A question about geodesics of hyperbolic space

Question

Let $ H^n $ be the the upper half space of $ R^n $ endowed with the conformal metric $ g=\frac{1}{x_{n}^{2}}|dz|^2 $ ($ |dz|^2 $ is the standard metric of $ R^n $). This space is the classical hyperbolic space and its Riemannian connection $ \overline{\nabla} $ satisfies

$$ \overline{\nabla}_{\partial_n}\partial_n =0 $$

where $ \partial_1, \ldots \partial_n $ is the standard frame of $ R^n $. Note that $ \overline{\nabla} $ is not the standard connection of $ R^n $.

Now it is easy to see that the smooth curve $ \gamma(t)= (0, \ldots, 0,t) $ , $ t >0 $ is a geodesic in $ H^n $. This curve is clearly defined for $ t>0 $ and it is not defined for all t in $ R $. Moreover it is well known that $ H^n $ is complete. This fact apparently contradicts the Hopf Rinow theorem (every geodesic in a complete space is defined for every $ t \in R $). What is the mistake in this argument?

Thanks

Answer 1

Your $\gamma(t)$ is actually not a geodesic. In fact, earlier you have a mistake in computing the Levi-Civitia connection. I'll use $X$ to denote $\partial_n$, just to save on typing. Via the formula on wikipedia, we have \begin{align*} \langle \overline{\nabla}_X X , X\rangle =& \frac{1}{2}\big(Xg(X,X) + Xg(X,X) - Xg(X,X)\\ &+ \quad g([X,X],X) - g([X,X],X) - g([X,X],X)\big)\\ \end{align*}

Since $X$ is a coordinate vector field, $[X,X] = 0$ so the last three terms vanish. Hence, we see that \begin{align*}\langle \overline{\nabla}_X X, X\rangle &= \frac{1}{2} Xg(X,X)\\ &= \frac{1}{2}\partial_n \frac{1}{x_n^2}\\ &= -\frac{1}{x_n^3}\end{align*}

In particular, $ \overline{\nabla}_{\partial_n} \partial_n\neq 0$.

This in turn affects the computation of $\Gamma^k_{ij}$ which affects the geodesic equation. The punchline is that your $\gamma$ has the same image as a geodesic, but is not constant speed. This can be seen since, for example, $O(n-1)$ acts on $\mathbb{H}^n$ by rotating about the image of $\gamma$ which is clearly an isometry. The fixed point set of this $O(n-1)$ action is thus a totally geodesic submanifold - that is, the image of $\gamma(t)$ is a geodesic.

On the other hand, the correct paramaterization is given by $\alpha(t) = (0,\ldots,0,e^t)$. Then this has the same image as $\gamma$, hence is a geodesic iff it's constant speed. But we have \begin{align*} \frac{d}{dt}\langle \alpha'(t),\alpha'(t)\rangle &= \frac{d}{dt} \langle e^t\partial_n, e^t\partial_n\rangle\\ &= \frac{d}{dt} e^{2t} \frac{1}{x_n^2}\\ &= \frac{d}{dt}e^{2t} \frac{1}{e^{2t}} \\ &= 0\end{align*} so the length squared, and hence the length, of $\alpha$ is constant.

Answer 2

Intuitively, the reason why there is no contradiction with Hopf-Rinow is because it takes "infinitely long" for the geodesic $\gamma$ you describe above to reach the boundary of the upper half plane. Remember that in the hyperbolic metric, distances get arbitrarily big as you get arbitrarily close to the boundary from the Euclidean perspective.

To be a little more precise. Let $\gamma(t) = (0, \ldots, t)$ be the geodesic described above, where $\gamma(0)$ is chosen to be an arbitrary point in the interior of the upper half plane. Then $\gamma(t)$ never hits the boundary for any value of $t$, because the boundary is "infinitely far away" in the hyperbolic metric.

In fact, not only does $\gamma(t)$ never hit the boundary, but the same argument shows that $\gamma(t)$ can never even be within a finite distance to the boundary. This shows that the metric completion of the upper half plane with respect to the hyperbolic metric can't include the boundary itself.

A way to get a concrete handle on this is the following. Take the sequence of points $p_n = (0, \ldots, 1/n)$ for $n = 1, 2, \ldots$ and calculate the distance between $p_1$ and $p_n$ (i.e. the length of the geodesic $\gamma$ connecting $p_1$ to $p_n$). You will see that this distance goes to infinity as $n \to \infty$.

Answer 3

The Hopf Rinow theorem is only valid in local (i.e. exponential) coordinates. The standard Riemannian metric you described is not the local metric, although it is extremely useful.