Timeline for Dimension of the moduli stack of vector bundles over a curve
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13 events
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| Nov 17, 2020 at 17:40 | comment | added | Martin Hurtado | Sorry for the following silly question about deformation theory but I am confused about which deformations is count by each $H^{i}$. I thought that $H^{2}(X,End(V))$ is the space of obstructions of $X$ and $H^{1}(X,End(V))$ is the space of $1-dimensional$ deformations of $X$. So the space that counts the $2$-dimensional deformations of $X$ is not $H^{2}(X,End(V))$ but $H^{2}(R \Gamma)$ (the second homology group of the tangent complex) ? Is super antiintuitive to me, it is like you have to add extensions (via cdga's) to measure something which is totally a deformation of the underived space | |
| Oct 6, 2020 at 9:26 | comment | added | Sam Gunningham | The point is that on a curve there is only $H^0$ and $H^1$, but on higher dimensional varieties there are higher cohomology groups. In general, the invariant that behaves well is the (negative) Euler characteristic $-\chi(End(V))$. You can think of this as the dimension of the tangent complex $R\Gamma(X;End(V))[1]$ at $V$ to the derived moduli stack of vector bundles. If we only consider the classical (i.e. underived) stack, then we only see the $H^0$ and $H^1$ term; the fact that there can be a varying $H^2$ term (on a surface, say) means that the stack is not smooth. | |
| Oct 6, 2020 at 2:53 | comment | added | Martin Hurtado | I meant $\dim(H^{1})$ is not constant when varying the vector bundle $V$ (not the point of $X$) | |
| Oct 5, 2020 at 18:56 | comment | added | Martin Hurtado | Is this non-smoothness (of the stack of vector bundles over higher dimensional varieties) the reason for the very limited literature on the topic?I have read about it as one of the motivations for derived stacks in Vezzosi-Toen works. | |
| Oct 5, 2020 at 18:53 | comment | added | Martin Hurtado | Also, for the case of a variety or scheme of dimension $>1$ it is known that the moduli stack of vector bundles exists but it is not smooth. The formula of the dimension of the stack should be also $\dim(H^{1})-\dim(H^{0})$ so I guess that what does not work there is $dim(H^{1})$ which is not constant when varying the point of $X$. In that case what does $dim(H^{1})$ represent? Not all the space of deformations but a part of it? If so, why does it happen with a surface but not with a curve? | |
| Oct 5, 2020 at 17:24 | comment | added | Martin Hurtado | Thank you. Your second comment seems to me quite useful. | |
| Oct 5, 2020 at 10:18 | comment | added | Sam Gunningham | 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. | |
| Oct 5, 2020 at 10:15 | comment | added | Sam Gunningham | 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. | |
| Oct 5, 2020 at 10:13 | comment | added | Sam Gunningham | 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. | |
| Oct 5, 2020 at 4:00 | history | edited | Martin Hurtado | CC BY-SA 4.0 |
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| Oct 5, 2020 at 3:57 | comment | added | Martin Hurtado | 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 | |
| Oct 5, 2020 at 3:33 | history | edited | Martin Hurtado | CC BY-SA 4.0 |
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| Oct 5, 2020 at 3:25 | history | asked | Martin Hurtado | CC BY-SA 4.0 |