On the classification of rational K-matrices

TL;DR

This paper classifies the residual symmetries of rational K-matrices using boundary Yang-Baxter equations, confirming previous assumptions and exploring non-invertible cases with non-reductive symmetry algebras.

Contribution

It provides a derivation of residual symmetries of rational K-matrices, confirming the existence of involutive automorphisms relating original and residual symmetry algebras.

Findings

Residual symmetries correspond to involutive automorphisms of the Lie algebra

Invertible K-matrices have reductive symmetry algebras

Non-invertible K-matrices have semi-direct sum symmetry algebras

Abstract

This paper presents a derivation of the possible residual symmetries of rational K-matrices which are invertible in the ''classical limit'' (the spectral parameter goes to infinity). This derivation uses only the boundary Yang-Baxter equation and the asymptotic expansions of the R-matrices. The result proves the previous assumption of the literature: if the original and the residual symmetry algebras are $\mathfrak{g}$ and $\mathfrak{h}$ then there exists a Lie-algebra involution of $\mathfrak{g}$ for which the invariant sub-algebra is $\mathfrak{h}$. In addition, we study some K-matrices which are not invertible in the ''classical limit''. It is shown that their symmetry algebra is not reductive but a semi-direct sum of reductive and solvable Lie-algebras.

Equations

$$\left[Y_A^{(i)},Y_B^{(i)}\right]$$

$$\left[{\bar{Y}}_{\bar{A}}^{(i)},{\bar{Y}}_{\bar{B}}^{(i)}\right]$$

$$C_{AB}^{(i)}={⟨Y_A^{(i)},Y_B^{(i)}⟩}_i{C}_{A\bar{B}}^{(i)}={⟨Y_A^{(i)},{\bar{Y}}_{\bar{B}}^{(i)}⟩}_i=0C_{\bar{A}\bar{B}}^{(i)}={⟨{\bar{Y}}_{\bar{A}}^{(i)},{\bar{Y}}_{\bar{B}}^{(i)}⟩}_i$$

$$\left[C^{(12)},{\rho }^{(1)}(X)\otimes 1+1\otimes {\rho }^{(2)}(X)\right]=0.$$

$$\left[\mathfrak{h},\mathfrak{h}\right]\subseteq \mathfrak{h}\left[\mathfrak{h},\mathfrak{f}\right]\subseteq \mathfrak{f}\left[\mathfrak{f},\mathfrak{f}\right]\subseteq \mathfrak{h}$$

$$R_{12}^{(12)}(u)R_{13}^{(13)}(u+v)R_{23}^{(23)}(v)=R_{23}^{(23)}(v)R_{13}^{(13)}(u+v)R_{12}^{(12)}(u)$$

$$R^{(ij)}(u)=1+\frac{1}{u}{C}^{(ij)}+\mathcal{O}(u^{-2}),$$

$$R_{12}^{(12)}(u)$$

$$R_{13}^{(13)}(u)$$

$${\rho }^{(1)}(X)K(u)-K(u){\rho }^{(2)}(X)+\left[{\rho }^{(B)}(X),K(u)\right]=0$$

$$R_{12}^{(11)}(u-v)K_{13}(u)R_{21}^{(12)}(u+v)K_{23}(v)=K_{23}(v)R_{12}^{(12)}(u+v)K_{13}(u)R_{21}^{(22)}(u-v).$$

$$K(u)=\kappa +\mathcal{O}(u^{-1}).$$

$$R_{12}^{(11)}\left(\frac{u-v}{x}\right)K_1\left(\frac{u}{x}\right)R_{21}^{(12)}\left(\frac{u+v}{x}\right)K_2\left(\frac{v}{x}\right)$$

$$K_2\left(\frac{v}{x}\right)R_{12}^{(12)}\left(\frac{u+v}{x}\right)K_1\left(\frac{u}{x}\right)R_{21}^{(22)}\left(\frac{u-v}{x}\right)$$

$$\frac{x}{u-v}\left(C^{(11)}{\kappa }_1{\kappa }_2-{\kappa }_1{\kappa }_2{C}^{(22)}\right)+\frac{x}{u+v}\left({\kappa }_1{C}^{(21)}{\kappa }_2-{\kappa }_2{C}^{(12)}{\kappa }_1\right)=\mathcal{O}(x^2),$$

$$C^{(11)}{\kappa }_1{\kappa }_2$$

$${\kappa }_1{C}^{(21)}{\kappa }_2$$

$$B^{AB}{Y}_A^{(1)}\otimes Y_B^{(1)}=B^{AB}\kappa Y_A^{(2)}{\kappa }^{-1}\otimes \kappa Y_B^{(2)}{\kappa }^{-1}=B^{AB}{\mathrm{Ad}}_{\kappa }(Y_A^{(2)})\otimes {\mathrm{Ad}}_{\kappa }(Y_B^{(2)})$$

$${\mathrm{Ad}}_{\kappa }(Y_A^{(2)})=M_A^B{Y}_B^{(1)}+N_A^{\bar{B}}{\bar{Y}}_{\bar{B}}^{(1)}.$$

$$B^{AB}{C}_{BC}^{(1)}{Y}_A^{(1)}=B^{AB}\left(M_A^D{Y}_D^{(1)}+N_A^{\bar{D}}{\bar{Y}}_{\bar{D}}^{(1)}\right)M_B^E{C}_{EC}^{(1)}.$$

$$Y_C^{(1)}=B^{AB}\left(M_A^D{Y}_D^{(1)}+N_A^{\bar{D}}{\bar{Y}}_{\bar{D}}^{(1)}\right)M_B^E{B}_{EC}.$$

$$\mathbf{B}{\mathbf{M}}^T{\mathbf{B}}^{-1}\mathbf{M}$$

$$\mathbf{B}{\mathbf{M}}^T{\mathbf{B}}^{-1}\mathbf{N}$$

$${\mathrm{Ad}}_{\kappa }(Y_A^{(2)})=M_A^B{Y}_B^{(1)}$$

$$B^{AB}{\mathrm{Ad}}_{\kappa }(Y_A^{(2)})\otimes Y_B^{(1)}=B^{AB}{Y}_A^{(1)}\otimes {\mathrm{Ad}}_{\kappa }(Y_B^{(2)}).$$

