Timeline for Index Theorem for the Twisted Dirac Operator
Current License: CC BY-SA 3.0
8 events
when toggle format | what | by | license | comment | |
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Apr 8, 2017 at 13:21 | vote | accept | Mtheorist | ||
Apr 8, 2017 at 13:19 | comment | added | Sebastian Goette | @Mtheorist No, it means, the number of pairs of $\psi_-$ and $\bar\psi_+$ zero modes matching up at the boundary as in (39.212) minus the number of pairs of $\bar\psi_-$ and $\psi_+$ zero modes matching up at the boundary. Without the boundary condition, there are infinitely many of each. For example take $\Sigma=D^2\subset\mathbb C$ and $E$ trivial. Then every holomorphic function on $D^2$ is a $\bar\!\partial_+=\partial_{\bar z}$ zero mode. | |
Apr 8, 2017 at 6:13 | comment | added | Mtheorist | So the RHS of equation 39.213 really means number of $\psi_-$ OR $\overline{\psi}_+$ zero modes minus number of $\overline{\psi}_-$ OR $\psi_+$ zero modes, correct? | |
Apr 8, 2017 at 4:30 | comment | added | Sebastian Goette | @Mtheorist I mean, consider $\operatorname{ind}(\not\!\partial^E_+-\not\!\partial^{E^*}_+)=\operatorname{ind}(\not\!\partial^E_+\oplus\not\!\partial^{E^*}_-)$. But please note that the individual operators $\not\!\partial^E_+$ and $\not\!\partial^{E^*}_-$ do not admit local elliptic boundary conditions. For each of them, you could use the nonlocal APS boundary conditions, which are too complicated here. But the operator $\mathcal D$ admits local boundary conditions like those of (39.212). By "local" I mean pointwise on $\partial\Sigma$. In particular, $\operatorname{ind}(D)$ does not exist. | |
Apr 7, 2017 at 7:20 | comment | added | Mtheorist | Thank you for your answer. I am sorry but I do not understand exactly what you mean by formal difference of the operators. Do you mean that $\textrm{Index }\mathcal{D}=\textrm{Index }\overline{D}-\textrm{Index }{D}=2\textrm{ Index }\overline{D}$? | |
Apr 7, 2017 at 3:27 | history | edited | Sebastian Goette | CC BY-SA 3.0 |
typo
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Apr 7, 2017 at 0:23 | comment | added | Bombyx mori | You meant "you can use", right? | |
Apr 7, 2017 at 0:11 | history | answered | Sebastian Goette | CC BY-SA 3.0 |