Mixed Products of Modified Greaves–Jing–Zhu Operators

Abstract

Let $\mathcal{Y}(z;t)$ be the modified Greaves–Jing–Zhu operator on the odd power-sum ring. We first point out that this operator can be obtained from the classical neutral operator by a simple diagonal change of variables. We then study products in which the two deformation parameters are not necessarily the same. For two parameters $t$ and $s$, we compute the scalar factor that appears in the mixed product. This factor has an explicit exponential form and, in a completed setting, can also be written as a quotient of infinite $t$-Pochhammer products. We also give a recurrence for its coefficients, a product formula for several mixed operators, and formulas for the coefficients obtained after applying the operators to $\mathbf{1}$.

A particularly simple case occurs when $s=t^{M}$. In this case the scalar factor becomes the finite quotient $(u;t)_{M}/(-u;t)_{M}$. Its coefficients are signed principal specializations of one-row Schur $Q$-functions. As a result, after removing the signs, these coefficients are nonnegative palindromic polynomials. We also give a Gaussian-binomial formula and a finite-order recurrence.

1 Introduction

Vertex operators provide a useful way to organize identities involving symmetric functions, Heisenberg algebras, and free-fermion representations. Greaves, Jing, and Zhu introduced a charged-fermion operator construction for the $t$-Schur functions and the associated $t$-Schur measure [1]. We adapted the same idea to the odd power-sum ring, which is the standard setting for Schur $Q$-functions [4]. The resulting modified operator has the same scalar factor as the classical neutral operator when both operators use the same parameter. Its modes generate shifted $t$-Schur functions and lead to a two-row formula, a Pfaffian Giambelli identity, a Cauchy identity, and a finite shifted Gessel formula.

The goal of this paper is more limited. We only study what happens when two modified operators with different parameters are multiplied. General linear transformations of vertex-operator presentations have been studied in a broader setting; see, for example, [6]. In the present case, the diagonal form of the modified Greaves–Jing–Zhu operator makes the mixed factor especially explicit.

Let

$$\rho_{t}(p_{n})=(1-t^{n})p_{n}\qquad(n\geq 1\text{ odd}).$$

Our first observation is

$$\mathcal{Y}(z;t)=\rho_{t}\Phi(z)\rho_{t}^{-1},\qquad\mathcal{Q}_{\lambda}(X;t)=Q_{\lambda}[X-tX],$$

where $\Phi(z)$ is the classical neutral operator. For two independent parameters, the mixed product has the form

$$\mathcal{Y}(z;t)\mathcal{Y}(w;s)=F_{t,s}(w/z):\mathcal{Y}(z;t)\mathcal{Y}(w;s):,$$ (1.1)

where

$$F_{t,s}(u)=\exp\!\left(-2\sumop\slimits@_{\begin{subarray}{c}n\geq 1\\ n\text{ odd}\end{subarray}}\frac{1-s^{n}}{1-t^{n}}\frac{u^{n}}{n}\right).$$ (1.2)

The specialization $s=t$ gives $(1-u)/(1+u)$, which is the usual neutral factor. In the $t$-adic completion, or analytically for $|t|<1$, this mixed factor may also be written as

$$F_{t,s}(u)=\frac{(u;t)_{\infty}(-su;t)_{\infty}}{(-u;t)_{\infty}(su;t)_{\infty}}.$$

We also derive a recurrence for the coefficients of $F_{t,s}(u)$ and use the mixed product formula to compute coefficients after applying the operators to $\mathbf{1}$.

The most concrete case is

$$s=t^{M},\qquad M\in\mathbb Z_{>0}.$$

Then the infinite product reduces to the finite product

$$F_{t,t^{M}}(u)=\prodop\slimits@_{j=0}^{M-1}\frac{1-t^{j}u}{1+t^{j}u}=\frac{(u;t)_{M}}{(-u;t)_{M}}.$$ (1.3)

If

$$F_{t,t^{M}}(u)=\sumop\slimits@_{m\geq 0}f_{m}^{[M]}(t)u^{m},$$

then

$$f_{m}^{[M]}(t)=(-1)^{m}q_{m}(1,t,\ldots,t^{M-1}).$$ (1.4)

Therefore $(-1)^{m}f_{m}^{[M]}(t)$ is a nonnegative palindromic polynomial. We give an explicit Gaussian-binomial formula and a finite recurrence of order $M$.

