# The 4-sphere does not admit dimension 2 foliations

I want to know if the 4-sphere admits dimension 2 foliations. I found the following theorem in a dissertation by Jonathan Bowden (this is googleable, but I won't link it because I don't know exactly what the copyright issues might be). Anyway, here it is:

A closed, oriented 4-manifold admits an oriented 2-plane distribution if and only if there exists a pair $K_+,K_-\in H^2(M)$ such that

$$\langle K_\pm^2,[M]\rangle = \pm 2\chi(M)+3\sigma(M)$$ $$K_\pm\equiv w_2(M)\mod 2$$

Since a requirement for a $n$-dimensional foliation is a 2-plane distribution, and a sphere has trivial middle homology, I think this implies the 4-sphere has no dimension 2 foliations. Am I correct?

EDIT: For any complex manifold $M$ we have the canonical line bundle $K_M=det_{\mathbb{C}}T^*_M$, and

$$K_M\cdot K_M=3\sigma (M)+2\chi(M)$$

So I'm not sure about the equivalence of these notations; I think that pairing over $[M]$ might be implied in the second one since if you represent these classes as forms you need to integrate over the manifold to get integers. But then again the 4-sphere is not a complex manifold, but maybe there is some connection between complex 2-manifolds and when they can be foliated in dimension 2?

EDIT': I have posted a followup: Examples of 2-dimensional foliations of a 4-sphere.

Assuming the validity of that statement, your reasoning looks good. I don't know the answer to your question about complex manifolds, but there is an easy way (if you know about characteristic classes) to show that $S^4$ does not admit a dimension 2 distribution (let alone foliation). Indeed the following generalization of the hairy ball theorem is true:
$TS^{2n}$ does not admit any nontrivial subbundles (i.e. $S^{2n}$ has no nontrivial distrubtions).
To see this, assume for contradiction that $E \subset TS^{2n}$ is a subbundle with $1 \le \text{rank } E \le 2n-1$. Then you can find a complement $E^\perp$ with $E\oplus E^\perp = TS^{2n}$ . The properties of the euler class gives $$e(TS^{2n}) = e(E)\cup e(E^\perp).$$ But $e(E) \in H^{\text{rank } E}(S^{2n};\mathbb Z) = 0$, giving $e(TS^{2n}) = 0$. This is a contradiction since $$2 = \chi(S^{2n}) = \int_{S^{2n}} e(TS^{2n}) = 0.$$
• I am obviously missing something. I thought a trivial distribution would be analogous to a trivial bundle - if one exists (of dimension $m$) we can always find a subspace of the tangent space at any point with that dimension. But it sounds like a trivial distribution always has dimension 0? – levitopher Jun 4 '13 at 5:34
• @levitopher Sorry for the confusion, by trivial I meant either the zero vector bundle or all of $TS^{2n}$. – Eric O. Korman Jun 4 '13 at 17:18