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

Let $X$ be a smooth complex algebraic variety, and $\varphi: \Gamma\curvearrowright X$ an action (by automorphisms) of a finite group $\Gamma$ on $X$.

Can we say that each irreducible component of the fix point set $X^{\Gamma}$ is smooth?

Here $X^{\Gamma}:=\{ x\in X\; | \; g(x):=\varphi_g (x)=x\;\forall g\in\Gamma\; \}$.

Remark: without loss of generality we can assume $X$ affine, since any $\Gamma$-variety can be covered by affine $\Gamma$-invariant open subsets.

share|improve this question
6  
I think this is a consequence of Cartan Lemma. In a neighborhood of a fixed point $p$ the action can be linearized, so locally analytically we may assume $X=\mathbb{C}^n$, $p=0$ and $G \subset \rm{GL}(n, \, \mathbb{C})$. Then in a neighborhood of $0$ the fixed locus is given by a union of linear subspaces and the claim follows. –  Francesco Polizzi Jun 10 at 19:08
1  
If $H$ is a linearly reductive affine group scheme over any field $k$ (e.g., any affine algebraic group with reductive identity component in char. 0, which includes any finite group, as well as any group of multiplicative type in any characteristic) and $Y$ is a separated $k$-scheme of equipped with an $H$-action then the functor $Y^H$ of $H$-fixed points is represented by a closed subscheme of $Y$ (that much doesn't use the linear reductivity) which moreover is smooth whenever $Y$ is smooth. In particular, irreducible components of $Y^H$ are always pairwise disjoint. –  user52824 Jun 11 at 12:27
    
@user52824: could you please also give a reference? –  Qfwfq Jun 11 at 19:03
1  
@Qfwfq: The case of finite groups with order invertible on the base is in Edixhoven's 1992 paper "Neron models and tame ramification" in Compositio 81 (see 3.1--3.4), and the general case is Prop. A.8.11 in the book "Pseudo-reductive groups" (which gives a proof in a more general setting over rings that generalizes Edixhoven's Prop. 3.4; various aspects simplify when working over a field). –  user52824 Jun 11 at 19:35
    
@user52824: thank you! –  Qfwfq Jun 11 at 22:48

1 Answer 1

up vote 8 down vote accepted

Yes. The reason is that locally around a fixed point, the action linearizes, i.e. is analytically equivalent to a linear action of $\Gamma $ onto a vector space -- so the fixed locus is locally isomorphic to a linear space. This fact, which follows easily from an averaging process, goes back (at least) to H. Cartan, in Algebraic geometry and topology (a symposium in honor of S. Lefschetz), pp. 90-102, Princeton University Press, 1957.

share|improve this answer
    
Thank you for the answer! –  Qfwfq Jun 10 at 19:19

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.