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Consider a compact Riemann surface $X$ of genus $\ge 2$, and consider the set $M$ of stable holomorphic vector bundles of rank $n$ and degree $d$ on $X$, up to isomorphism.

At that point, one states that one can describe a complex manifold $\mathcal{M}$, which as a set is in bijection with $M$.

My question is: The above statement is not really well-defined (it is only well-defined when given together with its proof); I would prefer a statement about existence of some complex manifold with some universal property. Could you specify such a property, or give a reference?

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$\mathcal{M}$ is what is called a coarse moduli space. In concrete terms, this means the following:

1) As a set, $\mathcal{M}$ can be viewed as the set of isomorphism classes of stable bundles (of rank $r$ and degree $d$):

2) Given any family of such vector bundles parametrized by an analytic space $S$ (that is, a vector bundle $\mathcal{E}$ on $S\times X$ such that for each $s\in S$, $\mathcal{E}_s:=\mathcal{E}_{|\{s\}\times X }$ is a stable bundle on $X$, of rank $r$ and degree $d$), the map $S\rightarrow \mathcal{M}$ given by $s\mapsto [\mathcal{E}_s]$ is holomorphic.

It is easy to see that this property characterizes $\mathcal{M}$.

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  • $\begingroup$ Thanks. 1) Do you have a reference? 2) Can I replace analytic spaces by complex manifolds and it will still characterize M? 3) It seems that there will not necessarily be a universal bundle on MxX? $\endgroup$
    – Sasha
    Commented Jan 18, 2016 at 9:19
  • $\begingroup$ 1) See this Wikipedia article and the references at the end. 2) Yes, because $\mathcal{M}$ is constructed as a quotient of a fine moduli space $Q$ by an algebraic group, and one can take $S=Q$ in the above answer. 3) There is a universal bundle if and only if $r$ and $d$ are coprime -- this is a result of Ramanan, The moduli spaces of vector bundles over an algebraic curve, Math. Ann. 200 (1973), 69–84. $\endgroup$
    – abx
    Commented Jan 18, 2016 at 10:36

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