Cohomology of foliations and closed forms along the leaves

Question

Let $M$ be a manifold equipped with a codimension one, transversely orientable, regular foliation $F \subset M$. Let $\alpha\in \Omega^k(M)$ be a differential form on $M$ that is not closed on $M$ but such that its restriction to every leaf of $F\subset M$ is closed, namely:

$\iota^{\ast}d\alpha = 0$

for every leaf immersion $\iota\colon \Sigma \hookrightarrow M$ with leaf $\Sigma$. Intuitively speaking $\alpha$ is a differential form on $M$ closed along the "leaves of a foliation" but not necessarily closed on $M$. Clearly, such $\alpha$ defines a cohomology class in the de Rham cohomology of every leaf. I was wondering if there was a more refined cohomology theory associated with the foliation $F$ for which such a differential form $\alpha$ would also define a canonical class.

Thanks.

Answer 1

I am hardly an expert on this topic, but here's a construction.

Let $\Omega^k:=\Omega^k(M)$, and let$$F\Omega^k=\{\omega\in\Omega^k:\omega(v_1,\dots,v_k)=0,\quad v_1,\dots,v_k\in T_x\Sigma,\ \Sigma\text{ is a leaf of }F\}.$$

This is a homogeneous ideal in $\Omega$ (if we define $F\Omega=\bigoplus _kF\Omega^k$), and because the distribution associated to $F$ is (obviously...) integrable, we have $\mathrm dF\Omega\subseteq F\Omega$, i.e. this is a differential ideal.

Define $F^p\Omega^q=F^p\Omega\cap \Omega^q$, where $F^p\Omega$ is the $p$th ideal power. Then $F^p\Omega\subseteq F^{p+1}\Omega$, so this is a filtration that is $\mathrm d$-compatible. Consider the spectral sequence $(E^{p,q}_r=E^{p,q}_r(F),d_r)$ associated to this filtration. If $m=\dim M$ and $n=\dim\Sigma$ ($\Sigma$ a typical leaf in $F$), then $E^{p,q}_r=0$ when $q>n$, so this is a first quadrant spectral sequence with $$ H^k(M,\mathbb R)\cong\bigoplus_{k=p+q}E^{p,q}_\infty,\quad E^{p,q}_\infty = E^{p,q}_r,\ r\ge n+1 $$.

By the definition, we have $E^{0,q}_0\cong\Omega^q/F\Omega^q$, $d_0:E^{0,q}_0\rightarrow E^{0,q+1}_0$ given by $d_0[\omega]=[\mathrm d\omega]$.

Let $\alpha\in\Omega^k$ be the form in the construction in the OP. Then $i^\ast\mathrm d\alpha=0$ means $\mathrm d\alpha\in F\Omega^{k+1}$, so the class $[\alpha]=\alpha\mod F\in E^{0,k}_0$ is $d_0$-closed, hence it determines an element $[[\alpha]]=[\alpha]\mod d_0\in E^{0,k}_1$.

The cohomology theory of the foliation is then essentially the study of the spectral sequence $E^{p,q}_r(F)$. The complex $(E^{0,\ast}_0,d_0)$ may then essentially be thought of as the "de Rham complex of the typical leaf with coefficients in functions on $M$", whereas the complexes $(E^{p,\ast}_0,d_0)$ for $p>0$ as being the "de Rham complexes of the typical leaf twisted by the bundle of $p$-forms transversal to the leaves".

The relationship between these and the de Rham complex on $M$ is then given by the convergence of the spectral sequence.

Answer 2

If $F$ is a regular foliation then we have a Lie algebroid on $M$ given as the (integrable) sub-bundle $L\subset T_M$ of vectors that are tangent to the leaves.

There is a Chevalley-Eilenberg complex $CE^\bullet(L)$ associated with this Lie algebroid, and the collection of pullbacks $\iota^*$ along leaves actually define a cochain map $$ r_F:DR^\bullet(M)=CE^\bullet(T_M)\to CE^\bullet(L). $$ Here $DR^\bullet(M)$ denotes the de Rham complex of $M$ (it coincides with the Chevalley-Eilenberg complex of the tangent Lie algebroid $T_M$). The condition $d\iota^*\alpha=\iota^*d\alpha=0$ tells you exactly that $r_F(\alpha)$ is closed in $CE^\bullet(L)$.

In other words, such an $\alpha$ defines a class in the Chevally-Eilenberg cohomology of $L$, a.k.a. leafwise de Rham cohomology for this particular kind of Lie algebroid.