Massey products on cycles of projective lines and trigonometric solutions of the Yang-Baxter equations

A. Polishchuk

Department of Mathematics, University of Oregon, Eugene, OR 97403

apolish@uoregon.edu

Summary. We show that a nondegenerate unitary solution r(u, v) of the associative Yang-Baxter equation (AYBE) for Mat(N, ℂ) (see [7]) with the Laurent series at u = 0 of the form r(u, v) = \frac{1⊗1}{u} + r_{0}(v) + … satisfies the quantum Yang-Baxter equation, provided the projection of r_{0}(v) to sl_{N} ⊗ sl_{N} has a period. We classify all such solutions of the AYBE extending the work of Schedler [8]. We also characterize solutions coming from triple Massey products in the derived category of coherent sheaves on cycles of projective lines.

Introduction

This paper is concerned with solutions of the associative Yang-Baxter equation (AYBE)

r^{12}(−u′, v)r^{13}(u+u′, v+v′)−r^{23}(u+u′, v′)r^{12}(u, v)+r^{13}(u, v+v′)r^{23}(u′, v′) = 0,

(1)

where r(u, v) is a meromorphic function of two complex variables (u, v) in a neighborhood of (0, 0) taking values in A ⊗ A, where A = Mat(N, ℂ) is the matrix algebra. Here we use the notation r^{12} = r ⊗ 1 ∈ A ⊗ A ⊗ A, etc. We will refer to a solution of (1) as an associative r-matrix. This equation was introduced in the above form in [7] in connection with triple Massey products for simple vector bundles on elliptic curves and their degenerations. It is usually coupled with the unitarity condition

r^{21}(−u, −v) = −r(u, v).  (2)

Note that the constant version of (1) was independently introduced in [1] in connection with the notion of infinitesimal bialgebra (where A can be any associative algebra). The AYBE is closely related to the classical Yang-Baxter equation (CYBE) with spectral parameter

[r^{12}(v), r^{13}(v + v′)] − [r^{23}(v′), r^{12}(v)] + [r^{13}(v + v′), r^{23}(v′)] = 0 (3)

for the Lie algebra sl_{N} (so r(v) takes values in sl_{N} ⊗ sl_{N}) and also with the quantum Yang-Baxter equation (QYBE) with spectral parameter

R^{12}(v)R^{13}(v + v′)R^{23}(v′) = R^{23}(v′)R^{13}(v + v′)R^{12}(v),  (4)

where R(v) takes values in A⊗A. In the seminal work [3] Belavin and Drinfeld made a thorough study of the CYBE for simple Lie algebras. In particular, they showed that all nondegenerate solutions are equivalent to either elliptic, trigonometric, or rational solutions, and gave a complete classification in the elliptic and trigonometric cases. In the present paper we extend some of their results and techniques to the AYBE. In addition, we show that often solutions of the AYBE are automatically solutions of the QYBE (for fixed u).

We will be mostly studying unitary solutions of the AYBE (i.e., solutions of (1) and (2)) that have the Laurent expansion at u = 0 of the form

r(u, v) = \frac{1 ⊗ 1}{u} + r_{0}(v) + ur_{1}(v) + …  (5)

It is easy to see that in this case r_{0}(v) is a solution of the CYBE. Hence, denoting by pr : Mat(N, ℂ) → sl_{N} the projection along ℂ · 1 we obtain that \overline{r}_{0}(v) = (pr ⊗ pr)r_{0}(v) is a solution of the CYBE for sl_{N}. We prove that if r(u, v) is nondegenerate (i.e., the tensor r(u, v) ∈ A ⊗ A is nondegenerate for generic (u, v)) then so is \overline{r}_{0}. Thus, \overline{r}_{0} falls within Belavin-Drinfeld classification. Furthermore, we show that if \overline{r}_{0} is either elliptic or trigonometric then r(u, v) is uniquely determined by \overline{r}_{0} up to certain natural transformations. The natural question raised in [7] is which solutions of the CYBE for sl_{N} extend to unitary solutions of the AYBE of the form (5). In [7] we showed that this is the case for all elliptic solutions and gave some examples with trigonometric solutions. In [8] Schedler studied further this question for trigonometric solutions of the CYBE of the form r_{0}(v) = \frac{r+e^{v}r^{21}}{1−e^{v}}, where r is a constant solution of the CYBE. He discovered that not all trigonometric solutions of the CYBE can be extended to solutions of the AYBE, and found a nice combinatorial structure that governs the situation (called associative BD triples). In this paper we complete the picture by giving the answer to the above question for arbitrary trigonometric solutions of the CYBE (see Theorem 0.2 below). We will also prove that every nondegenerate unitary solution r(u, v) of the AYBE with the Laurent expansion at u = 0 of the form (5) satisfies the QYBE with spectral parameter for fixed u, provided \overline{r}_{0}(v) either has a period (i.e., it is either elliptic or trigonometric) or has no infinitesimal symmetries (see Theorem 1.5). Thus, our work on extending trigonometric classical r-matrices (with spectral parameter) to solutions of the AYBE leads to explicit formulas for the corresponding quantum r-matrices. The connection with the QYBE was noticed before for elliptic solutions constructed in [7] (because they are given essentially by Belavin’s elliptic R-matrix) and also for those trigonometric solutions that are constructed in [8].

