# $\exp_q^{-1}(\exp_p(tX)) = \exp_q^{-1}(p)+t\Gamma_{p\to q}(X)+O(t^2)$ as $t\to 0$?

Let $$(M,g)$$ be a Riemannian manifold with induced metric $$d$$ and injectivity radius $$r>0$$. Let $$p, q$$ be two points in $$M$$ such that $$d(p,q). It is easy to see that $$p$$ and $$q$$ can be connected by an unique minimizing geodesic $$\gamma=\{\gamma(t):0\le t\le 1 \}$$ with $$\gamma(0)=p$$ and $$\gamma(1)=q$$. Denote by $$\Gamma_{p\to q}:T_p(M)\to T_q(M)$$ the parallel transport along the geodesic $$\gamma$$ from $$p$$ to $$q$$.

My question is: for a tangent vector $$X\in T_p(M)$$, does the following holds $$\exp_q^{-1}(\exp_p(tX)) = \exp_q^{-1}(p)+t\Gamma_{p\to q}(X)+O(t^2), \quad t\to 0?$$

Any comments or hints will be appreciated. TIA.

It does not hold : On $$2$$-dimensional unit sphere, consider two points of $$\frac{\pi}{2}$$ distance. And parallel vector field on the geodesic between two points is perpendicular to the geodesic (cf. Parallel transportion for Alexandrov space with curvature bounded below - Petrunin) :
When $$M=\mathbb{S}^2,\ |p-q|=\frac{\pi}{2}$$, then $$c$$ is unit speed geodesic from $$p$$ to $$q$$. Assume that $$X(t)$$ is a unit parallel vector field along $$c(t)$$ s.t. $$X(0)\perp c'(0)$$.
Then $$A=0,\ B=\exp_p^{-1}\ q,\ C=\exp_p^{-1}\ \exp_q\ tX(\pi/2)$$ is a triangle in $$T_pM$$ s.t. $$\angle\ BAC = t$$.
When $$A,\ B,\ C'$$ is triangle with $$\angle\ ABC'=\pi/2$$ and $$|B-C'|=t$$, then we find $$|C-C'|$$ : By cosine law, \begin{align*} |C-C'|^2&=t^2+(\pi\ \sin\ \frac{t}{2})^2-2t(\pi \ \sin\ \frac{t}{2})\cos\ \frac{t}{2} \\& =( t-\pi\ \sin\ \frac{t}{2})^2+ O(t^4) \\ |C-C'| &=t-\pi\ \sin\ \frac{t}{2}+ O(t^2)\end{align*}
• I think I understand your exmaple and it is really a good counterexample. But I derive the last two equations in your answer as $$|C-C'|^2 = (t- \frac{\pi}{2} \sin t)^2 + (\frac{\pi}{2} - \frac{\pi}{2} \cos t)^2 = t^2 - \pi t \sin t + \frac{\pi^2}{2} (1-\cos t) = (1-\pi+ \frac{\pi^2}{4}) t^2 + O(t^4),$$ $$|C-C'| = (1-\frac{\pi}{2}) t + O(t^2).$$ But, anyway, this negate that equality in my question. Thanks. – Q. Huang Jan 29 '19 at 6:11