I intend to solve the following exercise but I would like to have some help with understanding the ''big picture'':
Exercise. Describe a natural 1 to 1 correspondence between elements of $SO(3)$ and elements of $$T^1S^2=\{(p,v)\in \mathbb R^3 \times \mathbb R^3 | |p|=|v|=1 \text{ and } p \bot v \}$$
which can be thought of as the collection of all unit-length vectors $v$ tangent to all points $p$ of $S^2$.
At this point $SO(3)$ has been defined to be the set of all positions of a globe on a fixed stand. Since this seemed to be too sloppy to use I did some googling and found that $SO(3)$ is the set of all orthogonal matrices with determinant equal to $1$. So now my intention is to find a bijection from this set of matrices into $T^1S^2$.
But before I embark on this task I would like to understand what this exercise is really about. What exactly is $T^1S^2$? At first I thought it was the tangent bundle of $S^2$ but the tangent bundle has a slightly different definition in that its elements are equivalence classes whereas here we are talking about points in $\mathbb R^6$. Also, the notation is different: the tangent bundle would be denoted $TS^2$ but the notation used here is $T^1S^2$.
I am looking for some information about the meaning of this bijection in the exercise. I have no (very little) prior knowledge of differential geometry. Thank you.
From the description given, $T^1 S^2 \subset TS^2$ denotes the unit tangent bundle inside $TS^2$: each fiber $(T^1 S^2)_p$ is the unit sphere in $(TS^2)_p$. (The metric used is the one corresponding to the embedding $TS^2\subset \theta^3$, where $\theta\to S^2$ denotes the trivial bundle, coming from the usual embedding $S^2\subset \mathbb{R}^3$.)
For the bijection, note that each point $(p, v)\in T^1 S^2$ has $p$ and $v$ orthonormal. The matrix with columns $p, v, (p\times v)$ then lies in $SO(3)$.