$$M_C^D{Y}_D^{(1)}=B^{AB}{Y}_A^{(1)}{M}_B^E{B}_{EC}.$$

$$\mathbf{M}=\mathbf{B}{\mathbf{M}}^T{\mathbf{B}}^{-1},$$

$${\mathrm{Ad}}_{\kappa }({\rho }^{(2)}(X))={\rho }^{(1)}(\alpha (X))$$

$${\rho }^{(2)}={\mathrm{Ad}}_{{\kappa }^{-1}}\circ {\rho }^{(1)}\circ \alpha$$

$${\rho }^{(2)}={\mathrm{Ad}}_{{\left(V\kappa \right)}^{-1}}\circ \rho$$

$$\rho (\alpha (X))={\mathrm{Ad}}_{\kappa }(\rho (X))=\kappa \rho (X){\kappa }^{-1},$$

$$R_{12}(u-v)K_1(u)R_{21}(u+v)K_2(v)=K_2(v)R_{12}(u+v)K_1(u)R_{21}(u-v),$$

$$\rho (\alpha (X))={\mathrm{Ad}}_{\kappa }({\rho }_{cg}(X))=-\kappa \rho (X{)}^T{\kappa }^{-1},$$

Full text

Keywords: rational K-matrix, boundary Yang-Baxter equation, classification

1 Introduction

Integrable physical systems with boundaries can be defined by their R- and K-matrices [Skl88]. The R-matrices satisfy the famous Yang-Baxter equation and the K-matrix is the solution of the boundary Yang-Baxter equation (bYBE) for a given R-matrix. In this paper we give a derivation of the possible residual symmetries of the rational K-matrices. Some explicit forms of rational K-matrices (where there is no boundary degrees of freedom) were calculated for the defining representations of the matrix Lie-algebras [MS03]. Motivated by a classical argument, the derivation uses the following assumption. If the boundary breaks the bulk symmetry $G$ to $H$ then $G/H$ has to be a symmetric space, which means that there exists a Lie-group involution for which the subgroup $H$ is invariant. There are other direct calculations of K-matrices in the defining representation in [AAC*+*03, AAC*+*04].

It is well known that a special class of the ratioanal R-matrices are classified with the representations of Yangian algebras [Dri85, Loe16]. The introduction of boundaries into the theory of Yangians leads to a whole new class of the so-called reflection or twisted Yangian algebras [Ols92, MR02, GR16]. The twisted Yangians are in exact correspondence with the symmetric spaces $G/H$ which are given by a proper involution. The involution can be used to construct a co-ideal sub-algebra (which is isomorphic to the twisted Yangian) of the original Yangian. The representations of this co-ideal sub-algebra can be further used to calculate K-matrices [DMS01, Mac02].

In this paper we derive the assumption above - i.e. we prove for K-matrices (which are invertible in the ”classical limit”) that the residual symmetry algebra has to be the invariant sub-algebra of a Lie algebra involution. This derivation use only the boundary Yang-Baxter equation and the asymptotic expansion of the R- and K-matrices. We give an example for a K-matrix with non-invertible ”classical limit” and we describe its symmetry which is not a reductive Lie-algebra but a semi-direct sum of a solvable and a reductive algebra.

The paper is organized as follows. In section 2, we introduce the basic notations. Section 3 contains the main theorem and proofs which lead to the classification of possible K-matrices with invertible ”classical limit” when the boundary vector space is one dimensional. In this section, we also give an example for K-matrices with non-invertible ”classical limit” and describe its symmetry. In section 4, we extend the main theorem of section 3 to K-matrices with general finite dimensional boundary vector spaces.

2 Notations

Let $\mathfrak{g}$ be a complex simple Lie algebra with basis $\left\{X_{A}\right\}$ where $A=1,\dots,\mathrm{dim(\mathfrak{g})}$ and $\left[X_{A},X_{B}\right]=f_{AB}^{C}X_{C}$. If $\rho^{(i)}:\mathfrak{g}\to\mathrm{End}(\mathbb{C}^{d_{i}})$ is a faithful representation of $\mathfrak{g}$ then let $V^{(i)}:=\rho^{(i)}(\mathfrak{g})\subset\mathrm{End}(\mathbb{C}^{d_{i}})$. There is a non-degenerate invariant symmetric bilinear form (metric) $\left\langle\cdot,\cdot\right\rangle_{i}:\mathrm{End}(\mathbb{C}^{d_{i}})\otimes\mathrm{End}(\mathbb{C}^{d_{i}})\to\mathbb{C}$ for which $\left\langle Y,Z\right\rangle_{i}=\mathrm{Tr}\left(YZ\right)$. This metric can be used to define the orthogonal complement of $V^{(i)}$ in $\mathrm{End}(\mathbb{C}^{d_{i}})$: $\mathrm{End}(\mathbb{C}^{d_{i}})=V^{(i)}\oplus\bar{V}^{(i)}$ for which $\left\langle Y,\bar{Y}\right\rangle_{i}=0$ for all $Y\in V^{(i)}$ and $\bar{Y}\in\bar{V}^{(i)}$. We can choose a basis in $V^{(i)}$ and $\bar{V}^{(i)}$. The $\left\{Y_{A}^{(i)}=\rho^{(i)}(X_{A})\right\}$ is the obvious choice for $V^{(i)}$ and let $\left\{\bar{Y}_{\bar{A}}^{(i)}\right\}$ be a basis of $\bar{V}^{(i)}$ where $A=1,\dots,\mathrm{dim(\mathfrak{g})}$ and $\bar{A}=\mathrm{dim(\mathfrak{g})}+1,\dots,d^{2}$. The action of the commutator and the metric on the basis elemets can be written as

[TABLE]

and

[TABLE]

The metric $C_{AB}^{(i)}$ is proportional to the Killing form $B_{AB}=\mathrm{Tr}\left(\mathrm{ad}_{X_{A}}\circ\mathrm{ad}_{X_{B}}\right)$ i.e. $C_{AB}^{(i)}=c^{(i)}B_{AB}$. Let $B^{AB}$ be the inverse of $B_{AB}$ i.e. $B^{AB}B_{BC}=\delta_{C}^{A}$. If $\rho^{(1)}$ and $\rho^{(2)}$ are some representations of $\mathfrak{g}$ then we can define a useful matrix $C^{(12)}=B^{AB}\rho^{(1)}(X_{A})\otimes\rho^{(2)}(X_{B})$ which is invariant under the action of $\mathfrak{g}$:

[TABLE]

Let $\mathfrak{h}$ be a subalgebra of $\mathfrak{g}$ then ($\mathfrak{g}$,$\mathfrak{h}$) is a symmetric pair if there exists a Lie-algebra involution $\alpha$ for which $\alpha(X)=X$ for all $X\in\mathfrak{h}$. If $\mathfrak{f}$ is the orthogonal complement of $\mathfrak{h}$ in $\mathfrak{g}$ then $\mathfrak{g}=\mathfrak{h}\oplus\mathfrak{f}$ is a $\mathbb{Z}_{2}$ graded decomposition i.e.

