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.

This is intuitively clear, but I cannot solve this homework problem:

1) Let $(M,g)$ be a complete Riemannian manifold, let $c:[0,1]\to M$ be a continuous curve in $M$ such that $c(0)=p, c(1)=q$. Then prove that in the fixed-end-point homotopy class of $c$, there is a geodesic $\gamma$, i.e. there exists a geodesic $\gamma$ so that $\gamma$ is homotopic to $c$ with homotopy keeping the end points $p,q$ fixed.

2) My question: Assuming the above is true, is that geodesic $\gamma$ in the answer necessarily minimizing as well? I feel it should be.

I was thinking of using Hopf-Rinow theorem stating that geodesically complete is the same as metrically complete and starting with the contrary. But I got stuck.

share|improve this question

1 Answer 1

up vote 3 down vote accepted

Here is a sketch for (1). Let $L$ denote length, and $\gamma:[0,1]\to M$ some path with $\gamma(0)=p$ and $\gamma(1) = q$. Denote by $[\gamma]$ the class of paths in $M$ homotopic rel endpoints to $\gamma$. Let $L_0 =\inf_{[\gamma]}L(\gamma)$. Take a sequence of paths $\gamma_t$ with $L(\gamma_t)\to L_0$. Argue by completeness that the $\gamma_t$ must limit to a continuous path $\gamma_0$, then argue that $\gamma_0$ locally minimizes length.

The answer to (2) by the sketch above is "yes." However, geodesics do not necessarily minimize length in endpoint homotopy classes. Consider two nearby points on $S^2$. Then there are two geodesics between $p$ and $q$: the two arcs of the great circle through $p$ and $q$. The longer arc is a geodesic in the fixed-endpoint homotopy class of paths connecting $p$ and $q$, but it is maximal.

share|improve this answer
    
Thank you, at one point I had to parametrize the curves whose lengths converge to the infimum, according to arc length, then prove that they are pointwise bounded and equicontinuous, then apply Aezela-Ascoli theorem. Then use the fact that a continuous minimizer is indeed a geodesic. I hope that was correct way to do it. –  Mathmath Mar 9 '13 at 15:46
    
@Mathmath What do you mean by "continuous minimizer"? That sounds circular. –  Neal Mar 10 '13 at 4:39

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.