3.1.1 Generators and relations

For $i,j\in I=\mathbb {Z}/r\mathbb {Z}$ , we set

$$ \begin{align*} a_{i,j}&=\left\{\begin{array}{ll} 2 & j=i \\ -1 & j=i\pm 1 \\ 0 & \mbox{otherwise} \end{array}\right. \\[7pt] m_{i,j}&=\left\{\begin{array}{ll} \mp 1 & j=i\pm 1\\ 0 & \mbox{otherwise} \end{array}\right. \end{align*} $$

and we define

$$\begin{align*}g_{i,j}(z):=\frac{\mathfrak{q}^{a_{i,j}}z-1}{z-\mathfrak{q}^{a_{i,j}}}. \end{align*}$$

The quantum toroidal algebra $U_{s,u}(\ddot {\mathfrak {sl}}_r)$ is a unital associative $\mathbb {F}$ -algebra with generators

$$\begin{align*}\{e_{i,k},f_{i,k},\psi_{i,k},\psi_{i,0}^{-1},\gamma^{\pm\frac{1}{2}}, \mathfrak{q}^{\pm d_1},\mathfrak{q}^{\pm d_2}\}_{i\in I}^{k\in\mathbb{Z}}. \end{align*}$$

Its relations are described in terms of currents:

$$ \begin{align*} e_i(z) & :=\sum_{k\in\mathbb{Z}}e_{i,k}z^{-k}\\[7pt] f_i(z) & :=\sum_{k\in\mathbb{Z}}f_{i,k}z^{-k}\\[7pt] \psi_i^{\pm}(z) & :=\psi_{i,0}^{\pm 1}+\sum_{k>0}\psi_{i,\pm k}z^{\mp k}. \end{align*} $$

The relations are then

$$ \begin{align*} [\psi_i^{\pm}(z),\psi_j^{\pm}(w)]&=0,\,\gamma^{\pm\frac{1}{2}}\mbox{ are central},\\ \psi_{i,0}^{\pm1}\psi_{i,0}^{\mp1}&=\gamma^{\pm\frac{1}{2}}\gamma^{\mp\frac{1}{2}}=\mathfrak{q}^{\pm d_1}\mathfrak{q}^{\mp d_1}=\mathfrak{q}^{\pm d_2}\mathfrak{q}^{\mp d_2}=1,\\ \mathfrak{q}^{d_1}e_i(z)\mathfrak{q}^{-d_1}&=e_i(\mathfrak{q}^{-1} z),\, \mathfrak{q}^{d_1}f_i(z)\mathfrak{q}^{-d_1}=f_i(\mathfrak{q}^{-1} z),\, \mathfrak{q}^{d_1}\psi_i^{\pm}(z)\mathfrak{q}^{-d_1}=\psi_i^{\pm}(\mathfrak{q}^{-1} z),\\ \mathfrak{q}^{d_2}e_i(z)\mathfrak{q}^{-d_2}&=\mathfrak{q} e_i(z),\, \mathfrak{q}^{d_2}f_i(z)\mathfrak{q}^{-d_2}=\mathfrak{q}^{-1} f_i(z),\, \mathfrak{q}^{d_2}\psi_i^{\pm}(z)\mathfrak{q}^{-d_2}=\psi_i^{\pm}( z),\\ g_{i,j}(\gamma^{-1}\mathfrak{d}^{m_{i,j}}z/w)\psi_i^{+}(z)\psi_j^{-}(w)&=g_{i,j}(\gamma\mathfrak{d}^{m_{i,j}} z/w)\psi_j^{-}(w)\psi_i^{+}(z),\\ e_i(z)e_j(w)&=g_{i,j}(\mathfrak{d}^{m_{i,j}}z/w)e_j(w)e_i(z),\\ f_i(z)f_j(w)&=g_{i,j}(\mathfrak{d}^{m_{i,j}}z/w)^{-1}f_j(w)f_i(z),\\ (\mathfrak{q}-\mathfrak{q}^{-1})[e_i(z),f_j(w)]&=\delta_{i,j}\left(\delta(\gamma w/z)\psi_i^+(\gamma^{\frac{1}{2}}w)-\delta(\gamma z/w)\psi_i^-(\gamma^{\frac{1}{2}}z)\right),\\ \psi_i^{\pm}(z)e_j(w)&=g_{i,j}(\gamma^{\pm\frac{1}{2}}\mathfrak{d}^{m_{i,j}}z/w)e_j(w)\psi_i^{\pm}(z),\\ \psi_i^{\pm}(z)f_j(w)&=g_{i,j}(\gamma^{\mp\frac{1}{2}}\mathfrak{d}^{m_{i,j}}z/w)^{-1}f_j(w)\psi_i^{\pm}(z),\\ \mathrm{Sym}_{z_1,z_2}[e_i(z_1),[e_i(z_2),e_{i\pm1}(w)]_{\mathfrak{q}}]_{\mathfrak{q}^{-1}}&=0,\,[e_i(z),e_j(w)]=0\mbox{ for }j\not=i,i\pm1,\\ \mathrm{Sym}_{z_1,z_2}[f_i(z_1),[f_i(z_2),f_{i\pm1}(w)]_{\mathfrak{q}}]_{\mathfrak{q}^{-1}}&=0,\,[f_i(z),f_j(w)]=0\mbox{ for }j\not=i,i\pm1,\end{align*} $$

Here, $\delta (z)$ denotes the delta function

$$\begin{align*}\delta(z)=\sum_{k\in\mathbb{Z}}z^k\end{align*}$$

and for $v\in \mathbb {F}$ , $[a,b]_v=ab-v ba$ is the v-commutator. We will also work with elements $\{h_{i, k}\}_{i\in I}^{k\not = 0}$ defined by

(3.1) $$ \begin{align} \psi_i^{\pm}(z)=\psi_{i,0}^{\pm 1}\exp\left(\pm(\mathfrak{q}-\mathfrak{q}^{-1})\sum_{k>0}h_{i,\pm k}z^{\mp k}\right). \end{align} $$

Finally, we denote by

• • $'\ddot {U}$ the subalgebra obtained by dropping the generator $\mathfrak {q}^{d_1}$ ;

• • $\ddot {U}'$ the subalgebra obtained by dropping the generator $\mathfrak {q}^{d_2}$ ;

• • $'\ddot {U}'$ the subalgebra obtained by dropping both generators $\mathfrak {q}^{d_1}$ and $\mathfrak {q}^{d_2}$ .

3.1.2 Miki automorphism

We recall that $U_{\mathfrak {q},\mathfrak {d}}(\ddot {\mathfrak {sl}}_r)$ contains two copies of the quantum affine algebra $U_{\mathfrak {q}}(\dot {\mathfrak {sl}}_r)$ . The first, called the vertical copy, is generated by currents where $i\not =0$ . This copy is given in the new Drinfeld presentation. However, the second copy, called the horizontal copy, is generated by the constant terms of all the currents. This copy is given in the Drinfeld-Jimbo presentation. We do not go into detail on these two subalgebras as we will not need them in the sequel. However, we mention them because they give the ‘two loops’ of the quantum toroidal algebra. Let $\eta $ denote the $\mathbb {C}(\mathfrak {q})$ -linear antiautomorphism of $'\ddot {U}'$ defined by

(3.2) $$ \begin{align} \eta(\mathfrak{d})&=\mathfrak{d}^{-1}\nonumber\\ \eta(e_{i,k})&=e_{i,-k},\, \eta(f_{i,k})=f_{i,-k},\, \eta(h_{i,k})=-\gamma^kh_{i,-k},\\ \eta(\psi_{i,0})&=\psi_{i,0}^{-1},\,\eta(\gamma^{\frac{1}{2}})=\gamma^{\frac{1}{2}}.\nonumber \end{align} $$

The following beautiful result of Miki gives the ‘S-transformation’ of the torus:

3.1.3 Heisenberg subalgebras

Recall the generators $\{h_{i,n}\}_{i\in I}^{n\not =0}$ defined by (3.1). Together with $\gamma ^{\pm \frac {1}{2}}$ , these elements generate a rank r Heisenberg algebra. The relations are

(3.3) $$ \begin{align} [h_{i, n}, h_{j,n'}]=\delta_{n,-n'}\frac{(\gamma^n-\gamma^{-n})\mathfrak{d}^{-nm_{i,j}}[na_{i,j}]_{\mathfrak{q}}}{(\mathfrak{q}-\mathfrak{q}^{-1})n}\\ \gamma^{\frac{1}{2}}\mbox{ is central},\nonumber \end{align} $$

where $[n]_v$ is the usual quantum number:

$$\begin{align*}[n]_v=\frac{v^n-v^{-n}}{v-v^{-1}}. \end{align*}$$

We define dual elements $\{h_{i,n}^{\perp }\}_{i\in I}^{n\not =0}$ by

(3.4) $$ \begin{align} \begin{aligned} h_{i,n}^{\perp}&= \frac{\mathfrak{q}^{n}(\mathfrak{q}-\mathfrak{q}^{-1})n}{(1-\mathfrak{q}^{nr}\mathfrak{d}^{nr})(1-\mathfrak{q}^{nr}\mathfrak{d}^{-nr})[n]_{\mathfrak{q}}} \sum_{j,k=0}^{r-1} \mathfrak{q}^{n(j+k)}\mathfrak{d}^{n(j-k)}h_{i+j-k, n}\\ h_{i,-n}^{\perp}&= \frac{\mathfrak{q}^{n}(\mathfrak{q}-\mathfrak{q}^{-1})n}{(1-\mathfrak{q}^{nr}\mathfrak{d}^{nr})(1-\mathfrak{q}^{nr}\mathfrak{d}^{-nr})[n]_{\mathfrak{q}}} \sum_{j,k=0}^{r-1} \mathfrak{q}^{n(j+k)}\mathfrak{d}^{n(j-k)}h_{i-j+k,-n}. \end{aligned} \end{align} $$

Lemma 3.2. The elements $\{h_{i,n}^{\perp }\}$ are characterized by

(3.5) $$ \begin{align} [h_{i,n}^{\perp}, h_{j,-n'}]=[h_{j,n'}, h_{i,-n}^{\perp}]=\delta_{i,j}\delta_{n,n'}(\gamma^n-\gamma^{-n}) \end{align} $$

for $k>0$ .

Proof. Equations (3.5) obviously characterizes these elements. For $n>0$ , let $M_n$ be the matrix $r\times r$ matrix

$$\begin{align*}\left(M_n\right)_{ij}=\mathfrak{d}^{-nm_{i,j}}[na_{i,j}]_{\mathfrak{q}}. \end{align*}$$

We view the rows and coloumns as indexed by I. Equation (3.3) can be rephrased as

$$\begin{align*}[h_{i,n}, h_{j,-n}]= \left(M_{n}\right)_{ij}\frac{(\gamma^n-\gamma^{-n})}{(\mathfrak{q}-\mathfrak{q}^{-1})n}. \end{align*}$$

For any $r\times r$ matrix A (with rows and columns indexed by I), set

$$ \begin{align*} Ah_{i,n}=\sum_{k\in I} A_{ki}h_{k,n},\qquad\qquad\qquad Ah_{i,-n}=\sum_{k\in I}A_{ki}h_{k,-n}. \end{align*} $$

We then have for $n>0$ ,

$$ \begin{align*} \left[Ah_{i,n}, h_{j,-n}\right]&=(A^TM_n)_{ij}\frac{(\gamma^n-\gamma^{-n})}{(\mathfrak{q}-\mathfrak{q}^{-1})n},& \left[h_{j,n}, Ah_{i,-n}\right]&=\left(M_nA\right)_{ji}\frac{(\gamma^n-\gamma^{-n})}{(\mathfrak{q}-\mathfrak{q}^{-1})n}. \end{align*} $$

Thus, to obtain (3.5), we need to invert $M_n$ . To that end, we factorize $M_n$ :

$$ \begin{align*} M_n&=[n]_{\mathfrak{q}}\mathfrak{q}^{-n} \begin{pmatrix} (\mathfrak{q}^{2n}+1) & -\mathfrak{q}^n\mathfrak{d}^n & 0&\cdots & 0 & -\mathfrak{q}^n\mathfrak{d}^{-n}\\ -\mathfrak{q}^n\mathfrak{d}^{-n} & (\mathfrak{q}^{2n}+1) & -\mathfrak{d}^n & 0 &\cdots & 0\\ 0 & -\mathfrak{q}^n\mathfrak{d}^{-n} & (\mathfrak{q}^{2n}+1) & -\mathfrak{q}^n\mathfrak{d}&\ddots &0\\ \vdots & \ddots& \ddots& \ddots& \ddots & \vdots&\\ 0 & \cdots & 0 & -\mathfrak{q}^n\mathfrak{d}^{-n} & (\mathfrak{q}^{2n}+1) &-\mathfrak{q}^n\mathfrak{d}^n\\ -\mathfrak{q}^n\mathfrak{d}^n & 0 &\cdots & 0& -\mathfrak{q}^n\mathfrak{d}^{-n} & (\mathfrak{q}^{2n}+1) \end{pmatrix}\\ &=[n]_{\mathfrak{q}}\mathfrak{q}^{-n} \begin{pmatrix} 1 & 0&\cdots & 0 & -\mathfrak{q}^n\mathfrak{d}^{-n}\\ -\mathfrak{q}^n\mathfrak{d}^{-n} & 1 & 0 & & 0\\ 0 & -\mathfrak{q}^{n}\mathfrak{d}^{-n} & 1 & \ddots &0\\ \vdots & \ddots& \ddots& \ddots& 0&\\ 0 & \cdots & 0 & -\mathfrak{q}^{n}\mathfrak{d}^{-n} & 1 \\ \end{pmatrix} \begin{pmatrix} 1 & -\mathfrak{q}^n\mathfrak{d}^n & 0&\cdots & 0 \\ 0 & 1 & -\mathfrak{q}^n\mathfrak{d}^n & \ddots & \vdots\\ \vdots & \ddots & 1 & \ddots &0\\ 0 & \cdots & 0 & \ddots &-\mathfrak{q}^{n}\mathfrak{d}^n\\ -\mathfrak{q}^{n}\mathfrak{d}^n & 0 &\cdots & 0& 1 \end{pmatrix}. \end{align*} $$

Inverting the last two matrices, we obtain (3.4).

We denote by $\ddot {U}^0$ the subalgebra generated by $\{\gamma ^{\pm \frac {1}{2}}\}\cup \{h_{i, k}\}_{i\in I}^{k\not = 0}$ an call it the vertical Heisenberg subalgebra. In analogy with 3.1.2, we call $\varsigma (\ddot {U}^0)$ the horizontal Heisenberg subalgebra.

Remark 3.3. In [Reference Wen19], the author defines elements $\{b_{i,k}^{\perp }\}$ in terms of a pairing that is not used in this paper. By comparing the commutator (3.3) to the pairing in loc. cit., we have that

$$\begin{align*}h_{i,k}^{\perp}=-b_{i,k}^{\perp}. \end{align*}$$

3.2.1 Twisted group algebra

Recall that Q and $Q^\vee $ denote the $\mathfrak {sl}_r$ root and coroot lattices, respectively, with simple roots $\{\alpha _j\}_{j=1}^{r-1}$ , simple coroots $\{\alpha ^\vee _j\}_{j=1}^{r-1}$ , and canonical pairing $(-,-) : Q^\vee \times Q \to \mathbb {Z}$ :

$$\begin{align*}(\alpha_i^\vee,\alpha_j)=a_{i,j}. \end{align*}$$

Let P denote the $\mathfrak {sl}_r$ weight lattice and $\{\Lambda _p\}_{j=1}^{r-1}$ the fundamental weights. We will also need

$$\begin{align*}\alpha_0=-\sum_{j=1}^{r-1}\alpha_j,\quad \alpha^\vee_0=-\sum_{j=1}^{r-1}\alpha^\vee_j,\quad \Lambda_0:=0. \end{align*}$$

We have that $\{\alpha _2,\ldots ,\alpha _{r-1},\Lambda _{r-1}\}$ is a basis of P.

The twisted group algebra $\mathbb {F}\{P\}$ is the $\mathbb {F}$ -algebra generated by $\{e^{\alpha _j}\}_{j=2}^{r-1}\cup \{e^{\Lambda _{r-1}}\}$ satisfying the relations

$$ \begin{align*} e^{\alpha_i}e^{\alpha_j}&=(-1)^{\left( \alpha^\vee_i,\alpha_j\right)}e^{\alpha_j}e^{\alpha_i}\\ e^{\alpha_i}e^{\Lambda_{r-1}}&=(-1)^{\delta_{i,r-1}}e^{\Lambda_{r-1}}e^{\alpha_i}. \end{align*} $$

Given a general $\alpha \in P$ , we write $\alpha =\sum _{j=2}^{r-1}m_j\alpha _j+m_{r}\Lambda _{r-1}$ and then set

$$\begin{align*}e^{\alpha}=e^{m_2\alpha_2}\cdots e^{m_{r-1}\alpha_{r-1}}e^{m_r\Lambda_{r-1}}.\end{align*}$$

For example,

(3.6) $$ \begin{align} \begin{aligned} e^{\alpha_1}&=e^{-2\alpha_2}e^{-3\alpha_3}\cdots e^{-(r-1)\alpha_{r-1}}e^{r\Lambda_{r-1}}\\ e^{\alpha_0}&=e^{\alpha_2}e^{2\alpha_3}\cdots e^{(r-2)\alpha_{r-1}}e^{-r\Lambda_{r-1}}. \end{aligned} \end{align} $$

Define $\mathbb {F}\{Q\}$ to be the subalgebra of $\mathbb {F}\{P\}$ generated by $\{e^{\alpha _i}\}_{i=1}^{r-1}$ .

