# Analogue of roots of unity in n-sphere

The $$n$$-th roots of unity $$z_1,…,z_n$$ in $$\mathbb{C}\equiv \mathbb{R}^2$$ for $$n$$ prime have an interesting property: for $$0\leq p and $$u$$ a unit vector, the sum $$\sum_{k=1}^n \langle z_k,u\rangle ^p$$ does not depend on $$u$$. This can be shown by linearising powers of cosine that appear and use the property that for $$0, $$\sum_{k=1}^n z_k^p=0$$.

I would like to find an analogue set in $$\mathbb{R}^d$$ (or maybe it makes more sense in $$\mathbb{R}^{2d}\equiv \mathbb{C}^d$$) for some $$d\geq 2$$. It seems better to look for a finite set of rotation matrices $$R_1,...,R_m$$ on $$\mathbb{R}^d$$ such that $$\sum_{k=1}^m \langle R_k u,v\rangle^p$$ do not depend on the unit vectors $$u,v$$, for $$p, where $$n$$ can be arbitrarily large (any prime number?).

Any idea?

• Matrices corresponding to vertices of Platonic solids, maybe? Commented Jul 19 at 20:17
• Unfortunately thare are no such solids with arbitrarily many vertices Commented Jul 20 at 4:52
• In the second paragraph $m=n$, isn't it? Commented Jul 23 at 11:24
• Not necessarily, but it will depend on n Commented Jul 23 at 12:21

Most likely we will not have the luxury of getting this invariance for arbitrary $$n$$ when the dimensionality of the space is greater than two. In two dimensions the invariance stated in the problem is achieved through the geometric possibility of forming a regular polytope with any number of vertices $$\ge3$$. In three or more dimensions we are less lucky; only certain vertex counts give regular polytopes.

With this in mind consider the case of four points arranged at the vertices of a regular tetrahedron in three-dimensional space. Since the scalar product is invariant under rotation we may take the vertices as $$(\pm\sqrt{1/3},\pm\sqrt{1/3},\pm\sqrt{1/3})$$ where we select an even number of signs negative. We define our unit vector as $$(x,y,z)$$ with $$x^2+y^2+z^2=1$$ and seek to prove that the sum of powered scalar products is invariant for the prime exponents $$2$$ and $$3$$. We actually get

$$[(+\sqrt{1/3},+\sqrt{1/3},+\sqrt{1/3})\cdot(x,y,z)]^2+...+[(+\sqrt{1/3},-\sqrt{1/3},-\sqrt{1/3})\cdot(x,y,z)]^2=(4/3)(x^2+y^2+z^2)=4/3$$

but

$$[(+\sqrt{1/3},+\sqrt{1/3},+\sqrt{1/3})\cdot(x,y,z)]^3+...+[(+\sqrt{1/3},-\sqrt{1/3},-\sqrt{1/3})\cdot(x,y,z)]^3=(8/\sqrt3)(xyz)=\text{ not invariant}$$

The failure of the tetrahedral arrangement to achieve an invariant sum with exponents of $$3$$ may be tied to the theory of spherical harmonics. When we form the cubic sum, it will be some superposition of spherical harmonics of odd degree less than or equal to three. It must also conform with tetrahedral symmetry. Well, there is a degree three spherical harmonic with tetrahedral symmetry, which is $$xyz$$ — the same as the function we found in the summation above. There is, however, no tetrahedrally symmetry spherical harmonic of degree two, which accounts for successfully getting invariance with an exponent of two.

We can use this theory to assess the remaining four Platonic solids. Unlike the tetrahedron, these all possess a (proper) center of inversion, and only even-degree spherical harmonics conform with that. As we saw with the tetrahedral case, the only even prime ($$2$$) is too low a degree to generate a symmetry-conforming spherical harmonic, so the invariance requested by the problem is achieved for octahedral, cubic, icosahedral and dodecahedral arrangements of "roots" on a unit sphere.

But spherical harmonics with even composite degree may conform with the symmetry of the Platonic solids, so we cannot extend the invariance generally to composite exponents unless we stick with odd exponents. However, the icosahedral and dodecahedral arrangements do have a slight superiority over the other Platonic solids: there is no fourth-degree spherical harmonic conforming with fivefold rotational symmetry elements of these figures. Thus in addition to prime exponents and odd composite ones, the icosahedral and dodecahedral arragements give invariance with an exponent of four (but not six or in general higher even exponents).

• Well thanks! But I do not get everything. So you are saying that some Platonic solids can work until n=4 but not further right? Commented Jul 25 at 8:38
• Yes, for the icosahedron and dodecahedron up to $p=4$ and then all odd exponents after that. I directly tested the icosahedron for exponents of $4$ and $6$ and found as expected that the former works. But the latter does not.. Commented Jul 25 at 9:57
• Anyway, very nice findings! I imagine if it does not work with platonic solids, then nothing else would work... Commented Jul 25 at 11:58
• I also examined some 4-dimensional polytopes. The 5-cell works only with 2 among even exponents. The 24-cell is limited to 4 among even exponents. I do not expect the 120- or 600-cell to do better than the 24-cell, so four dimensional space appears no hetter than three. Commented Jul 25 at 21:47