# Number of connected components of an Automorphism group

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.

• 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.
– YCor
Commented Apr 17, 2017 at 11:21
• 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).
– YCor
Commented Apr 17, 2017 at 13:33
• 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. Commented Apr 17, 2017 at 15:27
• Relevant: mathoverflow.net/q/8812/89334 Commented Apr 18, 2017 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.
• 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... Commented Apr 17, 2017 at 14:46