Solutions to graded reflection equation of GL-type

Abstract

We list solutions of the graded reflection equation associated with the fundamental vector representation of a quantum supergroup of GL-type.

1 Introduction

There are two fundamental equations in the theory of quantum integrable models. The first one is the Yang–Baxter equation (YBE) which is a key ingredient of quantum inverse scattering method; it gave rise to quantum groups [1, 2]. The other is reflection equation (RE) also known as the boundary YBE [3,4,5]. It appears in models with open boundary conditions [6, 7] and is related with coideal subalgebras and quantum homogeneous spaces [8,9,10,11]. The theory of quantum symmetric pairs is developed in great detail in [12, 13], and the corresponding universal solutions to RE have been constructed in [14, 15].

Reflection equation associated with infinite dimensional quantum groups lead to solutions (K-matrices) depending on spectral parameter, in contrast with constant solutions, which are related to finite dimensional quantum groups. Classification of K-matrices is a fundamental problem, which is completely solved to date only for $\mathfrak {g}\mathfrak {l}(N)$: for the constant case in [11] and for those (invertible) with spectral parameter in [16]. Although the constant version of RE is of less demand in applications to physics, it plays a role in such a classification thanks to a baxterization procedure [17, 18]. It has been proved in [16] that every invertible K-matrix of the affine quantum group $U_q\bigl (\widehat{\mathfrak {g}\mathfrak {l}(N)}\bigr )$ can be obtained from a baxterized constant K-matrix of $U_q\bigl (\mathfrak {g}\mathfrak {l}(N)\bigr )$ via a family of transformations called equivalence. Thus, [16] is not just listing K-matrices with spectral parameter but clarifies their structure. In this context, classification of constant K-matrices becomes an important first step.

A supersymmetric version of RE was addressed in the literature with a focus on affine quantum supergroups and super-Yangians [19,20,21,22,23] and for special gradings of the underlying vector space. It is therefore interesting to obtain a classification of constant solutions to RE for an arbitrary grading, as a first step of the classification programme of [16]. We do this for the general linear quantum supergroups.

We give a complete classification of the solutions to RE of the aforementioned type. Our method is motivated by an earlier result of [11] for the non-graded case. It turns out that solutions of the graded RE are exactly even matrices that solve the non-graded RE. When it comes to non-degenerate K-matrices, solutions exist only for a special “symmetric” grading that is preserved by the longest element of the symmetric group acting on a weight basis of the defining representation.

In the next section, we give basics of the graded RE and prove its equivalence to a non-graded RE with an appropriate R-matrix. For even solutions, that is exactly the R-matrix satisfying the equivalent non-graded YBE equations. We use that correspondence and reduce the study to a non-graded RE. That allowed us to adapt the reasoning of [11] to the current exposition. We closely follow [11] making appropriate modifications as per grading, along with some additions and clarifications. The main result is stated in Sect. 3. The proof is arranged in a sequence of nine lemmas in Sect. 4. Baxterization of constant solutions to spectral parameter-dependent RE solutions is considered in Sect. 5.

2 $\mathbb {Z}_2$-graded Yang–Baxter and reflection equations

Let us recall basic definitions of graded algebras and modules.

An associative algebra $\mathcal {A}$ is called $\mathbb {Z}_2$-graded if it is presentable as sum of vector spaces $\mathcal {A}=\mathcal {A}_0\oplus \mathcal {A}_1$ satisfying $\mathcal {A}_i\mathcal {A}_j\subset \mathcal {A}_{i+j\mod 2}$. In particular, $\mathcal {A}_0$ is a subalgebra and $\mathcal {A}_1$ is a left/right $\mathcal {A}_0$-module. Elements of $\mathcal {A}_0$ are called even and elements of $\mathcal {A}_1$ are called odd. An $\mathcal {A}$-module V is called graded if it decomposes to a direct sum $V=V_0\oplus V_1$ such that $\mathcal {A}_i V_j\subset V_{i+j\mod 2}$.

An example is a complex vector space $V=\mathbb {C}^N$ with a basis $\{v_i\}_{i=1}^N$ equipped with an arbitrary parity function $\{1,\ldots ,N\}\rightarrow \{0,1\}$, $i\mapsto [i]=\deg (v_i)$. Then, $V_i=\textrm{Span}\{e_k\}_{[k]=i}$, $i=0,1$. It is a graded module over the algebra $\mathcal {A}=\textrm{End}(V)$ whose graded components are set to be $\mathcal {A}_i=\textrm{Span}\{e_{lk}\}_{[l]+[k]=i\mod 2}$, $i=0,1$. Here, $e_{lk}$ are the matrix units acting on the basis vectors by $e_{lk}v_j=\delta _{kj}v_l$.

Given two graded algebras $\mathcal {A}$ and $\mathcal {B}$, their tensor product $\mathcal {A}\otimes \mathcal {B}$ is a graded algebra too. The multiplication on homogeneous elements is defined by the rule

$$\begin{aligned} (a_1\otimes b_1)(a_2\otimes b_2)=(-1)^{\deg (a_2)\deg (b_1)}a_1 a_2\otimes b_1b_2. \end{aligned}$$

One has $(\mathcal {A}\otimes \mathcal {B})_0=(\mathcal {A}_0\otimes \mathcal {B}_0)\oplus (\mathcal {A}_1\otimes \mathcal {B}_1)$ and $(\mathcal {A}\otimes \mathcal {B})_1=(\mathcal {A}_1\otimes \mathcal {B}_0)\oplus (\mathcal {A}_0\otimes \mathcal {B}_1)$.