3.2.2 Vertex operators

The vertical Heisenberg subalgebra $\ddot {U}^0$ has a Fock representation $F_r$ defined as follows. Let $\ddot {U}^0_+$ denote the subalgebra generated by $\gamma ^{\frac {1}{2}}$ and $\{h_{i,k}\}_{i\in I}^{k>0}$ . $\ddot {U}^0_+$ has a one-dimensional representation $\mathbb {F}_{\mathfrak {q}}$ where $\gamma ^{\frac {1}{2}}$ acts by $\mathfrak {q}^{\frac {1}{2}}$ while $h_{i,k}$ acts by $0$ . $F_r$ is then the induced representation

$$\begin{align*}F_r:=\mathrm{Ind}_{\ddot{U}^0_+}^{\ddot{U}^0}\mathbb{F}_{\mathfrak{q}}\cong\mathbb{K}[h_{i,-k}]_{i\in I}^{k>0}. \end{align*}$$

The vertex representation is defined on the space $W:=F_r\otimes \mathbb {F}\{Q\}$ . For $v\otimes e^\alpha \in W$ where

$$ \begin{align*} v&=h_{i_1,-k_1}\cdots h_{i_N,-k_N} v_0\\ \alpha&=\sum_{j=1}^{r-1}m_j\alpha_j, \end{align*} $$

we define the operators $h_{i,k}$ , $e^\beta $ , $\partial _{\alpha _i}$ , $z^{H_{i,0}}$ and d by

(3.7) $$ \begin{align} \nonumber h_{i,k}(v\otimes e^\alpha )&:=(h_{i,k}v)\otimes e^\alpha ,\, e^\beta(v\otimes e^\alpha ):=v\otimes(e^\beta e^\alpha ),\\ \nonumber \partial_{\alpha_i}(v\otimes e^\alpha )&:=\left( \alpha^\vee_i,\alpha\right) v\otimes e^\alpha ,\\ z^{H_{i,0}}(v\otimes e^\alpha )&:=z^{\left( \alpha^\vee_i,\alpha\right)} \mathfrak{d}^{\frac{1}{2}\sum_{j=1}^{r-1}\left( \alpha^\vee_i,m_j\alpha_j\right) m_{i,j}}v\otimes e^\alpha ,\\ \nonumber d(v\otimes e^\alpha )&:=-\left(\dfrac{(\alpha,\alpha)}{2}+\sum_{i=1}^N k_i\right)v\otimes e^\alpha. \end{align} $$

Theorem 3.4 [Reference Saito16]

Let $\vec {c}=(c_0,\ldots , c_{r-1})\in (\mathbb {F}^{\times })^r$ . The following formulas endow W with an action of $\ddot {U}'$ :

$$ \begin{align*} \rho_{\vec{c}}(e_i(z))&=c_i\exp\left(\sum_{k>0}\frac{\mathfrak{q}^{-\frac{k}{2}}}{[k]_{\mathfrak{q}}}h_{i,-k}z^k\right)\\ &\times\exp\left(-\sum_{k>0}\frac{\mathfrak{q}^{-\frac{k}{2}}}{[k]_{\mathfrak{q}}}h_{i,k}z^{-k}\right)e^{\alpha_i} z^{1+H_{i,0}},\\ \rho_{\vec{c}}(f_i(z))&=\frac{(-1)^{r\delta_{i,0}}}{c_i}\exp\left(-\sum_{k>0}\frac{\mathfrak{q}^{\frac{k}{2}}}{[k]_{\mathfrak{q}}}h_{i,-k}z^k\right)\\ &\times\exp\left(\sum_{k>0}\frac{\mathfrak{q}^{\frac{k}{2}}}{[k]_{\mathfrak{q}}}h_{i,k}z^{-k}\right)e^{-\alpha_i} z^{1-H_{i,0}},\\\rho_{\vec{c}}(\psi_i^{\pm}(z))&=\exp\left(\pm(\mathfrak{q}-\mathfrak{q}^{-1})\sum_{k>0}h_{i,\pm k}z^{\mp k}\right)\mathfrak{q}^{\pm\partial_{\alpha_i}},\\ &\rho_{\vec{c}}(\gamma^{\frac{1}{2}})=\mathfrak{q}^{\frac{1}{2}},\,\rho_{\vec{c}}(\mathfrak{q}^{d_1})=\mathfrak{q}^{d}. \end{align*} $$

3.2.3 Embedding symmetric functions

We can let $\Lambda ^{I}$ act on $F_r$ via multiplication operators given by

(3.8) $$ \begin{align} p_k[X^{(i)}]\mapsto \frac{k}{[k]_{\mathfrak{q}}}h_{i,-k} \end{align} $$

for $k>0$ . To obtain an identification $W\cong \Lambda ^{I}\otimes \mathbb {F}\left\{ Q \right\}$ , we need to embed $\mathbb {K}$ into $\mathbb {F}$ :

(3.9) $$ \begin{align} q=\mathfrak{q}\mathfrak{d},\, t=\mathfrak{q}\mathfrak{d}^{-1}. \end{align} $$

Applying $\rho _{\vec {c}}$ to (3.5) sends $\gamma \mapsto \mathfrak {q}$ . Thus, as operators on $\Lambda ^{I}$ , we have the identification

$$\begin{align*}p_k[X^{(i)}]^{\perp}\mapsto kh_{i,k}^{\perp}. \end{align*}$$

Now consider transforming the formulas for $\rho _{\vec {c}}$ using matrix plethysms on $\{p_k[X^{(i)}]\}$ . We can obtain an isomorphic representation as long as we perform a corresponding transformation on $\{h_{i,k}\}$ to maintain the commutation relations, using (3.3) as a guide. First, we define $\rho _{\vec {c}}^+$ by performing the plethysm

$$\begin{align*}p_k[X^{(i)}]\mapsto \mathfrak{q}^{\frac{k}{2}}\left(p_k[X^{(i)}]-t^{-k}p_k[X^{(i-1)}]\right). \end{align*}$$

For $\rho _{\vec {c}}^+$ , we will only be interested in the currents $\{e_i(z)\}$ , although we have a representation for the entire algebra:

(3.10) $$ \begin{align} \begin{aligned} E_i(z):=\rho_{\vec{c}}^+(e_i(z))&=c_i\exp\left[\sum_{k>0}\left(p_{k}[X^{(i)}]-t^{-k}p_{k}[X^{(i-1)}]\right)\frac{z^k}{k}\right]\\ &\times\exp\left[\sum_{k>0}\left(-p_{k}[X^{(i)}]^{\perp}+q^{-k}p_k[X^{(i-1)}]^{\perp}\right)\frac{z^{-k}}{k}\right]e^{\alpha_i} z^{1+H_{i,0}}. \end{aligned} \end{align} $$

Similarly, we define $\rho _{c}^-$ by performing the plethysm

$$\begin{align*}p_k[X^{(i)}]\mapsto \mathfrak{q}^{-\frac{k}{2}}\left(t^{k}p_k[X^{(i)}]-p_k[X^{(i-1)}]\right). \end{align*}$$

Here, we will only be interested in the action of the currents $\{ f_i(z)\}$ :

(3.11) $$ \begin{align} \begin{aligned} F_i(z):=\rho_{\vec{c}}^-(f_i(z))&= \frac{(-1)^{r\delta_{i,0}}}{c_i}\exp\left[\sum_{k>0}\left(-t^{k}p_{k}[X^{(i)}]+p_{k}[X^{(i-1)}]\right)\frac{z^k}{k}\right]\\ &\times\exp\left[\sum_{k>0}\left(q^{k}p_{k}[X^{(i)}]^{\perp}-p_{k}[X^{(i-1)}]^{\perp}\right)\frac{z^{-k}}{k}\right]e^{-\alpha_i} z^{1-H_{i,0}}. \end{aligned} \end{align} $$

The following is a consequence of the main result of [Reference Wen19]:

3.2.4 Normal ordering

Later, we will make use of a particular expression for products of the currents $\{E_i(z)\}$ and $\{F_i(z)\}$ . We will need notation for an ordered product or composition of noncommuting operators $a_1,\ldots , a_m$ :

(3.12) $$ \begin{align} \begin{aligned} \overset{\curvearrowright}{\prod_{j=1}^m}a_j&:=a_1 a_2\cdots a_m\\ \overset{\curvearrowleft}{\prod_{j=1}^m}a_j&:=a_m a_{m-1}\cdots a_1 \end{aligned} \end{align} $$

Proposition 3.7. For $p\in I$ , we have

(3.13) $$ \begin{align} \begin{aligned} &\overset{\curvearrowright}{\prod_{a=1}^n}\,\overset{\curvearrowright}{\prod_{i=1}^{r}}E_{p+i}(z_{p+i,a})\\ &= \left( (-1)^{\frac{(r-2)(r-3)}{2}}\mathfrak{d}^{\frac{r}{2}-1}\prod_{i\in I} c_i\right)^n\\ &\times\prod_{1\le a<b\le n}\prod_{i\in I} \frac{\displaystyle\left(1-z_{i,b}/z_{i,a}\right)\left(1-q^{-1}t^{-1}z_{i,b}/z_{i,a}\right)}{\displaystyle\left(1-t^{-1}z_{i+1,b}/z_{i,a}\right)\left(1-q^{-1}z_{i-1,b}/z_{i,a}\right)}\\ &\times\prod_{a=1}^n \frac{z_{p,a}/z_{p+1,a}} {\left(1-q^{-1}z_{p,a}/z_{p+1,a}\right) \prod_{i\in I\setminus\{p+1\}} \left(1-t^{-1}z_{i,a}/z_{i-1,a}\right)}\\ &\times\prod_{i\in I}\exp\left(\sum_{a=1}^n\sum_{k>0}\left(p_{k}[X^{(i)}]-t^{-k}p_k[X^{(i-1)}]\right)\frac{z_{i,a}^{k}}{k}\right)\\ &\times\prod_{i\in I}\exp\left(\sum_{a=1}^n\sum_{k>0}\left(-p_{k}[X^{(i)}]^{\perp}+q^{-k}p_k[X^{(i-1)}]^{\perp}\right)\frac{z_{i,a}^{-k}}{k}\right) \prod_{i\in I}\prod_{a=1}^n z_{i,a}^{H_{i,0}}, \end{aligned} \end{align} $$

where all rational functions are Laurent series expanded assuming

(3.14) $$ \begin{align} |z_{i,a}|=1,\,|q|>1,\,|t|>1. \end{align} $$

For the F-currents, we have

(3.15) $$ \begin{align} &\overset{\curvearrowleft}{\prod_{a=1}^n}\,\overset{\curvearrowleft}{\prod_{i=1}^{r}}F_{p+i}(z_{p+i,a})\nonumber\\ &=\left( \frac{(-1)^{\frac{(r-2)(r-3)}{2}}}{\displaystyle\mathfrak{d}^{\frac{r}{2}-1} \prod_{i\in I} c_i}\right)^n\nonumber\\ &\times\prod_{1\le a<b\le n}\prod_{i\in I} \frac{\left(1-z_{i,a}/z_{i,b}\right)\left(1-qtz_{i,a}/z_{i,b}\right)}{\left(1-tz_{i-1,a}/z_{i,b}\right)\left(1-qz_{i+1,a}/z_{i,b}\right)}\\&\times \prod_{a=1}^n\frac{z_{p+1,a}/z_{p,a}}{\left(1-qz_{p+1,a}/z_{p,a}\right) \prod_{i\in I\setminus\{p\}} \left(1-tz_{i,a}/z_{i+1,a}\right)}\nonumber\\ &\times\prod_{i\in I}\exp\left(\sum_{a=1}^n\sum_{k>0}\left(-t^{k}p_{k}[X^{(i)}]+p_k[X^{(i-1)}]\right)\frac{z_{i,a}^{k}}{k}\right)\nonumber\\ &\times\prod_{i\in I}\exp\left(\sum_{a=1}^n\sum_{k>0}\left(q^{k}p_{k}[X^{(i)}]^{\perp}-p_k[X^{(i-1)}]^{\perp}\right)\frac{z_{i,a}^{-k}}{k}\right) \prod_{i\in I}\prod_{a=1}^n z_{i,a}^{-H_{i,0}},\nonumber \end{align} $$

where all rational functions are Laurent series expanded assuming

(3.16) $$ \begin{align} |z_{i,a}|=1,\, |q|< 1,\, |t|<1. \end{align} $$

Proof. The computation is standard. We will only go over the signs and powers of $\mathfrak {d}$ . The sign comes from the commutation of $\{e^{\alpha _i}\}$ ; in both cases, these factors simplify to $\pm e^0$ . For the E-currents, if $p=0$ , then by (3.6),

$$\begin{align*}e^{\alpha_1}= e^{r\Lambda_{r-1}}e^{-(r-1)\alpha_{r-1}}\cdots e^{-3\alpha_3}e^{-2\alpha_2}.\end{align*}$$

Thus,

$$\begin{align*}e^{\alpha_1}e^{\alpha_2}\cdots e^{\alpha_{r-1}}=(-1)^{\frac{(r-2)(r-3)}{2}}e^{r\Lambda_{r-1}}e^{-(r-2)\alpha_{r-1}}\cdots e^{-2\alpha_3}e^{-\alpha_2}. \end{align*}$$

However, if $p\not =0$ , we have

$$ \begin{align*} e^{\alpha_0}e^{\alpha_1}&= (-1)^{\frac{(r-1)(r-2)}{2}-1}e^{-\alpha_2}\cdots e^{-\alpha_{r-1}}\\ &=(-1)^{\frac{(r-1)(r-2)}{2}+r-3}e^{-\alpha_{r-1}}\cdots e^{-\alpha_{2}}\\ &= (-1)^{\frac{(r-2)(r-3)}{2}}e^{-\alpha_{r-1}}\cdots e^{-\alpha_{2}}, \end{align*} $$

which also leads to a sign of $(-1)^{\frac {(r-2)(r-3)}{2}}$ . For the F-currents, first consider the case $p=0$ .

$$ \begin{align*} e^{-\alpha_0}e^{-\alpha_{r-1}}\cdots e^{-\alpha_3}e^{-\alpha_2}&= (-1)^{r+\frac{(r-2)(r-3)}{2}}e^{-2\alpha_2}e^{-3\alpha_3}\cdots e^{-(r-1)\alpha_{-r-1}}e^{r\Lambda_{r-1}}\\ &= (-1)^{r+\frac{(r-2)(r-3)}{2}}e^{r\Lambda_{r-1}}e^{-(r-1)\alpha_{r-1}}\cdots e^{-3\alpha_2}e^{-2\alpha_2}. \end{align*} $$

If $p\not =0$ , then we use that

$$ \begin{align*} e^{-\alpha_1}e^{-\alpha_0}&=(-1)^re^{-\alpha_1}e^{r\Lambda_{r-1}}e^{-(r-2)\alpha_{r-1}}\cdots e^{-2\alpha_3}e^{-\alpha_2}\\ &= (-1)^{r+\frac{(r-2)(r-3)}{2}}e^{\alpha_2}e^{\alpha_3}\cdots e^{\alpha_{r-1}}. \end{align*} $$

Finally, note that $F_0(z)$ also has a sign of $(-1)^{r}$ . The power of $\mathfrak {d}$ comes from the interaction between $\{z^{\pm H_{i,0}}\}$ and $\{e^{\pm \alpha _{j}}\}$ . First observe that when considering $E_i(z_{i,a})$ and $E_{j}(z_{j,b})$ for $a\not =b$ , the powers of $\mathfrak {d}$ from $j=i-1$ and $j=i+1$ cancel out. When $a=b$ , there is a total power of $\mathfrak {d}^{\frac {r}{2}-1}$ . The case for $\{F_i(z)\}$ is similar but inverted.

3.3.1 Definition

In order to define the Fock representation, we will need some notation for partitions. For a partition $\lambda $ , let $\square =(a,b)\in D(\lambda )$ . We set the following:

1. 1. $\chi _\square =q^{a}t^{b}$ , the character of the box;

2. 2. $c_\square = b-a$ modulo r (its color);

3. 3. $d_i(\lambda )$ the number of elements of $D(\lambda )$ with content equivalent to i modulo r;

4. 4. $A_i(\lambda )$ and $R_i(\lambda )$ the addable and removable i-nodes of $\lambda $ , respectively.

Finally, we will abbreviate $a\equiv b\mbox { mod }r$ by simply $a\equiv b$ and use the Kronecker delta function $\delta _{a= b}:=\delta _{a-b,0}$ .

Let $v\in \mathbb {F}^{\times }$ . The Fock representation $\mathcal {F}(v)$ has a basis $\{|\lambda \rangle \}$ indexed by partitions.

Theorem 3.8 [Reference Feigin, Jimbo, Miwa and Mukhin5], cf. [Reference Wen19]

We can define a $'\ddot {U}$ -action $\tau _v$ on $\mathcal {F}(v)$ where the only nonzero matrix elements of the generators are

$$ \begin{gather*} \begin{aligned} \langle\lambda | e_i(z)|\lambda+\square\rangle&=\delta_{c_\square=i}(-\mathfrak{d})^{d_{i+1}(\lambda)}\delta\left( \frac{ z}{\chi_\square v}\right) \frac{\displaystyle\prod_{\blacksquare\in R_{i}(\lambda)}\left( \chi_\square-\mathfrak{q}^2\chi_\blacksquare \right)} {\displaystyle\prod_{\substack{\blacksquare\in A_{i}(\lambda)\\\blacksquare\not=\square}}\left(\chi_\square-\chi_\blacksquare\right)}\\ \langle\lambda+\square |f_i(z)|\lambda\rangle&=\delta_{c_\square=i}(-\mathfrak{d})^{-d_{i+1}(\lambda)}\delta\left(\frac{ z}{\chi_\square v}\right) \frac{\displaystyle\prod_{\substack{\blacksquare\in A_{i}(\lambda)\\\blacksquare\not=\square}}\left( \mathfrak{q}\chi_\square-\mathfrak{q}^{-1}\chi_\blacksquare \right)} {\displaystyle\prod_{\blacksquare\in R_{i}(\lambda)}\mathfrak{q}\left( \chi_\square-\chi_\blacksquare \right)}\\ \langle\lambda|\psi_i^{\pm}(z)|\lambda\rangle&= \prod_{\blacksquare\in A_{i}(\lambda)}\frac{\left(\mathfrak{q} z-\mathfrak{q}^{-1}\chi_{\blacksquare}v\right)}{\left( z-\chi_\blacksquare v\right)} \prod_{\blacksquare\in R_{i}(\lambda)}\frac{\left(\mathfrak{q}^{-1} z-\mathfrak{q}\chi_\blacksquare v\right)}{\left( z-\chi_\blacksquare v\right)}, \end{aligned}\\ \langle\lambda|\gamma^{\frac{1}{2}}|\lambda\rangle=1,\,\langle\lambda|\mathfrak{q}^{d_2}|\lambda\rangle=\mathfrak{q}^{-|\lambda|}. \end{gather*} $$

3.3.2 Tsymbaliuk isomorphism

The representation $\tau _v$ on $\mathcal {F}(v)$ has a cyclic vector $|\varnothing \rangle $ . However, $\rho _{\vec {c}}$ and $\rho _{\vec {c}}^{\pm }$ also have the natural cyclic vector $1\otimes 1\in F_r\otimes \mathbb {F}\{Q\}$ . The following theorem was proved by Tsymbaliuk:

Theorem 3.9 [Reference Tsymbaliuk17]

Let

(3.17) $$ \begin{align} v=(-1)^{\frac{(\ell-2)(\ell-3)}{2}}\frac{\mathfrak{q}\mathfrak{d}^{-\frac{\ell}{2}}}{c_0\cdots c_{\ell-1}}. \end{align} $$

The map of cyclic vectors

$$\begin{align*}\mathcal{F}(v)\ni|\varnothing\rangle\mapsto 1\otimes 1\in W \end{align*}$$

induces an isomorphism between the $'\ddot {U}'$ -module $\tau _v$ and the $\varsigma $ -twisted modules $\rho _{\vec {c}}\circ \varsigma $ , $\rho _{\vec {c}}^{\pm }\circ \varsigma $ .

The Tsymbaliuk isomorphism is defined only in terms of cyclic vectors. In light of Remark 3.6, the following result from [Reference Wen19] provides more detail on the Tsymbaliuk isomorphisms:

Theorem 3.10. The Tsymbaliuk isomorphisms (Theorem 3.9) between $\tau _v$ and $\rho _{\vec {c}}^{\pm }$ send

$$\begin{align*}\mathbb{F}|\lambda\rangle\to\mathbb{F} \left(P_{\lambda}[X^{\bullet};q,t^{-1}]\otimes e^{\kappa(\lambda)}\right).\end{align*}$$

Thus, we can study the eigenvalues of $\varsigma (\ddot {U}^0)$ on $P_{\lambda }$ by instead studying the eigenvalues of $\ddot {U}^0$ on the basis $\{|\lambda \rangle \}$ .

3.3.3 Infinite-variable eigenvalues

From the formulas in Theorem 3.8, we can see that

$$\begin{align*}\langle\lambda|\psi_{i,0}^{\pm 1}|\lambda\rangle= \mathfrak{q}^{\pm\left(|A_{i}(\lambda)|-|R_{i}(\lambda)|\right)}. \end{align*}$$

Therefore,

$$ \begin{align*} &\left\langle\lambda\left|\exp\left(\pm(\mathfrak{q}-\mathfrak{q}^{-1})\sum_{k>0}h_{i,\pm k}z^{\mp k}\right)\right|\lambda\right\rangle\\ &=\prod_{\blacksquare\in A_{i}(\lambda)}\frac{\mathfrak{q}^{\mp 1}\left(\mathfrak{q} z-\mathfrak{q}^{-1}\chi_{\blacksquare}v\right)}{\left( z-\chi_\blacksquare v\right)} \prod_{\blacksquare\in R_{i}(\lambda)}\frac{\mathfrak{q}^{\pm 1}\left(\mathfrak{q}^{-1} z-\mathfrak{q}\chi_\blacksquare v\right)}{\left( z-\chi_\blacksquare v\right)}\\ &= \exp\left[ \sum_{k>0}\left( \sum_{\blacksquare\in A_{i}(\lambda)}(1-\mathfrak{q}^{\mp 2k})\chi_{\blacksquare}^{\pm k} +\sum_{\blacksquare\in R_{i}(\lambda)}(1-\mathfrak{q}^{\pm 2k})\chi_\blacksquare^{\pm k}\right)\frac{v^{\pm k}z^{\mp k}}{k} \right]. \end{align*} $$

Taking logarithms, we see that for $k>0$ ,

(3.18) $$ \begin{align} \begin{aligned} \langle\lambda|h_{i,\pm k}|\lambda\rangle &= \frac{v^{\pm k}[ k]_{\mathfrak{q}}}{k}\left( \sum_{\blacksquare\in A_{i}(\lambda)}\mathfrak{q}^{\mp k}\chi_\blacksquare^{\pm k} -\sum_{\blacksquare\in R_{i}(\lambda)}\mathfrak{q}^{\pm k}\chi_\blacksquare^{\mp k}\right)\\ &= \frac{v^{\pm k}\mathfrak{q}^{\mp k}[ k]_{\mathfrak{q}}}{k}\left( \sum_{\blacksquare\in A_{i}(\lambda)}\chi_\blacksquare^{\pm k} - \sum_{\blacksquare\in R_{i}(\lambda)}(qt\chi_\blacksquare)^{\pm k}\right). \end{aligned} \end{align} $$

Using (3.18), we can try to piece together elements of $\ddot {U}^0$ whose eigenvalues are elementary symmetric functions in $\{q^{\pm \lambda _b}t^{\pm b}\}$ .

