On a Riemannian manifold (along a geodesic), is the relation *is conjugate to* transitive?

Question

Let $(M, g)$ be a complete Riemannian manifold.

Suppose $\gamma : \mathbb{R} \rightarrow M$ is a geodesic such that the instant $0$ is conjugate to both $a$ and $b$, where the numbers $a, b, 0$ are distinct.

Question: Does it follow that $a$ and $b$ are conjugate to each other?

I tried to prove this, but I couldn't. Then I tried finding a counterexample, only to realize that the only interesting example I know of when it comes to conjugate points is the sphere, which is rather special.

EDIT: Here's my definition of two instants being conjugate to each other along a geodesic $\gamma : \mathbb{R} \rightarrow M$.

The instants $t_0, t_1 \in \mathbb{R}$ are conjugate along $\gamma$ if and only if there exists a nontrivial (meaning not identically zero) Jacobi field along $\gamma$ vanishing at $t_0$ and $t_1$.

I'm at a loss. Any suggestions are welcome. Thanks!

Answer 1

No, this relation is not transitive. The simplest example I know is the product $M$ of two spheres $M_1, M_2$ of radii, say, $1$ and $1.1$. Now, take a geodesic $\gamma$ in $M$ which projects to nonconstant geodesics $\gamma_i$ in each factor. Then, lift Jacobi fields from $M_i$'s (along $\gamma_i$) to two Jacobi fields $J_1, J_2$ in $M$ along $\gamma$.

Answer 2

It is clear that the answer is yes if we use the same Jacobi field i.e. the Jacobi field vanishes twice at $a$ and at $b$ so the three points are conjugate pairwise. If you have two different variations of two different geodesics generating two different Jacobi fields $J_1$ and $J_2$ and they vanish at points $a$ and $b$ respectively, you have nothing to say about $a$ and $b$.