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Suppose $M$ is a smooth manifold of dimension $n$. Let $\gamma: [0,1] \mapsto M$ be a smooth curve on it. Assume $\dot{\gamma}(t)$ is never zero. Can we always find smooth vector fields $e_1(t), e_2(t), \ldots, e_n(t)$ along $\gamma$ such that they span $T_{\gamma(t)}$? In each coordinate neighborhood $U$ we can find such smooth frame but it seems we cannot simply glue them together. So if this is true, how to construct them? Thank you!

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    $\begingroup$ en.wikipedia.org/wiki/Parallel_transport $\endgroup$ – Max Jan 18 '18 at 20:47
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    $\begingroup$ All you really need is the fact that a vector bundle over the unit interval is always trivial. (This is a special case of the result that on a paracompact manifold homotopic maps pull back a bundle to isomorphic bundles.) $\endgroup$ – Ted Shifrin Jan 18 '18 at 21:07
  • $\begingroup$ @Ted, it is a safe bet that all that does not mean anything to anyone asking this question. $\endgroup$ – Mariano Suárez-Álvarez Jan 18 '18 at 21:37
  • $\begingroup$ @Mariano, yes, you're no doubt right, and I usually make this comment. Any approach is going to need some partitions of unity, I think. $\endgroup$ – Ted Shifrin Jan 18 '18 at 21:42
  • $\begingroup$ Lebesgue's lemma on [0,1] to get a partition fine enough such that each subinterval maps into a coordinate chart should be enough! (This is implicit in Ivo's answer, for example, and is in fact the non trivial part) $\endgroup$ – Mariano Suárez-Álvarez Jan 18 '18 at 21:50
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Yes. First equip $M$ with a Riemannian metric $\langle\cdot,\cdot\rangle$ (this is always possible, using for example partitions of unity). With it comes it's Levi-Civita connection, and a notion of parallelism, which we can use to give a relatively simple construction.

Given $e_1,\ldots,e_n \in T_{\gamma(0)}M$, there are unique parallel vector fields $E_1,\ldots,E_n$ along $\gamma$ such that $E_i(0) = e_i$. Since they're parallel, it holds that $\langle E_i(t), E_j(t)\rangle_{\gamma(t)} = \langle e_i,e_j\rangle_{\gamma(0)}$ for all $t \in [0,1]$, and so $(E_i(t))_{i=1}^n$ is a basis for $T_{\gamma(t)}M$ if $(e_i)_{i=1}^n$ is a basis for $T_{\gamma(0)}M$. The existence of these fields follows, at least locally, from solving a first-order differential equation in a coordinate chart. Since $\gamma([0,1])$ is compact, is can be covered by a finite quantity of coordinate chart, and the local solutions "glue" well. For details of the proof you can look at any Riemannian Geometry book (say, Lee's or do Carmo's will do).

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    $\begingroup$ So you started off by putting a Riemannian metric on $M$? It was only given to be smooth. $\endgroup$ – Ted Shifrin Jan 18 '18 at 21:00
  • $\begingroup$ Yes. You're right, that should be mentioned, I'll add an edit $\endgroup$ – Ivo Terek Jan 18 '18 at 21:03

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