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Let $Vect_{n}(C)$ the moduli stack of vector bundles $V$ of rank $n$ over a smooth curve $C$ of genus $g$. It is well known that $Vect_{n}(C)$ is a smooth stack of dimension $\dim(H^{0}(C,End(V)))-\dim(H^{1}(C,End(V)))=n^{2}(g-1)$

My concern is how to understand this dependence on the genus of the curve, in a clear geometrical way.

As $C$ is smooth the geometric and arithmetic genus match. The geometric genus $g=\dim(H^1(C,\mathcal{O}_{C}))$ equals the dimension of the space of regular differential $1$-forms on $C$. However I cannot figure out what is this space geomerically, I don't know how to draw such $1$-forms (In fact thir calculation is a bit involved in some cases (see The Construction of a Basis of Holomorphic Differential 1-forms for a given Planar Curve)).

If the dependence was "$\dim(Vect_{n}(C))=gn^{2}$" I could suppose intuitively that you have a different $n^{2}$-dimensional space over each of the "pieces of the curve" (supposing that each linear independent $1$-form was related somehow with each "piece of the curve" (#"pieces"$=g$)) but it is clear that this interpretation is totally wrong.

QUESTION: Could someone explain in a geometrical/intuitive way why the dimension of the stack depends on such a way on $g$?

(An aside question what does it mean when $C$ is an elliptic curve and consequently the dimension of the stack at is $0$?)

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  • $\begingroup$ I have just removed from there, I think I will get more interesting answers here, and even though the question may be a bit basic, I haven't found in the literature truly geometric explanations to the question $\endgroup$
    – Martin Hurtado
    Commented Oct 5, 2020 at 3:57
  • $\begingroup$ It might help to consider the case of line bundles. It sounds like you are more happy with with fact that $dim(Pic(C)) = g = \dim(H^1(\mathcal O))$. The moduli stack $Vect_1(C)$ is $Pic(C) \times BG_m$. The dimension of the stack is thus $g-1$, where the $-1$ is to account for the fact that line bundles have a 1-dimensional space of automorphisms. $\endgroup$
    – Sam Gunningham
    Commented Oct 5, 2020 at 10:13
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    $\begingroup$ In general, the formula for dimension of the moduli stack ($\dim(H^1) - \dim(H^0)$) at a point $V$ incorporates two things: the space of deformations of $V$ and the space of automorphisms of $V$. You could think that the former corresponds to the usual notion of dimension of a variety and the latter is a kind of stacky correction. For example, the statement that the dimension is 0 in genus 1 is saying that the amount of deformations is always balanced by the amount of automorphisms. $\endgroup$
    – Sam Gunningham
    Commented Oct 5, 2020 at 10:15
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    $\begingroup$ Note that for rank $>1$ vector bundles, the individual terms $dim(H^0)$ and $dim(H^1)$ can jump around as you vary the vector bundle, even though their difference remains constant. This is maybe an indication of why we need to include that stacky correction to get a nice meaninful notion of dimension. $\endgroup$
    – Sam Gunningham
    Commented Oct 5, 2020 at 10:18
  • $\begingroup$ Thank you. Your second comment seems to me quite useful. $\endgroup$
    – Martin Hurtado
    Commented Oct 5, 2020 at 17:24

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