Timeline for What is the equivariant cohomology of a group acting on itself by conjugation?
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| Mar 15, 2020 at 5:48 | history | edited | David Roberts♦ | CC BY-SA 4.0 |
Added links and more bibliographic info
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| Apr 14, 2010 at 18:37 | comment | added | Dan Ramras |
There's a difference between L(|BG|), the free loop space of the geometric realization of BG=pt/G, and the realization of the level-wise free loop space $|k \mapsto L(B_k G)|$ (here $B_k G = G^{k-1}$). When you consider free loops on the groupoid pt/G, won't its realization be the level-wise free loop space? When G is discrete, all level-wise loops are constant and the level-wise loop space is exactly |BG|, not L(|BG|). (When G is connected $L(|BG|) \simeq |k \mapsto L(B_k G)|$ by Theorem 12.3 in May's book math.uchicago.edu/~may/BOOKS/gils.pdf.)
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| Apr 14, 2010 at 0:08 | comment | added | David Ben-Zvi | Or finally there's a more "classical" proof of LBG=(G x EG)/G that Dan Freed explained to me - you write LBG as an associated bundle to the universal Omega BG=G bundle over BG (aka EG) -- you check that LBG is associated to the conjugation action of G on itself, and you're done. | |
| Apr 14, 2010 at 0:06 | comment | added | David Ben-Zvi | In any case that's overkill - we can present the stack of bundles as a groupoid by writing bundles as two trivial bundles on half-intervals with gluings on the overlap, and check it's equivalent to G/G (which of course is what you get if you think of it as a single G-bundle on an interval, with gluing of the two endpoints). Or alternatively, you can represent BG as the action groupoid of G on a point and use this to get a presentation of free loops into it as a topological groupoid (which will give you the same as the above in a different language). | |
| Apr 13, 2010 at 22:50 | comment | added | David Ben-Zvi | I'm not sure I understand the question. Are you asking for a presentation of the stack of bundles as a groupoid? I don't know one (other than a posteriori as G/G..) but the construction of a classifying space of a nerve for topological groupoids makes sense for arbitrary stacks. Up to getting sheaf/cosheaf confused it's just the global sections functor from stacks (thought of as sheaves of spaces or simplicial sets or groupoids) to spaces (or simplicial sets or higher groupoids). So a map of stacks gives a map of spaces. Does this seem reasonable? | |
| Apr 13, 2010 at 3:52 | comment | added | Dan Ramras |
I agree that a statement on stacks should be stronger. Unfortunately I don't think I understand the suggestion here. There is a map of stacks from G/G to the stack of G-bundles on $S^1$, given by sending an element of g to the bundle over $S^1$ formed by using g as clutching function (and conjugations induce explicit bundle isomorphisms). Now, if we view G/G as a topological groupoid, the classifying space of its nerve is (EG x G)/G. But I'm now sure how to view the stack of G-bundles over $S^1$ as a topological groupoid, so as to obtain a map of spaces from the above map of stacks.
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| Apr 8, 2010 at 13:45 | comment | added | David Ben-Zvi | Probably I'm missing something basic, but I would have thought the statement on topological stacks is strictly stronger than one on spaces, which is obtained from it by a geometric realization. So we consider the top. stack of G-bundles on S^1, to which there's an obvious map from the stack G/G, and which has an obvious (classifying) map to the free loop space of BG, and that's our equivalence, no? | |
| Apr 8, 2010 at 4:16 | comment | added | Dan Ramras | Sure, you can rephrase this in ways that make it sound like a tautology. But as far as I know, there is is no natural homotopy equivalence between these two objects, considered as topological spaces (not stacks). The proofs I've seen all make clever use of some third space, to which both of these spaces map (or vice versa). I don't think the abstract nonsense will help in writing down such a map, but I'd be quite interested to know if I'm wrong about that. I guess I'm saying that I think there's something more to the topological statement that just abstract nonsense. | |
| Apr 8, 2010 at 4:02 | comment | added | David Ben-Zvi | More geometrically, free loops in BG are the same as G-local systems on the circle (here G considered as a homotopy type). Those are classified (via their monodromy) as elements in G, up to conjugation, i.e. the quotient stack G/G (aka the Borel construction (G x EG)/G). | |
| Apr 8, 2010 at 3:27 | comment | added | David Ben-Zvi | Does this result deserve attribution? It seems to me equivalent to the fact that the inertia of the stack BG is the adjoint quotient stack G/G, which is almost definitional.. or more topologically, we know L(BG) is a Omega(BG)=G-bundle over BG, so comes from a G action on G, and we just need to check that it is the conjugation action.. | |
| Apr 8, 2010 at 0:48 | comment | added | Dan Ramras | I should say that some people attribute this result to a paper of Bokstedt-Hsiang-Madsen, but I've never really understood why. It's also sometimes said to be due to Waldhausen. And there's a proof for discrete groups in one of Dave Benson's books; Kate's argument is a clever extension of Benson's. | |
| Apr 8, 2010 at 0:46 | history | answered | Dan Ramras | CC BY-SA 2.5 |