In the special case when $\mathcal {A}=\mathcal {B}=\textrm{End}(V)$, there is a $\mathbb {Z}_2$-graded isomorphism

$$\begin{aligned} \varphi :\textrm{End}(V)\otimes \textrm{End}(V)\rightarrow \textrm{End}(V\otimes V), \quad e_{ij}\otimes e_{lk}\mapsto (-1)^{[j]([l]+[k])} e_{il,jk}, \end{aligned}$$

(1)

where $e_{il,jk}\in \textrm{End}(V\otimes V)$ are matrix units acting by $e_{il,jk}(v_{m}\otimes v_n)=\delta _{lm}\delta _{kn}v_{i}\otimes v_j$. Given $F=\sum _{ijlk}F_{ij,lk}e_{ij}\otimes e_{lk}\in \textrm{End}(V)\otimes \textrm{End}(V)$, we will write ${\tilde{F}}=\sum _{ijlk}{\tilde{F}}^{il}_{jk}e_{il,jk} \in \textrm{End}(V\otimes V)$ for its $\varphi$-image. The entries are related by ${\tilde{F}}^{il}_{jk}=(-1)^{[j]([l]+[k])}F_{ij,lk}$.

Suppose V is a graded vector space and $\mathcal {A}=\textrm{End}(V)$ is the corresponding graded matrix algebra. An invertible element $R\in \mathcal {A}\otimes \mathcal {A}$ is called an R-matrix if it satisfies Yang–Baxter equation

$$\begin{aligned} R_{12}R_{13}R_{23}=R_{23}R_{13}R_{12}, \end{aligned}$$

where the subscripts indicate the tensor factor in the graded tensor cube of $\textrm{End}(V)$. It is usually assumed that R is even. In terms of the matrix ${\tilde{R}}\in \textrm{End}(V\otimes V)$, the YBE reads

$$\begin{aligned} \sum _{a,b,c}{\tilde{R}}_{ca}^{lm}{\tilde{R}}_{ib}^{cn}{\tilde{R}}_{jk}^{ab} (-1)^{[a][b]+[n][a]}= \sum _{a,b,c}{\tilde{R}}_{bc}^{mn}\tilde{R}_{ak}^{lc}{\tilde{R}}_{ij}^{ab}(-1)^{[b][c]+[k][b]}. \end{aligned}$$

An even element $P=\sum _{i=1}^{n}(-1)^{[j]}e_{ij}\otimes e_{ji}\in \textrm{End}(V)\otimes \textrm{End}(V)$ is called graded permutation. It features

$$\begin{aligned} P(v\otimes w)=(-1)^{\deg (v)\deg (w)}w\otimes v \end{aligned}$$

for all homogeneous $v,w\in V$. Then, the operator $S=PR\in \textrm{End}(V)\otimes \textrm{End}(V)$ satisfies the braid relation

$$\begin{aligned} S_{12}S_{23}S_{12}=S_{23}S_{12}S_{23}. \end{aligned}$$

A matrix $A\in \textrm{End}(V)$ is said to satisfy the (graded) reflection equation if

$$\begin{aligned} SA_2S A_2= A_2 S A_2 S \end{aligned}$$

in $\textrm{End}(V)\otimes \textrm{End}(V)$. We will generally not assume that A is even.

It is known that graded YBE is equivalent to the non-graded YBE with the R-matrix with entries $\breve{R}^{ij}_{kl}=(-1)^{[i][j]+[j][k]+[k][l]}R_{ik,jl }$. Respectively, the graded braid equation with matrix S goes over to the non-graded braid equation with matrix elements $\breve{S}^{ij}_{kl}=\breve{R}^{ji}_{kl}$.

Proposition 2.1

The graded RE on $A\in \textrm{End}(V)$ is equivalent to the non-graded RE

$$\begin{aligned} {\hat{S}} A_2 {\hat{S}} A_2 =A_2 {\hat{S}} A_2 {\hat{S}}, \end{aligned}$$

(2)

with ${\hat{S}}^{ij}_{kl}=\breve{S}^{ij}_{kl}(-1)^{[i][j]+[k][l]}$, and ${\hat{S}}^{ij}_{kl}=\breve{S}^{ij}_{kl}$ under the assumption that A is even.

Proof

Apply the isomorphism $\varphi$ defined in (1) to the graded RE and get

$$\begin{aligned} {\tilde{S}}{\tilde{A}}_2{\tilde{S}} {\tilde{A}}_2={\tilde{A}}_2{\tilde{S}} \tilde{A}_2{\tilde{S}}. \end{aligned}$$

Observe that ${\tilde{S}}=\breve{S}$. Plugging in the entries of the tensor ${\tilde{A}}_2$ expressed through the matrix A, rewrite the equation as

$$\begin{aligned}{} & {} \sum _{a,b,c,d}{\tilde{S}}^{ij}_{ab}A_{bc}(-1)^{[a]([b]+[c])}\tilde{S}^{ac}_{md}(-1)^{[m]([d]+[n])} A_{dn}\\{} & {} \quad = \sum _{a,b,c,d}A_{ja}(-1)^{[i]([j]+[a])}\tilde{S}^{ia}_{bc} A_{cd}(-1)^{[b]([c]+[d])}{\tilde{S}}^{bd}_{mn}. \end{aligned}$$

