Timeline for Codimension of the complement of the stable locus
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
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| Sep 25, 2021 at 4:10 | comment | added | yors | @NicolasTholozan Thank you for your comment. I haven't seen this result before. Can you give a reference for this? | |
| Sep 25, 2021 at 4:06 | comment | added | yors | @Bernie Thank you for the reference. My understanding is that since the moduli space $\mathcal{M}$ is a normal variety and the stable locus is precisely the nonsingular locus, the codimension of the complement is at least $2$. Is it correct? | |
| Sep 24, 2021 at 21:54 | comment | added | Nicolast | Denote by $\mathcal M(r)$ the moduli space of semistable bundles of rank $r$ and degree $0$. Then $\mathcal M(r) \setminus \mathcal M^s(r)$ is essentially the union of $\mathcal M(k) \times \mathcal M(r-k)$ for $0<k<r$, which has codimension $2k (r-k)(g-1) -1$ if I'm not mistaken. | |
| Sep 24, 2021 at 19:21 | comment | added | Bernie | I think some of what you are looking for can be found in the proof of Theorem 1 in Section 4 of the paper " Moduli of Vector Bundles on a Compact Riemann Surface" by Narasimhan and Ramanan. | |
| Sep 24, 2021 at 13:53 | comment | added | yors | Thank you very much. I should shave mentioned that genus greater that 2 and for the parabolic case the genus is greater than 3. I have edited the post. | |
| Sep 24, 2021 at 13:53 | history | edited | yors | CC BY-SA 4.0 |
added 43 characters in body
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| Sep 24, 2021 at 12:27 | comment | added | Bernie | For g=2, r=2 and fixed determinant $\mathcal{O}_X$ the moduli space is isomorphic to $\mathbb{P}^3$ and the semistable locus is the singular Kummer surface associated to $X$, hence has codimension 1. This all explained in the papers by Narasimhan et. al. | |
| Sep 24, 2021 at 11:44 | history | asked | yors | CC BY-SA 4.0 |