2
$\begingroup$

Let $M$ be a smooth n dimensional manifold. The foliation values of $M$, denoted by $F(M)$, is defined as

\begin{equation} F(M)=\{ 1\leq k\leq n\mid \text{there exist an smooth $k$ dimensional foliation of M} \}\end{equation}

Question:

For $n\in \mathbb{N}$, What is $F(S^{n})$?

$\endgroup$
9
  • 2
    $\begingroup$ for $n$ even, it is $\{n\}$, because the Euler class implies that the tangent bundle is irreducible. For $n$ odd it always includes $1$. Not sure in about other $k$. $\endgroup$
    – Will Sawin
    Jan 18, 2014 at 23:11
  • 1
    $\begingroup$ Reeb constructed (in hs thesis) a codimension 1 foliation of the 3-sphere and from that H. B. Lawson, Jr, constructed codimension 1 foliations on spheres of dimensions $2^k+3$ in [Annals of Mathematics, Second Series, Vol. 94, No. 3 (Nov., 1971), pp. 494-503] So sometimes we do have $n-1$ in $F(S^n)$. Durfee generalized this to odd dimensional spheres. Thurston proved a couple of years later that in fact codimension 1 foliations exists in every closed manifold of Euler characteristic zero. $\endgroup$ Jan 19, 2014 at 2:37
  • 3
    $\begingroup$ Ah! Thurston himself tells of all this in a short paper in the Proc. of the ICM 74 in Vancouver. MathSciNet chasing is fun :-) See mathunion.org/ICM/ICM1974.1/Main/icm1974.1.0547.0550.ocr.pdf $\endgroup$ Jan 19, 2014 at 2:48
  • 1
    $\begingroup$ $S^{4n-1}$ gets a 3-dimensional foliation from the diagonal free action by $SU(2)$ (or unit quaternions). $\endgroup$
    – Ian Agol
    Jan 19, 2014 at 19:28
  • 2
    $\begingroup$ Also, $S^{15}$ is foliated by $S^7$'s, from the fibration $S^7\to S^{15}\to S^8$. $\endgroup$
    – Ian Agol
    Jan 19, 2014 at 20:05

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.