For $k\in \mathbb {Z}_{>0}$ and $p\in I$ , let us define

(3.19) $$ \begin{align} \hat{h}_{p,k}&:=\frac{1}{(1-t^{kr})}\sum_{i=0}^{r-1}t^{ k(i+1)}h_{p-i, k} \end{align} $$

(3.20) $$ \begin{align}\hspace{11.5pt} \hat{h}_{p,-k}&:= \frac{1}{(1-t^{- kr})}\sum_{i=0}^{r-1}t^{- k(i+1)}h_{p-i,- k}. \end{align} $$

Lemma 3.11. Assume $|t^{\pm 1}|<1$ (where ‘ $+$ ’ and ‘ $-$ ’ are separate cases). For $p\in I$ , we have

(3.21) $$ \begin{align} \begin{aligned} &\left\langle\lambda\left|\exp\left[-\sum_{k>0}\hat{h}_{p,\pm k} (-z)^{\mp k}{v^{\pm k}[ k]_{\mathfrak{q}}} \right]\right|\lambda\right\rangle\\ &= \exp\left[-\sum_{k>0}\left(\sum_{\substack{b>0\\b-\lambda_b\equiv p+1 }}q^{\pm k\lambda_b}t^{\pm kb}\right)\frac{(-z)^{\mp k}}{k}\right]\\ &= \prod_{\substack{b>0\\b-\lambda_b\equiv p+1 }}\left(1+q^{\pm\lambda_b}t^{\pm b}z^{\mp 1}\right), \end{aligned} \end{align} $$

where we set $\lambda _b=0$ for all $b>\ell (\lambda )$ .

Proof. Comparing (3.21) to (3.18), we need to establish the equality

(3.22) $$ \begin{align} \frac{1}{1-t^{\pm kr}}\sum_{i=0}^{r-1}t^{\pm k(i+1)}\left( \sum_{\blacksquare\in A_{p-i}(\lambda)}\chi_\blacksquare^{\pm k} - \sum_{\blacksquare\in R_{p-i}(\lambda)}(qt\chi_\blacksquare)^{\pm k}\right) =\left(\sum_{\substack{b>0\\b-\lambda_b\equiv p+1 }}q^{\pm k\lambda_b}t^{\pm kb}\right). \end{align} $$

We note that here, we consider $(1-t^{\pm kr})^{-1}$ as a geometric series. The summands on the right-hand side of (3.22) are $qt$ -shifts of the characters of color $p+1$ boxes that are the rightmost boxes in their row. We can account for these coordinates by starting at each addable box of $D(\lambda )$ , going straight up until we reach a box of color $p+1$ , then moving upwards by intervals of r, and ending the search once we are above the $qt$ -shift of the removable box above it. This is exactly what the left-hand side of (3.22) does. We illustrate this with Figure 2.

4.1.1 Shuffle product

We endow $\mathbf {S}$ with the shuffle product $\star $ , defined as follows. For $i,j\in I$ , we define the mixing terms:

$$\begin{align*}\omega_{i,j}(z,w):=\left\{\begin{array}{ll} \left(z-\mathfrak{q}^{2}w\right)^{-1}\left(z-w\right)^{-1} & \mbox{if }i=j\\ \left(\mathfrak{q} w-\mathfrak{d}^{-1}z\right) &\mbox{if }i+1=j\\ \left(z-\mathfrak{q}\mathfrak{d}^{-1} w\right) &\mbox{if }i-1=j\\ 1 &\mbox{otherwise.} \end{array}\right.\end{align*}$$

For $F\in \mathbf {S}_{k_{\bullet }}$ and $G\in \mathbf {S}_{l_{\bullet }}$ , let $F\star G\in \mathbf {S}_{k_{\bullet }+l_{\bullet }}$ be defined by

$$\begin{align*}F\star G:=\frac{1}{k_{\bullet}!l_{\bullet}!}\mathrm{Sym}_{k_{\bullet}+l_{\bullet}}\left[ F\left( \left\{ z_{i,a} \right\}_{i\in I}^{1\le a\le k_i} \right)G\left( \left\{ z_{j, b} \right\}_{j\in I}^{k_j<b\le k_j+l_j} \right) \prod_{i,j\in I}\prod_{\substack{1\le a\le k_i\\k_j<b\le k_j+l_j}}\omega_{i,j}(z_{i,a},z_{j,b}) \right], \end{align*}$$

where for $n_{\bullet }\in \left( \mathbb {Z}_{\ge 0} \right)^I$ ,

$$\begin{align*}n_{\bullet}!=\prod_{i\in I}n_i!=|\mathfrak{S}_{n_{\bullet}}| \end{align*}$$

and $\mathrm {Sym}_{n_{\bullet }}$ denotes the color symmetrization (i.e., the symmetrization over $\mathfrak {S}_{n_{\bullet }}$ ).

4.1.2 The shuffle algebra

Consider now for each $k_{\bullet }$ the subspace $\mathcal {S}_{k_{\bullet }}\subset \mathbf {S}_{k_{\bullet }}$ consisting of functions F satisfying the following two conditions:

1. 1. Pole conditions: F is of the form

   (4.1) $$ \begin{align} F=\frac{f(\{z_{i,r}\})}{\displaystyle \prod_{i\in I}\,\prod_{\substack{1\le r, r'\le k_i\\r\not= r'}}(z_{i,r}-\mathfrak{q}^2z_{i,r'})} \end{align} $$
   for a color-symmetric Laurent polynomial f.

2. 2. Wheel conditions: F has a well-defined finite limit when

   $$\begin{align*}\frac{z_{i,r_1}}{z_{i+\epsilon,s}}\rightarrow\mathfrak{q}\mathfrak{d}^\epsilon\mbox{ and }\frac{z_{i+\epsilon,s}}{z_{i,r_2}}\rightarrow\mathfrak{q}\mathfrak{d}^{-\epsilon}\end{align*}$$
   for any choice of i, $r_1$ , $r_2$ , s and $\epsilon $ , where $\epsilon \in \{\pm 1\}$ . This is equivalent to specifying that the Laurent polynomial f in the pole conditions evaluates to zero at
   $$ \begin{align*} \begin{aligned} z_{i,r_1}&=\mathfrak{q}\mathfrak{d}^{\epsilon}z_{i+\epsilon,s},& z_{i+\epsilon,s}&=\mathfrak{q}\mathfrak{d}^{-\epsilon}z_{i,r_2}. \end{aligned} \end{align*} $$

We set

$$\begin{align*}\mathcal{S}:=\bigoplus_{k_{\bullet}\in\left( \mathbb{Z}_{\ge 0} \right)^I}\mathcal{S}_{k_{\bullet}}. \end{align*}$$

The following is standard:

We call $(\mathcal {S},\star )$ the shuffle algebra of type $\hat {A}_{r-1}$ .

4.1.3 Relation to $U_{\mathfrak {q},\mathfrak {d}}(\ddot {\mathfrak {sl}_r})$

Let

• • $\ddot {U}^+\subset U_{\mathfrak {q},\mathfrak {d}}(\ddot {\mathfrak {sl}_r})$ be the subalgebra generated by $\left\{ e_{i}(z) \right\}_{i\in I}$ and

• • $\ddot {U}^-\subset U_{\mathfrak {q},\mathfrak {d}}(\ddot {\mathfrak {sl}_r})$ be the subalgebra generated by $\left\{ f_{i}(z) \right\}_{i\in I}$ .

Correspondingly, we set $\mathcal {S}^+:=\mathcal {S}$ and $\mathcal {S}^-:=\mathcal {S}^{op}$ . The following key structural result was proved by Neguţ:

Theorem 4.2 [Reference Neguţ11]

$\mathcal {S}^{\pm }$ is generated by $\{z_{i,1}^n\}_{i\in I}^{n\in \mathbb {Z}}$ and

$$ \begin{align*} \Psi_+(z_{i,1}^n)&= e_{i,n}\\ \Psi_-(z_{i,1}^n)&= f_{i,n} \end{align*} $$

induce algebra isomorphisms $\Psi _{\pm }:\mathcal {S}^{\pm }\rightarrow \ddot {U}^{\pm }$ .

Finally, note that the subalgebras $\ddot {U}^{\pm }$ are each closed under $\eta $ . We will need to understand how the antiautomorphism $\eta $ is manifested on the shuffle side:

Proposition 4.3. For $F\in \mathcal {S}^{\pm }_{k_{\bullet }}$ , define

$$\begin{align*}\eta_{\mathcal{S}}(F):= \left.F(z_{i,r}^{-1})\prod_{i\in I}\prod_{r=1}^{k_i}(-\mathfrak{d})^{k_{i+1}k_i} z_{i,r}^{k_{i+1}+k_{i-1}-2(k_i-1)}\right|_{\mathfrak{d}\mapsto\mathfrak{d}^{-1}}. \end{align*}$$

We have

(4.2) $$ \begin{align} \Psi_+^{-1}\eta\Psi_+(F)=\Psi_-^{-1}\eta\Psi_-(F)=\eta_{\mathcal{S}}(F). \end{align} $$

Proof. Equation (4.2) is true when $F=z_{i,1}^n$ is a generator. To see that it is a $\mathbb {C}(\mathfrak {q})$ -linear algebra antiautomorphism that inverts $\mathfrak {d}$ , we first observe that

(4.3) $$ \begin{align} \begin{aligned} z^{-2}w^{-2}\omega_{i,i}(z^{-1},w^{-1})\bigg|_{\mathfrak{d}\mapsto\mathfrak{d}^{-1}}&=\omega_{i,i}(w,z)\\ zw(-\mathfrak{d})\omega_{i,i+1}(z^{-1},w^{-1})\bigg|_{\mathfrak{d}\mapsto\mathfrak{d}^{-1}}&=\omega_{i+1,i}(w,z)\\ zw(-\mathfrak{d})\omega_{i+1,i}(z^{-1}, w^{-1})\bigg|_{\mathfrak{d}\mapsto\mathfrak{d}^{-1}}&=\omega_{i,i+1}(w,z). \end{aligned} \end{align} $$

Now, for $F\in \mathcal {S}_{k_{\bullet }}^+$ and $G\in \mathcal {S}_{l_{\bullet }}^+$ ,

(4.4) $$ \begin{align} \nonumber &\eta_{\mathcal{S}}(F\star G)\\ \nonumber &=\frac{1}{k_{\bullet}!l_{\bullet}!}\mathrm{Sym}\left[F\left(\{z_{i,a}^{-1}\}_{i\in I}^{1\le a\le k_i}\right) G\left(\{z_{j,b}^{-1}\}_{j\in I}^{k_j<b\le k_j+l_j}\right) \prod_{i,j\in I}\prod_{\substack{1\le a\le k_i\\k_j<b\le k_j+l_j}}\omega_{i,j}(z_{i,a}^{-1},z_{j,b}^{-1})\right]\\ &\quad\times \prod_{i\in I}\prod_{r=1}^{k_i+l_i}(-\mathfrak{d})^{(k_{i+1}+l_{i+1})(k_i+l_i)} z_{i,r}^{k_{i+1}+l_{i+1}+k_{i-1}+l_{i-1}-2(k_i+l_i-1)}\bigg|_{\mathfrak{d}\mapsto\mathfrak{d}^{-1}}. \end{align} $$

The monomial in (4.4) is color-symmetric, so we can move it inside the symmetrization. We can break up the exponents appearing in (4.4) as follows:

(4.5) $$ \begin{align} (k_{i+1}+l_{i+1})(k_i+l_i)&=k_{i+1}k_{i}+l_{i+1}l_i+\left[k_{i+1}l_i+k_il_{i+1}\right] \end{align} $$

(4.6) $$ \begin{align} k_{i+1}+l_{i+1}+k_{i-1}+l_{i-1}-2(k_i+l_i-1)&=k_{i+1}+k_{i-1}-2(k_i-1)+[l_{i+1}+l_{i-1}-2l_i] \end{align} $$

(4.7) $$ \begin{align} &=l_{i+1}+l_{i-1}-2(l_i-1)+[k_{i+1}+k_{i-1}-2k_i]. \end{align} $$

In (4.5), we will assign the bracketed summand to the mixing terms, $k_{i+1}k_{i}$ to F, and $l_{i+1}l_i$ to G. In a given summand of the symmetrization, if $z_{i,r}$ is assigned to F, then in (4.6), we assign the bracketed summand to the mixing terms and the rest to F. However, if $z_{i,r}$ is assigned to G, then in (4.7), we assign the bracketed summand to the mixing terms and the rest to G. Then, applying (4.3), we do indeed obtain

$$\begin{align*}\eta_{\mathcal{S}}(G)\star\eta_{\mathcal{S}}(F). \end{align*}$$

The case where $F,G\in \mathcal {S}^-$ is similar.

4.1.4 Shuffle presentation of horizontal Heisenberg elements

Recall the vertical Heisenberg elements (3.21) whose actions on $\mathcal {F}(v)$ are related to infinite-variable Macdonald operators. Previous work [Reference Wen19] gives us a better understanding of the action of $\varsigma ^{-1}$ on such elements. However, we need $\varsigma $ instead, and thus, we will apply the identity $\varsigma =\eta \varsigma ^{-1}\eta $ (cf. Theorem 3.1) and Proposition 4.3. To that end, recall the elements $\{\hat {h}_{p,\pm k}\}$ from (3.19) and (3.20). Observe that

(4.8) $$ \begin{align} \begin{aligned} &\varsigma\exp\left[-\sum_{k>0}\hat{h}_{p,\pm k}\frac{\mathfrak{q}^{\pm k} (-z)^{\mp k}}{[ k]_{\mathfrak{q}}} \right] \\ &=\varsigma\exp\left[-\sum_{k>0}\left(\frac{\sum_{i=0}^{r-1}t^{\pm k(i+1)}h_{p-i,\pm k}}{(1-t^{\pm kr})}\right)\frac{\mathfrak{q}^{\pm k} (-z)^{\mp k}}{[ k]_{\mathfrak{q}}} \right] \\ &=\eta\exp\left[\sum_{k>0}\left(\frac{\sum_{i=0}^{r-1}q^{\pm k(i+1)}\varsigma^{-1}(h_{p-i,\mp k})}{(1-q^{\pm kr})}\right)\frac{\mathfrak{q}^{\pm k} (-z)^{\mp k}}{[ k]_{\mathfrak{q}}} \right]\\ &= \eta\exp\left[(\mathfrak{q}-\mathfrak{q}^{-1})^{-1}\sum_{k>0}\left(\varsigma^{-1}(h_{p,\mp k}^{\perp})-t^{\pm k}\varsigma^{-1}(h_{p+1,\mp k}^{\perp})\right)\frac{q^{\pm k} (-z)^{\mp k}}{k} \right], \end{aligned} \end{align} $$

where in the last line, we use (3.3). Let $\delta =(1,\ldots , 1)\in \left( \mathbb {Z}_{\ge 0} \right)^I$ be the diagonal vector and consider the elements $\mathcal {E}_{p,n}^{\pm }\in \mathcal {S}^{\pm }$ given by

(4.9) $$ \begin{align} \begin{aligned} \mathcal{E}_{p,n}^+&:= \mathrm{Sym}_{n\delta}\left( \prod_{1\le a<b\le n}\left\{\frac{z_{p+1,a}-q^{-1}z_{p,b}}{z_{p+1,a}-tz_{p,b}}\prod_{i,j\in I}\omega_{i,j}\left( z_{i,a},z_{j,b} \right)\right\}\right.\\ &\ \quad\times \left.\prod_{a=1}^n\left\{\left( \frac{z_{0,a}}{z_{p,a}}-q^{-1}\frac{z_{0,a}}{z_{p+1,a}} \right)\prod_{i\in I}z_{i,a} \right\} \right)\\ \mathcal{E}_{p,n}^-&:= \mathrm{Sym}_{n\delta}\left( \prod_{1\le a<b\le n}\left\{\frac{z_{p+1,a}-q^{-1}z_{p,b}}{z_{p+1,a}-tz_{p,b}}\prod_{i,j\in I}\omega_{i,j}\left( z_{i,a},z_{j,b} \right)\right\}\right.\\ &\ \quad\times \left.\prod_{a=1}^n\left\{\left( q\frac{z_{p+1,a}}{z_{0,a}}-\frac{z_{p,a}}{z_{0,a}} \right)\prod_{i\in I}z_{i,a} \right\} \right). \end{aligned} \end{align} $$

By [Reference Wen19, Proposition 4.22], $\mathcal {E}_{p,n}^{\pm }\in \mathcal {S}^{\pm }$ .

Lemma 4.4. We have

$$ \begin{align*} \sum_{n=0}^\infty \frac{(-1)^{n}\mathfrak{q}^{n(r-1)}t^{-n}(1-q^{-1}t^{-1})^{nr}}{v^{-n}\prod_{a=1}^n(1-q^{-a}t^{-a})}\Psi_+\left( \mathcal{E}_{p,n}^+ \right)z^{-n} &= \varsigma\exp\left[-\sum_{k>0}\hat{h}_{p,-k}\frac{\mathfrak{q}^{-k} (-z)^{-k}}{v^{- k}[ k]_{\mathfrak{q}}} \right] \\ \sum_{n=0}^\infty \frac{(-1)^{nr-n}\mathfrak{d}^{-n(r-1)}t^n(1-qt)^{nr}}{v^{n}q^n\prod_{a=1}^n(1-q^{-a}t^{-a})}\Psi_-\left( \mathcal{E}_{p,n}^- \right)z^n &= \varsigma\exp\left[-\sum_{k>0}\hat{h}_{p,k}\frac{\mathfrak{q}^{ k} (-z)^{ k}}{v^{ k}[ k]_{\mathfrak{q}}} \right]. \end{align*} $$

Proof. In [Reference Wen19], it was shown that

$$ \begin{align*} &\exp\left[(\mathfrak{q}-\mathfrak{q}^{-1})^{-1}\sum_{k>0}\left(\varsigma^{-1}(h_{p, k}^{\perp})-t^{-k}\varsigma^{-1}(h_{p+1, k}^{\perp})\right)\frac{q^{-k} (-z)^{ k}}{k} \right]\\ &= \sum_{n=0}^\infty\frac{(-1)^{nr}(-q)^{-n}t^{nr}(1-q^{-1}t^{-1})^{nr}}{\mathfrak{q}^n\prod_{a=1}^n(1-q^{-a}t^{-a})}\Psi_+(\mathcal{H}_{p,n}^+)z^n \end{align*} $$

and

$$ \begin{align*} &\exp\left[(\mathfrak{q}-\mathfrak{q}^{-1})^{-1}\sum_{k>0}\left(\varsigma^{-1}(h_{p,- k}^{\perp})-t^{ k}\varsigma^{-1}(h_{p+1,- k}^{\perp})\right)\frac{q^{ k} (-z)^{- k}}{k} \right]\\ &= \sum_{n=0}^\infty\frac{(-q)^n(1-qt)^{nr}}{\mathfrak{q}^n\prod_{a=1}^n(1-q^{-a}t^{-a})}\Psi_-(\mathcal{H}_{p,n}^-)z^{-n}, \end{align*} $$

where

(4.10) $$ \begin{align} \begin{aligned} \mathcal{H}^+_{p,n}&= \mathrm{Sym}_{n\delta}\left( \prod_{1\le a<b\le n}\left\{\frac{t^{-1}z_{p+1,b}-z_{p,a}}{qz_{p+1,b}-z_{p,a}}\prod_{i,j\in I}\omega_{i,j}\left( z_{i,a},z_{j,b} \right)\right\}\right.\\ &\ \quad\times \left.\prod_{a=1}^n\left\{\left( \frac{z_{p,a}}{z_{0,a}}-t^{-1}\frac{z_{p+1,a}}{z_{0,a}} \right)\prod_{i\in I}z_{i,a} \right\} \right)\\ \mathcal{H}^-_{p,n}&= \mathrm{Sym}_{n\delta}\left( \prod_{1\le a<b\le n}\left\{\frac{t^{-1}z_{p+1,b}-z_{p,a}}{qz_{p+1,b}-z_{p,a}}\prod_{i,j\in I}\omega_{i,j}\left( z_{i,a},z_{j,b} \right)\right\}\right.\\ &\ \quad\times \left.\prod_{a=1}^n\left\{\left( t\frac{z_{0,a}}{z_{p+1,a}}-\frac{z_{0,a}}{z_{p,a}} \right)\prod_{i\in I}z_{i,a} \right\} \right). \end{aligned} \end{align} $$

It is helpful to recall Remark 3.3 when making comparisons with [Reference Wen19]. The result follows from applying Proposition 4.3 to (4.8). We note that the mixing terms in 4.10 contribute a power of $\mathfrak {d}^{-r n(n-1)}$ before inverting $\mathfrak {d}$ .

