It is well-known that any Lie group $G$ has $\pi_2(G)=0$: see this question. Is the same true for any compact (Hausdorff) topological group? Or even for locally compact ones? Maybe there is a way of showing it by expressing the group as an inverse limit of Lie groups?

It seems that it does not hold for all (Hausdorff) topological groups: see e.g. here (I did not go through the details...).

EDIT: according to a deep result by Malcev-Iwasawa, any connected locally compact group deformation retracts to a compact subgroup, so the locally compact case reduces to the compact one.

  • 3
    $\begingroup$ It’s a theorem of Browder that every H-space with finite homological dimension satisfies pi_2=0. $\endgroup$ – Dylan Wilson Dec 31 '17 at 14:34
  • 3
    $\begingroup$ So Browder's result comes from his paper "Torsion in H-Spaces", Th 6.11 for $G$ an arcwise connected H-space with finitely generated homology, nonvanishing in only a finite number of degrees. If that is of any use to you. $\endgroup$ – Tyrone Dec 31 '17 at 14:57
  • 6
    $\begingroup$ With no further hypotheses, topological groups can have the homotopy type of any loop space and so $\pi_2$ of a topological group can be any abelian group. Or, more parsimonously, the Eilenberg-MacLane spaces $B^2 A$, for $A$ any abelian group, have topological group (even topological abelian group) representatives. $\endgroup$ – Qiaochu Yuan Dec 31 '17 at 19:53
  • $\begingroup$ I wonder what happens to the homotopy groups of a topological group when you take its Bohr compactification... $\endgroup$ – Qiaochu Yuan Dec 31 '17 at 23:14
  • 1
    $\begingroup$ @QiaochuYuan: can you give a reference, or quick explanation, for how to see that $B^2A$ can have topological group models? $\endgroup$ – Peter LeFanu Lumsdaine Jan 6 '18 at 14:44

Your Answer

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

Browse other questions tagged or ask your own question.