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I have found this cute problem: Let's $M$ a 2-dimensional orientable manifold which has a non zero vector field $X$. Show that $M$ is always parallelizable. My idea was consider $p\in M$ so $X(p)\in T_pM$ is a non zero vector. Then I can complete $X(p)$ to a base of $T_pM$. I can't go further. Any ideas? Thanks

Basel J.
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Protagora1789
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  • How are you going to complete to a basis? If you do something completely different at nearby points, it won't be continuous. If you manage to do it in a smooth way, you are done. – Max Apr 14 '20 at 16:34
  • Hint: there are precisely 2 unit vectors which are perpendicular to $X(p)$. But using an orientation, you can always pick one in a consistent way. – Jason DeVito - on hiatus Apr 14 '20 at 17:28
  • Think carefully about how you're using the fact that $M$ is a $2$-dimensional manifold (and orientable). What's special to this situation? – Ted Shifrin Apr 14 '20 at 17:34
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    What is your definition of orientability? – Arctic Char Apr 16 '20 at 04:46
  • My definition of orientability is : a smooth atlas for $M$ is oriented if the transition functions $\phi_{ij}$ satisfies that $det(d(\phi_{ij})p)>0$ for all $p\in M$, where $d\phi{ij}$ is the differential. – Protagora1789 Apr 16 '20 at 13:33
  • @TedShifrin The fact that $M$ is a 2-dimensional manifold relates with the fact that $M$ can be embed in $\mathbb{R}^4$? – Protagora1789 Apr 16 '20 at 13:35
  • You should not care where this surface lives. Here’s a huge hint: Put a Riemannian metric on it, and think about a unit tangent vector orthogonal to $X(p)$ (for each $p$). – Ted Shifrin Apr 17 '20 at 06:41
  • if I put a riemannian metric $g$ on $M$ I have a sectionof $TM$ $g(p)$ for all $p\in M$. Thanks to $g$ i can found a ortogonal vector $n(p)$ orthogonal to $X(p)$ (as if it is a scalar product). But then i don't understand how i can found an orthogonal vector to $n(p)$ that is a vectori of $T_pM$ – Protagora1789 Apr 17 '20 at 12:52
  • @TedShifrin I've tried to put a riemannian metric on $M$, but I'm stucked. Can you give me another hint? – Protagora1789 Apr 18 '20 at 15:04
  • Putting a Riemannian metric on any manifold is a standard application of partitions of unity. Once you have one, there are two unit vectors $Y(p)$ in $T_pM$ orthogonal to $X(p)$. Use the orientation to pick that one so that $X(p),Y(p)$ is a positively-oriented basis. Now check that $Y$ is smooth and you have given a global trivialization of $TM$. Note that this argument works only for $n=2$. – Ted Shifrin Apr 18 '20 at 18:40
  • @TedShifrin It makes sense. One thing: how can I say that 'there are two unit vectors $Y(p)$ in $T_pM$ orthogonal to $X(p)$'? – Protagora1789 Apr 19 '20 at 13:25

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An orientable 2-dimensional manifold can be embedded into $\Bbb R^3$ so that it is a submanifold (see e.g. Wikipedia, or this question for the more trick non-compact cases). A submanifold $M$ of $\Bbb R^3$ has a well-defined normal vector $n(p)$ which depends continuously on $p\in M$.

The cross product $n(p)\times X(p)$ then also depends continuously on $p\in M$, is linearly independent to $X(p)$ and still lies in the tangent space. Therefore, it provides the desired second vector field.

M. Rumpy
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  • I know that every m-dimensional manifold can be embedded into $\mathbb{R}^{2m}$, so in this case I can say that $M$ can be embedded in $\mathbb{R}^4$ – Protagora1789 Apr 15 '20 at 08:01
  • @Protagora1789 In the answer I linked to a Wiki article that states that every compact orientable surface can be embedded into $\Bbb R^3$, which is also clear by the classification of such surfaces. I am uncertain about the non-compact case, but I cannot come up with a counter-example. – M. Rumpy Apr 15 '20 at 08:52
  • You're right. My concern was about the fact that M is not necessary compact, so i assume that we can only embed M into $\mathbb{R}^4$. – Protagora1789 Apr 15 '20 at 12:35
  • @Protagora1789 Because I don't know a better answer right now, do you know of a non-compact orientable surface that cannot be embedded into $\Bbb R^3$? – M. Rumpy Apr 17 '20 at 19:07
  • Actually no, but i don't know the answer. Maybe there is. Returning to your idea, it works if i consider $M$ embedded in $\mathbb{R}^4$? I mean, it's necessary to be in $\mathbb{R}^3$? – Protagora1789 Apr 17 '20 at 20:49
  • @Protagora1789 I see no immediate way to translate this approach to $\Bbb R^4$. There is no analogue to the cross product in four dimensions, and even if there were, we would need to explain why the approach does not work for e.g. the Klein bottle or the projective plane, which can be embedded into $\Bbb R^4$. – M. Rumpy Apr 18 '20 at 09:21
  • @Protagora1789 I thought this makes a good question, so I posted it. – M. Rumpy Apr 18 '20 at 10:45
  • The Klein bottle cannot be embedded in $\mathbb{R}^3$ neverthless it has dimension 2. – Protagora1789 Apr 18 '20 at 15:07
  • @Protagora1789 Yes, but it is compact and admits a non-vanishing tangent vector field. Any approach must explain why it does not work for this. – M. Rumpy Apr 18 '20 at 16:21
  • @Protagora1789 My answers to this question seem to suggest that being non-compact is not a problem. So my answer still holds. It seems to involve a non-trivial back box though. – M. Rumpy Apr 23 '20 at 20:08