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I’m trying to show that the (unique) geodesic joining two points $p, q$ in the hyperboloid model of the hyperbolic space is obtained from the intersection of the $2$-plane $E$ spanned by $p, q$ in $\mathbb R^{1,n}$ with the hyperboloid.

If I knew that the geodesic between two points is unique in the hyperbolic space then I could argue that if the geodesic joining $p, q$ is not in $E$, its reflection through $E$ would be another geodesic with the same endpoints, contradicting uniqueness.

However, there are examples of spaces where there is more than one geodesic connecting two points. For example, two antipodal points on a sphere may be connected through infinitely many geodesics.

Why is there only one geodesic between any two given points in hyperbolic space?

Rodrigo
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    Hint: Use the fact that geodesics in Riemannian manifolds are locally unique. – Moishe Kohan Apr 06 '24 at 19:18
  • @MoisheKohan are they locally unique in the sense that for every $x \in M$ there is a neighborhood of $x$ s.t. for every $y$ in this neighborhood the geodesic connecting $x$ and $y$ is unique? – Rodrigo Apr 06 '24 at 19:25
  • Even stronger: for any two points $p,q$ in that neighborhood $W$, the unit speed geodesic in $W$ connecting $p,q$ is unique. – Moishe Kohan Apr 06 '24 at 19:41

1 Answers1

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The real hyperbolic $n$-space is $$\mathbb H^n:=\{x \in \mathbb R^{n+1}: \langle x,x \rangle =-1 \text{ and }x_0>0\}$$ where $x=(x_0,x_1,\ldots,x_n)$ and the bilinear form is $$\langle x,y \rangle = -x_0y_0 + \sum_{k=1}^n x_k y_k.$$

The tangent space at $p\in \mathbb H^n$ is $T_p \mathbb H^n = p^\perp$.

Consider $p \in \mathbb H^n$ and a unit vector $v$ tangent to $\mathbb H^n$ at $p$. Let $\alpha(t)$ be a geodesic starting at $p$ with velocity $v$.

The fact that $\alpha(t)$ is a geodesic means that its acceleration is normal to $\mathbb H^n$, meaning that the curve has no tangential acceleration, in other words, $\alpha(t) \parallel\ddot \alpha(t)$.

There is a natural curve starting at $p$ with velocity $v$ and without tangential acceleration. $$\alpha(t) := \cosh(t)p +\sinh(t)v.$$

Note that

  • $\alpha(0)=p$,
  • $\dot \alpha(0)=v$,
  • $\ddot \alpha(t)=\alpha(t)$.

Therefore, $\alpha(t) = \cosh(t)p +\sinh(t)v$ is the geodesic starting at $p$ with velocity $v$ (it cannot exist another geodesic with this initial conditions because geodesics are described by a first-order ODE).

Observe $\alpha(\mathbb R) = (\mathbb R p + \mathbb R v)\cap \mathbb H^n$. With that, we conclude that geodesics are obtained as the intersection of two-dimensional subspaces of $\mathbb R^{n+1}$ with $\mathbb H^n$.

In your case, for two distinct points $p$ and $q$, the geodesic containing these points is $(\mathbb R p + \mathbb R q)\cap \mathbb H^n$ and it is unique because two linearly independent vectors $p,q$ define a unique two-dimensional subspaces of $\mathbb R^{n+1}$.

Hugo
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