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Let $G$ be a topological group, and $M$ a connected compact smooth manifold. I'm studying $$ \pi_0 (map (BG,M)). $$

For $G$ a finite group, we know that this is just a point by the Sullivan conjecture on maps from classifying spaces which was proven by Miller. (This does not require smoothness of $M$.)

On the other hand, if $G$ is an infinite discrete groups, this $\pi_0$ can be larger (take $G=\mathbb{Z}$ and $M=S^1$).

Question What happens if $G$ is a compact Lie group? Are there examples where this $\pi_0$ is more than a point?

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  • $\begingroup$ @YCor Yes, that's what I meant by "trivial". I have clarified this and made the question more precise. Also, I do indeed consider topological groups; I edited this as well. $\endgroup$
    – Alexander Körschgen
    Commented Dec 11, 2017 at 17:10

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You were right to single out Lie groups as potentially interesting. In [Topology 5 (1966), 241-243], Brayton Gray showed that the homotopy group of maps $[BS^1, S^3]$ was uncountable. Indeed, he showed that the subgroup of phantom maps -- maps null on every finite subcomplex - was uncountable. Then Alex Zabrodsky, in [Isreal J. Math. 58 (1987)], has a theorem that refines this: all maps in this case are phantom, and the group is isomorphic to $\hat Z/Z$.

More generally, there was a decade of work, after Miller's theorem, exploiting the Sullivan conjecture, and much of it was focused on understanding maps out of $BG$ for $G$ compact Lie. In particular, there are a number of papers identifying mapping spaces of the form $Map(BG,BH)$, for well chosen pairs of compact Lie groups. Now note that $\Omega Map(BG,BH) \simeq Map(BG,H)$ which is of the form you were asking about. Look up papers of Dwyer, Wilkerson, McClure, Oliver, and Lannes to get going in the literature.

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  • $\begingroup$ Thank you for the comprehensive answer. I would like to add that there is also a recent preprint by Rezk which computes $Map(BG,BH)$ for 1-truncated compact Lie groups $H$ (see arxiv.org/abs/1608.02999). However, this does not really help with this particular application because one gets $\pi_0 (Map (BG,H)) = \pi_0 (H) = \ast$ for such $H$. $\endgroup$
    – Alexander Körschgen
    Commented Dec 12, 2017 at 15:28

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