Let $\mathcal{X}$ be an algebraic stack of finite type over a (separably closed) field $ k$. Let's say that $\mathcal{X}$ has finite dimension $d \in \mathbb{Z}$. Is it still true that the number of irreducible components of dimension $d$ of $\mathcal{X}$ is the dimension of $H^{2d}_c(\mathcal{X},\bar{\mathbb{Q}}_{\ell})$ as in the case of schemes? (Here I'm referring to the lisse etale cohomology.)

If this is not true in general, does it hold for a suitable class of stacks like the ones of the form $[X/G]$ where $X$ is a $k$ scheme?


I will say yes, although the level of generality is a bit scary and I hope I am not missing some stacky subtlety. I just took the standard argument for schemes, stared at it, and couldn't see anything that wouldn't work in the general case.

Claim: if $X$ is an equidimensional finite type algebraic stack, and $d=\dim(X)$, then the rank of $H^{2d}_c(X)$ equals the number of irreducible components of $X$. [All coefficients are $\mathbf Q_\ell$ from now on.]

Proof: Consider first the case $X$ smooth. In this case by Poincaré duality, $H^{2d}_c(X) \cong H^0(X)^\vee$. But in the smooth case we necessarily have connected components = irreducible components, so it's fine.

In general choose $U \subset X$ smooth open with complement $Z$ of strictly smaller dimension. Then we have the long exact sequence $$ \ldots \to H^{k-1}_c(Z) \to H^{k}_c(U) \to H^{k}_c(X) \to H^k_c(Z) \to \ldots $$ and we win: $H^k_c(Z)$ vanishes for $k>2(d-1)$ by dimension reasons, so $H^{2d}_c(U)=H^{2d}_c(X)$.

  • $\begingroup$ For general stacks,how can you choose such an open dense substack? $\endgroup$ May 18 at 6:06
  • $\begingroup$ @Tommaso Choose $p:W \to X$ a smooth surjection from a scheme, and let $V \subset W$ be open dense smooth. Then $p(V)$ is an open dense smooth substack of $X$. $\endgroup$ May 18 at 8:58
  • $\begingroup$ If you call $R=W\times_{\mathcal{X}}W$ the open substacks of $X$ shouldn't be given by the $R$ invariant open subschemes of $W$? How you pick $V$ to be $R$ invariant in general? (I'm probably missing something basic, sorry! I'm really new to the stacky language) $\endgroup$ May 18 at 9:09
  • $\begingroup$ @Tommaso One can reason as follows: smooth morphisms are open, so if $V \subset W$ is an open subset then so is $p(V) \subset X$. $\endgroup$ May 18 at 17:45
  • $\begingroup$ Really thank you! Sorry with stacks I'm never sure if I'm missing something.. $\endgroup$ May 19 at 12:50

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