# Existence of complementary pairs of foliations on spheres

Let $M$ be an $n$-manifold, $0\leq k\leq n$. We define a $(k,n-k)$-bifoliation on $M$ to be a pair $(\mathscr{E},\mathscr{F})$ consisting of ($C^\infty$ nonsingular) foliations $\mathscr{E},\mathscr{F}$ on $M$ of dimension $k$ resp. $n-k$ which are complementary=transverse in the sense that $TM$ is the internal direct sum of $T\mathscr{E}$ and $T\mathscr{F}$.

I am interested in the case where $M$ is a sphere:

Q1. For which $k$ does $S^n$ admit a $(k,n-k)$-bifoliation?

Relevant facts and partial answers:

• Every $n$-manifold admits a unique $(0,n)$-bifoliation.
• If $(\mathscr{E},\mathscr{F})$ is a $(k,n-k)$-bifoliation, then $(\mathscr{F},\mathscr{E})$ is an $(n-k,k)$-bifoliation.
• If $S^n$ admits a $(k,n-k)$-bifoliation with $1\leq k\leq n-1$, then $n$ is odd: $TS^n$ does not split if $n$ is even.
• Every odd-dimensional $n$-manifold $M$ admits a $(1,n-1)$-bifoliation: It admits a nowhere vanishing vector field, hence an $(n-1)$-plane field. By W. Thurston: Existence of codimension-one foliations, it therefore admits an $(n-1)$-dimensional foliation $\mathscr{F}$. Any line subbundle $E\subseteq TM$ that is complementary to $T\mathscr{F}$ is the tangent bundle of a $1$-dimensional foliation.
• There exist $1$-dimensional foliations $\mathscr{E}$ on $S^3$ such that $S^3$ does not admit any $(1,2)$-bifoliation of the form $(\mathscr{E},\mathscr{F})$: see Tamura/Sato: On transverse foliations, §1, Example 2.
• $TS^n$ splits off a trivial sub vector bundle of rank $k$ if and only if $k\leq 2^p+8q-1$, where $p\in\{0,1,2,3\}$ and $q,r\in\mathbb{N}$ are determined by $n+1 = (2r+1) 2^p 16^q$; see e.g. here. If $TS^n$ splits off a vector bundle of rank $k\leq\frac{n}{2}$, then it splits off a trivial vector bundle of rank $k$; see Steenrod: The topology of fibre bundles, Theorem 27.16.
• Every homotopy class of $2$-plane fields on $S^n$ contains the tangent bundle of a foliation; see W. Thurston: The theory of foliations of codimension greater than one (Corollary 3).
• Let $k>\frac{n}{2}$ or $k=3$. Then $S^n$ admits a $k$-dimensional foliation if and only if $S^n$ admits a $k$-plane field: For the case $k>\frac{n}{2}$, see W. Thurston: The theory of foliations of codimension greater than one (Corollary 2). The obvious fibration $S^3\to S^{4m+3}\to\mathbb{HP}^m$ yields a $3$-dimensional foliation on $S^{4m+3}$.

Therefore the general problem Q1 raises the following questions (in particular about $k=2$):

Q2. Can you prove for some $n$ (necessarily $n=4m+3\geq7$) and some $k\in\{2,\dots,n-2\}$ that $S^n$ admits a $(k,n-k)$-bifoliation?

Q3. Can you prove, for some $k$ and some $n$-manifold $M$, that $M$ does not admit a $(k,n-k)$-bifoliation, but admits foliations $\mathscr{E},\mathscr{F}$ of dimensions $k$ resp. $n-k$ such that $T\mathscr{F}$ is homotopic to a sub vector bundle of $TM$ which is complementary to $T\mathscr{E}$? For instance, can you prove for some $m$ that $S^{4m+3}$ does not admit a $(2,4m+1)$-bifoliation?

Q4. What is known about (non-)existence of $k$-dimensional foliations on $S^n$ in the case $4 \leq k < \frac{n}{2}$?

In the definition of $(k,n-k)$-bifoliation, I assumed the foliations to be $C^\infty$. Can something interesting be said under other regularity assumptions, e.g. $C^1$?

• To me this looks more like a research project than like a mathoverflow question :-) – ThiKu Aug 9 '17 at 19:54
• @ThiKu: Answering Q1 completely would be a very ambitious research project, but asking for an answer (better than a simple "no") to Q2 or Q3 seems like a legitimate mathoverflow question to me, although it's maybe not an easy one. I am willing to accept an answer with a single example concerning Q2 or Q3. Also, I don't expect that the correct answer to Q4 is "nothing". – Marc Nardmann Aug 9 '17 at 22:38
• I had overlooked that Q4 has been asked before on mathoverflow. There is a 7-dimensional foliation on $S^{15}$: link. – Marc Nardmann Aug 10 '17 at 12:12
• @MarcNardmann What about $\mathbb{C}P^2$?Does it admit a 2 dimensional real foliation? If yes is there such a foliation whose leaves are totally geodesics immersed submanifolds of $\mathbb{C}P^2$ where the later is equiped with the Fubbini - study metric? – Ali Taghavi Oct 18 '17 at 6:04
• @Ali Taghavi: This is off-topic here, but anyway: For $M=\mathbb{CP}^2$, assume there exist $\mathbb{R}$-vector bundles $E_0,E_1$ of rank 2 with $TM=E_0\oplus E_1$. We have $c_1(TM)=3$ and $c_2(TM)=3$, thus $w_2(TM)=1$ and $w_4(TM)=1$; see Milnor/Stasheff 14.10, 14-B. Moreover, $w(E_i)=1+w_2(E_i)$ and $w(TM) = w(E_0)w(E_1)$, thus (in $H^{2k}(M;\mathbb{Z}_2)=\mathbb{Z}_2$) $1=w_2(TM)=w_2(E_0)+w_2(E_1)$ and $1=w_4(TM)=w_2(E_0)w_2(E_1)$, which is impossible. Hence $M$ does not even admit a 2-plane distribution. – Marc Nardmann Oct 19 '17 at 17:43