why is the spectrum of the schrödinger operator discrete?

Question

let (M,g) be a compact riemannian manifold. Then the spectrum of the Schrödinger operator $H=-\Delta +V$ with bounded potential V acting on $L^2(M)$ consists of discrete Eigenvalues $\lambda_1\leq\lambda_2\leq... $.

I ask myself, why this statement is true?

Most books are dealing with the case $M=\mathbb{R}^n$. One elementary result is (in the case of $L^2(\mathbb{R}^n)$), that if V is bounded from below and $V(x)\rightarrow \infty$ for $\|x\|\rightarrow \infty$ then the resolvent is compact.

But in my case I have no $\|x\|\rightarrow \infty$ ? I hope you can help me.

Answer 1

I don't know the result in the case of $\mathrm L^2(\mathbb R^n)$, but the condition you describe : $\mathrm V(x) \to \infty$ for $||x|| \to \infty$ is exactly saying that $\mathrm V$ is a proper map (i.e the inverse image of any compact is a compact). A continuous map $\mathrm V$ on a compact manifold is automatically proper and maybe it is sufficient to adapt the theorem you state for $\mathbb R^n$.

Answer 2

It is because of the Rellich lemma: $H^1$ is compactly embedded into $L^2$ on compact manifolds.