Is the Laplacian $-\Delta$ on a compact manifold an isomorphism?

Question

We know that for (a normal) domain $-\Delta:H^1_0(\Omega) \to H^{-1}(\Omega)$ is an isomorphism.

What is the corresponding result for the Laplace-Bulltrami operator or more generally a Laplacian operator on a manifold which has no boundary? What properties does it satisfy?

Answer 1

It holds also on sufficiently smooth Riemannian manifolds, and in fact for more general partial differential operators (even pseudo-differential operators.)

See for example Hebey, Nonlinear Analysis on Manifolds: Sobolev Spaces and Inequalities.

Answer 2

Once I had similar question. If you are given function $g$ on manifold $M$ which integral onver $M$ is nonzero. Than there is no solution of $\delta d f = g$. You can do this by contradiction. Assume that you have such a solution than: $$ 0\neq \int_M \star g = \int_M d{\star}df = \int_{\partial M} \star df = 0 $$ because $\partial M = 0$. This answer is from Ted Shifrin.

Actually there is nice physical interpretation for this result.

If you have heat equation with sources $\frac{d u}{dt}= \delta d u - g$ then its solution in infinite time tends to solution of $\delta d u = g$. But if integral of $g$ over $M$ is positive, than you pump energy to the system. So in infinite time you would pump infinite amount of energy to the system and there is "not enough space" on compact manifold to store that much energy or you do not have any boundary to consume that energy.