For two vector fields $X$ and $Y$ on a Riemannian manifold $M$ with metric $g$, we define
$$\langle X, Y \rangle_{L^2} = \int_M g_{ij}X^iY^j dV.$$
I have been unable to find a similar expression for two functions $f,g \in C_0(M)$. I am guessing that this will just be
$$\langle f, g \rangle = \int_M fgdV \tag{1}$$
Why do we not have to take the complex conjugate as in Euclidean space?
But if this definition is correct, consider the heat equation:
$$\partial_t u – \Delta u = 0$$
so that $u: M \rightarrow \mathbb{R}$. In these set of notes they state
$$\langle \Delta u, u\rangle_{L^2} = \int_M (\Delta u)u d\mu = -\int_M |\nabla u|^2 d\mu \leq 0 \tag{2}$$
where they take $d\mu$ to be the Riemannian measure.
Using (1), (2) is obtained from integrating by parts as one usually does on Euclidean space with $\mu$ the Lebesgue measure. Does this just mean we can always treat $dV$ (or $d\mu$ in the notation of the notes) to be some multiple of the Lebesgue measure? If so, what justifies this? If not, what justifies the integration by parts formula as in (2)?
Lastly, I have seen the integration by parts formula on a Riemannian manifold (for example, here) given as
$$\int_M \langle \nabla f, X \rangle = -\int_M f div X + \int_{\partial M} f \langle X, \nu \rangle.$$
But what is the relationship between this formula and the integration by parts of scalar functions as in (2)?
Best Answer
Because we’re dealing with real-valued stuff.
This makes no sense. There is no “Lebesgue measure” on $M$ in the sense of a non-zero translation-invariant measure (translations don’t even make sense in general). All we have is the Riemannian measure (we can call it the Riemann-Lebesgue measure if we wish, since its a measure arising in Riemannian geometry, and it’s defined on a $\sigma$-algebra on $M$ which is analogous to the Lebesgue $\sigma$-algebra on open sets of some $\Bbb{R}^n$, and in any chart domain, this measure pushes forward to a measure on an open subset of $\Bbb{R}^n$ which is absolutely continuous relative to the usual Lebesgue measure).
I think you’re forgetting the obvious fact that Laplacians $\Delta u$ are by definition the divergence of the gradient vector field. So, (2) really is a special case of the final quoted formula (please don’t memorize that… just derive it on the spot so you always know why it’s true). Integration by parts in 1D is really just a matter of doing a product rule in reverse, and applying the fundamental theorem of calculus. In higher dimensions, it is a matter of doing a variant of a product rule in reverse, and applying Stokes’ theorem or divergence theorem. For a slight generalization of your situation (also, writing $\nabla u$ to mean the vector field $\text{grad}_g(u)$): \begin{align} (\Delta u) v=(\text{div } \nabla u)v=\text{div}(v\,\nabla(u))-\langle\nabla v,\nabla u\rangle_g \end{align} Now, integrate both sides over $M$ using the measure $d\mu$. Now suppose either $M$ is compact and boundaryless, or that $u$ or $v$ decays sufficiently rapidly or some other assumption which allows you to invoke the divergence theorem and discard the resulting boundary term. This gives \begin{align} \int_M(\Delta u)v\,d\mu&=0-\int_{M}\langle\nabla v,\nabla u\rangle_g\,d\mu. \end{align} If you now take $v=u$, then you get \begin{align} \int_M(\Delta u)u\,d\mu&=-\int_M\|\nabla u\|_g^2\,d\mu\leq 0. \end{align}
Why did I redo the derivation? To emphasize that you shouldn’t be memorizing these things and that you should just do the integration-by-parts by-hand. Anyway, if you want to apply formulae, then take $f=u$ and $X=\nabla u$, and impose extra assumptions on $u$ or $M$ to discard the boundary term.