# When is the etale cohomology of $\mathrm{Sym}^n(X)$ isomorphic to the $\Sigma_n$-invariants in the étale cohomology of $X^n$?

Suppose $$X$$ is a smooth projective variety defined over an arbitrary algebraically closed field $$k$$, and consider the action of $$\Sigma_n$$ on the $$n$$-fold product $$X^n$$. Is it true that $$H_{\acute{e}t}^i(\mathrm{Sym}^n(X),\mathbb{Q}_\ell)\cong H_{\acute{e}t}^i(X^n,\mathbb{Q}_\ell)^{\Sigma_n}$$? In particular, what happens in the case where $$\operatorname{char} k=p>0$$?

In Grothendieck's Toh\^oku paper Sec. 5.2, he determines sufficient conditions to ensure that, for a topological space $$X$$ with a finite group $$G$$ acting on it (not necessarily faithfully), $$H^i(X/G,\mathcal{A})\cong H^i(X,\mathcal{A})^G$$ for a sheaf $$\mathcal{A}$$ (Cor. to Prop. 5.2.3). In characteristic zero, comparison theorems allow me to appeal to this result. In positive characteristic, if the variety lifts to characteristic zero, then I can make the same argument, but it seems like there ought to be a direct proof of this fact.

I am particularly interested in when $$X$$ is a surface, but would be happy to know of any general results (with references) similar to Grothendieck's result above.

• In general, there is a spectral sequence with $E_2$-page $\mathrm{H}_i(G;\mathrm{H}^i_\mathrm{et}(X,\mathcal{F}))$ abutting to $\mathrm{H}^i_\mathrm{et}(X/G,\mathcal{F})$. If $p>n$, then $|\Sigma_n|=n!$ is invertible in $k$, so the higher cohomology groups of $\Sigma_n$ vanish, and you get the desired result. – skd Sep 26 '18 at 15:48
• (That should be group cohomology, not group homology; sorry for the typo.) – skd Sep 26 '18 at 16:30
• @skd Your answer is basically right, but note that you have confused $p$, the characteristic of the ground field, with $\ell$, the characteristic of the coefficient group. – David E Speyer Sep 26 '18 at 18:02
• @skd I think a slightly more precise statement is that $E_2^{pq} = H^p(G,H^q_{et}(X,\mathcal F))$ converges to $H^{p+q}_{et}([X/G],\mathcal F)$, i.e. the spectral sequence will in general converge to the cohomology of the stack quotient. When $\mathcal F = \mathbb Q_\ell$, (or more generally a ring where the order of $G$ is invertible) the cohomology of the quotient stack coincides with the cohomology of the scheme-theoretic quotient $X/G$ (the coarse moduli space). – Dan Petersen Sep 26 '18 at 21:24
• @DanPetersen thanks for the helpful comment. Do you know of a reference that discusses the spectral sequence in the generality you give? I know it is in Milne's Etale Cohomology (III.2.20) if the action of $G$ on $X$ is free. What if the action is not free? – Sarah Frei Sep 26 '18 at 23:30

One can give a spectral-sequence free argument. Let $$X$$ be an algebraic variety and $$G$$ a finite group acting on $$X$$, acting freely on a dense open subset. Let us say that $$X$$ is quasi-projective so that $$X/G$$ exists as a scheme (rather than an algebraic space), but this is not essential. We have $$\pi \colon X \to X/G$$ which induces a pullback map $$\pi^\ast \colon H^\ast(X/G) \to H^\ast(X)$$ as well as a trace map $$\pi_\ast \colon H^\ast(X) \to H^\ast(X/G)$$. The trace map exists in great generality; if $$X$$ and $$X/G$$ are both smooth the trace can be defined simply as the Poincaré dual of the pushforward map in étale homology. This is not true in your case but your spaces are still rational homology manifolds so that Poincaré duality holds with $$\mathbb Q_\ell$$-coefficients. In any case, what makes it all work is that $$\pi_\ast \pi^\ast$$ is multiplication by $$\vert G\vert$$ on $$H^\ast(X/G)$$, and that $$\pi^\ast \pi_\ast$$ is the map $$x \mapsto \sum_{g \in G} g \cdot x$$ on $$H^\ast(X)$$. In particular, if $$\vert G\vert$$ is invertible in the coefficients of the cohomology then $$\pi_\ast \pi^\ast$$ is invertible and $$\pi^\ast \pi_\ast$$ is projection onto the $$G$$-invariants.
The following is a particular case of (SGA 4.3, XVII Th. 5.5.21) : Let $$X$$ be a quasi-projective scheme over an algebraically closed field $$k$$. Then for any $$n \geq 0$$ and any $$r \geq 1$$ we have $$R \Gamma_c(\mathrm{Sym}_{k}^n(X), \mathbb{Z}/r\mathbb{Z}) = L \Gamma^n ( R\Gamma_c(X, \mathbb{Z}/r\mathbb{Z})).$$ The functor $$L \Gamma^n$$ is the left derived functor of the non-additive functor $$\Gamma^n$$, which coincides on flat modules with the symmetric tensor'' functor. The $$R \Gamma_c$$ denotes higher direct image with compact supports.