Show that $\int_M\Delta f HdV=-\int_M\nabla f\cdot\nabla H dV $?

Question

$M$ is a compact Riemannian manifold,$f,H$ are functions on $M$,How to show that $\int_M\Delta f HdV=-\int_M\nabla f\cdot\nabla H dV $?Besides,if $M$ is not compact, is the equality right?

Beside, I want to know that, if $F,H$ are field on Riemannian manifold(example vector field), is $\int_M \nabla F\cdot HdV=-\int_M F\cdot \nabla HdV$ right?

In fact, I am fuzzy with there equality.I think $\cdot$ is Riemannian inner and no matter $f$ is function or field , $\nabla f$ should be treated as operator not gradient of $f$.But,in analysis, there are likely equality,treat them as gradient. So, I think they are same no matter treat it as gradient or operator.But I fail to understand the same.

Sorry for my poor English,I don't know whether precisely I describe my question .

Answer 1

Your statement follows from Stokes's theorem $\int \limits _{\partial M} \omega = \int \limits _M \Bbb d \omega$ and the fact that $M$ has no boundary (i.e. $\partial M = \emptyset$, so the left-hand side is $0$).

Note that $(\Delta f) H \ \Bbb d V = (\text{div} \ \nabla f) H \ \Bbb d V = \text{div} (H \nabla f) \ \Bbb d V - (\nabla f \cdot \nabla H) \ \Bbb d V$. If you define $\omega = i_{H \nabla f} (\Bbb d V)$ (the inner product of a field and a form or, if you prefer, the Lie derivative of $\Bbb d V$ along the field $H \nabla f$), then $\omega$ will be a $n-1$-form on $\partial M$ and it can be shown that $\Bbb d \omega = \text{div} (H \nabla f) \ \Bbb d V$. You have thus obtained that $\Bbb d \omega = (\Delta f) H \ \Bbb d V + (\nabla f \cdot \nabla H) \ \Bbb d V$. Plugging this into Stokes's theorem as described above, the integral of $\Bbb d \omega$ will be $0$, whence follows your formula.

If $M$ is not compact but either $H$ or $f$ has compact support, then the equality remains true. It may fail in the absence of compact support, though.

Answer 2

Not sure how the gradient of a specific function $f$ can be interpreted as an operator.

The integral of the divergence of a vector field on a compact manifold is zero. Apply this to the vector field $H\nabla f$ (scalar times vector).

But the divergence of $H\nabla f$ is equal to the inner product of $\nabla H$ with $\nabla f$ plus the product (as scalar fields) of $H$ with $\Delta f$.

On a noncompact manifold you need extra tricks to deal with the integral of the divergence of $H\nabla f$. For instance, on a manifold with boundary there is a boundary term involving the integral of the inner product of $H\nabla f$ with the normal to the boundary. Or you could add a hypothesis about $H$, $\nabla f$ and their derivatives becoming arbitrarily small outside compact subsets of the manifold.