Power-sum relation for arbitrary homogeneous symmetric polynomials

About 1 year old · traced to 2025

Let ${\bf x}^m=\{x_1,\ldots,x_m\}$ and ${\bf y}^m=\{y_1,\ldots,y_m\}$ . For an arbitrary homogeneous symmetric polynomial $S_n({\bf x}^m)$ of degree $n$ , write

S_n({\bf x}^m)=\sum_{\bf k}C_{n,\bf k}P_{n,\bf k}({\bf x}^m),\qquad P_{n,\bf k}({\bf x}^m)=\prod_{l=1}^n p_l({\bf x}^m)^{k_l},

where $\sum_{i=1}^n i k_i=n$ , and let $\pi({\bf x}^m)=\prod_{j=1}^m x_j$ . Define

V_n({\bf x}^m)=\frac{S_n({\bf x}^m)}{\pi({\bf x}^m)},\qquad U_n({\bf x}^m,{\bf y}^m)=V_n({\bf x}^m)-\sum_{i=1}^m y_i^{m-k-1}V_n({\bf s}_i^m),

where ${\bf s}_i^m$ has entries $s^m_{i,j}=y_i x_j-y_j x_i+x_i\delta_{ij}$ . Power-sum relation. The relation

U_n({\bf x}^m,{\bf y}^m)=0,

holds for $0\leq n\leq m-1$ . Moreover, at ${\bf y}^m={\bf 1}^m$ ,

U_n({\bf x}^m,{\bf 1}^m)=Y_{n-m}({\bf x}^m),\qquad n\geq m,

where $Y_{n-m}$ is homogeneous symmetric of degree $n-m$ and

Y_{n-m}({\bf x}^m)=\sum_{\bf k}C_{n,\bf k}Y_{n-m,\bf k}({\bf x}^m).

This packages the proposed relations for arbitrary power-sum expansions; the source gives no resolution status.

References

Primary source

Boris Y. Rubinstein, “A New Class of Relations for Homogeneous Symmetric Polynomials”, arXiv:2510.25749 (2025).

power-sums symmetric-polynomials

Posted 1 month ago

Progress summary

Refreshed 1 month ago

Claimed solved

A 2025 paper claims the relation works for every homogeneous symmetric polynomial, and a later posted argument claims a complete proof, but neither has been independently verified.

Rubinstein's 2025 preprint claims that the relations previously obtained for Bernoulli symmetric polynomials remain valid for arbitrary homogeneous symmetric polynomials, including the power-sum formulation considered here.

October 2025 general-validity claim

Rubinstein states that the relations are valid in the arbitrary-polynomial setting and derives consequences for Bernoulli numbers. The retrieved record does not independently verify the exact displayed formulation or establish a published proof.

Posted attempt

A posted argument claims a complete proof after correcting the exponent from $m-k-1$ to $m-n-1$ . It uses cancellation of residues at $x_i=0$ and $D_{ij}=0$ , then polynomiality, degree bounds, and symmetry; the attempt has not been independently verified.

Current status (as of August 2026): The general relation is claimed in Rubinstein's 2025 preprint and in a complete posted proof attempt, but independent verification is absent.

Sources

arxiv.org

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In the primary source, equations (25) and (27) use the exponent $m-n-1$ ; the displayed $m-k-1$ contains an unbound $k$ and must be corrected accordingly.

Let $S$ be any symmetric homogeneous polynomial of degree $n$ in $m$ variables. Set

D_{ij}=y_ix_j-y_jx_i, \qquad (s_i)_j= \begin{cases} x_i,&j=i,\\ D_{ij},&j\ne i, \end{cases} \qquad \pi(x)=\prod_i x_i,

and define

U= \frac{S(x)}{\pi(x)} - \sum_i \frac{y_i^{m-n-1}S(s_i)} {x_i\prod_{j\ne i}D_{ij}}.

Over the coefficient field generated by the $y_i$ , the only possible $x$ -poles are $x_i$ and $D_{ij}$ . On $x_i=0$ ,

S(s_i)=y_i^nS(x|_{x_i=0}), \qquad \prod_{j\ne i}D_{ij} =y_i^{m-1}\prod_{j\ne i}x_j,

so the residue of the $i$ -th summand cancels the residue of $S(x)/\pi(x)$ .

On $D_{ij}=0$ , write $c=y_j/y_i$ . Then

x_j=cx_i, \qquad D_{jr}=cD_{ir}, \qquad s_j=c\,\tau_{ij}(s_i),

and hence $S(s_j)=c^nS(s_i)$ . The two corresponding residues have ratio

-c^{(m-n-1)+n-(m-1)}=-1,

so they cancel. Therefore $U$ is a polynomial in $x$ , homogeneous of degree $n-m$ . In particular,

U=0\qquad(n<m).

For $n\ge m$ , put $d=n-m+1$ . Multiplication by $\pi(y)^d$ removes all explicit $y$ -denominators. With

H=\pi(x)\prod_{i<j}D_{ij}, \qquad C=\binom m2,

the residue cancellations show that the numerator over the common denominator $H$ is divisible by every pairwise nonassociated irreducible factor $x_i,D_{ij}$ . Thus

R=\pi(y)^dU\in\mathbb Q[x_1,\ldots,x_m,y_1,\ldots,y_m], \qquad U=\frac{R}{\pi(y)^d}.

Moreover,

\deg_x R=n-m.

The first term of $HR$ has $y$ -degree at most $md+C$ ; each remaining term has degree at most

(m-1)d+n+C-(m-1)=md+C.

Since $H$ is homogeneous of $y$ -degree $C$ , it follows that

\deg_yR\le md=m(n-m+1).

Simultaneously permuting the pairs $(x_i,y_i)$ leaves $U$ , $\pi(y)^d$ , and therefore $R$ invariant. Specializing $y_1=\cdots=y_m=1$ consequently gives a symmetric homogeneous polynomial

Y_{n-m}(x)=R(x,\mathbf1).

Finally, linearity in $S$ gives the asserted decomposition

S=\sum_{\mathbf k}C_{n,\mathbf k}P_{n,\mathbf k} \quad\Longrightarrow\quad Y_{n-m} =\sum_{\mathbf k}C_{n,\mathbf k}Y_{n-m,\mathbf k}.

This proves all four source assertions, including the nonspecialized rational form, the degree bound, and simultaneous-permutation symmetry.