Timeline for Relation between the de Rham and Hodge Laplacians on the Exterior Algebra
Current License: CC BY-SA 2.5
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 24, 2011 at 18:33 | vote | accept | Jean Delinez | ||
| Mar 24, 2011 at 18:31 | comment | added | Jean Delinez | Sorry, about that, superscripts in the wrong place. Yes, it still works out, but best to stick to convention. | |
| Mar 24, 2011 at 18:29 | history | edited | Jean Delinez | CC BY-SA 2.5 |
added 4 characters in body
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| Mar 24, 2011 at 18:28 | comment | added | diverietti | He does want superscripts, since what he calls Laplacian is not... He is writing the Dirac operator instead... | |
| Mar 24, 2011 at 18:24 | comment | added | José Figueroa-O'Farrill | (I don't think you want the superscript $2$ in your displayed formula.) On a Kähler manifold, the three laplacians associated with $d$, $\partial$ and $\bar\partial$ satisfy $$\Delta_d = 2 \Delta_{\partial} = 2 \Delta_{\bar\partial}$$ not just on functions, but also on forms. This is why on a compact Kähler manifold, one has the decomposition of the de Rham cohomology in terms of the Dolbeault cohomology groups. This can be found, e.g., in Well's Differential analysis on complex manifolds. | |
| Mar 24, 2011 at 18:22 | answer | added | diverietti | timeline score: 12 | |
| Mar 24, 2011 at 18:20 | comment | added | Donu Arapura | Yes, the identity holds for differential forms. This is what's behind the Hodge decomposition. See your favourite book on Kahler manifolds (Griffiths-Harris, Wells...) for an explanation. | |
| Mar 24, 2011 at 18:03 | history | asked | Jean Delinez | CC BY-SA 2.5 |