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It is not merely simple, it contains an explanation of why it works as a definition. (Un)flattening is an isometry, it doesn't change lengths of lines. If you

It is not merely simple, it contains an explanation of why it works as a definition. (Un)flattening is an isometry, it doesn't change lengths of lines. If you flatten a polyhedron and draw a straight line between two points, it is the shortest path between them (property of Euclidean space). Now fold the surface back, it remains the shortest path (and the geodesic).

deleted 6y ago
[deleted]
Someone 6y ago
The “Now fold the surface back, it remains the shortest path” part isn’t true, in general. A different unfolding of the polyhedron may yield a shorter path. See https://en.wikipedia.org/wiki/The_spider_and_the_fly_problem https://en.wikipedia.org/wiki/The_spider_and_the_fly_problem
- kazinator 6y ago
  Even if we are discussing a pair of points on adjacent polygons? Wouldn't the shortest path go across the one and only edge the polygons have in common? I'm looking at the diagrams on that Wikipedia page. It looks as if we require the path to visit certain polygons, then it is in fact the shorted path which visits those polygons. In both solutions, the path traced along the unfolded jacket of the box is in a straight line, obeying the principle we have been discussing. I don't think that what we are looking at there are simply two different "different unfoldings" of a polyhedron; they are different choices about which configuration of edges to separate in forming what is evidently called the net of the polyhedron: https://en.wikipedia.org/wiki/Net_(polyhedron) https://en.wikipedia.org/wiki/Net_(polyhedron) They are different "nettings", if you will. We can crack a polygon into more than one net, but a given net has just one unfolding. (If we choose a particular net of a polyhedron, can that have more than one shortest path between some two points?)
  - Someone 6y ago
    ”Even if we are discussing a pair of points on adjacent polygons?” Yes. Consider a pointy triangular wedge of two 1 by 100 rectangles and a third one that’s 1 by 1, and pick points on the two large rectangles that are close to the edge they share with the small one. The shortest path will visit all three rectangles. Also: I was only refuting the claim that such a line is the shortest line between the two points. I happened to know an (IMO) interesting counterexample, so I mentioned it. I think the conclusion that such a line is a geodesic is valid, but not because of the argument given.

deleted 6y ago

[deleted]

Someone 6y ago

The “Now fold the surface back, it remains the shortest path” part isn’t true, in general. A different unfolding of the polyhedron may yield a shorter path. See https://en.wikipedia.org/wiki/The_spider_and_the_fly_problem https://en.wikipedia.org/wiki/The_spider_and_the_fly_problem

kazinator 6y ago
Even if we are discussing a pair of points on adjacent polygons? Wouldn't the shortest path go across the one and only edge the polygons have in common? I'm looking at the diagrams on that Wikipedia page. It looks as if we require the path to visit certain polygons, then it is in fact the shorted path which visits those polygons. In both solutions, the path traced along the unfolded jacket of the box is in a straight line, obeying the principle we have been discussing. I don't think that what we are looking at there are simply two different "different unfoldings" of a polyhedron; they are different choices about which configuration of edges to separate in forming what is evidently called the net of the polyhedron: https://en.wikipedia.org/wiki/Net_(polyhedron) https://en.wikipedia.org/wiki/Net_(polyhedron) They are different "nettings", if you will. We can crack a polygon into more than one net, but a given net has just one unfolding. (If we choose a particular net of a polyhedron, can that have more than one shortest path between some two points?)
- Someone 6y ago
  ”Even if we are discussing a pair of points on adjacent polygons?” Yes. Consider a pointy triangular wedge of two 1 by 100 rectangles and a third one that’s 1 by 1, and pick points on the two large rectangles that are close to the edge they share with the small one. The shortest path will visit all three rectangles. Also: I was only refuting the claim that such a line is the shortest line between the two points. I happened to know an (IMO) interesting counterexample, so I mentioned it. I think the conclusion that such a line is a geodesic is valid, but not because of the argument given.