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Let $X$ be an algebraic smooth curve of genus $g$ over $\mathbb C$, and let $\mathcal M_X(r,d)$ (resp $\mathcal M_X^0(r)$) be the moduli space of vector bundle of rank $r$ and degree $d$ (resp. with trivial determinant) over $X$, we know that when $g=r=2$ and $d\equiv 0 \mod 2$ this space is smooth. So we ignore this case.

My question, as you knew from the title, is: What is the dimension of the singular locus of this moduli spaces?

Reference will be appreciated! thanks

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    $\begingroup$ I think that the answer is that the singular locus is the strictly semistable locus, assuming $g\ge 2$ and excluding the special case that you mention. The reason is that the moduli stack is smooth, the map from the moduli stack to the moduli space contracts the strictly semistable locus to smaller dimension, and finally the codimension of the strictly semistable locus in the stack is more than one, which shows that contracting it produces singularity. (The special thing about the case $g=r=2$ is that the codimension is one, which is why the contraction happens to give a smooth answer.) $\endgroup$
    – t3suji
    Oct 9, 2015 at 15:38
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    $\begingroup$ (You have to be slightly careful in the argument because the stack of semistable bundles is an Artin stack, but you can just ignore the bundles whose automorphism group jumps.) $\endgroup$
    – t3suji
    Oct 9, 2015 at 15:42
  • $\begingroup$ So, I minimised this dimension by considering the direct some morphism from cartisian product of two moduli spaces of vector bundles of smaller rank , say $\mathcal M_n\times\mathcal M_{r-n} \rightarrow \mathcal M_r$, so if I take the maximum of dimension of images of such maps, does this answer the question? $\endgroup$
    – Z.A.Z.Z
    Oct 10, 2015 at 11:17
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    $\begingroup$ Yes. (The maximum should be when $n=r/gcd(r,d)$, assuming of course that $gcd(r,d)\ne 1$.) $\endgroup$
    – t3suji
    Oct 10, 2015 at 13:51

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