Let $n$ be a nonnegative integer and $x\in S^n$ a point in the n-sphere.

Combining the map $\alpha\colon SO_{n+1}\longrightarrow S^n$ induced by matrix multiplication with $x$ and the connecting homomorphism of the fibration $SO_{n+1}\rightarrow SO_{n+2}\rightarrow S^{n+1}$, one obtains a group homomorphism:

$$\mathbb{Z}\cong\pi_{n+1}(S^{n+1})\rightarrow\pi_n(SO_{n+1})\rightarrow\pi_n(S^n)\cong \mathbb{Z},$$ which is multiplication with some $\lambda\in\mathbb{Z}$. The number $\lambda$ is (up to sign) independent of the chosen isomorphism and the point $x$.

I want to calculate $\lambda$. Can somebody give me a hint?


I finally found an answer to my question: $\lambda$ is up to sign always $2$ if $n$ is odd and $0$ if $n$ is even.

A sketch of a proof can be found in Levine's "Lectures on groups of homotopy spheres" on page 64.


I think the answer to your question will vary widely depending on $n$. For example, for $n=3,5$ and $7$ (mod 8), $\pi_{n-1}SO_{n}$ is known to be $\mathbb{Z}/2$, which is torsion (hence $\lambda=0$). There is a complete list of the homotopy groups $\pi_{n-1}SO_{n}$ in the book "Topology of Lie groups" by Mimura Toda.

  • $\begingroup$ I don't think that $\lambda$ will vary widely. I expect $\lambda$ up to sign depending only on $n$ being even or odd. $\endgroup$ – archipelago Jan 12 '15 at 17:59
  • $\begingroup$ @archipelago: Did you read my answer? I gave a large class of examples where $\lambda=0$. $\endgroup$ – user113529 Jan 12 '15 at 19:48
  • $\begingroup$ Yes. Those are all odd, so it fits to my expectation. $\endgroup$ – archipelago Jan 12 '15 at 19:53

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