Foliations of spheres 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?
 A: 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
A: 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.
