Timeline for When is $\mathrm{Aut}(X)\times \mathrm{Aut}(Y)$ of finite index in $\mathrm{Aut}(X\times Y)$?
Current License: CC BY-SA 3.0
12 events
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| Sep 4, 2013 at 5:00 | comment | added | Jérémy Blanc | 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. | |
| Sep 4, 2013 at 0:58 | history | edited | Ricardo Andrade | CC BY-SA 3.0 |
since answer was edited to impose uniform notation, I made the notation a bit more uniform (I approved the previous edit without noticing the incongruencies so I thought I should correct them)
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| S Sep 4, 2013 at 0:49 | history | suggested | José Hdz. Stgo. | CC BY-SA 3.0 |
homogenized notation
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| Sep 4, 2013 at 0:35 | review | Suggested edits | |||
| S Sep 4, 2013 at 0:49 | |||||
| Sep 3, 2013 at 15:50 | comment | added | Darius Math | @Jérémy Blanc . What if both $X$ and $Y$ are projective? Are then these two groups equal? | |
| Sep 3, 2013 at 15:31 | comment | added | Jérémy Blanc | 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. | |
| Sep 3, 2013 at 13:42 | comment | added | YCor | nb I should switch the words horizontal and vertical in the above sentence. | |
| Sep 3, 2013 at 13:11 | comment | added | YCor | I don't claim that there must be no nonconstant maps in both directions, but I claim that such a hypothesis must be done in Joe's argument. Indeed, as in the example I gave, you can expect an automorphism to preserve the horizontal foliation $(\{x\}\times Y)_{x\in X}$ but not the vertical foliation. | |
| Sep 3, 2013 at 12:30 | comment | added | Darius Math | 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? | |
| Sep 3, 2013 at 12:06 | comment | added | YCor | 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 $X\to Y\to X$. | |
| Sep 3, 2013 at 11:25 | comment | added | Darius Math | 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?) | |
| Sep 3, 2013 at 11:16 | history | answered | Joe Silverman | CC BY-SA 3.0 |