MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $F$ be a smooth foliation of a torus. Assume that $F$ can be mapped by a homeomorphism to an irrational-straight-line foliation $L$. Does it follow that $F$ can be mapped to $L$ by a diffeomorphism?

I am interested in 2 dimensional case and higher dimensional case.

share|cite|improve this question
up vote 6 down vote accepted

As far as I understand, the answer to the question is no, you can check this in Handbook of Dynamical Systems, Volume 1, Part 1 By Boris Hasselblatt, Anatole Katok, page 173. This question is identical to the following -- suppose we have a diffeo of S^1 toplogically conjugate to an irrational rotation, can we make this conjugation a diffeo? Here is the sitation from the book:

For smooth and analytic circle diffeomrophisms with extremely well approximable rotation number, the conjugacy to a roation and hence the invariant measure tend to be singular. Arnold's theorem exposed sigularity of the conjugacies as a generic phenomenon in typical one-parameters families of real-analityc maps

share|cite|improve this answer
I guess this means that the third step of my "proof" is broken. – Sam Nead Jul 11 '10 at 22:30

The second sentence is false. For an example, take the Reeb foliation of the annulus and glue the boundary components. See the first figure in this article for a picture.

share|cite|improve this answer
Second sentence is an assumption. BTW, it's for counterexample it's simpler to take foliation by circles. – Ainu Jul 11 '10 at 15:42
Thank you for editing your question - it is much clearer now. I'll think about it. – Sam Nead Jul 11 '10 at 16:25

After pondering your question, I believe that in low dimensions (less than four?) this sort of thing is true - namely if two smooth objects are homotopic then they are diffeotopic. Perhaps a reference for this can be found in Thurston's book - I don't have it at hand. I'll try to look for the reference tomorrow.

To actually give all of the details appears to be an extremely delicate exercise. Here is my attempt, with the last step basically omitted:

  • Use smoothness, irrationality, and closing lemma-type argument to find a simple closed curve $\alpha$ that is transverse to $F$.
  • Find a diffeotopy that turns $\alpha$ into a geodesic. Apply this diffeotopy to $F$.
  • The foliation $F$ now gives a diffeomorphism $f$ of $\alpha$, namely the first return map (check). Using a partition of unity, any diffeotopy of $\alpha$ can be used to give a diffeotopy of the two-torus, supported in a tubular neighborhood of $\alpha$. Since we can smoothly conjugate $f$ to be a rotation (this is the same problem down one dimension) there is a diffeotopy of $\alpha$ starting at the identity and ending at the conjugating map. So we may diffeotope $F$ to arrange that the first return map $f$ is the desired rotation.
  • In a neighborhood of $\alpha$, diffeotope $F$ to have the correct angle (computed using the rotation number of $f$). Similarly, we diffeotope $F$ to be straight in a tubular neighborhood of $\alpha$. (A "hair combing" diffeotopy.)
  • Pick any point $x \in \alpha$ and let $\gamma$ be the leaf of $F$ connecting $x$ to $f(x)$, inside the annulus $T - \alpha$. Using another partition of unity, diffeotope $\gamma$ to be straight. An additional diffeotopy ensures that $F$ is straight in a tubular neighborhood of $\gamma$.
  • Finally, we need to straighten $F$ inside of the box $B = T - (\alpha \cup \gamma)$, given that $F$ is straight with respect to the boundary. (This seems to be the heart of the matter - You can't use Alexander's trick and stay in the smooth category...)
share|cite|improve this answer
thank you – Ainu Jul 11 '10 at 23:24

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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