The polynomial-independence conjecture for normalized symmetric-group character values

Theorem. Let $\mathcal P_{\geq2}$ denote the partitions having no parts equal to $1$, and put

$$\mathcal R=\mathbb Q(N)[t_\mu:\mu\in\mathcal P_{\geq2}]. \tag{1}$$

For each $\mu$, let $T_\mu$ be the polynomial in single-cycle variables furnished by Theorem 2.4 of the source, with $T_{(k)}=t_{(k)}$. For a partition $\lambda\vdash n$, write

$$r_\mu(\lambda) =\frac{\chi^\lambda(\mu,1^{n-|\mu|})}{\chi^\lambda(1^n)} \qquad(n\geq|\mu|). \tag{2}$$

Then the ideal of all relations valid for every sufficiently large $n$ and every $\lambda\vdash n$ is exactly

$$\boxed{ \mathcal I =\left(t_\mu-T_\mu:\mu\in\mathcal P_{\geq2}\right) \subseteq\mathcal R. } \tag{3}$$

Here a rational coefficient is evaluated only away from its finitely many poles. In particular, the size and all single-cycle character ratios are jointly algebraically independent:

$$N,\quad t_{(2)},\quad t_{(3)},\quad t_{(4)},\quad\ldots. \tag{4}$$

The assertion continues to hold if, for any fixed number $K$ of these variables, the tested representations are restricted to partitions with at most $K$ rows.

Proof. For a Young diagram $\lambda$, define the shifted power sums

$$p_j(\lambda) =\sum_{i\geq1} \left[ \left(\lambda_i-i+\frac12\right)^j -\left(-i+\frac12\right)^j \right] \qquad(j\geq1). \tag{5}$$

Let $\mathcal A$ be the algebra of polynomial functions on Young diagrams. The classical Kerov--Olshanski theorem, in the explicit form of Ivanov--Olshanski, identifies

$$\mathcal A=\mathbb Q[p_1,p_2,p_3,\ldots], \qquad \deg p_j=j. \tag{6}$$

Define the central character of a single $j$-cycle by

$$P_j(\lambda) = \begin{cases} n^{\underline j}\,r_{(j)}(\lambda),&n=|\lambda|\geq j,\\ 0,&n<j, \end{cases} \qquad n^{\underline j}=n(n-1)\cdots(n-j+1). \tag{7}$$

For $j=1$, the character ratio is $1$, so $P_1(\lambda)=n$. Ivanov--Olshanski's triangular character formula gives

$$P_j=p_j+\text{a polynomial of canonical degree strictly less than }j. \tag{8}$$

Consequently, triangular change of generators in (6) yields

$$\boxed{ \mathcal A=\mathbb Q[P_1,P_2,P_3,\ldots], \qquad P_1(\lambda)=|\lambda|. } \tag{9}$$

We must strengthen ordinary algebraic independence to the eventual-size quantifier in the conjecture. Let

$$0\neq H\in\mathbb Q[P_1,\ldots,P_K], \qquad d=\deg_{\mathcal A}H. \tag{10}$$

Write its nonzero top homogeneous component as $h(p_1,\ldots,p_K)$. For any strictly decreasing positive real numbers $x_1>\cdots>x_K>0$, define

$$\lambda^{(a)} =\left(\lfloor ax_1\rfloor,\ldots,\lfloor ax_K\rfloor\right), \qquad a\longrightarrow\infty. \tag{11}$$

These are partitions for all sufficiently large $a$, their sizes tend to infinity, and (5) gives

$$p_j(\lambda^{(a)}) =a^j\sum_{i=1}^Kx_i^j+O(a^{j-1}). \tag{12}$$

Therefore

$$\lim_{a\to\infty} \frac{H(\lambda^{(a)})}{a^d} =h\left( \sum_{i=1}^Kx_i, \sum_{i=1}^Kx_i^2, \ldots, \sum_{i=1}^Kx_i^K \right). \tag{13}$$

If $H$ vanished on all sufficiently large partitions with at most $K$ rows, the right-hand side would vanish throughout the open cone $x_1>\cdots>x_K>0$. However, the Jacobian of its power-sum coordinates is

$$\det\left(jx_i^{j-1}\right)_{1\leq j,i\leq K} =K!\prod_{1\leq i<j\leq K}(x_j-x_i)\neq0. \tag{14}$$

Their image therefore contains an open set, and the polynomial $h$ would be identically zero, contradicting its definition. Thus no nonzero $H$ can vanish eventually, even on this bounded-row subfamily.

Now let

$$F(N,u_2,\ldots,u_K) \in\mathbb Q(N)[u_2,\ldots,u_K] \tag{15}$$

be a relation among the size and finitely many single-cycle ratios. Substitute

$$N=P_1, \qquad u_j=\frac{P_j}{P_1^{\underline j}} \quad(2\leq j\leq K). \tag{16}$$

Since $P_1,\ldots,P_K$ are algebraically independent by (9), this substitution embeds the polynomial ring in (15) into $\operatorname{Frac}(\mathcal A)$. If $F\neq0$, clearing the finitely many denominators produces a nonzero polynomial function

$$H =D(P_1) F\left( P_1, \frac{P_2}{P_1^{\underline2}}, \ldots, \frac{P_K}{P_1^{\underline K}} \right) \in\mathcal A, \qquad D\in\mathbb Q[N]\setminus\{0\}. \tag{17}$$

An eventual relation would make $H$ vanish on every sufficiently large diagram, contradicting (13)--(14). This proves (4).

Finally, define the reduction homomorphism

$$\operatorname{red}:\mathcal R \longrightarrow\mathbb Q(N)[t_{(2)},t_{(3)},\ldots], \qquad \operatorname{red}(t_\mu)=T_\mu. \tag{18}$$

Because $T_{(j)}=t_{(j)}$, this is a retraction. Successive substitution in the finitely many variables appearing in any polynomial gives

$$\ker(\operatorname{red}) =\left(t_\mu-T_\mu:\mu\in\mathcal P_{\geq2}\right). \tag{19}$$

The source's Theorem 2.4 shows that every generator on the right is an eventual character relation. Conversely, if $F\in\mathcal R$ is any eventual relation, then $\operatorname{red}(F)$ is an eventual relation involving only single cycles. By (17), it must be zero, so $F\in\ker(\operatorname{red})$. This proves (3).

For example, the double-transposition relation becomes

$$t_{(2,2)} = \frac{ N(N-1)t_{(2)}^2-4(N-2)t_{(3)}-2 }{(N-2)(N-3)}. \tag{20}$$

The same notation $T_\mu$ is reused in Example 2.7 of the source for the vanishing relation polynomial, rather than for the single-cycle expression. Throughout (3), (18), and (19), $T_\mu$ has its original Theorem 2.4 meaning, exactly as required by Conjecture 6.6.

Attribution. The precise conjecture and reduction polynomials are from T. Y. Lee, Relations between values and zeros of irreducible characters of symmetric groups, version 2, Theorem 2.4 and Conjecture 6.6. The classical free polynomial algebra and triangular character formula used in (6)--(9) are due to V. Ivanov and G. Olshanski, Kerov's central limit theorem for the Plancherel measure on Young diagrams, Symmetric Functions 2001 (2002), Propositions 1.5 and 3.4; see also the authors' full preprint. The eventual-size argument and reduction (13)--(19) apply those established results to prove the full stated conjecture, including the bounded-row strengthening.