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 don't know of a totally elementary proof of this result, but here is some context for it. More generally, an $n$-manifold $M$ (without boundary) has a classifying map $f : M \to BO(n)$ for its tangent bundle. Knowing when $M$ is parallelizable is equivalent to knowing when $f$ is null-homotopic. There is a general machine for doing this involving lifting $f$ higher and higher through the stages of the Whitehead tower of $BO(n)$, and it tells us that the complete set of obstructions to solving this problem are a set of cohomology classes in $H^k(M, \pi_k(BO(n)), k \le n$, each of which is well-defined provided that the previous one vanishes, such that $f$ is null-homotopic iff all of the classes vanish.
The construction of the first such class goes like this. Assume for simplicity that $M$ is connected. The first question is whether $f$ lifts to the universal cover of $BO(n)$, which is true iff $f$ induces the zero map on $\pi_1$ by standard covering space theory. Now, $\pi_1(BO(n)) \cong \mathbb{Z}_2$, and the induced map on $\pi_1$ gives a homomorphism $\pi_1(M) \to \mathbb{Z}_2$ which corresponds precisely to the first Stiefel-Whitney class $w_1$. This class vanishes iff $f$ lifts to the universal cover of $BO(n)$, which is $BSO(n)$, iff $M$ is orientable.
Now we want to try lifting $f$ to the $2$-connected cover of $BSO(n)$; this is analogous to the universal cover but involves killing $\pi_2(BSO(n)) \cong \mathbb{Z}_2$ (for $n \ge 3$) instead of $\pi_1$. Whether this is possible is controlled by the map
$$BSO(n) \to B^2 \mathbb{Z}_2$$
inducing an isomorphism on $\pi_2$. This is equivalently a universal characteristic class in $H^2(BSO(n), \mathbb{Z}_2)$ which turns out to be precisely the second Stiefel-Whitney class $w_2$. This class vanishes iff $f$ lifts to the $2$-connected cover of $BSO(n)$, which is $BSpin(n)$, iff $M$ has a spin structure.
The first surprise in this story is that (when $n \ge 3$) $BSpin(n)$ also turns out to be the $3$-connected cover; in other words, $\pi_3(BSpin(n)) = 0$, so the next step of this story involves $\pi_4$ and can be ignored for $3$-manifolds. If $M$ is a $3$-manifold, not necessarily compact, admitting both an orientation and a spin structure, then the classifying map of its tangent bundle lifts to a map $M \to BSpin(3)$, but since the latter is $3$-connected any such map is nullhomotopic. So:
A $3$-manifold, not necessarily closed, is parallelizable iff the first two Stiefel-Whitney classes $w_1, w_2$ vanish iff it admits an orientation and a spin structure.
To give an indication of the generality of this machinery, for $4$-manifolds the next step involves computing $\pi_4(BSpin(4)) \cong \mathbb{Z}^2$ and then lifting to the $4$-connected cover of $BSpin(4)$, which is called $BString(4)$. This says that the next obstruction to a $4$-manifold with an orientation and a spin structure being parallelizable is a pair of cohomology classes in $H^4(M, \mathbb{Z})$ which I believe turn out to be the Euler class $e$ and the first fractional Pontryagin class $\frac{p_1}{2}$ (a certain characteristic class of spin manifolds that when doubled gives the Pontryagin class $p_1$) respectively. Since $H^4(M, \mathbb{Z})$ is always torsion-free for a $4$-manifold, the conclusion is that
A $4$-manifold, not necessarily closed, is parallelizable iff the characteristic classes $w_1, w_2, e, p_1$ all vanish iff it admits an orientation, a spin structure, and a string structure.
And for $5$-manifolds and higher we really need to talk about $\frac{p_1}{2}$ and not just $p_1$.
The second surprise in this story is that for closed $3$-manifolds the condition that $w_2$ vanishes is redundant: it is implied by orientability by a standard computation with Wu classes. In other words, closed orientable $3$-manifolds automatically admit spin structures. I don't have a good intuitive explanation of this; it comes from a relationship between the Stiefel-Whitney classes, Steenrod operations, and Poincaré duality that I don't understand very well.
Best Answer
It's not about parallel transport but the ability to reasonably define when two vectors on different tangent spaces are parallel. The vector fields provide a natural basis on each tangent space and allows us to compare two tangent vectors by comparing their components in this basis. With this you can decide when two vectors are equal or parallel.
If you are familiar with bundles, parallelizability means that the tangent bundle is globally trivializable. This is just another way to say that you can describe tangent vectors irrespective of the base point. In other words, for any two points $x,y\in M$ there is a canonical linear isomorphism $F_{x,y}:T_xM\to T_yM$ which maps each $X_i(x)$ to $X_i(y)$. (This property defines $F_{x,y}$ uniquely.)
A connection on the tangent bundle gives a way to identify tangent spaces along a curve, but not between any two points. Parallel transport is defined along curves, not between points. Parallelizability in the sense of this question has nothing to do with connections. The global trivialization of the tangent bundle gives you a way to naturally translate tangent vectors between base points, but it is not parallel transport in the usual sense.
The definition of parallelizability involves no metric or connection, so parallel transport doesn't even have enough structure to work in. You can always get a metric by defining your frame to be orthonormal. This gives a "global inner product", but the vectors are not generally parallel transported with respect to this metric.
The term "parallel" in the context of the question's definition must be understood in the sense of parallel vectors in a general vector space, not in the sense of parallel transport along curves.