Power-sum relation for arbitrary homogeneous symmetric polynomials

From papers

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.

Progress summary

Open

An October 2025 preprint proposes the relation, but it remains a conditional conjecture rather than a verified solution.

The problem asks whether the proposed identity $U_n=0$ holds for $0\le n\le m-1$ , together with the stated specialization for larger degrees. A preprint dated October 2025 presents this general framework, but does not establish the identity unconditionally.

October 2025 preprint

The manuscript states the general power-sum formulation as Conjecture 6.1 and says it follows assuming earlier Conjectures 3.1 and 3.2. It therefore records a claimed extension to arbitrary homogeneous symmetric polynomials, not a proof; no counterexample, independent verification, referee report, withdrawal, or retraction was found. Its displayed definition also contains the unbound exponent $m-k-1$ .

Current status (as of August 2026): The relation is proposed in a 2025 preprint but remains unproved because its stated derivation depends on earlier unproved conjectures; no counterexample or verification is recorded.

Sources

arxiv.org

Sources & referencesView supporting material

Primary source

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

power-sums symmetric-polynomials

1 week ago

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.

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Shivam Patel · 1 day ago