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Explicitly construct a differentiable vector field $W$ in the torus.

Meridians of $T^2$ parameterized by arc length, for all $p \in T^2$, define $W (p)$ as the velocity vector of the meridian passing through $p$.

Maybe could help

$x=(R+r \cos \phi)\cos \theta$;

$y=(R+r \cos \phi)\sin \theta$;

$z=r \sin \phi$

Where $\theta, \phi \in [0, 2\pi[$ Thanks

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Is the first sentence the question and the rest is what you've tried? If it doesn't have to be nonvanishing you could easily make one on a coordinate chart, then multiply by a bump function to make it go to zero outside the chart. – Matt Jul 5 '12 at 15:51
Since you can represent $T^2 = \mathbb{R}^2/\mathbb{Z}^2$, just take a $\mathbb{Z}^2$-equivariant vector field on $\mathbb{R}^2$. – Neal Jul 5 '12 at 16:04
@Matt or just make it identically zero, can't be much more explicit than that... – user31373 Jul 5 '12 at 21:50
up vote 1 down vote accepted

I guess that the second question is actually a part of the exercise. Hint: the variable that parametrizes each meridian is $\phi$. Hence, the velocity vector of the meridian is the partial derivative with respect to $\phi$.

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