Given an element of $SO(n)$, it can be decomposed as the product of rotations through planes (two-dimensional subspaces of $\mathbb{R}^n$), for example, using Givens rotations.
However, because rotations don't commute, if we are only given a particular basis for $\mathbb{R}^n$, we can not necessarily generate any element of $SO(n)$ as a product of rotations through planes spanned by pairs of any two basis vectors. (See the paragraph beginning with "Notice that..." here.)
Analogously, knowledge of all sectional curvatures (i.e. of every 2-dimensional subspace of $\mathbb{R}^n$) completely determines the Riemannian curvature tensor (see here or here).
But if we are only given the sectional curvatures corresponding to planes spanned by pairs of two vectors from a particular basis, then we do not know enough to determine the entire Riemannian curvature tensor (as the answers to this MathOverflow question argue).
Question: Are these two facts related or connected on some level?
In other words, can we intuitively understand the reason why we need all sectional curvatures to determine the Riemannian curvature tensor, and not just those correspond to the planes spanned by a particular given basis, to be that "rotations don't commute"?
Seemingly there would likely be a strong connection between $SO(n)$ and the Riemannian curvature tensor, since both are generated by the inner product/Riemannian metric.
An intuitive "argument" or a pointer to a reference would suffice for an answer.
