Deformations of $\mathcal {W}$ algebras via quantum toroidal algebras

Abstract

We study the uniform description of deformed $\mathcal {W}$ algebras of type $\textsf {A}$ including the supersymmetric case in terms of the quantum toroidal ${\mathfrak {g}}{\mathfrak {l}}_1$ algebra ${{\mathcal {E}}}$. In particular, we recover the deformed affine Cartan matrices and the deformed integrals of motion. We introduce a comodule algebra $\mathcal {K}$ over ${{\mathcal {E}}}$ which gives a uniform construction of basic deformed $\mathcal {W}$ currents and screening operators in types $\textsf {B},\textsf {C},\textsf {D}$ including twisted and supersymmetric cases. We show that a completion of algebra $\mathcal {K}$ contains three commutative subalgebras. In particular, it allows us to obtain a commutative family of integrals of motion associated with affine Dynkin diagrams of all non-exceptional types except $\textsf {D}^{(2)}_{\ell +1}$. We also obtain in a uniform way deformed finite and affine Cartan matrices in all classical types together with a number of new examples, and discuss the corresponding screening operators.

Appendices

Appendix A: Proof of Theorem 4.11

In this Section we prove Theorem A. We have $C^2=\mu q_2$.

1.1 A.1: Commutativity $[{{\mathbf {I}}}_1,{{\mathbf {I}}}_2]=0$

As an illustration, let us verify the commutativity of ${{\mathbf {I}}}_1$ and ${{\mathbf {I}}}_2$.

We note that, by making use of the decomposition

$$\begin{aligned}&\omega _2(x)=q_2C^2f(x)\sigma _2(x)\sigma _2(x^{-1}),\quad \sigma _2(x)=(1-x)^3 \frac{(\mu x,\mu ^2 q_2x)_\infty }{(q_1^{-1}x,q_3^{-1}x)_\infty }, \end{aligned}$$

the integral (4.41) can be rewritten in terms of the currents (4.37) as

$$\begin{aligned} const.\ {{\mathbf {I}}}_n&= \int \!\!\cdots \!\!\int \varvec{E}(z_1,\ldots , z_n) \cdot \prod _{j\ne k}\sigma _2(z_k/z_j) \prod _{j=1}^n\frac{dz_j}{z_j}. \end{aligned}$$

Consider the products

$$\begin{aligned}&{{\mathbf {I}}}_1{{\mathbf {I}}}_2=\int \!\!\int \!\!\int \varvec{E}(z_1,z_2,z_3)\times f(z_2/z_1)^{-1}f(z_3/z_1)^{-1}\sigma _2(z_2/z_3)\sigma _2(z_3/z_2) \prod _{j=1}^3\frac{dz_j}{2\pi i z_j},\\&{{\mathbf {I}}}_2{{\mathbf {I}}}_1= \int \!\!\int \!\!\int \varvec{E}(z_1,z_2,z_3)\times f(z_1/z_2)^{-1}f(z_1/z_3)^{-1}\sigma _2(z_2/z_3)\sigma _2(z_3/z_2)\prod _{j=1}^3 \frac{dz_j}{2\pi i z_j}. \end{aligned}$$

The integral in ${{\mathbf {I}}}_1{{\mathbf {I}}}_2$ is initially defined for $|z_1|\gg |z_2|=|z_3|=1$, while in ${{\mathbf {I}}}_2{{\mathbf {I}}}_1$ it is defined for $|z_1|\ll |z_2|=|z_3|=1$. In both cases we move the contour for $z_1$ to the unit circle. Along the way we pick up residues at the poles $z_1=q_2^{-1}z_i, \mu ^{-1}q_2^{-1}z_i$ or $z_1=q_2z_i, \mu q_2 z_i$, $i=2,3$, respectively.

When all variables are on the unit circle, the two integrals coincide thanks to the identity (3.24).

Let us compare the residues at $z_1=q_2^{\pm 1}z_3$. We obtain respectively

$$\begin{aligned} J_1=\int \!\!\int _{|z_2|=|z_3|=1}\varvec{E}(z_2,q_2^{-1}z_3,z_3) f(q_2z_2/z_3)^{-1}\sigma _2(z_2/z_3)\sigma _2(z_3/z_2) \prod _{i=2}^3\frac{d z_i}{2\pi i z_i},\\ J_2=\int \!\!\int _{|z_2|=|z_3|=1}\varvec{E}(z_2,z_3,q_2z_3) f(q_2z_3/z_2)^{-1}\sigma _2(z_2/z_3)\sigma _2(z_3/z_2) \prod _{i=2}^3\frac{d z_i}{2\pi i z_i}. \end{aligned}$$

If we rename $z_3$ in $J_1$ to $q_2z_3$ (so that $q_2z_3$ is on the unit circle), then the two integrands become the same thanks to the identity

$$\begin{aligned} f(x)^{-1}\sigma _2(q_2^{-1}x)=\sigma _2(x)\frac{(1-x)^2(1-q_2^{-1}x)^2}{(1-\mu x)(1-\mu ^{-1}q_2^{-1}x)(1-q_1x)(1-q_3x)}. \end{aligned}$$

(A.1)

The integrand of $J_2$ has poles at (see Fig. 1)

$$\begin{aligned} z_3&=\mu ^mq_s^{-1}z_2\quad (s=1,3,\ m\ge 0);\\ z_3&=\mu ^{-m}q_sq_2^{-1}z_2\quad (s=1,3,\ m\ge 1);\\ z_3&=q_sz_2\quad (s=1,3). \end{aligned}$$

Among them the points $z_3=q_sz_2$ are inside the contour for $J_1$ (after renaming) and outside that for $J_2$. However these poles are actually absent due to the zero condition (4.38). Hence we have $J_1=J_2$.