An important input for our study of trigonometric solutions of the AYBE is the geometric picture with Massey products developed in [7] that involves considering simple vector bundles on elliptic curves and their rational degenerations. In loc. cit. we constructed all elliptic solutions in this way and some trigonometric solutions coming from simple vector bundles on the union of two projective lines glued at two points. In this paper we consider the case of bundles on a cycle of projective lines of arbitrary length. We compute explicitly corresponding solutions of the AYBE. Then we notice that similar formula make sense in a more general context and prove this by a direct calculation. The completeness of the obtained list of trigonometric solutions is then checked by combining the arguments of [8] with those of [3] (modified appropriately for the case of the AYBE). It is interesting that contrary to the initial expectation expressed in [7] not all trigonometric solutions of the AYBE can be obtained from the triple Massey products on cycles of projective lines (see Theorem 5.5). This makes us wonder whether there is some generalization of our geometric setup.

Another question that seems to be worth pursuing is the connection between the combinatorics of simple vector bundles on a cycle of projective lines X and the Belavin-Drinfeld combinatorics. Namely, the discrete type of a vector bundle on X is described by the splitting type on each component of X. As was observed in [4], Theorem 5.3, simplicity of a vector bundle corresponds to a certain combinatorial condition on these splitting types (see also Lemma 3.1). In this paper we show that this condition allows to associate with such a splitting type a Belavin-Drinfeld triple (or rather an enhanced combinatorial data described below). It seems that this connection might provide an additional insight on the problem of classifying discrete types of simple vector bundles on X.

In [6] Mudrov constructs solutions of the QYBE from certain algebraic data that should be viewed as associative analogues of Manin triples. Elsewhere we will show how solutions of the AYBE give rise to such data and will study the corresponding associative algebras that are related to both the classical and quantum side of the story.

Now let us present the combinatorial data on which our trigonometric solutions of the AYBE depend (generalizing Belavin-Drinfeld triples with associative structure considered in [8]). Let S be a finite set. To equip S with a cyclic order is the same as to fix a transitive cyclic permutation C_{0} : S → S. We denote by Γ_{C_{0}} := {(s, C_{0}(s)) | s ∈ S} the graph of C_{0}.

Definition 0.1. An associative BD-structure on a finite set S is given by a pair of transitive cyclic permutations C_{0}, C : S → S and a pair of proper subsets Γ_{1}, Γ_{2} ⊂ Γ_{C_{0}}, such that (C × C)(Γ_{1}) = Γ_{2}, where (C × C)(i, i′) = (C(i), C(i′)).

We can identify Γ_{C_{0}} with the set of vertices Γ of the affine Dynkin diagram Ã_{N−1}, where N = |S| (preserving the cyclic order). Then we get from the above structure a Belavin-Drinfeld triple (Γ_{1}, Γ_{2}, τ) for Ã_{N−1}, where the bijection τ : Γ_{1} → Γ_{2} is induced by C × C. It is clear that τ preserves the inner product. The nilpotency condition on τ is satisfied automatically. Indeed, choose (s_{1}, C_{0}(s_{1})) ∈ Γ_{S} \Γ_{1}. Then for every (s, C_{0}(s)) ∈ Γ_{1} there exists k ≥ 1 with C^{k}(s) = s_{1}, so that (C × C)^{k}(s, C_{0}(s)) ∉ Γ_{1}.

We extend the bijection τ to a bijection τ : P_{1} → P_{2} induced by C × C, where

P_{ι} = {(s, C_{0}^{k}(s)) | (s, C_{0}(s)) ∈ Γ_{ι}, (C_{0}(s), C_{0}^{2}(s)) ∈ Γ_{ι}, …, (C_{0}^{k−1}(s), C_{0}^{k}(s)) ∈ Γ_{ι}}, ι = 1, 2.

For a finite set S let us denote by A_{S} the algebra of endomorphisms of the ℂ-vector space with the basis (e_{i})_{i∈S}, so that A_{S} ≃ Mat(N, ℂ), where N = |S|. We denote by e_{ij} ∈ A_{S} the endomorphism defined by e_{ij}(e_{k}) = δ_{jk}e_{i}. We denote by 𝔥 ⊂ A_{S} the subalgebra of diagonal matrices (i.e., the span of (e_{ii})_{i∈S}). Now we can formulate our result about trigonometric solutions of the AYBE.

Theorem 0.2. (i) Let (C_{0}, C, Γ_{1}, Γ_{2}) be an associative BD-structure on a finite set S. Consider the A_{S} ⊗ A_{S}-valued function

r(u, v) = \frac{1}{1 − exp(−v)} ∑_{i} e_{ii} ⊗ e_{ii} + \frac{1}{exp(u) − 1} ∑_{0≤k<N,i} exp(\frac{ku}{N})e_{C^{k}(i),C^{k}(i)} ⊗ e_{ii}+

\frac{1}{exp(v) − 1} ∑_{0<m<N,j=C_{0}^{m}(i)} exp(\frac{mv}{N})e_{ij} ⊗ e_{ji}+

∑_{0<m<N,k≥1;j=C_{0}^{m}(i),τ^{k}(i,j)=(i′,j′)}[exp(−\frac{ku + mv}{N})e_{ji} ⊗ e_{i′j′} − exp(\frac{ku + mv}{N})e_{i′j′} ⊗ e_{ji}],

where i, i′, j, j′ denote elements of S, and the summation in the last sum is taken only over those (i, j) for which τ^{k} is defined on (i, j). Then r(u, v) satisfies (1) and (2). Furthermore, let us set

R(u, v) = ([exp(\frac{u}{2}) − exp(−\frac{u}{2})]^{−1} + [exp(\frac{v}{2} ) − exp(−\frac{v}{2} )]^{−1})^{−1} · r(u, v). (6)

