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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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    @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.2019-05-16

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