8

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 $N$.

  • I think your observation could be a good starting point for a proof. You know that spherical harmonics groups into irreducible finite dimensional representations under the action of $SO(3)$ (fix the positive index). Now take one of this fdim irrep $V$: "how many" harmonic polynomials of your form can you find in $V$? My question splits in 2 parts: understand how $SO(3) $ acts on your polynomials and how they group together; count dimension "representation-wise" and see if you get everything – Andrea Marino Dec 28 '20 at 00:30
  • @AndreaMarino Regarding your first comment: For arbitrary $N$, I'm not sure how to determine how many of those polynomials are linearly independent. Maybe it's easy, but I'm not seeing it right now. That seems like a prerequisite for using the representation theory of $SO(N)$ to prove that they form an (over)complete basis. – Chiral Anomaly Dec 28 '20 at 00:52
  • I thought you were in $N=3$ case, otherwise I think representation theory of $SO(N)$ becomes rather complicated, despite still tractable. Probably it is not the right approach. What about counting the dimension degreewise? The number of different ways you can choose the derivatives, recalling $\nabla_x \nabla_y = \nabla_y \nabla_x $, is $\binom{N+K-1}{K-1}$: this is the number of ways of distributing the $K$ derivatives among the $N$ components. – Andrea Marino Dec 28 '20 at 13:05
  • On the other hand, the harmonic homogenous polynomials of degree K are known to have degree $\binom{N+K-1}{K}-\binom{N+K-3}{K-2}$. This is weird: if my combinatorics above is right, then the dimension of harmonic polynomials grows as $N^K$ while the number of polynomials you have found grows as $N^{K-1}$, so we should be missing something... – Andrea Marino Dec 28 '20 at 13:15
  • 2
    @AndreaMarino In the polynomial that I wrote, contracting two of the indices (replacing $\nabla_a\nabla_b$ with $\nabla^2\equiv\sum_a\nabla_a\nabla_a$) gives zero, because $\nabla^2 r^{-(N-2)}=0$ for $r\neq 0$. So they're not all linearly independent, not even after accounting for index-permutation symmetries. I don't know what the number of such relations is, but in any case, this seems to make the discrepancy you noted even worse. Maybe that just means that my expression doesn't capture all of them, but I'm not sure. – Chiral Anomaly Dec 28 '20 at 13:59
  • 1
    We could close with finding a direct counterexample, but I would be personally more satisfied with understanding why we are not taking all of them. I think the reason is that these polynomials has a particular symmetry (aka lies in a subrep). Can we spot it? – Andrea Marino Dec 28 '20 at 14:05
  • You are probably talking about this approach to spherical harmonics. The explanation seems very good to me, however, I would just like to add a very non-rigorous comment that from the point of view of representation theory, this is a very natural way to think of spherical harmonics, since they basically are representations of SO(N), which are more or less symmetric functions. – user108687 Jan 21 '21 at 00:24

3 Answers3

5

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.

  • Thank you again for your response! I accepted a new answer that cites a reference which proves that every homogeneous harmonic polynomial can indeed be written as shown in the question. (Shubin's conjecture seems to makes a slightly stronger statement, which may still be unproven.) – Chiral Anomaly Nov 12 '21 at 02:32
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 $f$?

3

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.

TW Hu
  • 46
  • 2
  • The reference you cited is magnificent! I haven't finished working through the proof yet, but I've studied it enough to confirm that it does indeed answer my question completely (assuming the proof is sound). – Chiral Anomaly Nov 12 '21 at 02:27
  • 2
    I also prove Shubin’s Conjecture as well using theories in the same book. – TW Hu Nov 18 '21 at 20:29
  • Wow. My colleague Paul Bourdon would be happy to hear that. You should write this up and post a paper on arXiv. – Abdelmalek Abdesselam Nov 19 '21 at 10:53