Then R(u, v) satisfies the QYBE with spectral parameter (4) (for fixed u) and the unitarity condition

R(u, v)R^{21}(u, −v) = 1 ⊗ 1.  (7)

(ii) Assume that N > 1. Then every nondegenerate unitary solution of the AYBE for A = Mat(N, ℂ) with the Laurent expansion at u = 0 of the form (5), where \overline{r}_{0}(v) is a trigonometric solution of the CYBE for sl_{N}, is equal to

c exp(λuv) exp[u(1 ⊗ a) + v(b ⊗ 1)]r(cu, c′v) exp[−u(a ⊗ 1) − v(b ⊗ 1)], where r(u, v) is obtained from one of the solutions from (i) by applying an algebra isomorphism A_{S} ≃ A, λ, c and c′ are constants (c ≠ 0, c′ ≠ 0), and a, b ∈ 𝔥 are infinitesimal symmetries of r(u, v), i.e.,

[a ⊗ 1 + 1 ⊗ a, r(u, v)] = [b ⊗ 1 + 1 ⊗ b, r(u, v)] = 0.

Note that the complete list of scalar unitary solutions of the AYBE was obtained in Theorem 5 of [7]. The solution obtained from Theorem 0.2(i) in the case N = 1 coincides with the basic trigonometric solution from that list (up to changing v to −v).

We will also deduce the following result about solutions of the AYBE not depending on the variable u.

Theorem 0.3. Assume that N > 1. Let r(v) be a nondegenerate unitary solution of the AYBE for A = Mat(N, ℂ) not depending on the variable u. Then

r(v) = \overline{r}(v) + b ⊗ 1 + 1 ⊗ b + \frac{c · 1 ⊗ 1}{Nv},

where \overline{r}(v) is equivalent to a rational nondegenerate solution of the CYBE for sl_{N}, b ∈ sl_{N} is an infinitesimal symmetry of \overline{r}(v), c ∈ ℂ^{∗}. Also,

R(u, v) = (1 + \frac{cu}{v})^{−1} · (1 + ur(v))

is a unitary solution of the QYBE with spectral parameter for fixed u (hence, the same is true for \frac{v}{c} r(v) = lim_{u→∞} R(u, v)).

The case of nondegenerate unitary solutions of the AYBE not depending on v turns out to be much easier — in this case we get a complete list of solutions (see Proposition 1.1). Note that there are no constant nondegenerate solutions of the AYBE for A = Mat(N, ℂ) (unitary or not), as follows from Proposition 2.9 of [2].

The paper is organized as follows. In section 1 we discuss nondegeneracy conditions for solutions of the AYBE and show how to deduce the QYBE in Theorem 1.5. After recalling in section 2 the geometric setup leading to solutions of the AYBE, we calculate these solutions associated with simple vector bundles on cycles of projective lines in sections 3 and 4 (the result is given by formulas (32), (33)). Then in section 5 we consider associative BD-structures on completely ordered sets and classify such structures coming from simple vector bundles on cycles of projective lines (see Theorem 5.5). In section 6 we prove the first part of Theorem 0.2. In section 7 we establish a meromorphic continuation in v for a class of solutions of the AYBE and derive some additional information about these solutions. Finally, in section 8 we prove the second part of Theorem 0.2 and Theorem 0.3. Acknowledgment. I am grateful to Pavel Etingof for the crucial help with organizing my initial computations into a nice combinatorial pattern. I also thank him and Travis Schedler for useful comments on the first draft of the paper and the subsequent helpful discussions. Parts of this work were done while the author enjoyed the hospitality of the Max-Planck-Institute für Mathematik in Bonn and of the SISSA in Trieste. This work was partially supported by the NSF grant DMS-0601034.

1 The AYBE and the QYBE

Recall that we denote A = Mat(N, ℂ). Let r(u, v) be a meromorphic A ⊗ Afunction in a neighborhood of (0, 0). We say that r(u, v) is nondegenerate if the tensor r(u, v) is nondegenerate for generic (u, v).

We start by collecting some facts about nondegenerate unitary solutions of the AYBE. First, let us consider the case when r(u, v) does not depend on v. Then the AYBE reduces to

r^{12}(−u′)r^{13}(u + u′) − r^{23}(u + u′)r^{12}(u) + r^{13}(u)r^{23}(u′) = 0,  (8)

and the unitarity condition becomes r^{21}(−u) = −r(u).

Let us set P = ∑_{i,j} e_{ij} ⊗ e_{ji}.

Proposition 1.1. All nondegenerate unitary solutions of (8) have form

r(u) = (ϕ_{a}(cu) ⊗ id)(P),

where c ∈ ℂ^{∗}, a ∈ sl_{N}, ϕ_{a}(u) ∈ End(A) is the linear operator on A defined from the equation

uϕ_{a}(u)(X) + [a, ϕ_{a}(u)(X)] = X.

