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What about this: ListPlot3D[Flatten[data, 1], ColorFunction -> "LakeColors"] ?? It gives Have fun!


For any given theta, you have a curve in phi/omega space. Thus, one way to visualize this is to plot in the phi/omega plane over all theta: sol = Solve[{-Cos[φ] Sin[2 θ] Sin[π Ω]^2 + Sin[θ] Sin[φ] Sin[2 π Ω], -2 Sin[θ] Sin[π Ω] (Cos[φ] Cos[π Ω] + Cos[θ] Sin[φ] Sin[π Ω]), -Cos[θ]^2 - Cos[2 π Ω] Sin[θ]^2} == {0, 1, 0}, {θ, φ, Ω}]; ...


Because of that exponential in the definition of τ, the sampling of the function is tricky, and so trying to use ParametricPlot3D with enough PlotPoints, as we would have to, kills the kernel on my machine. So let's take an indirect approach by first plotting the function for δ = 1. For this, we use ParametricPlot: p1 = ParametricPlot[{E^-k BesselI[0, k], ...


It seems you don't need the other two equations, the second one by itself defines the curves: ContourPlot3D[ -2 Sin[θ] Sin[π Ω] (Cos[φ] Cos[π Ω] + Cos[θ] Sin[φ] Sin[π Ω]) == 0.99, {θ, 0, 2 π}, {φ, 0, 2 π}, {Ω, 0, 2}, PlotPoints -> 100, MaxRecursion -> 0] Another way to do it is to use Maxim Rytin's BoundaryStyle trick as described by Daniel ...

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