Timeline for Global Lichnerowicz Formula Proof (in the Kahler case)?
Current License: CC BY-SA 3.0
15 events
| when toggle format | what | by | license | comment | |
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| Mar 6, 2012 at 11:58 | comment | added | Deane Yang | It seems to me that you can't do this unless you have a global definition of scalar curvature. What's that? | |
| Mar 6, 2012 at 11:28 | answer | added | Alex_K | timeline score: 2 | |
| Mar 5, 2012 at 0:49 | comment | added | Otis Chodosh | @Paul, Gromov has something to say about that in ihes.fr/~gromov/PDF/positivecurvature.pdf, at least for the untwisted case. He makes the case that (1) scalar curvature is the only "natural" 0-th order operator on a Riem. mfld (he does this right in the beginning), and then on p 28 that the difference must be 0-th order, so it should not be surprising. I don't know if this is convincing at all, but it certainly is a cool point of view imo. | |
| Mar 4, 2012 at 17:33 | comment | added | Paul Siegel | If $S$ is any Clifford bundle then the Dirac operator is the composition $C^\infty(M,S) \to C^\infty(M, T^*M \otimes S) \to C^\infty(M, T M \otimes S) \to C^\infty(M,S)$ where the first map is the connection, the second is the metric, and the third is Clifford multiplication. Since $D$ is self-adjoint and only the connection differentiates anything, there should be an easy symbol argument which shows that $D^2 - \nabla^* \nabla$ is $0$th order. The mystery to me is why the difference is the twisting curvature of $\nabla$ plus the scalar curvature. | |
| Mar 4, 2012 at 11:27 | comment | added | Jean Delinez | @Tim: Yes, of course. It's adjusted now. | |
| Mar 4, 2012 at 11:26 | history | edited | Jean Delinez | CC BY-SA 3.0 |
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| Mar 3, 2012 at 22:42 | comment | added | Tim Perutz | Jean, the formula you give is for the spin Dirac operator on a spin manifold. However, the Dolbeault operator is an example of a spin-c Dirac operator $D_A$, associated with a connection $A$ in the determinant line bundle - in this case, the canonical line bundle for the complex manifold. The Lichnerowicz formula for such operators has another term, a multiple of $F_A$, viewed through Clifford multiplication as an endomorphism of the spinor bundle. | |
| Mar 3, 2012 at 20:22 | comment | added | shu | By the symbole argument and self adjointness, $D^2-\nabla^*\nabla$ is a 0 order operator on your manifold. The thing rest is to understand the curvature term. It is rather subtile. If you compare the diffusion of tow side, the thing is more or less like passing the Ito calculs to Stratonovich calculs. The later is coordinate free. | |
| Mar 3, 2012 at 20:08 | comment | added | shu | You can proof it by lifting in principal bundle which does not need a local coordinate system. But as Paul Siegel pointed out, there ...may not be a conceptual argument. | |
| Mar 3, 2012 at 18:50 | comment | added | Jean Delinez | No this is a global statement - $D$, $\nabla$, and $Sc$ are all module maps from $\Omega^\bullet$ to itself. There is no choice here of a local coordinate system, and I would like a proof that likewise makes no choice of local coordinate system. | |
| Mar 3, 2012 at 18:46 | comment | added | Deane Yang | The operator $\nabla^*\nabla$ is a self-adjoint positive elliptic operator operator, so its inverse is smoothing. Those are rather strong and useful properties. | |
| Mar 3, 2012 at 18:44 | comment | added | Deane Yang | Could one of you explain what you mean by a global proof? Isn't this a pointwise statement? | |
| Mar 3, 2012 at 17:13 | comment | added | Paul Siegel | As for (ii), Garding's inequality and Rellich's lemma together imply that any elliptic first order differential operator has compact resolvent. So the issue isn't really the Hilbert space properties of $\nabla^* \nabla$ but rather the way you intend to compute the symbol of $D$. | |
| Mar 3, 2012 at 16:58 | comment | added | Paul Siegel | I have asked a lot of knowledgeable people about (i), and nobody could point me to a satisfying global or conceptual argument. I think the physicists have some sort of heuristic explanation, but that's about it. | |
| Mar 3, 2012 at 16:24 | history | asked | Jean Delinez | CC BY-SA 3.0 |