# Projectivised tangent bundle of 2 sphere

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 $S^2$.
• Thanks for answering another of my questions. That sounds very straightforward when put like that, I wasn't thinking of the tangent space as sitting inside $\mathbb{R}^3$, which helps in this case. It would be hard to explicitly write down the action of a rotation about p though in $\theta$ and $\phi$ coordinates? Commented May 4, 2016 at 21:13
• Also, is $P(S^2)$ diffeomorphic to some other manifold? I know it has dimension 3 but that's about all I know about it. Commented May 4, 2016 at 21:20
• @Wooster: the unit tangent bundle of $S^2$ is acted on transitively and freely by $SO(3)$, so they're diffeomorphic, and $SO(3)$ is in turn diffeomorphic to $\mathbb{RP}^3$. But now I'm confused: the projective tangent bundle is a further quotient of this, which doesn't sound right... Commented May 5, 2016 at 1:26
• Yes, I am confused as well. Whilst this is intuitively obvious, I struggled to choose coordinates on $S^2$ such that the transformation of the tangent space is obvious? Commented May 5, 2016 at 10:52
• (+1) If I understand P(TS^{2})$, an element may be viewed as a pair of diametrically-opposite points in a fibre of$TS^{2}$. The$SO(3)$-action on the unit tangent bundle of the sphere passes to$P(TS^{2})$, but now a half turn about the axis through$p \in S^{2}$induces the identity map in the fibre at$p$(though of course doesn't induce the identity on$P(TS^{2})$; for that, a full turn of$S^{2}$is required). As an$S^{1}$-bundle over$S^{2}$,$P(TS^{2})$appears to have Euler class four, "the number of half-twists in a continuous line field on$S^{2}\$". Commented May 7, 2016 at 2:19