Real vector bundles on $S^{7}$

Question

Is it true that $\pi_{6}(O(n))=0$ for all n? Equivalently, are all real bundles on $S^{7}$ trivial?

Answer 1

Because of the covering space $SU(2) \to SO(3)$ given by the quaternions (cf. this document, for example), we know that the universal cover of $SO(3)$ is homeomorphic to $SU(2) \simeq S^3$.

This implies that $\pi_6 SO(3) \simeq \pi_6 S^3 \simeq \mathbb Z/12$ (according to that table).

So there is a nontrivial rank 3 bundle on $S^7$.

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Once the few remarks I made in my comment are clear, this article by Michel Kervaire gives the definitive answer (p. 162): $$\pi_6 SO(3) \simeq \mathbb Z/12\qquad \pi_6 SO(4) \simeq (\mathbb Z/12)^2\qquad \pi_6 SO(5) = \pi_6 SO(6) = \pi_6 SO(7) = 0.$$

Note that $\pi_6SO(4)$ also comes from the quaternions: indeed, they also give the universal cover $SU(2) \times SU(2) \to SO(4)$, so $\pi_6(SO(4)) \simeq (\pi_6(S^3))^2 \simeq (\mathbb Z/12)^2$.