4.2.1 Matrix elements

The following is a consequence of computations similar to those done for Proposition 3.7:

Proposition 4.6. For $v_1,v_2\in W$ , we have

(4.11) $$ \begin{align} \begin{aligned} \left\langle v_1\left| \overset{\curvearrowright}{\prod_{i=0}^{r-1}}\overset{\curvearrowright}{\prod_{a=1}^{k_i}} E_{i}(z_{i,a})\right|v_2\right\rangle &=\frac{f\left( \left\{ z_{i,a} \right\}_{i\in I}^{1\le a\le k_i} \right) \displaystyle\prod_{i\in I}\,\prod_{1\le a<b\le k_i}\left( z_{i,a}-z_{i,b} \right)\left( z_{i,a}-q^{-1}t^{-1}z_{i,b} \right)} {\displaystyle \prod_{\substack{1\le a\le k_0\\1\le b\le k_{r-1}}}\left( z_{0,a}-t^{-1}z_{r-1,b} \right) \prod_{i\in I\backslash\{r-1\}}\prod_{\substack{1\le a \le k_i\\1\le b\le k_{i+1}}}\left(z_{i,a}-q^{-1}z_{i+1,b}\right)} \end{aligned} \end{align} $$

for some Laurent polynomial f, where the rational functions are expanded into Laurent series assuming

(4.12) $$ \begin{align} |z_{i,a}|=1,\, |q|>1,\, |t|>1. \end{align} $$

However,

(4.13) $$ \begin{align} \begin{aligned} \left\langle v_1\left| \overset{\curvearrowleft}{\prod_{i=0}^{r-1}}\overset{\curvearrowleft}{\prod_{a=1}^{k_i}} F_{i}(z_{i,a})\right|v_2\right\rangle &=\frac{g\left( \left\{ z_{i,a} \right\}_{i\in I}^{1\le a\le k_i} \right) \displaystyle\prod_{i\in I}\,\prod_{1\le a<b\le k_i}\left( z_{i,b}-z_{i,a} \right)\left( z_{i,b}-qtz_{i,a} \right)} {\displaystyle \prod_{\substack{1\le a\le k_{r-1}\\1\le b\le k_{0}}}\left( z_{r-1,b}-tz_{0,a} \right) \prod_{i\in I\backslash\{0\}}\prod_{\substack{1\le b \le k_i\\1\le a\le k_{i-1}}}\left(z_{i,b}-qz_{i-1,a}\right)} \end{aligned} \end{align} $$

for some Laurent polynomial g, where the rational functions are now expanded into Laurent series assuming

(4.14) $$ \begin{align} |z_{i,a}|=1,\, |q|<1,\, |t|<1. \end{align} $$

Notice that $\omega _{i,i+1}(z_{i,a},z_{i+1,b})^{-1}$ and $\omega _{i,i-1}(z_{i,a},z_{i-1,b})^{-1}$ are rational functions that we can also expand according to (4.12) and (4.14). Thus, we can make sense of matrix elements of products of currents multiplied by these inverted mixing terms. We do not claim that such products yield well-defined series of operators – just that their matrix elements make sense. The following is a consequence of the toroidal relations:

Proposition 4.7. When computing matrix elements, we have the relations

(4.15) $$ \begin{align} \frac{E_{i}(z)E_i(w)}{\omega_{i,i}(z,w)}&=\frac{E_{i}(w)E_i(z)}{\omega_{i,i}(w,z)} \end{align} $$

(4.16) $$ \begin{align} \frac{E_{i}(z)E_{i+1}(w)}{\omega_{i,i+1}(z,w)}&= \frac{E_{i+1}(w)E_{i}(z)}{\omega_{i+1,i}(w,z)}\\ \frac{F_{i}(z)F_i(w)}{\omega_{i,i}(w,z)}&=\frac{E_{i}(w)E_i(z)}{\omega_{i,i}(z,w)}\nonumber\\ \nonumber \frac{F_{i}(z)F_{i+1}(w)}{\omega_{i+1,i}(w,z)}&= \frac{F_{i+1}(w)F_{i}(z)}{\omega_{i,i+1}(z,w)}. \end{align} $$

Proof. We will only prove the statements for $E_i(z)$ . Applying $\rho _{\vec {c}}^+$ to the relations from 3.1.1 yields

$$\begin{align*}E_i(z)E_j(w)=g_{i,j}(\mathfrak{d}^{m_{i,j}}z/w)E_j(w)E_i(z). \end{align*}$$

Strictly speaking, when unpacking this relation, we should clear denominators. We then obtain

(4.17) $$ \begin{align} \left(\frac{z}{w}-\mathfrak{q}^2\right)E_i(z)E_i(w)&=\left(\mathfrak{q}^2\frac{z}{w}-1\right) E_i(w)E_i(z) \end{align} $$

(4.18) $$ \begin{align} \left(\mathfrak{d}^{-1}\frac{z}{w}-\mathfrak{q}^{-1}\right)E_{i}(z)E_{i+1}(w)&=\left(\mathfrak{q}^{-1}\mathfrak{d}^{-1}\frac{z}{w}-1\right)E_{i+1}(w)E_i(z). \end{align} $$

Since $\omega _{i,i}(z,w)^{-1}=(z-\mathfrak {q}^2w)(z-w)$ , (4.17) directly yields (4.15). However, multiplying both sides of (4.18) by $-\mathfrak {q} w$ gives us

$$\begin{align*}\omega_{i+1,i}(w,z)E_i(z)E_{i+1}(w)=\omega_{i,i+1}(z,w)E_{i+1}(w)E_i(z). \end{align*}$$

This implies (4.16).

4.2.2 Constant term formula

For $F\in \mathcal {S}^+_{k_{\bullet }}$ and $G\in \mathcal {S}^-_{k_{\bullet }}$ , consider the rational functions

$$ \begin{align*} &\frac{\displaystyle F\times\left\langle v_1\left| \overset{\curvearrowright}{\prod_{i=0}^{r-1}}\overset{\curvearrowright}{\prod_{a=1}^{k_i}} E_{i}(z_{i,a})\right|v_2\right\rangle} {\displaystyle\left(\prod_{i\in I}\prod_{1\le a<a'\le k_i}\omega_{i,i}(z_{i,a},z_{i,a'})\right) \left(\prod_{0\le i< j\le r-1}\,\prod_{\substack{1\le a\le k_i\\1\le b\le k_j}}\omega_{i,j}(z_{i,a},z_{j,b})\right)}\\ &\frac{\displaystyle G\times\left\langle v_1\left| \overset{\curvearrowleft}{\prod_{i=0}^{r-1}}\overset{\curvearrowleft}{\prod_{a=1}^{k_i}} F_{i}(z_{i,a})\right|v_2\right\rangle} {\displaystyle\left(\prod_{i\in I}\prod_{1\le a<a'\le k_i}\omega_{i,i}(z_{i,a},z_{i,a'})\right) \left(\prod_{0\le i< j\le r-1}\,\prod_{\substack{1\le a\le k_i\\1\le b\le k_j}}\omega_{i,j}(z_{i,a},z_{j,b})\right)}. \end{align*} $$

We can expand these rational functions into Laurent series according to the assumptions (4.12) and (4.14), respectively. For any Laurent series, we denote by $\{-\}_0$ this operation of taking constant terms.

Lemma 4.8. For $F\in \mathcal {S}^+_{k_{\bullet }}$ and $G\in \mathcal {S}^-_{k_{\bullet }}$ , we have

(4.19) $$ \begin{align} \rho^+_{\vec{c}}(\Psi_+(F)) =\frac{1}{k_{\bullet}!}\left\{\frac{\displaystyle F\times\overset{\curvearrowright}{\prod_{i=0}^{r-1}}\,\overset{\curvearrowright}{\prod_{a=1}^{k_i}}E_i(z_{i,a})} {\displaystyle\left(\prod_{i\in I}\prod_{1\le a<a'\le k_i}\omega_{i,i}(z_{i,a},z_{i,a'})\right) \left(\prod_{0\le i< j\le r-1}\,\prod_{\substack{1\le a\le k_i\\1\le b\le k_j}}\omega_{i,j}(z_{i,a},z_{j,b})\right)}\right\}_0 \end{align} $$

where the right-hand side is expanded according to (4.12) and

(4.20) $$ \begin{align} \rho^-_{\vec{c}}(\Psi_-(G)) = \frac{1}{k_{\bullet}!}\left\{\frac{\displaystyle G\times\overset{\curvearrowleft}{\prod_{i=0}^{r-1}}\,\overset{\curvearrowleft}{\prod_{a=1}^{k_i}}F_i(z_{i,a})} {\displaystyle\left(\prod_{i\in I}\prod_{1\le a<a'\le k_i}\omega_{i,i}(z_{i,a},z_{i,a'})\right) \left(\prod_{0\le i< j\le r-1}\,\prod_{\substack{1\le a\le k_i\\1\le b\le k_j}}\omega_{i,j}(z_{i,a},z_{j,b})\right)}\right\}_0, \end{align} $$

where the right-hand side is expanded according to (4.14). In particular, the expressions on the right-hand side are well-defined operators on W.

Proof. A consequence of Theorem 4.2 and the toroidal relations is that $\mathcal {S}^{\pm }$ are both spanned by shuffle monomials

$$\begin{align*}z_{0,1}^{n(0,1)}\star z_{0,1}^{n(0,2)}\star\cdots\star z_{0,1}^{n(0,k_0)}\star z_{1,1}^{n(1,1)}\star\cdots\star z_{r-1,1}^{n(r-1,k_{r-1})} \end{align*}$$

since

$$ \begin{align*} \Psi_+\left( z_{0,1}^{n(0,1)}\star\cdots\star z_{r-1,1}^{n(r-1,k_{r-1})} \right)&= e_{0,n(0,1)}\cdots e_{r-1,n(r-1,k_{r-1})}\\ \Psi_-\left(z_{0,1}^{n(0,1)}\star\cdots\star z_{r-1,1}^{n(r-1,k_{r-1})} \right)&= f_{0,n(0,1)}\cdots f_{r-1,n(r-1,k_{r-1})}. \end{align*} $$

We will check that the matrix elements coincide for these monomials, from which the lemma follows. For the ‘ $+$ ’ case, the proposed formula gives us

$$ \begin{align*} &\frac{1}{k_{\bullet}!}\left\{ \displaystyle\mathrm{Sym}_{k_{\bullet}}\left(\left\{\prod_{i\in I}\prod_{a=1}^{k_i}z_{i,a}^{n(i,a)} \,\prod_{1\le a<a'\le k_i}\omega_{i,i}(z_{i,a},z_{i,a'})\right\} \left\{ \prod_{0\le i<j\le r-1}\,\prod_{\substack{1\le a\le k_i\\1\le b\le k_j}}\omega_{i,j}(z_{i,a},z_{j,b}) \right\}\right)\right.\\ &\times\left.\frac{\displaystyle\left\langle v_1\left| \overset{\curvearrowright}{\prod_{i=0}^{r-1}}\overset{\curvearrowright}{\prod_{a=1}^{k_i}} E_{i}(z_{i,a})\right|v_2\right\rangle} {\displaystyle\left(\prod_{i\in I}\prod_{1\le a<a'\le k_i}\omega_{i,i}(z_{i,a},z_{i,a'})\right) \left(\prod_{0\le i< j\le r-1}\,\prod_{\substack{1\le a\le k_i\\1\le b\le k_j}}\omega_{i,j}(z_{i,a},z_{j,b})\right)}\right\}_0. \end{align*} $$

Using (4.15) to swap variables, we can move both the matrix element and the mixing terms inside the symmetrization, where the mixing terms will all cancel out. Notice that taking the constant term is insensitive to the labeling of the variables, and thus, the constant terms of all the summands of the symmetrization are equal. The end result is

$$\begin{align*}\left\{\prod_{i\in I}\prod_{a=1}^{k_i}z_{i,a}^{n(i,a)} \left\langle v_1\left| \overset{\curvearrowright}{\prod_{i=0}^{r-1}}\overset{\curvearrowright}{\prod_{a=1}^{k_i}} E_{i}(z_{i,a})\right|v_2\right\rangle\right\}_0 =\left\langle v_1\left|\rho_{\vec{c}}^{+}\left(e_{0,n(0,1)}\cdots e_{r-1,n(r-1,k_{r-1})} \right)\right|v_2\right\rangle. \end{align*}$$

The ‘ $-$ ’ case is similar.

5.1.1 Finitized vertex operators

Recall that $\Lambda ^I_{N_{\bullet }}$ denotes the tensor product over $i\in I$ of rings of symmetric polynomials in $N_i$ variables and $\pi _{N_{\bullet }}: \Lambda ^I\to \Lambda ^I_{N_{\bullet }}$ is the natural projection. We will abuse notation and also denote the map $(\pi _{N_{\bullet }}\otimes 1):\Lambda ^I\otimes \mathbb {K}\{Q\}\rightarrow \Lambda ^I_{N_{\bullet }}\otimes \mathbb {K}\{Q\}$ by $\pi _{N_{\bullet }}$ . Recall Proposition 3.7.

Remark 5.1. The action (3.7) of the operator $z^{H_i,0}$ includes a power of $\mathfrak {d}$ . In Proposition 3.7 and throughout this paper, we will be working with products of currents that have an equal number of $E_i(z)$ for each $i\in I$ and likewise for $F_i(z)$ . In this setup, the powers of $\mathfrak {d}$ will cancel. Namely, because $m_{i,i\pm 1}=-m_{i\pm 2,i\pm 1}$ , we have that the power of $\mathfrak {d}$ from the action of $z_{i,a}^{H_{i,0}}$ will be canceled by those from the action of $z_{i\pm 2,a}^{H_{i\pm 2,0}}$ . Thus, we we will abuse notation and omit the $\mathfrak {d}$ from the action of $z^{H_{i,0}}$ . Applying the compatibility condition (2.9), this leaves

$$\begin{align*}z^{H_{i,0}}(e^\alpha)=z^{( \alpha_i^\vee,\alpha)}=z^{N_{i}-N_{i-1}}. \end{align*}$$

Proposition 5.2. Let $f\in \Lambda ^I$ be factored according to color:

$$\begin{align*}f= \prod_{i\in I} f_i[X^{(i)}], \end{align*}$$

where $f_i\in \Lambda $ for all $i\in I$ . For

(5.1) $$ \begin{align} |z|=1,\, |q|>1,\, |t|>1,\, |x_{l}^{(j)}|<1, \end{align} $$

the vertex operators from (3.13) act on f such that upon finitization, we have

(5.2) $$ \begin{align} \begin{aligned} &\pi_{N_{\bullet}} \left(\exp\left[ \sum_{k>0}\left( p_k[X^{(i)}]-t^{-k}p_k[X^{(i-1)}] \right) \frac{z^k}{k}\right]\right.\\ &\times\left.\exp\left[\sum_{k>0}\left( -p_k[X^{(i)}]^{\perp}+q^{-k}p_k[X^{(i-1)}]^{\perp} \right)\frac{z^{-k}}{k} \right]z^{H_{i,0}} (f\otimes e^\alpha)\right) \\ &= \frac{\displaystyle\prod_{l=1}^{N_{i-1}}\left(z^{-1}-t^{-1}x_{l}^{(i-1)}\right)}{\displaystyle\prod_{l=1}^{N_i} \left(z^{-1}-x_{l}^{(i)}\right)} \left( f_i\left[x_{\bullet}^{(i)}-z^{-1}\right]f_{i-1}\left[x_{\bullet}^{(i-1)}+q^{-1}z^{-1}\right] \prod_{\substack{j\in I\\j\not=i,i-1}} f_j\left[x_{\bullet}^{(j)}\right]\otimes e^\alpha\right). \end{aligned} \end{align} $$

However, for

$$ \begin{align*} |z|=1,\, |q|<1,\, |t|<1,\, |x_{l}^{(j)}|<1, \end{align*} $$

the vertex operators from (3.15) act as

(5.3) $$ \begin{align} \begin{aligned} &\pi_{N_{\bullet}} \left(\exp\left[ \sum_{k>0}\left( -t^kp_k[X^{(i)}]+p_k[X^{(i-1)}] \right) \frac{z^k}{k}\right]\right.\\ &\times \left.\exp\left[\sum_{k>0}\left( q^kp_k[X^{(i)}]^{\perp}-p_k[X^{(i-1)}]^{\perp} \right)\frac{z^{-k}}{k} \right]z^{-H_{i,0}} (f\otimes e^\alpha)\right) \\ &= \frac{\displaystyle\prod_{l=1}^{N_{i}}\left(z^{-1}-tx_{l}^{(i)}\right)} {\displaystyle\prod_{l=1}^{N_{i-1}} \left(z^{-1}-x_{l}^{(i-1)}\right)} \left( f_i\left[x_{\bullet}^{(i)}+qz^{-1}\right]f_{i-1}\left[x_{\bullet}^{(i-1)}-z^{-1}\right] \prod_{\substack{j\in I\\j\not=i,i-1}} f_j\left[x_{\bullet}^{(j)}\right]\otimes e^\alpha\right). \end{aligned} \end{align} $$

Proof. We will only consider (5.2) – the proof for (5.3) is similar. First consider the ‘left’ half of the vertex operator together with $z^{H_i,0}$ . We have

(5.4) $$ \begin{align} \nonumber &\pi_{N_{\bullet}} \left(\exp\left[ \sum_{k>0}\left( p_k[X^{(i)}]-t^{-k}p_k[X^{(i-1)}] \right) \frac{z^k}{k}\right]z^{H_{i,0}} (f\otimes e^\alpha)\right) \\ &= \frac{\displaystyle\prod_{l=1}^{N_{i-1}}\left(1-t^{-1}zx_{l}^{(i-1)}\right)}{\displaystyle\prod_{l=1}^{N_i} \left(1-zx_{l}^{(i)}\right)} z^{N_i-N_{i-1}} \left( \pi_{N_{\bullet}}(f)\otimes e^\alpha\right) = \frac{\displaystyle\prod_{l=1}^{N_{i-1}}\left(z^{-1}-t^{-1}x_{l}^{(i-1)}\right)}{\displaystyle\prod_{l=1}^{N_i} \left(z^{-1}-x_{l}^{(i)}\right)} \left( \pi_{N_{\bullet}}(f)\otimes e^\alpha\right). \end{align} $$

For this to hold, we will need to impose conditions on $|x_{l}^{(i)}|$ . Recall that we have the conditions (3.14) when working with $\{E_i(z)\}$ . We extend these to (5.1) for (5.4) to hold. Let us also point out that the compatibility condition (2.9) is used to obtain the factor $z^{N_i-N_{i-1}}$ after the first equality.

