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Consider the 2-dimensional torus $T^2=\mathbb{R}^2/\mathbb{Z}^2$, and a foliation on it (for example a foliation in circles, maybe the partition of the torus obtained form a Hopf-related map). I'm wondering if there are some condition on the foliation to be (the union of) the integral curves of a suitable vector field $X_\tau$ defined on the torus...

Note: One can clearly ask something more general (generic group action on a smooth manifold whose orbits are leaves of a given foliation), but I'm really dumb on making good (=well defined) questions so for the moment let's talk about a particular case.

Note 2: I'm not requiring much smoothness for $X_\tau$ just because I suspect that the answer will be "No if you suppose $X_\tau$ is not $C^k$-smooth with $k\ge k_0$".

Thanks a lot!

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    $\begingroup$ I am not quite sure I understand your question. Given a two dimensional manifold and a co-dimensional 1 $C^1$ foliation, the tangent spaces of your foliation define a 1-dimensional $C^0$ distribution in the tangent space of your ambient manifold. So are you asking just whether this distribution is generated by a (non-vanishing) vector field? $\endgroup$
    – Willie Wong
    Commented Sep 18, 2011 at 12:34
  • $\begingroup$ I'm wondering if there exists a vector field whose integral curves are precisely the leaves of the foliation... but, yes, as far as i know something about distributions you only restated this in a different way. $\endgroup$
    – fosco
    Commented Sep 18, 2011 at 14:18

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A foliation of a torus need not arise from a vector field, because its tangent line field may be non-orientable.

E.g., representing the torus by a square whose opposite sides are identified, think of a stack of arches (convex downward) filling up the square, except for the vertical sides -- which will constitute a single closed leaf.

In terms of $\mathbb T^2 = \mathbb R^2 / \mathbb Z^2$, this is represented by the real analytic tangent line field defined by

$$L(x,y) := \exp((x + 1/2) \pi i)$$

where antipodal points on the circle of angles are identified.

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  • $\begingroup$ Welcome to Math.SE, and thank you for this answer. I reformatted the formulas using LaTeX. $\endgroup$
    – user53153
    Commented Jan 7, 2013 at 22:55

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