This turns to the non-graded RE with braid matrix ${\tilde{S}}$ if A is even. For arbitrary A, multiplication of the equation by $(-1)^{[i][j]+[m][n]}$ proves the statement in general. $\square$

3 Classification of solutions

We fix a graded R matrix of the GL-type in the fundamental vector representation $V=\mathbb {C}^N$ with an arbitrary grading on the basis elements:

$$\begin{aligned} R=\sum _{i,j} e_{ii}\otimes e_{jj}q^{(-1)^{[i]}\delta _{ij}}+\omega \sum _{i>j}e_{ij}\otimes e_{ji}(-1)^{[j]}\in \textrm{End}(V)\otimes \textrm{End}(V), \end{aligned}$$

where $\omega$ stands for $=q-q^{-1}$. This matrix differs from the one in [24] in the sign of inequality in the second sum.

In what follows, we will work with the equivalent non-graded R-matrix and use the symbol $\otimes$ to denote the non-graded tensor product. In order to make double indexing more readable, we will use upper and lower indices. The matrix units $e_{ij}$ will be replaced with $e^i_j$.

The equivalent non-graded R-matrix is

$$\begin{aligned} \breve{R}=\sum _{i,j} (-1)^{[i][j]}q^{(-1)^{[i]}\delta _{ij}}e^{i}_j\otimes e^i_{j}+\omega \sum _{i<j}e^i_j\otimes e^j_i. \end{aligned}$$

It is convenient to represent the corresponding braid matrix in the form

$$\begin{aligned} {\hat{S}} =\breve{S}= & {} \sum _{i,j} (-1)^{[i][j]}q^{(-1)^i\delta _{ij}}e^j_i\otimes e^i_j +\omega \sum _{i<j}e^j_j\otimes e^i_i\> \nonumber \\= & {} \sum _{i,k} s_{ik}e^{ik}_{ik}+\sum _{i\not =j} (-1)^{[i][j]}e^{ij}_{ji},\quad \text{ where }\; s_{ik}= \left\{ \begin{array}{ll} \omega ,&{} \quad i <k,\\ (-1)^{i}q^{(-1)^i},&{} \quad i =k,\\ 0,&{} \quad i >k. \end{array} \right. \end{aligned}$$

The transposed matrix ${\hat{S}}_{21}$ is equivalent to the graded R-matrix from [24].

We denote by $[a,b]\subset \mathbb {Z}$ the intervals $\{k\in \mathbb {Z}| a\le k \le b\}$. Respectively, we use parentheses for the intervals defined by strict inequalities. Classification theorem is formulated with the use of the following data.

Definition 3.1

An admissible pair $(Y,\sigma )$ consists of an ordered subset $Y\subset I=[1,N]$ and a parity preserving strictly decreasing map $\sigma :Y \rightarrow I$ without stable points.

We distinguish two non-intersecting subsets $Y_+=\{i\in Y| i>\sigma (i)\}$ and $Y_-=\{i\in Y| i<\sigma (i)\}$; obviously $Y=Y_-\cup Y_+$. Denote $b_-=\max \{Y_-\cup \sigma (Y_+)\}$ and $b_+=\min \{Y_+\cup \sigma (Y_-)\}$. Because the $\sigma$ is decreasing, we have $b_-<b_+$. We adopt the convention $b_-=0$ and $b_+=N+1$ if $Y=\varnothing$.

Theorem 3.2

General solution to the graded RE of the GL-type is a matrix A of the form

$$\begin{aligned} A= & {} \sum _{i\in I} x_i e^i_i + \sum _{j\in Y}y_j e^{\sigma (j)}_j, \hspace{28pt} \end{aligned}$$

(3)

where

• $(Y,\sigma )$ is an admissible pair with

  $$\begin{aligned} Y= & {} [1, b_-]\cup [b_+, b_+ + b_- -1] , \end{aligned}$$

  (4)
  $$\begin{aligned} \sigma (i)= & {} b_+ + b_- -i \end{aligned}$$

  (5)

  for $b_-,b_+\in \mathbb {N}$ subject to the conditions $b_-<b_+$, $b_- + b_+\le N+1$ and

  $$\begin{aligned} x_i= & {} \left\{ \begin{array}{ll} \lambda +\mu ,&{} i\in [1,b_-],\\ \lambda ,&{} i\in (b_-,b_+),\\ 0,&{} i\in [b_+,N], \end{array} \right. \end{aligned}$$

  (6)
  $$\begin{aligned} y_i y_{\sigma (i)}= & {} -\lambda \mu \not =0 \end{aligned}$$

  (7)

  for $\lambda ,\mu \in \mathbb {C}$;

• $(Y,\sigma )$ is an admissible pair such that $Y \cap \sigma (Y)=\varnothing$ and

  $$\begin{aligned} x_i= & {} \left\{ \begin{array}{cc} \lambda ,&{} \quad i\in [1,b],\\ 0,&{} \quad i\in (b,N], \end{array} \right. \end{aligned}$$

  (8)
  $$\begin{aligned} y_i&\quad \not =&0 \end{aligned}$$

  (9)

  for $b\in [b_-,b_+)$ and $\lambda \in \mathbb {C}$.

As follows from the theorem, there are two classes of numerical RE matrices, those corresponding to $\sigma (Y)=Y$ and $\sigma (Y)\cap Y=\varnothing$. We call them solutions of Type 1 and 2, respectively.

