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Let $X$ be a smooth projective connected variety over the complex numbers with ample canonical bundle. If $X$ is generic and $\dim X \leq1$, the automorphism group of $X$ is trivial, see for instance

Why is a general curve automorphism-free?

This question is about generalizing this to arbitrary dimension. Let me be more precise.

Suppose that $X$ is "generic". Is the automorphism group of $X$ trivial?

This is probably true, and there are three approaches to this sketched in the above MO question. The first two might not be feasible.

  1. Use deformation theory, i.e., compute the tangent space at the moduli space, and use Lefschetz trace formula. Can somebody make this more precise in this case?

  2. Count parameters using Riemann-Hurwitz. This is going to be problematic in the higher-dimensional case, even though there is a Riemann-Hurwitz formula, I am not sure the dimension of the moduli space is explicitly known (as opposed to the one-dimensional case where it equals $3g-3$).

  3. Exhibit an $X$ as above with trivial automorphism group for any possible hilbert polynomial. In fact, the order of the automorphism group of $X$ is bounded (even explicitly) by a constant depending only on the Hilbert polynomial of $X$.

I think 3 is the most promising, but this would require me to come up with the following.

Let h be the hilbert polynomial of $X$. Then there exists a smooth projective connected variety $Y$ with ample canonical bundle and hilbert polynomial of the canonical bundle equal to $h$ such that Aut$(Y)$ is trivial.

So my problem is to do this for every occuring hilbert polynomial. Of course, writing down varieties $X$ as above with no automorphisms is not so difficult.

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You want to do this for every possible family? That is very hard to imagine. –  Angelo May 12 '13 at 7:57
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There exist varieties of general type with non-trivial automorphism groups and no deformations... –  ulrich May 12 '13 at 9:07
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I meant this: it is very hard to imagine this might be true, but if it were, it would probably be extremely difficult to prove. In any case, it seems that you are being given counterexamples. –  Angelo May 12 '13 at 9:42
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@Francesco: One does not need to consider fake projective planes. Any cocompact quotient of the complex 2-ball is rigid (by a theorem of Calabi and Vesentini) and of general type (by a theorem of Kodaira) so one can consider any non-trivial finite Galois cover of such a quotient. –  ulrich May 12 '13 at 14:46
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Don't curves of genus 2 already give a counterexample in dimension 1? –  Tom Graber May 13 '13 at 23:46
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2 Answers 2

up vote 12 down vote accepted

It seems to me that this is not true and that a counterexample can be constructed as follows.

Take a double cover $\alpha \colon X \longrightarrow A$ of an abelian surface $A$, branched over a smooth divisor $B \in |2 L|$, with $L$ very ample. We have $$K_X=\alpha^* L, \quad \alpha_* \mathcal{\omega}_X = \mathcal{\omega}_A \oplus \omega_A (L),$$ hence $$K_X^2 = 2L^2, \quad p_g(X) = 1+ h^0(A, L), \quad q(X)=2.$$ Than $X$ is a smooth surface of general type. Moreover, since $\alpha$ is a finite map, $X$ does not contract any curve, in particular $K_X$ is ample.

We have $q(X)=2$, so $\textrm{Alb}(X)$ is an abelian surface and, by the universal property of the Albanese map, the morphism $\alpha$ factors through $a \colon X \longrightarrow \textrm{Alb}(X)$.

But then, since $\deg \alpha =2$ and $X$ is not an abelian surface, it follows that the isogeny $\textrm{Alb}(X) \to A$ must be an isomorphism. Then the morphism $\alpha \colon X \longrightarrow A$ coincides the Albanese map of $X$.

By a result of Catanese (see A superficial working guide to deformation ands moduli, arXiv:1106.1368, Section 5) the degree of the Albanese map is a topological invariant. It follows that any deformation of $X$ is still a double cover of its Albanese variety.

Thus any surface $Y$ lying in the same connected component of the moduli space containing $X$ has non-trivial automorphism group, because the Albanese double cover induces a non-trivial involuton $\iota \colon Y \to Y$, and so $\mathbf{Z} /2 \mathbf{Z} \subset \textrm{Aut}(Y)$.

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I like this counterexample a lot. I do have some questions. 1. Is it really possible that $X$ is nonsingular. I thought $X$ would have cyclic quotient singularities above the singular locus of the branch locus. Or maybe this is where you use "degree two"? 2. This is not so important, but how do you show that your double cover really equals the albanese map. Is it not possibly necessary to compose with an isogeny of A to really get the albanese map? 3. Can you give rigid examples of your $X$, and examples where $X$ has infinitely many deformations? Thanks again! –  Jonathan May 12 '13 at 10:19
    
@Jonathan: 1. If you take the branch locus in a very ample linear system, by Bertini theorem the general element of the system will be smooth, hence the general cover will be smooth. 2. By the universal property of the Albanese map, the cover $\alpha \colon X \to A$ factors through the Albanese morphism $X \to \textrm{Alb}(X)$. Since $\deg \alpha =2$ and $X$ is of general type, the isogeny $\textrm{Alb}(X) \to A$ must be an isomorphism. 3. $X$ has non trivial deformations coming from the deformations of $A$ (and from those of the branch locus). –  Francesco Polizzi May 12 '13 at 21:34
    
I have added some further details –  Francesco Polizzi May 13 '13 at 13:09
    
thank you very much. –  Jonathan May 16 '13 at 7:37
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As Francesco points out, the claim is false when $\dim X>1$. The question is discussed in [B. Fantechi, R. Pardini, Automorphisms and moduli spaces of varieties with ample canonical class via deformations of abelian covers, Comm. Algebra 25 (1997), 1413-1441. math.AG/9410006] (see in particular Thm. 6.6).

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