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?


1 Answer 1


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, 2021 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, 2021 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, 2021 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, 2021 at 17:45
  • $\begingroup$ Really thank you! Sorry with stacks I'm never sure if I'm missing something.. $\endgroup$ May 19, 2021 at 12:50

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.