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.

  • 7
    $\begingroup$ Perhaps you can use the fact that $G$ acts on the Rees algebra $R(I)=\bigoplus_{m\ge 0}I_Z^m$ and that $\tilde{X}=Proj R(I)$.. $\endgroup$ – J.C. Ottem Feb 25 '13 at 20:21
  • $\begingroup$ 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. $\endgroup$ – Andrew Stout Feb 25 '13 at 20:23
  • $\begingroup$ edit: all $z \in Z$ $\endgroup$ – Andrew Stout Feb 25 '13 at 20:23
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    $\begingroup$ 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. $\endgroup$ – user5117 Feb 25 '13 at 20:43

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.

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    $\begingroup$ It works in fact for any group. Not necessarily algebraic... $\endgroup$ – Jérémy Blanc Feb 26 '13 at 23:53
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    $\begingroup$ I apologize since the post is quite old, but I don't understand some passages: which do you need to say that the inverse image of $Z$ is $G\times Z$ (i.e., where do you use it?)? isn't there a typo on the sentence "the morphism $G\times \tilde{X}\to \tilde{X}$ gives an action of $G$ to $X$: shouldn't there be written an action to $G$ on $\tilde{X}$? Thanks in advance for the patience. $\endgroup$ – Baobab Sep 30 '20 at 15:38

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