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 $J(0)=J(t)=0$.


2 Answers 2



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:


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.

  • 1
    $\begingroup$ What do you mean by "almost meet?" In general, the family of geodesics won't all intersect at $q$, but then in what sense do they almost meet? $\endgroup$
    – Mathmank
    Commented Feb 9, 2016 at 21:49

Let $p,q$ two points, and $c$ a path between them. The energy functional is a nice function on the space $\Omega_p^q$of paths from $p$ to $q$, and a geodesic $c$ is just a critical point of this functional. If you think of this $\Omega $ as a manifold, a Jacobi Field exists iff the second derivative $E"$ of $E$ is degenerate, and this Jacobi field is an element of the kernel of $E"$. A good way to produce this is to consider a one parameter family of geodesics $c_t$ such that at $t=0$, the geodesic is non degenerate with Morse index $i$ and at $t=t_0$ with Morse index $i+1$. Between, some geodesic must be degenerate. For instance, take a simple closed geodesic of positive index. Let $p$ be some point on $c$, and $c_t$ be the arc of $c$ between $p$ and a point at the distance $t$. If $t$ is small, the index is still $0$, and if $t=t_0$ is the length of $c$ the index is $1$, so somewhere in between there is a conjugate point to $p$

  • 1
    $\begingroup$ Thomas: Is $\Omega$ really a smooth manifold? $\endgroup$
    – C.F.G
    Commented Dec 17, 2021 at 7:36

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