Solutions to graded reflection equation of GL-type

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


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 gl(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 gl(N ) can be obtained from a baxterized constant K-matrix of U q gl(N ) 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 Section 3. The proof is arranged in a sequence of 9 lemmas in Section 4. Baxterization of constant solutions to spectral parameter dependent RE solutions is considered in Section 5.

Z 2 -graded Yang-Baxter and Reflection equations
Let us recall basic definitions of graded algebras and modules.
An associative algebra A is called Z 2 -graded if it is presentable as sum of vector spaces In particular, A 0 is a subalgebra and A 1 is a left/right A 0 -module.Elements of A 0 are called even and elements of A 1 are called odd.
An example is a complex vector space V = C N with a basis {v i } N i=1 equipped with an arbitrary parity function {1, . . ., N } → {0, 1}, i It is a graded module over the algebra A = End(V ) whose graded components are set to be Here e lk are the matrix units acting on the basis vectors by e lk v j = δ kj v l .
Given two graded algebras A and B their tensor product A ⊗ B is a graded algebra too.The multiplication on homogeneous elements is defined by the rule In the special case when A = B = End(V ), there is a Z 2 -graded isomorphism (1) ϕ : , where e il,jk ∈ End(V ⊗ V ) are matrix units acting by e il,jk we will write F = ijlk F il jk e il,jk ∈ End(V ⊗ V ) for its ϕ-image.The entries are related by F il jk = (−1) [j]([l]+[k]) F ij,lk .Suppose V is a graded vector space and A = End(V ) is the corresponding graded matrix algebra.An invertible element R ∈ A ⊗ A is called an R-matrix if it satisfies Yang-Baxter equation where the subscripts indicate the tensor factor in the graded tensor cube of End(V ).It is usually assumed that R is even.In terms of the matrix R ∈ End(V ⊗ V ), the YBE reads An even element P = n i=1 (−1) [j] e ij ⊗e ji ∈ End(V )⊗End(V ) is called graded permutation.It features A matrix A ∈ End(V ) is said to satisfy the (graded) reflection equation if in End(V ) ⊗ 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 Ȓij kl = (−1) Respectively, the graded braid equation with matrix S goes over to the non-graded braid equation with matrix elements Sij kl = Ȓji kl .
Proposition 2.1.The graded RE on A ∈ End(V ) is equivalent to the non-graded RE , and Ŝij kl = Sij kl under the assumption that A is even.
Proof.Apply the isomorphism ϕ defined in (1) to the graded RE and get S Ã2 S Ã2 = Ã2 S Ã2 S.
Observe that S = S. Plugging in the entries of the tensor Ã2 expressed through the matrix A, rewrite the equation as This turns to the non-graded RE with with braid matrix S if A is even.For arbitrary A, multiplication of the equation by (−1) proves the statement in general.

Classification of solutions
We fix a graded R matrix of the GL-type in the fundamental vector representation V = C N with an arbitrary grading on the the basis elements: where ω 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 ⊗ 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 It is convenient to represent the corresponding braid matrix in the form The transposed matrix Ŝ21 is equivalent to the graded R-matrix from [24].
We We distinguish two non-intersecting subsets Theorem 3.2.General solution to the graded RE of the GL-type is a matrix A of the form where As follows from the theorem, there are two classes of numerical RE matrices, those corresponding to σ(Y ) = Y and σ(Y ) ∩ Y = ∅.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 gl(N ).Different gradings impose different restricting conditions on A. So, invertible A (they belong to Type 1) are even only for a "symmetric" grading of the underlying vector space: or its opposite.