Fig. 1

Integration contours on the $z_3$-plane ($s=1,3$) for $J_1$, $J_2$

Next let us consider the residues at $z_1=(\mu q_2)^{\pm 1}z_3$. Similarly as above we obtain

$$\begin{aligned} J'_1&=\int \!\!\int _{|z_2|=|z_3|=1}\varvec{E}(z_2,\mu ^{-1}q_2^{-1}z_3,z_3) f(\mu q_2 z_2/z_3)^{-1}\sigma _2(z_2/z_3)\sigma _2(z_3/z_2) \prod _{i=2}^3\frac{d z_i}{2\pi i z_i},\\ J'_2&=\int \!\!\int _{|z_2|=|z_3|=1}\varvec{E}(z_2,z_3,\mu q_2 z_3) f(\mu q_2 z_3/z_2)^{-1}\sigma _2(z_2/z_3)\sigma _2(z_3/z_2) \prod _{i=2}^3\frac{d z_i}{2\pi i z_i}. \end{aligned}$$

We have another identity

$$\begin{aligned} f(x)^{-1}\sigma _2(\mu ^{-1}q_2^{-1}x)=\sigma _2(x)\frac{(1-x)^2(1-\mu ^{-1}q_2^{-1}x)^2}{(1-\mu ^{-1}q_1 x)(1-\mu ^{-1}q_3 x)(1-q_1x)(1-q_3x)}. \end{aligned}$$

(A.2)

After renaming $z_3\rightarrow \mu q_2 z_3$ in $J_1'$, the two integrands become the same. The integrand of $J_2'$ has poles at (see Fig. 2)

$$\begin{aligned} z_3&=\mu ^mq_s^{-1}z_2\quad (m\ge 0, s=1,3);\\ z_3&=\mu ^{-m}q_2^{-1}q_sz_2\quad (m\ge 1, s=1,3);\\ z_3&=q_s z_2,\ \mu ^{-1}q_sz_2\quad (s=1,3). \end{aligned}$$

The last ones $z_3=q_sz_2,\ \mu ^{-1}q_sz_2$ ($s=1,3$) are inside the contour for $J_1'$, while they are outside for $J_2'$. Using the zero conditions (4.38), (4.39), we conclude that $J_1'=J_2'$.$\square$

Fig. 2

Integration contours on the $z_3$-plane ($s=1,3$) for $J_1'$, $J_2'$

1.2 A.2: The general case

We consider the general case $[{{\mathbf {I}}}_m,{{\mathbf {I}}}_n]=0$. Call the integration variables $z_1,\ldots ,z_m$ for ${{\mathbf {I}}}_m$ and $w_1,\ldots ,w_n$ for ${{\mathbf {I}}}_n$. We proceed in the same way, rewriting ${{\mathbf {I}}}_m{{\mathbf {I}}}_n$ and ${{\mathbf {I}}}_n{{\mathbf {I}}}_m$ as integrals over the unit circle and picking residues with respect to some groups of variables.

First consider ${{\mathbf {I}}}_m{{\mathbf {I}}}_n$. In view of symmetry and zeros on the diagonal, it is sufficient to consider residues from

$$\begin{aligned}&z_i=q_2^{-1}w_i\quad (1\le i\le k),\\&z_i=\mu ^{-1} q_2^{-1}w_i\quad (k+1\le i\le k+l). \end{aligned}$$

The result is (we write only the integrand)

$$\begin{aligned}&J_1=\varvec{E}(\{q_2^{-1}w_i,w_i\}_{i=1}^k,\{\mu ^{-1}q_2^{-1}w_i,w_i\}_{i=k+1}^{k+l}, \{z_j\}_{j=k+l+1}^m,\{w_j\}_{j=k+l+1}^n) \times F_1G_1H_1, \end{aligned}$$

with

$$\begin{aligned} F_1&=\prod _{1\le i\ne j\le k}f(q_2w_j/w_i)^{-1}\sigma _2(w_j/w_i)^2 \prod _{k+1\le i\ne j\le k+l}f(\mu q_2w_j/w_i)^{-1}\sigma _2(w_j/w_i)^2\\&\quad \times \prod _{\begin{array}{c} 1\le i\le k\\ k+1\le j\le k+l \end{array}}f(q_2w_j/w_i)^{-1}\sigma _2(\mu ^{-1}w_j/w_i)\sigma _2(w_i/w_j)\\&\quad \prod _{\begin{array}{c} k+1\le i\le k+l\\ 1\le j\le k \end{array}}f(\mu q_2w_j/w_i)^{-1}\sigma _2(\mu w_j/w_i)\sigma _2(w_i/w_j) , \\ G_1&=\prod _{j=k+l+1}^{m} \Bigl (\prod _{1\le i\le k}f(w_i/z_j)^{-1}\sigma _2(q_2^{-1}w_i/z_j)\sigma _2(q_2z_j/w_i)\\&\quad \quad \prod _{k+1\le i\le k+l}\quad f(w_i/z_j)^{-1}\sigma _2(\mu ^{-1}q_2^{-1}w_i/z_j)\sigma _2(\mu q_2z_j/w_i)\Bigr )\\&\quad \times \prod _{j=k+l+1}^{n} \Bigl ( \prod _{1\le i\le k}f(q_2w_j/w_i)^{-1}\sigma _2(w_j/w_i)\sigma _2(w_i/w_j) \\&\quad \prod _{k+1\le i\le k+l}f(\mu q_2w_j/w_i)^{-1}\sigma _2(w_j/w_i)\sigma _2(w_i/w_j)\Bigr ) \\ H_1&=\prod _{\begin{array}{c} k+l+1\le i\le m\\ k+l+1\le j\le n \end{array}}f(w_j/z_i)^{-1}\\&\quad \prod _{k+l+1\le i\ne j\le m}\sigma _2(z_j/z_i) \prod _{k+l+1\le i\ne j\le n}\sigma _2(w_j/w_i). \end{aligned}$$

