Timeline for de Rahm Laplace operator on forms bounded
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 26, 2014 at 13:19 | answer | added | Matthias Ludewig | timeline score: 3 | |
| May 26, 2014 at 6:58 | comment | added | Willie Wong | "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. | |
| May 24, 2014 at 6:07 | comment | added | Luigi | By Hellinger-Toeplitz theorem it should be bounded. Isn't this right? | |
| May 24, 2014 at 4:26 | comment | added | Luigi | How can this be seen? Do you have any reference (textbook, paper, etc....)? | |
| May 23, 2014 at 21:16 | comment | added | José Figueroa-O'Farrill | It does if $M$ is compact. | |
| May 23, 2014 at 16:17 | comment | added | Luigi | Does this laplacian have a compact resolvent? | |
| May 23, 2014 at 16:05 | comment | added | Nate Eldredge | 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. | |
| May 23, 2014 at 15:54 | review | First posts | |||
| May 23, 2014 at 16:12 | |||||
| May 23, 2014 at 15:37 | history | asked | Luigi | CC BY-SA 3.0 |