The N=1 conjecture for character sums over Riordan partitions

A proof of the character-sum identity for every partition

We prove Hemmer–Straub–Westrem, Conjecture 6, numbered Conjecture 2.3 in the preprint. No restriction on the part sizes is needed.

Let $\lambda=(\lambda_1,\ldots,\lambda_r)\vdash n$. In the multiset $\mathrm{Ev}(\lambda)$, each occurrence of a part $a$ is independently replaced by either $2a$ or the pair $(a,a)$. Thus all $2^r$ choices are counted, including repetitions. Let $\mathcal R_3(2n)$ be the partitions of $2n$ with at most three parts, all even. Write

$$\begin{aligned} L_\lambda&=\sum_{\rho\in\mathrm{Ev}(\lambda)} (-1)^{\ell(\rho)} \sum_{\mu\in\mathcal R_3(2n)}\chi^\mu_\rho,\\ R_\lambda&=\sum_{\rho\in\mathrm{Ev}(\lambda)} \chi^{(n,n)}_\rho. \end{aligned}$$

Here $\chi^\mu_\rho$ is the irreducible symmetric-group character indexed by $\mu$, evaluated at cycle type $\rho$.

In fact, both sides have the following common value:

$$L_\lambda=R_\lambda =2^r\operatorname{CT}_t \left((1-t)\prod_{i=1}^r (t^{\lambda_i}+1+t^{-\lambda_i})\right),$$

where $\operatorname{CT}_t$ denotes the coefficient of $t^0$ in a Laurent polynomial.

The idea is to express the two character sums in three and two variables, respectively. A simple constant-term identity for three-variable Schur polynomials then makes the two expressions identical.

1. The two power-sum products

Write $p_a(x_1,\ldots,x_d)=\sum_{j=1}^d x_j^a$, and let $s_\mu$ denote the Schur polynomial. We use the standard character expansion

$$p_\rho(x_1,\ldots,x_d) =\sum_{\substack{\mu\vdash|\rho|\\ \ell(\mu)\leq d}} \chi^\mu_\rho\,s_\mu(x_1,\ldots,x_d).$$

This is Lemma 7 of the published paper. Specializing to $d$ variables preserves the coefficients of the Schur functions with at most $d$ parts; those with more parts vanish.

Define the symmetric functions

$$\begin{aligned} A_\lambda&=\prod_{i=1}^r (p_{\lambda_i}^{\,2}-p_{2\lambda_i}),\\ B_\lambda&=\prod_{i=1}^r (p_{\lambda_i}^{\,2}+p_{2\lambda_i}). \end{aligned}$$

Splitting a part contributes two parts and hence a positive sign; doubling it contributes one part and hence a negative sign. Expanding these products therefore gives, with exactly the required multiplicities,

$$\begin{aligned} A_\lambda&=\sum_{\rho\in\mathrm{Ev}(\lambda)} (-1)^{\ell(\rho)}p_\rho,\\ B_\lambda&=\sum_{\rho\in\mathrm{Ev}(\lambda)}p_\rho. \end{aligned}$$

Consequently, $L_\lambda$ is the sum of the coefficients of $s_\mu$ for $\mu\in\mathcal R_3(2n)$ in $A_\lambda(x,y,z)$, while $R_\lambda$ is the coefficient of $s_{(n,n)}$ in $B_\lambda(x,y)$.

Put

$$g_\lambda(x,y,z)=\prod_{i=1}^r (x^{\lambda_i}+y^{\lambda_i}+z^{\lambda_i}) =\sum_{\substack{\nu\vdash n\\ \ell(\nu)\leq3}}c_\nu s_\nu(x,y,z).$$

Since $p_a^2-p_{2a}=2((xy)^a+(xz)^a+(yz)^a)$ in three variables, we obtain

$$A_\lambda(x,y,z) =2^r(xyz)^n g_\lambda(x^{-1},y^{-1},z^{-1}).$$

The signed product identity is also the special case of Theorem 8 in the paper used here.

