I'm trying to understand how rotations act on the "projectivised" tangent bundle of the sphere.
Let $S^2$ be the two sphere and denote by $P(TS^2)$ the tangent bundle where each tangent space $T_xS^2$ is taken to be a projective vector space. I'm trying to show that given any $x$ and $y$ in $P(TS^2)$ there is a rotation of the sphere that maps $x$ to $y$.
If we use coordinates $(\theta,\phi)$ on the sphere then the bundle $P(TS^2)$ is locally $\{(\theta, \phi, [u:v]\}$ where $(u,v)$ are the projective coordinates corresponding to the basis $\frac{d}{d\theta},\frac{d}{d\phi}$.
We can write $x = (\theta_1,\phi_1, [u_1:v_1])$ and $y = (\theta_2,\phi_2, [u_2:v_2])$. Now it's clear that there is a rotation that takes the "manifold part" of $x$ and $y$ onto each other. Now intuitively I'd like to rotate about that point until the "tangent space parts" also match up.
I'm not sure if this is a good approach and I'm struggling to see how a rotation acts on the projective vector spaces. I'd also be interested in how one geometrically visualises such a projectivised tangent bundle - is there even a natural geometric interpretation in this case?
You can visualize this action very explicitly: a tangent vector to a point $p \in S^2$ is literally a little vector tangent to $S^2$ inside of $\mathbb{R}^3$, and rotation acts in the obvious way. The rotations around the axis through $p$ act transitively on unit tangent vectors at $p$ (and so act transitively on the projectivized tangent space at $p$), and rotations also act transitively on $S^2$.