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Let us start with metric on $\mathbb R^4=\mathbb R^2\times \mathbb R^2$. Defined by norm $\|{*}\|$ defined by $$\|(x,y)\|=\tfrac12(|x|+|y|),$$ $\|(x,y)\|=\int_0^1|t\cdot x+(1-t)\cdot y|\,dt,$$ where $|{ * }|$ denotes Euclidean norm on $\mathbb R^2$. This norm, is not strictly strongly convexand generically , so you have a lot of should expect many geodesics connectiong two between close points.

Now, your metric is the intrisic metric induced by on the hypersurface $\Sigma$ described by $|x-y|=1$. It often happens that one of the geodesic in $(\mathbb R^4,\|{*}\|)$ lies in $\Sigma$. For example the geodesic between segments $[(0,0),(0,1)]$ and $[(1,0),(0,0)]$.

Maybe this is a closed geodesic, but at least it is a closed line with 4 geodesic segments.

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Maybe it already makes you happy (?). If notSo, you have a Finsler metric on $\Sigma=\mathbb S^1\times \mathbb R^2$. The unit ball for some coordiantes $(u,v,w)$ of in the tangent plane can be described by is isometric the following inequality: $$\sqrt{u^2+w^2}+\sqrt{v^2+w^2}\le 2.$$ This intersection of the ball in the above norm with 3-dimensional subspace in one special direction. It seems that this intersection is strongly convex. The metric is smooth (sinse there is a transitive isometric group action on $\mathbb S^1\times \mathbb R^2$). It reamins to write differential equasion for geodesic; this should be in any book on Finsler geometry. (It should be pain, but it might help.)

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Let us start with metric on $\mathbb R^4=\mathbb R^2\times \mathbb R^2$. Defined by norm $\|{*}\|$ defined by $$\|(x,y)\|=\tfrac12(|x|+|y|),$$ where $|{ * }|$ denotes Euclidean norm. This norm, is not strictly convex and generically you have a lot of geodesics connectiong two points.

Now, your metric is the intrisic metric induced by on the hypersurface $\Sigma$ described by $|x-y|=1$. It often happens that one of the geodesic in $(\mathbb R^4,\|{*}\|)$ lies in $\Sigma$. For example the geodesic between segments $[(0,0),(0,1)]$ and $[(1,0),(0,0)]$.

Maybe this is a closed geodesic, but at least it is a closed line with 4 geodesic segments.

alt text

Maybe it already makes you happy already. (?). If not, you have a Finsler metric on $\Sigma=\mathbb S^1\times \mathbb R^2$. The unit ball for some coordiantes $(u,v,w)$ of the tangent plane can be described by the following inequality: $$\sqrt{u^2+w^2}+\sqrt{v^2+w^2}\le 2.$$ This ball is strongly convex. The metric is smooth (sinse there is a transitive isometric group action on $\mathbb S^1\times \mathbb R^2$). It reamins to write differential equasion for geodesic; this should be in any book on Finsler geometry.

show/hide this revision's text 3 added 254 characters in body

Let us start with metric on $\mathbb R^4=\mathbb R^2\times \mathbb R^2$. Defined by norm $\|{*}\|$ defined by $$\|(x,y)\|=\tfrac12(|x|+|y|),$$ where $|{ * }|$ denotes Euclidean norm. This norm, is not strictly convex and generically you have a lot of geodesics connectiong two points.

Now, your metric is the intrisic metric induced by on the hypersurface $\Sigma$ described by $|x-y|=1$. So It often happens that one of the geodesic in $(\mathbb R^4,\|{*}\|)$ lies in $\Sigma$. For example the geodesic between segments $[(0,0),(0,1)]$ and $[(1,0),(0,0)]$.

Maybe it is makes you happy already. If not, you have a Finsler metric on $\mathbb \Sigma=\mathbb S^1\times \mathbb R^2$. The unit ball for some coordiantes $(u,v,w)$ of the tangent plane can be described by the following inequality: $$\sqrt{u^2+w^2}+\sqrt{v^2+w^2}\le 2.$$ This ball is strongly convex. The metric is smooth (sinse there is a transitive isometric group action on $\mathbb S^1\times \mathbb R^2$). It reamins to write differential equasion for geodesic; this should be in any book on Finsler geometry.

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