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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.

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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.)

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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.

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