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.



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


Unfortunately I couldn't tell you a citable reference off hand. If I were to use this in a paper I wouldn't give a reference for this - sorry, I know that's not a very helpful thing to say.


