# Difference between parallel transport composed with exponential maps along two different geodesics starting at the same point?

I asked this question on math.stackexchange too: it's not a homework problem, but something that came to my mind while thinking of commutation:

https://math.stackexchange.com/questions/1356518/difference-between-exponential-maps-composed-with-parallel-transport-along-two-d

Let $(M,g)$ be a Riemannian manifold, and let $\gamma_{p,v}, \gamma_{p,w}$ be two geodesics starting from $p$ with directions/initial vectors $v,w$ respectively. Consider the two operations (to be formally stated in a moment):

I) Take the parallel transport $P^v$ of $w$ from time $0$ to $t$ along $\gamma_{p,v}$, then take exponential map at this point and travel for time $s$.

II)Take the parallel transport $P^w$ of $v$ from time $0$ to $s$ along $\gamma_{p,w}$, then take the exponential maps at this point and travel for time t.

My question is: how different are these two points for varying $s,t$?

To formalize things, consider $F(t,s):=exp_{exp_p(tv)}{(sP^v_{0,t}(w))}$ and $G(t,s)=exp_{exp_p(sw)}{(tP^w_{0,s}(v))}$.

Clearly, $F(t,s)$ and $G(t,s)$ are not necessarily equal, because if we fix $s$, $G$ is a geodesic but $F$ is not, unless the curature vanishes. But I'd like to know if we have results that estimate $||F(t,s)-G(t,s)||$.

• We have $F(t,0)=G(t,0)$ and $F(0,s)=G(0,s)$. So if you assume smoothness we have $d(F(t,s),G(t,s))=\mathcal{O}(st).$ Jul 13, 2015 at 12:56