Does a geodesic always extremize its path length? I've learned that a geodesic maximizes its proper time in Minkowski spacetime.
Is this still true in general curved spacetime?
In other words, does the geodesic equation give the globally extremal path from one point in spacetime to another. Assuming the points are time-like separated.
I would think that by the local equivalence theorem, the geodesic equation gives a locally extremal path. Can it proven to be globally extremal?
 A: Let $g$ be your spacetime metric. We can define the energy functional
$$E(\gamma) = \frac12\int_0^1g(\dot\gamma(t),\dot\gamma(t))dt$$
on curves $\gamma$ in your space. The Euler-Lagrange equations for this functional is the geodesic equation. Likewise we can define the length functional
$$\ell(\gamma) = \int_0^1\sqrt{g(\dot\gamma(t),\dot\gamma(t))}dt,$$
which maps a curve to its length. It is not hard to see that curves that extremize the former automatically extremize the latter, for details, see wikipedia. In particular, a geodesic always extremizes the path length. 
Depending on the topology of spacetime, it does not have to be a global extremal though. For example, on a cylinder $\mathbb R\times S^1$ with the Minkowski metric, consider the points $(0,P)$ and $(1,P)$ where $P$ is any point on the circle $S^1$. Then the curve $\gamma(t) = (t,P)$ gives a geodesic, but you get another one wrapping around the circle $n$ times for every $n$. To find these, just unroll the cylinder into $\mathbb R\times \mathbb R$ and draw straight lines, which we already know are geodesics in Minkowski space.
