## 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 2010 at 15:56

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.

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Needless to say, but just in case and since the OP has not made this clear: this will not work in indefinite signature, where Hopf-Rinow fails. There are incomplete homogeneous lorentzian manifolds, for example. – José Figueroa-O'Farrill Jul 8 2010 at 15:54

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.)

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 thank you to everybody. – kangdon Jul 11 2010 at 1:38