Next, from the ‘right’ half, we have

(5.5) $$ \begin{align} \nonumber &\pi_{N_{\bullet}}\left( \exp\left[\sum_{k>0}\left( -p_k[X^{(i)}]^{\perp}+q^{-k}p_k[X^{(i-1)}]^{\perp} \right)\frac{z^{-k}}{k} \right]\cdot f \right)\\ &= \pi_{N_{\bullet}}\left(f_i[X^{(i)}-z^{-1}]f_{i-1}[X^{(i-1)}+q^{-1}z^{-1}]\prod_{\substack{j\in I\\j\not=i,i-1}} f_j[X^{(j)}] \right)\\ \nonumber &= f_i\left[x_{\bullet}^{(i)}-z^{-1}\right]f_{i-1}\left[x_{\bullet}^{(i-1)}+q^{-1}z^{-1}\right] \prod_{\substack{j\in I\\j\not=i,i-1}} f_j\left[x_{\bullet}^{(j)}\right].\notag \end{align} $$

Here, (5.5) follows from checking on power sums $p_k[X^{(i)}]$ and $p_k[X^{(i-1)}]$ .

Remark 5.3. As in Proposition 5.2, when writing formulas involving vertex operators, we will express them in terms of functions that are factorizable according to color:

$$\begin{align*}f=\prod_{i\in I}f_i[X^{(i)}]. \end{align*}$$

Factorizable functions span $\Lambda ^I$ , so to define an operator, it suffices to consider its action on such functions. We can write our operators in terms of general f if we introduce colored plethystic notation. For instance, the terms at the bottom of (5.2) can be written as

$$ \begin{align*} &f_i\left[x_{\bullet}^{(i)}-z^{-1}\right]f_{i-1}\left[x_{\bullet}^{(i-1)}+q^{-1}z^{-1}\right] \prod_{\substack{j\in I\\j\not=i,i-1}} f_j\left[x_{\bullet}^{(j)}\right]\\ &=f\left[\left(x_{\bullet}^{(i)}-z^{-1}\right)+\left(x_{\bullet}^{(i-1)}+q^{-1}z^{-1}\right)+\sum_{\substack{j\in I\\j\not=i,i-1}}x_{\bullet}^{(j)}\right], \end{align*} $$

where the bottom denotes the image of $f[X_{N_{\bullet }}]$ under the ring map generated by

$$\begin{align*}p_n[x_{\bullet}^{(j)}]\mapsto \begin{cases} p_n[x_{\bullet}^{(i)}]-z^{-n} & j=i\\ p_n[x_{\bullet}^{(i-1)}]+q^{-n}z^{-n} & j=i-1\\ p_n[x_{\bullet}^{(j)}] & \mbox{otherwise} \end{cases}. \end{align*}$$

This notation can then be carried over to general f. However, the benefits of introducing this notation in our paper seemed marginal at best, so we have elected to making statements in terms of factorizable functions.

5.1.2 Applying the constant term formula

Our next goal is to obtain constant term formulas for the action of the shuffle elements $\mathcal {E}_{p,n}^{\pm }$ from (4.9). In light of Lemma 4.4, we will also incorporate the constants

$$ \begin{align*} c_{n}^+&:=\frac{(-1)^{n}\mathfrak{q}^{n(r-1)}t^{-n}(1-q^{-1}t^{-1})^{nr}}{v^{-n}\prod_{a=1}^n(1-q^{-a}t^{-a})},\quad c_{n}^-:=\frac{(-1)^{nr-n}\mathfrak{d}^{-n(r-1)}t^n(1-qt)^{nr}}{v^{n}q^n\prod_{a=1}^n(1-q^{-a}t^{-a})}, \end{align*} $$

where $v=(-1)^{\frac {(r-2)(r-3)}{2}}\mathfrak {q}\mathfrak {d}^{-\frac {r}{2}}(c_0\cdots c_{r-1})^{-1}$ .

Lemma 5.4. For any factorizable $f=\prod _{i\in I}f_i[X^{(i)}]\in \Lambda ^I$ , we have

(5.6) $$ \begin{align} \begin{aligned} &c_n^+ \pi_{N_{\bullet}}\bigg( (\rho_{\vec{c}}^+\circ\Psi_+)(\mathcal{E}_{p,n}^+)(f\otimes e^\alpha)\bigg)\\ &=\frac{(1-q^{-1}t^{-1})^{nr}}{t^{n}\prod_{a=1}^n(1-q^{-a}t^{-a})} \left\{\prod_{i\in I}\prod_{a=1}^n\prod_{l=1}^{N_i}\left(\frac{ z_{i+1,a}^{-1}-t^{-1}x_{l}^{(i)} }{ z_{i,a}^{-1}-x_{l}^{(i)} }\right)\right.\\ &\times\prod_{1\le a<b\le n} \left[ \frac{\left(1-z_{p+1,b}/z_{p+1,a}\right)\left(1-q^{-1}t^{-1}z_{p+1,b}/z_{p+1,a}\right)}{\left(1-tz_{p,b}/z_{p+1,a}\right)\left(1-t^{-1}z_{p+2,b}/z_{p+1,a}\right)}\right.\\ &\times\left.\prod_{i\in I\backslash\{p+1\}} \frac{\left(1-z_{i,b}/z_{i,a}\right)\left(1-q^{-1}t^{-1}z_{i,b}/z_{i,a}\right)}{\left(1-q^{-1}z_{i-1,b}/z_{i,a}\right)\left(1-t^{-1}z_{i+1,b}/z_{i,a}\right)}\right]\\ &\times\prod_{a=1}^n \left[ \left(\frac{z_{0,a}}{z_{p,a}}\right)\left(\frac{1}{1-t^{-1}z_{p+1,a}/z_{p,a}}\right)\right.\\ &\times\left. \prod_{i\in I\setminus\{p+1\}} \left(\frac{1}{1-t^{-1}z_{i,a}/z_{i-1,a}}\right) \left( \frac{1}{1-q^{-1}z_{i-1,a}/z_{i,a}} \right)\right]\\ &\left.\times \prod_{i\in I} f_i\left[ \sum_{l=1}^{N_i} x_{l}^{(i)}-\sum_{a=1}^nz_{i,a}^{-1}+q^{-1}\sum_{a=1}^n z_{i+1,a}^{-1} \right] \right\}_0\otimes e^\alpha \end{aligned} \end{align} $$

and

(5.7) $$ \begin{align} \begin{aligned} &c_n^-\pi_{N_{\bullet}}\bigg((\rho_{\vec{c}}^-\circ\Psi_-)(\mathcal{E}_{p,n}^-)(f\otimes e^\alpha)\bigg)\\ &=\frac{t^n(1-qt)^{nr}}{\prod_{k=1}^n(1-q^kt^k)} \left\{\prod_{i\in I}\prod_{a=1}^n\prod_{l=1}^{N_i}\left(\frac{ z_{i,a}^{-1}-tx_{l}^{(i)} }{ z_{i+1,a}^{-1}-x_{l}^{(i)} }\right)\right.\\ &\times\prod_{1\le a<b\le n}\left[ \frac{\left(1-z_{p,a}/z_{p,b}\right)\left(1-qtz_{p,a}/z_{p,b}\right)}{\left(1-t^{-1}z_{p+1,a}/z_{p,b}\right)\left(1-tz_{p-1,a}/z_{p,b}\right)}\right.\\ &\times\left. \prod_{i\in I\backslash\{p\}} \frac{\left(1-z_{i,a}/z_{i,b}\right)\left(1-qtz_{i,a}/z_{i,b}\right)}{\left(1-qz_{i+1,a}/z_{i,b}\right)\left(1-tz_{i-1,a}/z_{i,b}\right)}\right]\\ &\times \prod_{a=1}^n\left[ \left(\frac{z_{p,a}}{z_{0,a}}\right)\left(\frac{1}{1-tz_{p,a}/z_{p+1,a}}\right)\right.\\ &\times\left. \prod_{i\in I\setminus\{p\}} \left(\frac{1}{1-tz_{i,a}/z_{i+1,a}}\right) \left( \frac{1}{1-qz_{i+1,a}/z_{i,a}} \right) \right]\\ &\left.\times \prod_{i\in I} f_i\left[ \sum_{l=1}^{N_i} x_{l}^{(i)}+q\sum_{a=1}^n z_{i,a}^{-1}-\sum_{a=1}^n z_{i+1,a}^{-1} \right]\right\}_0\otimes e^\alpha. \end{aligned} \end{align} $$

Proof. Plugging in $\mathcal {E}_{p,n}^{\pm }$ into the formula from Lemma 4.8, we can use the toroidal relations and Proposition 4.7 to reorder the currents in alignment with Proposition 3.7. As in the proof of Lemma 4.8, we can use the toroidal relations to remove the symmetrizations in $\mathcal {E}_{p,n}^{\pm }$ . Taking the result for $\mathcal {E}_{p,n}^+$ , acting on $f\otimes e^\alpha $ and then applying $\pi _{N_{\bullet }}$ gives us

$$ \begin{align*} \begin{aligned} &\pi_{N_{\bullet}}\bigg( (\rho_{\vec{c}}^+\circ\Psi_+)(\mathcal{E}_{p,n}^+)(f\otimes e^\alpha)\bigg)\\ &=\left( (-1)^{\frac{(r-2)(r-3)}{2}}\mathfrak{d}^{\frac{r}{2}-1}\prod_{i\in I} c_i\right)^n \left\{\prod_{i\in I}\prod_{a=1}^n\prod_{l=1}^{N_i}\left(\frac{ z_{i+1,a}^{-1}-t^{-1}x_{l}^{(i)} }{ z_{i,a}^{-1}-x_{l}^{(i)} }\right)\right.\\ &\times\prod_{1\le a<b\le n} \left[ \frac{1-q^{-1}z_{p,b}/z_{p+1,a}}{1-tz_{p,b}/z_{p+1,a}} \prod_{i\in I} \frac{\left(1-z_{i,b}/z_{i,a}\right)\left(1-q^{-1}t^{-1}z_{i,b}/z_{i,a}\right)}{\left(1-t^{-1}z_{i+1,b}/z_{i,a}\right)\left(1-q^{-1}z_{i-1,b}/z_{i,a}\right)} \right]\\ &\times\prod_{a=1}^n \left[\left(\frac{z_{0,a}}{z_{p+1,a}}\right)\left(\frac{z_{p+1,a}}{\omega_{p+1,p}(z_{p+1,a},z_{p,a})}\right)\right.\\ &\times\left.\prod_{i\in I\setminus\{p+1\}} \frac{z_{i,a}}{\left(1-t^{-1}z_{i,a}/z_{i-1,a}\right)\omega_{i-1,i}(z_{i-1,a},z_{i,a})}\right]\\ &\times \prod_{i\in I}f_i\left[ \sum_{l=1}^{N_i}x_{l}^{(i)}-\sum_{a=1}^nz_{i,a}^{-1}+q^{-1}\sum_{a=1}^n z_{i+1,a}^{-1} \right]\Bigg\}_0\otimes e^\alpha, \end{aligned} \end{align*} $$

where all rational functions are expanded as Laurent series assuming

(5.8) $$ \begin{align} |z_{i,a}|=1,\, |x_{l}^{(j)}|<1,\, |t|>1,\, |q|>1. \end{align} $$

For $\mathcal {E}_{p,n}^-$ , we instead have

$$ \begin{align*} \begin{aligned} &\pi_{N_{\bullet}}\bigg((\rho_{\vec{c}}^-\circ\Psi_-)(\mathcal{E}_{p,n}^-) (f\otimes e^\alpha)\bigg)\\ &= \left( \frac{(-1)^{\frac{(r-2)(r-3)}{2}}}{\displaystyle\mathfrak{d}^{\frac{r}{2}-1} \prod_{i\in I} c_i}\right)^n \left\{\prod_{i\in I}\prod_{a=1}^n\prod_{l=1}^{N_i}\left(\frac{ z_{i,a}^{-1}-tx_{l}^{(i)} }{ z_{i+1,a}^{-1}-x_{l}^{(i)} }\right)\right.\\ &\times\prod_{1\le a<b\le n} \left[ \frac{q^{-1}-z_{p+1,a}/z_{p,b}}{t-z_{p+1,a}/z_{p,b}} \prod_{i\in I} \frac{\left(1-z_{i,a}/z_{i,b}\right)\left(1-qtz_{i,a}/z_{i,b}\right)}{\left(1-tz_{i-1,a}/z_{i,b}\right)\left(1-qz_{i+1,a}/z_{i,b}\right)} \right]\\ &\times \prod_{a=1}^n \left[ \left(\frac{z_{p+1,a}}{z_{0,a}}\right) \left(\frac{-z_{p,a}} {\omega_{p+1,p}(z_{p+1,a},z_{p,a})}\right)\right.\\ &\left.\times\prod_{i\in I\setminus\{p\}} \frac{z_{i,a}}{\left(1-tz_{i,a}/z_{i+1,a}\right)\omega_{i,i+1}(z_{i,a},z_{i+1,a})} \right]\\ &\left.\times \prod_{i\in I}f_i\left[ \sum_{l=1}^{N_i} x_{l}^{(i)}+q\sum_{a=1}^n z_{i,a}^{-1}-\sum_{a=1}^n z_{i+1,a}^{-1} \right]\right\}_0\otimes e^\alpha, \end{aligned} \end{align*} $$

where all rational functions are expanded into Laurent series assuming.

(5.9) $$ \begin{align} |z_{i,a}|=1,\, |x_{l}^{(j)}|<1,\, |q|<1,\, |t|<1. \end{align} $$

In both formulas, we are taking constant terms in the z-variables.

Finally, to obtain (5.6) and (5.7) from these formulas, we multiply through by $c_n^{\pm }$ and use $q=\mathfrak {q}\mathfrak {d}, t=\mathfrak {q}\mathfrak {d}^{-1}$ to write

$$ \begin{align*} \omega_{i,i+1}(z,w)&= \mathfrak{q} w-\mathfrak{d}^{-1}z=\mathfrak{d}^{-1}(qw-z)=\mathfrak{q}\left(w-q^{-1}z\right)\\ \omega_{i,i-1}(z,w)&= z-\mathfrak{q}\mathfrak{d}^{-1}w=z-tw.\\[-38pt] \end{align*} $$

Remark 5.5. Observe that the formulas in Lemma 5.4 are for symmetric functions in finitely many variables. To obtain constant term formulas for operators in infinitely many variables, we can apply Proposition 5.2. For example, starting from (5.6), we use (5.2) and replace

$$\begin{align*}\prod_{i\in I}\prod_{a=1}^n\prod_{l=1}^{N_i}\left(\frac{ z_{i+1,a}^{-1}-t^{-1}x_{l}^{(i)} }{ z_{i,a}^{-1}-x_{l}^{(i)} }\right) f_i\left[ \sum_{l=1}^{N_i} x_{l}^{(i)}-\sum_{a=1}^nz_{i,a}^{-1}+q^{-1}\sum_{a=1}^n z_{i+1,a}^{-1} \right] \otimes e^\alpha \end{align*}$$

with

$$ \begin{align*} &\exp\left[ \sum_{i\in I}\sum_{a=1}^n\left(\sum_{k>0}\left( p_k[X^{(i)}]-t^{-k}p_k[X^{(i-1)}] \right) \frac{z_{i,a}^k}{k}\right)\right]\\ &\times\exp\left[\sum_{i\in I}\sum_{a=1}^n\sum_{k>0}\left( -p_k[X^{(i)}]^{\perp}+q^{-k}p_k[X^{(i-1)}]^{\perp} \right)\frac{z_{i,a}^{-k}}{k} \right]z^{H_{i,0}} (f\otimes e^\alpha). \end{align*} $$

5.1.3 Integral formula

Regardless of f, the formulas obtained in 5.1.2 are constant terms of Laurent series expansions of some rational function. Note that all poles are simple except for the poles at zero possibly coming from the plethystic modifications done to f. Thus, it will be advantageous to invert all the z-variables: let

$$\begin{align*}w_{i,a}:=z_{i,a}^{-1} \end{align*}$$

and define the functions

(5.10) $$ \begin{align} g_{p,n}^+(w_{\bullet,\bullet},X_{N_{\bullet}}) &:=\frac{(-1)^{\frac{n(n-1)}{2}}(1-q^{-1}t^{-1})^{nr}}{t^{\frac{n(n+1)}{2}}\prod_{a=1}^n(1-q^{-a}t^{-a})} \prod_{i\in I}\prod_{a=1}^n\prod_{l=1}^{N_i}\left(\frac{ w_{i+1,a}-t^{-1}x_{l}^{(i)} }{ w_{i,a}-x_{l}^{(i)} }\right) \end{align} $$

(5.11) $$ \begin{align} &\times\prod_{1\le a<b\le n} \left[ \frac{\left(w_{p+1,b}-w_{p+1,a}\right)\left(w_{p+1,b}-q^{-1}t^{-1}w_{p+1,a}\right)} {\left(w_{p+1,a}-t^{-1}w_{p,b}\right)\left(w_{p+2,b}-t^{-1}w_{p+1,a}\right)}\right. \end{align} $$

(5.12) $$ \begin{align} &\times\left. \prod_{i\in I\backslash\{p+1\}} \frac{\left(w_{i,b}-w_{i,a}\right)\left(w_{i,b}-q^{-1}t^{-1}w_{i,a}\right)}{\left(w_{i-1,b}-q^{-1}w_{i,a}\right)\left(w_{i+1,b}-t^{-1}w_{i,a}\right)}\right] \end{align} $$

(5.13) $$ \begin{align} &\times\prod_{a=1}^n \left[ \left(\frac{w_{p,a}}{w_{0,a}}\right)\left(\frac{w_{p+1,a}}{w_{p+1,a}-t^{-1}w_{p,a}}\right)\right. \end{align} $$

(5.14) $$ \begin{align} &\times\left. \prod_{i\in I\setminus\{p+1\}} \left(\frac{w_{i,a}}{w_{i,a}-t^{-1}w_{i-1,a}}\right) \left( \frac{w_{i-1,a}}{w_{i-1,a}-q^{-1}w_{i,a}} \right)\right] \end{align} $$

and

(5.15) $$ \begin{align} g_{p,n}^-(w_{\bullet,\bullet},X_{N_{\bullet}}) &:= \frac{(-1)^{\frac{n(n-1)}{2}}t^{\frac{n(n+1)}{2}}(1-qt)^{nr}}{\prod_{a=1}^n(1-q^at^a)} \prod_{i\in I}\prod_{a=1}^n\prod_{l=1}^{N_i}\left(\frac{ w_{i,a}-tx_{l}^{(i)} }{ w_{i+1,a}-x_{l}^{(i)} }\right) \end{align} $$

(5.16) $$ \begin{align} &\times\prod_{1\le a<b\le n}\left[ \frac{\left(w_{p,a}-w_{p,b}\right)\left(w_{p,a}-qtw_{p,b}\right)}{\left(w_{p,b}-tw_{p+1,a}\right)\left(w_{p-1,a}-tw_{p,b}\right)}\right. \end{align} $$

(5.17) $$ \begin{align} &\times\left. \prod_{i\in I\backslash\{p\}} \frac{\left(w_{i,a}-w_{i,b}\right)\left(w_{i,a}-qtw_{i,b}\right)}{\left(w_{i+1,a}-qw_{i,b}\right)\left(w_{i-1,a}-tw_{i,b}\right)}\right] \end{align} $$

(5.18) $$ \begin{align} &\times \prod_{a=1}^n\left[ \left(\frac{w_{0,a}}{w_{p,a}-tw_{p+1,a}}\right) \prod_{i\in I\setminus\{p\}} \left(\frac{w_{i,a}}{w_{i,a}-tw_{i+1,a}}\right) \left( \frac{w_{i+1,a}}{w_{i+1,a}-qw_{i,a}} \right) \right]. \end{align} $$

Lemma 5.6. Let $f=\prod _{i\in I} f_i[X^{(i)}]\in \Lambda ^I$ be factorizable according to color. For the ‘+’ case, assume

$$\begin{align*}|x_{l}^{(i)}|<1,\, |q|>1,\, |t|>1. \end{align*}$$

We have

(5.19) $$ \begin{align} \nonumber &c_n^+ \pi_{N_{\bullet}}\bigg( (\rho_{\vec{c}}^+\circ\Psi_+)(\mathcal{E}_{p,n}^+)(f\otimes e^\alpha)\bigg)\\ &=\left(\underset{|w_{i,a}|=1}{\oint\cdots\oint}g_{p,n}^+(w_{\bullet,\bullet},X_{N_{\bullet}}) \prod_{i\in I} f_i\left[ \sum_{l=1}^{N_i} x_{l}^{(i)}-\sum_{a=1}^nw_{i,a}+\sum_{a=1}^{n}q^{-1}w_{i+1,a} \right] \prod_{a=1}^n \frac{dw_{i,a}}{2\pi\sqrt{-1}w_{i,a}} \right)\otimes e^\alpha, \end{align} $$

where we orient the unit circle $|w_{i,a}|=1$ counter-clockwise. In the ‘ $-$ ’ case, we now assume

$$\begin{align*}|x_{l}^{(i)}|<1,\, |q|<1,\, |t|<1. \end{align*}$$

We then have

(5.20) $$ \begin{align} \nonumber &c_n^- \pi_{N_{\bullet}}\bigg( (\rho_{\vec{c}}^-\circ\Psi_-)(\mathcal{E}_{p,n}^-)(f\otimes e^\alpha)\bigg)\\ &=\left(\underset{|w_{i,a}|=1}{\oint\cdots\oint}g_{p,n}^-(w_{\bullet,\bullet},X_{N_{\bullet}}) \prod_{i\in I} f_i\left[ \sum_{l=1}^{N_i} x_{l}^{(i)}+\sum_{a=1}^nqw_{i,a}-\sum_{a=1}^nw_{i+1,a} \right] \prod_{a=1}^n \frac{dw_{i,a}}{2\pi\sqrt{-1}w_{i,a}} \right)\otimes e^\alpha \end{align} $$

and also orient the unit circle counter-clockwise.

