark 1 Theorem 1 can be used to prove the DeLellis-Topping inequality.