# Automorphisms of Cartesian products

Let us consider the Cartesian product $X^r$, where $X$ is a smooth projective variety. There is a subgroup $Aut_{\Delta}(X^r)\subset Aut(X^r)$ of automorphisms of $X^r$ mapping a $k$-dimensional diagonal in $X^r$ isomorphically to a $k$-dimensional diagonal. In particular any automorphism in $Aut_{\Delta}(X^r)$ preserves the smallest diagonal $\Delta_{1,...,r}\subset X^r$.

If we consider the diagonal action of $Aut(X)$ on $X^r$ clearly we have $S_r\times Aut(X)\subseteq Aut_{\Delta}(X^r)$, where $S_r$ is the symmetric group. Under which hypothesis on $X$ (perhaps $X$ of general type) do we have $S_r\times Aut(X)= Aut_{\Delta}(X^r)$?

• You can act with different automorphism if X on each coordinate of X^r, and that is not in your subgroup. – Mariano Suárez-Álvarez Mar 5 '16 at 23:25
• In that way you do not preserv, for instance, the smallest diagonal $\Delta_{1,...,r}$. – japin Mar 6 '16 at 2:57

This is true under the assumption you suggest, $X$ of general type, when $X$ is a curve.

In the simplest case $r=2$, Corollary 3.9 of this paper of Catanese gives that $Aut(X^2)$ is the semidirect product of $Aut(X)^2$ with the $S_2$ exchanging the factors. Then by this complete description of $Aut(X^2)$ your equality follows immediately.

The corollary follows from the rigidity lemma 3.8 in the same page, which is stated for product of two curves, but whose proof has nothing to do with the number of components. In your case that gives, for $\dim X=1$, that $Aut(X^r)$ equals the semidirect product of $Aut(X)^r$ with the groups $S_r$ of the permutations of the factors. Again, your equality follows immediately.

It seems to me that the proof of that rigidity lemma can be extended with little work to varieties of general type of higher dimension, although I did not write the details.