Continuously Differentiable Functions of Dodecahedron Symmetry

Question

Which continuously differentiable functions $F(x,y,z)$ with closed form satisfy Dodecahedron Symmetry?

Dodecahedron Symmetry

1. xoy plane, $F(x,y,z) = F(x,y,-z)$
2. origin, $F(x,y,z) = F(-x,-y,-z)$
3. rotational, $F(x,y,z) = F(y,z,x)$
4. plane symmetry, $(φ+1)x + φy - z = 0$

where $φ=\frac{1+\sqrt{5}}{2}$ is golden ratio.

With this four symmetries, the function $F(x,y,z)$ in domain $\mathbb{R}^3$ can be mapped from $\{ F(x,y,z): x ≥ 0,y ≥ 0, z ≥ (φ+1)x + φy \}$

Recently, I noticed that three function families that satisfy Dodecahedron Symmetry (and of course their linear combinations).

i) constant function $F(x,y,z) = c$

ii) sphere $F(x,y,z) = x^2+y^2+z^2$

iii) 'golden ratio surface' (not sure whether we already have a name for it)

$F(x,y,z) = cos(x)cos(φy)+cos(y)cos(φz)+cos(z)cos(φx)$

The question is, besides these three families, is there any other continuously differentiable functions in $\mathbb{R}^3$ domain satisfying Dodecahedron Symmetry.

I have tried to replace $cos(x)$ to some other functions, while they will break the continuously differentiable property.

Pictures for golden ratio surface $F(x,y,z) = 0$.

Answer 1

The solution is rather surprising. The following function is more representative for this symmetric function family.

$F(x, y, z)=f(x+\phi y) + f(x-\phi y) + f(y+\phi z) + f(z-\phi x) + f(z+\phi x) + f(z-\phi x)$

with function $f$ being an even periodic function, e.g., $f=cos$. The magic does not come from function cos, but from the six $\phi$-planes, $x+\phi y=0, x-\phi y=0,...$, etc (refer to the vertices of Regular dodecahedron). This is because any such function $f$ can be approximated with linear combination of cos functions (Fourier Decomposition).

I am curious about the generalization of this symmetry function to higher dimensions.

Answer 2

There are many other functions with the desired symmetry. The general recipe (which is not specific to dodecahedral symmetries) is to take an arbitrary continuously differentiable function $f:\Bbb R^3\to\Bbb R$ and "force it to be symmetrical":

$$F(\vec r):=\sum_{g\in G} f(g\vec r),$$

where $G$ is the group of dodecahedral symmetries.