# Uncountably many 3 dimensional foliations on a 5 dimensional torus

This is a problem on a review for some upcoming quals:

Prove there are uncountably many 3-dimensional foliations on a 5-dimensional torus.

Unless I am looking at this wrong it seems like a pretty straight forward intuitive sort of problem. My mind first went to just the plain old torus $\mathbb{R}^2/\mathbb{Z}^2$. There are uncountably many 1-dimensional foliations. This can be seen by fixing a slope and taking lines through every point with the given slope. If the slope is rational the lines will eventually double back on themselves and form a loop. If they are irrational they will go on forever in both directions.

Similarily, for $\mathbb{R}^3/\mathbb{Z}^3$ pick a vector and then at every point pick a plane going through that point tangent to the vector. This gives uncountably many foliations of dimension 2.

For the problem given I suspect it's the same idea. In $\mathbb{R}^5$ there are uncountably many 3 dimensional planes. Line them all up facing the same direction, and in the quotient you will have a 3-dimensional foliation of $\mathbb{R}^5/\mathbb{Z}^5$.

I'm wondering if this is the way I want to look at this problem, or if there is a better way to view it.

Thanks :)

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An important question here is when do you count two foliations as distinct? The foliations of the 2-torus that you've described fall into two isomorphism classes up to homeomorphism of the torus, so I would only count two of them. If you don't mod out by surface homeomorphisms, you easily get uncountably many examples by taking a disk in the surface and applying any homeomorphism that preserves its boundary, introducing local kinks into the foliation but not altering it in an essential way. – Grumpy Parsnip Jul 22 '11 at 20:39
@Jim you make a very good point. I hadn't really thought of this to be honest. Looking back through the notes the professor wrote up for the class I see that he only briefly addressed foliations, and never talked about equivalence of them. He was also the person that wrote the practice exam. It seems what you described must be what he intended, as the way I was looking at it seems almost trivial. – MJoszef Jul 22 '11 at 21:02
@Jim We defined a k-dimensional foliation (on an n-dimensional manifold) as a collection of charts such that all the transition functions from $\mathbb{R}^k\times\mathbb{R}^{n-k}$ to $\mathbb{R}^k\times\mathbb{R}^{n-k}$ are of the form $(x,y)\rightarrow(f(x,y),g(y))$. I assume different foliations from this definition are ones with different charts. No I have to think about what this fits in with what you said though, and see if there is some equivalence. – MJoszef Jul 22 '11 at 21:03