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Jul 13, 2011 at 13:22 history edited Claudio Gorodski CC BY-SA 3.0
Added necessary assumption on theorem of Cartan.
Jul 13, 2011 at 13:20 comment added Claudio Gorodski Hugo: that's right, the Riemannian manifold must be simply-connected, thanks! A symmetric space of non-compact type is automatically simply-connected, and this can be seen from the argument in the "Edit".
Jun 26, 2011 at 16:18 comment added Hugo Chapdelaine Well I'm confused, say that you take $X=\mathbf{C}/\Lambda$ with $\Lambda=\mathbf{Z}+i\mathbf{Z}$ and you take the isometry $z\mapsto z+\frac{1}{2}$, then there is no fixed point! Are you assuming that your space is simply connected?
Jun 20, 2011 at 17:07 history edited Claudio Gorodski CC BY-SA 3.0
Added explanation regarding first question.
Jun 19, 2011 at 4:23 comment added Ben Wieland To answer the second question, you used the fact that $G/K$ is nonpostively curved. In the cast that $G$ is semisimple, this is a fairly easy computation using the Killing form. Moreover, this result answers the first question, that $G$ is diffeo to $K\times G/K$ and $G/K$ to $R^n$.
Jun 19, 2011 at 1:47 history answered Claudio Gorodski CC BY-SA 3.0