[TABLE]

We can choose a basis in $\mathfrak{h}$ and $\mathfrak{f}$: $\left\{X_{a}\right\}$ and $\left\{X_{\alpha}\right\}$ respectively where $a=1,\dots,\mathrm{dim(\mathfrak{h})}$ and $\alpha=\mathrm{dim}(\mathfrak{h})+1,\dots,\mathrm{dim(\mathfrak{g})}$. Using this basis, one can define $C^{(\mathfrak{h},12)}=B^{ab}\rho^{(1)}(X_{a})\otimes\rho^{(2)}(X_{b})$ and $C^{(\mathfrak{f},12)}=B^{\alpha\beta}\rho^{(1)}(X_{\alpha})\otimes\rho^{(2)}(X_{\beta})$ for which $C^{(12)}=C^{(\mathfrak{h},12)}+C^{(\mathfrak{f},12)}$. The quadratic Casimir of $\mathfrak{h}$ is $c^{(\mathfrak{h},1)}=B^{ab}\rho^{(1)}\left(X_{a}\right)\rho^{(1)}\left(X_{b}\right)$.

Let $\rho^{(B)}:\mathfrak{h}\to\mathrm{End}(\mathbb{C}^{d_{B}})$ be a representation of subalgebra $\mathfrak{h}$ then $C^{(\mathfrak{h},1B)}=B^{ab}\rho^{(1)}(X_{a})\otimes\rho^{(B)}(X_{b})$.

Let $W^{(i)}:=\rho^{(i)}(\mathfrak{h})\subset\mathrm{End}(\mathbb{C}^{d_{i}})$. Obviously $W^{(i)}\subseteq V^{(i)}$. The metric can be used to define the orthogonal complement of $W^{(i)}$ in $\mathrm{End}(\mathbb{C}^{d_{i}})$: $\mathrm{End}(\mathbb{C}^{d_{i}})=W^{(i)}\oplus\bar{W}^{(i)}$. We can choose bases $\left\{Y_{a}^{(i)}=\rho(X_{a})\right\}$ and $\left\{\bar{Y}_{\bar{a}}^{(i)}\right\}$ in $W^{(i)}$ and $\bar{W}^{(i)}$ respectively where $a=1,\dots,\mathrm{dim(\mathfrak{h})}$ and $\bar{a}=\mathrm{dim(\mathfrak{h})}+1,\dots,d^{2}$.

Definition 1**.**

The quasi-classical R-matrix in the representation $\rho^{(i)}\otimes\rho^{(j)}$ is a $R^{(ij)}(u):\mathbb{C}^{d_{i}}\otimes\mathbb{C}^{d_{j}}\to\mathbb{C}^{d_{i}}\otimes\mathbb{C}^{d_{j}}$ spectral parameter ($u\in\mathbb{C}$) dependent linear map which satisfy the Yang-Baxter equation

[TABLE]

and its asymptotic expansion is:

[TABLE]

where $\rho^{(i)}:\mathfrak{g}\to\mathrm{End}(\mathbb{C}^{d_{i}})$ are some representations for $i=1,2,3$ and

[TABLE]

Definition 2**.**

Let $R^{(ij)}(u)\in\mathrm{End}\left(\mathbb{C}^{d_{i}}\otimes\mathbb{C}^{d_{j}}\right)$ be a quasi classical R-matrix in the representation $\rho^{(i)}\otimes\rho^{(j)}$ and $\mathfrak{h}$ is a sub-algebra of $\mathfrak{g}$ where $i,j=1,2$. The map $K(u):\mathbb{C}^{d_{2}}\otimes\mathbb{C}^{d_{B}}\to\mathbb{C}^{d_{1}}\otimes\mathbb{C}^{d_{B}}$ is a $(\mathfrak{g},\mathfrak{h})$ symmetric K-matrix in the representation $\left(\rho^{(1)},\rho^{(2)}\right)$ if the following two conditions are satisfied.

• •

There exists a representation $\rho^{(B)}:\mathfrak{g}\to\mathrm{End}(\mathbb{C}^{d_{B}})$ for which

[TABLE]

for all $X\in\mathfrak{h}\subset\mathfrak{g}$.

• •

It satisfies the boundary Yang-Baxter equation (bYBE):

[TABLE]

The $\mathbb{C}^{d_{B}}$ is the boundary vector space.

Definition 3**.**

Let $K(u)$ be a $(\mathfrak{g},\mathfrak{h})$ symmetric K-matrix with the following asymptotic expansion

[TABLE]

The K-matrix $K(u)$ is called quasi-classical if $\kappa$ is invertible.

In the following we deal with quasi-classical K-matrices therefore $d_{1}=d_{2}=d$.

3 K-matrices with 1 dimensional boundary space

In this section we investigate K-matrices with one dimensional boundary space i.e. $d_{B}=1$.

3.1 Classification of possible quasi-classical K-matrices

Proposition 4**.**

If there exists a quasi classical K-matrix in the representation $\left(\rho^{(1)},\rho^{(2)}\right)$ then the map $\mathrm{Ad}_{\kappa}:\mathrm{End}(\mathbb{C}^{d})\to\mathrm{End}(\mathbb{C}^{d})$ is a bijection between $\rho^{(1)}(\mathfrak{g})$ and $\rho^{(2)}(\mathfrak{g})$ i.e. $\mathrm{Ad}_{\kappa}(\rho^{(2)}(X_{A}))=M_{A}^{B}\rho^{(1)}(X_{B})$ where $\mathbf{M}$ is invertible, and $\mathbf{M}^{2}=1$ ($\left(\mathbf{M}\right)_{A}^{B}=M_{A}^{B}$).