The paper is organized as follows. In section˜2 we review the odd power-sum ring and prove the diagonal-conjugation and plethystic formulas. In section˜3 we prove the general mixed product formula, its infinite-product form, its coefficient recurrence, and the corresponding coefficient formulas after applying the operators to $\mathbf{1}$. In section˜4 we specialize to $s=t^{M}$ and study the finite product and its coefficients.

2 The odd power-sum ring and diagonal conjugation

2.1 Strict partitions and Schur $Q$-functions

A strict partition is a finite sequence $\lambda=(\lambda_{1}>\cdots>\lambda_{\ell}>0)$. We write $|\lambda|=\lambda_{1}+\cdots+\lambda_{\ell}$ and $\ell(\lambda)=\ell$. Let $\mathrm{SP}(N)$ denote the set of strict partitions of $N$. Let $\mathrm{OP}(N)$ denote the set of partitions of $N$ all of whose parts are odd. Euler’s identity gives $|\mathrm{SP}(N)|=|\mathrm{OP}(N)|$.

Let

$$\Gamma=\mathbb{Q}[p_{1},p_{3},p_{5},\ldots]$$

be the odd power-sum ring, graded by $\deg p_{n}=n$. For a partition $\alpha=(1^{m_{1}}3^{m_{3}}\cdots)\in\mathrm{OP}(N)$, set

$$p_{\alpha}=\prodop\slimits@_{i}p_{\alpha_{i}},\qquad z_{\alpha}=\prodop\slimits@_{r\geq 1}r^{m_{r}}m_{r}!.$$

The elements $p_{\alpha}$, $\alpha\in\mathrm{OP}(N)$, form a basis of the homogeneous component _N.

We use the standard Schur $Q$ scalar product

$$\langle p_{\alpha},p_{\beta}\rangle=2^{-\ell(\alpha)}z_{\alpha}\,\delta_{\alpha\beta}.$$ (2.1)

The Schur $Q$-functions $Q_{\lambda}$, $\lambda\in\mathrm{SP}(N)$, form another basis of _N. We set

$$P_{\lambda}=2^{-\ell(\lambda)}Q_{\lambda}.$$

Then

$$\langle Q_{\lambda},P_{\mu}\rangle=\delta_{\lambda\mu},\qquad\langle Q_{\lambda},Q_{\mu}\rangle=2^{\ell(\lambda)}\delta_{\lambda\mu}.$$ (2.2)

We follow the standard conventions in [5, Chapter III, Section 8]; see also [3, 7].

The one-row functions $q_{r}=Q_{(r)}$ are determined by

$$Q_{X}(z):=\sumop\slimits@_{r\geq 0}q_{r}(X)z^{r}=\exp\!\left(2\sumop\slimits@_{\begin{subarray}{c}n\geq 1\\ n\text{ odd}\end{subarray}}\frac{p_{n}(X)}{n}z^{n}\right)=\prodop\slimits@_{i}\frac{1+x_{i}z}{1-x_{i}z}.$$ (2.3)

Here $q_{0}=1$ and $q_{r}=0$ for $r<0$. For $r\geq s\geq 0$, define

$$Q_{(r,s)}=q_{r}q_{s}+2\sumop\slimits@_{k=1}^{s}(-1)^{k}q_{r+k}q_{s-k}.$$ (2.4)

In particular, $Q_{(r,0)}=q_{r}$ and $Q_{(r,r)}=0$. If $\lambda$ has odd length, append a zero part. The Pfaffian Giambelli identity is

$$Q_{\lambda}=\operatorname{Pf}\bigl(Q_{(\lambda_{i},\lambda_{j})}\bigr).$$ (2.5)

For later use, define coefficients $\chi_{\alpha}^{\lambda}$ by the power-sum expansion

$$Q_{\lambda}=\sumop\slimits@_{\alpha\in\mathrm{OP}(|\lambda|)}\frac{2^{\ell(\alpha)}}{z_{\alpha}}\chi_{\alpha}^{\lambda}p_{\alpha}.$$ (2.6)

With the usual normalization, these are the spin-character coefficients. Their representation-theoretic interpretation will not be used below; we only need (2.6) and the orthogonality implied by (2.1).

