I am studying the proof of a theorem and I am stuck. It says that by integration by parts we get that:
For $g(t)$ a variation of Riemannian metrics wih $g'(0)=h,$
$$\int_{M} (-\Delta (tr h) + \delta^2 h)\;dV_g=0$$
where $\Delta$ is the rough Laplacian, and $\delta^2 h=\nabla^i \nabla^j h_{ij}$ is the double-divergence operator.
I don't understand how integration by parts imply that this integral is equal to zero.
Any help is appreciated!
Both of them are consequence of divergnece theorem: for any vector fields $X$ on $M$,
$$\int_M \text{div}(X) dV_g = 0. $$
(div is just $\delta$ in your question) The first term is zero as for all functions $f$,
$$\int_M \Delta f dV_g = \int_M \text{div} (\nabla f) dV_g = 0$$
while the second one, choose the vector fields to be
$$X_i = \nabla ^j h_{ij} $$
(Doing $F$/ $W$- Functional ^^?)