What are examples of parallelizable manifolds, and why does parallelizable correspond to $TM$ being trivial?

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 $M\mathbb{R}^n \to TM$.

I'll explain the relationship between trivialization of the tangent bundle and having $n$ linearly independent vector fields:

A vector field is a smooth map $X: M \rightarrow TM$ such that $X(p) \in TM_p$. That is, it's a smooth assignment of a tangent vector at $p$ to each point $p \in M$.

Suppose we have $n = \dim(M)$ vector fields $\{X_i\}_{i=1}^n$ which are linearly independent at every point, i.e. such that for all $p \in M$, the $n$ vectors $\{X_i(p)\}_{i=1}^n$ in $TM_p$ are linearly independent. Then we can define a map $f: M \times \mathbb{R}^n \rightarrow TM$ given by $f(p,\sum_{i=1}^n a_i e_i) = \sum_{i=1}^n a_i X_i(p) \in TM_p$.

This map is smooth because the $X_i$ are, and is a linear isomorphism when restricted to $\{p\}\times \mathbb{R}^n$ for any $p$. In fact, this map $f$ is a diffeomorphism: it's bijective, and we can compute its derivative well enough to know its derivative must be an isomorphism at every point, and thus by the inverse function theorem it's a diffeomorphism.

$$df_{p,v} = \left(\begin{array}{c|c} I_n & 0 \\ \hline * & A(p) \end{array}\right)$$

(In coordinates $\phi: U \rightarrow M$, $U \subset \mathbb{R}^n$ near $p$, giving a local trivialization $U \times \mathbb{R}^n \rightarrow TM$ via $(p,v) \mapsto d\phi_p(v) \in TM_p$. Here $A(p)$ is invertible and has columns given by the $X_i(p)$ expressed in the basis coming from $d\phi: TU_p (\cong \mathbb{R}^n) \rightarrow TM_p$.)

Conversely suppose we have a trivialization, i.e. a diffeomorphism $\Phi: M \times \mathbb{R}^n \rightarrow TM$ such that $\Phi(p,-): \{p\} \times \mathbb{R}^n \rightarrow TM_p$ is a linear isomorphism. Then we get vector fields by taking $X_i = \Phi(-,e_i): M \rightarrow TM$. The $n$ of these are linearly independent and are vector fields as described above.