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The Hopf fibration $S^1\rightarrow S^3\rightarrow S^2$ gives a decomposition of $S^3$ into 2-tori and to circles, so that the tori are foliated by circles of slope 1. If you take the region between two of this tori and identified by an homeomorphism that preserves the foliation you get a foliation by circles on the 3 torus, which I believe is equivalent to the foliation of $S^1$ on $S^1\times S^1\times S^1$ given by its product structure. Is this true?

Also if you remove the two circles of the decomposition I mentioned you get a fibration $S^1\rightarrow S^1\times S^1\times \mathbb R\rightarrow \ S^1\times \mathbb R$ which I believe is given by the multiplication in the first entry and projection in the other two. Is this also true?

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Yes, Hatcher's 3-manifolds notes are a short reference for this. But there's also the book on Seifert fiberings by Orlik that's much more elaborate on this topic. – Ryan Budney Jun 1 '13 at 20:33
The plural of torus is tori. – Paolo Ghiggini Jun 1 '13 at 20:46
up vote 4 down vote accepted

If you take the region between two of this tori and identified by an homeomorphism that preserves the foliation...

There is not a single isotopy class of (doubly orientation-preserving) homeomorphism that does this. Instead, there is a $\mathbb{Z}$-family of homeomorphisms, generated by a Dehn twist along a leaf of the foliation. For one of these you get the three-torus, but the others give Nil manifolds. But if you do get the three-torus, then the foliation is indeed standard.

To explain the comment "doubly orientation-preserving": A homeomorphism preserving the standard foliation of the two-torus by circles can

  1. reverse the orientation of the surface and/or
  2. reverse the orientation of the leaves.

You'll want to preserve both to get the $\mathbb{Z}$-family, generated by a Dehn twist.

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