Proof.

At first, we use the bYBE

[TABLE]

in the $x\to 0$ limit. The first non trivial term is:

[TABLE]

where we used the expansions (1),(3). This is the classical boundary Yang-Baxter equation (cbYBE). The above equation is equivalent to two constraints on $\kappa$:

[TABLE]

Equation (6) can be written in the following form:

[TABLE]

From this, we can derive that $\mathrm{Ad}_{\kappa}$ is a bijection between $\rho^{(1)}(\mathfrak{g})$* and $\rho^{(2)}(\mathfrak{g})$*. The action of $\mathrm{Ad}_{\kappa}$ on $\rho^{(2)}(\mathfrak{g})$ can be written as follows

[TABLE]

Appling the operator $1\otimes\left\langle Y_{C}^{(1)},\cdot\right\rangle_{1}$ on (8) we obtain that

[TABLE]

Since $C_{AB}^{(1)}=c^{(1)}B_{AB}$

[TABLE]

From this, we obtain two constrains for $\mathbf{M}$ and $\mathbf{N}$:

[TABLE]

It follows from the first that $\mathbf{M}$ is invertible and $\mathbf{M}^{-1}=\mathbf{B}\mathbf{M}^{T}\mathbf{\mathbf{B}}^{-1}$. Using this in the second equation, we obtain that $\mathbf{N}=0$. Therefore

[TABLE]

which implies that $\mathrm{Ad}_{\kappa}$ is a bijection between $\rho^{(1)}(\mathfrak{g})$ and $\rho^{(2)}(\mathfrak{g})$.

The equation (7) can be written in the following form:

[TABLE]

Using (9) and applying $1\otimes\left\langle Y_{C}^{(1)},\cdot\right\rangle_{1}$

[TABLE]

Therefore

[TABLE]

but we have seen above that $\mathbf{M}^{-1}=\mathbf{B}\mathbf{M}^{T}\mathbf{B}^{-1}$ therefore $\mathbf{M}^{2}=1$. ∎

Corollary 5**.**

If there exists a quasi classical K-matrix in the representation $\left(\rho^{(1)},\rho^{(2)}\right)$ then there exists a Lie algebra involution $\alpha:\mathfrak{g}\to\mathfrak{g}$, $\alpha^{2}=\mathrm{id}_{\mathfrak{g}}$ for which $\mathrm{Ad}_{\kappa}(\rho^{(2)}(X))=\rho^{(1)}(\alpha(X))$.

Proof.

Since $\mathrm{Ad}_{\kappa}(\rho^{(2)}(X))\in\rho^{(1)}(\mathfrak{g})$ for all $X\in\mathfrak{g}$ and $\rho^{(1)}$ is faithful, the map $\alpha=\left(\rho^{(1)}\right)^{-1}\circ\mathrm{Ad}_{\kappa}\circ\rho^{(2)}:\mathfrak{g}\to\mathfrak{g}$ exists and it is a Lie algebra automorphism. From (9) we obtain that $\alpha(X_{A})=M_{A}^{B}X_{B}$. We also saw that $\mathbf{M}^{2}=1$, therefore $\alpha^{2}=\mathrm{id}_{\mathfrak{g}}$ i.e. $\alpha$ is a Lie algebra involution.

From the definition of $\alpha$ we obtain that $\alpha(X)=\left(\rho^{(1)}\right)^{-1}\left(\mathrm{Ad}_{\kappa}\left(\rho^{(2)}(X)\right)\right)$ therefore

[TABLE]

for all $X\in\mathfrak{g}$. ∎

Remark 6*.*

The equation (10) can be written as

[TABLE]

Therefore if we choose an arbitrary representation $\rho^{(1)}$ and a Lie algebra involution $\alpha$ then the equation above fixes $\rho^{(2)}$. Let us choose $\rho^{(1)}=\rho$ then there are two possibilities for $\rho^{(2)}$.

1. (i)

There exists $V\in\mathrm{Aut}(\mathbb{C}^{d})$ such that $\rho\circ\alpha=\mathrm{Ad}_{V^{-1}}\circ\rho$. This $V$ exists for all inner and some outer involutions. Using $V$, the equation (10) reads as

[TABLE]

which means $\rho$ and $\rho^{(2)}$ are equivalent representations, it is therefore advisable to choose a basis where $\rho^{(2)}=\rho$. In this basis

[TABLE]

Therefore these K-matrices belong to the usual untwisted boundary Yang-Baxter equation:

[TABLE]

where $R(u)$ is the R-matrix in the $\rho\otimes\rho$ representation. 2. (ii)

There is no $V\in\mathrm{Aut}(\mathbb{C}^{d})$ such that $\rho\circ\alpha=\mathrm{Ad}_{V^{-1}}\circ\rho$ which is true for some outer involutions. These belong to the $\mathbb{Z}_{2}$ automorphisms of the Dynkin-diagrams of the Lie-algebras, e.g. in the case of the $A_{n}$ series, these connect the representations to their contra-gradient representations, therefore $\rho^{(2)}=\rho_{cg}$ i.e.

[TABLE]

for all $X\in\mathfrak{g}$. Therefore these K-matrices belong to the twisted boundary Yang-Baxter equation:

[TABLE]

where $\bar{R}(u)$ is the crossed R-matrix of $R(u)$:

[TABLE]

where $\Gamma$ is the crossing parameter.

Theorem 7**.**

Let $K(u)$ be a quasi classical $(\mathfrak{g},\mathfrak{h})$ symmetric K-matrix in the representation $\left(\rho^{(1)},\rho^{(2)}\right)$ then $(\mathfrak{g},\mathfrak{h})$ is a symmetric pair.

Proof.

From the previous corollary there exists a Lie algebra involution $\alpha$, for which* $\mathrm{Ad}_{\kappa}(\rho^{(2)}(X))=\rho^{(1)}(\alpha(X))$.* This involution can be used for a $\mathbb{Z}_{2}$ graded decomposition: $\mathfrak{g}=\mathfrak{h}_{0}\oplus\mathfrak{f}$ where $\alpha(X^{(+)})=+X^{(+)}$ and $\alpha(X^{(-)})=-X^{(-)}$ for all $X^{(+)}\in\mathfrak{h}_{0}$ and $X^{(-)}\in\mathfrak{f}$. Therefore $(\mathfrak{g},\mathfrak{h}_{0})$ is a symmetric pair.

