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....)?


  • 3
    $\begingroup$ 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. $\endgroup$ May 23, 2014 at 16:05
  • $\begingroup$ Does this laplacian have a compact resolvent? $\endgroup$
    – Luigi
    May 23, 2014 at 16:17
  • $\begingroup$ It does if $M$ is compact. $\endgroup$ May 23, 2014 at 21:16
  • $\begingroup$ How can this be seen? Do you have any reference (textbook, paper, etc....)? $\endgroup$
    – Luigi
    May 24, 2014 at 4:26
  • 1
    $\begingroup$ "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. $\endgroup$ May 26, 2014 at 6:58

1 Answer 1


Differential operators are never bounded unless they are of order zero.

Standard references are Berline, Getzler, Vergne, "Heat Kernels and Dirac Operators" and Gilkey, "Invariance Theory, The Heat Equation and the Atiyah-Singer Index Theorem".

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.


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