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I'll be using homological grading throughout this question.

Let $G$ be a compact connected lie group. The following isomorphisms are classical and can be proven using several methods:

$$H^{\bullet}(BG;\mathbb{R}) \cong Sym^{\bullet}(\mathfrak{g}^*[2])^G$$

$$H^{\bullet}(G; \mathbb{R}) \cong Sym^{\bullet}(\mathfrak{g}^*[1] )^G$$

Where on the RHS we take invariants w.r.t. the adjoint action of $G$ on the lie algebra. Replacing $G$ with its complexification (which is the complex points $\mathbb{G} (\mathbb{C})$ of a connected semi-simple complex algebraic group) doesn't change the cohomology so let us do this. Next let's interpret the graded rings above as (derived if you like) prestacks (functors $CommAlg_{\mathbb{C}} \to \infty$-groupoids). Translating to this language we get the following (which is just a different way to notate the above):

$$SpecH^{\bullet}(B\mathbb{G}(\mathbb{C});\mathbb{C}) \cong \mathfrak{g}_{\mathbb{C}}[-2] \text{ //}G \cong \mathbb{T}_{B\mathbb{G}}[-1] \text{ //} \mathbb{G}$$

$$Spec H^{\bullet}(\mathbb{G}(\mathbb{C}); \mathbb{C}) \cong \mathfrak{g}_{\mathbb{C}}[-1] \text{ //}G \cong \mathbb{T}_{B\mathbb{G}} \text{ //} \mathbb{G}$$

I find it weird that both sides of the equation naturally take as input a connected semisimple complex algebraic group $\mathbb{G}$ but the isomorphism itself is of a transcendental nature (by passing through the real points of the compact real form as a topological group). This mysterious relationship does not manifest itself at the level of the proofs of these statement (at least in the proofs I know). It would be nice to know whether these interpretations can be turned to proofs (at least giving one of the results given the other one):

Question 1: Is their a higher mechanism at work here? Phrased differently: Are these isomorphisms a formal consequences of the formalism of DAG together with some basic facts about groups?

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  • $\begingroup$ In standard formulations of DAG, rings of functions on affine schemes are required to be connective, i.e. concentrated in positive homological degrees. However, these cohomology rings are coconnective. $\endgroup$ – Justin Campbell Dec 17 '17 at 7:02
  • $\begingroup$ @JustinCampbell Agreed. Changed schemes to prestacks. $\endgroup$ – Saal Hardali Dec 17 '17 at 7:21
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    $\begingroup$ Further to Justin's comment, your enhancements are exclusively in the stacky, rather than derived, direction, so DAG won't tell you anything more than AG here. Your expressions are almost descriptions of the de Rham stacks for $BG$ and $G$, slightly garbled by passage to cohomology. $\endgroup$ – Jon Pridham Dec 17 '17 at 9:52
  • $\begingroup$ @JonPridham Agreed. I guess it's a matter of terminology. Some people would say "stacky" for them is deriving "one side of the coin" and replacing rings with derived rings is deriving the "other side". I tried not to make my interpretations too rigid to make room for the possibility that there's a slightly different interpretation which can be explained more formally (for instance maybe my convention for grading is wrong etc...). $\endgroup$ – Saal Hardali Dec 17 '17 at 9:57
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    $\begingroup$ Note the statement is very false for G eg unipotent.. so can't be all that formal, it really is a statement about reductive groups $\endgroup$ – David Ben-Zvi Dec 17 '17 at 17:52
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I don't know of a DAG mechanism that implies these statements, any more formally than the standard proofs -- i.e., we reduce to a maximal torus, where the statement follows from the shape of the cohomology of the circle.

There are higher mechanisms though of which this statement is a hint. On one side you have topology (of BG), on the other side (derived) algebraic geometry -- suggesting that rather than something formal this is a shadow of mirror symmetry. Indeed the statement is [essentially equivalent to] the mirror symmetry between the G-equivariant A-model on a point and the B-model into a shift of the Cartan mod Weyl group. That is probably a natural setting for this question.

Or if you prefer, the statement can be seen (at the risk of tremendous overkill) as a feature of geometric Langlands (roughly the case of ${\mathbf P}^1$), specifically of the wonderful derived Geometric Satake theorem of Bezrukavnikov-Finkelberg https://arxiv.org/abs/0707.3799 (though of course the basic statements above are used in the proof!) Namely there's an equivalence of monoidal dg categories

$$D_{LG_+}(LG/LG_+) \simeq D_{coh}({\mathfrak g}^{\vee,\ast}[2]/G^\vee)$$

between equivariant sheaves on the affine Grassmannian for $G$ and equivariant coherent sheaves on the coadjoint representation of the Langlands dual. Your statement is an "affinization" of this -- describing modules for the endomorphisms of the unit (or natural coefficient rings) on both sides -- on the left there's the equivariant category

$$D_{LG_+}(pt)=D_G(pt)=D(BG)$$

and on the right there's the affinization of the coadjoint quotient,

$${\frak g}^{\vee,\ast}[2]//G^\vee \simeq {\frak h}^{\vee,\ast}[2]//W \simeq {\frak h}[2]//W\simeq {\frak g}[2]//G.$$

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I believe you can pass from one statement to the other through Koszul duality and formality. More precisely, the cochains of the classifying space and the chains of the group are Koszul dual. So once you know that the cochains are quasisomorphic to the $G$ invariants of a polynomial ring, you get that the chains are quasisomorphic to the $G$ invariants of an exterior algebra.

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