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>



Broken geodesic in Finsler polyhedral space
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.


