Let be $M$ a $n-$dimensional manifold. A distribution $D$ on $M$ is an assignment of subspace $D\_m \subset T\_mM$, for all $m\in M$.


A distribution $D$ on $M$ is said to be locally constant if for every $m\in M$ there is an open neighbourhood $U$ of $m$ such that $dim (D\_u)=k$ for all $u\in U$.


Is there any theorem which say that every involutive distribution is locally constant? 


I think to use the Frobenius theorem which say that an involutive distribution is integrable, so the dimension of $D$ is constant on each integral manifold, but the problem is that those integral manifold are not necessary open.



Sufficient condition for a distribution to be locally constant
Here is a counterexample to such a theorem. Choose a point $x \in M$, and take $D\_x = \{0\}$ and $D\_m = T\_m M$ for $m \neq x$. Then $D$ is involutive (in the sense that for any vector fields $X, Y$ in $D$, the bracket $[X,Y]$ is also in $D$), but not locally constant. Check it locally:
$$[\sum\_i f\_i \partial\_i,\sum\_i g\_j \partial\_j] = \sum\_{i,j} (f\_i (\partial\_i g\_j) \partial\_j - g\_j (\partial\_j f\_i) \partial\_i).$$
If $f\_i, g\_j$ are all zero at $x$, then the right-hand side is also zero at $x$.


Note that one of the assumptions of the Frobenius theorem is that the distribution is locally constant. 


Moreover, people usually include the "locally constant" condition in the definition of a distribution. Without this condition, it's usually called a *generalized distribution*.


