9
$\begingroup$

A smooth "singular" distribution $D\subseteq TM$ on an $n$-dimensional manifold $M$ is integrable if it is tangent to immersed submanifolds $N_\alpha$ that are disjoint and cover $M$. If dim$D=k$ constant, then the distribution is regular and the dimension of the submanifolds are all the same. In this case we have Frobenius theorem, stating that the distribution is integrable if and only if it is involutive ($X,Y\in D\Rightarrow [X,Y]\in D$). An equivalent conditions (Frobenius condition) is phrased in terms of $n-k$ independent differential forms $\omega_i\in\Omega^1(M)$ with $1\leq i \leq n-k$ that vanish on $D$, i.e. $\langle\omega_i,X\rangle=0 \;\forall i\forall X\in D$. So $D$ is integrable if and only if there are $1$-forms $\{\alpha^i_j\}_{i,j=1}^{n-k}$ such that \begin{equation} d\omega_i=\sum_j \alpha^j_i\wedge\omega_j \end{equation} and for co-dimension one foliations the condition reduces to $d\omega\wedge \omega=0$.

Question: Is there a similar condition for singular, but nevertheless integrable distributions? For the dimensions to work out there would then have to be a set of $n-k$ one forms, where $k=\min_{x\in M}\dim D_x$, that are not linearly independent. Is anyone aware of a reference or a conditon similar to one above?

$\endgroup$
1

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.