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Consider the complex algebraic moduli space $X:=\mathcal A_g^n$ of ppav's of dimension $g$ with some high enough level $n$ structure (so that it represents the corresponding functor).

Can one compute the dimension of the infinitesimal space of deformations to $X$ (as a function of $g$ and the level)?

If $V$ is a smooth projective variety, then I'm basically asking about $\dim H^1(V,T_V)$. I presume that, as $X$ is not projective, the dimension of its infinitesimal deformation space is given by $\dim H^1(\overline X,T_{\overline X}(log D))$, where $\overline X$ is a good compactification of $X$ and $D$ is its boundary divisor. [Edit: This is wrong as is explained below by Jason Starr. The space of infinitesimal deformations of $X$ as a $\mathbb C$-scheme is $\mathrm{Ext}^1(\Omega_{X/\mathbb C}, \mathcal O_X)$.

Computing the space $\mathrm{Ext}^1(\Omega_{X/\mathbb C}, \mathcal O_X)$ explicitly will require writing down $\Omega_{X/\mathbb C}$. Can this be done by pulling-back the relative cotangent bundle from the universal family along the zero section?

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  • $\begingroup$ For any finite type scheme $V$ over a field $k$, whether or not it is smooth, projective, etc., the vector space of first order deformations of $V$ as a $k$-scheme equals $\text{Ext}^1_{\mathcal{O}_V}(\Omega_{V/k},\mathcal{O}_V)$. $\endgroup$
    – Jason Starr
    Commented Feb 15, 2016 at 15:05
  • $\begingroup$ @JasonStarr Thank you for your comment. If $V$ is log-canonically polarized with Hilbert polynomial $h$, will the dimension of $Ext^1(\Omega, \mathcal O)$ be equal to the dimension of the tangent space to the object $[V]$ in the stack of log-canonically polarized varieties with Hilbert polynomial $h$? Or is this $Ext^1$ possibly bigger than this tangent space? $\endgroup$
    – Christian
    Commented Feb 15, 2016 at 15:16
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    $\begingroup$ Christian: I would advise you to write down one example. For instance, if $\overline{V}$ is a smooth, projective curve of genus $g>1$, if $D$ is an effective Cartier divisor of degree $1$ on $\overline{V}$, and if $V$ is the open complement of $D$, then $\text{Ext}^1_{\mathcal{O}_V}(\Omega_{V/k},\mathcal{O}_V)$ is the zero vector space, while $H^1(\overline{V},T_{\overline{V}/k}(\text{log}\ D))$ is nonzero. $\endgroup$
    – Jason Starr
    Commented Feb 15, 2016 at 15:22
  • $\begingroup$ @JasonStarr Thank you again. It seems that my intuition was completely wrong. $\endgroup$
    – Christian
    Commented Feb 15, 2016 at 16:14
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    $\begingroup$ I believe $T_X$ is the same as $\operatorname{Sym}^2 $ of the relative tangent bundle of the universal family. $\endgroup$
    – Will Sawin
    Commented Feb 15, 2016 at 16:35

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