# Broken geodesic in Finsler polyhedral space

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

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.

• Construction of $F$ is nice. – Hee Kwon Lee Nov 21 '18 at 11:12

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 $$q$$.

• This is an example s.t. $\{y\geq1\}$ has a Euclidean norm. It is desired by me. Thank you. – Hee Kwon Lee Dec 2 '18 at 15:46