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 equation ( 2) organized into a sequence of lemmas.Explicitly (2) reads 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: 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 i j ⊗ e i j implies that A is even.Equations ( 12) results from comparing terms with e i m ⊗ e i n and e m i ⊗ e n i .Equation ( 13) with i = j is obtained from e i j ⊗ e m n .For i = j, it coincides with (15) modulo third line of (14).The former comes from e i n ⊗ e m i and the latter from e i i ⊗ e n m .The sign factor in ( 15) is reduced from (−1) because A is even.The 1st line of (14) comes from e i m ⊗ e i n while the 2nd from e m j ⊗ e n j .Equation (16) results from e m i ⊗ e i m .All other equations arising from (2) are fulfilled in virtue of this system.
Lemma 4.2.If A a solution to equation (2), then it can be presented in the form (3), where (Y, σ) is an admissible pair, x i = A i i for i ∈ I, and Proof.By virtue of equations ( 12), the matrix A has at most one non-zero off-diagonal entry in every row and every column.Indices of such rows form a subset Y in I, and the offdiagonal entries may be written as stable points.This is equivalent to equation (12).The map σ is decreasing, that is encoded in equation (13).It preserves parity because A is even.
Lemma 4.3.The system ( 14) is equivalent to by virtue of (13).They imply Proof.The proof is similar to non-graded case because subsystem ( 14) turns out to be the same: it is equivalent to (13) and substituting it into the above system we reduce it to (17).Setting 17) leads to x i y i = 0 and therefore x i = 0. Similarly, the assumption (17) to y i x σ(i) = 0 and hence x σ(i) = 0.So x i = 0 for all i ∈ Y + ∪ σ(Y − ) and i = b + in particular.
where m = i = n = m.A n σ(m) = 0, which is fulfilled as well, due to A n σ(m) = 0.It remains to study the case n = σ(m).Under this hypothesis, the term A n i A i m vanishes.Indeed, since i = m, one has A i m = 0 ⇒ i = σ(m).But this contradicts the condition i = n = σ(m).In terms of the variables x i and y i , equation ( 18) then reads Depending on allocation of the indices i, m, and σ(m), this equation splits into the following four implications.
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 ≥ b + .Equations ( 20) and ( 21) are thus satisfied as well.Equation (19) . Applying Lemma 4.3, we find x i = x b− , in either cases.
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 ∈ Y , because the sum turns into Therefore the first term vanishes and we come to We should study the situation i < m.Equation (17) holds if σ(i) < m because two possibly non-zero terms A ν i with ν = i and ν = σ(i) are not in the range of summation ν m.We may think that m ≤ σ(i); then These requirements are fulfilled, by Lemma 4.4.
It remains to satisfy equation (16), in order to complete the proof of Theorem 3.2.Let Y 0 be the subset in Y , such that σ restricted to Y 0 is involutive.By Lemma 4.4, either σ(i) ∈ Y or i and σ(i) belong to Y 0 simultaneously.Equation ( 16) brakes down into four equations Everywhere i < m, see (16).
Thus we complete the proof of Theorem 3.2

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 C m+n : 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.

satisfies the braid relation S 12 S
23 S 12 = S 23 S 12 S 23 .
denote by [a, b] ⊂ Z the intervals {k ∈ Z|a ≤ k ≤ 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, σ) consists of an ordered subset Y ⊂ I = [1, N ] and a parity preserving strictly decreasing map σ : Y → I without stable points.
the σ is decreasing, we have b − < b + .We adopt the convention b − = 0 and b

Lemma 4 . 4 .( 1 )
With Lemma 4.3 taken into account, equation (15) is equivalent to the following two assertions.For any m ∈ Y , either σ 2 (m) = m or σ(m) ∈ Y .(2) x i = x b− and x j = 0 whenever i ≤ b − and j ≥ b + .Proof.First note that if m ∈ Y , equation (15) holds identically.So we can assume m ∈ Y and rewrite (15) as and A σ(m) m = y m = 0, we obtain A n σ(m) = 0. Since n = σ(m) = i and n = m, this means either σ(m) ∈ Y or σ 2 (m) = m.Assuming i = σ(m), we come to the equation s iσ(m) A σ(m) m
a solution of Type 2 from Theorem 3.2.Proof.If Y 0 = ∅, then Y ∩ σ(Y ) = ∅ by Lemma 4.4, Statement 1. Equation (16) is reduced to (23).It is satisfied if and only if x m = λ, i ≤ b, and x m = 0, b < i, for some λ ∈ C and b ∈ Z. Comparing this with Lemma 4.4, Statement 2, we come to a solution of Type 2 with b ∈ [b − , b + ).
The first author (D.A.) is thankful to the Deanship of Scientific Research at University of Bisha for the financial support through the Scholarship Program of the University.Declarations