Timeline for Spin Representation
Current License: CC BY-SA 3.0
7 events
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| Dec 22, 2011 at 19:35 | comment | added | Paul Siegel | A bundle on $M^n$ which is locally isomorphic to the Clifford algebra $\mathbb{C}_n$ which also comes equipped with local supersymmetries as above is the same thing as the choice of a spin$^c$ structure on $M$, and in the even dimensional case one can use the supersymmetries to reduce the Clifford algebra down to the ordinary spin representation. So your question comes down to: why must we remember the supersymmetry data? Generally the answer is that if you forget them you end up solving a trivial index problem. This is most transparent in the K-homology proof. | |
| Dec 22, 2011 at 19:24 | comment | added | Paul Siegel | @Kofi: The short answer is that $S$ is a direct summand of any complex representation of the Clifford algebra, so the spinor bundle . The long answer is more subtle. The basic issue is that the index of an operator is determined not just by the operator itself but by whatever "super-symmetric" data happens to be lying around. Specifically a local ONB $\{e_i\}$ acts by right multiplication on the bundle of Clifford algebras as supersymmetries (odd, self-adjoint, squares to $-1$), and these supersymmetries anti-commute with the Dirac operator. | |
| Dec 22, 2011 at 8:49 | comment | added | Matthias Ludewig | Though I must say, I always wondered, why you want to have S as small as possible. In every book it just says you do (and you inserted this phrase in your exposé as well), but I never really understood why this would be important. Especially since some Manifolds don't admit a spin structure, but you can always look at the bundle of Clifford algebras and the Dirac operator living there. | |
| Dec 22, 2011 at 3:23 | comment | added | Paul Siegel | Quite right - thanks for the corrections! I edited my answer to account for the former. | |
| Dec 22, 2011 at 3:21 | history | edited | Paul Siegel | CC BY-SA 3.0 |
added 10 characters in body; deleted 2 characters in body
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| Dec 21, 2011 at 13:25 | comment | added | Jan Jitse Venselaar | Small nitpick: surely you mean $ c_i c_j + c_j c_i =0 $ Also, Dirac actually wanted to find such an operator whose square was the D'Alembert operator, the Laplacian in space-time, so with one +-sign (the time direction) and three minus-signs (space directions), so the $c_i$ generate the Clifford algebra $Cl_{3,1}$ (or $Cl_{1,3}$, I always forget the signs). | |
| Dec 21, 2011 at 8:05 | history | answered | Paul Siegel | CC BY-SA 3.0 |