Let us assume that $X\in\mathfrak{h}$. This implies that $K(u)\rho^{(2)}(X)=\rho^{(1)}(X)K(u)$ (see equation (2)) which reads as $\kappa\rho^{(2)}(X)=\rho^{(1)}(X)\kappa$ in the asymtotic limit. Therefore $\mathrm{Ad}_{\kappa}(\rho^{(2)}(X))=\rho^{(1)}(X)$ which implies that $\alpha(X)=X$ i.e $X\in\mathfrak{h}_{0}$ therefore $\mathfrak{h}\subseteq\mathfrak{h}_{0}$.

Now, we take the bYBE in the $v\to\infty$ limit:

[TABLE]

which implies that

[TABLE]

Therefore

[TABLE]

which is equivalent to

[TABLE]

for all $X\in\mathfrak{h}_{0}$ which implies that $\mathfrak{h}_{0}\subseteq\mathfrak{h}$. We have seen previously that $\mathfrak{h}\subseteq\mathfrak{h}_{0}$ therefore $\mathfrak{h}_{0}=\mathfrak{h}$ i.e. $\left(\mathfrak{g},\mathfrak{h}\right)$ is a symmetric pair. ∎

Theorem 8**.**

The $(\mathfrak{g},\mathfrak{h})$ symmetric K-matrix in the representation $\left(\rho^{(1)},\rho^{(2)}\right)$ is unique up to a multiplicative scalar function if $\mathfrak{h}$ is semi-simple, $\rho^{(1)}$ is irreducible and the matrix $\kappa$ is fixed. If $\mathfrak{h}$ is not semi-simple but reductive then the K-matrix may have a free parameter.

Remark 9*.*

The proof of this theorem can be found in [BS19] for a special case when $\mathfrak{g}=\mathfrak{sl}(n)$ and $\rho^{(1)}$ and $\rho^{(2)}$ are the defining representations. The authors of that paper uses the Sklyanin determinant but we do not use it in the following modified proof.

Proof.

Let the asymptotic expansion of $K(u)$ be

[TABLE]

where $\kappa$ is a fixed. The bYBE is invariant under the scalar multiplication: $K(u)\to c(u)K(u)$. We can fix this freedom by

[TABLE]

Let us assume that we have two K-matrices satisfying the bYBE with the above normalization, and these agree up to order $r-1$. The difference of the two K-matrices are

[TABLE]

From the normalization (11) we obtain that

[TABLE]

Substituting $K$ and $\tilde{K}$ into bYBE (4) and subtracting them from each other, the leading non-trivial terms at lowest order in $x$ are the following:

[TABLE]

The spectral parameter dependent functions

[TABLE]

are linearly independent for $r>1$ and linearly dependent for $r=1$:

[TABLE]

For $r>1$ we have 4 constrains. Let us see the first one:

[TABLE]

or in an equivalent form:

[TABLE]

Using (6), this can be written as

[TABLE]

i.e.

[TABLE]

which implies that

[TABLE]

for all $X\in\mathfrak{g}$. Since $\rho^{(1)}$ is irreducible, $\delta k^{(r)}\kappa^{-1}$ has to be proportional to the identity but from (12) we can see that $\delta k^{(r)}$ has to vanish.

For $r=1$ we have three equations

[TABLE]

We can see that equation (14) follows from (13) and (15), therefore we only have to deal with these two. Let us start with equation (13). Multiplying by $\kappa_{1}^{-1}\kappa_{2}^{-1}$ from the right:

[TABLE]

Using (6) we obtain that

[TABLE]

i.e.

[TABLE]

Using the basis in $\mathrm{End}(\mathbb{C}^{d})$ related to $\rho^{(1)}$: $\delta k^{(1)}\kappa^{-1}=Z^{A}Y_{A}^{(1)}+Z^{\bar{A}}\bar{Y}_{\bar{A}}^{(1)}$, we obtain that

[TABLE]

Applying the operator $1\otimes\left\langle Y_{D}^{(1)},\cdot\right\rangle_{1}$we obtain that

[TABLE]

which implies that

[TABLE]

for all $X\in\mathfrak{g}$. Because of $\rho^{(1)}$ is irreducible, $Z^{\bar{C}}\bar{Y}_{\bar{C}}^{(1)}=c1$ where $c\in\mathbb{C}$, therefore $\delta k^{(1)}\kappa^{-1}=Z^{A}Y_{A}^{(1)}+c1$. From (12), we can obtain that $c=0$ which implies that $\delta k^{(1)}\kappa^{-1}\in\rho^{(1)}(\mathfrak{g})$.

Let us continue with (15)

[TABLE]

Using (6) and (7) we can obtain that

[TABLE]

which implies that

[TABLE]

Using (10), we obtain that

[TABLE]

therefore

[TABLE]

for all $X\in\mathfrak{h}$. Since $\delta k^{(1)}\kappa^{-1}\in\rho^{(1)}(\mathfrak{g})$, $\delta k^{(1)}=0$ if $\mathfrak{h}$ is semi-simple. For reductive $\mathfrak{h}$, the $\delta k^{(1)}\kappa^{-1}$ is an element of the center of $\mathfrak{h}$ which is a one dimensional subspace for the $(\mathfrak{g},\mathfrak{h})$ symmetric pair (see the classification of symmetric pairs [Hel79]), therefore $K(u)$ may has a free parameter. ∎

3.2 Comments on the classical limit

The reason we use the ”quasi classical” name is the following [Dri85]. The YBE does not fix the norm of the R-matrix and the spectral parameter (if $R(u)$ a solution then $c(u)R(xu)$ is also a solution for all $x\in\mathbb{C}$ and any complex function $c(u)$) therefore one can redefine the spectral parameter as $u\to u/\hbar$ for which

[TABLE]

where $r(u)$ is the classical R-matrix

[TABLE]

which satisfies the classical Yang-Baxter equation

[TABLE]

which is the order $\hbar^{2}$ term of the YBE. Scaling the spectral parameter in the K-matrix similarly, we obtain that

[TABLE]

The first non-trivial term of the bYBE in $\hbar$ reads as

[TABLE]

which is the classical boundary Yang-Baxter equation (cbYBE) which was investigated in Proposition 4.

