# de Rahm Laplace operator on forms bounded

Let $M$ be a closed differentiable manifold. Let $E^{p}(M)$ be the vector space of $p$-forms on $M$ equipped with the $L^{2}$-inner product $(\alpha, \beta) = \int_{M}\alpha \wedge \star \beta$. The Laplace de Rahm operator is then defined by $\Delta = d^{*}d + dd^{*} : E^{p}(M) \rightarrow E^{p}(M)$. All these definitions are as in the ordinary Hodge theory. Is the operator $\Delta : E^{p}(M) \rightarrow E^{p}(M)$ bounded with respect to the $L^{2}$-inner product? I have not found any literature which treats these kind of questions. Is there any reference concerning functional analytical properties of this operator (spectrum etc....)?

Luigi

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No, differential operators are hardly ever bounded on $L^2$. For an explicit counterexample you could take $p=0$ so that $\Delta$ is the ordinary Laplacian. It's easy to find a function $f$ with $\|f\|_2$ small but $\|\Delta f\|_2$ large. –  Nate Eldredge May 23 at 16:05
Does this laplacian have a compact resolvent? –  Luigi May 23 at 16:17
It does if $M$ is compact. –  José Figueroa-O'Farrill May 23 at 21:16
How can this be seen? Do you have any reference (textbook, paper, etc....)? –  Luigi May 24 at 4:26
"By Hellinger-Toeplitz it should..." What is "it"? If you mean the Hodge Laplacian, it is not everywhere defined on $L^2$. It is everywhere defined on $E^p$ if you mean it to be the vector space of, say, smooth $p$-forms, but then $E^p$ with the $L^2$ product is not complete, and so not a Hilbert space. –  Willie Wong May 26 at 6:58

Your operator is an example of a general Laplace-type operator, which is a differential operator $L$ acting on a vector bundle $\mathcal{V}$ whose principal symbol is given by $$\sigma(L, \xi) = - |\xi|^2 \cdot \mathrm{id},$$ where $\mathrm{id}$ denotes the identity endomorphism of your bundle.
All these operators (if they are symmetric on smooth sections with respect to the $L^2$ inner product) have in common that they are unbounded on $L^2(M, \mathcal{V})$ and self-adjoint on the domain $H^2(M, \mathcal{V})$ (the Sobolev space of order $2$), given that your manifold is compact. Furthermore, they have a compact resolvent.
If the manifold is non-compact, the resolvent will usually not be compact, but in good cases (there are curvature assumption to be made, if I am not mistaken, and assumptions on the potential if you have one), the operator is still self-adjoint on $H^2(M, \mathcal{V})$ and has spectrum bounded from below.