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?

  • $\begingroup$ I think the quotient is a point, i.e. there is only one orbit. At least, this seems to be the case for $n=2$ and the standard riemannian metric. $\endgroup$ Jan 2, 2014 at 11:57
  • $\begingroup$ Maybe I'm dense, but I'm not quite sure what your equivalence relation is. Specifically, do you mean they are equivalent if the geodesic $\gamma$ with $\gamma'(0)=v$ passes through $x'$ at some time $t_0$ and has $\gamma'(t_0)=v'$, or are you asking wether there is some great circle containing both $x,x'$ such that the parallel transport of $v$ along the geodesic that is the great circle takes on the value $v'$? The first is an equivalence relation and the quotient is most certainly a Grassmannian for the standard metric. The second seems to contain only one orbit. $\endgroup$ Jan 2, 2014 at 12:07
  • $\begingroup$ @Olivier Bégassat : I meant the first definition. $\endgroup$
    – D. Thomine
    Jan 2, 2014 at 12:19

1 Answer 1


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

  • $\begingroup$ Yes, I realized that a couple minutes ago. Actually, it is not the Grassmannian, but the oriented Grassmannian, because there a two oriented great circles in any 2-dimensional subspace. Anyway, thank you ! $\endgroup$
    – D. Thomine
    Jan 2, 2014 at 12:18
  • $\begingroup$ Oh, I missed that you wanted oriented great circles. My sloppiness! $\endgroup$ Jan 2, 2014 at 13:04

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