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Hello!

If a compact Lie group $K$ acts smoothly on a smooth manifold $M$, then the set $M^K$ of fixed points under this action is a smooth submanifold of $M$. This is proved for example in Duistermaat's book on Lie groups, using the Bochner Linearization Theorem.

I am interested in knowing if some variant of this statement is also true in algebraic geometry. In other words: can one describe a class of algebraic groups where the fixed points for an arbitrary action on a smooth variety is again smooth?

Thank you!

Hanno

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2 Answers 2

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Linearly reductive groups, that is, linear algebraic groups, say over a field, in which the functor of invariants from finite dimensional representations to vector spaces is exact. I am not sure this is in the literature in this generality, but it is not so hard to prove with a formal scheme argument.

Also, I would conjecture that this is optimal, that is, given a non linearly reductive algebraic group, one can find a smooth variety on which this acts, such that the fixed point locus is not smooth.

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    $\begingroup$ A reference is Birger Iversen, A fixed point formula for action of tori on algebraic varieties, Invent. Math., 1972. $\endgroup$ Mar 2, 2012 at 16:15
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    $\begingroup$ @DanPetersen's reference, clickably: Iversen - A fixed point formula for action of tori on algebraic varieties (MSN). Despite the title, some of the results handle arbitrary linearly reductive groups. (Note that linearly reductive = reductive in characteristic 0.) $\endgroup$
    – LSpice
    Feb 11, 2020 at 15:39
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Angelo's conjectural statement is proven in Fogarty, J.; Norman, P. "A fixed-point characterization of linearly reductive groups." MR0485896

(not enough reputation to post this as comment, sorry)

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    $\begingroup$ These sorts of answers that usefully supplement another answer (even if not directly addressing the OP's question) have typically been accepted by the community as worth keeping around, and so I think we can let it stand, if only for the sake of getting the poster to the 50-point threshold (so that next time he/she can comment instead). $\endgroup$
    – Todd Trimble
    Feb 11, 2020 at 20:09

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