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.

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?

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 $3g-3$).

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.

everypossible family? That is very hard to imagine. $\endgroup$ – Angelo May 12 '13 at 7:57