Pad $\nu$ with zeros to three parts. The Schur alternant formula gives the usual complementary-partition identity

$$(xyz)^n s_\nu(x^{-1},y^{-1},z^{-1}) =s_{(n-\nu_3,\ n-\nu_2,\ n-\nu_1)}(x,y,z).$$

All parts on the right are nonnegative because $\nu_1\leq n$. The resulting partitions have size $2n$. Those with all parts even correspond precisely to $\nu_i\equiv n\pmod2$ for all three indices. Partitions $\mu$ with $\mu_1>n$ do not occur in this expansion and have coefficient zero. Thus

$$\frac{L_\lambda}{2^r} =\sum_{\substack{\nu\vdash n,\ \ell(\nu)\leq3\\ \nu_1\equiv\nu_2\equiv\nu_3\equiv n\pmod2}}c_\nu.$$

2. A constant term that selects the parity

For every partition $\nu$ with at most three parts, we claim

$$\operatorname{CT}_t\bigl((1-t)s_\nu(t,1,t^{-1})\bigr) = \begin{cases} 1,&\nu_1,\nu_2,\nu_3\text{ have the same parity},\\ 0,&\text{otherwise}. \end{cases}$$

Here and below missing parts are zero. We give a direct calculation of this identity.

Set $c=\nu_3$, $a=\nu_1-c$, and $b=\nu_2-c$, so $a\geq b\geq0$. Removing the factor $(xyz)^c$ does not change the specialization at $(t,1,t^{-1})$. The three-variable alternant formula yields

$$(1-t)s_{(a,b,0)}(t,1,t^{-1}) =-\frac{t^2N_{a,b}(t)}{(1-t)^2(1+t)},$$

where

$$\begin{aligned} N_{a,b}(t)={}&t^{a+2}-t^{a-b+1}-t^{b+1}\\ &+t^{b-a-1}+t^{-b-1}-t^{-a-2}. \end{aligned}$$

Expand the denominator at $t=0$:

$$\frac1{(1-t)^2(1+t)} =\frac1{(1-t)(1-t^2)} =\sum_{k\geq0}q_k t^k, \qquad q_k=\left\lfloor\frac{k}{2}\right\rfloor+1.$$

Set $q_k=0$ for $k<0$. The first three monomials of $t^2N_{a,b}(t)$ have positive exponent. Taking the constant term of the remaining three gives

$$\begin{aligned} \operatorname{CT}_t\bigl((1-t)s_{(a,b,0)}(t,1,t^{-1})\bigr) &=q_a-q_{a-b-1}-q_{b-1}\\ &=\left\lfloor\frac a2\right\rfloor+1 -\left\lceil\frac{a-b}{2}\right\rceil -\left\lceil\frac b2\right\rceil. \end{aligned}$$

This equals $1$ exactly when both $a$ and $b$ are even, and equals $0$ in the other three parity cases. That is precisely the claimed condition on $\nu$. The convention for negative indices includes $b=0$ and $a=b$, so the calculation also covers partitions with fewer than three parts.

Since $\nu_1+\nu_2+\nu_3=n$, having three parts of the same parity is equivalent to all three being congruent to $n$ modulo $2$. Applying the identity term by term to $g_\lambda$ now gives

$$\frac{L_\lambda}{2^r} =\operatorname{CT}_t\bigl((1-t)g_\lambda(t,1,t^{-1})\bigr).$$

3. The rectangular character gives the same constant term

For a homogeneous symmetric polynomial $F(x,y)$ of degree $2n$, the two-variable Schur formula implies

$$[s_{(n,n)}]F =[x^ny^n]\bigl((1-y/x)F(x,y)\bigr).$$

Indeed, for a partition $(u,v)$ of $2n$,

$$(1-y/x)s_{(u,v)}(x,y) =x^uy^v-x^{v-1}y^{u+1}.$$

The monomial $x^ny^n$ occurs in this expression only for $(u,v)=(n,n)$, with coefficient $1$.

In two variables, $p_a^2+p_{2a}=2(x^{2a}+x^ay^a+y^{2a})$. Hence

$$B_\lambda(x,y) =2^r(xy)^n\prod_{i=1}^r \bigl((x/y)^{\lambda_i}+1+(y/x)^{\lambda_i}\bigr).$$

Dividing by $(xy)^n$ and putting $t=y/x$ turns the coefficient extraction into a constant term. Therefore,

$$\begin{aligned} \frac{R_\lambda}{2^r} &=\operatorname{CT}_t \left((1-t)\prod_{i=1}^r (t^{\lambda_i}+1+t^{-\lambda_i})\right)\\ &=\operatorname{CT}_t\bigl((1-t)g_\lambda(t,1,t^{-1})\bigr) =\frac{L_\lambda}{2^r}. \end{aligned}$$

This proves the conjecture for every partition $\lambda$, including arbitrary mixtures and repetitions of part sizes.