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I am currently writing a paper about the Hodge theorem for an algebraic topology course. The specific formulation I am proving can be stated thus. Let $M$ be a compact, orientable n-dimensional Riemannian manifold. We use $\Delta^kM$ to denote the vector space of elliptic differential $k$-forms, i.e the differential $k$-forms $\alpha \in \Omega^kM$ such that $\Delta \alpha = 0$ where $\Delta = dd^* + d^*d$ is the Laplace Beltrami operator. We use $H^kM$ to denote the $k$-th de Rham cohomology of $M$. The Hodge thoerem states that: $$ \Delta^kM \simeq H^kM $$

A big part of the proof of this theorem is showing that the following equality holds: $$ \Omega^kM = \Delta^kM \oplus \Delta(\Omega^kM) $$

This equality in follows from the following facts;

a) $\Omega^kM$ sits inside some Sobolev space $H$ where $\Delta$ is well defined.

b) Since $\Delta$ is a linear self-adjoint operator on $H$, we have that: $$H = Ker(\Delta) \oplus Im(\Delta^*) = Ker(\Delta) \oplus Im(\Delta)$$ This means that any $\alpha \in \Omega^kM$ can be expressed as $\alpha = \beta + \Delta\gamma$ where $\beta, \gamma \in H$.

c) Now, $\beta$ is a harmonic form and it is not difficult to show that this implies that it is closed (i.e that $d\beta = 0$). This implies that it is smooth, so $\beta \in \Omega^kM$. Thus, $\Delta\gamma$ must be smooth. By elliptic regularity, this implies that $\gamma$ is smooth, leading us to the result that $\gamma \in \Omega^kM$ and thus that $$ \Omega^kM = \Delta^kM \oplus \Delta(\Omega^kM) $$

I have done a fair bit of snooping, but I have not found a clear, short explanation as to what the Sobolev space $H$ looks like and how it constructed. So I guess the best formulation of my question would be:

How does one construct a Sobolev space $H$ containing the space of differential k-forms $\Omega^kM$ of a closed Riemannian manifold $M$ such that the differential Laplace-Beltrami operator $\Delta$ is well-defined?

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Um it would not be easier. They're all very busy doing research, and I've already asked Melrose and my topology professor for some clarification. I am definitely speaking with others though. Please keep in mind that my grasp of this material is incomplete, the course I'm taking is at the undergraduate level and somewhat unrelated to this stuff. Would you briefly point out precisely where I went wrong above so that I can understand the material properly? Thanks! –  Julian Chaidez May 16 '13 at 23:45
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Maybe Section 2.1.4 of the notes below might help www3.nd.edu/~lnicolae/ind-thm.pdf –  Liviu Nicolaescu May 16 '13 at 23:59
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Actually, the only detail that you didn't get right in the summary is that the Laplacian is not an operator from $H$ to $H$, where $H$ is a fixed Sobolev space. The Sobolev space you want to work with is $W^{2,k}$, where the norm of $\omega$ is the sum of the $L^2$ norms of the derivatives of $\omega$ up to order $k$. Then the Laplacian is a bounded linear map from $W^{2,k}$ to $W^{2,k-2}$. You construct the Sobolev space as the completion of smooth forms on $M$. –  Deane Yang May 17 '13 at 2:12
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And you shouldn't be shy about interrupting these busy professors. No matter how much research they're doing, they have a responsibility to teach their students and answer questions. You can tell them I said so. –  Deane Yang May 17 '13 at 2:14
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There is a very nice book where this is (somewhat unexpectedly) REALLY nicely done: Schrödinger Operators: With Applications to Quantum Mechanics and Global Geometry by Barry Simon et al. As a bonus, they have a good exposition of the relation between Hodge theory and Morse theory that was first uncovered by Witten. –  alvarezpaiva May 17 '13 at 7:36

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