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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.

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  • $\begingroup$ I can't think of an example of distribution that arises "in nature", for which we don't immediately see whether its subspaces have constant dimension, but for which we eventually come to that conclusion by using some theorem or some indirect information. $\endgroup$
    – Ben McKay
    Commented Nov 25, 2017 at 20:24
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    $\begingroup$ If you want to add smoothness as a hypothesis, you need to formulate it carefully. The example below (vector fields vanishing at the origin) is "smooth" in the sense that it is locally the span of smooth vector fields, for example the span of the linear vector fields, in any system of local coordinates. "Smoothness" in the sense of a smooth section of a Grassmann bundle will not allow jumping of the dimensions of the fibers. $\endgroup$
    – Ben McKay
    Commented Nov 27, 2017 at 19:01

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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.

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  • $\begingroup$ Thank you very much for your answer and sorry for the delay. I forgot to mention that here the distribution is supposed to be smooth, as I see your conter example is not smooth ... I am looking forward for your answer. $\endgroup$
    – Antema Kely
    Commented Nov 27, 2017 at 18:08
  • $\begingroup$ What do you mean by "smooth"? My distribution is generated by smooth vector fields $r^2 \partial_i$, where $r^2 = \sum_i x_i^2$. Can you define a "smooth distribution"? $\endgroup$
    – Piotr
    Commented Nov 28, 2017 at 19:02

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