# [GIF] Loop (Geodesics in quadrilateral space)

Posted 6 years ago
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| LoopAs I've mentioned before, there's an identification between the Stiefel manifold $V_2(\mathbb{R}^4)$ -- which consists of $4 \times 2$ matrices with orthonormal columns -- and the moduli space of planar quadrilaterals. Of course, geodesics in the Stiefel manifold are easy to describe: given a starting matrix and an ending matrix, just rotate the first column of the starting matrix towards the first column of the ending matrix, while simultaneously rotating the second column of the starting matrix towards the second column of the ending matrix. If the plane spanned by the columns of the starting matrix is perpendicular to the plane spanned by the columns of the ending matrix, then this geodesic will be closed (periodic).Anyway, the above GIF shows 9 points from the closed geodesic connecting the standard square to a rotated square, with the angle of rotation being a function of time. (Since 9 is odd, you can't actually see the target square; obviously this is not very useful for understanding the geodesic, but in my opinion is more aesthetically pleasing. Change n to 2 if you just want to see the starting and ending squares.)Here's the source code: ToReal[z_] := {Re[z], Im[z]}; ToComplex[{x_, y_}] := x + I y; FrameToEdges[frame_] := ToReal[ToComplex[#]^2] & /@ Transpose[frame]; FrameToVertices[frame_] := Accumulate[FrameToEdges[frame]]; DualFourFrame[frame_] := Orthogonalize[NullSpace[frame]]; Manipulate[Module[{fr, dfr, dfrrot, cols, ?}, fr = {{1/Sqrt, 1/2, 0, 1/2}, {0, 1/2, 1/Sqrt, -1/2}}; dfr = {{0, -1/2, 1/Sqrt, 1/2}, {-1/Sqrt, 1/2, 0, 1/2}}; cols = RGBColor /@ {"#E5FCC2", "#547980"}; dfrrot = dfr.Transpose[RotationMatrix[?, dfr]]; ? = ?/ 2 + ?/2 (1/2 - 1/2 Cos[? (s - Floor[s])] + Floor[s]); n = 9; Graphics[{FaceForm[None], EdgeForm[Directive[JoinForm["Round"], cols[], Thickness[.01]]], Table[Module[{verts}, verts = FrameToVertices[Cos[t] fr + Sin[t] dfrrot]; Polygon[-Mean[verts] + # & /@ verts]], {t, 0, ? - ?/n, ?/n}]}, ImageSize -> {540, 540}, PlotRange -> 1/Sqrt, Background -> cols[]]], {s, 0, 2}] Answer
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Posted 6 years ago
 Nice visualization.It reminds me that I was recently thinking about the moduli space of quadrilaterals, but from the point of view of edge lengths, where for each edge length sequence the moduli parameter is the diagonal. This is a considerably simpler, more Euclidean moduli space, but not as col as this one. The reason I was thinking about this was for a mentorship where a student is working on Thurston's algorithm, and I was thinking there must be a simpler way to go from edge lengths to a planar embedding.The other thing it reminds me of is the general issue of visualizing higher dimensional stuff. On the one hand there is DimensionReduce. There are animations like this one, with well chosen paths like geodesics. In complex systems types of things, we want to know how simple or complex something is, and the extra dimensions add to that challenge.Some comments on the code. It is better form (this is just my opinion) to have only one Module, and when you are doing dynamics, it can't hurt to use DynamicModule instead. Code which is static can either go in front or be part of the Initialization option of Manipulate. DynamicModule[{fr, dfr, dfrrot, cols, \[Theta], verts}, fr = {{1/Sqrt, 1/2, 0, 1/2}, {0, 1/2, 1/Sqrt, -1/2}}; dfr = {{0, -1/2, 1/Sqrt, 1/2}, {-1/Sqrt, 1/2, 0, 1/2}}; cols = RGBColor /@ {"#E5FCC2", "#547980"}; dfrrot = dfr.Transpose[RotationMatrix[\[Theta], dfr]]; Manipulate[...]] Answer
Posted 6 years ago
 It reminds me that I was recently thinking about the moduli space of quadrilaterals, but from the point of view of edge lengths, where for each edge length sequence the moduli parameter is the diagonal. This is a considerably simpler, more Euclidean moduli space, but not as col as this one. Yes, fixed edgelength spaces are a somewhat different beast, though in the case of polygons in 3-space they arise as a symplectic reduction of the Grassmannian spaces I'm using (see Hausmann and Knutson's paper). Somewhat surprisingly, planar polygons are actually harder to deal with than polygons in space, at least if you want to really nail the probability measure (or the volume form). I don't think what you said about quadrilaterals is quite right: if the edgelengths are fixed and you choose a diagonal length, there are usually more than one planar quadrilateral with those edgelengths and that diagonal length.I am curious though: what do you mean by Thurston's algorithm? I'm aware of some of his work (e.g., this paper) on constructing polyhedra, but not of anything relevant to planar polygons, so I'd be very happy to learn more.As for my code, it's certainly kind of gross and I have to admit that I'd never even heard of DynamicModule before. I clearly need to play around with it and learn a bit more about how it works, because just copying your suggested code (and making a couple of necessary modifications) resulted in huge CPU usage. Answer
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