Broken geodesic in Finsler polyhedral space

Question

Here we assume that all norms has only one geodesic, i.e. locally minimizing, between any two points.

Example : In $\mathbb{R}^2$, a line $y=kx,\ k>0$ divides $\mathbb{R}^2$ into two regions. We define norms $\|\ \|_U,\ \|\ \|_L$ on upper, lower regions, respectively, where $\| (1,k)\|_U=\|(1,k)\|_L$. In further, if $S_U,\ S_L$ are unit sphere wrt these norms, then assume that these spheres are invariant under the map $T(x,y)=(-x,y)$. Then in the glued space, the line $x=0$ is a geodesic.

Question (Observation 3 in reference) : Define Euclidean norm on $\{(x,y)|y\geq 1\}$ and $\{ (x,y)| y\leq -1\}$. And we define a norm $\| \ \|_M$ on $\{ (x,y)| -1\leq y\leq 1\}$ s.t. $S_M$ is $T$-invariant.

Then geodesic between $p=(0,2)$ and $ q=(0,-2)$ in the glued space can be a broken line ?

Since $p, \ q$ are in a vertical line, then $\{0\}\times [-2,2]$ is a unique geodesic. Am I wrong ?

[Add] Euclidean polyhedral space with locally unique geodesic has a globalization. Here question is related to that of Finsler olyhedral space with locally unique geodesic.

[Add] Define a norm on $\{(x,y)|y\geq 1\}$ which is $T$-invariant, strict, smooth and close to $\|\ \|_\infty$.

And define a norm on $\{ (x,y)|y\leq 1\}$ which is $T$-invariant, strict and close to $\|\ \|_1$. If its unit sphere $S$ passes $(0,1)$ and $(x,y)\in S$ implies $y\leq 1$, then assume that $S$ is not smooth at $(0,1),\ (0,-1)$ only.

Then in glued space, geodesic segment between $(0,0)$ and $ (0,1)$ has at least two extensions that are geodesics.

Reference : Polyhedral Finsler spaces with locally unique geodesics - Burago and Ivanov https://arxiv.org/abs/1210.5286

Answer 1

I don't follow quite the argument in the reference but maybe the following helps:

On the one hand assuming strict convexity of the inner metric there can be at most one geodesic (see (1) below). On the other hand, if you assume $T$-invariance then whenever $\gamma$ is geodesic connecting $p$ and $q$ then also $T(\gamma)$ is a geodesic connecting $p$ and $q$ as both points are fixed by $T$. Combined this means that strict convexity together with $T$-invariance implies "no broken geodesics" as only the $y$-axis is fixed by $T$. Note that it is easy to see that without strict convexity in the $y$-direction there will be broken geodesics conecting $p$ and $q$.

(1) By stricty convexity of the norms (one may replace the Euclidean norms by some other norms) the following function is strictly convex in as a function on $\mathbb{R}^2$ $$ F:(x,y) \mapsto d_U((0,2),(x,1)) + d_M((x,1),(y,-1))+d_L((y,-1),(0,-2)). $$

To see convexity note that the line $x\mapsto (x,1)$ is a geodesic w.r.t. to both the upper metric and the middle one. Similarly for $x\mapsto (y,-1)$. Now strict convexity of the norms implies that all three metrics are Busemann convex, i.e. each term in the sum is separately convex. Finally, strict convexity of $F$ follows from the fact that the distance $d(\cdot,r)$ of a strictly convex norm is not strictly convex only along a geodesics containing the point $r$.

To conclude the claim (1) just observe that the minimum of $F$ equals the distance of the points $p$ and $q$. By strict convexity (of $F$) this means there is at most one such value.

Answer 2

I am one of the authors of the reference in question. Perhaps there is a confusion between "geodesics" and "minimizing geodesics". A minimizing geodesic between $p$ and $q$ is unique, as Martin Kell explained. However, a geodesic (which is just locally minimizing) may be non-unique.

For example, define $\|\cdot\|_M$ on $\{-1\le y\le 1\}$ by $$ \|v\|_M=(1-\varepsilon)\|v\|_1+ \varepsilon\|v\|_2 , \quad v\in\mathbb R^2, $$ where $\varepsilon=\frac 1{1000}$. The metrics on $\{y\ge 1\}$ and $\{y\le-1\}$ are standard Euclidean.

Consider the (unique) shortest path from the point $r=(\frac1{10},0)$ to $p=(0,2)$. One easily sees that it is a broken line composed of a vertical segment from $r$ to $(\frac 1{10},1)$ and a segment from $(\frac 1{10},1)$ to $p$. Similarly, the shortest path from $r$ to $q=(0,-2)$ starts as a vertical segment from $r$ down to the point $(\frac 1{10},-1)$. These two shortest paths join smoothly at $r$ and thus form a (non-minimizing) geodesic between $p$ and $q$.