Why do only fixed points contribute to the Euler characteristic?

Question

Let $G$ be an algebraic group with zero Euler characteristic, acting on a variety $X$ (over $\mathbb C$). I read some time ago that then the Euler characteristic of $X$ can be computed as $$\chi(X)=\chi(X^G),$$ where $X^G$ is the locus of fixed points. What is a proof of this fact?

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Attempt: one can decompose $X$ as a disjoint union of $X^G$ and the set $$S=\coprod G.x,$$ the union being on nontrivial orbits $G.x\neq\{x\}$. If I show that $\chi(G.x)=0$ for every orbit appearing in the decomposition of $S$, we are done. But I can't. I initially thought that each such orbit is the base of a locally trivial fibration $$G\times\{x\}\to G.x$$ with fiber $G_x$, the stabilizer of the point $x$ (The map being just the restriction of the action $G\times X\to X$).

First, I am not sure this is true in the context of actions by algebraic groups. Second: even if it were true, it would just tell me (using that $\chi(G)=0$) that $$0=\chi(G)=\chi(G.x)\chi(\textrm{fiber})=\chi(G.x)\chi(G_x),$$ but this does not quite allow me to concude that $\chi(G.x)=0$, because $\chi(G_x)$ might be zero as well.

Can anyone help me figuring out how to fix the argument, or provide a proof?

Thank you!

Answer 1

See [BB], in particular Corollary 2:

Suppose $G=\mathbb{G}_m$ acts on an algebraic scheme $X.$ Then $\chi(X) = \chi (X^G).$

I'm not certain what is precisely meant by an algebraic scheme in that paper, but it probably includes your notion of a variety. Bialynicki-Birula's result is for any algebraically closed field.

If you ever yearn for the same result for any reduced and irreducible scheme over an arbitrary field, see Theorem 1.3 of [HU].

You may also be interested in this blog post by Brendan Pawlowski, which gives a proof sketch of the following:

If a torus $T$ acts smoothly on a compact smooth manifold $M$ with finitely many fixed points, then $|M^T| = \chi(M).$

What I wrote above are the most general results that I am aware of. More usually (in my particular experience) one needs to use information specific to the situation at hand to push through the argument in some way. The end of user160609's answer sketches one approach.

Here are some facts that may help you in a particular situation you encounter.

See section 3.34 of [HA] for an introduction to cohomology with compact supports. The main things you need to know about it are:

1. If $X$ is a compact space then $H_c^{\bullet}(X) = H^{\bullet}(X).$

2. $H^i_c(\mathbb{R}^n)= 0$ if $i\neq n$ and has rank $1$ if $i=n.$

3. Cohomology with compact supports is not homotopy invariant.

4. If $X$ is homeomorphic to $Y$ then $H^{\bullet}_c(X) \cong H^{\bullet}_c(Y).$

5. Let $U\subseteq X$ be an open subset and $C= X\setminus U.$ There is a long exact sequence: $$ \cdots \to H^i_c(U) \to H^i_c (X) \to H^i_c (C) \to H^{i+1}_c (U) \to \cdots$$

6. For complex algebraic varieties, $\chi_c(X) = \chi(X).$

Proof: If $X$ is a compact space then every singular simplex trivially satisfies the extra condition of compact support, so 1) follows. Property 2) can be found in Section 3.34 of [HA]. Since $\mathbb{R}$ is homotopy equivalent to a point, this shows that cohomology with compact supports is not homotopy invariant. Recall that $X$ is homotopy equivalent to $Y$ if there are continuous maps $f:X\to Y, g:Y\to X$ such that $f\circ g$ is homotopic to the identity on $Y$ and $g\circ f$ is homotopic to the identity on $X.$ The root of the problem is that given a continuous map $X\to Y,$ the induced map on their cohomology does not always restrict to their cohomology with compact support (an example of this is the constant map from $\mathbb{R}^n$ to $\{ 0 \}$). This would be true if the preimage of every compact set in $Y$ was guaranteed to be compact, which is the case when the map is a homeomorphism. The rest of the usual proof that singular cohomology is homotopy invariant goes through and property 4) is still true.

Property 5) is proved is the special case of 7.8 (page 186) of [IV] where the chain map is the identity.

Property 6) is proved on pages 141-142 of [FU].

If $Y$ is locally closed in $X$ then $\chi_c(X) = \chi_c(X\setminus Y) + \chi_c(Y).$

Proof : Suppose $U$ is an open subset of $X$ and $C=X\setminus U.$ Using exactness and the Rank-Nullity theorem at every point in the long exact sequence for cohomology with compact supports (property 5 above) gives $$\sum_{i=0}^{\infty} (-1)^i \dim H^i_c(X) = \sum_{i=0}^{\infty} (-1)^i \left(\dim H^i_c (C) +\dim H^i_c (U)\right),$$ which gives $\chi_c(X) = \chi_c(U) + \chi_c(C).$ We repeatedly apply this property. The closure $\overline{Y}$ of $Y$ is closed, so $\chi_c(X) = \chi_c(X\setminus \overline{Y}) + \chi_c(\overline{Y}).$ Since $Y$ is locally closed (open in its closure) we have $\chi_c(\overline{Y}) = \chi_c(\overline{Y}\setminus Y) + \chi_c(Y).$ Since $X\setminus \overline{Y}$ is open in $X\setminus Y$ we have $\chi_c(X\setminus Y) = \chi_c( X\setminus \overline{Y} ) + \chi_c( \overline{Y}\setminus Y ).$ Combining these gives the result.

