# Action of the group of isometries on a manifold

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

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The result you want is not true in indefinite signature, so you need to assume that you are working with positive-definite riemannian manifolds. (Perhaps this is implicit in your tag, but a growing number of people use riemannian geometry to include also the indefinite case.) –  José Figueroa-O'Farrill Jul 8 '10 at 15:56

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.
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.)