Let $X$ be a smooth quasi-projective irreducible variety over the field of complex numbers $\mathbb{C}$. We denote by $\mathrm{Aut}(X)$ the group of algebraic automorphisms of $X$. Moreover, for a variety $V$, we call a map $V \to \mathrm{Aut}(X)$ a morphism, if the induced map $V \times X \to X$ is a morphism of varieties. The connected component of $\mathrm{Aut}(X)$ of the neutral element $e \in \mathrm{Aut}(X)$ we define by $$ \mathrm{Aut}(X)^\circ = \left\{ g \in \mathrm{Aut}(X) \Big| \begin{array}{l}  \, \textrm{$\exists$ an irreducible variety $V$ and a morphism} \\ \textrm{$V \to \mathrm{Aut}(X)$ s.t. the image contains $g$ and $e$} \end{array} \right\} \, . $$ This notion goes back to Ramanujam, see [Ram64]. Clearly, $\mathrm{Aut}(X)^\circ$ is a normal subgroup of $\mathrm{Aut}(X)$.

I am interested in the size of the group $Q(X) = \mathrm{Aut}(X) / \mathrm{Aut}(X)^\circ$. In case $X$ is projective, then $Q(X)$ is countable. Also in case $X$ is affine, $Q(X)$ is countable. My question is, whether this is true in general, i.e. whether for all smooth irreducible quasi-projective variety $X$, the group $Q(X)$ is countable. Every proof, counter-example or textbook reference would be perfect.

[Ram64] Ramanujam, C.P., A note on automorphism groups of algebraic varieties, Math. Ann. 156, 25-33 (1964). ZBL0121.16103.

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    $\begingroup$ If I'm optimistic, I'd expect that there's a reasonable notion of degree such that automorphisms of degree $\le n$ form a variety (or something not so far from a variety) and hence would map to only a finite subset of $Q(X)$, which would entail the countability result. In case $X$ is affine it's easy to check working with coordinate rings. $\endgroup$ – YCor Apr 17 '17 at 11:21
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    $\begingroup$ I've found several claims by H. Kraft that the automorphism group of an affine variety is an ind-group (which implies this countability result), with reference to a paper in preparation with Furter. I don't know if some unexpected difficulty explains that the paper is not available now, nor if there is any reason not to expect that this holds in more generality, namely for arbitrary quasi-projective varieties (which certainly would be more technical to prove). $\endgroup$ – YCor Apr 17 '17 at 13:33
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    $\begingroup$ I have a certain confidence in the claim that the automorphism group of an affine variety is an ind-group, though I have never checked the details. However, I am 100% sure that it is an ind-variety, which is enogh for the required countability result. I happened to explain this recently in mathoverflow.net/a/266803/89334. It seems reasonable to extend this to arbitrary varieties. $\endgroup$ – Uri Bader Apr 17 '17 at 15:27
  • $\begingroup$ Relevant: mathoverflow.net/q/8812/89334 $\endgroup$ – Uri Bader Apr 18 '17 at 21:20

Let $i:X\to \overline{X}$ be any dense open immersion of $X$ in a projective scheme. For every morphism $\phi:X\to X$, denote by $\overline{\Gamma}_\phi\subset \overline{X}\times_{\text{Spec}(\mathbb{C})}\overline{X}$ the closure of the graph of $\phi$. For every choice of ample invertible sheaf $\mathcal{L}$ on $\overline{X}\times_{\text{Spec}(\mathbb{C})}\overline{X}$, there is an associated Hilbert polynomial $P(t)\in \mathbb{Q}[t]$ of $\overline{\Gamma}_\phi$ with respect to $\mathcal{L}$. For every choice of $P(t)$, denote by $\text{Hilb}^{P(t)}$ the associated Hilbert scheme parameterizing closed subschemes $Z\subset \overline{X}\times_{\text{Spec}(\mathbb{C})}\overline{X}$ with Hilbert polynomial $P(t)$. There is an open subscheme $U$ that parameterizes those $Z$ such that both projections $Z \cap (X\times_{\text{Spec}(\mathbb{C})} X) \to X$ are isomorphisms. As an open subscheme of a projective scheme, $U$ has only finitely many irreducible components. Also there are only countably many possible numerical polynomials $P(t)$. Thus, the group $Q(X)$ is finite or countably infinite.

  • $\begingroup$ Several things seem to be unclear to me: First of all, why does there exist such an open subscheme $U$ of the Hilbert scheme $\mathrm{Hilb}^{P(t)}$? And secondly, assume, I take an irreducible component of $U$, call it $U_1$. Then we need that the map $U_1 \times X \to X$ is a morphism of varieties, which is not clear to me... $\endgroup$ – Anonymous Apr 17 '17 at 14:46
  • $\begingroup$ I edited my answer. $\endgroup$ – Jason Starr Apr 17 '17 at 14:55
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    $\begingroup$ The convention that finite sets are countable is so convenient! $\endgroup$ – YCor Apr 17 '17 at 18:39
  • $\begingroup$ It's not important, and was not a correction, but I mean it's convenient because you can just say "countable" instead of emphasizing the artificial distinction "finite or XXX" where XXX is anything you mean to be "infinite countable". $\endgroup$ – YCor Apr 17 '17 at 19:28

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