# Existence of minimizing geodesic in each fixed-end-point homotopy class in a complete manifold?

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.

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## 1 Answer

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.

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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