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?



Irreducible components of an algebraic stack
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 $H^{2d}\_c(U)=H^{2d}\_c(X)$.