For ${{\mathbf {I}}}_n{{\mathbf {I}}}_m$ we work similarly, picking poles at

$$\begin{aligned}&z_i=q_2w_i\quad (1\le i\le k),\\&z_i=\mu q_2 w_i\quad (k+1\le i\le k+l), \end{aligned}$$

ending up with

$$\begin{aligned}&J_2=\varvec{E}(\{q_2 w_i,w_i\}_{i=1}^k,\{\mu q_2 w_i,w_i\}_{i=k+1}^{k+l}, \{z_j\}_{j=k+l+1}^m,\{w_j\}_{j=k+l+1}^n) \times F_2G_2H_2, \end{aligned}$$

where

$$\begin{aligned} F_2&=\prod _{1\le i\ne j\le k}f(q_2w_j/w_i)^{-1}\sigma _2(w_j/w_i)^2\\&\quad \prod _{k+1\le i\ne j\le k+l}f(\mu q_2w_j/w_i)^{-1}\sigma _2(w_j/w_i)^2\\&\quad \times \prod _{\begin{array}{c} 1\le i\le k\\ k+1\le j\le k+l \end{array}}f(q_2w_i/w_j)^{-1}\sigma _2(w_i/w_j)\sigma _2(\mu w_j/w_i)\\&\quad \prod _{\begin{array}{c} k+1\le i\le k+l\\ 1\le j\le k \end{array}}f(\mu q_2w_j/w_i)^{-1}\sigma _2(w_j/w_i)\sigma _2(\mu ^{-1} w_i/w_j) , \\ G_2&=\quad \prod _{i=k+l+1}^{m}\quad \Bigl (\prod _{1\le j\le k}f(z_i/w_j)^{-1}\sigma _2(q_2w_j/z_i)\sigma _2(q_2^{-1}z_i/w_j)\quad \\&\quad \prod _{k+1\le i\le k+l}\quad f(z_i/w_j)^{-1}\sigma _2(\mu q_2w_j/z_i)\sigma _2(\mu ^{-1}q_2^{-1}z_i/w_j)\Bigr )\\&\quad \times \prod _{j=k+l+1}^{n} \Bigl ( \prod _{1\le i\le k}f(q_2w_i/w_j)^{-1}\sigma _2(w_i/w_j)\sigma _2(w_j/w_i) \\&\quad \prod _{k+1\le i\le k+l}f(\mu q_2w_i/w_j)^{-1}\sigma _2(w_i/w_j)\sigma _2(w_j/w_i)\Bigr ), \\ H_2&=\prod _{\begin{array}{c} k+l+1\le i\le m\\ k+l+1\le j\le n \end{array}}f(z_i/w_j)^{-1}\\&\quad \prod _{k+l+1\le i\ne j\le m}\sigma _2(z_j/z_i) \prod _{k+l+1\le i\ne j\le n}\sigma _2(w_j/w_i). \end{aligned}$$

Using the identities (A.1), (A.2) one can check that

$$\begin{aligned} J_1\bigr |_{w_i\rightarrow q_2w_i(1\le i\le k),\ w_j\rightarrow \mu q_2 w_j\ (k+1\le j\le k+l)}=J_2. \end{aligned}$$

The contours can be chosen to be the unit circle by the same mechanism observed above. It remains to show that under symmetrization with respect to $\{z_i\}_{i=k+l+1}^m$ and $\{w_j\}_{j=k+l+1}^n$ we have

$$\begin{aligned} \mathop {{\mathrm {Sym}}}H_1=\mathop {{\mathrm {Sym}}}H_2. \end{aligned}$$

This reduces to identity (3.24). The proof of Theorem 4.11 is now complete.$\square$

Appendix B: K matrices

It is well known that the Hopf algebra ${{\mathcal {E}}}$ is equipped with the universal R matrix, which gives rise to an intertwiner of ${{\mathcal {E}}}$ modules

$$\begin{aligned} {\check{\textsf {R}}}(u_1/u_2):{\mathcal {F}}_2(u_1)\otimes {\mathcal {F}}_2(u_2)\rightarrow {\mathcal {F}}_2(u_2)\otimes {\mathcal {F}}_2(u_1). \end{aligned}$$

(B.1)

Since ${\check{\textsf {R}}}(u_1/u_2)$ commutes with the diagonal Heisenberg $\Delta h_r$, it is written in terms of a single boson $\{\textsf {a}^\textsf {A}_r\}$ entering the root current $A(z)=\,\,:\Lambda _1(s_2z)\Lambda _2(s_2z)^{-1}$. Exhibiting the dependence on parameters we shall write

$$\begin{aligned} {\check{\textsf {R}}}(u_1/u_2)={\check{\textsf {R}}}(u_1/u_2;q_1,q_2,q_3,\{\textsf {a}^\textsf {A}_r\}). \end{aligned}$$

Proposition B.1

Consider a $\mathcal {K}$ module ${\mathcal {F}}_2(u)\otimes {\mathcal {F}}^X_c$ ($X=\textsf {B},\textsf {CD}$), and let $A(z)=\,\,:e^{\sum _{r\in {{\mathbb {Z}}}}\textsf {a}_rz^{-r}}$ be the root current associated with E(z). Then there exists an intertwiner of $\mathcal {K}$ modules

$$\begin{aligned}&\textsf {K}(u):{\mathcal {F}}_2(u)\otimes {\mathcal {F}}^{X}_c\rightarrow {\mathcal {F}}_2(u^{-1})\otimes {\mathcal {F}}^{X}_c\, \end{aligned}$$

in the following cases.

$$\begin{aligned} \hbox { type } \textsf {B}&\quad (X=\textsf {B}, c=3):&\textsf {K}(u)={\check{\textsf {R}}}(u;q_1,q_2q_3^{1/2},q_3^{1/2},\{\textsf {a}_r\}), \end{aligned}$$

(B.2)

$$\begin{aligned} \hbox { type } \textsf {C}&\quad (X=\textsf {CD}, c=3):&\textsf {K}(u)={\check{\textsf {R}}}(u^2;q_1,q_2q_3^{-1},q_3^2,\{\textsf {a}_r\}), \end{aligned}$$

(B.3)

$$\begin{aligned} \hbox { type } \textsf {D}&\quad (X=\textsf {CD}, c=2):&\textsf {K}(u)=(-1)^{\textsf {N}}. \end{aligned}$$

(B.4)

In the last line, $\textsf {N}$ denotes the number operator $\sum _{r>0}\nu _r^{-1}\textsf {a}_{-r}\textsf {a}_r$, $\nu _r=[\textsf {a}_r,\textsf {a}_{-r}]$. Note that $\textsf {K}(u)$ is independent of u in this case.