Proof. Upon making the substitution $w_{i,a}:=z_{i,a}^{-1}$ , the right-hand side of (5.6) is equal to

(5.21) $$ \begin{align} \left\{g_{p,n}^+(w_{\bullet,\bullet},X_{N_{\bullet}})\prod_{i\in I}f_i\left[\sum_{l=1}^{N_i} x_{l}^{(i)}-\sum_{a=1}^nw_{i,a}+\sum_{a=1}^nq^{-1}w_{i+1,a} \right]\right\}_0\otimes e^\alpha. \end{align} $$

Now, all the poles appearing in (5.21) are simple. Similarly, the right-hand side of (5.7) becomes

(5.22) $$ \begin{align} \left\{g_{p,n}^-(w_{\bullet,\bullet},X_{N_{\bullet}})\prod_{i\in I}f_i\left[ \sum_{l=1}^{N_i}x_{l}^{(i)}+\sum_{a=1}^nqw_{i,a}- \sum_{a=1}^nw_{i+1,a} \right]\right\}_0\otimes e^\alpha. \end{align} $$

In case ‘ $\pm $ ’, the integrands are given by series in the $x^{(i)}_l$ and $q^{\mp 1}$ , $t^{\mp 1}$ , with coefficients which are Laurent polynomials in the $w_{i,a}$ . Under the given assumptions, these series converge uniformly absolutely on the integration cycle, and thus, we can exchange the order of summation and integration. This turns the integrals (5.19) and (5.20) into the constant term formulas (5.21) and (5.22), respectively.

5.1.4 Cyclic-shift operators

To describe the results of our computation, we need to introduce some difference operators that also permute variables. As before, let $X_{N_{\bullet }}=\{x_{l}^{(i)}\}_{i\in I}^{1\le l\le N_i}$ denote our set of variables compatible with our r-core via (2.9). Define a shift pattern of $X_{N_{\bullet }}$ to be a subset of $X_{N_{\bullet }}$ that contains no more than one variable of each color. A shift pattern contains color $p\in I$ if it contains a variable of color p. Let $Sh_p(X_{N_{\bullet }})$ denote the set of all shift patterns containing color p.

For a shift pattern ${\underline {J}}$ , let $J\subset I$ denote the colors of the variables in ${\underline {J}}$ . We denote the variables in ${\underline {J}}$ by $x_{{\underline {J}}}^{(i)}$ , so ${\underline {J}}=\{ x_{{\underline {J}}}^{(i)} \}_{i\in J}$ . To $\underline {J}$ we associate the following:

1. 1. Gap labels: For $i\in I$ , let $i^{\vartriangle }\in J$ be first element greater than or equal to i in the cyclic order. Similarly, let $i^{\triangledown }\in J$ be the first element less than or equal to i in the cyclic order. We stipulate that $0\le i^{\vartriangle }-i, i-i^{\triangledown }\le r-1$ . With this set, we define

   $$ \begin{align*} x_{{\underline{J}}^\vartriangle}^{(i)}&=q^{(i-i^{\vartriangle})}x_{{\underline{J}}}^{(i^\vartriangle)}\\ x_{{\underline{J}}^\triangledown}^{(i)}&=q^{(i-i^{\triangledown})}x_{{\underline{J}}}^{(i^\triangledown)}. \end{align*} $$

   To clarify, $x_{{\underline {J}}^\vartriangle }^{(i)}=x_{{\underline {J}}^\triangledown }^{(i)}=x_{{\underline {J}}}^{(i)}$ if $i\in J$ . Thus, while ${\underline {J}}$ gives a list of variables colored by $J\subset I$ , we ‘fill in the gaps’ for values $i\in I\backslash J$ with certain q-shifts of the elements of ${\underline {J}}$ . Note that the q-shifts are negative for $x_{{\underline {J}}^\vartriangle }^{(i)}$ and positive for $x_{{\underline {J}}^\triangledown }^{(i)}$ .

2. 2. A cyclic-shift operator: For $i\in J$ , let $i^{\blacktriangledown }\in J$ be the first element strictly less than i in the cyclic order. We set $1\le i-i^\blacktriangledown \le r$ , where r occurs if and only if $|J|=\{i\}$ . We then define the operator $T_{{\underline {J}}}$ on $\mathbb {K}[X_{N_{\bullet }}]$ as the algebra map induced by

   $$ \begin{align*} T_{{\underline{J}}}(x_{l}^{(i)})&= \left\{ \begin{array}{ll} q^{(i-i^\blacktriangledown)}x_{{\underline{J}}}^{(i^\blacktriangledown)} & \text{if } i\in J \text{ and } x_{l}=x_{{\underline{J}}}^{(i)} \\ x_{l}^{(i)}&\mbox{otherwise.} \end{array} \right. \end{align*} $$

   Note that this q-shift is positive. If we let $i^{\blacktriangle }\in J$ be the first element strictly greater than i in the cyclic order, then observe that

   $$ \begin{align*} T^{-1}_{{\underline{J}}}(x_{l}^{(i)})&= \left\{ \begin{array}{ll} q^{(i-i^\blacktriangle)}x_{{\underline{J}}}^{(i^\blacktriangle)} & \text{if } i\in J \text{ and } x_{l}^{(i)}=x_{{\underline{J}}}^{(i)} \\ x_{l}^{(i)}&\mbox{otherwise}, \end{array} \right. \end{align*} $$

   where as before, we view $1\le i^{\blacktriangle }-i\le r$ . Finally, we note the following: for $i\in J$ ,

   (5.23) $$ \begin{align} \begin{aligned} T_{\underline{J}}(x_{{\underline{J}}}^{(i)})&= qx_{{\underline{J}}^\triangledown}^{(i-1)}\\ T_{\underline{J}}^{-1}(x_{{\underline{J}}}^{(i)})&= q^{-1}x_{{\underline{J}}^\vartriangle}^{(i+1)}. \end{aligned} \end{align} $$

The cyclic-shift operators $T_{{\underline {J}}}^{\pm 1}$ will arise when evaluating the integrals of Lemma 5.6 by iterated residues. For later use, and to clarify this relationship, we record the following:

Lemma 5.8. For any $f=\prod _i f_i[X^{(i)}]\in \Lambda ^I$ , ${\underline {J}}\in Sh_p(X_{N_{\bullet }})$ , define the following evaluations on a set of auxilliary variables $\{w_i\}_{i\in I}$ :

$$ \begin{align*} \mathrm{ev}_{{\underline{J}}}^+ : \ \text{for } i=p,p+1,\dotsc,p-1 \text{ cyclic order,} \ w_i &\mapsto \begin{cases} x^{(i)}_{{\underline{J}}} & \text{if } i\in J\\ q^{-1}w_{i+1} & \text{if } i\in I\setminus J \end{cases}\\ \mathrm{ev}_{{\underline{J}}}^- : \ \text{for } i=p+1,p,\dotsc,p+2 \text{ in reverse cyclic order,} \ w_i &\mapsto \begin{cases} x^{(i-1)}_{{\underline{J}}} & \text{if } i-1\in J\\ qw_{i-1} & \text{if } i-1\in I\setminus J. \end{cases} \end{align*} $$

We then have

(5.24) $$ \begin{align} \begin{aligned} \mathrm{ev}_{{\underline{J}}}^+(w_i)&= x_{{\underline{J}}^\vartriangle}^{(i)},& \mathrm{ev}_{{\underline{J}}}^-(w_i)&= x_{{\underline{J}}^\triangledown}^{(i)} \end{aligned} \end{align} $$

and

(5.25) $$ \begin{align} T_{{\underline{J}}}^{-1}f\left[X_{N_{\bullet}}\right] &= \mathrm{ev}_{{\underline{J}}}^+\left( \prod_{i\in I} f_i\left[ \sum_{l=1}^{N_i} x_{l}^{(i)}-w_{i}+q^{-1}w_{i+1} \right] \right) \end{align} $$

(5.26) $$ \begin{align} T_{{\underline{J}}}f\left[X_{N_{\bullet}}\right] &= \mathrm{ev}_{{\underline{J}}}^-\left( \prod_{i\in I} f_i\left[ \sum_{l=1}^{N_i} x_{l}^{(i)}+qw_{i}-w_{i+1} \right] \right). \end{align} $$

Proof. The equations (5.24) follow from the definitions. Equipped with that, the right-hand side of (5.25) becomes

$$\begin{align*}\prod_{i\in I}f_i\left[\sum_{l=1}^{N_i} x_i^{(i)}-x_{{\underline{J}}^\vartriangle}^{(i)}+q^{-1}x_{{\underline{J}}^\vartriangle}^{(i+1)}\right] \end{align*}$$

If $i\in I\backslash J$ , then $x_{{\underline {J}}^\vartriangle }^{(i)}=q^{-1}x_{{\underline {J}}^\vartriangle }^{(i+1)}$ and so $f_i$ is unchanged. However, if $i\in J$ , then $x_{{\underline {J}}^\vartriangle }^{(i)}=x_{{\underline {J}}}^{(i)}$ and we obtain $T_{{\underline {J}}}^{-1}f_i$ by (5.23). The case of (5.26) is similar.

Example 5.9. For instance, suppose $r=3$ , $p=0$ , and ${\underline {J}}=\{x^{(0)}_1,x^{(2)}_1\}$ . Then the right-hand side of (5.25) is

$$ \begin{align*}\mathrm{ev}_{{\underline{J}}}^+\left(f_{0}\left[ \sum_{l=1}^{N_{0}} x_{l}^{(0)}-w_0+q^{-1}w_{1} \right] f_1\left[ \sum_{l=1}^{N_1} x_{l}^{(1)}-w_{1}+q^{-1}w_{2} \right] f_{2}\left[ \sum_{l=1}^{N_{2}} x_{l}^{(2)}-w_{2}+q^{-1}w_0 \right]\right) \end{align*} $$

with $\mathrm {ev}_{{\underline {J}}}^+$ given by evaluating $w_0\mapsto x^{(0)}_1,w_1\mapsto q^{-1}w_2,w_2\mapsto x^{(2)}_1$ in this order. The result is

$$ \begin{align*}f_{0}\left[ \sum_{l=1}^{N_{0}} x_{l}^{(0)}-x^{(0)}_1+q^{-2}x^{(2)}_1 \right] f_1\left[ \sum_{l=1}^{N_1} x_{l}^{(1)}+0\right] f_{2}\left[ \sum_{l=1}^{N_{2}} x_{l}^{(2)}-x^{(2)}_1+q^{-1}x^{(0)}_1\right] = T_{{\underline{J}}}^{-1}f\left[X_{N_{\bullet}}\right]. \end{align*} $$

We will also make use of n-tuples of shift patterns. For such an n-tuple $\mathbf {\underline {J}}=(\underline {J}_1,\ldots , \underline {J}_n)$ and $0\le k\le n$ , we denote

$$ \begin{align*} |\mathbf{\underline{J}}|&=\underline{J}_1\cup\cdots\cup\underline{J}_n\subset X_{N_{\bullet}}\\ |\mathbf{\underline{J}}|_{\le k}&=\underline{J}_1\cup\cdots\cup\underline{J}_{k}\subset X_{N_{\bullet}}\\ |\mathbf{\underline{J}}|_{\ge k}&=\underline{J}_k\cup\cdots\cup\underline{J}_{n}\subset X_{N_{\bullet}}. \end{align*} $$

If $\mathbf {\underline {J}}$ is an n-tuple of shift patterns all containing color p, we say $\mathbf {\underline {J}}$ is p-distinct if the p-colored variables $x_{\underline {J}_k}^{(p)}$ are all distinct. Let $Sh_p^{[n]}(X_{N_{\bullet }})$ denote the set of all p-distinct n-tuples of shift patterns containing color p.

5.2.1 The t-poles

First consider the ‘ $+$ ’ case. Here, we begin with the residue $w_{p,1}=t^{-1}w_{p-1,1}$ . Let us group together the factors

$$\begin{align*}\frac{w_{p,1}w_{p-1,1}}{w_{p,1}\left( w_{p-1,1}-q^{-1}w_{p,1} \right)\left( w_{p,1}-t^{-1}w_{p-1,1} \right) \displaystyle \prod_{l=1}^{N_p}\left( w_{p,1}-x_{l}^{(p)} \right)} \prod_{l=1}^{N_{p-1}}\frac{\left( w_{p,1}-t^{-1}x_{l}^{(p-1)} \right)} {\left( w_{p-1,1}-x_{l}^{(p-1)} \right)}. \end{align*}$$

Upon taking taking the residue, this becomes

$$\begin{align*}\frac{t^{-N_{p-1}}}{\left(1-q^{-1}t^{-1}\right) \displaystyle \prod_{l=1}^{N_p}\left( t^{-1}w_{p-1,1}-x_{l}^{(p)} \right)}. \end{align*}$$

Because of the additional restriction $|t|\gg 1$ , the poles above will be outside the unit circle $|w_{p-1,1}|=1$ .

This pattern persists as we continue downwards in cyclic order until we reach $w_{p+1,1}$ . Here, we have

$$ \begin{align*} &\left. \frac{w_{p+1,1}w_{p,1}}{w_{p+1,1}w_{0,1} \left( w_{p+1,1}-t^{-1}w_{p,1} \right)} \right|_{\substack{w_{0,1}\mapsto t^{p+1-r}w_{p+1,1}\\ w_{p,1}\mapsto t^{-(r-1)}w_{p+1,1}}} \prod_{l=1}^{N_{p}} \frac{\left( w_{p+1,1}-t^{-1}x_{l}^{(p)} \right)} {\left( t^{-r+1}w_{p+1,1}-x_{l}^{(p)} \right)}\\ &= \frac{t^{-p}}{1-t^{-r}} \cdot\frac{1}{w_{p+1,1}}\prod_{l=1}^{N_{p}} \frac{\left( w_{p+1,1}-t^{-1}x_{l}^{(p)} \right)} {\left( t^{-r+1}w_{p+1,1}-x_{l}^{(p)} \right)}. \end{align*} $$

The only pole here is the simple pole at $w_{p+1,1}=0$ . After taking this residue, (5.29) becomes just $f\left[ X_{N_{\bullet }} \right]$ . Bringing in the front matter in (5.27), we are left with

$$ \begin{align*} \frac{t^{-p-1-|N_{\bullet}|}} {1- t^{-r } } f\left[ X_{N_{\bullet}} \right]. \end{align*} $$

Here, we recall that $N_{\bullet }=(N_0,\ldots , N_{r-1})$ records the number of x-variables and $|N_{\bullet }|=\sum _{i\in I}N_i$ .

For the ‘ $-$ ’ case, recall that we begin at $w_{p+1,1}$ and take the residue $w_{p+1,1}=tw_{p+2}$ . We group together the factors

$$\begin{align*}\frac{w_{p+1,1}w_{p+2,1}}{w_{p+1,1}\left( w_{p+2,1}-qw_{p+1,1} \right)\left( w_{p+1,1}-tw_{p+2,1} \right) \displaystyle \prod_{l=1}^{N_p}\left( w_{p+1,1}-x_{l}^{(p)} \right)} \prod_{l=1}^{N_{p+1}} \frac{\left( w_{p+1,1}-tx_{l}^{(p+1)} \right)} {\left( w_{p+1,2}-x_{l}^{(p+1)} \right)}, \end{align*}$$

which upon taking the residue becomes

$$\begin{align*}\frac{t^{N_{p+1}}}{\left( 1-qt \right) \displaystyle \prod_{l=1}^{N_p}\left( tw_{p+2,1}-x_{l}^{(p)} \right)}. \end{align*}$$

The remaining poles above lie outside the unit circle $|w_{p+2,1}|=1$ because we have assumed $|t|\ll 1$ . We continue upwards in cyclic order, yielding similar calculations until we arrive at $w_{p,1}$ . Here, we have the factors

$$ \begin{align*} &\left.\frac{w_{0,1}} {w_{p,1}\left( w_{p,1}-tw_{p+1,1} \right)}\right|_{\substack{w_{0,1}\mapsto t^p w_{p,1}\\w_{p+1,1}\mapsto t^{r-1}w_{p,1}}} \prod_{l=1}^{N_p} \frac{\left( w_{p,1}-tx_{l}^{(p)} \right)} {\left( t^{r-1}w_{p,1}-x_{l}^{(p)} \right)}\\ &= \frac{t^p}{1-t^r} \cdot \frac{1}{w_{p,1}} \prod_{l=1}^{N_p} \frac{\left( w_{p,1}-tx_{l}^{(p)} \right)} {\left( t^{r-1}w_{p,1}-x_{l}^{(p)} \right)}. \end{align*} $$

The only pole within the unit circle $|w_{p,1}|=1$ is $w_{p,1}=0$ . After taking this residue, the final result (after including the front matter) is

$$ \begin{align*} \frac{t^{p+1+|N_{\bullet}|}}{1-t^r}f\left[ X_{N_{\bullet}} \right]. \end{align*} $$

5.2.2 The x-poles

We will first work out the ‘ $+$ ’ case. Thus, we have taken the residue of $w_{p,1}$ at the pole $w_{p,1}=x_{l}^{(p)}$ for some $1\le l \le N_p$ . This variable $x_{l}^{(p)}$ will be an element of a shift pattern ${\underline {J}}$ . Therefore, we call it $x_{{\underline {J}}}^{(p)}$ . It will be advantageous to now group together the factors

$$\begin{align*}\frac{\displaystyle w_{p,1}w_{p+1,1}} {w_{0,1}w_{p,1}\left( w_{p+1,1}-t^{-1}w_{p,1} \right)} \prod_{l=1}^{N_{p}} \frac{\left( w_{p+1,1}-t^{-1}x_{l}^{(p)} \right)} {\left( w_{p,1}-x_{l}^{(p)} \right)}. \end{align*}$$

After taking the residue, we leave behind

$$\begin{align*}\frac{w_{p+1,1}}{w_{0,1}} \prod_{\substack{l=1\\x_{l}^{(p)}\not=x_{{\underline{J}}}^{(p)}}}^{N_{p}} \frac{\left( w_{p+1,1}-t^{-1}x_{1}^{(p)} \right)} {\left( x_{{\underline{J}}}^{(p)}-x_{l}^{(p)} \right)}. \end{align*}$$

Next, we consider $w_{p+1,1}$ . We group together the factors

$$\begin{align*}\frac{ w_{p+1,1}w_{p+2,1}} {w_{p+1,1} \left( w_{p+2,1}-t^{-1}w_{p+1,1} \right) \underbrace{\left( w_{p+1,1}-q^{-1}w_{p+2,1} \right)}_{(1)} } \prod_{l=1}^{N_{p+1}} \frac{\left( w_{p+2,1}-t^{-1}x_{l}^{(p+1)} \right)} {\underbrace{\left( w_{p+1,1}-x_{l}^{(p+1)} \right)}_{(2)}}. \end{align*}$$

The only (nonremovable) poles within the unit circle $|w_{p+1,1}|=1$ are marked (1) and (2). We thus have two cases:

1. 1. Residue at $w_{p+1,1}=q^{-1}w_{p+2,1}$ : In this case, $\left( w_{p+2,1}-t^{-1}w_{p+1,1} \right)$ cancels with a $w_{p+2,1}$ in the numerator, leaving behind

   $$\begin{align*}\frac{1} {\left(1 -q^{-1}t^{-1} \right)} \left. \prod_{l=1}^{N_{p+1}} \frac{ \left( w_{p+2,1}-t^{-1}x_{l}^{(p+1)} \right)} {\left( w_{p+1,1}-x_{l}^{(p+1)} \right)}\right|_{w_{p+1,1}\mapsto q^{-1}w_{p+2,1}}. \end{align*}$$

   Because $|q|\gg 1$ , the poles above lie outside the unit circle $|w_{p+2,1}|=1$ .

