TITLE: Action of the group of isometries on a manifold
QUESTION [6 upvotes]: Hi guys,
I am able to prove that any symmetric manifold is complete (Consider a local geodesic and use the symmetry to flip it, effectively doubling the length of the geodesic, ad infinitum). I want to use a similar procedure to prove that a manifold whose isometries act transitively is complete, i.e there is always an isometry which maps the start point of a local geodesic to its end point, preserving the geodesic. I am, however, unable to ensure that it is not `rotated' in the process, i.e I want the pushforward of the initial tangent, by the isometry, to be the final tangent, ensuring the resultant doubled geodesic is smooth.
My Lie group theory is a bit scratchy but I assume there is a method which allows me to construct the correct pushforward using only transitivity.
Any ideas would be great,
regards,
MK

REPLY [2 votes]: Your original approach for extending the geodesic is in fact possible, after a slight modification. (Credint: I heard this Idea from Eran Assaf). 
(You do not pushforward the original geodesic, but instead you pushforward another geodesic which starts at $p$).
Let $p \in M$. It's known there exists a neighbourhood (of $p$) $V$ in $M$, and $\epsilon > 0$ such that for every $q \in V \, , w \in T_qM , |w|<\epsilon$  the (unique) geodesic which starts at $q$ with velocity $w$ is defined on $(-2,2)$.
Let $c$ be geodesic:  $c(0)=p,|\dot c(0)|<\epsilon$. By the above statement, we can assume $c$ is defined on $(-2,2)$,So we can define $q = c(1)$. 
By assumption there is an isometry $f:M \to M$ such that $f(p)=q$.
Now let $\gamma$ be the geodesic satisfying: $\gamma(0)=p,\dot \gamma(0)=(df_p)^{-1}(\dot c(1))$. Since $f$ is an isometry: $|\dot \gamma(0)|=|\dot c(1)|\overset{ c \, \text{geodesic} }{=}|\dot c(0)|<\epsilon$. Hence $\gamma$ is also defined on $(-2,2)$.
Now consider the curve $\beta = f \circ \gamma$: 
$\beta(0) = f(\gamma(0))=f(p)=q=c(1)$
$\dot \beta(0) = df_p(\dot \gamma(0))=\dot c(1)$
Since $f$ is an isometry and $\gamma$ is a geodesic, $\beta$ is also a geodesic. Hence, $\beta$ is the unique geodesic starting at $q = c(1)$ with velocity $\dot c(1)$. But so is $\tilde c(t)=c(t+1)$, hence they are equal, whenever both are defined. 
So, we can now concatenate, to obtain a geodsic $\alpha:(-2,3) \to M$,
$\alpha(t) =
\begin{cases}
c(t),  & t \in (-2,2)  \\
\beta(t-1), & t \in (-1,3)
\end{cases}$
$\alpha$ is well-defined, and a geodesic. Now we may repeat this procedure ad infinitum.