Is every harmonic polynomial a linear combination of these?

Question

In $N$-dimensional space, we can show by direct calculation that the polynomial $$ r^{2K+N-2}\nabla_a\nabla_b\nabla_c\cdots \frac{1}{r^{N-2}} \hspace{1cm} \text{(with $K$ derivatives)} $$ is harmonic (annihilated by the Laplacian $\nabla^2$), where $\nabla_n$ is the partial derivative with respect to the $n$th coordinate and $r$ is the distance from the origin.

Is every homogeneous harmonic polynomial of degree $K$ a linear combination of these? If so, how do we prove this? If not, what is a counterexample?

Motive: I'm studying physics, and this seems like a much nicer way to approach the theory of spherical harmonics (just divide this polynomial by $r^K$ to get a spherical harmonic) compared to the typical physics-textbook approach using spherical coordinates, but it's not obvious to me that all spherical harmonics are linear combinations of these for an arbitrary number of dimensions $N$.

Answer 1

I notice that $$ r^{2K+N−2} P(\partial) \frac{1}{r^{N−2}} = K(P(\partial) \frac{1}{r^{N−2}}) $$ Where $K(f)$ is Kelvin transform of the unit sphere and $P(\partial)$ is a term of derivatives. By Theorem 5.25 in Harmonic Function Theory. Your statement is true.

Prove of Shubin’s Conjecture Using Corollary 5.20 in the same book. For dimension N>2, if $p$ is a homogeneous harmonic function of degree $m$. Then we have $$ P= K(P(\partial) \frac{1}{r^{N−2}})/c_m=r^{2m+N−2} \frac{1}{c_m} P(\partial) \frac{1}{r^{N−2}}. $$ Where $c_m$ is a constant define as $\prod_{k=1}^{m} (4-N-2k)$. The uniqueness is follow by theorem 5.18.

Answer 2

This is a theorem of Maxwell and Sylvester, see for example https://arxiv.org/abs/math-ph/0408046 , https://arxiv.org/abs/0805.1904 and https://arxiv.org/abs/astro-ph/0412231 but, it is in 3d only.

In higher dimension, this is the object of a conjecture of Shubin, see page 10 of https://arxiv.org/abs/0704.1174

Great question BTW, related to some interesting and beautiful mathematics.

Answer 3

I think there is an error in my calculations in my comments. The following is not a complete answer but it gives an hint on how to proceed.

Define $P(N,K)$ to be the space of homogeneous polynomials of degree $K$ in $N$ variables. It has dimension $\binom{N+K-1}{N-1}=\binom{N+K-1}{K}$ the way of distributing the degree $K$ to the $N$ variables. The space $H(N,K) \subset P(N,K)$ is the subspace of harmonic polynomials. One can show that $\dim H(N, K) = \dim P(N, K) - \dim P(N, K-2) $, because the direct sum of the small spaces is isomorphic to the big space.

Note that the expression you wrote is a map from homogeneous polynomials of degree $K$ to harmonic polynomials of degree $K$ by linear extension. For example, in the case $N=5, K=2$ we have

$$ xy +yz \mapsto r^7\nabla_x \nabla_y \frac{1}{r^3} + r^7 \nabla_y \nabla_z \frac{1}{r^3} $$

Let us call this map $\rho: P(N, K) \to H(N, K) $. Now the relationship you wrote in the comments $\sum_a \nabla_a \nabla_a \frac{1}{r^{\bullet}} = 0$ rewrites as $ \rho(r^2) =0$, where $ r^2 = x_1^2 +\ldots + x_N^2$ is the "norm" polynomial.

We consider the map $r^2: P(N, K-2) \to P(N, K) $ that multiplies a polynomial by $r^2$. By a similar argument, we get that the composition of the two maps

$$ P(N, K-2) \underset{r^2}{\to} P(N, K) \underset{\rho}{\to} H(N, K) $$

is zero. Also, the dimension of the first and the third space sum to the dimension of the middle space. This means that the second map is surjective if and only if the kernel is the image of the first map. Here I can't go on. In other words, we have to show that the only relations that can occur are the ones that we mentioned: contracting two indices and permuting derivatives. I think it is remarkable to mention that $P(N, K) $ writes as the direct sum of $H(N, K) $ and the image of the first map, $r^2 P(N, K-2) $; the thesis is thus equivalent to the fact that every harmonic polynomial $f$ is uniquely expressible as $\rho(f_v) $ for some harmonic polynomial $f_v$. Is this $f$ itself, or a scalar multiple? Which harmonic polynomial we could expect to be canonically associated to any $f$?