Irreducible components of an algebraic stack

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 $$H^{2d}_c(U)=H^{2d}_c(X)$$.

• For general stacks,how can you choose such an open dense substack? May 18, 2021 at 6:06
• @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 $X$. May 18, 2021 at 8:58
• 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) May 18, 2021 at 9:09
• @Tommaso One can reason as follows: smooth morphisms are open, so if $V \subset W$ is an open subset then so is $p(V) \subset X$. May 18, 2021 at 17:45
• Really thank you! Sorry with stacks I'm never sure if I'm missing something.. May 19, 2021 at 12:50