2. 2. Residue at $w_{p+1,1}=x_{l}^{(p+1)}=:x_{{\underline {J}}}^{(p+1)}$ : Here, $\left( w_{p+2,1}-t^{-1}w_{p+1,1} \right)$ cancels with a factor in the numerator, leaving behind

   (5.33) $$ \begin{align} \frac{w_{p+2,1}}{\left(x_{{\underline{J}}}^{(p+1)}-q^{-1}w_{p+2,1}\right)} \prod_{\substack{l=1\\x_{l}^{(p+1)}\not=x_{{\underline{J}}}^{(p+1)}}}^{N_{p+1}} \frac{\left( w_{p+2,1}-t^{-1}x_{1}^{(p+1)} \right)} {\left( x_{{\underline{J}}}^{(p+1)}-x_{l}^{(p+1)} \right)}. \end{align} $$

   Again, because $|q|\gg 1$ , the first pole above lies outside the unit circle $|w_{p+2,1}|=1$ .

This pattern and dichotomy for residues continues upwards in cyclic order. The x-variables in the type (2) residues constitute a shift pattern ${\underline {J}}$ , and our gap labels $x_{{\underline {J}}^\vartriangle }^{(i)}$ incorporate the q-shifts from the type (1) residues. Therefore, $w_{i,1}$ is always evaluated at $x_{{\underline {J}}^\vartriangle }^{(i)}$ . Finally, observe that by Lemma 5.8, (5.29) becomes $T_{{\underline {J}}}^{-1}f\left[ X_{N_{\bullet }} \right]$ . The end result is $t^{-|N_{\bullet }|}D_{p,1}^*(q,t^{-1})f\left[ X_{N_{\bullet }} \right]$ .

The ‘ $-$ ’ case is similar. Our first variable is $w_{p+1,1}$ , for which we take the residue at $w_{p+1,1}=x_{l}^{(p)}=:x_{{\underline {J}}}^{(p)}$ . We consider the factors

$$\begin{align*}\left(\frac{w_{0,1}}{w_{p+1,1}} \right) \frac{1}{\left(w_{p,1}-tw_{p+1,1}\right)} \prod_{l=1}^{N_p} \frac{\left(w_{p,1}-tx_{l}^{(p)} \right)} {\left( w_{p+1,1}-x_{l}^{(p)} \right)}. \end{align*}$$

After taking the residue, the pole from $(w_{p,1}-tw_{p+1,1})$ cancels with a factor in the numerator, leaving behind

$$\begin{align*}\frac{w_{0,1}}{x_{{\underline{J}}}^{(p)}} \prod_{\substack{l=1\\x_{l}^{(p)}\not=x_{{\underline{J}}}^{(p)}}}^{N_p} \frac{\left(w_{p,1}-tx_{l}^{(p)} \right)} {\left( x_{{\underline{J}}}^{(p)}-x_{l}^{(p)} \right)}. \end{align*}$$

We now proceed downward in cyclic order. For each $w_{i,1}$ , we consider the factors

$$\begin{align*}\frac{w_{i,1}w_{i-1,1}} {w_{i,1}\left( w_{i-1,1}-tw_{i,1} \right)\underbrace{\left( w_{i,1}-qw_{i-1,1} \right)}_{(1)}} \prod_{l=1}^{N_{i-1}} \frac{\left( w_{i-1,1}-tx_{l}^{(i-1)} \right)} {\underbrace{\left( w_{i,1}-x_{l}^{(i-1)} \right)}_{(2)}}. \end{align*}$$

Because $\left( w_{i,1}-tw_{i+1,1} \right)$ has been canceled at this point, the only poles within the unit circle $|w_{i,1}|=1$ are those marked (1) and (2). The analysis is as before:

1. 1. Residue at $w_{i,1}=qw_{i-1,1}$ : This leaves behind

   $$\begin{align*}\frac{1}{(1-qt)} \prod_{l=1}^{N_{i-1}} \left. \frac{\left( w_{i-1,1}-tx_{l}^{(i-1)} \right)} {\left( w_{i,1}-x_{l}^{(i-1)} \right)}\right|_{w_{i,1}\mapsto qw_{i-1,1}}. \end{align*}$$

2. 2. Residue at $w_{i,1}=x_{l}^{(i-1)}=:x_{{\underline {J}}}^{(i-1)}$ : The leftovers are now

   (5.34) $$ \begin{align} \frac{w_{i-1,1}}{\left( x_{i-1,{\underline{J}}}-qw_{i-1,1} \right)} \prod_{\substack{l=1\\x_{l}^{(i-1)}\not=x_{{\underline{J}}}^{(i-1)}}}^{N_{i-1}} \frac{\left( w_{i-1,1}-tx_{l}^{(i-1)} \right)} {\left( x_{{\underline{J}}}^{(i-1)}-x_{l}^{(i-1)} \right)}. \end{align} $$

The x-variables where we have taken residues constitute a shift pattern ${\underline {J}}$ , and $w_{i,1}$ is always evaluated at $x_{{\underline {J}}^\triangledown }^{(i-1)}$ . Again, by Lemma 5.8, (5.32) becomes $T_{{\underline {J}}}f\left[ X_{N_{\bullet }} \right]$ . Here, we obtain $t^{|N_{\bullet }|}D_{p,1}(q,t^{-1})f\left[ X_{N_{\bullet }} \right]$ .

5.2.3 Degree one eigenfunction equation

Finally, we enhance Proposition 5.10 by obtaining eigenfunction equations for $D_{p,1}^*(q,t)$ and $D_{p,1}(q,t)$ for generic values of the parameters.

Theorem 5.11. For generic values of q, t,

(5.35) $$ \begin{align} D_{p,1}^*\left( X_{N_{\bullet}};q,t \right)P_{\lambda}[X_{N_{\bullet}};q,t] &=\left(\sum_{\substack{b=1\\ b-\lambda_b\equiv p+1}}^{|N_{\bullet}|} q^{-\lambda_b}t^{-|N_{\bullet}|+b}\right) P_{\lambda}[X_{N_{\bullet}};q,t] \end{align} $$

(5.36) $$ \begin{align} D_{p,1}\left( X_{N_{\bullet}};q,t \right)P_{\lambda}[X_{N_{\bullet}};q,t] &=\left(\sum_{\substack{b=1\\ b-\lambda_b\equiv p+1}}^{|N_{\bullet}|} q^{\lambda_b}t^{|N_{\bullet}|-b}\right) P_{\lambda}[X_{N_{\bullet}};q,t]. \end{align} $$

Proof. We will only consider the ‘ $+$ ’ case – the ‘ $-$ ’ case is similar. Combining Lemma 3.11 and Proposition 5.10, we have for $\lambda \in \mathbb {Y}$ with $\kappa (\lambda )=\alpha $ and $|\mathrm {quot}(\lambda )|\le |N_{\bullet }|$ ,

$$ \begin{align*} \left(t^{-|N_{\bullet}|}D_{p,1}^*(X_{N_{\bullet}};q,t^{-1})+\frac{t^{-p-1-|N_{\bullet}|}}{1-t^{-r}}\right)P_{\lambda}\left[ X_{N_{\bullet}}; q,t^{-1} \right] &= \left(\sum_{\substack{b>0\\ b-\lambda_b\equiv p+1}} q^{-\lambda_b}t^{-b}\right) P_{\lambda}[X_{N_{\bullet}}; q,t^{-1}], \end{align*} $$

where we assume $|q|\gg 1$ , $|t|\gg 1$ and $|x_{i,l}|<1$ . Even here, it is essential that $|t|\gg 1$ as we are working with series in $t^{-1}$ . We can do away with this once we notice that since $|N_{\bullet }|$ is divisible by r (Proposition 2.7) and $\ell (\lambda )\le |N_{\bullet }|$ ,

(5.37) $$ \begin{align} \begin{aligned} \sum_{\substack{b>0\\ b-\lambda_b\equiv p+1}} q^{-\lambda_b}t^{-b} &= \left(\sum_{k=0}^\infty t^{-p-1-|N_{\bullet}|-rk}\right)+ \left(\sum_{\substack{b=1\\ b-\lambda_b\equiv p+1}}^{|N_{\bullet}|} q^{-\lambda_b}t^{-b}\right)\\[8pt] &=\frac{t^{-p-1-|N_{\bullet}|}}{1-t^{-r}}+\left(\sum_{\substack{b=1\\ b-\lambda_b\equiv p+1}}^{|N_{\bullet}|} q^{-\lambda_b}t^{-b}\right). \end{aligned} \end{align} $$

Here, we have split off the terms corresponding to rows above height $|N_{\bullet }|$ . Thus, (5.36) holds under our conditions on $|q|$ , $|t|$ and $|x_{l}^{(i)}|$ .

Finally, we address the genericity of parameters. The equations (5.35) and (5.36) are equalities of rational functions in the space $(X_{N_{\bullet }}, q,t)$ . We have established them over an analytic open subset of $(X_{N_{\bullet }}, q,t)$ . After subtracting one side to the other, this is equivalent saying a rational function is zero on a codimension zero subspace, and thus, it must be zero.

The eigenvalues of $\left\{ D_{p,1}(X_{N_{\bullet }};q,t) \right\}_{p\in I}$ on $\left\{ P_{\lambda }[X_{N_{\bullet }};q,t] \right\}$ are nondegenerate. Therefore, we have the following:

Example 5.13. Let $r=3$ , $p=1$ , $N_{\bullet }=(2,1,0)$ and $\lambda =(3,1,1)$ . In this case, $\lambda $ is a $3$ -core and so

$$\begin{align*}P_{\lambda}[X_{N_{\bullet}};q,t]=1.\end{align*}$$

There are three shift patterns containing $p=1$ :

$$ \begin{align*} {\underline{J}}_1&=\{x_1^{(1)}\}\\ {\underline{J}}_2&=\{x_1^{(0)},x_1^{(1)}\}\\ {\underline{J}}_3&=\{x_2^{(0)},x_1^{(1)}\}. \end{align*} $$

The operator $D_{1,1}(X_{N_{\bullet }};q,t)$ is then

(5.38) $$ \begin{align} D_{1,1}(X_{N_{\bullet}};q,t)&=q\left(\frac{qtx_1^{(1)}-x_1^{(0)}}{q^2x_1^{(1)}-x_1^{(0)}}\right)\left(\frac{qtx_1^{(1)}-x_2^{(0)}}{q^2x_1^{(1)}-x_2^{(0)}}\right)T_{{\underline{J}}_1} \end{align} $$

(5.39) $$ \begin{align} &+(1-qt^{-1})q\left(\frac{qtx_1^{(1)}-x_2^{(0)}}{x_1^{(0)}-x_2^{(0)}}\right)\left(\frac{qtx_1^{(1)}}{x_1^{(0)}-q^2x_1^{(1)}}\right)T_{{\underline{J}}_2} \end{align} $$

(5.40) $$ \begin{align} &+(1-qt^{-1})q\left(\frac{qtx_1^{(1)}-x_1^{(0)}}{x_2^{(0)}-x_1^{(0)}}\right)\left(\frac{qtx_1^{(1)}}{x_2^{(0)}-q^2x_1^{(1)}}\right)T_{{\underline{J}}_3}. \end{align} $$

The cyclic-shift operators act trivially on $P_{\lambda }(X_{N_{\bullet }};q,t)$ . Consolidating (5.39) and (5.40) gets us

(5.41)

However, (5.38) becomes

(5.42) $$ \begin{align} \nonumber &q\left(\frac{qtx_1^{(1)}-x_1^{(0)}}{q^2x_1^{(1)}-x_1^{(0)}}\right)\left(\frac{qtx_1^{(1)}-x_2^{(0)}}{q^2x_1^{(1)}-x_2^{(0)}}\right)\\ &=q\left\{ \frac{q^2t^2x_1^{(1)}x_1^{(1)}-qtx_1^{(1)}x_2^{(0)}-qtx_1^{(0)}x_1^{(1)}+x_1^{(0)}x_2^{(0)}} {(x_1^{(0)}-q^2x_1^{(1)})(x_2^{(0)}-q^2x_1^{(1)})} \right\}. \end{align} $$

Combining (5.41) and (5.42) gets us

$$ \begin{align*} D_{1,1}(X_{N_{\bullet}};q,t)P_{\lambda}[X_{N_{\bullet}};q,t]&= q\left\{ \frac{q^4x_1^{(1)}x_1^{(1)}-q^2x_1^{(1)}x_2^{(0)}-q^2x_1^{(0)}x_1^{(1)}+x_1^{(0)}x_2^{(0)}} {(x_1^{(0)}-q^2x_1^{(1)})(x_2^{(0)}-q^2x_1^{(1)})} \right\}\\ &=q\frac{(x_1^{(0)}-q^2x_1^{(1)})(x_2^{(0)}-q^2x_1^{(1)})}{(x_1^{(0)}-q^2x_1^{(1)})(x_2^{(0)}-q^2x_1^{(1)})}\\ &=qP_{\lambda}[X_{N_{\bullet}};q,t]. \end{align*} $$

Example 5.14. Let $r=2$ , $p=0$ , $N_{\bullet }=(1,1)$ , and $\lambda =(1,1)$ . Here,

$$\begin{align*}P_{\lambda}[X_{N_{\bullet}};q,t]=x_1^{(1)}.\end{align*}$$

There are two shift patterns containing $0$ :

$$ \begin{align*} {\underline{J}}_1&=\{x_1^{(0)}\}\\ {\underline{J}}_2&=\{x_1^{(0)},x_1^{(1)}\}. \end{align*} $$

We then have

$$ \begin{align*} D_{0,1}(X_{N_{\bullet}};q,t)&= q\left( \frac{tx_1^{(0)}-x_1^{(1)}}{qx_1^{(0)}-x_1^{(1)}} \right)T_{{\underline{J}}_1} +(1-qt^{-1}) \frac{x_1^{(1)}}{x_1^{(0)}} \left( \frac{tx_1^{(0)}}{x_1^{(1)}-qx_1^{(0)}} \right)T_{{\underline{J}}_2}. \end{align*} $$

Observe that

$$ \begin{align*} T_{{\underline{J}}_1}x_1^{(1)}&=x_1^{(1)}\\ T_{{\underline{J}}_2}x_1^{(1)}&=qx_1^{(0)}. \end{align*} $$

Altogether then,

$$ \begin{align*} D_{0,1}(X_{N_{\bullet}};q,t)P_{\lambda}[X_{N_{\bullet}};q,t] &=q\left( \frac{tx_1^{(0)}-x_1^{(1)}}{qx_1^{(0)}-x_1^{(1)}} \right)x_1^{(1)} +(1-qt^{-1}) \frac{x_1^{(1)}}{x_1^{(0)}} \left( \frac{tx_1^{(0)}}{x_1^{(1)}-qx_1^{(0)}} \right)qx_1^{(0)}\\ &=qx_1^{(1)}\left(\frac{tx_1^{(0)}-x_1^{(1)}-(t-q)x_1^{(0)}}{qx_1^{(0)}-x_1^{(1)}}\right)\\ &=qx_1^{(1)}\left(\frac{qx_1^{(0)}-x_1^{(1)}}{qx_1^{(0)}-x_1^{(1)}}\right)\\ &=qP_{\lambda}[X_{N_{\bullet}};q,t]. \end{align*} $$

5.3.1 Only x-poles

In both the ‘ $+$ ’ and ‘ $-$ ’ cases, each of the n variables $\{ x_{l_a}^{(p)} \}_{a=1}^n$ will become part of a shift pattern containing p, so we set $x_{{\underline {J}}_a}^{(p)}:=x_{l_a}^{(p)}$ . Furthermore, as these variables must be distinct, we have that the tuple $\mathbf {{\underline {J}}}:=\left( {\underline {J}}_1,\ldots ,{\underline {J}}_n \right)$ will be p-distinct. After taking these residues, we will proceed as in 5.2.2 for a specific value of a.

First consider the ‘ $+$ ’ case. To see the effect of taking the residues $w_{p,b}=x_{{\underline {J}}_b}^{(p)}$ , we group together the factors

$$ \begin{align*} &\left(\frac{w_{p,b}}{w_{0,b}}\right)\frac{w_{p+1,b}}{w_{p,b}\left( w_{p+1,b}-t^{-1}w_{p,b} \right)} \prod_{l=1}^{N_p} \frac{\left( w_{p+1,b}-t^{-1}x_{l}^{(p)} \right)} {\displaystyle\left( w_{p,b}-x_{l}^{(p)} \right)}\\ &\times \frac{1} {\displaystyle \prod_{b<c} \left( w_{p+1,b}-t^{-1}w_{p,c} \right)} \prod_{a<b}\frac{ \displaystyle \left( w_{p,b}-w_{p,a} \right)\left( w_{p+1,b}-q^{-1}t^{-1}w_{p+1,a} \right)} {\left( w_{p+1,a}-t^{-1}w_{p,b} \right)}. \end{align*} $$

Upon taking residues, this becomes

(5.45) $$ \begin{align} \left(\frac{w_{p+1,b}}{w_{0,b}}\right) \frac{\displaystyle \prod_{\substack{l=1\\x_{l}^{(p)}\not\in|\mathbf{{\underline{J}}}|}}^{N_p} \left( w_{p+1,b}-t^{-1}x_{l}^{(p)} \right)} {\displaystyle \prod_{\substack{l=1\\x_{l}^{(p)}\not\in|\mathbf{{\underline{J}}}|_{\le b}}}^{N_p} \left( x_{{\underline{J}}_b}^{(p)}-x_{l}^{(p)} \right)} \underbrace{\prod_{a<b}\left( w_{p+1,b}-q^{-1}t^{-1}w_{p+1,a} \right)}_{(\dagger)}. \end{align} $$

The next variable we consider is $w_{p+1,1}$ . Notice that we have canceled the poles $\left( w_{p+1,1}-t^{-1}w_{p,b} \right)$ for all $b\ge 1$ , and consequently, the only two kinds of poles within the unit circle $|w_{p+1,1}|=1$ are as before in 5.2.2. We group together the factors

$$ \begin{align*} &\frac{w_{p+1,1}w_{p+2,1}} {w_{p+1,1}\left( w_{p+2,1}-t^{-1}w_{p+1,1} \right)\left( w_{p+1,1}-q^{-1}w_{p+2,1} \right)}\\ &\times \underbrace{\prod_{1<b} \frac{ \left( w_{p+1,b}-w_{p+1,1} \right)\left(w_{p+2,b}-q^{-1}t^{-1}w_{p+2,1} \right)} {\left( w_{p+2,b}-t^{-1}w_{p+1,1} \right)\left( w_{p+1,b}-q^{-1}w_{p+2,1} \right)}}_{(*)} \underbrace{\prod_{b=1}^n\prod_{l=1}^{N_{p+1}} \frac{\left( w_{p+2,b}-t^{-1}x_{l}^{(p+1)} \right)} {\left( w_{p+1,b}-x_{l}^{(p+1)} \right)}}_{(**)}. \end{align*} $$

The residues are

1. 1. Residue at $w_{p+1,1}=q^{-1}w_{p+2,1}$ : In this case, the factors in $(*)$ cancel out, leaving behind

   $$\begin{align*}\frac{1}{(1-q^{-1}t^{-1})} \left.\prod_{b=1}^n\prod_{l=1}^{N_{p+1}} \frac{\left( w_{p+2,b}-t^{-1}x_{l}^{(p+1)} \right)} {\left( w_{p+1,b}-x_{l}^{(p+1)} \right)}\right|_{w_{p+1,1}\mapsto q^{-1}w_{p+2,1}}. \end{align*}$$

   As in 5.2.2, $w_{p+1,1}$ will ultimately be evaluated at $x^{(p+1)}_{{\underline {J}}_1^\vartriangle }$ , and the poles above lie outside the unit circle $|w_{p+2,1}|=1$ because $|q|\gg 1$ .

