i have the following problem: Let $(M,g)$ be a compact Riemannian manifold with metric $g$ and $\nabla$ be the Levi-Civita Connection. Denote by $G(M) =${$\gamma: \mathbb{R} \rightarrow M | \gamma \text{ is a geodesic } $} the space of geodesics on $M$ with respect to $\nabla$. Has $G(M)$ then the structure of a manifold? And if yes does it have finite dimension. If no what could one do to make it a manifold with finite dimension ? Or can you give me some reference to read ? I hope for answers and thanks in advance.


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    $\begingroup$ The existence and uniqueness theorem for ODEs tells you a geodesic is uniquely determined by a tangent vector. So your space of geodesics is basically just the tangent bundle -- differentiate $\gamma$ at time $t=0$, $\gamma'(0)$. That's the map from your space of geodesics to the tangent bundle. Is this a homework problem? $\endgroup$ Feb 7, 2012 at 8:04
  • $\begingroup$ no this is not homework ... not at all $\endgroup$
    – william
    Feb 7, 2012 at 8:43
  • $\begingroup$ Why do you want to equip the space of geodesics with a manifold structure? Does Ryan's suggestion do what you want? $\endgroup$
    – Yemon Choi
    Feb 7, 2012 at 9:31

4 Answers 4


Dear Marco,

If you take unparameterized geodesics (the quotient of the unit co-sphere bundle by the action of the geodesic flow), here is an answer:

The space of geodesics of a Riemannian or even Finsler manifold of dimension $n$ is itself a manifold of dimension $2n - 2$ in a variety of special, albeit important, cases. For example, the space of geodesics of any rank-one symmetric space (Euclidean space, hyperbolic space, the sphere, real, complex, quaternionic spaces, the Cayley plane, complex hyperbolic space) is a manifold. Other less standard examples are Zoll manifolds, Hadamard manifolds or any convex neighborhood of a Riemannian or Finsler space.

When the space of geodesics is a manifold, it carries a natural symplectic structure inherited by symplectic reduction from the unit co-sphere bundle of the Riemannian or Finsler metric. Families of geodesics normal to immersed submanifolds are immersed Lagrangian submanifolds in the space of geodesics (a result that dates back to Hamilton's Theory of systems of rays, arguably the first paper on symplectic geometry). In particular, points in the manifold correspond to Lagrangian spheres in the space of geodesics (i.e. all geodesics passing through a point is a Lagrangian sphere). Using these spheres and the symplectic structure you can easily reconstruct the metric (see the paper Symplectic geometry and Hilbert's fourth problem, Journal Diff. Geom. Volume 69, Number 2 (2005) for this and as reference to everything else I'm writing here).

To make things concrete here are some examples:

$\bullet$ The space of geodesics of Euclidean or hyperbolic n-space is symplectomorphic to the cotangent of the $n-1$-sphere.

$\bullet$ The space of geodesics of the $n$-sphere or the $n$-dimensional real projective space is symplectomorphic to a complex quadric = the Grassmannian of oriented two-planes in ${\mathbb R}^{n+1}$

$\bullet$ The space of geodesics of the complex projective space ${\mathbb CP}^n$ is symplectomorphic to the space of 1-2 flags (point-line) in ${\mathbb CP}^n$

The space of geodesics may inherit additional structure from the structure of the manifold: for example Hitchin looks at the complex structure of the space of geodesics of ${\mathbb R}^3$ to study monopoles and minimal surfaces in Monopoles and geodesics, Comm. Math. Phys. Volume 83, Number 4 (1982), 579-602.

If you want additional references go to my JDG paper or my 1995 thesis (The symplectic geometry of spaces of geodesics) that you will find on the web.

  • 1
    $\begingroup$ Welcome to MO, Juan Carlos! $\endgroup$
    – Todd Trimble
    Feb 7, 2012 at 19:28
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    $\begingroup$ Another useful reference is chapter 2 of Besse's book Manifolds all of whose geodesics are closed. $\endgroup$
    – Dan Fox
    Feb 8, 2012 at 8:20

Marco, do you mean the space of (constant-length) parametrized geodesics or the space of geodesics. In the first case you would get the tangent bundle, as Ryan said. The second case is in general more complicated, and I think you won't get a nice manifold structure on that space in general. Nice would mean for example that the projection from the tangent bundle (space of constant lenght parametrized geodesics) to the space of geodesics is a submersion. This is not the case for a flat torus. More generally, this is not the case when there exist a geodesic $\gamma$ such that the closure of the image of its tangent curve $\gamma'$ is not the image itself. I am not sure about this, but I guess that this will always be the case on compact manifolds such that not all geodesics are closed.

But if you assume for example that you have negative curvature on a simply connected manifold, you get a nice manifold structure, as you may see from the theory of Jacobi fields for geodesics.


For simply connected pseudo-Riemannian space of constant curvature or a rank one Riemannian symmetric space (other than the octonion hyperbolic plane) the answer is yes, and can be found at http://arxiv.org/abs/0911.2602

  • $\begingroup$ The URL seems incomplete. $\endgroup$ Feb 7, 2012 at 18:25
  • $\begingroup$ The link is fixed now. $\endgroup$ Feb 8, 2012 at 12:15

To expand Ryan Budney's comment, the geodesics of $(M,g)$ are the projection on the base $M$ of the integral curves for the vector field $S_g$ on $TM.$

$S_g$ is the unique vector field on $TM$ which is at the same time:

special (i.e. it represents a $2^\textrm{nd}$-order edo on $M$, or equivalently its integral curves are the tangential lifting of their projection on the base) and

horizontal (i.e. it is a section of the horizontal distribution on $TM$.)

Because $S_g$ is a spray (i.e. $\mathcal{L}_Z S_g=S_g,$ where $Z$ is the Euler vector filed), it is called the geodesic spray of $(M,g).$

It can be realized even that $S_g$ is the hamiltonian vector field of the kinetic energy $K_g:v\in TM\to\tfrac{1}{2}g(v,v)\in\mathbb{R}$ with respect to the pull-back through $g^\flat$ of the canonical symplectic form on $T^\ast M.$

So $S_g$ preserves the sphere bundles, and we get that if $M$ is compact then $S_g$ is complete.

Through the flow of $S_g,$ its integral curves (which are the tangential lifting of their projection on the base) are indentified with $TM$.

  • $\begingroup$ so, what is then the dimension of these manifold. is it the dimension of the cotangent bundle? $\endgroup$
    – william
    Feb 7, 2012 at 17:20
  • $\begingroup$ is this an arguement why the space of geodesics has a manifold structure ? $\endgroup$
    – william
    Feb 7, 2012 at 17:22
  • $\begingroup$ do we then have a diffeomorphism from $TM$ to $G(M)$ ? $\endgroup$
    – william
    Feb 7, 2012 at 18:01
  • $\begingroup$ @Marco: I have just pointed out the bijection between the parametrized geodesics, appearing in your question, and the maximal integral curves of the geodesic spray, which are in correspondence one-to-one with the initial condition, i.e. the points of $TM.$ Another question in to give a manifold structure to the space orbits of the geodesic flow, (such as for the flow of an arbitrary vector fields), and this alternate problem is tackled by the other answers. $\endgroup$
    – agt
    Feb 7, 2012 at 18:10
  • $\begingroup$ Through a bijection f:X→Y, any smooth structure on X is transported on a smooth structure on Y, so that $f$ becomes a diffeo $\endgroup$
    – agt
    Feb 7, 2012 at 18:15

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