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Suppose $X$ is a nowhere vanishing vector field on the 2-torus that preserves the standard area element $\mu=d\theta\wedge d\zeta$. By area preservation, $$ i_X\mu=dh+ad\theta+bd\zeta, $$ for some smooth function $h$ and constants $a,b\in\mathbb{R}$. Is there a diffeomorphism $\phi$ of the 2-torus such that $\phi^*X$ is a (re-scaling of) the constant vector field $Y=b\frac{\partial}{\partial\theta}-a\frac{\partial}{\partial\zeta}$?

Based on the discussion in this old paper by T. Saito

http://projecteuclid.org/DPubS/Repository/1.0/Disseminate?view=body&id=pdf_1&handle=euclid.jmsj/1261734857

it seems like the answer definitely could be yes, but I'm having a hard time proving it myself or finding a reference that addresses the question.

Progress so far

-When one of the constants, say $b$, is zero (note that both cannot be zero) the answer is yes. In this case one such $\phi$ is $\phi^{-1}(\theta,\zeta)=(\theta+\frac{1}{a}h(\theta,\zeta),\zeta)$.

-When the maximum value of $|\partial h/\partial\theta|^2+|\partial h/\partial\zeta|^2$ is less than $a^2+b^2$, then you can use Moser's trick (nice discussion of it here http://concretenonsense.wordpress.com/2009/09/03/symplectic-geometry-ii/) to prove the answer is yes. In particular, you can show that $dh+ad\theta+bd\zeta$ is strongly isotopic to $a d\theta+bd\zeta$.

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1 Answer 1

The orbits of the flow by the vector field $X$ forms a foliation $\mathcal{F}_X$ of $T^2$. There is a transverse measure to the foliation: for a curve $\sigma$ transverse to $\mathcal{F}_X$, define the measure of $\sigma $ to be $\int_\sigma i_X\mu$. Since the vector field $X$ and 2-form $\mu$ are preserved by the flow by $X$, this measure is invariant under the flow by $X$, and in fact by transverse isotopy to $\mathcal{F}_X$ rel endpoints.

I think it's well-known that a measured foliation $\mathcal{F}_X$ is homeomorphic to a foliation by lines of a fixed slope (in your case, it would be slope $a/b$). However, I don't know what regularity one can choose for this homeomorphism, in particular, is it a diffeomorphism?

Addendum: For one description of measure foliations, you can have a look at A primer on mapping class groups. However, I think the point I'm making is fairly simple. Consider a simple closed curve $\sigma$ transverse to $\mathcal{F}_X$. The curve $\sigma$ has a measure (absolutely continuous with respect to Lebesgue measure) coming from the transverse measure to $\mathcal{F}_X$. Every leaf of $\mathcal{F}_X$ must meet $\sigma$, and cutting $T^2$ along $\sigma$ gives an annulus, with a foliation consisting of intervals connecting both sides. This must be a product foliation, and the identification of opposite sides rotates $\sigma$ by some fraction $\alpha$ (since the flow from one side to the other preserves transverse measure). So take a Euclidean annulus $A$ with the same area at $T^2$, and with geodesic boundary components of the same length as the measure of $\sigma$. There is a canonical way to connect opposite sides by orthogonal lines, so glue opposite sides by an $\alpha$-fraction rotation of the circle. Then I think there is homeomorphism sending this torus to the original $T^2$, and sending the foliation by lines orthogonal to $\partial A$ to $\mathcal{F}_X$. Being a bit more careful, I think one modify this map to be an area-preserving map, but I haven't thought this through carefully.

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Thanks very much for your answer. It's pretty slow going for me as I now look through the literature in order to find a more detailed description of why this is true. Could you possibly point me to a reference, maybe a textbook? –  Josh Burby Nov 5 '12 at 5:51
    
The answer I gave I don't think really answers your question, so you could uncheck it and wait for a better answer. –  Ian Agol Nov 5 '12 at 16:22

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