In Tu's "An Introduction to Manifolds", exercise 16.8, we are supposed to prove that $M$ has a trivial tangent bundle $\Leftrightarrow$ M admits a smooth frame $\{X_1, \dots, X_n\}$ (i.e., smooth vector fields $X_1, \dots, X_n$ such that $\{X_{1_p}, \dots, X_{n_p}\}$ is a basis for $T_pM$ for every $p\in M$).
When I read this, I concluded that $M$ is orientable $\Rightarrow$ M admits a smooth frame $\{X_1, \dots, X_n\}$ $\Rightarrow M$ has trivial tangent bundle. However, I've learned that $\mathbb{S}^2$ is an orientable manifold which has no trivial tangent bundle.
What am I missing here?
Best Answer
I am referring to your discussion with @melomm in the comments to the question. The right definition of orientability involving vector fields is the following. You have to specify an orientation of the tangent space $T_xM$ for all $x\in M$, which are compatible in the sense that for each $x\in M$, there is an open neighborhood $U$ of $x\in M$ and there are local vector fields $X_1,\dots X_n\in\mathfrak X(U)$ such that for each $y\in U$ the values $X_1(y),\dots,X_n(y)$ define a positively oriented basis for $T_yM$. Of course, this implies that $TM$ is trivial over $U$, but $TM$ is locally trivial anyway.
To see that this is equivalent to the definition via oriented atlasses roughly goes as follows. It easily follows from the defintions that for a connected local chart $(U_\alpha,\phi_\alpha)$ the choice of an orientation of $T_xM$ for one $x\in U_\alpha$ gives rise to a compatible choice for all $y\in U_\alpha$. If you have an oriented atlas than you can extend this orientation to all points in $M$. Conversely, if you have compatible orientations the charts inducing this given orientation form an oriented atlas.