Example for infinitely many points with more than one minimizing geodesic to a point?

Question

Question:

Does there exist a Riemannian manifold, with a point $p \in M$, and infinitely many points $q \in M$ such that there is more than one minimizing geodesic from $p$ to $q$?

Edit:

As demonstrated in Jack Lee's answer, one can construct many exmaples in the following way:

Take $X$ to be a manifold which has a pair of points $p,q$, with more than one minimizing geodesic connecting them. Take $Y$ to be any geodesically convex (Riemannian) manifold. Then $X \times Y$ satisfies the requirement:

Indeed, let $\alpha,\beta$ be two different geodesics in $X$ from $p$ to $q$.

Fix $y_0 \in Y$, and let $y \in Y$ be arbitrary. Let $\gamma_y$ be a minimizing geodesic in $Y$ from $y_0$ to $y$. Then $\alpha \times \gamma_Y,\beta \times \gamma_Y$ are minimizing from $(p,y_0)$ to $(q,y)$.

Hence, if $Y$ is positive-dimensional (hence infinite), we are done.

"Open" question: Are there examples which are not products? (This is probably hard, I am not even sure what obstructions exist for a manifold to be a topological product of manifolds)

---

Note that for any $p$, the set $$\{q \in M \,| \, \text{there is more than one minimizing geodesic from $p$ to $q$} \}$$

is of measure zero.

Indeed, let $M$ be a connected Riemannian manifold, and let $p \in M$.

The distance function from $p$, $d_p$ is $1$-Lipschitz, hence (by Rademacher's theorem) differentiable almost everywhere.

It is easy to see that if there are (at least) two different length minimizing geodesics from $p$ to $q$, then $d_p$ is not differentiable at $q$. (We have two "natural candidates" for the gradients).

Answer 1

Take $M$ to be the following cylinder in $\mathbb R^3$: $$ M = \{(x,y,z): x^2 + y^2=1\}. $$ Then let $p$ be the point $(1,0,0)\in M$. If $q$ is any point of the form $(-1,0,z)$, then there are two minimizing geodesics from $p$ to $q$.

Answer 2

I will introduce two examples

(1) Consider a torus in $\mathbb{R}^3$

(2) Consider two dimensional regular triangle $T$ in $\mathbb{R}^2\subset X=\mathbb{R}^3$ If $U$ is a suitable tubular neighborhood of $T$ in $X$, then consider $\partial U$ which is homeomorphic to $S^2$

There are three points $p_i$ in $\partial U$ whose Gaussian curvature attains local maximum. Then cut locus of $p_1$ ${\rm Cut}\ (p_1)$ is a curve $c:[0,1] \rightarrow \partial U$ between $p_2$ and $p_3$ And interior points $c(t),\ 0multiplicity $2$ i.e. there are exactly two minimizing geodesics from $p_1$ to $c(t)$

(3) (As far as I know) Generally, in Riemannian manifold $M$, if ${\rm Cut}\ (p)$ is not point set, then points in ${\rm Cut}\ (p)$ of multiplicity $\geq 2$ are dense

(4) Another highdimensional example is $\mathbb{C}P^2$

Answer 3

Between two points $(p,q)$ on any surface of revolution there are indefinitely many geodesic trajectories possible.

Just as you can throw a stone between two points $(p,q)$ situated at different heights choosing a different parabola at different angle of slope or angle of attack.

In a boundary value problem after specifying $(p,q)$ we use a shoot-through trial and error numerical procedure that satisfies the differential equation action.

That said however, there is a minimum that can be chosen from the set of all possible geodesics that minimize length, time or any chosen object function from among them.

The first minimization respects only the phenomenon, the second chooses minimum of all possibilities from a set satisfying them.