However, we saw that if the residual symmetry is not semi-simple then the K-matrix may have a free parameter ($K(u)\to K(u,a)$ where $a$ is the free parameter). These free parameter dependent solutions really exist [Gom18b]. The asymptotic expansion of these K-matrices can be written as

[TABLE]

where $X_{0}$ is the central element of $\mathfrak{h}$ and

[TABLE]

The above parameter can become an $\hbar$ dependent function: $a\to a(\hbar)$. Using a proper function $a(\hbar)$ the K-matrix can be written as

[TABLE]

which means that the classical limit of the K-matrix can be spectral parameter dependent. Using this $\hbar$ expansion, the first non-trivial term of the bYBE can be written as

[TABLE]

which is the classical boundary Yang-Baxter equation for spectral parameter dependent $\kappa$-matrix. In [Gom18a], there was derived some solutions of this equation in the defining representations of the matrix Lie-algebras. These solutions can be matched to the parameter dependent solutions of the bYBE.

One can classify the solutions of the general cbYBE

[TABLE]

if we assume that $\tilde{\kappa}(u)$ is invertable in the asymtotic limit i.e

[TABLE]

where $\kappa$ is invertable. This $\kappa$ also satisfies the spectral parameter independent cbYBE:

[TABLE]

We have seen at the proof of Proposition 4 and Corollary 5 that if $\kappa$ satisfies the equation above then it defines an involution $\alpha$ for which $\rho^{(1)}(\alpha(X))=\mathrm{Ad}_{\kappa}(\rho^{(2)}(X))$. Let $\mathfrak{h}$ be the invariant sub-algebra of $\alpha$. We can fix the normalization by

[TABLE]

Proposition 10**.**

If $\mathfrak{h}$ is semi-simple Lie-algebra then $\tilde{\kappa}(u)=\kappa$.**

Proof.

Let the asymptotic expansion of $\tilde{\kappa}(u)$ be

[TABLE]

Substituting this into

[TABLE]

the order $x^{r+1}$ term is

[TABLE]

Using the same argument as in the proof of Theorem 8, one can prove that $k^{(r)}=0$ which implies that $\tilde{\kappa}(u)=\kappa$. ∎

Proposition 11**.**

If $\mathfrak{h}$ is reductive Lie-algebra then the asymptotic expansion of $\tilde{\kappa}(u)$ has to be

[TABLE]

where $X_{0}$ is a central element of $\mathfrak{h}$ for which $\left\langle\rho(X_{0}),\rho(X_{0})\right\rangle=1$ and $a_{0}\in\mathbb{C}$. If we fix $a_{0}$ then $\tilde{\kappa}(u$) is unique.**

Proof.

Putting

[TABLE]

into equation (20) we obtain the equation (21) for $r=1$. Using the same argument as the proof of Theorem 8, one can prove that $k^{(1)}\kappa^{-1}$ has to be a central element of $\mathfrak{h}$ which means that $k^{(1)}=a_{0}\rho(X_{0})\kappa$.

The proof of the universality is the same as for $K(u)$. ∎

Corollary 12**.**

Let $\tilde{\kappa}(u,a_{0}$) be the $\kappa$-matrix with a fixed parameter $a_{0}$ and $\tilde{\kappa}(u)=\tilde{\kappa}(u,1)$ then $\tilde{\kappa}(u,a_{0})=\tilde{\kappa}(u/a_{0},1)=\tilde{\kappa}(u/a_{0})$.**

We can connect the parameter $a_{0}$ of (22) to the function $a(\hbar)$. From equations (16),(17) and (22) we can obtain that

[TABLE]

In summary, if the residual symmetry algebra is semi-simple then the classical $\kappa$-matrix is fixed by the symmetry, and the quantum correction to K-matrix is fixed by this $\kappa$ and the bYBE (up to a normalization)

[TABLE]

However, if the residual symmetry algebra is not semi-simple but reductive then the classical $\kappa$-matrix is not totally fixed by the symmetry since it has a free parameter $a_{0}$, and the K-matrix has a dynamical parameter $a(\hbar)$ which is not fixed by the $\kappa(a_{0}u)$ and the bYBE

[TABLE]

3.3 Not quasi classical K-matrices

There also exist non quasi-classical solutions of the bYBE. If $\mathfrak{g}=\mathfrak{sl}(n)$ and $\rho^{(1)}=\rho^{(2)}=\rho$, where $\rho$ is the defining representation then

[TABLE]

is a solution for $\kappa^{2}=0$ i.e. $K(u)$ is not quasi classical. This K-matrix is satisfies the unitary condition:

[TABLE]

The residual symmetry algebra is not reductive but a semi-direct sum of a solvable and a reductive Lie-algebras

[TABLE]

where $\mathfrak{h}_{s}$ is solvable and $\mathfrak{h}_{r}$ is reductive. Since

[TABLE]

the above sum is semi-direct. With a suitable base change, $\kappa$ can always be brought into Jordan canonical form:

[TABLE]

where there are $k$ non-trivial block. The reductive sub-algebra is $\mathfrak{h}_{r}=\mathfrak{sl}(n-2k)$. The solvable part can be written in a direct sum of subspaces: $\mathfrak{h}_{s}=\mathfrak{h}_{2}+\mathfrak{h}_{D}+\mathfrak{h}_{+}+\mathfrak{h}_{-}$ where

[TABLE]

where $E_{i,j}$s are the elementary matrices: $\left(E_{i,j}\right)_{ab}=\delta_{ia}\delta_{jb}$. The Lie-bracket of these are the following:

[TABLE]

Therefore $\left[\mathfrak{h}_{s},\mathfrak{h}_{s}\right]:=\mathfrak{h}_{1}=\mathfrak{h}_{2}+\mathfrak{h}_{+}+\mathfrak{h}_{-}$ and $\left[\mathfrak{h}_{1},\mathfrak{h}_{1}\right]=\mathfrak{h}_{2}$ which is a commutative Lie-algebra.

