Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Hi friends,

Let $X$ be a projective variety over a field $k$ of characteristic zero. Assume that $X$ comes with the action of a finite group $G$. Now let $Z$ be a closed subvariety stable under the action of $G$. Let $\pi: \tilde{X} \to X$ be the blow-up of $X$ along $Z$.

How can one extend the action of $G$ from $X$ to $\tilde{X}$?

Any help will be appreciated.

share|improve this question
7  
Perhaps you can use the fact that $G$ acts on the Rees algebra $R(I)=\bigoplus_{m\ge 0}I_Z^m$ and that $\tilde{X}=Proj R(I)$.. –  J.C. Ottem Feb 25 '13 at 20:21
    
isn't a blow-up a proper birational morphism? meaning that on $\bar{X} - \pi^{-1}(Z)$ the group action must be induced canonically via the isomorphism $\bar{X} - \pi^{-1}(Z) \cong X $. Thus, you could extend by acting on fibers in $Z$ $g\pi^{-1}(z) = \pi^{-1}(gz)$ for all $g\in G$ and all $Z\in Z$. well, this is a start anyway. –  Andrew Stout Feb 25 '13 at 20:23
    
edit: all $z \in Z$ –  Andrew Stout Feb 25 '13 at 20:23
1  
I guess it depends what kind of answer you want. J.C. Ottem's answer is the correct one, but more simplemindedly, and geometrically, if Z is smooth: Think of the exceptional divisor E as the projectivised normal bundle of Z in X. The differential of the map corresponding to $g \in G$ gives a linear aut. of the normal bundle, which then descends to E. –  Artie Prendergast-Smith Feb 25 '13 at 20:43

1 Answer 1

Yes you can extend the action. One can prove this using the universal property of blow-ups (see Hartshorne Corollary II.7.15).

As $Z$ is invariant under the action of $G$, the inverse image of $Z$ with respect to the morphism $G \times X \to X$ is $G \times Z$. Therefore on applying the universal property of blow-ups to this morphism we obtain a morphism $G \times \widetilde{X} \to \widetilde{X}$. Now, by assumption this morphism satisifies the identities $$(gh)x = g(hx), \quad ex = x$$ for all $x$ in $\widetilde{X}\setminus E$, where $E$ denotes the exceptional divisor of the blow-up. However since any two morphisms which are equal on an open dense subset must be equal on the whole space, we see that these identities hold for all $x$ in $\widetilde{X}$, i.e. the morphism $G \times \widetilde{X} \to \widetilde{X}$ gives an action of $G$ on $X$.

Note that in this argument we did not use the fact that $G$ was finite, it works for any algebraic group.

share|improve this answer
3  
It works in fact for any group. Not necessarily algebraic... –  Jérémy Blanc Feb 26 '13 at 23:53

Your Answer

 
discard

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.