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I am studying Riemannian Geometry in my own. And I was going through the proof of Hopf Rinow from the book of Carmo . I have a few doubts if one could help.

  1. In one side it has been proved that if $exp_p$ is defined on all of $T_pM$ then there exists a length minimizing geodesics between $p$ and $q$. Since geodesics are locally length minimizing by definition, I am not clear how to get exactly the idea how far it minimizes.

  2. In this direction of the proof which is essentially same in this pdf https://www.mathi.uni-heidelberg.de/~lee/Sven05.pdf

where we are exactly using the assumption that $exp_p$ is defined on all of $T_pM$?

  1. Here in the proof we are taking $x_0=exp_p(\delta v)$ and I guess it's coming from the fact that every geodesics starting from $p$ in normal nbd of the from $ exp_p(sv)$ for some unit tangent vector in $T_pM$.

However in Prop.5.11 of Lee's book in normal nbd geodesics from $p$ is given by $\gamma_V(t)=(tV^1,..,tV^n)$ . So I am very much confused about what is exactly the expression of geodesics starting from the point $p$ in normal neighborhoods .

Any help would definitely help my understanding in the subject.

User11111
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1 Answers1

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For the Hopf-Rinow theorem, we want to find a distance minimizing geodesic $\gamma$ from $p$ to $q$. The idea is:

  1. Start with a normal neighbourhood $exp_p(B_{\delta}(0))$ of $p$; in this neighbourhood the geodesics from $p$ are of the form $exp_p(sv)$ for some $v$ in $T_p(S)$ with $|v| = 1$ and $s < \delta$.
  2. Find a point $x_0$ on the boundary of $exp_p(B_{\delta}(0))$ that minimizes $d(x, q)$ amongst all $x$ on the boundary of $exp_p(B_{\delta}(0))$. Then our desired $\gamma$ starts with the geodesic from $p$ to $x_0$.
  3. Repeat this process from $x_0$ until we get to $q$. At each step, we prove that the geodesic segment added in fact coincides with $\gamma$.

Why do we need $exp_p$ to be defined on all of $T_p(S)$? Well, if our surface was not complete, then this process of extending $\gamma$ iteratively might terminate at a point where $\gamma$ is not defined. But since $\gamma$ is defined for all real numbers, we do not get a contradiction.

The Special One
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