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$\newcommand{\ga}{\gamma}$ $\newcommand{\al}{\alpha}$

I would like to find an example for a Riemannian manifold, that has two conjugate points $p,q$ with only one connecting geodesic between them.

(This is the geodesic they are conjugate along)

Explanation:

Consider a parametrized family of geodesics starting from a fixed point $p$, i.e:

$\ga_s(t)=\ga(t,s), \ga_s(0)=\ga_0(0)=p$ where for each fixed $s$ , the path $t \to \ga_s(t)$ is a geodesic in $M$.

Then $J(t)= \frac{\partial \ga}{\partial s}(t,0)$ is a Jacobi field, along the geodesic $\ga_0$.

Moreover, every Jacobi field can be realized from such a variation of geodesics.

By definition, if $p,q$ are conjugate along some geodesic $\al$, there exsits a nonzero Jacobi field along $\ga$ that vanishes at $p,q$. This means there is some variation $\ga(t,s)$ of $\al$ ($\ga_0=\al$) where $J(t)= \frac{\partial \ga}{\partial s}(t,0)$.

Assume $\al(t_0)=q$. Then $0=J(t_0)= \frac{\partial \ga}{\partial s}(t_0,0)$, so one can say that "$\gamma_s(1)$, is the point $q$ only up to first order in $s$", but we cannot conclude there exists an $s \neq 0$ such that $\ga_s(t_0)=q$.

(Of course, if we knew that $\ga_s(t_0)=q$ for all $s \in (\epsilon,\epsilon)$ this would imply $J(t_0)=0$ but not vice-versa).

In the language of wikipdeia:

"Therefore, if two points are conjugate, it is not necessary that there exist two distinct geodesics joining them"

Asaf Shachar
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    I don't see how that is possible. Conjugate points arise from a variation of geodesics. – Aloizio Macedo Jan 10 '16 at 20:40
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    Your comment is relevant, but my question makes sense. Please read my modified question. – Asaf Shachar Jan 10 '16 at 21:59
  • Maybe I am confused but doesn't it follow from Hopf-Rinow that this is only possible on a manifold which is not complete? For a non-complete example, consider the sphere $S^2$ with first fundamental form. Now remove a single point from the equator. The resulting manifold has an example of such points. – russoo Sep 20 '24 at 11:57

1 Answers1

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An example in here should do it: http://arxiv.org/pdf/math/0211091.pdf. Look on the first three pages or so. Basically a paraboloid is an example. Pick $p$ and travel along the meridian. If you track the minimizing geodesics joining $p$ to the point you're meeting along your travels, you'll see at first there's only one and then at some point that single minimal geodesic bifurcates into two. The bifurcation point is what you're looking for. Maybe it's easier to imagine the cone $z^2 = x^2 + y^2$ as a singular example of this bifurcation phenomenon. There the bifurcation point is easily identified as the vertex.

The whole paper I linked to is devoted to an in-depth analysis of when this phenomenon occurs.

Nick Strehlke
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    Thank you, this answer was extremely helpful! (and the original question too) For some reason, Petersen's book Riemannian geometry doesn't provide an explicit example (at least I haven't found one). – Michał Miśkiewicz Jan 27 '21 at 13:32
  • In the paper you linked, the author claimed: "Using the Implicit Function Theorem, it follows immediately from the above Definition that if $\gamma(t_0)$ is a bifurcation point for $\gamma$, then necessarily $\gamma(t_0)$ must be conjugate to $\gamma(a)$ along $\gamma$." How to use the implicit function theorem to get the claim? – HeroZhang001 Oct 29 '24 at 13:28