Theorem 3.2 asserts that solutions of graded RE are exactly even matrices from [11] solving the non-graded RE of $\mathfrak {g}\mathfrak {l}(N)$. Different gradings impose different restricting conditions on A. So, invertible A (they belong to Type 1) is even only for a “symmetric” grading of the underlying vector space:

$\left( \underbrace{0, \ldots , 0}_\ell ;\underbrace{1,\ldots ,1}_{N-2\ell },\underbrace{0, \ldots , 0}_\ell \right) .$

or its opposite.

4 Proof of the classification theorem

The proof of Theorem (3.2) formulated in the previous section is combinatorial. It is a result of the direct analysis of Eq.  (2) organized into a sequence of lemmas. Explicitly (2) reads

$$\begin{aligned} {\hat{S}}A_2{\hat{S}}A_2= & {} \sum _{i,\beta ,\alpha ,\nu } s_{i\beta }s_{i\nu } A^{\nu }_\beta A^\alpha _{\nu } \; e^i_i\otimes e^\beta _\alpha + \sum _{{\begin{array}{c} {i,j,\beta ,\alpha }\\ {j\not = i} \end{array}}} (-1)^{[i][j]}s_{i\alpha }A^{j}_\alpha A^\beta _i \; e^i_j\otimes e^\alpha _\beta \end{aligned}$$

(10)

$$\begin{aligned}+ & {} \sum _{{\begin{array}{c} {i,\beta ,\alpha ,\nu }\\ {i\not = \beta } \end{array}}} (-1)^{[i][\beta ]} s_{i\nu } A^{\nu }_\beta A^\alpha _{\nu } \; e^\beta _i\otimes e^i_\alpha + \sum _{{\begin{array}{c} {i,j,\beta ,\alpha }\\ {j\not = i \not = \alpha } \end{array}}} (-1)^{[i]([\alpha ]+[j])} A^{j}_\alpha A^\beta _i \; e^\alpha _j\otimes e^i_\beta ,\nonumber \\ A_2{\hat{S}}A_2{\hat{S}}= & {} \sum _{i,\beta ,\alpha ,\nu } s_{i\alpha }s_{i\nu } A^{\nu }_\beta A^\alpha _{\nu } \; e^i_i\otimes e^\beta _\alpha + \sum _{{\begin{array}{c} {i,j,\beta ,\alpha }\\ {j\not = i} \end{array}}} (-1)^{[i][j]} s_{j\beta }A^{j}_\alpha A^\beta _i \; e^i_j\otimes e^\alpha _\beta \nonumber \\+ & {} \sum _{{\begin{array}{c} {i,\beta ,\alpha ,\nu }\\ {i\not = \alpha } \end{array}}}(-1)^{[i][\alpha ]} s_{i\nu } A^{\nu }_\beta A^\alpha _{\nu } \; e^i_\alpha \otimes e^\beta _i + \sum _{{\begin{array}{c} {i,j,\beta ,\alpha }\\ {i\not = j \not = \beta } \end{array}}} (-1)^{[i]([\beta ]+[j])}A^{j}_\alpha A^\beta _i \; e^i_\beta \otimes e^\alpha _j. \nonumber \\ \end{aligned}$$

(11)

Comparison of similar terms gives rise to the system of quadratic equations on the matrix elements $A^i_j$.

Lemma 4.1

A general solution to (2) is an even matrix A satisfying the following system of equations:

$$\begin{aligned} \left\{ \begin{array}{rrl} A^m_i A^n_i &{}=&{}0, \\ A^i_m A^i_n &{}=&{}0, \end{array} \right. \hspace{44pt} \quad m\not = n \not = i \not = m , \hspace{20pt} \end{aligned}$$

(12)

$$\begin{aligned} A^n_i A^j_m \hspace{8pt}=\hspace{8pt}0, \hspace{36pt} \quad \left\{ \begin{array}{l} j\not = m \not = n\not =i,\\ (m-i)(n-j)<0, \end{array} \right. \end{aligned}$$

(13)

$$\begin{aligned} \left\{ \begin{array}{rrll} ((-1)^{[i]})q^{(-1)^{[i]})} - s_{mi}) A^i_i A^m_i &{}=&{} \sum _{\nu }s_{i\nu } A^{\nu }_i A^m_{\nu },&{} i \not = m , \\ ((-1)^{[i]})q^{(-1)^{[i]})} - s_{mi}) A^i_i A^i_m &{}=&{} \sum _{\nu }s_{i\nu } A^{\nu }_m A^i_{\nu },&{} i \not = m , \\ 0&{}=&{}\sum _{\nu }s_{i\nu } A^{\nu }_m A^n_{\nu } ,&{} (m-i)(n-i)<0, \end{array} \right. \end{aligned}$$

(14)

$$\begin{aligned} (-1)^{[i][n]}A^n_i A^i_m - \sum _{\nu }s_{i\nu } A^{\nu }_m A^n_{\nu }= (s_{ni} - s_{im}) A^i_i A^n_m , \quad m\not = i \not = n \not = m , \hspace{8pt} \end{aligned}$$

(15)

$$\begin{aligned} \omega A^m_m A^i_i = \sum _{\nu }s_{i\nu } A^{\nu }_m A^m_{\nu } - \sum _{\nu }s_{m\nu } A^{\nu }_i A^i_{\nu } , \quad i< m. \end{aligned}$$

(16)

Using a chance, we point out a typo in $s_{mi}$ (order of indices) in (14) admitted in [11]. This presentational flaw did not affect Lemma 3.3 because it was based on the correct expressions.

