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Let $X$ be a smooth projective variety over the complex numbers. Denote by $Aut(X)$ its automorphism group, by $Aut(X)_0$ the connected component of the identity, and by $G$ the quotient $Aut(X)/Aut(X)_0$.

The pull-back gives a representation $$ \rho \colon G \to GL(H^{1,1}(X,\mathbb{C})) $$

I look for examples of varieties $X$ such that the kernel of $\rho$ is not a finite group.

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  • $\begingroup$ I'm sure you know this, but maybe worth mentioning for passers-by: such examples must have $b_1 \neq 0$. $\endgroup$
    – Lazzaro Campeotti
    Mar 29, 2016 at 14:10
  • $\begingroup$ @potentially dense: why? $\endgroup$
    – Francesco Polizzi
    Mar 29, 2016 at 14:23
  • $\begingroup$ Maybe I'm missing something trivial, but is it not $\mathbb{P}^2$ such an example? In this case the automorphisms are the projectivities, and these send the class of a line (i.e., the generator of $H^{1, \,1} (\mathbb{P}^2) \cong \mathbb{Z}$) to itself. $\endgroup$
    – Francesco Polizzi
    Mar 29, 2016 at 14:32
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    $\begingroup$ @FrancescoPolizzi: If $b_1=0$ then the first Chern class map is injective. So $\operatorname{ker} \rho$ is a subgroup of the kernel of the map $G \rightarrow GL(Pic(X))$. But the group of automorphisms fixing the class of an ample line bundle is a closed subgroup of $PGL(N)$ for some $N$, hence has finitely many connected components. $\endgroup$
    – Lazzaro Campeotti
    Mar 29, 2016 at 14:34
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    $\begingroup$ Wait, I think I got a little muddled about torsion in my previous comment. I really want the first Chern class map $\operatorname{Pic} X \rightarrow H^2(X,\mathbf C)$ to be injective, so my condition should be modified appropriately. $\endgroup$
    – Lazzaro Campeotti
    Mar 29, 2016 at 14:40

2 Answers 2

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The group of connected components of the subgroup of $Aut(X)$ fixing $H^{11}(X)$ is finite. This follows from the main result of

Polarized Varieties, Fields of Moduli and Generalized Kummer Varieties of Polarized Abelian Varieties, T. Matsusaka, American Journal of Mathematics, Vol. 80, No. 1 (Jan., 1958), pp. 45-82

I quote from the mathscinet review:

The main result (section 3) is as follows. Let $U$ be a projectively embeddable non-singular variety; then there is a maximal connected algebraic group $G$ of automorphisms of $U$. Moreover, $G$ is of finite index in the group $G′$ of those automorphisms of $U$ which, for a given projective embedding of $U$, carry a hyperplane section $H$ into a cycle numerically equivalent to $H$.

Since the group of numerical equivalence classes of divisors embeds in $H^{11}$, the group of automorphisms preserving $H^{11}$ is a subgroup of the group $G'$ above. It is clearly a closed algebraic subgroup, so it likewise has finitely many components.

I must admit, on a quick read, I also endorse the last sentences of the review:

Unfortunately, the author's style, added to the intrinsic difficulties of the subject, makes it exceedingly hard to check the accuracy of many technical details which seem essential for the validity of the proofs; this is all the more to be regretted, since his results are so valuable and important. It is to be hoped that he, or someone else, will some day give a completely lucid and completely convincing exposition of these topics.

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    $\begingroup$ This result was later proved more generally for compact Kahler manifolds, independently by A. Fujiki and D. Lieberman around 1978. See their papers in Inventiones and LNM respectively. It is enough to consider automorphism which fix one Kahler class. $\endgroup$
    – YangMills
    May 6, 2016 at 0:23
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Edit: This is just a long comment.

Let $X$ be a smooth projective variety.

If $\omega_X$ is ample, then $\mathrm{Aut}(X)$ is finite, so that $G$ is finite. So you see that we should avoid $\omega_X$ being ample.

If $\omega_X^\vee$ is ample, then $\mathrm{Aut}(X)$ is not necessarily finite (e.g., $X=\mathbb P^n$), but it's (the $\mathbb C$-points of) an affine finite type group scheme. In particular, its group of connected components is finite. In particular, $G$ is finite here as well.

So, as you're looking for a variety with $G$ infinite, and we should clearly avoid $\omega_X$ being ample or anti-ample, my guess is that looking at CY-variety might work. [Edit: I now realize that this never works. If $X$ is a CY manifold then the kernel is finite.]

Now, as you know, if $X$ is a curve of genus one (resp. a K3 surface), your representation $\rho$ will have finite kernel (resp. trivial kernel). So, we will have to start considering CY-threefolds.

Now, there are examples of CY-threefolds with infinite automorphism group. The first example that comes to mind is provided by the variety "X" in http://arxiv.org/pdf/1306.1590v3.pdf (also mentioned previously on MO: Can a rigid CY threefold have infinitely many automorphisms).

EDIT (thanks for David Speyer for the comments):

My first guess would be that the associated representation has infinite kernel in this case. Unfortunately, it does not. In fact, you will also have to look further than Calabi-Yau manifolds, as Prop. 2.4 in http://arxiv.org/pdf/1206.1649v3.pdf shows that the kernel of $\rho$ is always finite in this case as well.

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  • $\begingroup$ I don't think this example works. Let $\omega$ be a primitive third root of unity, and let $R = \mathbb{Z}[\omega]$. The example there is the following: Let $A$ be an abelian $3$-fold with automorphisms by $SL_3(R)$. Let $Z$ be the center of $SL_3(R)$, a cyclic group of order $3$. Then $X$ is a crepant resolution of $A/Z$, which thus inherits an action of $SL_3(R)/Z$. (continued) $\endgroup$
    – David E Speyer
    Mar 30, 2016 at 5:14
  • $\begingroup$ Let $V$ be the obvious $SL_3(R)$ rep on $\mathbb{C}^3$. Then $H^{10}(A) \cong V$, $H^{01}(A) \cong \bar{V}$ and $H^{11}(A) \cong V \otimes \bar{V}$. Since $Z$ acts trivially on $H^{11}(A)$, classes in $H^{11}$ should descend to $H^{2}(A/Z)$ and should then pass to $H^{11}(X)$. (I am not absolutely certain about this, because of the intermediate singular variety $A/Z$ in the argument.) If I am right, this means there is a copy of $V \otimes \bar{V}$ inside $H^2(X)$. But $SL_3(R)/Z$ acts without kernel on $V \otimes \bar{V}$. $\endgroup$
    – David E Speyer
    Mar 30, 2016 at 5:16
  • $\begingroup$ I guess I should read the paper. $H^2$ is one dimensional? But all of the forms in $H^{11}(A)$ are $Z$-invariant, so they certainly descend to the smooth locus of $A/Z$, and I would have thought they extend to the resolution since it is crepant. $\endgroup$
    – David E Speyer
    Mar 30, 2016 at 5:30
  • $\begingroup$ I don't see a formula for $H^2$ in the paper you link or in the MO post. How are you computing it? $\endgroup$
    – David E Speyer
    Mar 30, 2016 at 5:33

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