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For me, from the definition of a geodesic:

Let $\gamma : t \rightarrow \gamma(t)$, $t \in I$, be a curve in a manifold $M$. The curve $\gamma$ is called a geodesic if the family of tangent vectors $\dot\gamma(t)$ is parallel with respect to $\gamma$.

it is not obvious, that for any two given points in a connected manifold $M$ there is a geodesic passing through them. Is this so?

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I guess $M$ is assumed to be connected. – Berci Dec 11 '12 at 11:35
You are right, I fixed this. – Hans Stricker Dec 11 '12 at 11:35
up vote 4 down vote accepted

No. Think about $M=\mathbb{R}^2-\{(0,0)\}$ with respect to the standard metric. Then all geodesics in $M$ are straight line. Now consider the points $p=(1,0)$ and $q=(-1,0)$ in $M$. There is no straight line joining $p$ and $q$ because $(0,0)$ is not in $M$.

In fact, $M$ is called geodesicaly complete if for any two given points $p$ and $q$ in $M$, there is a geodesic joining $p$ and $q$. And Hopf-Rinow Theorem says that $M$ is geodesicaly complete if and only if $M$ is complete as a metric space. See here. So the above example $\mathbb{R}^2-\{(0,0)\}$ is not complete as a metric space, as we know from calculus.

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