# [GIF] Fall Out (Rotating circle on the projective plane)

Posted 2 years ago
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| Fall OutThis was a result of various experiments mapping from the sphere to the plane. In this case, I'm taking a circle of disks on the sphere, then mapping to the plane by the following map: each point on the sphere (except those on the equator) is sent to the point on the $z=1$ plane lying on the same line through the origin (this map arises as a way of identifying [most of] the projective plane with an actual plane). Then, apply the inversion in the unit circle $z \mapsto \frac{z}{|z|^2}$. The circle on the sphere is the orbit of the point $p = (0,1/2,\sqrt{3}/2)$ under rotations around $(\cos s, 0 \sin s)$. Here $s$ is treated as the time parameter and varies from $0$ to $\pi$. Here's the code: inversion[p_] := p/Norm[p]^2; With[{n = 141, d = .01, p = {0, 1/2, Sqrt/2}, b = NullSpace[{N[{0, 1/2, Sqrt/2}]}], cols = {Black, GrayLevel[.95]}}, Manipulate[ Graphics[ {PointSize[.01], cols[], Polygon /@ Table[inversion[#1[[1 ;; 2]]/#1[]] &[RotationMatrix[t, {Cos[s], 0, Sin[s]}].(Cos[d] p + Sin[d] (Cos[θ] b[] + Sin[θ] b[]))], {t, 0., 2 π, 2 π/n}, {θ, 0., 2 π - 2 π/20, 2 π/20}]}, PlotRange -> 4, ImageSize -> 540, Background -> cols[[-1]]], {s, 0., π}]] Answer - Congratulations! This post is now a Staff Pick as distinguished by a badge on your profile! Thank you, keep it coming! Answer