parallelizable spheres

Let \(M = S^1\), then the tangent bundle yields a rank 1 vector bundle \begin{align*} {\mathbf{R}}\to TM \to M \end{align*} We can find a nonzero global section: the vector field \begin{align*} s(x, y) := -y\,dx+ x\,dy\quad\in \Gamma( T{\mathbf{R}}^2 ) \end{align*} restricted to \(S^1\):

Let \(M = S^2\), which is associated to the rank 2 vector bundle \begin{align*}{\mathbf{R}}^2 \to TM \to M\end{align*}

Then \(TM\) is trivial iff there are 2 independent global sections. Since there is a zero section, a second independent section must be everywhere-nonzero - however, this would be a nowhere vanishing vector field on \(S^2\), which by the Hairy Ball theorem does not exist.

Any such a vector field would allow a homotopy between the identity and the antipodal map on \(S^2\). But this is a contradiction: the identity map \(\operatorname{id}_{S^n}\) has degree 1 but the degree of the antipodal map \(S^n\to S^n\) has degree \((-1)^{n+1}\), since it is the composition of \(n+1\) reflections, all of which have degree \((-1)\), and the degree is a group morphism \(([S^n, S^n], \circ) \to ({\mathbf{Z}}, \cdot)\).