line bundles and spheres

Question

Suppose the tangent bundle of $S^{2n+1}$ has two trivial line subbundles. Are those real line bundles isomorphic? Does this imply that the orthogonal complements are isomorphic?

Answer 1

The group $\pi_n O(k)$ is in canonical bijection with vector bundles over $S^{n+1}$ of rank $k$ by what's known as the clutching construction. The effect of the inclusion homomorphism $\pi_k O(n) \to \pi_k O(n+1)$ in this language is to take the direct sum with the trivial bundle. Your question, then, asks what the kernel of $\pi_{2n} O(2n) \to \pi_{2n} O(2n+1)$ is.

Because our map $O(2n) \to O(2n+1)$ fits into a fiber sequence $$O(2n) \to O(2n+1) \to S^{2n},$$ the homomorphism of interest fits into a sequence $$\pi_{2n+1} O(2n) \to \pi_{2n+1} O(2n+1) \to \pi_{2n+1}(S^{2n}) \to \pi_{2n} O(2n) \to \pi_{2n} O(2n+1).$$ In particular, the kernel of this map is a quotient of $\Bbb Z/2 = \pi_{2n+1}(S^{2n})$ for $n>1$.

I claim that this cyclic group is $\Bbb Z/2$ for $n \geq 4$ and $n=2$, and is zero for $n=1,3$. That is to say, there are exactly two isomorphism classes of $(2n)$-plane bundles which embed as subbundles of $TS^{2n+1}$ for all $n$ except $1,3$.

I only know how to do this in a roundabout way: this paper of Kervaire lists off $\pi_{k} O(k)$ and $\pi_k O(k+1)$ in a range $k \geq 8$, and in all cases the groups are such that any map $\pi_{2n} O(2n) \to \pi_{2n} O(2n+1)$ necessarily has kernel once $n \geq 4$.

All that is left is the low-dimensional cases.

For $n=0$ there is of course nothing to say.

For $n=1$ we have explicitly that $\pi_2 O(2) = 0$, so the kernel is automatically zero.

For $n=2$ we can exploit the exceptional isomorphisms to get $\pi_4 O(4) = (\Bbb Z/2)^2$ and $\pi_4 O(5) = \pi_4 \text{Sp}(2) = \pi_4 \text{Sp}(1) = \pi_4 S^3 = \Bbb Z/2$. So the kernel must be $\Bbb Z/2$.

For $n=3$ we have following the exceptional isomorphisms that $\pi_6 O(6) \cong \pi_6 U(4)$. This is in the stable range, where we necessarily have $\pi_6 U(4) \cong \pi_6 U = \pi_0 U = 0$.

Thus for almost all odd-dimensional spheres, there is exactly one non-trivial bundle whose sum is isomorphic to the tangent bundle.

Note that this is not computing the number of non-isomorphic $2n$-plane distributions in $TS^{2n+1}$. That is different.

Edited to fix my typo. Thank you to Jason DeVito for pointing it out.