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If I have a compact connected Lie group $G$ and a (relatively nice) simply-connected topological abelian group $A$, when is it the case that a given $f\colon BG \to BA$ loops to a (continuous) homomorphism $G\to A$? This seems like it can't always be true, and also should be some kind of classical result, but I'm not sure where to look.

David Roberts
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    Should that be "when is $BG\to BA$ the delooping of a homomorphism"? – Mark Grant Jun 21 '19 at 06:37
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    A more usual way of posing the question is: when does an $E_1$-homomorphism come from a map of topological groups? I don't know the answer though, but it is not always – Denis Nardin Jun 21 '19 at 06:42
  • @DenisNardin yes, that's another way I was thinking of it. – David Roberts Jun 21 '19 at 06:50
  • @MarkGrant yes, sorry, my mistake. Edited the title, too. – David Roberts Jun 21 '19 at 07:04
  • $E_1$ or $A_∞$ (they are the same thing) control associativity. $E_∞$ is the commutative one. The other $E_n$ interpolate between the two (e.g. $E_2$ gives braided multiplication). $H_∞$ is the "shadow" of $E_∞$ in the homotopy category – Denis Nardin Jun 21 '19 at 07:13
  • Does this have anything to do with similar question "when does a morphism of stacks $BG\rightarrow BH$ is coming from a morphism of Lie groups $G\rightarrow H$"??? – Praphulla Koushik Jun 21 '19 at 07:13
  • @Denis ok. And no, $G$ is not abelian. – David Roberts Jun 21 '19 at 07:14
  • @PraphullaKoushik no, nothing at all. That answer is easy in comparison: they all do. – David Roberts Jun 21 '19 at 07:14
  • Oh. Ok ok :) @DavidRoberts – Praphulla Koushik Jun 21 '19 at 07:16
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    Unfortunately I think that this question is quite sensitive to $A$ and not just to $BA$. For example, $A$ does not have any homomorphisms from $G$ if it does not have any homomorphisms from $S^1$, which requires that it have elements of finite order $(g^n = 1)$; however, any classifying space $BA$ is equivalent to a classifying space $BA'$ where $A'$ has no finite-order elements. – Tyler Lawson Jun 21 '19 at 08:00
  • @Tyler hmm, interesting. It may be my hopes are misplaced, then. – David Roberts Jun 21 '19 at 08:07
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    I'll point to https://mathoverflow.net/questions/156408/equivariant-classifying-spaces-from-classifying-spaces?rq=1, which addresses a related question: what criteria imply that $\mathrm{Hom}(G,H)\to \mathrm{Map}_*(BG,BH)$ is a weak homotpy equivalence? The answer there says nothing about your case I think, but perhaps the methods used could be helpful. I don't know. – Charles Rezk Jul 05 '19 at 19:38

1 Answers1

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If $G$ is a compact connected topological group and $A$ is a locally compact abelian topological group, then for any map $f:BG\to BA$ the looped map $\Omega f:\Omega BG\to \Omega BA$ is homotopically equivalent to a homomorphism $\phi:G\to A$. This follows from the main result of

Scheffer, Wladimiro, Maps between topological groups that are homotopic to homomorphisms, Proc. Am. Math. Soc. 33, 562-567 (1972). ZBL0236.22008.

I'm not sure if this answers your question, however, which seems to be about when $f$ is in the image of the classifying space functor $B:\mathsf{TopGrp}\to\mathsf{Top}$.

Mark Grant
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  • Thanks, you are indeed answering the intended question, but unfortunately I can't assume $A$ is locally compact! It is in my example the geometric realisation of a simplicial abelian Lie group.... But this is a good answer in any case. – David Roberts Jun 21 '19 at 07:51
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    Reformulation of the question: is $\hom(G,A) \to \hom_{A_\infty}(G,A)$ surjective on $\pi_0$? (The target is $A_\infty$-homomorphisms $G \to A$). One might be able to use Boardman-Vogt obstruction theory to study this, by replacing $G$ with a cofibrant $A_\infty$-space. – John Klein Jun 21 '19 at 12:28
  • I will accept this in the absence of any other suggestions, because it shows how subtle the question is, and how much it will depend on the topology of $A$. – David Roberts Jul 22 '19 at 12:07