I'm reading Do Carmo Riemannian geometry. Given $M$ riemann manifold, it defines the conjugate locus of $p\in M$ as the set of first conjugate points to $p$, for all the geodesics that start at $p$. I was thinking: is it true that given a jacobi field (along a geodesic that starts at $p$), which admits at least a conjugate point to $p$, has a first conjugate point? can't happen that there's a succession of distinct conjugate point to $p$?

I've read some other related questions/papers, some says that conjugate points are in general discrete, some uses Morse index theorem, ... quite confusing. Is there a way to prove it using Do carmo theory? or is it something out of purpose for Do carmo book?

In general, my idea was this: let $\gamma: [0,a]\to M$ be geodesic with $\gamma(0)=p$ and $\gamma'(0)=v$. Consider $J$ jacobi field along $\gamma$ with $J(0)=0$ and $D_tJ(0)=w$. I know that can be written as $J(t)=(d_{tv}exp_p)(tw)$. Locally $exp_p$ is a diffeomorphism, so for $t$ near $0$ there aren't conjugate points. Consider $A$ the set of $t\neq0$ such that $J$ vanish and let $\bar{t}$ be the $\inf{A}$, then for continuity $J$ vanishes at $\bar{t}$ and $\bar{t}$ is the first conjugate point.

I don't know if my idea is okay or if it can be trivially simplified, but if so, is it in general true that each jacobi field has discrete finite number of vanishing point (i.e conjugate points) that I read somewhere?

  • 1
    $\begingroup$ Your proof is correct. $\endgroup$
    – Deane
    Commented May 11 at 16:06

1 Answer 1


Your proof is fine. Here is how you prove the general statement that zeros of (non-vanishing) Jacobi fields are isolated.

Let $J$ be a non-vanishing Jacobi field along a geodesic $\gamma$ and suppose $J(t_0) = 0$. Then $D_t J(t_0) \ne 0$, since otherwise $J \equiv 0$. Indeed, you can see that $\bar J(t) = J(t_0 - t)$ is a Jacobi field along the reversed geodesic $\bar{\gamma}$, with initial condition $\bar J(0) = J(t_0) = 0$ and $D_t\bar J(0) = - D_t J(t_0) = 0$. As a consequence of the uniqueness and existence of Jacobi fields, we have $\bar J(t) = 0$, which forces $J \equiv 0$.

Now, fix some parallel frame, say, $e_1(t), \cdots, e_n(t)$ along $\gamma$ near $t_0$, and write $J(t) = \sum_{i=1}^n J^i(t) e_i(t)$. It follows that $D_tJ(t) = \sum_{i=1}^n \dot J^i(t) e_i(t)$, as $D_t e_i = 0$. Then, for some $1 \le i_0 \le n$, we have $J^{i_0}(t_0) = 0$ but $\dot J^{i_0}(t_0) \ne 0$. Hence, $J^{i_0}$ is locally injective, and so $J^{i_0} \ne 0$ in a deleted neighborhood of $t_0$. Consequently zeros of $J$ are isolated.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .