I have read that the curvature is invariant under isometry in the general, though I'm urrently reading up a little on symmetric spaces and I've seen that we have
$\nabla R=0$ iff the space is locally symmetric.
But isn't this condition just the statement that curvature is invarient under parallel transport, which is an isometry?
I'm guessing that not all spaces are locally symmetric, but can't seem to figure out where I'm making the mistake.
Best Answer
What is true is that if $f : M \to M$ is an isometry, , $f^*g = g$, and then $f^*\nabla = \nabla$ and consequently this leads to $f^*R = R$. So an isometry of $M$ leaves unvariant the curvature.
But this has nothing to do with parallel transport: say $\gamma : I \to M$ is a smooth path and let $P_t$ be the parallel transport from $\gamma(0)$ to $\gamma(t)$. Then: $$ P_t : \left(T_{\gamma(0)}M,g_{\gamma(0)}\right) \to \left(T_{\gamma(t)}M,g_{\gamma(t)}\right) $$ is a linear isometry. It is not an isometry of $M$.
In fact, on a connected riemannian manifold, any two points can be joined by a smooth path and thus, there are always parallel transport between them. But they may not even exist an isometry of $M$ sending one point to another given one in the general case. This would imply that $M$ is two-point homogeneous and this has huge consequences on the curvature.