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Let $\mathcal{M}$ denote the moduli space of semistable vector bundles of fixed rank and degree over a compact Riemann surface $X$. Let $\mathcal{M}^s \subset \mathcal{M}$ be the moduli of stable vector bundles.

What can we say about the codimension of the complement $\mathcal{M} - \mathcal{M}^s$ (assuming genus $>2$)? Is it always $\geq 2$?

What can we say in case of moduli of parabolic bundles (assuming genus $>3$)? Are there any references?

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    $\begingroup$ 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. $\endgroup$
    – Bernie
    Commented Sep 24, 2021 at 12:27
  • $\begingroup$ 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. $\endgroup$
    – yors
    Commented Sep 24, 2021 at 13:53
  • $\begingroup$ 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. $\endgroup$
    – Bernie
    Commented Sep 24, 2021 at 19:21
  • $\begingroup$ 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. $\endgroup$
    – Nicolast
    Commented Sep 24, 2021 at 21:54
  • $\begingroup$ @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? $\endgroup$
    – yors
    Commented Sep 25, 2021 at 4:06

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