Holonomy of H^{n}

Question

I am trying to show that $Hol_{p}(H^{n})=SO(n)$. I know that Iso$_{p}=SO(n)$. From here can I conclude that $Hol_{p}(H^{n})=SO(n)$? For $S^{2}$ if we have two vectors $u,v$ at north-pole $N$ then let $\gamma , \sigma $ be geodesics from $N$ with velocity $u,v$ and intersect the equator at $p,q$. Then if we parallelly translate $u$ along $\gamma$, then along equator and then along $\sigma$ we get $v$. A similar argument works for $S^{n}$. Is there a similar argument for $H^n$?

Thank you in advance.

Answer 1

Start by looking at the case $n=2$. Then, by considering holonomies along various geodesic triangles with one vertex at $p$, you see that every rotation about $p$ is realized (this is very similar to the sphere case: You use the fact that each nondegenerate triangle has nonzero angle deficit). Therefore, the holonomy group of $H^2$ is $SO(2)$. Now, use this fact for the general $n$ to conclude that any rotation about a codimension 2 subspace is in the holonomy group. Lastly, use the fact that $SO(n)$ is generated by rotations of the above type (normal form of orthogonal transformations).