14
$\begingroup$

Suppose $X$ is a smooth projective variety over a field $k$, with ample canonical bundle. If $\operatorname{char}(k)=0$ or $\operatorname{char}(k)>\dim(X)$ and $X$ lifts to $W_2(k)$ (thanks Olivier Benoist!), it's not hard to see that $X$ has finite automorphism group.

Proof. Let $X\hookrightarrow \mathbb{P}\Gamma(X, \omega_X^{\otimes n})^\vee$ be a closed embedding. Then the induced map $\operatorname{Aut}(X)\hookrightarrow PGL(\Gamma(X, \omega_X^{\otimes n})^\vee)$ is a closed embedding, so $\operatorname{Aut}(X)$ is finite type. But the tangent space to $\operatorname{Aut}(X)$ at the identity is $\Gamma(X, T_X)$, which is trivial by Kodaira vanishing. So $\operatorname{Aut}(X)$ is a $0$-dimensional finite type group scheme, hence finite. $\blacksquare$

(An essentially identical argument avoiding the representability of $\operatorname{Aut}(X)$ may be found here.)

This proof breaks down for $\operatorname{char}(k)$ small: $\Gamma(X, T_X)$ need not be trivial. But I don't know of an example of a canonically polarized variety with infinite (thus necessarily positive-dimensional) automorphism group in small characteristic.

Is the automorphism group of a smooth variety with ample canonical bundle always finite, even if $\operatorname{char}(k)$ is small?

Improve this question
$\endgroup$
10
  • 1
    $\begingroup$ How do you know that the map ${\rm{Aut}}_{X/k} \rightarrow {\rm{PGL}}(\Gamma(X,\omega^{\otimes n})^{\vee})$ has closed image (e.g., why is it finite type?)? It is certainly not sufficient that it has trivial kernel. One can make plenty of maps from a locally finite type $k$-group to a finite type $k$-group with trivial kernel. $\endgroup$
    – user76758
    Commented Dec 21, 2013 at 1:19
  • $\begingroup$ The proof in your link is also flawed; why should $v$ there have anything to do with infinitesimal automorphisms of $X$ if it does not arise from the tangent space to the automorphism scheme (but just that of the closure of its image)? $\endgroup$
    – user76758
    Commented Dec 21, 2013 at 1:23
  • 2
    $\begingroup$ @user76758 : no it is OK : consider the closed subgroup scheme $G\subset PGL_{N+1}$ consisting of automorphisms preserving $X$. There are natural morphisms of group schemes in both directions : $G\to Aut(X)$ by restricting the action to $X$, and $Aut(X)\to G$ using the induced action on pluricanonical forms. Using the fact that $X\to\mathbb{P}^N$ is an embedding, you can see that these two maps are inverse to each other. $\endgroup$
    – Olivier Benoist
    Commented Dec 21, 2013 at 1:32
  • $\begingroup$ @DanielLitt : I am worried about your statement that you will be able to apply Kodaira vanishing if $p>dim(X)$. Are you sure you're not mistaken ? $\endgroup$
    – Olivier Benoist
    Commented Dec 21, 2013 at 1:34
  • 4
    $\begingroup$ @DanielLitt : to apply Deligne-Illusie, you also need to know that $X$ lifts to the second Witt vectors of $k$. $\endgroup$
    – Olivier Benoist
    Commented Dec 21, 2013 at 2:05

1 Answer 1

Reset to default
17
$\begingroup$

A much more general result is proven in Martin-Deschamps, Lewin-Menegaux : Applications rationnelles séparables dominantes sur une variété de type général. Théorème 2 : if $X$ and $Y$ are smooth and proper, and if $Y$ is of general type, the set of dominant separable rational maps from $X$ to $Y$ is finite ! What you want follows immediately: it is their Corollaire 4, and they attribute it to Matsumura.

Improve this answer
$\endgroup$
1
  • $\begingroup$ Perfect! I'd seen Kobayashi's analytic argument for this over $\mathbb{C}$, but hadn't been able to find an algebraic argument. $\endgroup$
    – Daniel Litt
    Commented Dec 21, 2013 at 2:03

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .