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I am looking for an efficient way to compute the homotopy groups, as well as morphisms between them, of certain matrix groups and compact symmetric spaces. To be specific, I want to determine the long exact sequence of the form

$$ \cdots \to \pi_{n+1}[H] \to \pi_{n+1}[G] \to \pi_{n+1}[G/H] \to \pi_{n}[H] \to \cdots$$

where $(G,H)$ runs over the ten cases $$ (O(n)\times O(n),O(n)_{diag}),\; (O(m+n),O(m)\times O(n)),\; (O(2n),U(n)),\; (U(n)\times U(n),U(n)_{diag}),\; (U(m+n),U(m)\times U(n)),\; (U(n)/O(n)),\; (U(2n),Sp(n)),\; (Sp(n)\times Sp(n),Sp(n)_{diag}),\; (Sp(n),U(n)),\; (Sp(m+n),Sp(m)\times Sp(n)). $$

That is, my goal is to determine what the homotopy groups are (especially in the unstable range) and what are the group homomorphisms.

Among these, I already know pretty well the real and complex Grassmannians $O(m+n)/O(m)\times O(n),\; U(m+n)/U(m)\times U(n).$ Also, the cases containing diagonal action seems easy since the relevant $G-$ bundles are trivial, as Aphelli commented.

However, I am not familiar with the other five cases.

Is there a systematic way to work this out? Any help would be greatly appreciated.

Hyeongmuk LIM
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  • Isn’t the case where $H=\ast$ tautological? – Aphelli Sep 07 '23 at 12:44
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    How much of the long exact sequence do you want? The (unstable) homotopy groups of these spaces are not completely known. – JHF Sep 07 '23 at 14:50
  • @JHF Thank you for letting me know that. I would like to compute $\pi_n[X]$ for $n=1$ up to $n=4$. It would be preferable if a few higher homotopy groups could be computed. – Hyeongmuk LIM Sep 08 '23 at 01:06
  • @Aphelli Thank you for pointing out that. I corrected those cases to the form $(G\times G,G)$ where $G$ acts diagonally. – Hyeongmuk LIM Sep 08 '23 at 03:24
  • @JHF Should I also restrict the size of matrices in $O(m), U(m), Sp(m)$? Is restricting to $\pi_{1-4}$ not sufficient to make the problem doable? – Hyeongmuk LIM Sep 08 '23 at 03:27
  • The fibration $G_{diag} \rightarrow G\times G \rightarrow G\times G/G_{diag}$ is trivial, isn’t it? Isn’t it isomorphic to $G \overset{in_1}{\rightarrow} G \times G \overset{p_2}{\rightarrow} G$? – Aphelli Sep 08 '23 at 06:30
  • @Aphelli By the middle morphism $G\times G\to G\times G: (a,b) \mapsto (a,a^{-1}b)$, right? So they were in fact easy cases. Thank you for your help! – Hyeongmuk LIM Sep 08 '23 at 06:43

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