4 K-matrices with general boundary space

In this section we investigate K-matrices with $d_{B}>1$.

Definition 13**.**

Let $K(u)=E_{ij}\otimes\Psi^{ij}(u)\in\mathrm{End}(\mathbb{C}^{d}\otimes\mathbb{C}^{d_{B}})$ where $E_{ij}$ are the elementary matrices of $\mathrm{End}(\mathbb{C}^{d})$. The K-matrix $K(u)$ is irreducible if there is no proper invariant subspace $\mathbb{C}^{d_{0}}\subset\mathbb{C}^{d_{B}}$ of $\Psi^{ij}(u)$ for all $i,j$.

Corollary 14**.**

If $K(u)\in\mathrm{End}(\mathbb{C}^{d}\otimes\mathbb{C}^{d_{B}})$ is an irreducible K-matrix then $\tilde{\kappa}=\kappa\otimes 1$ where $\kappa\in\mathrm{End}(\mathbb{C}^{d})$.

Proof.

The bYBE in the $v\to\infty$ limit reads as

[TABLE]

We can write the K-matrix and the matrix $\tilde{\kappa}$ as

[TABLE]

where $E_{ij}$s are the elementary matrices of $\mathrm{End}(\mathbb{C}^{d})$ and $\Psi^{ij}(u),\psi^{ij}\in\mathrm{End}(\mathbb{C}^{d_{B}})$ for all $i,j$. Putting this into the equation above, we obtain that

[TABLE]

for all $i,j,k,l$. Since $K(u)$ is irreducible, the matrices $\psi^{kl}$ have to be proportional to the identity therefore $\tilde{\kappa}=\kappa\otimes 1$. ∎

Proposition 15**.**

Let $K(u)\in\mathrm{End}(\mathbb{C}^{d}\otimes\mathbb{C}^{d_{B}})$ be an irreducible K-matrix in the representation $\left(\rho^{(1)},\rho^{(2)}\right)$ then there exists a Lie-algebra involution $\alpha:\mathfrak{g}\to\mathfrak{g}$, $\alpha^{2}=\mathrm{id}_{\mathfrak{g}}$ for which $\mathrm{Ad}_{\kappa}(\rho^{(2)}(X))=\rho^{(1)}(\alpha(X))$.

Proof.

From the previous corollary, we know that the leading order of an irreducible K-matrix is $\kappa\otimes 1$ therefore the leading non-trivial term of the bYBE

[TABLE]

is the same as (5). Therefore the proof of this proposition is the same as it was for the scalar boundary. ∎

Lemma 16**.**

Let $K(u)\in\mathrm{End}(\mathbb{C}^{d}\otimes\mathbb{C}^{d_{B}})$ be an irreducible K-matrix in the representation $\left(\rho^{(1)},\rho^{(2)}\right)$ and $\alpha$ is the Lie-algebra involution for which $\mathrm{Ad}_{\kappa}(\rho^{(2)}(X))=\rho^{(1)}(\alpha(X))$ then the asymptotic expansion of $K(u)$ has to be

[TABLE]

where $\rho^{(B)}:\mathfrak{h}\to\mathrm{End}(\mathbb{C}^{d_{B}})$ is a representation of $\mathfrak{h}$ which is the invariant subalgebra of $\alpha$ and $D\in\mathrm{End}(\mathbb{C}^{d})$ for which $\left[D,\rho^{(1)}\left(\mathfrak{g}\right)\right]=0$.

Proof.

Let the asymptotic expansions be

[TABLE]

We can substitute this to (23) and the next to the leading order in $x$ is the following:

[TABLE]

where

[TABLE]

Multipling equation (25) by $uv(u-v)^{2}(u+v)^{2}$, we obtain that

[TABLE]

Let us start with the $u^{4}$ term:

[TABLE]

Multipling this by $\kappa_{1}^{-1}\kappa_{2}^{-1}$ from the right.

[TABLE]

Since

[TABLE]

we obtain that

[TABLE]

Let us use the following notation:

[TABLE]

where $Z^{a},\bar{Z}^{\bar{a}}\in\mathrm{End}(\mathbb{C}^{d_{B}})$. Substituting this into (27) we obtain that

[TABLE]

If we apply $\left\langle Y_{d}^{(1)},\cdot\right\rangle_{1}\otimes\left\langle Y_{e}^{(1)},\cdot\right\rangle_{1}\otimes\mathrm{id}$ then we get

[TABLE]

where we used that $C_{ab}^{(1)}=c^{(1)}B_{ab}$. Using the definition

[TABLE]

we can obtain that

[TABLE]

which implies that there exists a representation ($\rho^{(B)}$) of $\mathfrak{h}$ for which $\rho^{(B)}(X_{a})=Z_{a}$.

Appling $\left\langle\bar{Y}_{\bar{d}}^{(1)},\cdot\right\rangle_{1}\otimes\left\langle Y_{e}^{(1)},\cdot\right\rangle_{1}\otimes\mathrm{id}$ to (28) we obtain that

[TABLE]

and using $\left\langle\bar{Y}_{\bar{d}}^{(1)},\cdot\right\rangle_{1}\otimes\left\langle\bar{Y}_{\bar{e}}^{(1)},\cdot\right\rangle_{1}\otimes\mathrm{id}$, we get

[TABLE]

Using these, $\bar{Y}_{\bar{a}}^{(1)}\otimes\bar{Z}^{\bar{a}}$ can be written as $\bar{Y}_{\bar{a}}^{(1)}\otimes\bar{Z}^{\bar{a}}=y^{i}\otimes U_{i}$ where $\left[U_{i},\rho^{(B)}(X)\right]=0$, $\left[U_{i},U_{j}\right]=0$, $\left\langle U_{i},U_{j}\right\rangle_{B}=\delta_{ij}$ for all $X\in\mathfrak{h}$ and $y^{i}\in\rho^{(1)}(\mathfrak{f})\oplus\bar{V}^{(1)}$. Using $\left\langle Y_{d}^{(1)},\cdot\right\rangle\otimes\mathrm{id}\otimes\left\langle U_{i},\cdot\right\rangle$ on (28), we obtain that

[TABLE]

From the $u^{3}v$ and $uv^{3}$ terms of (26) we can obtain that

[TABLE]

