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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.

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    $\begingroup$ 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. $\endgroup$
    – Francesco Polizzi
    Commented Jun 10, 2014 at 19:08
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    $\begingroup$ 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. $\endgroup$
    – user27920
    Commented Jun 11, 2014 at 12:27
  • $\begingroup$ @user52824: could you please also give a reference? $\endgroup$
    – Qfwfq
    Commented Jun 11, 2014 at 19:03
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    $\begingroup$ @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). $\endgroup$
    – user27920
    Commented Jun 11, 2014 at 19:35

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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.

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