# why is the spectrum of the schrödinger operator discrete?

let (M,g) be a compact riemannian manifold. Then the spectrum of the Schrödinger opartor $H=-\Delta +V$ 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.

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Here you have something related, hope it helps. –  Giuseppe Negro Feb 12 '13 at 14:53
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 $\mathbb R^n$.
Because it is continuous and $\rm M$ is compact ? –  Damien L Feb 13 '13 at 10:11