2.2 The modified neutral operator

Let $t$ be an indeterminate and work over $\mathbb{Q}(t)$. For positive odd $n$, set

$$a_{-n}=p_{n},\qquad a_{n}=\frac{n}{2}\frac{\partial}{\partial p_{n}}.$$ (2.7)

Then

$$[a_{m},a_{n}]=\frac{m}{2}\delta_{m,-n}\qquad(m,n\text{ odd}).$$ (2.8)

The classical neutral operator is

$$\Phi(z)=\exp\!\left(\sumop\slimits@_{\begin{subarray}{c}n\geq 1\\ n\text{ odd}\end{subarray}}\frac{2}{n}a_{-n}z^{n}\right)\exp\!\left(-\sumop\slimits@_{\begin{subarray}{c}n\geq 1\\ n\text{ odd}\end{subarray}}\frac{2}{n}a_{n}z^{-n}\right).$$ (2.9)

Its Fourier expansion is

$$\Phi(z)=\sumop\slimits@_{m\in\mathbb Z}\phi_{m}z^{-m}.$$

Following [4], define the modified operator

$$\mathcal{Y}(z;t)=\exp\!\left(\sumop\slimits@_{\begin{subarray}{c}n\geq 1\\ n\text{ odd}\end{subarray}}\frac{2(1-t^{n})}{n}a_{-n}z^{n}\right)\exp\!\left(-\sumop\slimits@_{\begin{subarray}{c}n\geq 1\\ n\text{ odd}\end{subarray}}\frac{2}{n(1-t^{n})}a_{n}z^{-n}\right).$$ (2.10)

In terms of derivatives, the second exponential equals

$$\exp\!\left(-\sumop\slimits@_{\begin{subarray}{c}n\geq 1\\ n\text{ odd}\end{subarray}}\frac{1}{1-t^{n}}\frac{\partial}{\partial p_{n}}z^{-n}\right).$$

Write

$$\mathcal{Y}(z;t)=\sumop\slimits@_{m\in\mathbb Z}\mathcal{Y}_{m}(t)z^{-m}.$$ (2.11)

For a strict partition $\lambda=(\lambda_{1}>\cdots>\lambda_{\ell}>0)$, set

$$\mathcal{Q}_{\lambda}(X;t)=\mathcal{Y}_{-\lambda_{1}}(t)\cdots\mathcal{Y}_{-\lambda_{\ell}}(t)\mathbf{1}.$$ (2.12)

2.3 Diagonal conjugation

Define the graded algebra automorphism

$$\rho_{t}:{}_{\mathbb{Q}(t)}\longrightarrow{}_{\mathbb{Q}(t)},\qquad\rho_{t}(p_{n})=(1-t^{n})p_{n}\quad(n\text{ odd}).$$ (2.13)

Its inverse is given by $\rho_{t}^{-1}(p_{n})=(1-t^{n})^{-1}p_{n}$.

Proposition 2.1 (Diagonal conjugation).

The modified operator is a conjugate of the classical neutral operator:

$$\mathcal{Y}(z;t)=\rho_{t}\Phi(z)\rho_{t}^{-1}.$$ (2.14)

Consequently,

$$\mathcal{Q}_{\lambda}(X;t)=\rho_{t}Q_{\lambda}(X)=Q_{\lambda}[X-tX].$$ (2.15)

Proof.

Because $a_{-n}$ is multiplication by $p_{n}$,

$$\rho_{t}a_{-n}\rho_{t}^{-1}=(1-t^{n})a_{-n}.$$

For $f\in{}_{\mathbb{Q}(t)}$,

$$\rho_{t}\frac{\partial}{\partial p_{n}}\rho_{t}^{-1}f=\frac{1}{1-t^{n}}\frac{\partial f}{\partial p_{n}}.$$

Hence

$$\rho_{t}a_{n}\rho_{t}^{-1}=\frac{1}{1-t^{n}}a_{n}.$$

Conjugating the two exponentials in (2.9) gives exactly (2.10), proving (2.14).

Since $\rho_{t}^{-1}\mathbf{1}=\mathbf{1}$, the mode expansion of (2.14) gives

$$\mathcal{Q}_{\lambda}(X;t) =\rho_{t}\phi_{-\lambda_{1}}\rho_{t}^{-1}\cdots\rho_{t}\phi_{-\lambda_{\ell}}\rho_{t}^{-1}\mathbf{1}$$

$$=\rho_{t}\phi_{-\lambda_{1}}\cdots\phi_{-\lambda_{\ell}}\mathbf{1}=\rho_{t}Q_{\lambda}(X).$$

In plethystic notation, $p_{n}[X-tX]=(1-t^{n})p_{n}[X]$, which proves the final equality in (2.15). ∎

Corollary 2.2.

