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.

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    $\begingroup$ Fix an ample class $L$ in Neron-Severi group of $X$. The subgroup of automorphisms sending $L$ to itself is of finite type. So the real question is: what is the image of $Aut(X)$ in $Aut(NS(X))$? Is that finitely generated/presented? These automorphisms permute ample classes, so if the semigroup of ample classes is f.g. (which happens rarely), we are OK. $\endgroup$
    – VA.
    Apr 22, 2010 at 17:40
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    $\begingroup$ For minimal surfaces a result of Dolgachev says that (possibly over the complex numbers only) that the image of $\mathrm{Aut}(X)$ in $\mathrm{Aut}(K_X^\perp)$ (the orthogonal complement of the canonical class) is a quotient of a subgroup of finite index of the full automorphism group of that lattice. Hence it is at least finitely generated. The normal subgroup by which one takes the quotient is the subgroup generated by reflections in nodal curves. $\endgroup$ Apr 22, 2010 at 17:53
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    $\begingroup$ Right. One can look at other features of NS(X) that are preserved. For example for Fanos, the closure of the ample cone is finitely generated, so that leads to the proof. CYs seem like maybe the hardest case. Such a natural question... Must be known, I hope someone answers. $\endgroup$
    – VA.
    Apr 22, 2010 at 21:32
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    $\begingroup$ VA, Mumford gave a colloq. talk here today, and after the dinner I mentioned the question. He was intrigued, and said he'd never heard anything about a result in that direction. Mazur wrote a paper with some theoretical arguments for Tate-Shaf. sets which required finite presentation hypotheses on Aut-scheme. In that part he acknowledged assistance from Gabber, to the extent of saying that for some result Gabber weakened the hypothesis to finite generation...but not eliminated it! So seems Gabber thought about it without success. If it is known, I will then be amazed (and very happy). $\endgroup$
    – BCnrd
    Apr 23, 2010 at 3:53
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    $\begingroup$ "Mazur wrote a paper ... " On the passage from local to global in number theory. Bull. Amer. Math. Soc. (N.S.) 29 (1993), no. 1, 14--50. $\endgroup$ Apr 23, 2010 at 5:15

2 Answers 2


I wanted to add some things to the comments I had already made but the list of comments have become very large and the comments I have already made are becoming more and more difficult to follow so I'll put everything (including the things I have already said) here instead even though it is not an answer to the question.

Let us first consider the case of a minimal surface $X$ (by minimal I mean $K_X$ nef). Dolgachev (Dolgachev: Reflection groups in algebraic geometry is a good reference even though the proof is only referenced there not given) gave a kind of structure theorem for the image $A_X$ of $\mathrm{Aut}(X)$ in $\mathrm{Aut}(S_X)$, where $S_X$ is the orthogonal complement of $K_X$ in $\mathrm{NS}(X)$ modulo torsion. His result says that there is a normal subgroup $W_X$ of $\mathrm{Aut}(S_X)$ generated by reflections in $-2$-curves and the group $P_X$ generated by $A_X$ and $W_X$ is a semi-direct product and of finite index in $\mathrm{Aut}(S_X)$. Note that it is possible to have $W_X=\{e\}$ and then $A_X$ itself is of finite index and hence an arithmetic (and thus finitely presented). It is also possible to have $W_X$ of finite index and then $A_X$ is finite (and thus finitely presented). However, there are intermediate cases where both $A_X$ and $W_X$ are infinite. Still $A_X$ is a quotient of $P_X$ and hence is finitely generated. I do not know if it is always finitely presented. Borcherds (Coxeter groups, Lorentzian lattices, and $K3$ surfaces. Internat. Math. Res. Notices 1998) gives examples where it is (and where it is even nicer) but also examples where it is finitely generated but not arithmetic.


I now realise that finite presentation is always true: For that we only need to show that $W_X$ is normally generated in $\mathrm{Aut}(X)$ by a finite number of elements and for that it is enough to show the same thing for $\mathrm{Aut}(S_X)$. We know that $W_X$ is generated by reflections in $-2$-elements. There are however only a finite number of conjugacy classes $-2$-elements. For this it is, by standard lattice theoretic arguments, enough to prove that there are only a finite number of isomorphism classes of orthogonal complements. However, the discriminant of such a complement is bounded in terms of the rank and discriminant of $S_X$ and there are only a finite number of forms of bounded rank and discriminant.


