How do I see that $\mathbb{RP}^4$ and $\mathbb{RP}^6$ do not admit fields of tangent $2$-planes?

Question

A manifold $M$ is said to admit a field of tangent $k$-planes if its tangent bundle admits a subbundle of dimension $k$. How do I see that $\mathbb{RP}^4$ and $\mathbb{RP}^6$ do not admit fields of tangent $2$-planes?

Answer 1

It has no subbundles at all. If it did, pulling back along the projection map $p: S^{2n} \to \Bbb{RP}^{2n}$ would get you a nontrivial subbundle of $TS^{2n}$, which would get you a splitting $\xi \oplus \eta = TS^{2n}$. Now, note that beacuse $H^1(S^{2n}) = 0$, and hence that $w_1(\xi) = 0$ and $\xi$ is orientable (and the same for $\eta$). Now looking at Euler classes, $$0=e(\xi)e(\eta) = e(\xi \oplus \eta) = e(TS^{2n})=2.$$ The first equality is because the cohomology of $S^{2n}$ is zero in degrees less than $2n$.