For $\lambda\in\mathrm{SP}(N)$,

$$\mathcal{Q}_{\lambda}(X;t)=\sumop\slimits@_{\alpha\in\mathrm{OP}(N)}\frac{2^{\ell(\alpha)}}{z_{\alpha}}\chi_{\alpha}^{\lambda}\prodop\slimits@_{a\in\alpha}(1-t^{a})p_{\alpha}(X).$$ (2.16)

In particular, $\mathcal{Q}_{\lambda}(X;t)$ is polynomial in $t$ with coefficients in .

Proof.

Apply $\rho_{t}$ to (2.6) and use

$$\rho_{t}(p_{\alpha})=\prodop\slimits@_{a\in\alpha}(1-t^{a})p_{\alpha}.$$

∎

Remark 2.3.

The operator (2.10) itself cannot be specialized directly at $t=1$ because its second exponential contains $(1-t^{n})^{-1}$. The functions in (2.16), however, are polynomial in $t$. For every nonempty strict partition $\lambda$,

$$\mathcal{Q}_{\lambda}(X;1)=0.$$

Thus the specialization of the functions exists even though the specialization of the operator does not.

Remark 2.4.

Every odd power sum is primitive:

$$\Delta(p_{n})=p_{n}\otimes 1+1\otimes p_{n}.$$

It follows that $\rho_{t}$ is a graded Hopf algebra automorphism over $\mathbb{Q}(t)$. In particular, same-parameter product and coproduct identities for the $\mathcal{Q}_{\lambda}(X;t)$ are transported directly from the classical Schur $Q$ theory. The new part of the story appears when two different parameters are compared.

3 Mixed operator products

We now work over $\mathbb{K}=\mathbb{Q}(t,s)$. Write

$$A_{t}(z)=\sumop\slimits@_{\begin{subarray}{c}n\geq 1\\ n\text{ odd}\end{subarray}}\frac{2(1-t^{n})}{n}a_{-n}z^{n},\qquad B_{t}(z)=-\sumop\slimits@_{\begin{subarray}{c}n\geq 1\\ n\text{ odd}\end{subarray}}\frac{2}{n(1-t^{n})}a_{n}z^{-n}.$$ (3.1)

Then

$$\mathcal{Y}(z;t)=e^{A_{t}(z)}e^{B_{t}(z)}.$$

We put the $A$-parts to the left and the $B$-parts to the right, and write

$$:\mathcal{Y}(z;t)\mathcal{Y}(w;s):=e^{A_{t}(z)+A_{s}(w)}e^{B_{t}(z)+B_{s}(w)}.$$ (3.2)

All scalar factors below are expanded as formal power series in $w/z$.

3.1 The mixed operator product expansion

Theorem 3.1 (Mixed operator product).

For independent parameters $t$ and $s$,

$$\mathcal{Y}(z;t)\mathcal{Y}(w;s)=F_{t,s}(w/z):\mathcal{Y}(z;t)\mathcal{Y}(w;s):,$$ (3.3)

where

$$F_{t,s}(u)=\exp\!\left(-2\sumop\slimits@_{\begin{subarray}{c}n\geq 1\\ n\text{ odd}\end{subarray}}\frac{1-s^{n}}{1-t^{n}}\frac{u^{n}}{n}\right)\in\mathbb{K}[[u]].$$ (3.4)

When $s=t$,

$$F_{t,t}(u)=\frac{1-u}{1+u}.$$ (3.5)

Proof.

The $A$-parts commute with each other, and the $B$-parts also commute with each other. By (2.8),

$$=-\sumop\slimits@_{\begin{subarray}{c}m,n\geq 1\\ m,n\text{ odd}\end{subarray}}\frac{2}{m(1-t^{m})}\frac{2(1-s^{n})}{n}[a_{m},a_{-n}]z^{-m}w^{n}$$

$$=-2\sumop\slimits@_{\begin{subarray}{c}n\geq 1\\ n\text{ odd}\end{subarray}}\frac{1-s^{n}}{1-t^{n}}\frac{(w/z)^{n}}{n}.$$

This commutator is a scalar. Therefore

$$e^{B_{t}(z)}e^{A_{s}(w)}=e^{[B_{t}(z),A_{s}(w)]}e^{A_{s}(w)}e^{B_{t}(z)}.$$

Substituting this into $e^{A_{t}(z)}e^{B_{t}(z)}e^{A_{s}(w)}e^{B_{s}(w)}$ proves (3.3) and (3.4).

