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This should be completely straightforward as every book I have read on the Hopf-Rinow theorem states this is "obvious". But I can find the life of me to justify this.

Geodesically complete means every geodesic $\gamma$ extends to all of time $\mathbb{R}$. How is that related to the velocity $\gamma'$ (which hence relates to exp)? If the curve extends all time, then are they implying the speed also extends all time? Does that even make sense since speed is constant for a geodesic, so why would that be attached to the time parameter?

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Let $v \in T_p M$, $\gamma$ a geodesic that has speed $v$ at $\gamma(0)=p$.

The exponential map defined on $p$ takes vectors $v$ from $T_p M$ and considers the geodesic that has that speed at the point $p$. This map does the following, $v \to \gamma(1)$ whenever it is defined the point $\gamma(1)$. Geometrically what the map is doing is taking a geodesic and riding it until it has elapsed a unit of time. So if the surface is geodesically complete there is no reason for it to stop. That means $\gamma(1)$ is always defined and so the domain of the exponential map is all of $T_p M$. Conversely if for all points $p$ and all speed $v \in T_p M$ the exponential map is defined it means the surface is geodesically complete.

This shows why being geodesically complete in the sense that the domain of a geodesic is all $\mathbb R$ is exactly the same that the exponential map is defined on all the points of the surface and for all the speeds in the tangent space to the point.

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    $\begingroup$ The exponential map takes velocity vectors as you say but it's image is not a vector of the tangent space. It's image is a point in the surface. The main idea of the exponential map is to parametrize a surface near a point $p$. That means it's image are points in the surface near $p$. $\endgroup$ – Leo Lerena Feb 26 at 6:34
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    $\begingroup$ Yes, it was a typo, sorry about that. Which question you mean. I tried to show the equivalence you wrote on the title of the question. About the question "If the curve extends all time, then are they implying the speed also extends all time?". The speed is constant as you said. What they are implying is that the exponential map can be defined on all of $T_p M$ as opposed to an open subset of it. $\endgroup$ – Leo Lerena Feb 26 at 6:45
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    $\begingroup$ Slightly abuse from my part. It takes a velocity vector right. There is a unique geodesic at that point which possess that velocity, so in a sense for each $v \in T_p M$ you have a unique geodesic $\gamma$ which has $\gamma'(0)=v$ and $\gamma(0)=p$. I can add this to the answer if I was not clear enough. $\endgroup$ – Leo Lerena Feb 26 at 6:57
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    $\begingroup$ The existence and uniqueness of that geodesic relies on the existence and uniqueness of the solution of a differential equation, which is a local theorem. $\endgroup$ – Leo Lerena Feb 26 at 6:59
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    $\begingroup$ @Leo: by your discussion in chat, it only implies that if we assume $D_p = T_p(M)$, then geodesics starting at $p$ are defined on an interval containing $[0,1]$, how does it imply that they're indeed defined on all of $\mathbb{R}$? $\endgroup$ – Mojojojo Apr 18 at 16:01
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Let $(M,g)$ a Riemannian manifold. Recall that for all $p\in M$ and $v \in T_pM$ there exists some $\varepsilon > 0$ and a $C^\infty$curve $\gamma(t)\colon (-\varepsilon, \varepsilon)\to M$ such that $\gamma$ is the unique geodesic passing through $p$ at time zero with velocity $\dot{\gamma}(0) = v$. The fact that the exponential map is defined for some $v \in T_pM$ has nothing to do with existence of a geodesic through $p$ with velocity $v$ (this always happens!), but with the fact that $\gamma(1)$ is defined ($\varepsilon > 1$). One has to be careful in thinking that a reparameterization would change the speed of a geodesic. A much careful treatment is given in the first five pages of Chapter 3 in do Carmo's book

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  • $\begingroup$ "but with the fact that $\gamma(1)$ is defined ($\varepsilon > 1$" I am certain this is what keeps me from understanding this. If past $\varepsilon > 1$ we still get the geodesic (via completeness), you are implying this is where it shows the equivalence? But like you say I could just take any $(p,v_p)$ and I would get a unique geodesic, so where is completeness coming in? $\endgroup$ – Hawk Feb 28 at 1:30
  • $\begingroup$ "But like you say I could just take any (𝑝,𝑣𝑝) and I would get a unique geodesic, so where is completeness coming in?" such geodesic might not be defined at $1$, thus, how do you evaluate the exponential map? $\endgroup$ – Juan Diego Rojas Feb 28 at 1:33
  • $\begingroup$ Can we work with an example? Let's just take $\mathbb{R} - 0$? $\endgroup$ – Hawk Feb 28 at 8:50
  • $\begingroup$ Yes. Let $p=(-1,0)$ what is $exp_p(1,0)$? $\endgroup$ – Juan Diego Rojas Mar 3 at 1:14

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