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The 30 straight edges of an icosahedron (with constant Euclidean vertex to vertex distance, and constant sphere center to vertex distance) have normal curvatures $\kappa_n=0$ in radial planes. They span and tessellate 20 equilateral triangles of Gauss curvature $K=0$. We try to find parametrization of the surface in other cases of icosahedral symmetry.

When $K=1$, we have a sphere with all $\kappa_n=1$.

How should the normal curvatures change so that the surface has:

  • $K=2$? I imagine spiky surfaces somewhat like the 'Mathematica' logo and/or bulgy ovaloids.
  • $K=-1$?

Thanks in advance for a solution or other suggestions.

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  • $\begingroup$ The only case of spherical symmetry is the sphere of constant curvature. Do you mean icosahedron symmetry? $\endgroup$
    – Alexandre Eremenko
    Commented Feb 2, 2023 at 13:29

1 Answer 1

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There is an equilateral triangle of constant curvature with three equal angles, and the angle can be arbitrary $\alpha>0$. By Gauss-Bonnet, the constant curvature $K$ has the same sign as $3\alpha-\pi$. Now consider the conformal map of the flat equilateral triangle onto this equilateral one, with any given $K$ and any angle consistent with the above restriction. Reflections according to the icosahedron group will give you the developing map of a surface with conic singularities (at the vertices), icosahedral symmetry, and given curvature. This surface can be embedded to $R^3$ if $\alpha\leq 2\pi/5$.

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  • $\begingroup$ So in particular what is 3d parametrization for a chosen $\alpha$ so that $K=-1?$ $\endgroup$
    – Narasimham
    Commented Feb 3, 2023 at 12:18
  • $\begingroup$ I did not work it out explicitly. $\endgroup$
    – Alexandre Eremenko
    Commented Feb 3, 2023 at 13:33

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