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