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I am trying to understand the 6-functor formalism of sheaves on topological stacks. As explained in this answer, there is a 6-functor formalism of sheaves for locally compact Hausdorff spaces, which can be extended to stacks (see Scholze's notes, in particular Thm 7.4 and Thm 4.20). Since the setup is quite abstract (to me at least), I have trouble actually computing things.

Concretely, I am interested in how to compute $f_!$ for $f \colon BG \to *$ with $G$ a (compact) Lie group, and in particular, I would like to know what $f_! \underline{\mathbb{Q}}$ is when $G = \text{SU}(2)$. It seems crucial to view $BG$ as a stack rather than the classifying space '$EG/G$' as $f_!$ does not exist for $f \colon EG/G \to *$.

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Such computations are indeed not formal; and virtually impossible if one tries to compute directly from the definitions.

The starting point here is that $f$ is actually cohomologically smooth. More precisely, $g: \ast\to BG$ is cohomologically smooth, as it pulls back to the smooth map $G\to \ast$; and thus $BG\to \ast$ is cohomologically smooth-locally cohomologically smooth (as the composite $\ast\to BG\to \ast$ is evidently smooth), and hence cohomologically smooth.

Thus, $f^!$ is $f^\ast$ up to twist. The next task is to identify the twist. Let's first figure out the degree. Note that $g^! f^!(1)=1$ and $g^! f^!(1) = g^\ast f^!(1)\otimes g^!(1)$. This means that the shifts in $f^!$ and $g^!$ must cancel. But $g^!$ shifts by $\mathrm{dim}(G)$ to the left, so $f^!(1)$ must sit in cohomological degree $\mathrm{dim}(G)$. In that degree, it gives a $1$-dimensional representation of $G$ on a $\mathbb Q$-vector space, but as $G$ is connected, it cannot act on it, so $f^!(1)\cong 1[-\mathrm{dim}(G)]$.

(In general, computing dualizing complexes is best done via degeneration to the normal cone, as I learned from Clausen. This works in virtually all situations.)

Now $f_!$ is the left adjoint of $f^!$. As $f^!\cong f^\ast[-\mathrm{dim}(G)]$, we learn that $f_!$ is $f_\natural[\mathrm{dim}(G)]$ where $f_\natural$ denotes the left adjoint of $f^\ast$. In classical language, $f_\natural$ computes group homology. Thus, $f_!\mathbb Q$ is the group homology of $G$ shifted by $\mathrm{dim}(G)$ to the left.

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  • $\begingroup$ I only know the group cohomology of SU(2), being $f_*Q=H^*(BSU(2),Q)=\bigoplus_{n\ge 0}Q[-4n]$. Would it be okay to say $f_!Q=Hom(f_!Q,Q)^\vee=Hom(Q,f^!Q)^\vee=Hom(Q,f^*Q[-\dim G])^\vee=Hom(Q,Q[-\dim G])^\vee=Hom(f^*Q,Q[-\dim G])^\vee=Hom(Q,f_*Q[-\dim G])^\vee=(f_*Q)^\vee[\dim G]$ resulting in $\bigoplus_{n\ge 0} Q[4n+3]$? I am not sure whether these duals work with infinite-dimensional vector spaces. Also, it seems weird that $f_!$ produces something unbounded to the left, is this a result of passing to stacks? $\endgroup$
    – jessetvogel
    Commented May 23 at 10:55
  • $\begingroup$ Yes, for formal reasons cohomology is dual to homology, and if it is finite-dimensional in each degree, you can reverse this and compute homology as the dual of cohomology. And yes, that $f_!$ is unbounded to the left is a stacky phenomenon. By the way, as the diagonal of $f$ is proper (if $G$ is a compact Lie group), one gets for formal reasons a natural map $f_!\to f_\ast$. This recovers the "norm maps" used in equivariant homotopy theory. The cofiber of the norm map yields "Tate cohomology" which often has some nice periodicity; this can nicely be observed here. $\endgroup$
    – Peter Scholze
    Commented May 23 at 14:35

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