Answer by Sergei Ivanov for Geodesics in $\mathbb{R}^2 \times \mathbb{S}^1$...
In addition to Anton Petrunin's answer, here is a trick to simplify (and in some sense solve) the geodesic equation.Since the metric has three-dimensional group of isometries (generated by rigid...
View ArticleAnswer by Jean-Marc Schlenker for Geodesics in $\mathbb{R}^2 \times...
A useful keyword for this problem is the Wasserstein distance, see wikipedia. I believe that this Wasserstein distance, for $p=1$, provides a variant of the distance you're considering but for...
View ArticleAnswer by j.c. for Geodesics in $\mathbb{R}^2 \times \mathbb{S}^1$ under...
What follows are just some illustrations, not a full answer; please refer to Anton Petrunin's answer for a nice description of the 4-dimensional geometry that the original question is embedded in....
View ArticleAnswer by Anton Petrunin for Geodesics in $\mathbb{R}^2 \times \mathbb{S}^1$...
Let us start with the metric on $\mathbb R^4=\mathbb R^2\times \mathbb R^2$defined by the norm $\|{*}\|$ defined by$$\|(x,y)\|=\int_0^1|t\cdot x+(1-t)\cdot y|\,dt,$$ where $|{ * }|$ denotes the...
View ArticleGeodesics in $\mathbb{R}^2 \times \mathbb{S}^1$ under "segment" metric
Represent the position of a unit-length, oriented segment $s$ in the planeby the location $a$ of its basepoint andan orientation $\theta$: $s = (a,\theta)$. So $s$ can beviewed as a point in...
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