# The homotopy type of the mapping space $Map_{B\rho}(BS^1,BG)$? for $G$ a compact Lie group

Given a homomorphism $\rho:S^1\rightarrow G$ with $G$ a compact Lie group there is an induced map of classifying spaces $B\rho:BS^1\rightarrow BG$. What is known about the homotopy type of the mapping space $Map_{B\rho}(BS^1,BG)$?

Integrally the space will be a mess of phantom maps. On the other hand, after rationalisation, $BG_\mathbb{Q}$ will be a product of even dimensional Eilenberg-Mac Lane Spaces and the mapping space too will become a similar product after one uses Thom's theorem. Is there an understandable middle ground?

I have read Notbohm's work and understand that if $C(\rho)\subseteq G$ denotes the centraliser of $\rho$ in $G$, then there is a homomorphism $C(\rho)\times S^1\rightarrow G$ and the adjoint of the map that classifies this homomorphism, $e_{\rho}:BC(\rho)\rightarrow Map_{B\rho}(BS^1,BG)$, induces an isomorphism on mod p homology for any prime. Is this map then not a homotopy equivalence after completion at any prime?

As an example of my lack of a thorough understanding of these concepts Take $\rho: S^1=U(1)\rightarrow U(2)$ to be the canonical inclusion. Then

$BC(\rho)=BT=BS^1\times BS^1$

classifies a maximal torus. But on the other hand

$Map_{B\rho}(BS^1,BU(2))_\mathbb{Q}\simeq Map_{B\rho}(BS^1,BU(2)_\mathbb{Q})\simeq K(\mathbb{Q},2)\times K(\mathbb{Q},2)\times K(\mathbb{Q},4)$

So from this I don't understand how the map $e_\rho$ can be an equivalence of any sort.

Edit: The result from Notbohm I quote is to be found in his paper "Maps Between Classifying Spaces" Mathematische Zeitschrift, 1991, Volume 207, Number 1, Page 153.

• I suspect that you are going wrong by equating $Map_f(X,Y)_{\mathbb Q}$ with $Map(X,Y_{\mathbb Q})$. – Tom Goodwillie Apr 7 '16 at 16:55
• I have thought of that. My reference is, of course, Hilton, Mislin and Roitberg (theorem 3.11) and I realise that $\mathbb{C}P^{\infty}$ isn't a finite complex. But then I get $Map_{B\rho}(BS^1,BU(2))^{\wedge}_p\simeq (BS^1)^{\wedge}_p\times (BS^1)^{\wedge}_p$. I identify the equivalence with the toral inclusion and get from the evaluation fibration that $Map^*_{B\rho}(BS^1,BU(2))^{\wedge}_p\simeq (S^2)^{\wedge}_p$. I'm pretty certain that none of this can be true. I apologise about my ignorance of both localization and completion. – Tyrone Apr 7 '16 at 17:13
• I'm not sure what it is that's actually bothering you. – Charles Rezk Apr 8 '16 at 12:04
• Mainly my questions is what is the homotopy type of $Map_{B\rho}(BS^1,BG)$? Either in integral, local, rational or complete homotopy theory. The example I bring up of $Map_{B\rho}(BS^1,BU(2))$ seems to be one the the easier cases to understand and happens to have applications in some of my research. – Tyrone Apr 8 '16 at 14:48