Proof. Let us write r(u, v) in the form r(u) = (ϕ(u)⊗id)(e), where e ∈ A^{∗} ⊗A is the canonical element, ϕ(u) : A^{∗} → A is an operator, nondegenerate for generic u. Now set B(u)(X, Y) = (X, ϕ(u)^{−1}(Y)) for X, Y ∈ A. It is easy to see that the equation (8) together with the unitarity condition are equivalent to the following equations on B(u):

B(−u)(XY, Z) + B(−u′)(Y Z, X) + B(u + u′)(ZX, Y) = 0,  (9)
B(u)(X, Y) + B(−u)(Y, X) = 0.  (10)

Substituting Z = 1 in the first equation we find

B(−u)(XY, 1) + (B(u + u′) − B(u′))(X, Y) = 0, i.e.,
B(u + u′)(X, Y) = ξ(u)(XY) + B(u′)(X, Y),

where ξ(u)(X) = −B(−u)(X, 1). Exchanging u and u′ we get that C(X, Y) = B(u)(X, Y) − ξ(u)(XY) does not depend on u. Substituting B(u)(X, Y) = ξ(u)(XY) + C(X, Y) into the previous equation we get

ξ(u + u′) = ξ(u) + ξ(u′),

hence, ξ(u) = u·ξ for some ξ ∈ A^{∗}. Now substituting B(u)(X, Y) = u·ξ(XY)+ C(X, Y) into (10) we derive that ξ(XY) = ξ(Y X) and C is skew-symmetric. Therefore, ξ = c · tr. Finally, equation (9) reduces to the equation

C(XY, Z) + C(Y Z, X) + C(ZX, Y) = 0.

Together with the skew-symmetry of C this implies that C(X, 1) = C(1, X) = 0 and the restriction of C to sl_{N} × sl_{N} is a 2-cocycle. Hence, C(X, Y) = l(XY − Y X) for some linear functional l on sl_{N}. Conversely, for C of this form the above equation is satisfied. Thus, all solutions of (9) and (10) are given by

B(u)(X, Y) = cu tr(X, Y) + l(XY − Y X),

where c ∈ ℂ^{∗} and l is a linear functional on sl_{N}. Let us identify A with A^{∗} using the metric tr(XY). Then we can view ϕ(u) as an operator from A to A such that B(u)(X, Y) = tr(Xϕ(u)^{−1}(Y)). Representing the functional l in the form l(X) = − tr(Xa) we obtain the formula

ϕ(u)^{−1}(Y) = cuY + [a, Y].
⊓ ⊔

Remark 1.2. It is easy to see that ϕ_{a}(u) (and hence the corresponding associative r-matrix) always has a pole at u = 0 with order equal to the maximal k such that there exists X ∈ A with ad^{k}(a)(X) = 0 and ad^{k−1}(a)(X) ≠ 0. Indeed, ϕ_{a}(u) cannot be regular at u = 0 since this would give [a, ϕ_{a}(0)(1)] = 1. Let

ϕ_{a}(u) = \frac{ψ_{−k}}{u^{k}} + \frac{ψ_{−k+1}}{u^{k−1}} + …

be the Laurent expansion of ϕ_{a}(u). Then we have

ψ_{i−1} + ad(a) ∘ ψ_{i} = 0

for i ≠ 0 and

ψ_{−1} + ad(a) ∘ ψ_{0} = id .

Decomposing End(A) into generalized eigenspaces of the operator ψ ↦ ad(a)∘ ψ we see that ψ_{−1} is the component of id ∈ End(A) corresponding to the zero eigenvalue. This immediately implies our claim. For example, if a is semisimple then ϕ_{a}(u) has a simple pole at u = 0. More precisely, taking diagonal matrix a = ∑_{i} a_{i}e_{ii} we get the associative r-matrix

r(u) = ∑_{ij} \frac{1}{u + a_{i} − a_{j}} e_{ij} ⊗ e_{ji}.

The proofs of the next two results are parallel to those of Propositions 2.2 and 2.1 in [3], respectively.

Lemma 1.3. Let r(u, v) be a nondegenerate unitary solution of the AYBE. Assume that r(u, v) does not have a pole at v = 0. Then r(u, 0) is still nondegenerate, and hence has the form described in Proposition 1.1.

Proof. Let us fix v_{0} such that r(u, v) does not have a pole at v = v_{0} and r(u, v_{0}) is nondegenerate for generic u. Then we can define a meromorphic function ϕ(u, v) with values in End_{ℂ}(A) by the condition

(ϕ(u, v) ⊗ id)(r(u, v_{0})) = r(u, v).

We claim that this function satisfies the identity

ϕ(u + u′, v)(XY) = ϕ(u, v)(X)ϕ(u′, v)(Y),  (11)

where X, Y ∈ A. Indeed, since r(u, v) does not have a pole at v = 0, substituting v′ = 0 in (1) we get

r^{12}(−u′, v)r^{13}(u + u′, v) = r^{23}(u + u′, 0)r^{12}(u, v) − r^{13}(u, v)r^{23}(u′, 0).

Note that the right-hand side is obtained by applying ϕ(u, v) ⊗ id ⊗ id to the right-hand side for v = v_{0}. Applying the above equation for v = v_{0} we deduce that it is equal to

(ϕ(u, v) ⊗ id ⊗ id)(r^{12}(−u′, v_{0})r^{13}(u + u′, v_{0})).

On the other hand, the left-hand side can be rewritten as

[(ϕ(−u′, v) ⊗ id)r(−u′, v_{0})]^{12}[(ϕ(u + u′, v) ⊗ id)r(u + u′, v)]^{13}.

Thus, if we write r(u, v_{0}) = ∑ K^{α}(u) ⊗ e_{α}, where e_{α} is a basis of A, then we derive

ϕ(−u′, v)(K^{α}(−u′))ϕ(u + u′, v)(K^{β}(u + u′)) = ϕ(u, v)(K^{α}(−u′)K^{β}(u + u′)).

By nondegeneracy of r(u, v_{0}) this implies (11). Taking Y = 1 in this equation we obtain

ϕ(u + u′, v)(X) = ϕ(u, v)(X)ϕ(u′, v)(1).  (12)

Similarly, we deduce that

ϕ(u + u′, v)(Y) = ϕ(u′, v)(1)ϕ(u, v)(Y).

