Timeline for Hodge decomposition of smooth n-forms: is it an isomorphism of topological vector spaces?
Current License: CC BY-SA 4.0
18 events
| when toggle format | what | by | license | comment | |
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| May 15 at 4:01 | comment | added | Pedro Lauridsen Ribeiro | Thanks, I do appreciate that. Glad you got nice answers from Stefan and Tobias. | |
| May 14 at 23:49 | vote | accept | David Roberts♦ | ||
| May 14 at 23:49 | comment | added | David Roberts♦ | @PedroLauridsenRibeiro sorry to hear that, and thanks for your early input on this! it gave me hope. I held off accepting an answer in case you still wanted to write one, but there you go ^_^ | |
| May 14 at 22:13 | comment | added | Pedro Lauridsen Ribeiro | Well... My computer decided to crash today. I've managed to fix it just now, but then Stefan and Tobias had already beat me to it. Please feel free to acknowledge whatever answer you feel more appropriate. | |
| May 14 at 10:27 | history | became hot network question | |||
| May 14 at 9:20 | answer | added | Stefan Waldmann | timeline score: 16 | |
| May 14 at 8:12 | answer | added | Tobias Diez | timeline score: 11 | |
| May 14 at 6:45 | comment | added | Pedro Lauridsen Ribeiro | Diferential operators of positive order and smooth coefficients are indeed unbounded with respect to the $L^2$ (Hilbert space) norm, but are certainly bounded with respect to the $C^\infty$ (Fréchet space) topology as given by the $L^2$ Sobolev norms of all non-negative orders. I'll compile the answer by tomorrow at the latest. | |
| May 14 at 5:45 | comment | added | David Roberts♦ | @PedroLauridsenRibeiro thanks! Can you compile those comments into an answer? I was doubting the $C^\infty$-continuity of $d$ because of something I'd seen saying differential operators are unbounded. | |
| May 14 at 4:53 | comment | added | Pedro Lauridsen Ribeiro | The remarks in my last comment above put together should yield that the harmonic decomposition is a topological isomorphism of Fréchet spaces, even though I don't really think Warner addresses this question explicitly. | |
| May 14 at 4:50 | comment | added | Pedro Lauridsen Ribeiro | @DavidRoberts Also, $L^2$ orthogonality works because continuity of $d$, $\delta$ and $\Delta$ as well as elliptic regularity of $\Delta$ are expressed in terms of $L^2$ Sobolev norms on $\Omega^n(M)$. These norms also generate the Fréchet topology of $\Omega^n(M)$ thanks to the Sobolev imbedding theorem. Finally, recall that the direct sum of the last two direct summands of the harmonic decomposition equals the image of $\Delta$ in $\Omega^n(M)$ - in other words, the equation $\Delta\omega=\alpha$ has a solution $\omega\in\Omega^n(M)$ iff $\alpha$ is $L^2$ orthogonal to $\mathcal{H}^n(M)$. | |
| May 14 at 4:41 | comment | added | Pedro Lauridsen Ribeiro | Regarding your latter points, recall that every de Rham cohomology class on $M$ has a unique (smooth) harmonic representative, so the dimension of the space of smooth harmonic $n$-forms on $M$ (which is finite by the Hodge theorem, btw) equals the $n$-th Betti number of $M$ and thus is a topological invariant. Since the direct summand $d\Omega^{n-1}(M)$ doesn't change from one harmonic decomposition to another, from that it should be straightforward to obtain an isomorphism between different harmonic decompositions. | |
| May 14 at 4:20 | comment | added | David Roberts♦ | @PedroLauridsenRibeiro does it address continuity of the decomposition in the Fréchet $C^\infty$-topology? | |
| May 14 at 4:03 | comment | added | Pedro Lauridsen Ribeiro | The whole last chapter of F. W. Warner, Foundations of Differentiable Manifolds and Lie Groups (Springer-Verlag, 1983) is dedicated to the proof of the Hodge theorem, including all the necessary prerequisites. It doesn't use pseudodifferential machinery to do so. | |
| May 14 at 2:35 | history | edited | David Roberts♦ | CC BY-SA 4.0 |
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| May 13 at 22:59 | comment | added | David Roberts♦ | Thanks! Do you have a detailed reference of this? | |
| May 13 at 21:47 | comment | added | Branimir Ćaćić | Quick comment: any account proving Hodge decomposition via elliptic regularity of the Laplace-Beltrami operator $\Delta := (d+\delta)^2 = d\delta+\delta d$ (and related theory of pseudodifferential operators) will necessarily cover pretty much all points in your first paragraph. | |
| May 13 at 11:49 | history | asked | David Roberts♦ | CC BY-SA 4.0 |