2. 2. Residue at $w_{p+1,1}=x_{l}^{(p+1)}=:x^{(p+1)}_{{\underline {J}}_1}$ : Here, the factors in $(*)$ cancel with those in $(**)$ containing $x^{(p+1)}_{{\underline {J}}_1}$ . We are left with

   (5.46) $$ \begin{align} \nonumber&\frac{w_{p+2,1}}{\left( x^{(p+1)}_{{\underline{J}}_1}-q^{-1}w_{p+2,1} \right)} \prod_{\substack{l=1\\x^{(p+1)}_{l}\not=x^{(p+1)}_{{\underline{J}}_1}}}^{N_{p+1}} \frac{\left( w_{p+2,1}-t^{-1}x^{(p+1)}_{l} \right)} {\left( x_{p+1,{\underline{J}}_1}-x^{(p+1)}_{l} \right)}\\ &\times\prod_{1<b} \left\{\frac{\left( w_{p+2,b}-q^{-1}t^{-1}w_{p+2,1} \right)} {\left( w_{p+1,b}-q^{-1}w_{p+2,1} \right)} \prod_{\substack{l=1\\x^{(p+1)}_{l}\not=x^{(p+1)}_{{\underline{J}}_1}}}^{N_{p+1}} \frac{\left( w_{p+2,b}-t^{-1}x^{(p+1)}_{l} \right)} {\left(w_{p+1,b}-x^{(p+1)}_{l}\right)}\right\}. \end{align} $$

   Because $|q|\gg 1$ , the pole $\left( x^{(p+1)}_{{\underline {J}}_1}-q^{-1}w_{p+2,1} \right)$ lies outside the unit circle $|w_{p+2,1}|=1$ . Our key organizational trick here is that when $w_{p+2,1}$ is ultimately evaluated at $x^{(p+2)}_{{\underline {J}}_1^\vartriangle }$ , then we can use (5.23) to write (5.46) as

   $$\begin{align*}T_{{\underline{J}}_1}^{-1}\left( \prod_{1<b}^n\prod_{l=1}^{N_{p+1}} \frac{\left( w_{p+2,b}-t^{-1}x^{(p+1)}_{l} \right)} {\left( w_{p+1,b}-x^{(p+1)}_{l} \right)} \right) \end{align*}$$
   since $T_{{\underline {J}}_1}$ will only affect $x^{(p+1)}_{{\underline {J}}_1}$ .

This pattern continues upwards in cyclic order for the variables $w_{i,1}$ . The x-variables where we take residues give a shift pattern ${\underline {J}}_1$ containing p, and $w_{i,1}$ is evaluated at $x^{(i)}_{{\underline {J}}_1^\vartriangle }$ . In (5.45), the term in $(\dagger )$ for $a=1$ can be rewritten as $\left( w_{p+1,b}-t^{-1}T_{{\underline {J}}_1}^{-1}x^{(p)}_{{\underline {J}}_1} \right)$ . Finally, we note that by Lemma 5.8, these residues result in $T_{{\underline {J}}_1}^{-1}$ applied to $f\left[ X_{N_{\bullet }} \right]$ . Thus, we can rewrite the result after taking the residues for $a=1$ as

$$ \begin{align*} &\frac{(-1)^{\frac{n(n-1)}{2}}t^{-\frac{n(n+1)}{2}}(1-q^{-1}t^{-1})^{r(n-1)}}{\prod_{a=1}^n(1-q^{-a}t^{-a})} \sum_{{\underline{J}}_1\in Sh_p}(1-q^{-1}t^{-1})^{|{\underline{J}}_1|}\left(\frac{x^{(p+1)}_{{\underline{J}}_1^\vartriangle}}{x^{(0)}_{{\underline{J}}_1^\vartriangle}}\right) \\ &\times \frac{\displaystyle \prod_{\substack{l=1\\x^{(p)}_{l}\not\in|\mathbf{{\underline{J}}}|_{\ge 1}}}^{N_p}\left( x^{(p+1)}_{{\underline{J}}_1^\vartriangle}-t^{-1}x^{(p)}_{l} \right)} {\displaystyle \prod_{\substack{l=1\\x^{(p)}_{l}\not\in|\mathbf{{\underline{J}}}|_{\le 1}}}^{N_p}\left( x^{(p)}_{{\underline{J}}_1}-x^{(p)}_{l} \right)} \left( \prod_{\substack{i\in I\\i\not= p}}\, \prod_{\substack{l=1\\x^{(i)}_{l}\not=x^{(i)}_{{\underline{J}}_1^\vartriangle}}}^{N_i}\frac{\left(x^{(i+1)}_{{\underline{J}}_1^\vartriangle}-t^{-1}x^{(i)}_{l}\right)} {\left( x^{(i)}_{{\underline{J}}_1^\vartriangle}-x^{(i)}_{l} \right)} \right) \left( \prod_{i\in J_1\backslash \left\{ p \right\}} \frac{q T_{{\underline{J}}_1}^{-1}x^{(i)}_{{\underline{J}}_1}} {\left( x^{(i)}_{{\underline{J}}_1}-T_{{\underline{J}}_1}^{-1}x^{(i)}_{{\underline{J}}_1} \right)} \right)\\&\times \underset{|w_{i,a}|=1}{\oint\cdots\oint} T_{{\underline{J}}_1}^{-1}\left( \prod_{a=1}^{n}\left\{ \left( w_{p+1,a}-t^{-1}x^{(p)}_{{\underline{J}}_1} \right) \prod_{1< a<b\le n} \left( w_{p+1,b}-q^{-1}t^{-1}w^{(p+1)}_{a} \right) \right.\right.\\ &\times \left. \frac{\displaystyle \prod_{\substack{l=1\\ x^{(p)}_{l}\not\in|\mathbf{{\underline{J}}}|}}^{N_p}\left( w_{p+1,a}-t^{-1}x^{(p)}_{l} \right)} {\displaystyle \prod^{N_p}_{\substack{l=1\\x^{(p)}_{l}\not\in |\mathbf{{\underline{J}}}|_{\le a}}}\left( x^{(p)}_{{\underline{J}}_a}-x^{(p)}_{l} \right)} \prod_{i\in I\backslash\left\{ p \right\}}\prod_{l=1}^{N_{i}} \frac{\displaystyle \left( w_{i+1,a}-t^{-1}x^{(i)}_{l} \right)}{\displaystyle\left( w_{i,a}-x^{(i)}_{l} \right)}\right\}\\ &\times\prod_{1< a<b\le n} \prod_{i\in I\backslash\{p\}} \frac{(w_{i,b}-w_{i,a})(w_{i+1,b}-q^{-1}t^{-1}w_{i+1,a})}{(w_{i+1,b}-t^{-1}w_{i,a})(w_{i,b}-q^{-1}w_{i+1,a})}\\ &\times\prod_{a=2}^{n} \left\{ \left(\frac{w_{p+1,a}}{w_{0,a}}\right) \prod_{i\in I\setminus\{p\}} \frac{w_{i,a}w_{i+1,a}}{\left(w_{i+1,a}-t^{-1}w_{i,a}\right)\left(w_{i,a}-q^{-1}w_{i+1,a}\right)} \right\}\\ &\left. \prod_{i\in I}f_i\left[ \sum_{l=1}^{N_i}x_{l}^{(i)}-\sum_{a=2}^n w_{i, a}+\sum_{a=2}^nq^{-1} w_{i+1,a} \right]\right)\prod_{i\in I\backslash\{p\}}\prod_{a=2}^n\frac{dw_{i,a}}{2\pi\sqrt{-1}w_{i,a}}. \end{align*} $$

We have written this so that we can repeat the calculation for $a=1$ for general a in increasing order. Note that as we do this, we can rewrite factors in $(\dagger )$ of (5.45) in terms of $T_{{\underline {J}}_a}^{-1}x_{p,{\underline {J}}_a}$ using (5.23). The end result of the residue calculation is

$$\begin{align*}t^{-n|N_{\bullet}|}D_{p,n}^*(X_{N_{\bullet}}; q,t^{-1})f\left[ X_{N_{\bullet}} \right].\end{align*}$$

The ‘ $-$ ’ case is similar. We begin by taking residues of $\{w_{p+1,\bullet }\}$ and then start instead at $x^{(p)}_{n}$ . Afterwards, we continue downwards in cyclic order until we have taken constant terms of all variables with $a=n$ . We then continue downwards in a. The end result is then

$$\begin{align*}t^{n|N_{\bullet}|}D_{p,n}(X_{N_{\bullet}};q,t^{-1})f\left[ X_{N_{\bullet}} \right]. \end{align*}$$

5.3.2 Mixed poles

In the case where there are t-poles, our goal is to show that the result is a linear combination of the lower order operators applied to $f\left[ X_{N_{\bullet }} \right]$ : $D_{p,k}^*(X_{N_{\bullet }}; q,t^{-1})$ in the ‘ $+$ ’ case and $D_{p,k}(X_{N_{\bullet }}; q,t^{-1})$ in the ‘ $-$ ’ case, where $k<n$ . Unlike in the case of $n=1$ , we will not try to compute the coefficients of this linear combination – we will compute them indirectly in 5.4. As in all the previous cases, the initial residues force a string of other residues, and we will first compute these strings that start from the initial t-poles. Once these variables are evaluated, the remaining terms will evaluate like 5.3.1.

In the ‘ $+$ ’ case, let $1\le b_1^p\le n$ be any index where the residue for $w_{p, b_{1}^p}$ is taken at a t-pole. Denote this pole by $w_{p,b_1^p}=t^{-1}w_{p-1,b_1^{p-1}}$ . In contrast to our previous calculations, we will not always cancel out factors but rather remark on why taking residues at certain poles will result in zero. The poles contributing within the unit circle $|w_{p-1,b_1^{p-1}}|=1$ are as follows.

1. 1. $(w_{p-1,b_1^{p-1}}-q^{-1}w_{p,a} )$ for $a\ge b_1^{p-1}$ : If $a<b_1^p$ , then the factor $( w_{p,b_1^p}-q^{-1}t^{-1}w_{p,a } )$ in the numerator of (5.12) becomes zero when taking this residue. If $a=b_1^{p-1}=b_1^p$ , then this is a pole at $0$ , which cancels with the extra factor of $w_{p-1,b_1^{p-1}}$ as in 5.2.1.

2. 2. $( w_{p-1,b_1^{p-1}}-x^{(p-1)}_{l} )$ : The factor $( w_{p,b_1^p}-t^{-1}x^{(p-1)}_{l} )$ in the numerator of (5.10) will evaluate to zero.

3. 3. $( w_{p-1,b_1^{p-1}}-t^{-1}w_{p-2,a} )$ for $a\le b_1^{p-1}$ : These poles possibly yield nonzero residues.

Taking a residue of the third kind, we evaluate $w_{p-1,b_1^{p-1}}=t^{-1}w_{p-2,b_1^{p-2}}$ for some $b_1^{p-2}\le b_1^{p-1}$ .

This pattern continues downwards in cyclic order, picking out variables $w_{i, b_1^i}$ where $b_1^p\ge b_1^{p-1}\ge \cdots \ge b_1^{p+1}$ . At $w_{p+1,b_1^{p+1}}$ , the pole of type (3) becomes

1. (3’) $(w_{p+1,b_1^{p+1}}-t^{-1}w_{p,a})$ for all a: If $w_{p,a}$ is evaluated at an x-variable $x^{(p)}_{l}$ , then as in 5.3.1, the factor $(w_{p+1,b_1^{p+1}}-t^{-1}x^{(p)}_{l})$ will evaluate to zero upon taking this residue. Thus, only the case where $w_{p,a}$ is evaluated at a t-pole yields a nonzero residue. For $a=b_1^p$ , this is a pole at $w_{p+1,b_1^{p+1}}=0$ . If $b_{1}^{p+2}=b_1^{p+1}$ , then because of the analogue of case (1), there are no extra powers of $w_{p+1,b_1^{p+1}}$ to cancel this pole.

If we take the residue in (3’) at $w_{p,a}$ evaluated at a t-pole but $a\not =b_1^p$ , then we set $b_2^p:=a$ . Letting the t-pole be $(w_{p,b_2^p}-t^{-1}w_{p-1,b_2^{p-1}})$ for $b_2^{p-1}\le b_2^p$ , the process is similar to as before. There is just one alteration to the poles of type (3):

1. (3”) $(w_{i, b_2^{i}}-t^{-1}w_{i-1,b_1^{i-1}})$ : This is a pole at $0$ , which cancels with the factor $\pm (w_{i, b_2^i}- w_{i, b_1^i})$ in the numerator of (5.11).

Thus, we avoid variables that we have already evaluated. Note that at first glance, the product of factors in (5.11) and (5.12) involving $w_{i, b_2^i}$ and $w_{j,b_1^j}$ may contribute a pole at $0$ , but in fact, their products have total degree zero and thus become a constant. There is an outlier case of $(w_{p+1,b_1^{p+1}}-t^{-1}w_{p,b_{2}^p})$ , which has been removed when we take residues, but this can be replaced with $(w_{p+1,b_1^{p+1}}-t^{-1}w_{p,b_{1}^p})$ to restore the degree zero balance. We continue like this to new indices $\{b_3^i\}_{i\in I}, \left\{ b_{4}^i \right\}_{i\in I},$ etc. until either there are no more nonzero residues or we finally take the residue at $0$ of $z_{p+1, b_k^{p+1}}$ for some final value k.

For $1\le m<m'\le k$ , we note that as in the $(m,m')=(1,2)$ case, the product of the binomials in (5.11) and (5.12) involving one variable from $\{w_{i,b_m^i}\}_{i\in I}$ and another variable from $\{w_{j, b_{m'}^j}\}_{i\in I}$ has degree zero provided we make the same adjustment for $i=p+1$ and $m'=m+1$ . Thus, these factors turn into a constant. To consider binomials involving only $\{w_{i,b_m^i}\}_{i\in I}$ for one value of m, we note that when we take the residues, we remove

$$ \begin{align*} &\frac{1}{w_{i,b_m^i}-t^{-1}w_{i+1,b_m^{i+1}}} &&\mbox{for }i\not=p+1,\\ &\frac{1}{w_{p+1,b_m^{p+1}}-t^{-1}w_{p,b_{m+1}^{p}}} &&\mbox{for }1\le m<k. \end{align*} $$

There is a leftover power of $w_{i,b_{m}^i}$ for $i\not =p$ from (5.13) and (5.14), and as discussed in the pole of type (1) above, these are only absorbed when $b_{m}^{i+1}=b_{m}^i$ . These unabsorbed powers turn the entire integral zero when we take the final residue $w_{p+1,b_k^{p+1}}=0$ . Thus, we only need to consider the case where for each m,

$$\begin{align*}b_m^p=b_m^{p-1}=\cdots=b_m^{p+1}=:b_m.\end{align*}$$

In this case, all factors only involving $\{w_{i,b_m}\}_{i\in I}^{1\le m\le k}$ leave behind a constant. Evidently, the corresponding terms in (5.13) and (5.14) disappear. The terms involving $w_{i,b_m}$ and an x-variable in (5.10) leave behind a power of t when we cancel

$$\begin{align*}\prod_{m=1}^k\prod_{i\in I}\prod_{l=1}^{N_i}\frac{\left( w_{i+1,b_{m}}-t^{-1}x^{(i)}_{l} \right)} {\left( w_{i,b_{m}}-x^{(i)}_{l} \right)}. \end{align*}$$

Finally, the product of terms in (5.11) involving any index $1\le a\le n$ and $b_m$ leave behind a constant when we evaluate $w_{i,b_m}=0$ for all $i\in I$ . The remaining factors are a scalar multiple of the calculation for $\mathcal {E}_{p,n-k}$ . The ‘ $-$ ’ case is analyzed similarly.

5.4.1 Spectral variables

Letting $\lambda $ vary over partitions with $\mathrm {core}(\lambda )$ compatible with $N_{\bullet }$ and $\ell (\lambda )\le |N_{\bullet }|$ , we note that by Proposition 2.7, the stabilized eigenvalues

$$\begin{align*}e_n\left[\sum_{\substack{b=1\\b-\lambda_b\equiv p+1}}^\infty q^{-\lambda_b}t^{-b}\right] \mbox{ and } e_n\left[\sum_{\substack{b=1\\b-\lambda_b\equiv p+1}}^\infty q^{\lambda_b} t^{b}\right] \end{align*}$$

depend only on the $N_p$ values of b, where $1\le b\le |N_{\bullet }|$ and $b-\lambda _b=p+1$ . We define the color p spectral variables $\{ s^{(p)}_{ a} \}_{a=1}^{N_p}$ by setting

$$\begin{align*}s^{(p)}_{a}=q^{\lambda_{b_a}}t^{b_a}, \end{align*}$$

where $1\le b_a\le |N_{\bullet }|$ is the ath number where $b_a-\lambda _{b_a}\equiv p+1$ . Using these variables, we can rewrite

$$ \begin{align*} e_n\left[\sum_{\substack{b=1\\b-\lambda_b\equiv p+1}}^\infty q^{-\lambda_b}t^{-b}\right] &=e_n\left[\sum_{k=0}^{\infty}t^{|N_{\bullet}|-p-1-kr}+\sum_{\substack{b=1\\b-\lambda_b\equiv p+1}}^{|N_{\bullet}|} q^{-\lambda_b}t^{-b}\right]\\ &=e_n\left[\frac{t^{-|N_{\bullet}|-p-1}}{1-t^{-nr}}+\sum_{a=1}^{N_p}\left(s^{(p)}_{a}\right)^{-1} \right], \end{align*} $$

where $|t|\gg 1$ . Here, we have split off the parts above row $|N_{\bullet }|$ as in (5.37). Similarly,

$$ \begin{align*} e_n\left[\sum_{\substack{b=1\\b-\lambda_b\equiv p+1}}^\infty q^{\lambda_b} t^{b}\right] &=e_n\left[\sum_{k=0}^\infty t^{|N_{\bullet}|+p+1+kr}+\sum_{\substack{b=1\\b-\lambda_b\equiv p+1}}^{|N_{\bullet}|} q^{\lambda_b} t^{b}\right]\\ &=e_n\left[ \frac{t^{|N_{\bullet}|+p+1}}{1-t^{nr}}+\sum_{a=1}^{N_p}s^{(p)}_{a} \right], \end{align*} $$

where $|t|\ll 1$ . The following is but a slight alteration of Lemma 3.2 from [Reference Feigin, Hashizume, Hoshino, Shiraishi and Yanagida4]:

Lemma 5.19. For $|t|\gg 1$ , we have

(5.49) $$ \begin{align} e_n\left[\frac{t^{-|N_{\bullet}|-p-1}}{1-t^{-nr}}+\sum_{a=1}^{N_p}\left(s^{(p)}_{a}\right)^{-1} \right]=\sum_{k=0}^n\frac{t^{-n|N_{\bullet}|-(n-k)(p+1)-r\binom{n-k}{2}}}{\displaystyle\prod_{l=1}^{n-k}(1-t^{-rl})} e_k\left[ \sum_{a=1}^{N_p}t^{|N_{\bullet}|}\left(s^{(p)}_{a}\right)^{-1} \right], \end{align} $$

while for $|t|\ll 1$ , we have

(5.50) $$ \begin{align} e_n\left[\frac{t^{|N_{\bullet}|+p+1}}{1-t^{nr}}+\sum_{a=1}^{N_p}s^{(p)}_{a} \right]=\sum_{k=0}^n\frac{t^{n|N_{\bullet}|+(n-k)(p+1)+r\binom{n-k}{2}}}{\displaystyle\prod_{l=1}^{n-k}(1-t^{rl})} e_k\left[ \sum_{a=1}^{N_p}t^{-|N_{\bullet}|}s^{(p)}_{a} \right]. \end{align} $$