[Math] How to understand conjugate points on a Riemannian manifold

Question

I'm having trouble grasping what it means for two points to be conjugate on a Riemannian manifold. Could someone provide a geometric or intuitive explanation for this?

For clarification: given a geodesic $\gamma: [0,a] \to M$, the point $\gamma(t)$ is conjugate to $p=\gamma(0)$ if there exists a Jacobi field $J$, not identically zero, along $\gamma$ such that $J(0)=J(t)=0$.

Answer 1

$\newcommand{\ga}{\gamma}$

Jacobi fields measure how geodesics that start at the same point but with different velocities spread apart from each other.

Let $p \in M, v,w \in T_PM$. We consider a $1$-parameter family of geodesics:

$$\ga_s(t)=\exp_p(t(v+sw)).$$

Recall that $\ga(t)=\exp_p(tv)$ is the unique geodesic passing through $p$ (at $t=0$) with velocity $v$. Thus, $\ga_s$ is the unique geodesic passing through $p$ (at $t=0$) with velocity $v+sw$. (We are slowly changing the initial velocity of our geodesic).

$J(t)=\frac{\partial \ga_s(t)}{\partial s}|_{s=0} \in T_{\exp_p(tv)}M=T_{\ga(t)}M$ is a vector field along the initial geodesic $\ga$.

Note that $J(0)=0$. (since all our geodesics start at the same point $p$ at $t=0$). Moreover, every Jacobi field along $\ga$ satisfying $J(0)=0$ can arise from such a family of geodesics.

$\|J(t)\|$ measures the rate of "spreading apart" of the geodesics at time $t$, when their initial velocity at $p$ change.

Hence, if $q=\ga(t_0)$ is conjugate to $\ga(0)=p$ along $\ga$, this means that there is a continuous family of geodesics starting from $p$ which "almost" meet at $q$. (They meet at $q$ only "up to first order").

Clearly, if $\ga_s(t_0)=q$ for every $s$ (that is all the geodesics in the family meet at $q$ at time $t_0$) then $J(t_0)=0$, but the reverse implication is false in general.