Upper bound of derivative of exponential map

Question

We know that for any simply connected surface $M$,whose Gaussian curvature $K\leq 0$, for any $p\in M$, $exp_p: T_pM\to M$ is diffeomorphism.

We know that for any $v\in T_pM$ and $w\in T_v(T_pM)\simeq T_pM$, $$|d(exp_p)_vw|\geq |w|$$

That is we have $$\||d(exp_p)_v\|_{op}\geq 1$$

Question: Can we find an upper bound of $\||d(exp_p)_v\|_{op}$ for all $v\in T_pM$ for particular $p$ in particular $M$. Or this is not atall possible for every $p$ and $M$.

If question is of level math.stackexchange, please suggest me.

Answer 1

To get an upper bound of the kind you seek in general, you need a lower bound on $K$. Thus, for example, if you know that $K\ge -c^2$ on your surface, then you get $$ \|\mathrm{d}(\exp_p)_v\|_{op} \le \frac{\sinh\bigl(c\,|v|\bigr)}{c\,|v|} \simeq 1 + \frac{\bigl(c\,|v|\bigr)^2}{3!} + \frac{\bigl(c\,|v|\bigr)^4}{5!} +\cdots. $$ This follows by a standard application of Sturm comparison on geodesics leaving $p$, and is usually treated in a first course in differential geometry, at least at the graduate level.

A more precise estimate is possible if you know that $-b^2\ge K\ge -c^2$ for some constants $0\le b\le c$, for then one has $$ \frac{\sinh\bigl(b\,|v|\bigr)}{b\,|v|} \le\|\mathrm{d}(\exp_p)_v\|_{op} \le \frac{\sinh\bigl(c\,|v|\bigr)}{c\,|v|}. $$ Moreover, one knows that, in this case, $\mathrm{d}(\exp_p)_v$ has two singular values, $\sigma_1\equiv1$ and $\sigma_2(p,v) = \|\mathrm{d}(\exp_p)_v\|_{op}\ge 1$. This is special for surfaces, though.