Let $\mathrm{Aut}(X)$ denote the group of biholomorphic autmorphisms of the (noncompact) complex manifold $X$. If $X$ and $Y$ are two (noncompact) complex manifolds, then $\mathrm{Aut}(X)$ and $\mathrm{Aut}(Y)$ and $\mathrm{Aut}(X)\times \mathrm{Aut}(Y)$ are subgroups of $\mathrm{Aut}(X\times Y)$. My question is: Are there reasonable conditions, under which $\mathrm{Aut}(X)\times \mathrm{Aut}(Y)$ is a subgroup of finite index in $\mathrm{Aut}(X\times Y)$? Of course I am interested in the nontrivial cases, i.e. when $\mathrm{Aut}(X\times Y)$ itself is not a finite group. You may also assume that $X$ and $Y$ are algebraic manifolds.

1$\begingroup$ It might be useful to look for entire curves. If $X$ is a ball and $Y=\mathbb{C}^n$, then all entire curves in $X \times Y$ lie in $x_0 \times Y$ fibers, and automorphisms must take entire curves to entire curves. The same idea for rational curves in compact complex manifolds. $\endgroup$ – Ben McKay Sep 3 '13 at 12:34
Easiest case that I know is when $X$ and $Y$ are nonisogenous abelian varieties. If you want infinite automorphism group $\DeclareMathOperator{\Aut}{Aut} \Aut(X\times Y)$, you'll need at least one of them to have complex multiplication with endomorphism rings $\DeclareMathOperator{\End}{End} \End(X)$ or $\End(Y)$ having infinite unit group, but that's easy enough to arrange.
More generally, won't it be true that if $\sigma\in\Aut(X\times Y)$ does not come from $\Aut(X)\times\Aut(Y)$, then you get a nonconstant map $X\to Y$ via $X\xrightarrow{i\times y_0} X\times Y\xrightarrow{\sigma}X\times Y\xrightarrow{p_2}Y$? So if there are no nonconstant maps from $X$ to $Y$, you'll have $\Aut(X)\times\Aut(Y)=\Aut(X\times Y)$. (I'm pretty sure that this is right if $X$ and $Y$ are projective, not entirely sure about the noncompact case.)

$\begingroup$ Thanks! by the last sentence do you mean that $Aut(X\times Y)=Aut(X)\times Aut(Y)$ always holds? (at least for projectve varieties?) $\endgroup$ – Darius Math Sep 3 '13 at 11:25

3$\begingroup$ No, the last argument assumes there are no nonconstant maps $X\to Y$. If $X,Y$ are abelian varieties and $f:X\to Y$ is a nonconstant regular map, then $(x,y)\to (x,y+f(x))$ is certainly not a product of automorphisms. Also I think that in the argument you really need that there are no nonconstant regular maps in both directions $X\to Y\to X$. $\endgroup$ – YCor Sep 3 '13 at 12:06

1$\begingroup$ Yves. What is your argument that there must be no nonconstant maps in both directions? if only we have no nonconstant maps $X\rightarrow Y$ which problem can arise? $\endgroup$ – Darius Math Sep 3 '13 at 12:30

1$\begingroup$ Yves is right, take $X=\mathbb{P}^1$, $Y=\mathbb{A}^1$, there is no nonconstant map $X\to Y$ but $Aut(X\times Y)$ is of infinite dimension, and $Aut(X)\times Aut(Y)$ is only of finite dimension. $\endgroup$ – Jérémy Blanc Sep 3 '13 at 15:31

1$\begingroup$ The question is independent of the fact that $X$ and $Y$ are projective or not. If there is no nonconstant morphism $X\to Y$ and $Y\to X$, then $Aut(X\times Y)=Aut(X)\times Aut(Y)$, since both fibrations are invariant. However, one direction is not enough: let $X$ be an elliptic curve and $Y$ a projective curve with a morphism $Y\to X$, then $Aut(X\times Y)\not=Aut(X)\times Aut(Y)$ because you can act by translation, given by the morphism. $\endgroup$ – Jérémy Blanc Sep 4 '13 at 5:00
If $X$ or $Y$ is $\mathbb{D}^n$ (the unit complex ball) or $\mathcal{T}^n$ (Teichmuller space), then $Aut(X\times Y)$ will be finite index in $Aut(X)\times Aut(Y)$. This follows by considering the Kobayashi (pseudo)metric. On a hyperbolic domain, the Kobayashi metric is a Finsler metric. On a product $X\times Y$, the Kobayashi metric is pointwise the maximum of the two metrics by a theorem of Royden. This metric is a nondegenerate Finsler metric on these examples. The product structure will then be seen pointwise in the Finsler norm, so any holomorphic biautomorphism will have to locally preserve the product structure (this follows from Royden's theorem for the Teichmuller metric case, which is the Kobayashi metric of Teichmuller space). I think the same will hold for $X$ or $Y$ products of these metrics.