Comparing these equation we see that ϕ(u′, v)(1) commutes with ϕ(u, v)(X) for any X ∈ A. Using nondegeneracy of r(u, v) we derive that ϕ(u, v)(1) = f(u, v) · 1 for some scalar meromorphic function f(u, v). Furthermore, we should have

f(u + u′, v) = f(u, v)f(u′, v),

which implies that f(u, v) = exp(g(v)u) for some function g(v) holomorphic near v = 0. Next, from (12) we obtain that exp(−g(v)u)ϕ(u, v) does not depend on u. Thus, all solutions of (12) have form ϕ(u, v) = exp(g(v)u)ψ(v), where for every v ψ(v) is an algebra automorphism of A or zero. By our assumption ϕ(u, v) does not have a pole at v = 0. Therefore, ψ(v) is holomorphic near v = 0. Now we use the fact that every algebra automorphism of A is inner, and hence has determinant equal to 1 (it is enough to check this for the conjugation with a diagonalizable matrix). Since, ψ(v_{0}) = id this implies that det ψ(v) = 1 identically. Therefore, det ψ(0) = 1 and ϕ(u, 0) is invertible. ⊓ ⊔

Lemma 1.4. Let r(u, v) be a nondegenerate unitary solution of the AYBE. Assume that r(u, v) has a pole at v = 0. Then this pole is simple and lim_{v→0} vr(u, v) = cP for some nonzero constant c.

Proof. Let r(u, v) = \frac{θ(u)}{v^{k}} + \frac{η(u)}{v^{k−1}} + … be the Laurent expansion of r(u, v) near v = 0. Considering the polar parts as v′ → 0 (resp., v → 0) in (1) we get

−θ^{23}(u + u′)r^{12}(u, v) + r^{13}(u, v)θ^{23}(u′) = 0,  (13)
θ^{12}(−u′)r^{13}(u + u′, v′) − r^{23}(u + u′, v′)θ^{12}(u) = 0.  (14)

Let V ⊂ A be the minimal subspace such that θ(u) ∈ V ⊗ A (for all u where θ(u) is defined). Then we have r^{13}(u, v)θ^{23}(u′) ∈ A ⊗ V ⊗ A. Hence, from (13) we get θ^{23}(u + u′)r^{12}(u, v) ∈ A ⊗ V ⊗ A. This implies that r^{12}(u, v) ∈ A ⊗ A_{1}, where

A_{1} = {a ∈ A : (a ⊗ 1)θ(u) ∈ V ⊗ A for all u}.

By nondegeneracy we get A_{1} = A, hence AV ⊂ V. Similarly, using (14) we derive that V A ⊂ V. Thus, V is a nonzero two-sided ideal in A, so we have V = A. Now let us prove that the order of pole k cannot be greater that 1. Indeed, assuming that k > 1 and considering the coefficient with v^{1−k} in the expansion of (1) near v = 0 we get

η^{12}(−u′)r^{13}(u + u′, v′) − r^{23}(u + u′, v′)η^{12}(u) + θ^{12}(−u′)\frac{∂r^{13}}{∂v} (u + u′, v′) = 0.

Now looking at polar parts at v′ = 0 we get θ^{12}(−u′)θ^{13}(u + u′) = 0 which contradicts to the equality V = A established above. Therefore, k = 1. Now let us look at (13) again. Let us fix u and consider the subspace

A(u) = {x ∈ A : θ(u + u′)(x ⊗ 1) = (1 ⊗ x)θ(u′) for all u′}.

Then from (13) we get that r(u, v) ∈ A⊗A(u). By nondegeneracy this implies that A(u) = A for generic u, so we get an identity

θ(u + u′)(x ⊗ 1) = (1 ⊗ x)θ(u′)

for all x ∈ A. Taking x = 1 we see that θ(u) = θ is constant. Finally, any tensor θ ∈ A ⊗ A with the property θ(x ⊗ 1) = (1 ⊗ x)θ is proportional to P. ⊓ ⊔

Recall that if r(u, v) is a solution of the AYBE with the Laurent expansion at u = 0 of the form (5) then r_{0}(v) is a unitary solution of the CYBE (see proof of Lemma 1.2 in [7], or Lemma 2.9 of [8]). The same is true for \overline{r}_{0}(v) = (pr ⊗ pr)(r_{0}(v)) ∈ sl_{N} ⊗ sl_{N}. We will show below that the nondegeneracy of r(u, v) implies that \overline{r}_{0}(v) is also nondegenerate, hence it is either elliptic, trigonometric, or rational. The first two cases are distinguished from the third by the condition that \overline{r}_{0}(v) is periodic with respect to v ↦ v + p for some p ∈ ℂ^{∗}.

Recall that by an infinitesimal symmetry of an A⊗A-valued function f(x) we mean an element a ∈ A such that [a ⊗ 1 + 1 ⊗ a, f(x)] = 0 for all x.

Theorem 1.5. Let r(u, v) be a nondegenerate unitary solution of the AYBE with the Laurent expansion at u = 0 of the form (5), and let \overline{r}_{0}(v) = (pr ⊗ pr)(r_{0}(v)). Then

(i) \overline{r}_{0}(v) is a nondegenerate unitary solution of the CYBE. (ii) The following conditions are equivalent:

(a) r(u, v) satisfies the QYBE (4) in v (for fixed u);

(b) the product r(u, v)r(−u, v) is a scalar multiple of 1 ⊗ 1;

(c) \frac{d}{dv} (r_{0}(v) − \overline{r}_{0}(v)) is a scalar multiple of 1 ⊗ 1.

