Foliations of spheres

Question

Reading a paper, I came across the statement that there are no transversely orientable codimension-one foliations on even dimensional spheres $\mathbb{S}^{2n}$. Could anyone explain why?

Answer 1

Thurston has shown that if a closed manifold has a codimension one foliation, its Euler characteristic is zero. But the Euler number of the $2n$-sphere is $2$.

Existence of Codimension-One Foliations

W. P. Thurston

Annals of Mathematics

Second Series, Vol. 104, No. 2 (Sep., 1976), pp. 249-268

Answer 2

If I am not misunderstanding, even dimensional spheres don't have any foliations where the leaves have positive dimension, except the trivial case where the whole sphere is a leaf.

To see this, suppose $\mathcal{F}$ is a foliation of $S^{2n}$ and let $\mathcal{V}\subseteq TS^{2n}$ denote the sub-bundle of $TS^{2n}$ consisting of vectors tangent to leaves in $\mathcal{F}$.

Since vector sub-bundles always have complements, we get a decomposition $TS^{2n} = \mathcal{V}\oplus \mathcal{V}'$.

Now, compute the Euler class. We have $e(TS^{2n}) = 2\in H^{2n}(S^{2n})$, as Tsemo mentions, so $2 = e(\mathcal{V})\cup e(\mathcal{V}')$. Since $H^k(S^{2n})$ is non-trivial only when $k = 0$ or $2n$, it follows that $\mathcal{V}$ has rank $0$ or $2n$. That is, the leaves are either points, or full dimension.