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Let $$P_k(x_1\ldots x_n)=\sum_{\lvert \alpha\rvert=k} c_\alpha x_1^{\alpha_1}\ldots x_n^{\alpha_n}, \qquad (x_1\ldots x_n)\in \mathbb{R}^n$$ be a homogeneous polynomial of degree $k$. Assume that this polynomial is zonal, meaning that its restriction to the unit sphere $\mathbb{S}^{n-1}$ is a function of the last variable $x_n$ only. (If the unit sphere is identified with the Earth, then we are requiring this polynomial to be constant on parallels.)

Is it true that $$P_k(x_1\ldots x_n)=p_k(x_n), \qquad \forall (x_1\ldots x_n)\in\mathbb{S}^{n-1}$$ where $p_k$ is a polynomial of a single variable of degree $k$?

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I guess I got it. The answer is partly affirmative, I am posting it here for reference.

We can write $P_k$ as a polynomial in $x_n$ having as coefficients polynomials in $x_1\ldots x_{n-1}$: $$ P_k(x_1\ldots x_n)=\sum_{j=0}^k x_n^j a_{k-j}(x_1\ldots x_{n-1}). $$ Because of the zonal symmetry, the coefficients $a_{k-j}$ must depend only on the value of $x_1^2+\dots +x_{n-1}^2$ (the parallel). Since they also need to be polynomials, the only possibility is that $$ a_{k-j}=\begin{cases} c_j(x_1^2+\dots+x_{n-1}^2)^\frac{j}{2}, & j\ \text{even} \\ 0, & j\ \text{odd} \end{cases}$$ And so $$ P_k(x_1\ldots x_n)=c_kx_n^k+ c_{k-2}(x_1^2+\dots+x_{n-1}^2)x_n^{k-2}+c_{k-4}(x_1^2+\dots+x_{n-1}^2)^2x_n^{k-4}\dots$$ Setting $x_1^2+\dots+x_n^2=1$, we obtain the desired polynomial $p_k$: $$p_k(x_n)=c_k x_n^k +c_{k-2}(1-x_n^2)x_n^{k-2} + c_{k-4}(1-x_n^2)^2x_n^{k-4} +\dots$$ WARNING! This polynomial needs not be of degree $k$. It is of lower degree if $c_k$ vanishes. This coefficient can be recovered quickly by evaluating $P_k$ at the North Pole: $$c_k=P_k(0\ldots 0, 1).$$

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  • $\begingroup$ I would say $$ a_{k-j}=\begin{cases} c_j(x_1^2+\dots+x_{n-1}^2)^\frac{k-j}{2}, & k-j\ \text{even} \\ 0, & k-j\ \text{odd} \end{cases}$$ and $p_k$ is of lower degree if $c_k = c_{k-2} + c_{k-4} + \cdots$ $\endgroup$ – inquisitor Dec 1 '17 at 17:03
  • $\begingroup$ @inquisitor: Thank you for your commentary. I am keeping the post as is since I forgot what was the rationale behind my choice of indices, and I don't have the motivation to come back to this old question now. $\endgroup$ – Giuseppe Negro Dec 2 '17 at 18:26

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