Some good $($introductory$)$ sources, in general, for all things smooth manifolds:
- Topology from the Differentiable Viewpoint, by Milnor
- Differential Topology, by Guillemin-Pollack
- Differential Forms and Applications, by Do Carmo
- A Comprehensive Introduction to Differential Geometry, Vol. 1, by Spivak
- Introduction to Smooth Manifolds, by Lee
- Foundations of Differentiable Manifolds and Lie Groups, by Warner
- Brian Conrad's Differential Geometry Notes
1.
Easy examples of parallelizable manifolds are $\mathbb{R}^n$, the tori $\mathbb{R}^n/\mathbb{Z}^n$ $($the points are cosets of the additive subgroup $\mathbb{Z}^n$ of $\mathbb{R}^n$, and the charts are open subsets of $\mathbb{R}^m$ that contain at most one point in each equivalence class, with each point mapping to its equivalence class$)$, and the argument used here to show $SO(n,\mathbb{R})$ is parallelizable $($$SO(n)$ is parallelizable$)$ actually shows that any Lie group is parallelizable. $($If $X_i$ are linearly independent tangent vectors at the identity, then for any paths $\gamma_i(t)$ with $\gamma_i'(0) = X_i$ we note that the paths $g\gamma(t)$ are smooth, and we will say that $V_i(g)$ is the equivalence class of $g\gamma_i'(0)$. Then each $V_i$ is a smooth vector field with no zeros, and the vector fields $V_i$ parallelize our Lie group.$)$
There are a lot of obstructions to parallelizability. One is the Euler characteristic. It turns out that any surface with nonzero Euler characteristic cannot have even a single smooth non-vanishing vector field. With algebraic topology one can show that a closed manifold with nonzero Euler characteristic cannot have a non-vanishing vector field. This includes genus $g$ surfaces for any $g \neq 1$. Another obstruction to paralleizability is non-orientability: every parallelizable manifold is orientable. Perhaps this is easier to see: given a parallelizable manifold with vector fields $X_1, \dots, X_n$ forming a basis for the tangent space at each point, we can always reverse the orientation of a connected chart so that they form a positively oriented basis at some point, and by continuity, at all points in each chart $($since determinants and the vector fields $X_i$ are continuous, and our charts are connected$)$. Then the transition maps will always have positive determinant, since the basis $X_1, \dots, X_n$ at each point is positively oriented at all points in all charts, and gets taken to itself via transition maps. So for example, the Klein bottle, Möbius band, $\mathbb{P}^n(\mathbb{R})$ for even $n > 1$ cannot be parallelized.
2.
Suppose the tangent bundle of $M$ is trivial; then our diffeomorphism $\rho^{-1}: M \times \mathbb{R}^n \to TM$ gives us vector fields $\rho^{-1}(M \times \{e_i\})$, $1 \le i \le n$ where $e_i$ are the standard basis coordinates. By hypothesis, the vectors $\rho^{-1}(p, e_i)$ are smooth functions of $p$ in local coordinates and are linearly independent at each $p \in M$. Conversely, suppose we have $n$ vector fields that are linearly independent at each point; then the map that takes each point $(p, \xi)$, $\xi = (\xi_1, \dots, \xi_n) \in \mathbb{R}^n$, to the tangent vector $\sum \xi_i X_i(p)$ is easily checked to be bijective, smooth, and have derivative of rank $2n$ at all points, and is thus a diffeomorphism $M\mathbb{R}^n \to TM$.
The distinction to be made is that a differentiable structure is a choice of maximal smooth atlas $\mathcal A$, but two different choices $\mathcal A$ and $\mathcal A'$ can lead to isomorphic smooth structures. As an example, the canonical smooth structure $\mathcal A$ on $\mathbb R$ that contains the smooth function ${\rm id}:\mathbb R\longrightarrow \mathbb R$ is isomorphic to the smooth structure $\mathcal A'$ that contains the smooth function $x\mapsto x^3$, although $\mathcal A'\neq \mathcal A$. Thus, although a manifold admits uncountably many different smooth structures, it may have finitely many isomorphism classes of such structures.
Best Answer
On $\mathbb{R}^n$ there is the obvious basis $e_1,e_2,\dots,e_n$. Since $TM\cong M\times\mathbb{R}^n$ you have the $p\mapsto e_i$ for all $p$ defines a smooth section $X_i$ of $TM$. The $X_1,\dots,X_n$ defines a frame at every point.
Conversely, if you have global frame field $X_1,\dots,X_n$, then for $v\in T_pM$ we have $v=v^iX_i(p)$, and you send this $v$ to $(p,(v^i))\in M\times\mathbb{R}^n$. Check it works.