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I'd like to ask how to compute homotopy groups of quotient groups, whose homotopy groups I already know. I found this answer, but I don't understand how to derive the homotopy group of $\pi_n (G/H)$ using the long exact sequence.


In general, if I know $\pi_n(G)$ and $\pi_n(H)$, can I compute $\pi_n(G/H)$?

  • What if $\pi_n(G)$ is trivial?
  • Or $\pi_n(H)= \mathbb{Z}_N$?
  • What about a combination of the two?
  • Any special cases that are easy?
  • What about $n=1$?

I know these are many questions so to clarify, in principal I'd like the most general answer but if one is not known, then any examples where a computation can be done is acceptable.

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    $\begingroup$ I assume that by "group" you mean "topological group"? That setup is way too broad. Even very simple examples behave nontrivially under quotients, even when both $G$ and $H$ have trivial homotopy groups, e.g. $G=\mathbb{R}$ and $H=\mathbb{Z}$. The long exact sequence is a very weak connection, doesn't really determine $\pi_n(G/H)$. It does tell you something, but not much. $\endgroup$
    – freakish
    Jun 16, 2020 at 14:01
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    $\begingroup$ @freakish That's fair, I do realize my question is too general, which is why I included simpler cases hoping that any of them would have an answer. I'm mostly thinking of Lie groups. One of the things I'm interested in is $\pi_n (G/Z(G))$, where $Z(G)$ is the center of the group and $n$ is small like 1,2 or 3. A simple example is $G=SU(2)$, $Z(G)= \mathbb{Z}_2$, then $G/H = SO(3)$ and $\pi_1(G/H) = \pi_1(SO(3) ) = \mathbb{Z}_2$. I wonder how well this generalizes to $SU(N)$, for example, and other Lie groups. I guess I asked the question in a broader way hoping to learn more. $\endgroup$
    – Stratiev
    Jun 16, 2020 at 14:42
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    $\begingroup$ For $H=Z(G)$ that's indeed more interesting. $\endgroup$
    – freakish
    Jun 16, 2020 at 14:43

2 Answers 2

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For the cases you're interested in $H$ is discrete which simplifies the situation a lot. In this case most of the long exact sequence in homotopy vanishes and we get that $G$ and $G/H$ have the same $\pi_n, n \ge 2$, so the only thing left to do is analyze $\pi_1$. Here the long exact sequence ends

$$1 \to \pi_1(G) \to \pi_1(G/H) \to H \to 1$$

so we get that $\pi_1(G/H)$ is an extension of $H$ by $\pi_1(G)$. Furthermore in all the examples you've named $\pi_1(G)$ vanishes so we just get that $\pi_1(G/H)$ is $H$.

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  • $\begingroup$ Thank you. Can you clarify this statement please " In this case most of the long exact sequence in homotopy vanishes and we get that G and G/H have the same πn,n≥2... " ? How can I see this? If the explanation is too lenghty, can you direct me to a reference? $\endgroup$
    – Stratiev
    Aug 7, 2020 at 21:50
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    $\begingroup$ @Stratiev: this is a general sort of tiny lemma you get used to early on in long exact sequence computations. The bits of the long exact sequence go $\dots \pi_n(H) \to \pi_n(G) \to \pi_n(G/H) \to \pi_{n-1}(H) \to \dots$ and since $\pi_n(H)$ vanishes for $n \ge 1$ the LES breaks up into a bunch of short exact sequences which just say that the map $\pi_n(G) \to \pi_n(G/H)$ is an isomorphism for $n \ge 2$. $\endgroup$ Aug 7, 2020 at 22:02
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It turns out that the long exact sequence associated to the fibration $H \rightarrow G \rightarrow G/H$ is actually the same as the long exact sequence of relative homotopy groups associated to the pair $(G,H)$. On top of being somewhat surprising (since we usually want to compare the relative homotopy groups to the topological, not group quotient), this actually can be useful in certain situations.

For example, we want to show that $G/H$ is weakly equivalent to the classifying space of some functor, we can show that the relative homotopy groups of $(G,H)$ classify the functor applied to the spheres.

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