MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $\mathrm{Aut}(X)$ denote the group of biholomorphic autmorphisms of the (non-compact) complex manifold $X$. If $X$ and $Y$ are two (non-compact) 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 non-trivial 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.

share|cite|improve this question
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. – Ben McKay Sep 3 '13 at 12:34

Easiest case that I know is when $X$ and $Y$ are non-isogenous 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 non-constant 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 non-constant 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.)

share|cite|improve this answer
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?) – Darius Math Sep 3 '13 at 11:25
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 $X\to Y\to X$. – YCor Sep 3 '13 at 12:06
Yves. What is your argument that there must be no non-constant maps in both directions? if only we have no non-constant maps $X\rightarrow Y$ which problem can arise? – Darius Math Sep 3 '13 at 12:30
Yves is right, take $X=\mathbb{P}^1$, $Y=\mathbb{A}^1$, there is no non-constant map $X\to Y$ but $Aut(X\times Y)$ is of infinite dimension, and $Aut(X)\times Aut(Y)$ is only of finite dimension. – Jérémy Blanc Sep 3 '13 at 15:31
The question is independent of the fact that $X$ and $Y$ are projective or not. If there is no non-constant 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. – 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 non-degenerate Finsler metric on these examples. The product structure will then be seen pointwise in the Finsler norm, so any holomorphic bi-automorphism 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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.