Subject: LBG
From: "Nicholas J. Kuhn"
Date: Fri, 03 Nov 2006 14:32:41 -0500
Regarding the fact that EG x_G G_conj = Map(S1,BG) (=LBG) ...
1. This has been well known at least since the mid 1980's ... one
reference would be: MR0814144 (88e:18007) Burghelea, Dan The cyclic
homology of the group rings. Comment. Math. Helv. 60 (1985), no. 3,
354--365.
2. The simplest proof goes as follows ...
There is a pushout square:
S0 --> I
| |
v v
I -----> S1.
This induces, for all spaces X, a pullback square
LX --> X^I
| |
v v
X^I -> X x X,
which rewrites as a homotopy pullback
LX --> X
| |
v v
X --> X x X,
where the two maps X --> X x X are the diagonal map `D'.
Now let X = BG, and replace the vertical D by the equivalent fibration
E(GxG)_{GxG} (GxG/D(G)) --> B(GxG).
One gets that LBG = EGx_G(GxG/D(G)). But (GxG/D(G)) = G_conj. QED
3. Another way of constructing a map from the Borel construction to the
loop space goes as follows ...
Given any two topological groups G and H, Hom(H,G) is a G-space via
conjugation. Considering a G-space as a category in the usual way
(objects are the G-space X, and morphisms are X x G, ...) , the evaluation
map
H x Hom(H,G) --> G induces
BH x BHom(H,G) --> BG, and thus
EGx_GHom(H,G) = BHom(H,G) --> Map(BH,BG).
Now let H = Z, the integers. Ha!
(This is always an equivalence if H and G are discrete. When H = a finite
p-group, and G is compact Lie, this is the map studied by Dywer-Zabrodsky,
Lannes, ...)
Nick Kuhn
> >> 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