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?](https://mathoverflow.net/questions/54994/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.



Proving that a generic variety with ample canonical bundle has no automorphisms
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 $\mathbf{Z} /2 \mathbf{Z} \subset \textrm{Aut}(Y)$.


