TITLE: Finiteness property of automorphism scheme
QUESTION [34 upvotes]: Some time ago I mentioned a certain open question in an MO answer, and Pete Clark suggesting posting the question on its own.  OK, so here it is:
First, the setup. Let $X$ be a projective scheme over a field $k$.  By Grothendieck, there is a locally finite type $k$-scheme $A = {\rm{Aut}}_ {X/k}$ representing the functor assigning to any $k$-scheme $T$ the group of $T$-automorphisms of $X_T$.  (Artin proved a related result with projectivity relaxed to properness, even allowing $X$ to be an algebraic space.)  The construction uses Hilbert schemes, so at most countably many geometric connected components can occur. 
In some cases the automorphism scheme is connected (such as for projective space, when the automorphism scheme is ${\rm{PGL}}_n$), and in other cases the geometric component group $\pi_0(A) = (A/A^0)(\overline{k})$ can be infinite.  For the latter, a nice example is $X = E \times E$ for an elliptic curve $E$ without complex multiplication over $\overline{k}$; in this case $A$ is an extension of ${\rm{GL}}_ 2(\mathbf{Z})$ by $E \times E$, so $\pi_0(A) = {\rm{GL}}_ 2(\mathbf{Z})$.  This latter group is finitely presented. 
Question: is the geometric component group $\pi_0(A)$ of the automorphism scheme $A$ of a projective $k$-scheme $X$ always finitely generated?  Finitely presented? And with projectivity relaxed to properness, and "scheme" relaxed to "algebraic space"?
Feel free to assume $X$ is smooth and $k = \mathbf{C}$, since I believe that even this case is completely wide open.
Remark: Let me mention one reason one might care (apart from the innate appeal, say out of analogy with finite generation of Neron-Severi groups in the general proper case).  If trying to study finiteness questions for $k$-forms of $X$ (say for fppf topology, which amounts to projective $k$-schemes $X'$ so that $X'_K = X_K$ for a finite extension $K/k$), then the language of ${\rm{H}}^1(k, {\rm{Aut}}_{X/k})$ is convenient.  To get hands on that, the Galois cohomology of the geometric component group intervenes.  So it is useful to know if that group is finitely generated, or even finitely presented.

REPLY [17 votes]: The answer is negative even for smooth projective varieties over $\mathbb C$: a counterexample is constructed in http://arxiv.org/abs/1609.06391.  The example is a smooth, uniruled sixfold of Picard rank somewhere in the high 30s.
Here's the basic idea.  One should try to find a variety $X$ whose automorphism group is something like a free group. It's possible to rig things so that there's a point $x$ on $X$ whose stabilizer in $\operatorname{Aut}(X)$ is a non-finitely generated subgroup; free groups have lots of these.  If you blow up $x$, the automorphisms of $X$ that lift to the blow-up are precisely the ones that fix $x$.  One also needs to check that the blow-up doesn't have any automorphisms other than the ones lifted from $X$, but this isn't too hard.  Things don't work out quite this cleanly in the linked example, but this is the strategy.
There's also the issue of how to actually prove the automorphism group isn't finitely generated.  We've arranged that the "obvious" automorphisms we can write down are a non-finitely generated group, but we might worry that there are other automorphisms we don't know about.  Maybe we actually just found a non-finitely generated subgroup of some larger finitely generated group.  The solution is to arrange that there's a rational curve $C$ on $X$ and prove that every automorphism of $X$ must fix $C$, and in fact restrict to $C$ as a map of the form $z \mapsto z+c$.
This means we have a restriction map $\rho : \operatorname{Aut}(X) \to \operatorname{Aut}(C)$, and the image lies in the abelian subgroup of translations fixing $\infty$.  We construct explicit automorphisms $\mu_n$ (for every $n$) whose restrictions to $C$ are the maps $z \mapsto z + 1/3^n$.  The means the image of $\rho$ is an abelian group containing $\mathbb Z[1/3]$, so it's not finitely generated.  This means that $\operatorname{Aut}(X)$ isn't finitely generated either, though I don't know exactly what it is.  If I had to guess, it's probably free on countably many generators.
(I apologize that this is very vague, but the paper is not so long, and kind of fun!)