Proof

First note that, by extracting the diagonal Heisenberg, the intertwining relation (B.1) for type $\textsf {A}$ reduces to

$$\begin{aligned} {\check{\textsf {R}}}(u_1/u_2)\bigl (u_1\Lambda ^\textsf {A}_+(z)+u_2\Lambda ^\textsf {A}_-(z)\bigr ) =\bigl (u_2\Lambda ^\textsf {A}_+(z)+u_1\Lambda ^\textsf {A}_-(z)\bigr ){\check{\textsf {R}}}(u_1/u_2), \end{aligned}$$

where $\Lambda ^\textsf {A}_\pm (z)$ are vertex operators in $\{\textsf {a}^\textsf {A}_r\}_{r\ne 0}$ such that

$$\begin{aligned}&\Lambda ^\textsf {A}_-(z)=\,\,:\Lambda ^\textsf {A}_+(q_2z)^{-1}, \end{aligned}$$

(B.5)

$$\begin{aligned}&{\mathcal {C}}\bigl (\Lambda ^\textsf {A}_+(z),\Lambda ^\textsf {A}_+(w)\bigr )=-\frac{(1-q_3)(1-q_1)}{1+q_2}q_2. \end{aligned}$$

(B.6)

For type $\textsf {C}$ and type $\textsf {D}$ we proceed the same way as in type $\textsf {A}$ to obtain the reduced intertwining relation

$$\begin{aligned}&\textsf {K}(u)\bigl (u \Lambda _+(z)+u^{-1}\Lambda _-(z)\bigr )=\bigl (u^{-1}\Lambda _+(z)+u\Lambda _-(z)\bigr )\textsf {K}(u), \end{aligned}$$

where $\Lambda _\pm (z)$ are vertex operators in $\{\textsf {a}_r\}_{r\ne 0}$ such that

$$\begin{aligned}&\Lambda _-(z)=\,\,:\Lambda _+(q_2q_c^{-1}z)^{-1}. \end{aligned}$$

(B.7)

For type $\textsf {C}$ ($c=3$), we have further

$$\begin{aligned} {\mathcal {C}}\bigl (\Lambda _+(z),\Lambda _+(w)\bigr )=-\frac{(1-q^2_3)(1-q_1)}{1+q_2q_3^{-1}}q_2q_3^{-1}. \end{aligned}$$

(B.8)

Comparing (B.7), (B.8) with (B.5), (B.6), we obtain (B.3).

For type $\textsf {D}$ ($c=2$), (B.7) becomes $\Lambda _-(z)=\,\,:\Lambda _+(z)^{-1}$, so that the intertwining relation reduces further to $\textsf {K}(u)\textsf {a}_r=-\textsf {a}_r\textsf {K}(u)$ for all $r\ne 0$. This leads to the solution (B.4).

For type $\textsf {B}$, the reduced intertwining relation involves three terms

$$\begin{aligned} \textsf {K}(u)\bigl (u\Lambda _{++}(z)+k\Lambda _0(z)+u^{-1}\Lambda _{--}(z)\bigr ) =\bigl (u^{-1}\Lambda _{++}(z)+k\Lambda _0(z)+u\Lambda _{--}(z)\bigr )\textsf {K}(u), \end{aligned}$$

which corresponds to the qq character of the three dimensional representation of $U_q\widehat{{\mathfrak {s}}{\mathfrak {l}}}_2$. One can reduce it further to intertwining relation for the two dimensional one

$$\begin{aligned}&\textsf {K}(u)\bigl (u^{1/2} \Lambda _+(z)+u^{-1/2}\Lambda _-(z)\bigr )=\bigl (u^{-1/2}\Lambda _+(z)+u^{1/2}\Lambda _-(z)\bigr )\textsf {K}(u)\, \end{aligned}$$

by introducing $\Lambda _\pm (z)$ such that $\Lambda _{\pm \pm }(z)=\,\,:\Lambda _\pm (s_3^{1/2}z)\Lambda _\pm (s_3^{-1/2}z)$, $\Lambda _0(z)=\,\,:\Lambda _+(s_3^{1/2}z)\Lambda _-(s_3^{-1/2}z)$. The rest is similar to type $\textsf {C}$. $\square$

On tensor products (4.9), (4.10), we write the intertwiners as ${\check{\textsf {R}}}_{i,i+1}(u_i/u_{i+1})$ ($i=1,\ldots ,\ell -1$) indicating the tensor components where they act non-trivially. For the intertwiners involving boundary Fock modules we write as $\textsf {K}_\ell (u_\ell )$. The standard argument (with $\ell =2$) leads to the boundary Yang–Baxter equation

$$\begin{aligned}&{\check{\textsf {R}}}_{1,2}(u_1/u_2)\textsf {K}_2(u_1){\check{\textsf {R}}}_{1,2}(u_1u_2)\textsf {K}_2(u_2) =\textsf {K}_2(u_2){\check{\textsf {R}}}_{1,2}(u_1u_2)\textsf {K}_2(u_1){\check{\textsf {R}}}_{1,2}(u_1/u_2). \end{aligned}$$

(B.9)

As noted above, the K matrix $\textsf {K}_\ell$ of type $\textsf {D}$ is independent of $u_\ell$ and satisfies $\textsf {K}^2_\ell =1$. Comparing the intertwining relation with the definition of the currents $A_{\ell -1}(z)$ (4.15) and $A_\ell (z)$ (4.18), we see that

$$\begin{aligned} \textsf {K}_\ell A_{\ell -1}(z)=A_\ell (z) \textsf {K}_\ell . \end{aligned}$$

Though the zeroth node of the Dynkin diagram is not associated with boundary Fock modules, one can consider K matrices depending only on $A_0(z)$ and satisfying the intertwining relations, for example for type $\textsf {C}$

$$\begin{aligned} \textsf {K}_1(u_1)\bigl (\varvec{\Lambda }_{{{\bar{1}}}}(z)+\varvec{\Lambda }_1(\mu z)\bigr )= \bigl (u_1^2\varvec{\Lambda }_{{{\bar{1}}}}(z)+u_1^{-2}\varvec{\Lambda }_1(\mu z)\bigr )\textsf {K}_1(u_1). \end{aligned}$$

Appendix C: The library of Cartan matrices

1.1 C.1: Conventions

The matrix of contractions ${{\hat{B}}}$ is the deformed version of the symmetrized Cartan matrix. We will give a list of explicit deformed Cartan matrices ${{\hat{C}}}$ of low rank which can be used to write all others as explained in “Appendix C.4”.

