Subject: Re: two postings
From: Johannes Ebert
Date: Fri, 3 Nov 2006 10:18:29 +0100 (CET)
Hello,
for connected groups, the answer is as follows. If G is connected, then
LBG=B(LG); LG= continuous loops in G. The map EG \times_G G^ad
\to B LG arises in the following way: Consider the space PG of all maps F
: R \to G; such that f(x+ 2\pi )f(x)^-1 = const. There is a map PG \to G
by sending f to f(2 \pi) f(0)^-1. This is an LGB- principal bundle
(perhaps one needs some point-set-topology conditions on G at this point).
Moreover, G acts on PG from the left and PG \to G^ad is G-equivariant.
Thus we have an LG-principal bundle on EG \times_G G^ad, which has a
classifying map EG \times_G G^ad \to BLG. This is the desired map. To see
that it is a homotopy equivalence, one compares the two fibrations
G \to EG \times_G G^ad \to BG
and
B (\Omega G) \to B LG \to BG.
The principal bundle PG \to G has a (nonequivariant) reduction of the
structural group to \Omega G (loops based at the identity), which gives us
a map G \to B \Omega G. Moreover, the two G-principal bundles on EG
\times_G G^ad induced by the compositions
EG \times_G G^ad \to B LG \to BG
and EG \times_G G^ad \to BG
are isomorphic, and we have a comparison diagram between both fibrations,
with the identity on BG.
If G is connected, then B \Omega G \sim G, and the above is seen to be an
equivalence. In the end, the long exact homotopy sequence and the 5-lemma
shows that EG \times_G G^ad is an equivalence.
If G is not connected, then a similar construction should work.
Bets regards,
Johannes Ebert
On Thu, 2 Nov 2006, Don Davis wrote:
>> Two postings: A question and a conference update...............DMD
>> ______________________________________________________
>>
>> Subject: for toplist
>> From: "Claude Schochet"
>> Date: Thu, 2 Nov 2006 14:20:36 -0500
>>
>> I am told that the following result is "well-known". Does anybody know
a
>> reference?
>>
>> Prop: Let $G$ be a topological group. Then there is a natural
equivalence
>> \[
>> EG \times _G G^{ad} \simeq F(S1, BG)
>> \]
>> where $F(S1, BG)$ is the free loop space on $BG$ and $G^{ad}$ indicates
>> $G$ acting on itself via the adjoint action.
>> Thanks!
>>
>> Claude Schochet
>> Math Dept, Wayne State Univ.
>> Detroit, MI 48202
>> claude@math.wayne.edu
>> __________________________________________________________