$M$ a complete Riemannian manifold with nonpositive sectional curvature. Show that $|d(\exp_p)_v(w)| \geq |w|$ for all $v \in T_p M$, and all $w \in T_v(T_p M)$.


Based on the hints I've gotten, here is my attempted solution.

We will compare $M$ with $\tilde{M}=T_p M \approx T_v T_p M$ via the Rauch comparison theorem. Let $\gamma:[0,1] \rightarrow M$ be a geodesic and let $\tilde{\gamma}: [0,1] \rightarrow \tilde{M}$ be a comparison geodesic (which means that $\gamma$ and $\tilde{\gamma}$ have the same speed). Note that $\tilde{\gamma}$ also has no conjugate points because curvature zero. Write $\gamma(0)=p$ and $\gamma'(0)=v$.

Let $J(t)$ be the geodesic along $\gamma$ with $J(0)=0$ and $J'(0)=w$; then $J(t)$ can be written


so we see that $J(1)=d(exp_p)tv(tw)$.

The Rauch comparison theorem gives us that, for a Jacobi field $\tilde{J}$ along $\tilde{\gamma}$ with

  1. $\tilde{J}(0)=J(0)=0$
  2. $|\tilde{J}'(0)|=|J'(0)|$

  3. $\langle \tilde{J}'(0),\gamma'(0) \rangle = \langle J'(0), \gamma'(0) \rangle$

we have $|\tilde{J}| \leq |J|$, so if we had a Jacobi field $\tilde{J}$ along $\tilde{\gamma}$ with $\tilde{J}(1)=w$ we'd be done.

There is a theorem that says if there is no conjugate points along $\tilde{\gamma}$ then there is a unique Jacobi field $\tilde{J}$ along $\tilde{\gamma}$ with $\tilde{J}(0)=0$ and $\tilde{J}(1)=w$ but I am unsure how how to get that the highlighted conditions (2) and (3) above hold.

  • $\begingroup$ Compare $M$ with $\tilde{M}=T_pM.$ $\endgroup$ – Frieder Jäckel Aug 31 '18 at 14:50
  • 1
    $\begingroup$ Hint: $d(\exp_p)_v(w)$ is equal to $J(1)$, where $J$ is a Jacobi field along the geodesic $t\mapsto \exp_p(tv)$ that vanishes at $t=0$. $\endgroup$ – Jack Lee Aug 31 '18 at 18:48
  • $\begingroup$ $\tilde{M}=T_pM$ has constant metric, so the curvature tensor vanishes and the covariant derivative along curves is the usual derivative. Therefore $\tilde{J}(t)=tw$ is the Jacobifield (Uniqueness!) along $t\to tv$ with $\tilde{J}(0)=0$ and $\tilde{J}‘(0)=w.$ $\endgroup$ – Frieder Jäckel Sep 4 '18 at 19:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.