The deformed symmetrized Cartan matrix ${{\hat{B}}}$ and the deformed Cartan matrix ${{\hat{C}}}$ are related by a diagonal matrix ${{\hat{D}}}$, namely ${{\hat{B}}}={{\hat{D}}}{{\hat{C}}}$, where the diagonal entries $d_i$ of ${{\hat{B}}}$ and the diagonal entries of ${{\hat{C}}}$ are given as follows.

The nodes of type $\textsf {A}$ (corresponding to ${\mathcal {F}}_{c_i}\otimes {\mathcal {F}}_{c_{i+1}}$).

The nodes of type $\textsf {B}$ (corresponding to ${\mathcal {F}}_{c_{\ell }}\otimes {\mathcal {F}}^{\textsf {B}}_{c_{\ell +1}}$).

The nodes of type $\textsf {C}$ (corresponding to ${\mathcal {F}}_{c_\ell }\otimes {\mathcal {F}}^{\textsf {C}\textsf {D}}_{c_{\ell +1}}$, $c_{\ell +1}\ne c_{\ell }$).

The nodes of type $\textsf {D}$ (corresponding to ${\mathcal {F}}_{c_{\ell -1}}\otimes {\mathcal {F}}_{c_\ell }\otimes {\mathcal {F}}^{\textsf {C}\textsf {D}}_{c_{\ell +1}}$, $c_{\ell +1}=c_{\ell }$).

Affine nodes (corresponding to dressing in non $\textsf {A}$ types) are the same as $\textsf {B}, \textsf {C},\textsf {D}$ nodes after the change

$$\begin{aligned} c_i\rightarrow c_{\ell +1-i}, \qquad d_i\rightarrow d_{\ell -i},\qquad C_{i,i}\rightarrow C_{\ell -i,\ell -i}. \end{aligned}$$

The deformed affine Cartan matrices we obtain have “local" form: $C_{i,j}=0$ if $|i-j|$ is sufficiently large. Moreover, the non-zero terms stabilize with $\ell$ increased. We give here a number of explicit deformed Cartan matrices of low rank which can be used to write all others as explained in “Appendix C.4”.

We use the following notation. We write $X(c_0;c_1,\dots ,c_{\ell -1};c_{\ell +1})Y$ for the choice of the module and the dressing and give the $(\ell +1)\times (\ell +1)$ matrix ${{\hat{C}}}$. Here Y can be $\textsf {B},\textsf {C}$, or $\textsf {D}$ depending on which boundary module we consider ${\mathcal {F}}_{c_{\ell +1}}^{\textsf {B}}$ or ${\mathcal {F}}_{c_{\ell +1}}^{\textsf {C}\textsf {D}}$. As always, in the latter case we choose $\textsf {D}$ if $c_{\ell +1}=c_\ell$ and $\textsf {C}$ otherwise. Similarly X can be $\textsf {B},\textsf {C}$, or $\textsf {D}$ depending on which dressing we choose. Namely $C^2/\mu =q_{c_0}^{-1/2}$ corresponds to $X=\textsf {B}$ while $C^2/\mu =q_{c_0}$ corresponds to $X=\textsf {C}$, if $c_0\ne c_1$ and to $X=\textsf {D}$ if $c_0=c_1$.

We skip writing X and $c_0$ in finite types. In addition we also skip $c_{\ell +1}$ and Y in $\textsf {A}$ type.

1.2 C.2: Finite types

Type $\textsf {A}$.

$$\begin{aligned} (2,2,2)&\left( \begin{matrix} s_2+s_2^{-1} &{}\quad -1\\ -1&{}\quad s_2+s_2^{-1} \end{matrix}\right) \\ (1,2,3)&\left( \begin{matrix} t_3 &{}\quad t_1 \\ t_3 &{}\quad t_1 \end{matrix}\right) \qquad \\ (1,2,1)&\left( \begin{matrix} t_3 &{}\quad t_1 \\ t_1 &{}\quad t_3 \end{matrix}\right) \qquad \\ (2,2,1)&\qquad \left( \begin{matrix} s_2+s_2^{-1} &{}\quad -1 \\ t_1 &{}\quad t_3 \end{matrix}\right) \ \\ (1,2,2)&\left( \begin{matrix} t_3 &{}\quad t_1 \\ -1 &{}\quad s_2+s_2^{-1} \end{matrix}\right) \ \end{aligned}$$

Type $\textsf {B}$.

$$\begin{aligned} (2,2;2)\textsf {B}&\begin{pmatrix} s_2+s_2^{-1}&{}\quad -1\\ -s_2^{1/2}-s_2^{-1/2}&{}\quad s_2^{3/2}+s_2^{-3/2}\\ \end{pmatrix}\quad \\ (2,2;3)\textsf {B}&\begin{pmatrix} s_2+s_2^{-1}&{}\quad -1\\ -s_3^{1/2}-s_3^{-1/2}&{}\quad s_1^{1/2}s_2^{-1/2}+s_1^{-1/2}s_2^{1/2} \\ \end{pmatrix} \\ (1,2;2)\textsf {B}&\begin{pmatrix} t_3&{}\quad t_1\\ -s_2^{1/2}-s_2^{-1/2}&{}\quad s_2^{3/2}+s_2^{-3/2}\\ \end{pmatrix}\quad \\ (1,2;1)\textsf {B}&\begin{pmatrix} t_3&{}\quad t_1\\ -s_1^{1/2}-s_1^{-1/2}&{}\quad s_3^{1/2}s_2^{-1/2}+s_3^{-1/2}s_2^{1/2} \\ \end{pmatrix} \\ (1,2;3)\textsf {B}&\begin{pmatrix} t_3&{}\quad t_1\\ -s_3^{1/2}-s_3^{-1/2}&{}\quad s_1^{1/2}s_2^{-1/2}+s_1^{-1/2}s_2^{1/2} \\ \end{pmatrix} \end{aligned}$$

Type $\textsf {C}$.    