(d) (pr ⊗ pr ⊗ pr)[\overline{r}_{0}^{12}(v)\overline{r}_{0}^{13}(v + v′) − \overline{r}_{0}^{23}(v′)\overline{r}_{0}^{12}(v) + \overline{r}_{0}^{13}(v + v′)\overline{r}_{0}^{23}(v′)] = 0. (iii) The equivalent conditions in (ii) hold when \overline{r}_{0}(v) either admits a period or has no infinitesimal symmetries in sl_{N}.

Remarks 1.6. 1. In fact, our proof shows that equivalent conditions in (ii) hold under the weaker assumption that the system

[\overline{r}_{0}(v), a^{1} + a^{2}] = [\overline{r_{0}}(v), b^{1} + b^{2} + va^{1}] = [b, a] = 0

on a, b ∈ sl_{N} implies that a = 0.

2. Note that the implication (b) ⟹ (a) in part (ii) of the theorem holds for any unitary solution of the AYBE (as follows easily from Lemma 1.8 below). It is plausible that one can check condition (b) in other situations than those considered in the above theorem. For example, we have nondegenerate unitary solutions of the AYBE of the form

r(u, v) = \frac{ω}{u^{n}} + \frac{P}{v} ,

where n ≥ 1, ω ∈ A ⊗ A satisfies ω^{12}ω^{13} = 0 and ω^{21} = (−1)^{n−1}ω. It is easy to see that these solutions satisfy r(u, v)r(−u, v) = 1 ⊗ 1/v^{2}, so they also satisfy the QYBE. On the other hand, the solutions of the AYBE constructed in Proposition 1.1 do not satisfy the QYBE in general.

Lemma 1.7. Assume that N > 1. Let r(u, v) be a nondegenerate unitary solution of the AYBE with the Laurent expansion at u = 0 of the form (5). Then r(u, v) has a simple pole at v = 0 with the residue c · P, where c ∈ ℂ^{∗}.

Proof. By Lemma 1.4 we only have to rule out the possibility that r(u, v) has no pole at v = 0. Assume this is the case. Then r(u, 0) is the solution of (8) that has a simple pole at u = 0 with the residue 1 ⊗ 1. Let ϕ(u) : A → A be the linear operator such that r(u, 0) = (ϕ(u) ⊗ id)(P). Then ϕ(u) has the Laurent expansion at u = 0 of the form

ϕ(u)(X) = \frac{tr(X) · 1}{u} + ψ(X) + …

for some operator ψ : A → A. By Proposition 1.1 we have

cuϕ(u)(X) + [a, ϕ(u)(X)] = X

for some c ∈ ℂ^{∗} and a ∈ sl_{N}. Considering the constant terms of the expansions at u = 0 we get

c tr(X) · 1 + [a, ψ(X)] = X.

It follows that [a, ψ(X)] = X for all X ∈ sl_{N}. Hence, the operator pr ψ|_{sl_{N}} : sl_{N} → sl_{N} is invertible. Taking in the above equality X ∈ sl_{N} such that pr ψ(X) = a we derive that a = 0 which leads to a contradiction. ⊓ ⊔

The next two lemmas constitute the core of the proof of Theorem 1.5.

Lemma 1.8. For a triple of variables u_{1}, u_{2}, u_{3} (resp., v_{1}, v_{2}, v_{3}) set u_{ij} = u_{i} − u_{j} (resp., v_{ij} = v_{i} − v_{j}). Then for every unitary solution of the AYBE one has

r^{12}(u_{12}, v_{12})r^{13}(u_{23}, v_{13})r^{23}(u_{12}, v_{23}) − r^{23}(u_{23}, v_{23})r^{13}(u_{12}, v_{13})r^{12}(u_{23}, v_{12}) =

s^{23}(u_{23}, v_{23})r^{13}(u_{13}, v_{13}) − r^{13}(u_{13}, v_{13})s^{23}(u_{21}, v_{23}) =

r^{13}(u_{13}, v_{13})s^{12}(u_{32}, v_{12}) − s^{12}(u_{12}, v_{12})r^{13}(u_{13}, v_{13}),

where s(u, v) = r(u, v)r(−u, v).

Proof. In the following proof we will use the short-hand notation r^{ij}(u) for r^{ij}(u, v_{ij}). The AYBE can be rewritten as

r^{12}(u_{12})r^{13}(u_{23}) − r^{23}(u_{23})r^{12}(u_{13}) + r^{13}(u_{13})r^{23}(u_{21}) = 0.  (15)

On the other hand, switching indices 1 and 2 and using the unitarity condition we obtain

r^{23}(u_{23})r^{13}(u_{12}) − r^{12}(u_{12})r^{23}(u_{13}) + r^{13}(u_{13})r^{12}(u_{32}) = 0.  (16)

Multiplying (16) with r^{12}(u_{23}) on the right we get

576  A. Polishchuk

r^{23}(u_{23})r^{13}(u_{12})r^{12}(u_{23}) − r^{12}(u_{12})r^{23}(u_{13})r^{12}(u_{23}) + r^{13}(u_{13})s^{12}(u_{32}) = 0. On the other hand, switching u_{1} and u_{2} in (15) and multiplying the obtained equation with r^{12}(u_{12}) on the left we obtain s^{12}(u_{12})r^{13}(u_{13}) − r^{12}(u_{12})r^{23}(u_{13})r^{12}(u_{23}) + r^{12}(u_{12})r^{13}(u_{23})r^{23}(u_{12}) = 0. Taking the difference between these cubic equations gives r^{23}(u_{23})r^{13}(u_{12})r^{12}(u_{23})−r^{12}(u_{12})r^{13}(u_{23})r^{23}(u_{12}) = s^{12}(u_{12})r^{13}(u_{13})−r^{13}(u_{13})s^{12}(u_{32}). The other half of the required equation is obtained by switching the indices 1 and 3 and using the unitarity condition. ⊓ ⊔

