A question about geodesics of hyperbolic space 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 &gt;0 $ is a geodesic in $ H^n $. This curve is clearly defined for $ t&gt;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
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 =&amp; \frac{1}{2}\big(Xg(X,X) + Xg(X,X) - Xg(X,X)\\ &amp;+ \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 &amp;= \frac{1}{2} Xg(X,X)\\ &amp;= \frac{1}{2}\partial_n \frac{1}{x_n^2}\\ &amp;= -\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 &amp;= \frac{d}{dt} \langle e^t\partial_n, e^t\partial_n\rangle\\ &amp;= \frac{d}{dt} e^{2t} \frac{1}{x_n^2}\\ &amp;= \frac{d}{dt}e^{2t} \frac{1}{e^{2t}} \\ &amp;= 0\end{align*} so the length squared, and hence the length, of $\alpha$ is constant.
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.