$$\begin{aligned} (2,2;3)\textsf {C}&\begin{pmatrix} s_2+s_2^{-1}&{}\quad -s_3-s_3^{-1}\\ -1&{}\quad s_2s_3^{-1}+s_2^{-1}s_3 \\ \end{pmatrix} \\ (1,2;1)\textsf {C}&\begin{pmatrix} t_3&{}\quad s^2_1-s_1^{-2}\\ -1&{}\quad s_2s_1^{-1}+s_2^{-1}s_1 \\ \end{pmatrix}\ \ \\ (1,2;3)\textsf {C}&\begin{pmatrix} t_3&{}\quad -t_1(s_3+s_3^{-1})\\ -1&{}\quad s_2s_3^{-1}+s_2^{-1}s_3 \\ \end{pmatrix}\ \ \end{aligned}$$

Type $\textsf {D}$.    

$$\begin{aligned} (2,2,2;2)\textsf {D}&\begin{pmatrix} s_2+s_2^{-1}&{}\quad -1 &{}\quad -1\\ -1&{}\quad s_2+s_2^{-1}&{}\quad 0\\ -1&{}\quad 0&{}\quad s_2+s_2^{-1}\\ \end{pmatrix} \qquad \ \ \\ (1,2,2;2)\textsf {D}&\begin{pmatrix} t_3&{}\quad t_1 &{}\quad t_1\\ -1&{}\quad s_2+s_2^{-1}&{}\quad 0\\ -1&{}\quad 0&{}\quad s_2+s_2^{-1}\\ \end{pmatrix} \qquad \ \ \ \ \\ (2,1,2;2)\textsf {D}&\begin{pmatrix} t_3&{}\quad t_2&{}\quad t_2\\ t_2&{}\quad t_3&{}\quad s_1s_2^{-1}-s_1^{-1}s_2\\ t_2&{}\quad s_1s_2^{-1}-s_1^{-1}s_2&{}\quad t_3\\ \end{pmatrix}\ \ \ \ \\ (1,1,2;2)\textsf {D}&\begin{pmatrix} s_1+s_1^{-1}&{}\quad -1 &{}\quad -1\\ t_2&{}\quad t_3&{}\quad s_1s_2^{-1}-s_1^{-1}s_2\\ t_2&{}\quad s_1s_2^{-1}-s_1^{-1}s_2&{}\quad t_3\\ \end{pmatrix} \\ (3,1,2;2)\textsf {D}&\begin{pmatrix} t_2&{}\quad t_3&{}\quad t_3\\ t_2&{}\quad t_3 &{}\quad s_1s_2^{-1}-s_1^{-1}s_2\\ t_2&{}\quad s_1s_2^{-1}-s_1^{-1}s_2&{}\quad t_3\\ \end{pmatrix}\ \ \ \ \end{aligned}$$

1.3 C.3: Affine types

$2\times 2$ cases.

$\textsf {B}-\textsf {B}$ types.

$$\begin{aligned} \textsf {B}(2;2;2)\textsf {B}&&\begin{pmatrix} s_2^{3/2}+s_2^{-3/2}&{}\quad -s_2^{1/2}-s_2^{-1/2}\\ -s_2^{1/2}-s_2^{-1/2}&{}\quad s_2^{3/2}+s_2^{-3/2}\\ \end{pmatrix}\qquad \ \ \\ \textsf {B}(2;2;1)\textsf {B}&&\begin{pmatrix} s_2^{3/2}+s_2^{-3/2}&{}\quad -s_2^{1/2}-s_2^{-1/2}\\ -s^{1/2}_1-s_1^{-1/2}&{}\quad s_2^{1/2}s_3^{-1/2}+s_2^{-1/2}s_3^{1/2}\\ \end{pmatrix}\quad \ \\ \textsf {B}(1;2;1)\textsf {B}&&\begin{pmatrix} s_2^{1/2}s_3^{-1/2}+s_2^{-1/2}s_3^{1/2}&{}\quad -s^{1/2}_1-s_1^{-1/2}\\ -s^{1/2}_1-s_1^{-1/2}&{}\quad s_2^{1/2}s_3^{-1/2}+s_2^{-1/2}s_3^{1/2}\\ \end{pmatrix} \\ \textsf {B}(1;2;3)\textsf {B}&&\begin{pmatrix} s_2^{1/2}s_3^{-1/2}+s_2^{-1/2}s_3^{1/2}&{}\quad -s^{1/2}_1-s_1^{-1/2}\\ -s^{1/2}_3-s_3^{-1/2}&{}\quad s_2^{1/2}s_1^{-1/2}+s_2^{-1/2}s_1^{1/2}\\ \end{pmatrix} \end{aligned}$$

$\textsf {B}-\textsf {C}$ types.

$$\begin{aligned} \textsf {B}(2;2;1)\textsf {C}&&\begin{pmatrix} s_2^{3/2}+s_2^{-3/2}&{}\quad -(s_1+s_1^{-1})(s_2^{1/2}+s_2^{1/2})\\ -1&{}\quad s_1s^{-1}_2+s^{-1}_1s_2\\ \end{pmatrix}\qquad \\ \textsf {B}(1;2;1)\textsf {C}&&\begin{pmatrix} s_2^{1/2}s_3^{-1/2}+s_2^{-1/2}s_3^{1/2}&{}\quad -(s_1+s_1^{-1})(s^{1/2}_1+s_1^{-1/2})\\ -1&{}\quad s_1s^{-1}_2+s^{-1}_1s_2\\ \end{pmatrix} \\ \textsf {B}(3;2;1)\textsf {C}&&\begin{pmatrix} s_1^{1/2}s_2^{-1/2}+s_1^{-1/2}s_2^{1/2}&{}\quad -(s_1+s_1^{-1})(s^{1/2}_3+s_3^{-1/2})\\ -1&{}\quad s_1s^{-1}_2+s^{-1}_1s_2\\ \end{pmatrix} \end{aligned}$$

$\textsf {C}-\textsf {C}$ types.

$$\begin{aligned} \textsf {C}(1;2;1)\textsf {C}&&\begin{pmatrix} s_1s^{-1}_2+s_1^{-1}s_2&{}\quad -s_1-s_1^{-1}\\ -s_1-s_1^{-1}&{}\quad s_1s^{-1}_2+s_1s_2^{-1}\\ \end{pmatrix}\qquad \\ \textsf {C}(1;2;3)\textsf {C}&&\begin{pmatrix} s_1s^{-1}_2+s_1^{-1}s_2&{}\quad -s_3-s_3^{-1}\\ -s_1-s_1^{-1}&{}\quad s_3s^{-1}_2+s_3s_2^{-1}\\ \end{pmatrix}\qquad \end{aligned}$$

$3\times 3$ cases.