Lemma 1.9. Let r(u, v) be a unitary solution of the AYBE with the Laurent expansion (5) at u = 0. Assume also that r(u, v) has a simple pole at v = 0 with the residue cP. Then one has

s(u, v) = r(u, v)r(−u, v) = a ⊗ 1 + 1 ⊗ a + (f(u) + g(v)) · 1 ⊗ 1 with

g(v) = −\frac{c}{N} (tr ⊗ tr)(\frac{dr_{0}(v)}{dv}),

f(u) = \frac{1}{N} tr μ(\frac{∂r(u, 0)}{∂u}),

a = pr μ(\frac{∂r(u, 0)}{∂u}),

where μ : A ⊗ A → A denotes the product. Furthermore, a ∈ sl_{N} is an infinitesimal symmetry of r(u, v), and if we write

r_{0}(v) = \overline{r}_{0}(v) + α(v) ⊗ 1 − 1 ⊗ α(−v) + h(v) · 1 ⊗ 1,

where \overline{r}_{0}(v) ∈ sl_{N} ⊗ sl_{N} and α(v) ∈ sl_{N}, then

α(v) = α(0) − \frac{v}{cN} a.

Proof. Let us write r(u, v) = \frac{cP}{v} + r̃(u, v), where r̃(u, v) does not have a pole at v = 0. Then we can rewrite the AYBE as follows (where v_{ij} = v_{i} − v_{j}): r^{13}(u, v_{13})r^{23}(−u + h, v_{23}) = r^{23}(h, v_{23})r^{12}(u, v_{12}) − r^{12}(u − h, v_{12})r^{13}(h, v_{13}) = r^{23}(h, v_{23})r^{12}(u, v_{12}) − r^{12}(u, v_{12})r^{13}(h, v_{13}) + [r^{12}(u, v_{12}) − r^{12}(u − h, v_{12})]r^{13}(h, v_{13}) =

\frac{r^{23}(h, v_{23}) − r^{23}(h, v_{13})}{v_{12}}cP^{12} + [r^{23}(h, v_{23})r^{12}(u, v_{12}) − r^{12}(u, v_{12})r^{13}(h, v_{13})]+

[r^{12}(u, v_{12}) − r^{12}(u − h, v_{12})]r^{13}(h, v_{13}).

Passing to the limit v_{2} → v_{1} we derive

r^{13}(u, v)r^{23}(−u + h, v) = −\frac{∂r^{23}}{∂v} (h, v)cP^{12} + [r^{23}(h, v)r^{12}(u, 0) − r^{12}(u, 0)r^{13}(h, v)]+

[r^{12}(u, 0) − r^{12}(u − h, 0)]r^{13}(h, v).

Next, we are going to apply the operator μ ⊗ id : A ⊗ A ⊗ A → A ⊗ A, where μ is the product on A. We use the following easy observations:

(μ⊗id)(x^{13}y^{23}) = xy, (μ⊗id)(x^{23}y^{12}−y^{12}x^{13}) = 0, (μ⊗id)(x^{23}P^{12}) = 1⊗tr_{1}(x),

where x, y ∈ A ⊗ A, tr_{1} = tr ⊗ id : A ⊗ A → A (the last property follows from the identity ∑_{ij} e_{ij}ae_{ji} = tr(a) · 1 for a ∈ A). Thus, applying μ ⊗ id to the above equation we get

r(u, v)r(−u+h, v) = −c·1⊗tr_{1}(\frac{∂r}{∂v} (h, v))+(μ⊗id) ([r^{12}(u, 0) − r^{12}(u − h, 0)]r^{13}(h, v)) .

Finally, taking the limit h → 0 we derive

s(u, v) = −c · 1 ⊗ tr_{1}(\frac{dr_{0}(v)}{dv}) + μ(\frac{∂r(u, 0)}{∂u}) ⊗ 1,  (17)

where we used the equalities \frac{∂r(0,v)}{∂v} = \frac{dr_{0}(v)}{dv} and \frac{∂r̃(u,v)}{∂u} = \frac{∂r(u,v)}{∂u}. Hence, we can write s(u, v) in the form

s(u, v) = a(u) ⊗ 1 + 1 ⊗ b(v) + (f(u) + g(v))1 ⊗ 1,

where a(u) and b(v) take values in sl_{N}, and

b(v) = −c pr tr_{1}(\frac{dr_{0}(v)}{dv}).

The unitarity condition on r(u, v) implies that s^{21}(−u, −v) = s(u, v). This immediately gives the required form of s(u, v) with some a ∈ sl_{N}, as well as the formulas for g(v), f(u), a and α(v). The fact that a is an infinitesimal symmetry of r(u, v) follows from the second equality in the identity of Lemma 1.8. ⊓ ⊔

Lemma 1.10. Let r_{0}(v) ∈ A ⊗ A be a unitary solution of the CYBE of the form

r_{0}(v) = \overline{r}_{0}(v) + α(v) ⊗ 1 − 1 ⊗ α(−v) + h(v) · 1 ⊗ 1,

where \overline{r}_{0}(v) ∈ sl_{N} ⊗ sl_{N} and α(v) ∈ sl_{N}. Then

[\overline{r}_{0}(v − v′), α(v) ⊗ 1 + 1 ⊗ α(v′)] = [α(v), α(v′)] = 0.

