Timeline for Index of linearized operator for symplectic vortex equations
Current License: CC BY-SA 3.0
8 events
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| Apr 9, 2017 at 3:56 | comment | added | Sebastian Goette | So, you are studying the operator (23)? For the Cauchy-Riemann operator, there are typically no local boundary conditions. Also, I don't see anything else you could pair this with to get local boundary conditions as in the other question. Is there a reason that you need local boundary conditions? If you like, we can continue the discussion by email. | |
| Apr 8, 2017 at 17:07 | comment | added | Mtheorist | Thank you for your answer and the reference. Am I right to say that since the Dolbeault operator does not admit local boundary conditions, then we cannot compute the index of the linearized operator in reference [1] for an open Riemann surface, since the other part of the linearized operator is a Cauchy-Riemann operator, as shown on page 28? (I was hoping to use the local boundary conditions which you discussed in mathoverflow.net/questions/266520/…) | |
| Apr 8, 2017 at 13:13 | comment | added | Sebastian Goette | The Euler operator admits local boundary conditions, and you can choose either absolute or relative ones (as for cohomology). The Dolbeault operator does not, so now you should stick to the Euler operator. Because the boundary is odd-dimensional, it has Euler number 0, and the index is still given by the formula above both for absolute and for relative boundary conditions. If you are interested, read the intro to Atiyah, Patodi, Singer, Spectral asymmetry and Riemannian geometry I, Math. Proc. Camb. Phil. Soc. 77 (1975), 43-69. | |
| Apr 8, 2017 at 6:36 | comment | added | Mtheorist | May I know if the answer above and your additional comment hold in the case where the Riemann surface has boundaries? | |
| Apr 8, 2017 at 6:22 | vote | accept | Mtheorist | ||
| Apr 8, 2017 at 5:11 | history | edited | Sebastian Goette | CC BY-SA 3.0 |
missing factor added
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| Apr 8, 2017 at 5:07 | comment | added | Sebastian Goette | Maybe your explanation works, too, but for Riemann-Roch, you only consider the $\bar\partial$ part of $d$, so you have a different operator. You split $\Omega^1=\Omega^{0,1}\oplus\Omega^{1,0}$ and consider the sum of $\bar\partial^*$ and $\partial^*$, and then you would end up with the same formula. | |
| Apr 8, 2017 at 5:03 | history | answered | Sebastian Goette | CC BY-SA 3.0 |