Action of the group of isometries on a manifold

Question

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

Answer 1

By the Hopf-Rinow theorem, you only have to prove that the manifold is a complete metric space. By homogeneity, the injectivity radius is bounded from below by a uniform positive constant. Using this and the compacity of balls whose radius is smaller than the injectivity radius of their center it is easy to check the convergence of Cauchy sequences.

Another way to do this is to interpret the bound on injectivity radius, $r$ say, in term of geodesic extension: a geodesic $\gamma$ defined on $[a,b]$ can be extended to a geodesic defined on $(a-r,b+r)$. From this the conclusion follows.

Answer 2

It is easy to see that any metrically homogeneous, locally compact, metric space, $X$, is complete. If $p$ is some point of $X$ then, by local compactness, for some $\epsilon > 0$, the closed $\epsilon$-ball about $p$ is compact and hence complete. Then, by metric homogeneity, the closed $\epsilon$-ball about every point is complete. Then, if $x_n$ is a Cauchy sequence in X, eventually the $x_n$ are all within $\epsilon/2$ of eachother, and so by the triangle inequality they eventually lie in a closed $\epsilon$ ball about one of them. qed (Of course, the same argument shows more generally that if a metrically homogeneous metric space has one point with a complete neighborhood then it is complete.)

Answer 3

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.