Is $SO(n)$ diffeomorphic to $SO(n-1) \times S^{n-1}$

Question

There is a fibration $SO(n-1) \mapsto SO(n) \mapsto S^{n-1}$, from basically taking the first column of the matrix in $\mathbb{R}^n$. Is this fibration trivializable?

Answer 1

$SO(n)$ is the orthonormal frame bundle of $S^{n-1}$. A trivialization of this bundle would in particular imply that $S^{n-1}$ is parallelizable (i.e., has trivial tangent bundle), so this happens precisely for $n=2$, $4$, and $8$.

Answer 2

If $n=3$, the Hopf fibration is the composition $S^3=Spin(3)\rightarrow SO(3)\rightarrow S^2$, so if $SO(3)\rightarrow S^2$ is trivial, so the hopf fibration will be flat and this is not true.

https://en.wikipedia.org/wiki/Hopf_fibration#Geometric_interpretation_using_rotations