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While studying a unique continuation property of some determinants, I encountered the following problem.

Let $p_{ij}$ be real or complex harmonic polynomials (with homogeneous real and imaginary parts) in $\mathbb{R^n}$ for $i, j = 1, 2$ such that $p_{11} p_{22} = p_{12} p_{21}$. The question is to classify all such polynomials.

For $n = 2$ and the real case, I was able to prove by hand that up to constants $\{p_{11}, p_{22}\} = \{p_{12}, p_{21}\}$. For the complex case I believe that (up to constants and conjugation) we must have $p_{ij} = z^{n_{ij}}$ and $n_{11} + n_{22} = n_{12} + n_{21}$.

For $n > 2$ and the complex case, one class of examples is given by $p_{ij}(x) = \big(\sum_i c_i x_i \big)^{n_{ij}}$ with $n_{11} + n_{22} = n_{12} + n_{21}$ and $\sum_i c_i^2 = 0$. Is this the only possibility (up to trivial things)? Do we have "uniqueness" in the real case, as for $n = 2$?


Some related questions/approaches:

As the restriction of harmonic polys to the sphere $S^{n - 1}$ give spherical harmonics (eigenfunctions of $\Delta_{S^{n-1}}$), algebraic identities for these could be useful. For $n = 3$, these are well known and express $Y_l^m \cdot Y_{l'}^{m'}$ as a linear combination of $Y_{l''}^{m''}$ for $|l - l'| \leq l'' \leq l + l'$ and $m'' = m + m'$ ("selection principles") using Clebsch-Gordan coefficients. It is claimed that this expansion follows from the representation theory of $SO(3)$ since $H_l \otimes H_{l'} \cong \oplus H_{l''}$ in the given range, where $H_k$ is the irrep of $SO(3)$ on spherical harmonics.

  1. I don't see this, i.e. how to relate the rep theory with the algebraic identity for $Y_l^m \cdot Y_{l'}^{m'}$? Any textbook I can look this up?

  2. Are there similar expansions for products of two spherical harmonics for $n>3$? What are the selection principles for these?

  3. In the real case and if $n = 3$, by using the selection principles we expect to get $|n_{11} - n_{22}| = |n_{12} - n_{21}|$ ($n_{ij}$ is the degree of $p_{ij}$), and so $\{n_{11}, n_{22}\} = \{n_{12}, n_{21}\}$, which is a good sign.

Any ideas would be greatly appreciated!

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  • $\begingroup$ Depends on what you mean by "up to trivial things". For instance, $$[(x+iy)^2(z-it)^6+(x+iy)^4(z-it)^4+(x+iy)^6(z-it)^2][(x+iy)^2-(z-it)^2]=[(x+iy)^2(z-it)^2][(x+iy)^6-(z-it)^6]$$. Is it still within the trivial range from your example? $\endgroup$
    – fedja
    Dec 14, 2017 at 2:06
  • $\begingroup$ Sorry for not replying earlier -- nope, that's not in the trivial range as the factors are not of the form $(\sum_i c_i x_i)^k$ as above! Thanks! That's right, powers of holomorphic functions are holomorphic and products of coordinate independent functions are also harmonic, so the complex case is a bit wilder... I figured out the point 1. above on my own (it's hidden in fisica.net/quantica/…), but I still don't know what happens when $p_{ij}$ are real polynomials. Maybe one can use a computer to check it in some small cases? $\endgroup$
    – Ceka
    Jan 11, 2018 at 16:47

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