Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
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
add comment

1 Answer

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.