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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.

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The classifying space of a not necessarily connected topological group is analysed in my paper with Chris Spencer "$\cal G$-groupoids, crossed modules and the fundamental groupoid of a topological group", Proc. Kon. Ned. Akad. v. Wet. 7 (1976) 296-302. Can this be linked with the other material you give? There are related ideas in my paper with O. Mucuk, "Covering groups of non-connected topological groups revisited", Math. Proc. Camb. Phil. Soc, 115 (1994) 97-110. – Ronnie Brown Oct 6 '13 at 9:41
@RonnieBrown: I just downloaded your papers. However, I am rather slow and it will take me some time to understand if I can make a useful connection between them and the world I am coming from. The notions I am speaking of are very nicely explained in Bernstein-Lunts' book `Equivariant sheaves and functors'. It can be downloaded at It's #51. Further, the relevant chapter is #4. In particular, you may want to look at pg. 93 and 103-105. – Reladenine Vakalwe Oct 6 '13 at 13:10
I have the dual problem! IN fact I put this type of question, i.e. relation to triangulated categories, as Problem 16.1.26 in our book on "Nonabelian algebraic topology". Incidentally, the download has separate page numbers for the chapters, so 4 has only 58 pages. – Ronnie Brown Oct 6 '13 at 14:01
@RonnieBrown: The pg number references are to the actual pages in the book. Sorry for not clarifying earlier. Ch. 4 starts at pg 68 and ends at 125. Everything up to pg 93 is just the algebra of dg-algebras. Pg. 93 (section head is DG-modules and sheaves of topological spaces') explains why one might care. The natural subcategory' of $D(X)$ alluded to in the middle of the page is the subcategory generated by the constant sheaf. If you wish I can add additional comments about how I think about the equivariant derived category of a point. But I'd be surprised if any of it were news to you! – Reladenine Vakalwe Oct 6 '13 at 15:03
up vote 3 down vote accepted

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

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Oliver: I think you have to be quite careful in using Wolfgang's Ind Res proposition at the upgraded dg-level in this way. Not saying that the argument doesn't work, just that you have to be rather careful. Anyway, thanks for letting me know, I will ask Wolfgang for details. – Reladenine Vakalwe Oct 7 '13 at 1:33
Yes, you are right, and this being "careful" is the whole point of the diploma thesis mentioned above. If you really need this thesis I can scan it and send it to you via eMail. – Oliver Straser Oct 7 '13 at 6:17
Besides that, the methods described above can be used also in different settings: For example, let $N$ be an algebraic group, with $N_0=T$ is a torus. If $N$ acts on some variety, such that the action restricted to $N_e$ makes $X$ to a toric variety, then you can proof by similar methods that $D^b_N(X)$ is equivariantly formal. (Note that formality of $D_{N_e}(X)$ was shown already by Lunts) – Oliver Straser Oct 7 '13 at 6:23
I would be quite grateful if you could send me a copy via email. You were also supposed to send me a copy of your work on HC-modules. It's been a year now! – Reladenine Vakalwe Oct 7 '13 at 19:38

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