Proof

The proof is done by a direct analysis of which we indicate the key steps. Comparison of coefficients before $e_j^i\otimes e_j^i$ implies that A is even. Equation (12) results from comparing terms with $e^i_m\otimes e^i_n$ and $e_i^m\otimes e_i^n$. Equation (13) with $i\not =j$ is obtained from $e^i_j\otimes e^m_n$. For $i=j$, it coincides with (15) modulo third line of (14). The former comes from $e^i_n\otimes e_i^m$ and the latter from $e^i_i\otimes e_m^n$. The sign factor in (15) is reduced from $(-1)^{[m][i]+[m][n]+[i][n]}$ because A is even. The first line of (14) comes from $e^i_m\otimes e^i_n$, while the second from $e_j^m\otimes e_j^n$. Equation (16) results from $e_i^m\otimes e_m^i$. All other equations arising from (2) are fulfilled in virtue of this system. $\square$

   The next lemma accounts for Eqs. (12) and (13).

Lemma 4.2

If A a solution to Eq. (2), then it can be presented in the form (3), where $(Y,\sigma )$ is an admissible pair, $x_i=A^i_i$ for $i\in I$, and $y_i=A^{\sigma (i)}_i\not = 0$ for $i\in Y$.

Proof

By virtue of Eq. (12), the matrix A has at most one nonzero off-diagonal entry in every row and every column. Indices of such rows form a subset Y in I, and the off-diagonal entries may be written as $y_i=A_i^{\sigma (i)}\not = 0$, $i\in Y$, for some bijection $\sigma :Y\rightarrow I$ with no stable points. This is equivalent to Eq. (12). The map $\sigma$ is decreasing, that is encoded in Eq. (13). It preserves parity because A is even. $\square$

Lemma 4.3

The system (14) is equivalent to

$$\begin{aligned} \sum _{\nu \ge \max (i,m)}A^\nu _i A^m_\nu =0,\quad i\not = m, \end{aligned}$$

(17)

by virtue of (13). They imply $x_i=0$ for $i = b_+$.

Proof

The proof is similar to non-graded case because subsystem (14) turns out to be the same: It is equivalent to

$$\begin{aligned} \left\{ \begin{array}{rrll} 0&{}=&{} \sum _{i<\nu }A^{\nu }_i A^m_{\nu },&{} \quad m>i,\\ 0&{}=&{} \sum _{i\le \nu } A^{\nu }_i A^m_{\nu },&{} \quad i>m,\\ 0&{}=&{} \sum _{m<\nu }A^{\nu }_i A^m_{\nu },&{} \quad i>m,\\ 0&{}=&{} \sum _{m\le \nu } A^{\nu }_i A^m_{\nu },&{} \quad m > i, \\ 0&{}=&{}\sum _{k\le \nu }A^{\nu }_i A^m_{\nu } ,&{} \quad (i-k)(m-k)<0, \end{array} \right. \end{aligned}$$

Setting $i=j$ in (13) and substituting it into the above system, we reduce it to (17). Setting $m =\sigma (i) < i \in Y_+$ in (17) leads to $x_i y_i =0$, and therefore, $x_i=0$. Similarly, the assumption $m=\sigma (i)>i \in Y_-$ reduces (17) to $y_i x_{\sigma (i)}=0$, and hence, $x_{\sigma (i)}=0$. So $x_i=0$ for all $i\in Y_+\cup \sigma (Y_-)$ and $i=b_+$ in particular. $\square$

Lemma 4.4

With Lemma 4.3 taken into account, Eq. (15) is equivalent to the following two assertions.

• (1) For any $m\in Y$, either $\sigma ^2(m)=m$ or $\sigma (m)\not \in Y$.

• (2) $x_i = x_{b_-}$ and $x_j = 0$ whenever $i\le b_-$ and $j\ge b_+$.

Proof

First note that if $m\not \in Y$, Eq. (15) holds identically. So we can assume $m \in Y$ and rewrite (15) as

$$\begin{aligned} (-1)^{[i][n]}A^n_i A^i_m - s_{im} A^{m}_m A^n_{m}- s_{i\sigma (m)} A^{\sigma (m)}_m A^n_{\sigma (m)}= (s_{ni} - s_{im}) A^i_i A^n_m, \end{aligned}$$

(18)

where $m\not =i\not =n \not =m$.

Supposing $n\not =\sigma (m)$, we find $(-1)^{[i][n]}A^n_i A^i_m - s_{i\sigma (m)} A^{\sigma (m)}_m A^n_{\sigma (m)}= 0.$ If $i=\sigma (m)$, then, having in mind $s_{ii}=(-1)^{[i]}q^{(-1)^{[i]}}\not =\pm 1$ and $A^{\sigma (m)}_m=y_m\not =0$, we obtain $A^n_{\sigma (m)}=0$. Since $n\not =\sigma (m)=i$ and $n\not =m$, this means either ${\sigma (m)}\not \in Y$ or ${\sigma ^2(m)}=m$. Assuming $i\not =\sigma (m)$, we come to the equation $s_{i\sigma (m)} A^{\sigma (m)}_m A^n_{\sigma (m)}=0$, which is fulfilled as well, due to $A^n_{\sigma (m)}=0$.

