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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 $F(S^{n})$?

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    $\begingroup$ 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
    Commented Jan 18, 2014 at 23:11
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    $\begingroup$ 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. $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Jan 19, 2014 at 2:37
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    $\begingroup$ 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 $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Jan 19, 2014 at 2:48
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    $\begingroup$ $S^{4n-1}$ gets a 3-dimensional foliation from the diagonal free action by $SU(2)$ (or unit quaternions). $\endgroup$
    – Ian Agol
    Commented Jan 19, 2014 at 19:28
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    $\begingroup$ Also, $S^{15}$ is foliated by $S^7$'s, from the fibration $S^7\to S^{15}\to S^8$. $\endgroup$
    – Ian Agol
    Commented Jan 19, 2014 at 20:05

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