The Platonic solids has the most symmetric for discrete rotation in 3-D space. They are symmetry in all the faces, lines, and vertexes. The Spherical harmonics also has discrete rotational symmetry on the z-axis for all order. However, for small order, the spherical harmonics also has other symmetric axis.

For example,

$\displaystyle Y_{32}(\theta, \phi) = \sqrt{\frac{105}{32\pi}} \cos\theta \sin^2\theta \exp(2i\phi)$

The spherical plot and spherical contour plot for $1 + Y_{32}$ are

The plot clearly shows the tetrahedron’s symmetry. In fact, the tetrahedron, hexahedron (cubic), and octahedron share the same symmetry.

The tetrahedron inscribe itself.

The hexahedron and octahedron inscribe each other.

The dodecaherdo and icosahedron also inscribe each other.

We can systematically construct the tetrahedron, hexahedron (octahedron), and dodecahedron.

The tetrahedron has 120 degree rotation symmetry around a vertex. And we can guess, $Y_{33}$ should be a suitable candidate. Thus, we can form the tetrahedron from

$f(\theta, \phi) = Y_{30}(\theta,\phi) + Y_{33}(\theta, \phi)$

However, at angle $\tan(\theta/2) = \sqrt{2}, \phi = \pi/3$, the value of $f$ is less then $\theta = \phi = 0$.  To restore the symmetry,

$\displaystyle f_t(\theta, \phi) = Y_{30}(\theta,\phi) + \sqrt{\frac{8}{5}} Y_{33}(\theta, \phi)$

The plots for $f_t$ are show below.

Similarly, we can from the octahedron using $Y_{4m}$ with a coefficient.

$\displaystyle f_h(\theta, \phi) = Y_{40}(\theta,\phi) + \sqrt{\frac{10}{7}} Y_{44}(\theta, \phi)$

For dodecahedron, the use of $Y_{5,5}$ seem to be the logical choice. However, it turns out it cannot has the correct symmetry because $Y_{50}$ is not “even”.

$\displaystyle f_d(\theta, \phi) = Y_{60}(\theta,\phi) + \sqrt{\frac{28}{11}} Y_{65}(\theta, \phi)$

The Mathematica code for generating the graph is

GraphicsGrid[{{
SphericalPlot3D[1 + f(θ,φ), {θ,  0, π}, {φ, 0, 2 π}],
ParametricPlot3D[{Cos[φ] Sin[θ], Sin[φ] Sin[θ], Cos[θ]}, {θ,  0, π}, {φ, 0, 2 π},
ColorFunction -> Function[{x, y, z, u, v},  Hue[Abs[1 + f(u,v)]]],
ColorFunctionScaling -> False, Mesh -> None, PlotPoints -> 200]
}}, ImageSize -> 600]

The code for calculating the coefficient is

Solve[
SphericalHarmonicY[n, 0, k, 0] + a SphericalHarmonicY[n, n, k, 0]
== SphericalHarmonicY[n, 0, 0, 0]
, a]

Where $n$ is the order of the spherical harmonic, $k$ is the angle for the next face or vertex for fixed $\theta = 0$.

Tetraherdon $k = 2 \tan^{-1}(\sqrt{2})$.

Octaherdon $k = \pi/2$.

Dodecahedron $k = 2 \tan^{-1}( (1+\sqrt{5})/2)$.