5
$\begingroup$

Let us have an algebraic action by a finite group G on a complex projective variety $X=\bigcup\limits_{i=1}^N X_i$, whose irreducible components $X_i$ are all smooth and of the same dimension $d$, and whose singular cohomology $H^*(X)$ is generated by algebraic cycles (hence supported in even degrees). Note that variety X itself is NOT smooth (unless N=1).

Is there a fixed-points type theorem that would give us that $$\text{rank}(H^*(X^G))=\text{rank}(H^*(X))?$$

NB I have EDITED the question.

$\endgroup$
4
  • $\begingroup$ mathoverflow.net/questions/171525/… $\endgroup$
    – SWS
    Apr 16 '19 at 22:14
  • $\begingroup$ I am aware that it 1,2,3 holds when the whole variety is smooth. Here I am just saying that its components are smooth, not the variety itself - hence the reason to challenge those statements. $\endgroup$
    – Filip92
    Apr 16 '19 at 22:34
  • 4
    $\begingroup$ @Filip92: 1. does not hold even for smooth projective varieties: a translation on an abelian variety has no fixed points. $\endgroup$ Apr 16 '19 at 22:40
  • $\begingroup$ @SándorKovács I completely agree. However, I was having in mind examples with non-zero Euler characteristic (as written in my edited question). $\endgroup$
    – Filip92
    May 2 '19 at 0:24
8
$\begingroup$

Setup

Let $G$ be a finite group acting on a smooth projective variety $X$, and let $$ \rho: G \times X \to X $$ be the action morphism. For any $g \in G$ let $\rho_g$ denote the composition $$ X \simeq \{g\} \times X \subset G \times X \xrightarrow{\rho} X $$

The fixed point locus $X^G \subset X$ is a closed subscheme

(possibly empty):

Edit incorporating Peter McNamara's comment: $X^G$ is always closed when $G$ is an affine algebraic group (link in comments).

When $G$ is finite there's an elementary argument: note that for any $g \in G$, the $g$-fixed points $X^g$ fit into the cartesian diagram $\require{AMScd}$ \begin{CD} X^g @>>> X \\ @VVV @V \mathrm{id} \times \rho_g VV \\ X @> \Delta >> X \times X \end{CD}

Since $\Delta$ (and $\mathrm{id} \times \rho_{g}$ for that matter) are closed immersions and closed immersions are compatible with base change, $X^{g} \to X$ is a closed immersion. Since $X^{G} = \bigcap_{g \in G} X^{g}$, $X^{G}$ is also a closed subscheme of $X$.

Remark: This heavily relies on the fact that $G$ is a finite (or at least discrete) group, which is fine since that was the question, but I'd be interested in whether/when/how-to-show $X^{G}$ is closed in the case where $G$ is, say, a positive dimensional linearly reductive affine group scheme acting on $X$.

where $\Delta$ is the diagonal. As Sándor pointed out in a comment it definitely can happen that $X^{G}$ is empty, e.g. if $X$ is an abelian variety and $G \subset X$ is a finite subgroup acting by translations.

The fixed point locus is smooth

This is more difficult to prove -- an analytic argument due to Cartan appears in Algebraic geometry and topology, and an alternative approach is Luna's étale slice theorem.

The idea behind both approaches is to get local enough that the local geometry at a fixed point $x \in X^G$ is modeled by the tangent space $T_{X,x}$ with it's linear $G$--action (if $\rho_{g}: X \to X$ is the action of $g \in G$ on $X$, then $g$ acts on $T_{X,x}$ via $d \rho_{g}: T_{X, x} \to T_{X, \rho_{g}(x)} = T_{X, x}$). One then proves that $T_{X^{G}, x} = T_{X,x}^{G}$, the $G$-invariant subspace.

I won't attempt to go into further detail.

The fixed locus can have arbitrary dimension:

Let $\rho: G \times V \to V$ be a linear representation of a finite group $G$. Then $\mathbb{P}(V)$ is a smooth projective variety with an induced $G$-action $$ \bar{\rho}: G \times \mathbb{P}(V) \to \mathbb{P}(V), \, \, \text{ where } \bar{\rho}[v] = [\rho(v)] $$ Observe that for a non-0 vector $v \in V$, $[v] \in \mathbb{P}(V)^{G}$ if and only if for each $g \in G$ there's a scalar $\lambda_{g} \in \mathbb{C}^{\times}$ so that $\rho_{g}(v) = \lambda_{g} v$. Evidently this occurs if and only if $v$ lies in the isotypical summand of a character (1-dimensional representation) $L$ of $V$.

This observation together with some representation theory of finite groups yields examples where $\dim X^{G}$ takes arbitrary values in $0, \dots, \dim X$.

Criteria for $X^{G}$ to be non-empty

There are various ``fixed point theorems'' which guarantee the existence of, well, fixed points. See for instance the Lefschetz fixed point theorem and the holomorphic Lefschetz fixed point theorem. The later is especially powerful and shows for instance that if $$ H^{i}(X, \mathscr{O}_{X}) = 0 \,\, \text{ for } i > 0 $$ (for example, if $X$ is Fano or even just rationally connected) then $X^{G} \neq \emptyset$.

$\endgroup$
3
  • $\begingroup$ Your answer is in great details, thank you for that. However, it is still only for a smooth projective variety, whereas my setup includes (and in general is) rather singular. $\endgroup$
    – Filip92
    May 2 '19 at 0:01
  • 2
    $\begingroup$ The fixed point locus X^G is always a closed subscheme when G is affine. See mathoverflow.net/a/14876 (which ultimately references Prop A.8.10 of Conrad-Gabber-Prasad's book "Pseudo-Reductive Groups"). $\endgroup$ May 2 '19 at 1:17
  • 1
    $\begingroup$ @Filip92 at least if $G$ is finite (and maybe more generally, don't know right off the top of my head) you could lift the action of $G$ to the normalization $\coprod_i X_i \to \bigcup_i X_i = X$. Based on your assumptions the normalization is smooth, so anything we know in the smooth projective case would apply to the $G$-action on $\coprod_i X_i$. The trick would be to see what (if anything) fixed points/cohomology calculations on the normalization tell us about those of $X$ $\endgroup$
    – cgodfrey
    May 2 '19 at 7:12
6
$\begingroup$

The answer to the edited question is no. Take a double covering $\pi :X\rightarrow \mathbb{P}^2$ branched along a smooth quartic curve $C$ (so $X$ is a Del Pezzo surface), and $G=\langle \sigma \rangle$, where $\sigma $ is the involution such that $\pi \circ\sigma =\pi $. Then $\operatorname{rk} H^*(X)=10$, but $\operatorname{rk} H^*(X^G)=\operatorname{rk} H^*(C)=8$.

$\endgroup$
2
  • $\begingroup$ Thanks! Maybe I am missing something but, in my equality, it is rather $\text{rk }(H^*(X))$ instead of $\text{rk }(H^*(X)^G)$ which you've written. $\endgroup$
    – Filip92
    May 2 '19 at 11:07
  • 2
    $\begingroup$ Oh, sorry. But my example still works, since $\operatorname{rk}H^*(X)=10 $. I have edited my answer. $\endgroup$
    – abx
    May 2 '19 at 11:44
1
$\begingroup$

There is actually a simple answer. Pick X to be two $\mathbb{C}P^1$'s intersecting transversally, and a $\mathbb{Z}/2$ action that swaps two spheres. Then $\text{rk } H^*(X)=3$ whereas $\text{rk } H^*(X^G) = 1.$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.