A further step would be to blow up points of $X$ (still assumed minimal). As $X$ is the unique minimal model any automorphism of the blowing up is given by an automorphism of $X$ that permutes the blown up points (and the subgroup fixing the points is commensurable with the full automorphism group). In the case of abelian or hyperelliptic surfaces blowing up just one point is pointless as it just serves to kill off the connected component of $\mathrm{Aut}(X)$ so in that case the first interesting case is blowing up two points.

Consider the case of blowing up two points when $X$ is abelian. So we have two points on $X$ one of which we can assume is $0$ and the other we'll call $x$. An automorphism of $X$ that fixes both of these points will be en automorphism of $X$ as abelian variety that fixes pointwise the closed subgroup $A$ generated by $x$. The group fixing $x$ will then have finite index in the the group fixing $A$ pointwise. For any abelian subvariety $A$ of $X$, the subgroup of $\mathrm{Aut}(X)$ fixing all the points of $A$ is an arithmetic subgroup (in a not necessarily semi-simple group) and in particular is finitely presented.

The same argument works for abelian varieties of any dimension. There one of course also has the option of blowing up positive dimensional varieties, assume $S$ is a smooth closed subvariety. This time the automorphism group is the subgroup of automorphisms $X$ that fixes $S$. We thus get an induced action on $S$ and the kernel of that action has the same structure as before. Unless I am mistaken, the automorphisms of $S$ that extend to $X$ are of finite index in $\mathrm{Aut}(S)$ (look at $\mathrm{Alb}(S) \rightarrow X$ and split it up to isogeny). Hence the finite generation etc for the blowing up is reduced to finite generation for $S$ (and conversely for $X$ replaced by $\mathrm{Alb}(S)$).

Consider now the case of $X$ still minimal but non-abelian or hyper-elliptic and look at blowing up of one point $x$. For a general point of $X$ (in the sense of being outside a countable number of proper subvarieties) the automorphism group is trivial and hence finitely generated. The situation for arbitrary $x$ seems unclear but one thread of the discussion started concerning itself with whether for a general $X$ there is a characterisation (up to commensurability) of $\mathrm{Aut}(X)$ similar to the minimal case: Look at all automorphisms of the integral cohomology of $X$ that preserves multiplicative structure, Hodge structure, Chern classes (of the tangent bundle) and effective cones (spanned by effective cycles). Is the image of the automorphism group of $X$ of finite index in this group? I think the answer is no (and I hope that what I present here is a proof). For that we need to recall some facts on Seshadri constants (Lazarsfeld: Positivity in Algebraic Geometry, I is my reference). Given a point $x$ the Seshadri constants $\epsilon(L;x)$ for $L$ nef (but also for $L$ restricted to be ample) determine (and are determined by) the nef cone of the blowing up at $x$; $L-rE$ is nef precisely when $0\leq r\leq \epsilon(L;x)$. Switching tack, there is a subset $U$ of $X$ which is the intersection of a countable number of open non-empty subsets of $X$ such that $\epsilon(L;x)$ is constant on $U$ for all ample $L$. Indeed, $\epsilon(L;x)$ can be expressed (loc. cit.: 5.1.17) in terms of whether or not $kL$ separates $s$-jets at $x$ and for fixed $k$ and $s$ the separation is true on an open subset.

The conclusion is that there is a $U$ which is the intersection of a countable number of non-empty open subsets for which the nef cone of the blowing up of $X$ at $x$ is independent of $x$ when one expresses it in the decomposition $\mathrm{NS}(X)\bigoplus\mathrm Z E$. If we assume now that $K_X$ is numerically trivial we have that the first Chern class of the tangent bundle of the blowup of $X$ at some $x$ equals $E$ (up to torsion) and hence the group above will preserve the decomposition $\mathrm{NS}(X)\bigoplus\mathrm Z E$ and fix $E$ so come from an automorphism of $\mathrm{NS}(X)$. The only further condition we put on it is that it preserve the nef cone but for $x\in U$ this cone is independent of $x$. As we can further arrange it so that $x\in U \implies \varphi(x)\in U$ (as $\mathrm{Aut}(X)$ is countable) we get that all elements of $\mathrm{Aut}(X)$ give structure preserving automorphisms of the cohomology of the blowup of $X$ at $x$. However, as observed before, at the price of shrinking $U$ we can assume that that the automorphism group of the blowing up is trivial. Hence, if we let $X$ be for instance a K3-surface with infinite automorphism group we get an example.


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!)


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