Multiplying this with $\kappa_{1}^{-1}\kappa_{2}^{-1}$ from the right, we get

[TABLE]

The first term can be written as

[TABLE]

Using the definition $\bar{Y}_{\bar{a}}^{(1)}\otimes\bar{Z}^{\bar{a}}=y^{i}\otimes U_{i}=\tilde{y}^{i}\otimes U_{i}+\frac{1}{2}c^{(\mathfrak{h},1)}\otimes 1$ the $\tilde{k}^{(1)}$ can be written as

[TABLE]

Substituting this into the equation above, we obtain that

[TABLE]

Using $\left\langle Y_{\gamma}^{(1)},\cdot\right\rangle\otimes\mathrm{id}\otimes\left\langle U_{i},\cdot\right\rangle$, we get

[TABLE]

From (31) and (32) we can see that $\left[\rho^{(1)}(X),\tilde{y}^{i}\right]=0$ for all $X\in\mathfrak{g}$ therefore $\tilde{y}^{i}=d^{ij}D_{j}$ where $\left[D_{j},\rho^{(1)}(X)\right]=0$ for all $X\in\mathfrak{g}$, $\left\langle D_{i},D_{j}\right\rangle_{1}=\delta_{ij}$ and $D_{i}\in\rho^{(1)}(\mathfrak{f})\oplus\bar{V}^{(1)}$.

In summary,

[TABLE]

where $U^{i}=d^{ji}U_{j}$.

In the following we prove that $U^{i}$ has to be proportional to $1$. For this, we use the higher order terms of (23). Let the asymptotic expansions be

[TABLE]

where $C^{(0)(ij)}=1\otimes 1$, $C^{(1)(ij)}=C^{(ij)}$, $k^{(0)}=\kappa\otimes 1$.The $x^{r+1}$ order terms of (23) read as

[TABLE]

where $a+b+c+d=r+1$. Because of $k^{(0)}=\kappa\otimes 1$, the $a=r+1$ and $b=r+1$ terms are trivial therefore $a,b\leqq r$. Multipling these by $u^{r}v^{r}(u-v)^{r+1}(u+v)^{r+1}$, the spectral parameter dependencies are

[TABLE]

Expanding the brackets, the highest order term in $u$ is $u^{2r+1+b}v^{r-b}$. We concentrate on the $u^{3r+1}$ terms i.e. when $b=r$. From this we can obtain the following equation:

[TABLE]

Multiply this by $\kappa_{1}^{-1}\kappa_{2}^{-1}$ from the right, we get

[TABLE]

Using (33) and applying $\left\langle D_{i},\cdot\right\rangle\otimes\mathrm{id}\otimes\mathrm{id}$, we obtain that

[TABLE]

Since this is true for all $r$, the equation (35) implies that

[TABLE]

If $K(u)$ is irreducible then $U^{i}=e^{i}1$ therefore

[TABLE]

where $D=e^{i}D_{i}$ and we wanted to prove this. ∎

Theorem 17**.**

Let $K(u)$ be a quasi classical, irreducible $(\mathfrak{g},\mathfrak{h})$ symmetric K-matrix in the representation $\left(\rho^{(1)},\rho^{(2)}\right)$ then $\left(\mathfrak{g},\mathfrak{h}\right)$ is a symmetric pair.

Proof.

Similarly to the scalar boundary case, the leading order of $K(u)$ defines a Lie-algebra involution $\alpha$, for which* $\mathrm{Ad}_{\kappa}(\rho^{(2)}(X))=\rho^{(1)}(\alpha(X))$.* This involution can be used for a $\mathbb{Z}_{2}$ graded decomposition: $\mathfrak{g}=\mathfrak{h}_{0}\oplus\mathfrak{f}$ where $\alpha(X^{(+)})=+X^{(+)}$ and $\alpha(X^{(-)})=-X^{(-)}$ for all $X^{(+)}\in\mathfrak{h}_{0}$ and $X^{(-)}\in\mathfrak{f}$ therefore $\left(\mathfrak{g},\mathfrak{h}_{0}\right)$ is a symmetric pair.

Let us assume that $X\in\mathfrak{h}$ which implies that there exists a representation $\tilde{\rho}$ of $\mathfrak{h}$ such that

[TABLE]

Going to the $u\to\infty$ limit, we obtain that $\kappa\rho^{(2)}(X)=\rho^{(1)}(X)\kappa$ i.e. $\mathrm{Ad}_{\kappa}(\rho^{(2)}(X))=\rho^{(1)}(X)$ which means $\alpha(X)=X$ i.e $X\in\mathfrak{h}_{0}$ therefore $\mathfrak{h}\subseteq\mathfrak{h}_{0}$.

Now, we take the $v\to\infty$ limit of the bYBE:

[TABLE]

Multipling this by $\kappa_{2}^{-1}$ from the right and using (36), we obtain that

[TABLE]

which is equivalent to

[TABLE]

for all $X\in\mathfrak{h}_{0}$ which implies that $\mathfrak{h}_{0}\subseteq\mathfrak{h}$. We have seen previously that $\mathfrak{h}\subseteq\mathfrak{h}_{0}$ therefore $\mathfrak{h}_{0}=\mathfrak{h}$ i.e. $\left(\mathfrak{g},\mathfrak{h}\right)$ is a symmetric pair. ∎

5 Conclusion

In this paper we derived directly from the boundary Yang-Baxter equation that the possible residual symmetry algebras of the quasi-classical K-matrices have to be invariant sub-algebras of Lie-algebra involutions. It was also proved that if the boundary vector space is one dimensional then these K-matrices are universal (up to a normalization) when the residual sub-algebra is semi-simple, and for the non semi-simple ones the K-matrices have a free parameter. In the following it might be interesting to try to generalize these statements and proofs to the trigonometric cases where similar classification seemed to be present [RV16, NR18].

In addition, a not quasi-classical K-matrix was briefly examined. We have seen that its residual symmetry algebra is a semi-direct sum of a reductive and a solvable Lie-algebra. For further work, the classification of these may also be interesting. It is possible that they also play a role in the classification of the integrable initial states of spin chains [PVCP19a, PVCP19b, PPV19].

Acknowledgment

I thank Zoltán Bajnok for the useful discussions and for reading the manuscript. The work was supported by the NKFIH 116505 Grant.