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I'm working through Smooth Manifolds by Lee and came to problem 19-8, where you're asked to prove that the level sets of a submersion form a foliation of the domain. However when I try to solve this problem it seems like the assumption that the map is a submersion is not needed, ie. one need only assume the map is of constant rank. Lee's problems are usually carefully written to use minimal hypotheses, so I'm not sure if I'm wrong.

Let $F: M \rightarrow N$ be a map of constant rank $r$, with $m = dim(M)$ and $n = dim(N)$. The non-empty level sets are a collection $\mathcal{F}$ of disjoint immersed submanifolds of codimension $r$ whose union is $M$. To show that $\mathcal{F}$ is a foliation, I have to show that each $p \in M$ is contained in a flat chart for $\mathcal{F}$:

A smooth chart $(U,\phi )$ for $M$ is said to be flat for $\mathcal{F}$ if $\phi (U)$ is a cube in $\mathbb{R}^n$, and each submanifold in $\mathcal{F}$ intersects $U$ in either the empty set or a countable union of k-dimensional slices

The definition of a k-slice is

If $U$ is an open subset of $\mathbb{R}^n$ and $k \in \{0,...,n\}$, a k-dimensional slice of U (or simply a k-slice) is any subset of the form

$$S =\{ {(x^1,...,x^k,x^{k+1},...x^n) \in U: x^{k+1} = c^{k+1}, x^n = c^n } \}$$

for some constants $c^{k+1},...,c^n$.

By the rank theorem, for each $p \in M$ there is a coordinate representation of $F$ in a neighborhood $U$ of $p$ with the form $F(x^1,...,x^r,x^{r+1},...,x^m) = (x^1,...,x^r,0,...,0)$. We can assume $U$ is a cube, shrinking it if necessary.

Every submanifold in $\mathcal{F}$ is of the form $F^{-1}(q), q \in N$. There are two possibilities.

  • If $q \notin F(U)$, then $F^{-1}(q) \cap U$ is empty.

  • If $q \in F(U)$, then $q = F(a^1,...,a^m)$ for $(a^1,...,a^m) \in U$, and by the coordinate representation of $F$, we have that $F^{-1}(q)$ is the (m-r)-slice $\{ {(x^1,...,x^r,x^{r+1},...x^n) \in U : x^1=a^1,...,x^r=a^r} \}$

Thus I've shown that each submanifold in $\mathcal{F}$ intersects $U$ in a single (m-r)-slice, so $U$ is flat for $\mathcal{F}$. Is this correct?

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    $\begingroup$ Yes, it's correct. As a challenge, come up with an interesting foliation that comes from a map of constant rank but not from a submersion. If you can do so I'll bounty you 200 points. $\endgroup$
    – user98602
    Apr 5, 2016 at 2:35
  • $\begingroup$ @MikeMiller, locally I can reduce to the case of a submersion. You have some global scenario in mind? $\endgroup$ Apr 5, 2016 at 3:13
  • $\begingroup$ @Ted Yup, such a thing would make me care about constant rank foliations. $\endgroup$
    – user98602
    Apr 5, 2016 at 3:17
  • $\begingroup$ Mobius bandy thing?! $\endgroup$ Mar 20, 2018 at 3:49

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