If $s=t$, the ratio in (3.4) is $1$. Since

$$2\sumop\slimits@_{\begin{subarray}{c}n\geq 1\\ n\text{ odd}\end{subarray}}\frac{u^{n}}{n}=\log\frac{1+u}{1-u},$$

we obtain (3.5). ∎

Corollary 3.2 (Several mixed operators).

For parameters $t_{1},\ldots,t_{r}$,

$$\mathcal{Y}(z_{1};t_{1})\cdots\mathcal{Y}(z_{r};t_{r})=\prodop\slimits@_{1\leq i<j\leq r}F_{t_{i},t_{j}}(z_{j}/z_{i}):\mathcal{Y}(z_{1};t_{1})\cdots\mathcal{Y}(z_{r};t_{r}):.$$ (3.6)

Each factor is expanded in nonnegative powers of $z_{j}/z_{i}$.

Proof.

Move the $B$-part of the $i$th operator through the $A$-parts of the operators to its right. Each ordered pair $i<j$ contributes exactly the scalar factor computed in theorem˜3.1. ∎

3.2 Infinite-product form and coefficient recurrence

For $m\geq 0$, let

$$(a;t)_{m}=\prodop\slimits@_{j=0}^{m-1}(1-at^{j}),\qquad(a;t)_{\infty}=\prodop\slimits@_{j\geq 0}(1-at^{j}).$$

Proposition 3.3.

Analytically for $|t|<1$, or formally in the $t$-adic completion,

$$F_{t,s}(u)=\frac{(u;t)_{\infty}(-su;t)_{\infty}}{(-u;t)_{\infty}(su;t)_{\infty}}.$$ (3.7)

Proof.

Use

$$\frac{1-s^{n}}{1-t^{n}}=\sumop\slimits@_{j\geq 0}t^{jn}-\sumop\slimits@_{j\geq 0}(st^{j})^{n}$$

in (3.4). For each variable $x$,

$$\exp\!\left(-2\sumop\slimits@_{n\text{ odd}}\frac{x^{n}}{n}\right)=\frac{1-x}{1+x}.$$

It follows that

$$F_{t,s}(u)=\prodop\slimits@_{j\geq 0}\frac{(1-t^{j}u)(1+st^{j}u)}{(1+t^{j}u)(1-st^{j}u)},$$

which is (3.7). ∎

Write

$$F_{t,s}(u)=\sumop\slimits@_{m\geq 0}f_{m}(t,s)u^{m},\qquad f_{0}(t,s)=1.$$ (3.8)

Proposition 3.4 (Coefficient recurrence).

For $m\geq 1$,

$$mf_{m}(t,s)=-2\sumop\slimits@_{\begin{subarray}{c}1\leq j\leq m\\ j\text{ odd}\end{subarray}}\frac{1-s^{j}}{1-t^{j}}f_{m-j}(t,s).$$ (3.9)

Proof.

Logarithmically differentiate (3.4):

$$uF^{\prime}_{t,s}(u)=-2F_{t,s}(u)\sumop\slimits@_{j\text{ odd}}\frac{1-s^{j}}{1-t^{j}}u^{j}.$$

Comparing the coefficient of $u^{m}$ gives (3.9). ∎

3.3 Coefficients after applying the operators to $\mathbf{1}$

Define the modified one-row series

$$\mathsf{H}_{t}(X;z):=\mathcal{Y}(z;t)\mathbf{1}=\exp\!\left(2\sumop\slimits@_{\begin{subarray}{c}n\geq 1\\ n\text{ odd}\end{subarray}}\frac{1-t^{n}}{n}p_{n}(X)z^{n}\right)=\sumop\slimits@_{r\geq 0}q_{r}(X;t)z^{r}.$$ (3.10)

Thus $q_{r}(X;t)=\mathcal{Y}_{-r}(t)\mathbf{1}$.

