10
$\begingroup$

Let $T=S^1\times S^1 \times S^1$ be the 3-torus. Let us denote by $\alpha$ the isotopy class of the loop $S^1\times pt\times pt$ and let $\mathcal F_\alpha$ be the set of all smooth oriented foliations $F$ of $T$ by circles such that every leaf of $F$ lies in the class $\alpha$ (in particular $F$ yields a smooth principal $S^1$-bundle over the 2-torus having $T$ as total space).

Question: Is $\mathcal F_\alpha$ connected?

$\endgroup$
  • $\begingroup$ if the $S^1$ fibers tangle in any way, then I suspect they are not homotopic to the "trivial" foliation, $S^1 \times pt$. Is that right? $\endgroup$ – john mangual Mar 9 '17 at 16:19
  • 3
    $\begingroup$ A: Yes. Such foliations are Seifert fiberings, Using things like Alexander's theorem you can classify the the Seifert fiberings of manifolds that admit them, even up to isotopy. For example, take a look in Hatcher's 3-manifolds notes. $\endgroup$ – Ryan Budney Mar 9 '17 at 16:31
  • 3
    $\begingroup$ The fact that circle foliations are Seifert fibrations (for arbitrary 3-manifold, even a noncompact one, and even without smoothness assumption) is due to D.B.A.Epstein "Periodic flows on 3-manifolds" Annals of Math, 1972. $\endgroup$ – Misha Mar 9 '17 at 21:16
  • $\begingroup$ @johnmangual: I edited the question, so that now $\alpha$ is the isotopy class of the loop and not the homotopy class (maybe they are different and this could answer your point). $\endgroup$ – Gabriele Benedetti Mar 10 '17 at 19:16
  • $\begingroup$ @RyanBudney: Thanks a lot for the reference. I had a look at Hatcher's notes. For him $T^3$ is the exceptional case $M_1$ on p.37, for which he does not seem to provide the isotopy classification (but maybe I am wrong). I also found Theorem 5.2 (originally due to Waldhausen) in these notes by Jankins and Neumann math.columbia.edu/~neumann/preprints/… which asserts that except some cases ($T^3$ is case iii) two homeomorphic Seifert fiberings are actually isotopic. $\endgroup$ – Gabriele Benedetti Mar 10 '17 at 19:33
1
$\begingroup$

If I understood correctly it is enough to show that there exists a surface of section isotopic to $pt \times S^1 \times S^1$. After isotopy, you can assume that that surface is indeed the section and all circles of the foliation have the same length (see the paper D.B.A.Epstein "Periodic flows on 3-manifolds" Annals of Math, 1972. cited by Misha above). This allows to construct the desired diffeotopy to the canonical one.

To show that such a section exists, there exists an averaging trick going (at least) back to Fuller (see Fuller, On the surface of section and periodic trajectories, American J. of Math 1965) that allows to construct a surface of section in any homology class which is 'transverse' to the directions spanned by periodic orbits.

I think this is enough for what you are looking for.

$\endgroup$

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.