Tangential behavior of Riemannian exponential - MathOverflow

Tangential behavior of Riemannian exponential

2011-06-24T17:45:35Z

I wonder if the following holds in an arbitrary Riemannian manifold $M$:

assume $x\in M$, $h\in T_x M$, do we have for $u\in T_x M$ exponentiable (if necessary of small enough norm) that:

$$\lim_{t\to 0} \frac{d(\exp_{\exp_x u}t\tau(h), \exp_x (u+th))}{t}=0$$ where $\tau(h)$ is the parallel transport of $h$ along $[0,1]\rightarrow M, t\mapsto \exp_x tu$.

If it doesn't hold in general, then which geometrical conditions are sufficient for it to hold, and do we have at least in general that for any $x\in M$, $h\in T_x M$: $$\lim_{t\to 0, u\to 0} \frac{d(\exp_{\exp_x u}t\tau(h), \exp_x (u+th))}{t}=0$$

Answer by Sergei Ivanov for Tangential behavior of Riemannian exponential

2011-06-24T18:40:12Z

The second identity is always true because both arguments of $d$ are smooth functions of $u$ and $t$ and they coincide when $u=0$.

The first one holds true for all $u$ and $h$ only if the metric is flat. Indeed, the l.h.s. is the length of the difference of the initial velocity vectors of two curves $t\mapsto \exp_{\exp_x u} t\tau(h)$ and $t\mapsto\exp_x(u+th)$. The first velocity vector is $\tau(h)$, so its length equals $|h|$. The second one is the derivative of $\exp_x$ at $u$ along $h$. If its length equals $|h|$ for all $u$ and $h$, then $\exp_p$ is a Riemannian isometry (by definition), so the metric is flat.