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I am not a mathematician, but am interested in computing this homotopy group. I have two questions related to this.

  1. How can I use the long exact sequence to compute this homotopy group? Part of it would look like $\pi_n\left(\mathrm{SO}(N) \right) \rightarrow \pi_n\left(\mathrm{SO}(2N) \right) \rightarrow \pi_n\left(\mathrm{SO}(2N)/\mathrm{SO}(N)\right)$, I think. How can I use this to compute the homotopy group of interest?

  2. Can I compute this homotopy group by finding a homomorphism between $\mathrm{SO}(2N)/\mathrm{SO}(N)$ and $\mathrm{SO}(2N)/\mathrm{U}(N)$, to show that $\pi_d\left(\mathrm{SO}(2N)/\mathrm{SO}(N) \right)$ is isomorphic to $\pi_d\left(\mathrm{SO}(2N)/\mathrm{U}(N) \right)$? This relates to the following proofwiki page: https://proofwiki.org/wiki/Homotopy_Group_is_Homeomorphism_Invariant.

Impossibear
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    $SO(2N)/SO(N)$ and $SO(2N)/U(N)$ are not homeomorphic and they do not have the same homotopy groups. – Michael Albanese Feb 13 '22 at 17:03
  • @MichaelAlbanese: Thank you for the comment. Do you have a reference (textbook or research paper) for the homotopy groups of $\mathrm{SO}(2N)/\mathrm{SO}(N)$? – Impossibear Feb 14 '22 at 13:49
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    Not that I know of. You just have to extract as much information as your can from the long exact sequence. By the way, the sequence in your post is incorrect. The sequence is $\dots \to \pi_{n+1}(SO(2N)/SO(N)) \to \pi_n(SO(N)) \to \pi_n(SO(2N)) \to \pi_n(SO(2N)/SO(N)) \to \pi_{n-1}(SO(N)) \to \dots$ – Michael Albanese Feb 14 '22 at 17:00
  • What regime of parameters $N$ and $d$ are you thinking of? As $N$ gets bigger, the lower homotopy groups stabilize. – AHusain Feb 14 '22 at 22:37
  • @AHusain: I am most interested in $d=1$,$2$,$3$. I am most interested in general $N$, but even large $N$ (such as the regime you mention) would be useful. Or, conversely, $N=3$ could also be useful. – Impossibear Feb 15 '22 at 10:26

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