$\textsf {B}-\textsf {D}$ types.

$$\begin{aligned} \textsf {B}(2;2,2;2)\textsf {D}&\quad \begin{pmatrix} s_2^{3/2}+s_2^{-3/2}&{}\quad -s_2^{1/2}-s_2^{-1/2} &{}\quad -s_2^{1/2}-s_2^{-1/2} \\ -1 &{}\quad s_2+s_2^{-1}&{}\quad 0\\ -1 &{}\quad 0 &{}\quad s_2+s_2^{-1}\\ \end{pmatrix}\qquad \\ \textsf {B}(1;2,2;2)\textsf {D}&\quad \begin{pmatrix} s_2^{1/2}s_3^{-1/2}+s_2^{-1/2}s_3^{1/2}&{}\quad -s_1^{1/2}-s_1^{-1/2} &{}\quad -s_1^{1/2}-s_1^{-1/2} \\ -1 &{}\quad s_2+s_2^{-1}&{}\quad 0\\ -1 &{}\quad 0 &{}\quad s_2+s_2^{-1}\\ \end{pmatrix} \\ \textsf {B}(2;1,2;2)\textsf {D}&\quad \begin{pmatrix} s_1^{1/2}s_3^{-1/2}+s_1^{-1/2}s_3^{1/2}&{}\quad -s_2^{1/2}-s_2^{-1/2} &{}\quad -s_2^{1/2}-s_2^{-1/2}\\ t_2 &{}\quad t_3 &{}\quad s_1s_2^{-1}-s_1^{-1}s_2\\ t_2 &{}\quad s_1s_2^{-1}-s_1^{-1}s_2 &{}\quad t_3\\ \end{pmatrix} \\ \textsf {B}(1;1,2;2)\textsf {D}&\quad \begin{pmatrix} s_1^{3/2}+s_1^{-3/2}&{}\quad -s_1^{1/2}-s_2^{-1/2} &{}\quad -s_1^{1/2}-s_1^{-1/2} \\ t_2 &{}\quad t_3 &{}\quad s_1s_2^{-1}-s_1^{-1}s_2\\ t_2 &{}\quad s_1s_2^{-1}-s_1^{-1}s_2 &{}\quad t_3 \\ \end{pmatrix}\qquad \\ \textsf {B}(1;3,2;2)\textsf {D}&\quad \begin{pmatrix} s_2^{1/2}s_3^{-1/2}+s_2^{-1/2}s_3^{1/2}&{}\quad -s_1^{1/2}-s_1^{-1/2} &{}\quad -s_1^{1/2}-s_1^{-1/2} \\ t_2 &{}\quad t_1 &{}\quad s_3s_2^{-1}-s_3^{-1}s_2\\ t_2 &{}\quad s_3s_2^{-1}-s_3^{-1}s_2 &{}\quad t_1\\ \end{pmatrix} \end{aligned}$$

$\textsf {C}-\textsf {D}$ types.

$$\begin{aligned}&\textsf {C}(1;2,2;2)\textsf {D}\quad \qquad \begin{pmatrix} s_1s_2^{-1}+s^{-1}_1s_2&{}\quad -1 &{}\quad -1 \\ -s_1-s_1^{-1} &{}\quad s_2+s_2^{-1}&{}\quad 0\\ -s_1-s_1^{-1} &{}\quad 0 &{}\quad s_2+s_2^{-1}\\ \end{pmatrix}\qquad \qquad \\&\textsf {C}(1;3,2;2)\textsf {D}\quad \qquad \begin{pmatrix} s_1s_3^{-1}+s^{-1}_1s_3&{}\quad -1 &{}\quad -1 \\ (s_1+s_1^{-1})t_2 &{}\quad t_1 &{}\quad s_2^{-1}s_3-s_2s_3^{-1}\\ (s_1+s_1^{-1})t_2 &{}\quad s_2^{-1}s_3-s_2s_3^{-1} &{}\quad t_1 \\ \end{pmatrix} \\&\textsf {C}(2;1,2;2)\textsf {D}\quad \qquad \begin{pmatrix} s_1s_2^{-1}+s^{-1}_1s_2&{}\quad -1 &{}\quad -1 \\ s^2_2-s_2^{-2} &{}\quad t_3 &{}\quad s_2^{-1}s_1-s_2s_1^{-1}\\ s^2_2-s_2^{-2} &{}\quad s_2^{-1}s_1-s_2s_1^{-1} &{}\quad t_3\\ \end{pmatrix}\quad \end{aligned}$$

$\textsf {D}-\textsf {D}$ types.

$$\begin{aligned} \textsf {D}(2;2,2;2)\textsf {D}&\quad \begin{pmatrix} s_2+s_2^{-1}&{}\quad 0 &{}\quad -s_2-s_2^{-1} \\ 0 &{}\quad s_2+s_2^{-1} &{}\quad 0\\ -s_2-s_2^{-1} &{}\quad 0 &{}\quad s_2+s_2^{-1}\\ \end{pmatrix}\qquad \quad \\ \textsf {D}(1;1,2;2)\textsf {D}&\quad \begin{pmatrix} t_3&{}\quad s_1^{-1}s_2-s_1s_2^{-1} &{}\quad -t_3 \\ s_1^{-1}s_2-s_1s_2^{-1} &{}\quad t_3 &{}\quad -s_1^{-1}s_2+s_1s_2^{-1}\\ -t_3 &{}\quad -s_1^{-1}s_2+s_1s_2^{-1} &{}\quad t_3\\ \end{pmatrix} \end{aligned}$$

$4\times 4$ cases   

$\textsf {D}-\textsf {D}$ types.