It remains to study the case $n =\sigma (m)$. Under this hypothesis, the term $A^n_i A^i_m$ vanishes. Indeed, since $i\not = m$, one has $A^i_m \not =0 \Rightarrow i= \sigma (m)$. But this contradicts the condition $i\not = n =\sigma (m)$. In terms of the variables $x_i$ and $y_i$, Eq. (18) then reads

$$\begin{aligned} - s_{im} x_m y_m-s_{i\sigma (m)} y_m x_{\sigma (m)}= (s_{\sigma (m)i} - s_{im})x_i y_m, \end{aligned}$$

where $m\not =i\not =\sigma (m)$.

Depending on allocation of the indices i, m, and $\sigma (m)$, this equation splits into the following four implications.

$$\begin{aligned} i<m\quad \text{ and } \quad i<\sigma (m)&\Longrightarrow&x_i=x_m+x_{\sigma (m)} , \end{aligned}$$

(19)

$$\begin{aligned} i<m\quad \text{ and } \quad i>\sigma (m)&\Longrightarrow&x_m=0 , \end{aligned}$$

(20)

$$\begin{aligned} i>m\quad \text{ and } \quad i<\sigma (m)&\Longrightarrow&x_{\sigma (m)}=0 , \end{aligned}$$

(21)

$$\begin{aligned} i>m\quad \text{ and } \quad i>\sigma (m)&\Longrightarrow&x_i=0 . \end{aligned}$$

(22)

Recall that the index m is assumed to be from Y. Equation (22) is equivalent to $x_i = 0$ for $i>b_+$. By Lemma 4.3, this is also true for $i\ge b_+$. Equations (20) and (21) are thus satisfied as well. Equation (19) states $x_i = x_{b_-}+x_{\sigma (b_-)}$ if $i<b_- \in Y_-$ or $x_i = x_{b_-}+x_{\sigma ^{-1}(b_-)}$ if $i<b_-\in \sigma (Y_+)$. Applying Lemma 4.3, we find $x_i = x_{b_-}$, in either cases. $\square$

Remark 4.5

One can see that the proof of this lemma is almost literally the same as in the non-graded case.

Lemma 4.6

Equation (17) is fulfilled by virtue of Lemma 4.4.

Proof

Consider the case $m<i$. Equation (17) holds if $i\not \in Y$, because the sum turns into $A^i_iA^m_i$. So we may assume $i \in Y$ and distinguish two cases: $i\in Y_-$ and $i\in Y_+$. Assumption $i<\sigma (i)$ leads to $A^i_i A^m_i + A^{\sigma (i)}_i A^m_{\sigma (i)}=0$. The equality $m={\sigma (i)}$ is impossible, since otherwise $m={\sigma (i)}<i<{\sigma (i)}$. Therefore, the first term vanishes, and we come to $A^m_{\sigma (i)}=0$, $m\not = i$. This condition is satisfied, by Lemma 4.4, Statement 1. The case $i\in Y_+$ results in $x_i A^m_i=0$. Setting $m =\sigma (i)$, we come to $x_i =0$. Once $i\in Y_+\Rightarrow i\ge b_+$, we encounter a particular case of Lemma 4.4, Statement 2.

We should study the situation $i<m$. Equation (17) holds if $\sigma (i)<m$ because two possibly nonzero terms $A_i^\nu$ with $\nu =i$ and $\nu =\sigma (i)$ are not in the range of summation $\nu \geqslant m$.

We may think that $m\le \sigma (i)$; then,

$$\begin{aligned} \begin{array}{llrll} i<m=\sigma (i)&{} \Rightarrow &{} A^m_m = 0&{} \Rightarrow &{} x_{\sigma (i)}=0,\\ i<m<\sigma (i)&{} \Rightarrow &{} A^m_{\sigma (i)} = 0&{} \Rightarrow &{} \sigma (i)\not \in Y. \end{array} \end{aligned}$$

These requirements are fulfilled, by Lemma 4.4. $\square$

It remains to satisfy Eq. (16), in order to complete the proof of Theorem 3.2. Let $Y_0$ be the subset in Y, such that $\sigma$ restricted to $Y_0$ is involutive. By Lemma 4.4, either $\sigma (i) \not \in Y$ or i and $\sigma (i)$ belong to $Y_0$ simultaneously. Equation (16) brakes down into four equations

$$\begin{aligned} x_m(x_i-x_m)= & {} 0 , \hspace{134pt} i\not \in Y_0,\> m\not \in Y_0, \end{aligned}$$

(23)

$$\begin{aligned} \omega x_m(x_i-x_m)= & {} s_{i\sigma (m)} y_m y_{\sigma (m)},\quad \hspace{66pt} i \not \in Y_0,\> m \in Y_0, \end{aligned}$$

(24)

$$\begin{aligned} \omega x_m(x_i-x_m)= & {} -s_{m\sigma (i)} y_i y_{\sigma (i)} ,\quad \hspace{67pt} i \in Y_0,\> m \not \in Y_0, \end{aligned}$$

(25)

$$\begin{aligned} \omega x_m(x_i-x_m)= & {} s_{i\sigma (m)} y_m y_{\sigma (m)} - s_{m\sigma (i)} y_i y_{\sigma (i)} ,\quad i \in Y_0,\> m \in Y_0. \end{aligned}$$

(26)

Everywhere $i<m$, see (16).

