9
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ @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 Dec 11 '17 at 17:10
19
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 Dec 12 '17 at 15:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.