$$\begin{aligned} \textsf {D}(2;2,2,2;2)\textsf {D}&\quad \begin{pmatrix} s_2+s^{-1}_2&{}\quad 0 &{}\quad -1 &{}\quad -1 \\ 0&{}\quad s_2+s^{-1}_2 &{}\quad -1 &{}\quad -1 \\ -1 &{}\quad -1 &{}\quad s_2+s^{-1}_2 &{}\quad 0 \\ -1 &{}\quad -1 &{}\quad 0 &{}\quad s_2+s^{-1}_2 \\ \end{pmatrix}\qquad \qquad \quad \\ \textsf {D}(2;2,1,2;2)\textsf {D}&\quad \begin{pmatrix} t_3 &{}\quad s_1s_2^{-1}-s_1^{-1}s_2 &{}\quad t_2 &{}\quad t_2 \\ s_1s_2^{-1}-s_1^{-1}s_2&{}\quad t_3 &{}\quad t_2 &{}\quad t_2 \\ t_2 &{}\quad t_2 &{}\quad t_3 &{}\quad s_1s_2^{-1}-s_1^{-1}s_2 \\ t_2 &{}\quad t_2 &{}\quad s_1s_2^{-1}-s_1^{-1}s_2 &{}\quad t_3 \\ \end{pmatrix} \\ \textsf {D}(1;1,2,2;2)\textsf {D}&\quad \begin{pmatrix} t_3&{}\quad s_1^{-1}s_2-s_1s_2^{-1} &{}\quad t_1 &{}\quad t_1 \\ s_1^{-1}s_2-s_1s_2^{-1}&{}\quad t_3 &{}\quad t_1 &{}\quad t_1 \\ -1 &{}\quad -1 &{}\quad s_2+s^{-1}_2 &{}\quad 0 \\ -1 &{}\quad -1 &{}\quad 0 &{}\quad s_2+s^{-1}_2 \\ \end{pmatrix}\qquad \\ \textsf {D}(1;1,3,2;2)\textsf {D}&\quad \begin{pmatrix} t_2 &{}\quad s_1^{-1}s_3-s_1s_3^{-1} &{}\quad t_1 &{}\quad t_1 \\ s_1^{-1}s_3-s_1s_3^{-1}&{}\quad t_2 &{}\quad t_1 &{}\quad t_1 \\ t_2 &{}\quad t_2 &{}\quad t_1 &{}\quad s_2^{-1}s_3-s_2s_3^{-1} \\ t_2 &{}\quad t_2 &{}\quad s_2^{-1}s_3-s_2s_3^{-1} &{}\quad t_1 \\ \end{pmatrix} \end{aligned}$$

1.4 C.4: General case

In “Appendix C.2” we gave several deformed Cartan matrices of finite type, and in “Appendix C.3” several deformed Cartan matrices of affine type. In fact “Appendix C.2” contains all maximal submatrices of stable deformed affine Cartan matrices whose upper right and bottom left corner elements are both non zero. “Appendix C.3” contains all non-stable deformed affine Cartan matrices appearing in our construction.

It is important to keep in mind that we immediately obtain many more examples by permuting $s_1,s_2,s_3$. Another set of examples is obtained by reading the data from right to left. More precisely, the matrices ${{\hat{C}}}$ corresponding to $X(c_0;c_1,\dots ,c_{\ell };c_{\ell +1})Y$ and to $Y(c_{\ell +1};c_{\ell },\dots ,c_1;c_0)X$ are related by conjugation by the matrix $(\delta _{i+j,\ell })_{i,j=0}^\ell$.

The matrices listed in “Appendices C.2 and C.3” allow us to write a deformed affine Cartan matrix corresponding to arbitrary data $X(c_0;c_1,\dots ,c_{\ell };c_{\ell +1})Y$, since for larger $\ell$ the deformed Cartan matrices stabilize. One has to follow the following procedure.

First, search the affine examples, keeping in mind the symmetries. If the matrix is found, stop. If the matrix is not in the list, conclude that ${{\hat{C}}}$ is stable, that is $C_{0,\ell }=C_{\ell ,0}=0$.

Second, find the listed matrix of the finite type and of the largest size corresponding to right most colors: $(c_{\ell -i},\dots ,c_{\ell };c_{\ell +1})Y$ (it is $2\times 2$ for all cases except $Y=\textsf {D}$ when it is $3\times 3$). That gives the right bottom submatrix of ${{\hat{C}}}$.

Third, repeat for left most colors, that is, find the listed matrix of the finite type and of the largest size corresponding to $X(c_0;c_1,\dots ,c_i)$. This gives the left upper submatrix.

The rest nonzero entries of the matrix are recovered from matrices of type $\textsf {A}$.

The final result is the superposition of matrices which are linked via diagonal entries.

For example, the matrix ${{\hat{C}}}$ corresponding to $\textsf {B}(2;3,1;3)\textsf {C}$ is a superposition of the last matrix of type $\textsf {B}$ in our list and of the second matrix of type $\textsf {C}$ with necessary symmetries (the $C_{1,1}$ entry $t_2$ is common):

$$\begin{aligned} \begin{pmatrix} s_2^{1/2}s_3+s_2^{-1/2}s_3^{-1} &{}\quad -s_2^{1/2}-s_2^{-1/2} &{}\quad 0 \\ t_1 &{}\quad t_2 &{}\quad s_3^2-s_3^{-2} \\ 0 &{}\quad -1 &{}\quad s_2s_3^{-1}+s_2^{-1}s_3 \\ \end{pmatrix}. \end{aligned}$$

Similarly, matrix ${{\hat{C}}}$ corresponding to $\textsf {B}(2;3,2,1;3)\textsf {C}$ is a superposition of three: the fourth on type $\textsf {B}$ list, he second on type $\textsf {A}$ list and the last on type $\textsf {C}$ list. The first two share a common $C_{1,1}$ entry $t_1$ and the last two share a common $C_{2,2}$ entry $t_3$.

$$\begin{aligned} \begin{pmatrix} s_2^{1/2}s_3+s_2^{-1/2}s_3^{-1} &{}\quad -s_3^{1/2}-s_3^{-1/2} &{}\quad 0 &{}\quad 0 \\ t_3 &{}\quad t_1 &{}\quad t_3 &{}\quad 0 \\ 0 &{}\quad t_1 &{}\quad t_3 &{}\quad -t_2(s_3+s_3^{-1})\\ 0 &{}\quad 0&{}\quad -1 &{}\quad s_1s_3^{-1}+s_1^{-1}s_3\\ \end{pmatrix}. \end{aligned}$$