8
$\begingroup$

I'm trying to understand why manifolds with negative sectional curvature not have conjugate points. In fact for me it is sufficient to understand it for surfaces, but of course i'd be interested in the general case.

many regards, and thanks in advance, Leo

$\endgroup$

2 Answers 2

13
$\begingroup$

Let $\gamma:[0,\epsilon]\rightarrow M$ be a geodesic curve, $\epsilon>0$ and $J$ be a Jacobi vector field along $\gamma$ with $J(0)=J(\epsilon)=0$.

I remind you that :

  1. $J$ being a Jacobi vector field means that $\dfrac{D^2}{dt^2}J+R(\gamma',J)\gamma'=0$ with the usual notation for the covariant derivative $D$ and $R$ the Riemannian curvature tensor.
  2. $\left[\|X\|^2\|Y\|^2-\langle X,Y\rangle^2\right] \kappa(X,Y)=\langle R(X,Y)X,Y\rangle$ for any $X,Y\in TM$ with $\kappa$ the sectional curvature.

Now look at the map $\phi:t\mapsto \|J(t)\|^2$. Its second derivative is easy to compute and using the two remarks you get : $$\begin{align} \phi''(t) & = \langle \dfrac{D^2}{dt^2}J,J\rangle + \langle \dfrac{D}{dt}J,\dfrac{D}{dt}J\rangle\\ & = -\langle R(\gamma',J)\gamma',J\rangle + \|\dfrac{D}{dt}J\|^2 \\ & = -\left[\|X\|^2\|Y\|^2-\langle X,Y\rangle^2\right]\kappa(\gamma',J) + \|\dfrac{D}{dt}J\|^2.\end{align}$$

In particular, if all sectional curvatures are non-positive then $\phi''\geq 0$. So $\phi$ is a convex map and since $\phi(0)=\phi(\epsilon)=0$, $\phi(t)=0$ for any $t\in [0,\epsilon]$. It follows that $J$ has to be trivial.

So we can't have conjugate points in $M$.

$\endgroup$
2
  • $\begingroup$ thanks a lot Bebop for the very nice answer. So the argument goes: 1. the Jacobi field has $\frac{d^2}{dt^2}||J(t)||^2>0$, i.e. the length squared is a convex function on $[0,\epsilon]$ 2. $||J(0)||^2=||J(\epsilon)||^2$ implies that J is trivial. But why do we require $J(0)=J(\epsilon)$? Is that the condition for conjugate points? $\endgroup$
    – Marlo
    Apr 24, 2013 at 23:04
  • $\begingroup$ I forgot to write that $J(0)=J(\epsilon)=0$, otherwise the conclusion would be that $\phi$ is constant. But don't you know the definition of conjugate points ? This is all about your question ? Go on wikipedia to get the full definition. $\endgroup$
    – Bebop
    Apr 24, 2013 at 23:19
4
$\begingroup$

This is very late, but to provide an alternative proof:

The setup for the Rauch Comparison Theorem is very wordy, but the claim follows immediately from comparison with $\mathbb{R}^m$.

Assume that $M$ is a Riemannian manifold with nonpositive sectional curvature, that $p\in M$, and that $\gamma:\left[0,T\right]\to M$ is any unit speed geodesic originating at $p$. Let $m=dim(M)$. Take any Jacobian vector field $J$ along $\gamma$ with $J(0)=0$ and $||J'(0)||>0$.

Let $\tilde\gamma(t)=(t,0,0,...,0)$ be a unit-speed geodesic in $\mathbb{R}^m$. Let $\tilde{J}$ be the Jacobian vector field along $\tilde\gamma$ satisfying $\tilde{J}(0)=0$, $\tilde{J}'(0)=(<\gamma'(0),J'(0)>,(||J'(0)||^2-<\gamma'(0),J'(0)>^2)^\frac{1}{2},0,0,...,0)$. Then $J(0)=0=\tilde{J}(0)$, $||\tilde{J}'(0)||=||J'(0)||$, and $<\tilde{\gamma}'(0),\tilde{J}'(0)>=<\gamma(0),J'(0)>$.

For all $t\in\left[0,T\right]$, $max(K(\gamma'(t),X))\leq0=min(\tilde{K}(\tilde{\gamma}'(t),\tilde{X}))$ and $dim(\mathbb{R}^m)\leq dim(M)$ so, by the Rauch Comparison Theorem, $||J(t)||\geq||\tilde{J}(t)||$ for all $t\in\left[0,T\right]$. In particular, since $\mathbb{R}^m$ has no conjugate points, $\tilde{J}'(0)\not=0$ guarantees that $||\tilde{J}||>0$ for all positive $t$, hence $||J(t)||>0$ for all positive $t$, thus $p$ has no conjugate points along $\gamma$.

Since $p$ and $\gamma$ were arbitrary, $M$ has no conjugate points.

$\endgroup$
1
  • $\begingroup$ Thanks @Pepper! That's also a nice way of showing it :) $\endgroup$
    – Marlo
    Dec 25, 2018 at 17:59

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.