For $r,m\geq 0$, define the mixed two-row coefficient

$$\mathcal{Q}_{r,m}^{t,s}(X):=[z^{r}w^{m}]\mathcal{Y}(z;t)\mathcal{Y}(w;s)\mathbf{1}=\mathcal{Y}_{-r}(t)\mathcal{Y}_{-m}(s)\mathbf{1}.$$ (3.11)

When $t\neq s$, these coefficients are not skew-symmetric in $r$ and $m$, so we do not regard them as a strict-partition basis.

Proposition 3.5.

For $r,m\geq 0$,

$$\mathcal{Q}_{r,m}^{t,s}(X)=\sumop\slimits@_{k=0}^{m}f_{k}(t,s)q_{r+k}(X;t)q_{m-k}(X;s).$$ (3.12)

For $s=t$, this becomes

$$\mathcal{Q}_{r,m}^{t,t}(X)=q_{r}(X;t)q_{m}(X;t)+2\sumop\slimits@_{k=1}^{m}(-1)^{k}q_{r+k}(X;t)q_{m-k}(X;t).$$ (3.13)

Proof.

Applying (3.3) to $\mathbf{1}$ gives

$$\mathcal{Y}(z;t)\mathcal{Y}(w;s)\mathbf{1}=F_{t,s}(w/z)\mathsf{H}_{t}(X;z)\mathsf{H}_{s}(X;w).$$

Substitute (3.8) and (3.10). To obtain $z^{r}w^{m}$, the three indices must satisfy

$$z^{-k}z^{r+k}=z^{r},\qquad w^{k}w^{m-k}=w^{m},$$

which gives (3.12). If $s=t$, then

$$\frac{1-u}{1+u}=1+2\sumop\slimits@_{k\geq 1}(-1)^{k}u^{k},$$

so (3.13) follows. ∎

4 Cyclotomic mixed products

Fix a positive integer $M$ and define the finite plethystic alphabet

$$A_{M}(t)=1+t+\cdots+t^{M-1}.$$ (4.1)

We call a polynomial $f(t)$ palindromic of darga $d$ if $f(t)=t^{d}f(t^{-1})$; this convention permits zero coefficients at both ends of the ambient degree interval $[0,d]$. For every $n\geq 1$,

$$p_{n}[A_{M}(t)]=1+t^{n}+\cdots+t^{(M-1)n}=\frac{1-t^{Mn}}{1-t^{n}}.$$ (4.2)

4.1 Finite mixed scalar factors

Theorem 4.1 (Cyclotomic mixed product).

For $M\geq 1$,

$$\mathcal{Y}(z;t)\mathcal{Y}(w;t^{M})=\frac{(w/z;t)_{M}}{(-w/z;t)_{M}}:\mathcal{Y}(z;t)\mathcal{Y}(w;t^{M}):.$$ (4.3)

Equivalently,

$$F_{t,t^{M}}(u)=\prodop\slimits@_{j=0}^{M-1}\frac{1-t^{j}u}{1+t^{j}u}=\frac{(u;t)_{M}}{(-u;t)_{M}}.$$ (4.4)

Proof.

By (3.4) and (4.2),

$$\log F_{t,t^{M}}(u) =-2\sumop\slimits@_{n\text{ odd}}\left(\sumop\slimits@_{j=0}^{M-1}t^{jn}\right)\frac{u^{n}}{n}$$

$$=\sumop\slimits@_{j=0}^{M-1}\log\frac{1-t^{j}u}{1+t^{j}u}.$$

Exponentiating gives (4.4), and (4.3) follows from theorem˜3.1. ∎

Write

$$F_{t,t^{M}}(u)=\sumop\slimits@_{m\geq 0}f_{m}^{[M]}(t)u^{m}.$$ (4.5)

Proposition 4.2 (Principal specialization).

For every $m\geq 0$,

$$f_{m}^{[M]}(t)=(-1)^{m}q_{m}(1,t,\ldots,t^{M-1}).$$ (4.6)

In particular,

$$(-1)^{m}f_{m}^{[M]}(t)\in\mathbb{N}[t].$$ (4.7)

Moreover,

$$(-1)^{m}f_{m}^{[M]}(t)=\sumop\slimits@_{a=0}^{\min(M,m)}t^{\binom{a}{2}}\genfrac{[}{]}{0.0pt}{}{M}{a}_{t}\genfrac{[}{]}{0.0pt}{}{M+m-a-1}{m-a}_{t}.$$ (4.8)

Proof.