Lemma 4.7

Suppose $Y_0\not =\varnothing$. Then, $Y=Y_0$ and, moreover,

$$\begin{aligned} Y_-=\{1,\ldots ,b_-\} ,\quad Y_+=\{b_+,\ldots ,b_+ +b_- -1\}, \end{aligned}$$

(27)

$$\begin{aligned} \sigma (i) = b_+ + b_- -i. \end{aligned}$$

(28)

Proof

Let $k \in Y_0 \cap Y_-$ and $\sigma (k)\in Y_0 \cap Y_+$ . Suppose either $l \in Y_-\backslash Y_0$ or $l<\min (Y_-)$. The assumption $l<k$ contradicts Eq. (24) if one sets $i=l$, $m=k$. The inequality $k<l$ does not agree with (25) if one sets $i=k$, $m=l$. In both cases, one uses Lemma 4.4, Statement 2, and gets $0=y_k y_{\sigma (k)}$; that is impossible since $y_k\not =0$ for all $k\in Y$, by definition of Y. Therefore, $Y_-\subset Y_0$ and $\min (Y_-)=1$. Assume now $l \in Y_+\backslash Y_0$. Then, l cannot exceed $\sigma (k)$, because otherwise $\sigma (l) <\sigma ^2(k) = k \Rightarrow \sigma (l)\in Y_0\Rightarrow l\in Y_0$, an absurd. The only possibility is $l<\sigma (k)$. But this again contradicts Eq. (25) if one sets $i=k$ and $m=l$.

Let us prove that $Y_+$ is the integer interval $[b_+,b_++b_--1]$. Suppose the opposite, then there are $i\in Y_-$ and $m\not \in Y_+$ such that $b_+<m<\sigma (i)\in Y_+$. Equation (2.6) turns to $0=-s_{m\sigma (i)}y_iy_{\sigma (i)}=-\omega y_iy_{\sigma (i)},$ which is impossible. $\square$

Lemma 4.8

Suppose $Y_0\not =\varnothing$. Then, there is $a\in \mathbb {C}$ such that $y_i y_{\sigma (i)} = a$ for all $i\in Y$. If $(b_-,b_+)\not =\varnothing$, then $x_m=\lambda$ for all $m \in (b_-,b_+)$, where $\lambda$ is a solution to the quadratic equation $\lambda (\lambda -x_{b_-})=a$.

Proof

By Lemmas 4.4 and 4.7, Eq. (26) is fulfilled if and only if the product $y_i y_{\sigma (i)}$ does not depend on $i\in Y$ and is equal to some $a \in \mathbb {C}$. If $(b_-,b_+)\not =\varnothing$, Eq. (25) suggests $x_m(x_m-x_{b_-})=a$ as soon as $m\in (b_-,b_+)$. It is easy to see that Eq. (23) is equivalent to $x_m=x_i=\lambda$ for $i,m\in (b_-,b_+)$ and some $\lambda \in \mathbb {C}$. $\square$

It is convenient to introduce the parameterization $x_{b_-}=\lambda +\mu$, $a=-\mu \lambda$. Then, $x_m = \lambda + \mu$ for $m\le b_-$, $x_m= \lambda$ for $b_-<m<b_+$, and $y_i y_{\sigma (i)}=-\lambda \mu$. This parameterization makes sense even if $(b_-,b_+)=\varnothing$. The scalars $\lambda$ and $\mu$ will have the meaning of eigenvalues of the matrix A.

Lemma 4.9

Let A be a matrix satisfying Eqs. (12)–(15) and $(Y,\sigma )$ the corresponding admissible pair. Then, Eq. (16) gives rise to the alternative

• $Y_0=Y$. A is a solution of Type 1 from Theorem 3.2.

• $Y_0=\varnothing$. A is a solution of Type 2 from Theorem 3.2.

Proof

If $Y_0=\varnothing$, then $Y\cap \sigma (Y)=\varnothing$ by Lemma 4.4, Statement 1. Equation (16) is reduced to (23). It is satisfied if and only if $x_m=\lambda$, $i\le b$, and $x_m=0$, $b< i$, for some $\lambda \in \mathbb {C}$ and $b\in \mathbb {Z}$. Comparing this with Lemma 4.4, Statement 2, we come to a solution of Type 2 with $b\in [b_-,b_+)$.

Suppose $Y_0\not = \varnothing$. Then, Eq. (16) implies Lemmas 4.7 and 4.8. Conversely, these lemmas ensure (23)–(26), which are equivalent to (16). Altogether, this gives a solution of Type 1. $\square$

Thus, we complete the proof of Theorem 3.2

5 Baxterization

In the previous sections, we dealt with so-called constant RE. Its version with spectral parameter important for applications was addressed in a number of papers, e.g. [21, 22]. Note that consideration was restricted to the following two gradings (standard and symmetric) of the underlying vector space $\mathbb {C}^{m+n}$:

$$\begin{aligned}{} & {} [i]= \left\{ \begin{array}{ccc} 0, &{} \quad i\leqslant m\\ 1, &{} \quad m<i\leqslant m+n \end{array} \right. , \\{} & {} [i]= \left\{ \begin{array}{cccc} 0, &{} \quad i\leqslant k &{}\text{ or } &{} k+n<i\\ 1, &{} \quad k<i\leqslant k+n \end{array} \right. , \quad \text{ with } \quad m=2k. \end{aligned}$$

The full list of solutions in the standard grading is given [22].

