Let $X$ and $Y$ be (smooth projective connected) varieties over $\mathbf{C}$.

Let $\pi:X\to Y$ be a finite surjective flat morphism.

Does this induce (by base change) a map $\mathrm{Aut}(Y) \to \mathrm{Aut}(X)$?

I think it does. Given an automorphism $\sigma:Y\to Y$, the base change via $\pi:X\to Y$ gives an automorphism of $X$.

My real question is as follows:

Is $\mathrm{Aut}(Y)\to \mathrm{Aut}(X)$ injective?

If not, under which hypotheses is $\mathrm{Aut}(Y)\to \mathrm{Aut}(X)$ injective? Does $\pi$ etale do the trick?

What if $\dim X=\dim Y =1$?


I don't think that there is any (canonical) map $Aut(Y) \to Aut(X)$ of the kind you presume exists.

E.g. if $X$ is a curve of genus $g \geq 2$ and $Y$ is $\mathbb P^1$, then $Aut(X)$ is finite (often trivial), while $Aut(Y)$ equals $PGL_2(\mathbb C)$, which is simple. What is the map that you have in mind? (In any case, whatever it it is, it won't be injective.)

  • $\begingroup$ Let me first motivate my question. We know that, for each hyperbolic curve $X$, the inequality $\# \mathrm{Aut}(X) \leq 84(g-1)$ holds. Now, forgetting for a moment that we already know this, suppose that $X$ is a curve for which this inequality holds. If $Y$ is hyperbolic of genus $g$ (same genus as $X$!) and $Y\to X$ is a finite morphism, can we conclude that the same inequality holds for $\# \mathrm{Aut} (Y)$? So, back to the question. Let $\pi:Y\to X$ be finite (I reversed $X$ and $Y$...). Let $\sigma:X\to X$ be an autom. We can make a Cartesian diagram and base change. This gives a $\endgroup$ – Harry Jun 18 '12 at 14:28
  • $\begingroup$ morphism $Y\times_{X,\sigma} X\to Y$. I probably did something stupid and identified $Y\times_{X,\sigma} X$ with $Y$ and said this base changed morphism is an automorphism. This would be then the map I had in mind. $\endgroup$ – Harry Jun 18 '12 at 14:29
  • $\begingroup$ @Harry: Dear Harry, certainly the fibre product you write down will be isomorphic to $Y$ as a variety, but it won't usually be isomorphic to $Y$ as an $X$-variety, i.e. we can't find an isomorphism of it with $Y$ which lies over the automorphism $\sigma$ of $X$. As the simplest examples show (higher genus curves mapping to $\mathbb P^1$, higher genus curves mapping to genus one curves, etc.), automorphisms typically don't lift through finite flat maps. Regards, $\endgroup$ – Matt E Jun 18 '12 at 15:06

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