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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?

Wooster
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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 $S^2$.

Qiaochu Yuan
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  • 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? – Wooster May 04 '16 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. – Wooster May 04 '16 at 21:20
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    @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... – Qiaochu Yuan May 05 '16 at 01:26
  • 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? – Wooster May 05 '16 at 10:52
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    (+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 $S^{2}$". – Andrew D. Hwang May 07 '16 at 02:19
  • @AndrewD.Hwang The unit tangent bundle of $ \mathbb{R}P^2 $ is also a circle bundle over $ S^2 $ with Euler class four. See https://math.stackexchange.com/questions/4369088/orientability-and-unit-tangent-bundle-of-surfaces?rq=1 Is there perhaps an canonical identification of $ P(TS^2) $ with $ UT(\mathbb{R}P^2) $? – Ian Gershon Teixeira Feb 19 '22 at 13:38