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Given a topological space $X$ with a group action $ \varphi : G \times X \to X$, we can construct a group extension with $\pi_1(X)$ as follows:

Take the universal cover $\pi : \tilde X \to X$ and consider the set of lifts:

$$L_1= \{ (\tilde f,g) \in \text{Homeo}(\tilde X)\times G : \pi\circ \tilde f = \varphi(g) \circ \pi\} $$

By identifying the associated deck transformation with an element of $\pi_1(X)$ then we get a group extension: $$ 0 \to \pi_1(X) \to L_1 \to G \to 0 $$

Question: Does there exist a similar extension for higher homotopy groups?

My idea would be to use the $n^{th}$ stage $X_n$ of the Whitehead tower, since this is the $n$-connected version of the universal cover. This stage sits in a fibration with fiber $K(\pi_n(X), n-1)$. I am wondering if we perform the same construction as $L_1$, except we lift to $X_n$ instead of $\tilde X$, if we get a group extension:

$$ 0 \to K(\pi_n(X),n-1) \to L_n \to G \to 0 $$

Idea: Maybe one way to do this would be to consider the loop space $\Omega^{n-1}X$, since we know that $\pi_1(\Omega^{n-1}X)=\pi_n(X)$. Then maybe a similar argument to the universal cover extension could be used? However this would seem to require inducing the $G$-action on $X$ to a $G$-action on $\Omega^{n-1}X$. I am confused about how to do this if the $G$-action is free, and so does not preserve the basepoint of the loop space.

Edit: fixed $L_1$ definition

FirstAid
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  • Could you please give a clear definition of $L_1$. In this current form I think some thing is missing. – Ali Taghavi Sep 11 '24 at 16:11
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    Neat question! It looks like the compositions are backwards in $L_1$. Also, I suppose you have to argue that $L_1 \to G$ is surjective -- a priori, maybe there is some $g$ such that there is no automorphism of the universal cover which descends to the action of $g$ on $X$. – hunter Sep 11 '24 at 16:22
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    @AliTaghavi I think what OP means is: an element of $L_1$ is an element of $G$ (viewed as a homeomorphism of $X$) together with a homeomorphism of $\widetilde{X}$ lifting it. (Symbolically, this is what you get from what OP wrote if you swap both the orders of composition, and interpret $\phi(g)$ to mean the homeomorphism $x \mapsto \phi(g, x)$ from $X$ to itself). – hunter Sep 11 '24 at 16:24
  • @hunter Thanks! I think if $G$ is connected then surjectivity could be argued using the homotopy lifting property. That is, we could consider a homotopy from any $g$-action to the identity, and lift this to the universal cover – FirstAid Sep 11 '24 at 16:39
  • should there be a $\pi_1$ or something in the first term of the proposed sequence? (like the fundamental group of the eilenberg maclane space?) – hunter Sep 11 '24 at 16:52
  • oh or do you just mean $\pi_n(X)$? – hunter Sep 11 '24 at 16:54
  • @hunter I am not entirely sure. I guess the question boils down to the existence of some higher-homotopy analogue of a deck transformation. I am hoping that maybe this can be formalized using the natural group structure on the Eilenberg-Maclane space as described in https://math.stackexchange.com/questions/45556/group-structure-on-eilenberg-maclane-spaces – FirstAid Sep 11 '24 at 17:02
  • @FirstAid got it, didn't know about that group structure! If something like this is going to be true, then $G$ has to act on the abelianization of that group, is there a natural reason to think that? – hunter Sep 11 '24 at 17:27
  • (I guess that group is abelian.) – hunter Sep 11 '24 at 17:47
  • @hunter I think a sufficient condition for the extension is that there should be an inclusion of $K(\pi_n(X), n-1) \subset \text{Homeo}(X_n)$ by the abelian group action on the fibers . Maybe this is enough to conclude that such an extension exists? – FirstAid Sep 11 '24 at 17:57

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