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?
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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 $2n$.
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To consider Euler classes, don't you need $\xi$ and $\eta$ to be orientable subbundles? Are we guaranteed orientability? – Ted Shifrin Dec 15 '15 at 22:26
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@TedShifrin: Good point, I should have said this explicitly. I'll edit my answer. – Dec 15 '15 at 22:26
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Why is the pullback of a non-trivial subbundle on $\mathbb{RP}^{2n}$ by $p$ necessarily non-trivial? – Michael Albanese Dec 17 '15 at 01:35
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@MichaelAlbanese: I'm not sure what (non)triviality of a subbundle of $TM$ means. Note that in this answer all I need is that the subbundle of $TS^{2n}$ exist. – Dec 17 '15 at 01:36
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@Michael: Since this question was marked a duplicate, but this answer is different than the one on the other question, should I port it over? – Dec 17 '15 at 03:22
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I think you can leave it here. Both questions will be linked, so anyone who is interested can see the answers on either version. I'm not sure what the official site policy is though. – Michael Albanese Dec 17 '15 at 03:24
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Ok, just checking for some input. Thanks. – Dec 17 '15 at 03:25