Existence of smooth frame along a curve on the manifold

Question

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!

Answer 1

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).