The tangent bundle of spheres as a fiber bundle

468 Views Asked by At

Let $\mathbb{S}_n$ be the sphere of dimension $n \geq 2$, with a Riemannian metric of constant curvature. Let $X_n := T^1 \mathbb{S}_n$ be its unit tangent bundle.

I say that two points $(x, v)$ and $(x', v')$ of $X_n$ are equivalent, and write $(x, v) \sim (x', v')$, if they are on the same orbit of the geodesic flow, or (equivalent definition) if they lie in the same oriented great circle.

This gives me a projection $\pi : \ X_n \to X_{n / \sim}$, which should be a fiber bundle by circles. The space $X_{n / \sim}$ seems well-behaved: it is obviously a topological space, but it is also homogeneous (there is a transitive action of $SO_{n+1}$ on it), and can be given a manifold structure.

What does $X_{n / \sim}$ look like?

1

There are 1 best solutions below

2
On BEST ANSWER

Great circles are parametrized by two-dimensional subspaces of $\Bbb R^{n+1}$. So your quotient space is the Grassmannian $G(2,n+1)$.