6
$\begingroup$

In Toen's Affine Stacks, he computes that $\mathcal{O}(B\mathbb{G}_a) = k[\epsilon]$ with $|\epsilon| = 1$ and trivial differential (where here $\mathcal{O}$ is computed in a derived sense, and we choose dg algebras as a model for derived rings). To prove this, he explicitly constructs a cosimplicial ring whose $\text{Spec}$ is $B\mathbb{G}_a$, verifying this by showing they satisfy the same universal property.

I'm wondering if there's a more explicit way to do this calculation. In particular, I would like to compute $\mathcal{O}(BG)$ for various linear algebraic $G$ (unipotent, solvable, semisimple...). I would also like to compute $\mathcal{O}(G/G)$ (the adjoint action).

Here's an idea for an approach, which might be totally wrong. These stacks are Artin stacks, so in particular one can write $BG$ as a colimit of a simplicial diagram in schemes. Then by definition, $\mathcal{O}(BG)$ is a limit of cosimplicial diagram in algebras. Here I'm not exactly sure what I'm doing, but I want to say I can apply some kind of (dual) monoidal Dold-Kan to realize this limit in dg algebras. I think (not sure) one only gets a commutative algebra structure up to homotopy here, but at least taking cohomology, this ends up just being the Hochschild complex for computing the ("rational") cohomology of the (rational) trivial representation of $G$.

Using this, I think one computes that $\mathcal{O}(B\mathbb{G}_m) = k$ (but haven't verified the details), and Jantzen's book (Representations of Algebraic Groups) verifies Toen's result for $\mathbb{G}_a$. Very quickly though, this computation becomes difficult. Are there general results/computations known for unipotent, solvable, or reductive $G$, and can one compute $\mathcal{O}(G/G)$ in a similar way (e.g. using the rational cohomology of $k[G]$ under the adjoint action)? Do you have a preferred way to think about this?

$\endgroup$
  • $\begingroup$ I found this answer: mathoverflow.net/questions/32498/… which resolves the question for reductive $G$, and the answer is $\mathcal{O}(BG) = k$ and $\mathcal{O}(G/G) = k[G]$. $\endgroup$ – Harrison Chen Sep 11 '15 at 4:34
  • $\begingroup$ Note that I should say, in the above $k$ is a field of characteristic zero. $\endgroup$ – Harrison Chen Sep 11 '15 at 4:35
  • 3
    $\begingroup$ As you say, $\mathcal{O}(BG) = C^\bullet(G, k)$. For a unipotent group $C^\bullet(G, k)\cong C^\bullet(\mathfrak{g}, k)$, which is complicated. $\endgroup$ – Pavel Safronov Sep 11 '15 at 13:41
  • $\begingroup$ I guess, Proposition II.4.11 in Jantzen's book also says $C^\bullet(P, k) \simeq k$ for $P$ parabolic for a field $k$ of any characteristic. In characteristic zero it follows from Borel-Weil-Bott by identifying the category of $G$-equivariant sheaves on $G/B$ with the category of rational $B$-representations. $\endgroup$ – Harrison Chen Sep 28 '15 at 6:26
3
$\begingroup$

For a smooth groupoid $G_1\rightrightarrows G_0$ , here is a way to compute derived global sections $\mathcal O(X)$ of $\mathcal O_X$, where $X=[G_0/G_1]$.

Observe that $X$ is the homotopy colimit of the nerve $G_\bullet$ of the groupoid. As you said then $\mathcal O(X)$ is the homotopy limit of $\mathcal O(G_\bullet)$. This homotopy limit can indeed be computed via cosimplicial Dold-Kan.

This is particular tells you that, as Pavel wrote in his comment, $\mathcal O(BG)=C^*(G,k)$.

Considering the action groupoid $G\times G\rightrightarrows G$ for the conjugation action you get (as suggested in your question) that $\mathcal O(G/G)=C^*(G,\mathcal O(G))$, where the action of $G$ on $\mathcal O(G)$ comes from the conjugation on $G$.

For a reductive $G$, we know that $C^*(G,V)=V^G$ for any $G$-module $V$. Hence $\mathcal O(BG)=k$ and $\mathcal O(G/G)=\mathcal O(G)^G$ is the algebra of class functions.

As Pavel said in his comment, $\mathcal O(B\mathbb{G}_a)=C^*(\mathbb{G}_a,k)=C^*(Lie(\mathbb{G}_a),k)$. Then $Lie(\mathbb{G}_a)=Free(k)$ is the free Lie algebra generated by $k$. Now for any perfect complex $V$, $C^*(Free(V),k)$ is quasi-isomorphic to the square-zero extension $k\oplus V^*[-1]$. One thus gets back that $\mathcal O(B\mathbb{G}_a)=k[\epsilon]$.

$\endgroup$
  • $\begingroup$ I think the only nontrivial thing here is to show that $\mathcal{O}(B\mathbb{G}_a)\cong C^\bullet(\mathbb{G}_a, k)\cong C^\bullet(Lie(\mathbb{G}_a), k)=k[\epsilon]$ is a quasi-isomorphism of cdgas. The problem of course is that $C^\bullet(\mathbb{G}_a, k)$ is a totalization of a cosimplicial commutative algebra and is naively just a dga. $\endgroup$ – Pavel Safronov Jul 6 '17 at 9:18

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.