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Which sequences of adjacent edges of a polyhedron could be considered to be a geodesic? The edges of a face most surely will not, but the "equator" of the octahedron eventually will. But for what reasons? How do the defining property of a geodesic - having zero geodesic curvature - apply to a sequence of edges?

(One crude guess: any sequence of edges that pairwise don't share a face? What does this have to do with curvature?)

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Perhaps: Nudge the geodesic to one side so it does not lie exactly on the edge. Then it has a unique continuation past the vertex. Now nudge it to the other side, and get a different continuation. Then if you go exactly along the edge, when you hit the vertex, any exiting direction that lies between the two above should qualify as a geodesic. (For sufficiently small angular defect, this won't include all the edges. It might not even include any.) –  Rahul Jan 30 '13 at 22:54
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3 Answers

It so happens I drew the "equator of a dodecahedron" for one of my papers, so I can't resist including it here:
           Dodecahedron equator

Two points I'd like to make. First, a geodesic is a curve that has $\le \pi$ surface to each side at every point. This is Alexandrov's definition, and it is the right way to think of geodesics on polyhedra. He and Pogorelov called these quasigeodesics (Alexandrov and Zalgaller, Intrinsic Geometry of Surfaces, 1967, p.16; Pogorelov, Extrinsic Geometry of Convex Surfaces, 1973, p.28).

Second, if you allow a doubly covered polygon as a polyhedron of zero volume (as Alexandrov did), then indeed the edges of a face could be a geodesic: consider a doubly covered square, for example. Then the edges of one square face have $\pi$ at all interior-edge points to either side, and $\pi/2$ to either side at the four corners.

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Then there are no geodesics through a vertex with negative angular defect (on a nonconvex polyhedron)? –  Rahul Jan 31 '13 at 1:51
    
Yes, that is correct. Another way to think of it: Such a curve could not be unfolded straight. There is always too much surface to one side or the other. –  Joseph O'Rourke Jan 31 '13 at 1:52
    
I had thought about the following "definition" of a geodesic: it has exactly $2\pi$ curvature to each side. Is this somehow related to Alexandrov's definition? (Do you eventually have a reference for the latter?) –  Hans Stricker Jan 31 '13 at 8:19
    
I discuss this on p.373 of my book Geometric Folding Algorithms. I'd add references above. –  Joseph O'Rourke Jan 31 '13 at 11:46
    
@Hans: Your definition works for "pure geodesics." They can never go through a vertex. Alexandrov's quasigeodesics can go through vertices. Theorems about geodesics on smooth surfaces often extend to quasigeodesics on polyhedral surfaces. –  Joseph O'Rourke Jan 31 '13 at 12:41
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Some thoughts are given by Konrad Polthier and Markus Schmies, Straightest geodesics on polyhedral surfaces. Form the abstract:

Geodesic curves are the fundamental concept in geometry to generalize the idea of straight lines to curved surfaces and arbitrary manifolds. On polyhedral surfaces we introduce the notion of discrete geodesic curvature of curves and define straightest geodesics. This allows a unique solution of the initial value problem for geodesics, and therefore a unique movement in a given tangential direction, a property not available in the well-known concept of locally shortest geodesics.

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Thank you very much for the hint. I have had a look into this paper, but when you take a look at the figure on page 8, you will see, that the authors are not so much interested in "curves" restricted to sequences of edges. That had been my impression, so I didn't keep that track. Should I? –  Hans Stricker Jan 31 '13 at 0:26
    
You got $7$ pages farther than I did. I just read the abstract and thought the paper might be relevant, but the final judgement on that question can only be yours. –  Gerry Myerson Jan 31 '13 at 1:27
    
Polthier/Schmies seem to consider geodesics on polyhedral (= non-smooth) surfaces (with an emphasis on "surface") like Joseph (in his answer above) and Alexandrov do. Maybe I should have made clearer that I am specifically interested in geodesics as sequences of edges. (Somehow these two approaches may meet.) –  Hans Stricker Feb 2 '13 at 14:52
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There seems to be an almost trivial definition (which relies on a specific realization of a polyhedron): if the polyhedron is convex and inscribed into a sphere and the central projections of the edges onto the sphere sum up to one great arc or circle, then the edges are geodesic. (Note that each single edge is projected onto a single great arc.)

Due to this definition, the obvious "equators" of the regular octahedron are geodesic, but also the edges of any face of any inscribable polyhedron can be geodesic (in a specific realization). On the other side, some sequences of edges cannot be - in no realization of the polyhedron - geodesic due to this definition, e.g. the zig-zag-sequence around the "equator" of the dodecahedron (?).

It remains open, whether there is a more combinatorial definition of geodesics, say in a polyhedral graph.

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