Formality of classifying spaces (for not necessarily connected groups)

Question

As should be evident from the title this question has a similar flavor to:

Formality of classifying spaces

However, unlike Geordie's question, I will be working with torsion free coefficients (say the complex numbers). The `torsion' in the current question is going to be from a different source.

Let $G$ be a reductive linear algebraic group (over $\mathbb{C}$) say. Relatively general formal considerations show that the $G$-equivariant derived category of a point is equivalent to the dg-derived category of a dg-algebra (I am going to ignore finiteness issues).

For instance, if $G$ is connected (so that $BG$ is simply connected), the dg-algebra is the algebra associated to the De Rham complex of $BG$ (let's be friendly and ignore finiteness issues again). In this case the $G$-equivariant derived category of a point is the subcategory of $D(BG)$ generated by the constant sheaf. The latter is equivalent to the derived category of the De Rham complex. Further, since $H^*(BG)$ is a polynomial ring (I am still in the $G$ is connected case) generated in even degree, it is easy to see that this dg-algebra is formal (the point is that the De Rham complex is (super)commutative before even passing to cohomology, so you can very naively construct your quasi-isomorphism by hand).

Ok, that's pretty nice. Now let's look at $G$ not connected. Then the $G$-equivariant derived category of a point is still governed by a dg-algebra, but it's a bit nastier. Essentially we want the subcategory of $D(BG)$ generated by local systems, and since $BG$ isn't simply connected anymore we are going to have non-trivial local systems. The dg-algebra we are after is the $Ext$-algebra of the sum of all the irreducible local systems. Now we can certainly compute the cohomology of this dg-algebra: it is the $H^*(BG^0)$-twisted group algebra of $G/G^0 = \pi_1(BG)$. Added later: to see this, loop the fibration $BG^0\hookrightarrow BG \twoheadrightarrow B(G/G^0)$.

The question: is the dg-algebra governing $D(BG)$ (for $G$ not necessarily connected) formal? If so, is it possible to deduce this formality from the connected case?

I am quite happy assuming that $G/G^0$ is abelian.

Answer 1

The answer is yes $D^b_G(X)$ is equivariantly formal. The result has been proved in a diploma thesis written under the supervision of Wolfgang Soergel. (Unfortunately it is not available electronically).

The proof goes roughly as follows. Let $\pi:BG_0 \to BG$ the quotient map. Let $\mathcal{L}$ be the sum of simple perverse sheaves on $D_G(pt)$ and $\mathcal{L}_0$ the sum of simple perverse sheaves on $D_{G_0}(pt)$. Let $I_{\mathcal{L}}$ and $I_{\mathcal{L}_0}$ (note that $I_{\mathcal{L}_0}$ is a direct summand of $\pi^*(I_{\mathcal{L}})$). The corresponding injective resolutions i.e. $D^b_G(pt)\cong D^b(End(I_{\mathcal{L}}))$ ...

You want to show that the $dg$-algebra $End(I_{\mathcal{L}})$ is formal. Of course it would also be fine to show that $End(\pi_*\pi^*I_{\mathcal{L}})$ is formal. Now by [Soe01] Theorem 2.4.2 there exists an isomorphism $$End(\pi_*\pi^*I_{\mathcal{L}})\cong End( I_{\mathcal{L}_0})\otimes \mathbb{C}[G/G_0] $$ Be carefull: The multiplication of $End( I_{\mathcal{L}_0})\otimes \mathbb{C}[G/G_0]$ has some twist corresponding to the action of $G/G_0$ on $End( I_{\mathcal{L}_0})$ Anyway, one can show that the "classical" quasi-isomorphism of $End(I_{\mathcal{L}_0})$ to $H^*(...)$ is equivariant with respect to the $G/G_0$ action on so induces the quasi-isomorphism you are looking for.

[Soe01] Langlands' Philosophy and Koszul Duality http://home.mathematik.uni-freiburg.de/soergel/PReprints/langlands.ps