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