2
$\begingroup$

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

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Construction of $F$ is nice. $\endgroup$ – Hee Kwon Lee Nov 21 '18 at 11:12
3
$\begingroup$

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 $q$.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.