Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
1  
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
1  
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
2  
@Matt or just make it identically zero, can't be much more explicit than that... –  user31373 Jul 5 '12 at 21:50

1 Answer 1

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 $\phi$.

share|improve this answer

Your Answer

 
discard

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.