Let $T S^2$ denote the tangent bundle of the 2-sphere $S^2 = \{x \in \mathbb{R}^3 : \|x\|=1\}$. In this paper, Fodor proves that $T S^2$ is parallelizable, using machinery that I do not understand --- see Theorems 2.5 and 3.2 therein.
Question: A simple way to convince someone that a manifold is parallelizable is to present explicit formulas for a global frame. What is an explicit global frame for $T S^2$?
Thanks for the help!!
Recall: A smooth manifold $M$ is parallelizable if it admits a smooth global frame. It is a theorem that $M$ is parallelizable if and only if $T M$ is trivial as a vector bundle (i.e., splits as a product).
For the problem at hand, can one give an explicit diffeomorphism between $T (T S^2)$ and $T S^2 \times \mathbb{R}^4$?
Note: I am asking if $T S^2$ as a manifold is parallelizable. I am not asking whether $S^2$ is parallelizable (indeed, it is not).
Attempt: Spent some time trying to explicitly build a global frame, without success. Here is a brief description.
We have $$T S^2 = \{(x, v) \in \mathbb{R}^3 \times \mathbb{R}^3 : \|x\| = 1, x^\top v = 0\},$$ which is a 4-dimensional manifold. Its tangent bundle is $$T (T S^2) = \{(x, v, \dot x, \dot v) \in (\mathbb{R}^3)^4 : \|x\|=1, x^\top v = 0, x^\top \dot{x} = 0, v^\top \dot{x} + x^\top \dot{v} = 0\}.$$ We want to pick 4 nonvanishing vector fields $V_i, i = 1, ..., 4$, on $T S^2$, so that at every $(x,v)$, $\{V_1(x,v), \ldots, V_4(x,v)\}$ are linearly independent.
Let's start with the easier task of building a nonvanishing vector field $V$ on $T S^2$. Let $\times$ denote the cross product in $\mathbb{R}^3$. Fix two linearly independent unit vectors $u, w \in \mathbb{R}^3$, and define $$V(x,v) = (w \times x, u \times x - v^\top (w \times x) \cdot x).$$ One easily checks that $V(x, v) \in T_{(x,v)} T S^2$, and that $V$ is smooth and nonvanishing everywhere (details omitted).
In fact, one can check that if you take generic unit vectors $u_1, u_2, w_1, w_2 \in \mathbb{R}^3$, then the two vector fields $V_1, V_2$ are always linearly independent (defining $V_1, V_2$ as above).
Unfortunately, this construction doesn't seem to work with 3 or more vector fields $V_i$, constructed as above. For example, if you choose $u_1, u_2, u_3, w_1, w_2, w_3 \in \mathbb{R}^3$ randomly then you can numerically check that there always seems to exist an $(x,v)$ so that $\{V_1(x,v), V_2(x,v), V_3(x,v)\}$ are linearly dependent.
Perhaps one can pick the vectors $u_1, u_2, u_3, w_1, w_2, w_3$ more carefully than randomly? But it is not clear to me how to do this.