# Homotopy groups of the orthogonal group

I'm interested in knowing what $n$-dimensional vector bundles on the $n$-sphere look like, or equivalently in determining $\pi_{n-1}(SO(n))$; here's a case that I haven't been able to solve.

Let $n \equiv 1 \mod 8, n >1$. The fiber sequence $SO(n) \to SO(n+1) \to S^n$ yields a long exact sequence $$\pi_n S^n \to \pi_{n-1} SO(n) \to \pi_{n-1} SO(n+1) \to \pi_{n-1} S^n = 0.$$

The group $\pi_{n-1}(SO(n+1))$ is in the stable range and is equal to $\pi_{n-1}(SO) = \mathbb{Z}/2$ by Bott periodicity. The image of the generator of $\pi_n S^n$ is the class of the tangent bundle in $\pi_{n-1} SO(n)$, and this is annihilated by 2 (because $2$ is in the image of $\pi_n SO(n+1) \to \pi_n S^n$) and no less (as $S^n$ is not parallelizable). Consequently, we have a short exact sequence $$0 \to \mathbb{Z}/2 \to \pi_{n-1} SO(n) \to \mathbb{Z}/2 \to 0.$$

Does this split?

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Yes, this is in Corollary 6.14 of Mimura-Toda's "Topology of Lie groups". – Fernando Muro Mar 9 '12 at 23:58
Thanks! This answers the question. – Akhil Mathew Mar 10 '12 at 21:55
You're welcome :-) – Fernando Muro Mar 14 '12 at 11:05
To perhaps save someone else the trouble of searching the chapters, this is Cor. 6.14 in Ch. IV, p. 217. – Ulrik Buchholtz Mar 28 at 15:12