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.  


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).


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.  

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