By (2.3),

$$\sumop\slimits@_{m\geq 0}q_{m}(1,t,\ldots,t^{M-1})z^{m}=\prodop\slimits@_{j=0}^{M-1}\frac{1+t^{j}z}{1-t^{j}z}.$$

Substitute $z=-u$ and compare with (4.4); this proves (4.6) and (4.7).

For the explicit formula, use the finite and reciprocal $q$-binomial expansions

$$\prodop\slimits@_{j=0}^{M-1}(1+t^{j}u)=\sumop\slimits@_{a=0}^{M}t^{\binom{a}{2}}\genfrac{[}{]}{0.0pt}{}{M}{a}_{t}u^{a}$$

and

$$\prodop\slimits@_{j=0}^{M-1}\frac{1}{1-t^{j}u}=\sumop\slimits@_{b\geq 0}\genfrac{[}{]}{0.0pt}{}{M+b-1}{b}_{t}u^{b}.$$

The coefficient of $u^{m}$ in their product is the right side of (4.8). ∎

Corollary 4.3 (Reciprocity for scalar factors).

For every $m\geq 0$,

$$f_{m}^{[M]}(t)=t^{(M-1)m}f_{m}^{[M]}(t^{-1}).$$ (4.9)

Thus $(-1)^{m}f_{m}^{[M]}(t)$ is palindromic with darga $(M-1)m$.

Proof.

As multisets,

$$A_{M}(t)=t^{M-1}A_{M}(t^{-1}).$$

Since $q_{m}$ is homogeneous of degree $m$,

$$q_{m}[A_{M}(t)]=t^{(M-1)m}q_{m}[A_{M}(t^{-1})].$$

Apply (4.6). ∎

Proposition 4.4 (Finite-order recurrence).

Set

$$e_{j}^{[M]}(t):=e_{j}(1,t,\ldots,t^{M-1})=t^{\binom{j}{2}}\genfrac{[}{]}{0.0pt}{}{M}{j}_{t},\qquad 0\leq j\leq M.$$ (4.10)

Then the coefficients in (4.5) satisfy

$$\sumop\slimits@_{j=0}^{\min(M,m)}e_{j}^{[M]}(t)f_{m-j}^{[M]}(t)=\begin{cases}(-1)^{m}e_{m}^{[M]}(t),&0\leq m\leq M,\\ 0,&m>M.\end{cases}$$ (4.11)

In particular, for $m>M$,

$$f_{m}^{[M]}(t)=-\sumop\slimits@_{j=1}^{M}e_{j}^{[M]}(t)f_{m-j}^{[M]}(t).$$ (4.12)

Proof.

The numerator and denominator in (4.4) expand as

$$(u;t)_{M}=\sumop\slimits@_{j=0}^{M}(-1)^{j}e_{j}^{[M]}(t)u^{j},\qquad(-u;t)_{M}=\sumop\slimits@_{j=0}^{M}e_{j}^{[M]}(t)u^{j}.$$

Multiplying

$$(-u;t)_{M}\sumop\slimits@_{m\geq 0}f_{m}^{[M]}(t)u^{m}=(u;t)_{M}$$

and comparing the coefficient of $u^{m}$ gives (4.11). The homogeneous recurrence follows when $m>M$. ∎

Corollary 4.5 (Cyclotomic mixed coefficients).

For $r,m\geq 0$,

$$\mathcal{Q}_{r,m}^{t,t^{M}}(X)=\sumop\slimits@_{k=0}^{m}(-1)^{k}q_{k}(1,t,\ldots,t^{M-1})q_{r+k}(X;t)q_{m-k}(X;t^{M}).$$ (4.13)

Proof.

Combine proposition˜3.5 with (4.6). ∎

Example 4.6 (The case $M=2$).

For $M=2$,

$$F_{t,t^{2}}(u)=\frac{(1-u)(1-tu)}{(1+u)(1+tu)}.$$

Here

$$e_{1}^{[2]}(t)=1+t,\qquad e_{2}^{[2]}(t)=t,$$

and

$$f_{0}^{[2]}(t) =1,$$

$$f_{1}^{[2]}(t) =-2(1+t),$$

$$f_{2}^{[2]}(t) =2(1+t)^{2},$$

$$f_{3}^{[2]}(t) =-2(1+t)(1+t+t^{2}).$$

For $m>2$, the coefficients satisfy

$$f_{m}^{[2]}(t)=-(1+t)f_{m-1}^{[2]}(t)-tf_{m-2}^{[2]}(t).$$