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Suppose we have a (smooth) manifold $M$, and an integrable smooth distribution $\Delta$ on $M$. Somewhere, I read that we can define a natural map $\pi:M \rightarrow \frac{M}{\Delta}$.

First of all, what should I understand by $\frac{M}{\Delta}$? I guess that with the answer to this question I would be able to know also who is $\pi$.

For what I've also read, this quotient at least locally would be a manifold (locally because it seems like globally it could fail to be Hausdorff) but I don't know how to prove this, since I don't even know what is this quotient.

Any idea will be useful.

Paco Pacotilla
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1 Answers1

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The quotient is just the set of leaves of the foliation, i.e. the set of maximal integral manifolds of the distribution $\Delta$. The map $\pi$ is the obvious one, sending a point to the leaf it is an element of.

Eric Wofsey
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  • So, by why you're saying, $\frac{M}{\Delta}$ should have the same dimension as a manifold as the distribution $\Delta$, but this is not true. There's any other interpretation in which this quotient has dimension $n-dim \Delta$? Because it should be this way, according to the text I'm reading. – Paco Pacotilla Jan 17 '16 at 15:15
  • No, the dimension of $M/\Delta$ is $n-\dim\Delta$, at least if you take the quotient locally so that it is a manifold. Locally you can choose a coordinate chart so that $M\cong\mathbb{R}^n$ and the leaves are just the sets ${x}\times\mathbb{R}^{\dim\Delta}$ for $x\in\mathbb{R}^{n-\dim\Delta}$, and then the quotient can identified with the set of such $x$. – Eric Wofsey Jan 17 '16 at 17:56
  • Thanks @Eric, I was confused cause you first said that the quotient is the set of leaves, and I think you were trying to say that any leaf goes to a point, it's all clear now. – Paco Pacotilla Jan 17 '16 at 18:27