There is still a problem of clarifying the structure of solutions with spectral parameter in an arbitrary grading. In the non-graded case, every solution with spectral parameter is obtained from constant via a baxterization procedure [17, 18] up to equivalence see, [16]. This equivalence is given by a family of transformations described in [16], Lemma 3.3. There is a question if a similar fact holds true in an arbitrary grading. Then, the list of constant solutions given in this paper lays a base for such a classification. The second step (baxterization) is discussed next.

Recall that a spectral dependent version of the YBE is

$$\begin{aligned} S_{23}(x)S_{12}(xy)S_{23}(y) = S_{12}(y)S_{23}(xy)S_{12}(x), \quad x,y\in \mathbb {C}^\times , \end{aligned}$$

where S is regarded as a function $\mathbb {C}\rightarrow \textrm{End}(V)\otimes \textrm{End}(V)$ of (nonzero) spectral parameter. It is clearly equivalent to a non-graded YBE

$$\begin{aligned} \breve{S}_{23}(x)\breve{S}_{12}(xy)\breve{S}_{23}(y) = \breve{S}_{12}(y)\breve{S}_{23}(xy)\breve{S}_{12}(x). \end{aligned}$$

(29)

In the GL-case, the constant braid matrix $\breve{S}$ satisfies the Hecke condition

$$\begin{aligned} \breve{S}^2=\omega \breve{S} +1. \end{aligned}$$

Then, $\breve{S}(x)=\breve{S}-x^{-1}\breve{S}^{-1}$ solves (29), cf. e.g. [25, 26]. A baxterization procedure of [17] makes a constant RE-matrix A a solution to RE with spectral parameter:

$$\begin{aligned} \breve{S}\left( x/y\right) A_2(x)\breve{S}(xy)A_2(y) = A_2(y)\breve{S}(xy)A_2(x)\breve{S}\left( x/y\right) . \end{aligned}$$

The minimal polynomial of the matrix A from Theorem 3.2 is either quadratic or cubic:

$$\begin{aligned} (A-\lambda )(A-\mu ), \quad A(A-\lambda )(A-\mu ). \end{aligned}$$

A matrix with cubic minimal polynomial is of Type 1. Then, the A(x) is, respectively, either

$$\begin{aligned} A(x)=A-\frac{x^{-1}(\xi x-\lambda -\mu )}{x-x^{-1}} \end{aligned}$$

or

$$\begin{aligned} A(x)=A^2+(\xi x-\lambda -\mu )A-\frac{x^{-1}(x^2\xi ^2-\xi x(\lambda +\mu )+\lambda \mu )}{x-x^{-1}}, \end{aligned}$$

where $\xi$ is an arbitrary parameter.

A universal baxterization formula independent of the minimal polynomial of A can be found in [18]. It was used for Hecke algebraic formulation of open spin chains in [27]. In that setting, the matrix A is replaced with the so-called universal matrix $\mathcal {K}$ of the quantum general linear supergroup, $U_q(\mathfrak {g})$. The Hecke algebra admits a representation in $V^{\otimes n}\otimes \mathcal {A}$, where $V=\mathbb {C}^N$ is the natural $U_q(\mathfrak {g})$-module and $\mathcal {A}$ a certain $U_q(\mathfrak {g})$-algebra. This representation takes $\mathcal {K}$ to a matrix K whose entries generate $\mathcal {A}$. The assignment $K\mapsto A$ defines a one-dimensional representation of $\mathcal {A}$.

The algebra $\mathcal {A}$ can be realized as an invariant subalgebra in $U_q(\mathfrak {g})$ (under the adjoint action) generated by the entries of $K=L_-^{-1}L_+$, where the quantum Lax operators $L_\pm \subset \textrm{End}(V)\otimes U_q(\mathfrak {g})$ are expressed through the universal R-matrix $\mathcal {R}$ as $L_+=(\pi \otimes \text{ id})(\mathcal {R})$ as $L_-=(\pi \otimes \text{ id})(\mathcal {R}_{21}^{-1})$. This factorization of K implies that non-trivial one-dimensional representations of $\mathcal {A}$ cannot be restricted from $U_q(\mathfrak {g})$. A one-dimensional representation of $U_q(\mathfrak {g})$ assigns $\pm c\in \mathbb {C}$ with an arbitrary sign to natural generators of the Cartan subalgebra (diagonal entries of $L^\pm$) and kills the root vector generators. It takes $L_-^{-1}L_+$ to a scalar matrix A, as the only non-vanishing diagonal entries in $L_\pm ^{\pm 1}$ produce $c^2$ when squared. In other words, non-scalar solutions to RE cannot be obtained from one-dimensional representations of $U_q(\mathfrak {g})$. This of course applies to the standard quantum supergroup.

In the non-graded case, semi-simple invertible constant solutions to RE classify quantum symmetric pairs via a so-called “sandwich” construction introduced in [9] and systematically studied in [4]. Entries of the matrix $L_-^{-1}AL_+$ generate a left coideal subalgebra $\mathcal {B}\subset U_q(\mathfrak {g})$ that centralizes A. The dual sandwich $T^{-1}AT$, where T is the RTT-matrix of coordinate functions on the quantum group, satisfies RE. Its entries generate the subalgebra of $\mathcal {B}$-invariants in the Hopf dual to $U_q(\mathfrak {g})$.