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Suppose I have some closed $4$-manifold $X$ and a codimension-two foliation $\mathcal{F}$, as well as a closed surface $\Sigma$ of nonnegative self-intersection that is everywhere transverse to $\mathcal{F}$.

Then what kind of restrictions are there on the foliation $\mathcal{F}$? This question gives some answers in the case where $X$ is a complex surface and $\mathcal{F}$ is holomorphic, but I'm more interested in what happens in the real case.

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    $\begingroup$ For the special case that $\Sigma$ is the fiber of a surface bundle structure for $X$, the foliations you ask about are completely classified. That probably barely scatches the surface of the problem though. $\endgroup$
    – guest
    Commented Aug 19, 2020 at 0:37
  • $\begingroup$ What is the classification in this case? $\endgroup$
    – Rohil Prasad
    Commented Aug 19, 2020 at 0:51
  • $\begingroup$ Foliations transverse to the fibering of a fiber bundle with compact fiber, in any dimension and codimension, correspond to conjugacy classes of representations of $\pi_1(B)$, into $Homeo(F)$. Where $B$ is the base space and $F$ is the fiber. Let $\rho$ be such a representation, and let $\tilde{B}$ be the universal cover of $B$ and let $G = \{(\alpha, \rho(\alpha)\}$ where $\alpha$ ranges over $\pi_1(B)$. To construct the foliation, mod out $\tilde{B} \times F$ by $G$. The leaves descend from the product foliation of $\tilde{B} \times \{point\}$, and leafwise cover $B$ under bundle projection. $\endgroup$
    – guest
    Commented Aug 19, 2020 at 1:48

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In this real case, there are few restrictions. Indeed, choose $\Sigma\subset X$ such that $X$ admits a smooth 2-plane field $\xi$ (not necessarily integrable) transverse to $\Sigma$. Then, it is easy to perturb slightly $\xi$ to make it integrable on a small neighborhood of $\Sigma$. Then, by a theorem of Thurston (Commentarii 1974), $\xi$, being of real dimension $2$, can be homotoped rel. $\Sigma$ to become integrable everywhere. You can even begin with extending $\xi$ to a partial foliation of your choice over any regular subset of $X$. So, the possibilities are enormous.

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