Here's an example of how these facts help us to compute the Euler Characteristic. Even if you haven't studied toric varieties, you'll be able to see what type of specific facts we need to prove to push the computation through.

If $X=X(\Delta)$ is a toric variety then $\chi_c(X)$ is the number of nondegenerate cones in $\Delta.$

Proof : First we index the cones of $\Delta$ as $\sigma_1, \sigma_2, \ldots, \sigma_{|\Delta|}$ such that their dimensions are nondecreasing. Let $\mathcal{O}_i$ be the orbit associated to the cone $\sigma_i.$ A fact about toric varieties is that we have a decomposition of $X=X(\Delta)$ into finitely many disjoint locally closed subvarieties: $\displaystyle X= \bigsqcup_{i=1}^{|\Delta|} \mathcal{O}_{i}.$ A theorem about the orbits of the torus action further implies that $\overline{\mathcal{O}_{k+1}}$ is disjoint from $\bigcup_{i=1}^{k} \mathcal{O}_{i}$ since $\sigma_{k+1}$ can not be a face of any of the cones that come before in the sequence. This implies that $\overline{\mathcal{O}_{k+1}},$ the closure of $\mathcal{O}_{k+1}$ in $X,$ is also the closure of $\mathcal{O}_{k+1}$ in the subspace $X\setminus \bigcup_{i=1}^k \mathcal{O}_{i}.$ Therefore $\mathcal{O}_{k+1}$ is still locally closed even when considered as a subset of $X\setminus \bigcup_{i=1}^k \mathcal{O}_{\sigma_i}.$ Applying the result that if $Y$ is locally closed in $X$ then $\chi_c(X) = \chi_c(X\setminus Y)+\chi_c(Y)$ repeatedly gives

$$ \chi_c(X) = \chi_c( \mathcal{O}_1) + \chi_c (X\setminus \mathcal{O}_1) = \sum_{i=1}^2 \chi_c( \mathcal{O}_i) + \chi_c (X\setminus \bigcup_{i=1}^2 \mathcal{O}_i ) = \ \cdots \ = \sum_{i=1}^{|\Delta|} \chi_c(\mathcal{O}_i).$$

Another fact about the orbits is that $\mathcal{O}_i \cong (\mathbb{C}^*)^{\operatorname{codim}(\sigma_i)}.$ Using the facts about cohomology with compact supoorts and the Kunneth formula lets you compute $\chi_c( (\mathbb{C}^*)^n)$ which finishes the computation.

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[BB] A. Bialynicki-Birula, On fixed point schemes of actions of multiplicative and additive groups. Topology 12 (1973), 99–103.

[FU] W. Fulton, Introduction to Toric Varieties, Annals of Mathematics Studies, vol. 131, Princeton University Press, Princeton, New Jersey, 1993, The William H. Roever Lectures in Geometry.

[HA] A. Hatcher, Algebraic topology. URL: http://www.math.cornell.edu/~hatcher/AT/AT.pdf

[HU] W. Hu, On additive invariants of actions of additive and multiplicative groups. arxiv/0912.0563

[IV] B. Iversen, Cohomology of Sheaves, Universitext Series, Springer-Verlag, Berlin, 1986.

Answer 2

The basic idea is that if the circle $S^1$ acts on a space, then the orbit of any non-fixed point will be a copy of the circle, so the space decomposes as a union of circles, and the fixed points.

Since circles have Euler char. $0$, the Euler char. will then be equal to just the Euler char. of the fixed point set.

Here the intuition being used is that the Euler char. of a union is the sum of the Euler chars. Of course this is not naively correct for infinite unions, so to make the above argument rigorous, one has to work a little harder (and possibly, even probably, add some hypotheses on the action and on the space being acted on) to make the above proof idea into a rigorous argument. But hopefully it at least adds to the plausibility of the statement, at least in the case of a circle (of of a torus, in the algebraic setting).

For other groups the idea is similar; one can try to reduce to the case of a circle, for example, by considering maximal tori or related objects, or find other, related, variants of the above argument.

Added later: Let's suppose $X$ is a variety, and $G = \mathbb G_m$. If a point $x$ is not fixed, then the stabilizer of $x$ is a finite subgroup of $\mathbb G_m$; thus equals $\mu_n$ for some $n$. Now if we write $X_n$ to denote the set of $x$ with stabilizer $\mu_n$, then $X_n$ is locally closed, and so (I think) the $X_n$, together with $X^G$, give a decomposition of $X$ into finitely many locally closed pieces.

Now each $X_n$ has a free action of $G/H_n$, with is again a copy of $\mathbb G_m$, so one is reduced to proving that if $\mathbb G_m$ acts freely on a variety $X$, then $X$ has Euler char. zero. One idea would be to prove that $X$ locally admits a slice to its orbits, so that locally $ X = G \times T$ for some slice $T$ (and then multiplicativity of Euler chars. gives the result). There are results of this kind in some contexts (e.g. Luna's slice theorem).