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Give examples of vector fields on the $n$-dimensional torus.

What I have done:

on $S^1$ it's easy to give one example with perpendicular vectors of length $1$ rotating in one direction, and another example in the other direction. How many different vector fields are there on $S^1$? I know they are $X =$ $f$ $\cdot$ $d\over dx$ thus they should be an infinite vector space over real numbers.

on $T^2$ I think we can draw the square and draw arrows in just 1 direction to get a vector field, for example all the vertical arrows of length $1$. Again, how many different vector field are there on $T^2$?

on $T^n$ I think we can use that $T^n = S^1 \times \cdots \times S^1$ for $n$ times, maybe something like $X =$ $f_1$ $\cdot$ $d\over d\theta_1$ $+ \cdots +$ $f_n$ $\cdot$ $d\over d\theta_n$, but I don't know!

Michael Albanese
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Maffred
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1 Answers1

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It depends on what you mean by "how many". If you are really counting each one as different, then there is always an infinite number of vector fields.

However, each manifold you mention is parallelizable. It follows that they must have $n$ vector fields which are linearly independent at every point of the manifold. Note however that your question can't be properly addressed in general. For example, I don't know how one could make sense (in a non-trivial way) of the question "How many vector fields are there on $S^2$?".

$S^1$ is clearly parallelizable (you can just stack the tangent spaces vertically on the circle), and the product of parallelizable manifolds is parallelizable. Hence, $T^n$ is parallelizable.

With respect to your specific questions: Yes, that gives a vector field in $S^1$. Note that in the "parallelization" this is just the "constant" unit vector upwards, or downwards, depending on orientation.

The vector field which you propose in $T^2$ is also another example of a vector field. Note that in the standard embedding of $T^2 \hookrightarrow \mathbb{R}^3$, this vector field is a "flow of water on the surface of the pipe" given by the torus.

Aloizio Macedo
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  • Thank you! When you say that there are $n$ vector fields, do you mean that they generate a vector space over R of dimension $n$? I'm confused since I read that every vector field is given by the derivations multiplied each one by a function, this seems so big! – Maffred Jan 08 '16 at 03:28
  • No, I mean that there exists $n$ vector fields that, when you look at any point of the manifold, the vectors (of the vector fields) on that point are linearly independent. Note that those vector fields don't form a base for the space of vector fields at all. – Aloizio Macedo Jan 08 '16 at 03:44
  • Ok! Even if we only use vectors of length $1$? Is then true that they form a basis? – Maffred Jan 08 '16 at 03:57
  • No. Forget $S^1$ or $T^n$. Think about $\mathbb{R}^n$. Try to find a finite basis for the vector space of vector fields on $\mathbb{R}^n$? Do you see the problem? – Aloizio Macedo Jan 08 '16 at 04:01
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    Of course there are much much more! – Maffred Jan 08 '16 at 04:10
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    Oh wait, $R^n$ is parallelizable right? – Maffred Jan 08 '16 at 04:12
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    Ok I got it, thanks! – Maffred Jan 08 '16 at 04:13