In particular, if α(v) depends linearly on v, i.e., α(v) = b + v · a, then

[\overline{r}_{0}(v), a ⊗ 1 + 1 ⊗ a] = [\overline{r}_{0}(v), b ⊗ 1 + 1 ⊗ b + va ⊗ 1] = [b, a] = 0.

Proof. Applying pr ⊗ pr ⊗ pr to both sides of the CYBE we see that \overline{r}_{0} itself satisfies the CYBE. Taking this into account the equation can be rewritten as

[\overline{r}_{0}^{12}(v_{12}), α^{1}(v_{13}) + α^{2}(v_{23})] + [α^{1}(v_{12}), α^{1}(v_{13})] + c.p.(1, 2, 3) = 0,

where v_{ij} = v_{i} − v_{j} (the omitted terms are obtained by cyclically permuting 1, 2, 3). Applying the operator pr ⊗ pr ⊗ id gives

[\overline{r}_{0}^{12}(v_{12}), α^{1}(v_{13}) + α^{2}(v_{23})] = 0.

Now returning to the above equality and applying pr ⊗ id ⊗ id we derive that [α(v), α(v′)] = 0. ⊓ ⊔

Proof of Theorem 1.5. (i) In the case N = 1 the statement is vacuous, so we can assume that N > 1. By Lemma 1.7, r_{0}(v) has a simple pole at v = 0 with the residue cP, where c ∈ ℂ^{∗}. Projecting to sl_{N} we deduce that \overline{r}_{0}(v) is nondegenerate. (ii) By Lemma 1.8, r(u, v) satisfies the QYBE iff

s^{23}(u, v_{23})r^{13}(2u, v_{13}) = r^{13}(2u, v_{13})s^{23}(−u, v_{23}).

Using the formula for s(u, v) from Lemma 1.9 we see that this is equivalent to the equality

[r(u, v), 1 ⊗ a] = 0

which is equivalent to a = 0 by the nondegeneracy of r(u, v). Note that by Lemma 1.9, both conditions (b) and (c) are also equivalent to the equality a = 0. It remains to show the equivalence of (d) with this equality. To this end we use the identity

r_{0}^{12}(v)r_{0}^{13}(v+v′)−r_{0}^{23}(v′)r_{0}^{12}(v)+r_{0}^{13}(v+v′)r_{0}^{23}(v′) = r_{1}^{12}(v)+r_{1}^{13}(v+v′)+r_{1}^{23}(v′)

(18)

deduced by substituting the Laurent expansions in the first variable into (1). Let us denote the expression in the left-hand-side of (18) by AY BE[r_{0}](v, v′). Using the relation between r_{0}(v) and \overline{r}_{0}(v) from Lemma 1.9 we obtain

− cN · (pr ⊗ pr ⊗ pr) (AY BE[r_{0}](v, v′) − AY BE[\overline{r}_{0}](v, v′)) =
v\overline{r}_{0}^{12}(v)a^{3} + v′\overline{r}_{0}^{23}(v′)a^{1} + (v + v′)\overline{r}_{0}^{13}(v + v′)a^{2},

where a^{1} = a⊗1⊗1, etc. Note that (18) implies that (pr ⊗ pr ⊗ pr)AY BE[r_{0}](v, v′) = 0. Therefore, it suffices to prove that the equation

v\overline{r}_{0}^{12}(v)a^{3} + v′\overline{r}_{0}^{23}(v′)a^{1} + (v + v′)\overline{r}_{0}^{13}(v + v′)a^{2} = 0

on a ∈ sl_{N} implies that a = 0. Passing to the limit as v → 0 and v′ → 0 we deduce from the above equality that

(pr ⊗ pr ⊗ pr)[P^{12}a^{3} + P^{23}a^{1} + P^{13}a^{2}] = 0.

Let a = ∑ a_{ij}e_{ij}. Looking at the coefficient with e_{ij} ⊗ e_{ji} ⊗ e_{ij} we deduce that a_{ij} = 0 for i ≠ j. Finally, looking at the projection to e_{12} ⊗ e_{21} ⊗ sl_{N} we deduce that a_{ii} does not depend on i, hence a = 0. ⊓ ⊔

(iii) It suffices to prove that under our assumptions the infinitesimal symmetry a ∈ sl_{N} appearing in Lemma 1.9 is equal to zero. We only have to consider the case when \overline{r}_{0} has a period, i.e., \overline{r}_{0}(v + p) = \overline{r}_{0}(v) for some p ∈ ℂ^{∗}. By Lemma 1.10, it remains to check that the equation

[\overline{r}_{0}(v), b ⊗ 1 + 1 ⊗ b + va ⊗ 1] = 0

on a, b ∈ sl_{N} implies that a = 0. From the periodicity of \overline{r}_{0} we derive that

[\overline{r}_{0}(v), a ⊗ 1] = 0.

By the nondegeneracy of \overline{r}_{0}, it follows that a = 0. ⊓ ⊔

2 Solutions of the AYBE associated with simple vector bundles on degenerations of elliptic curves

Now let us review how solutions of the AYBE arise from geometric structures on elliptic curves and their degenerations. Let X be a nodal projective curve over ℂ of arithmetic genus 1 such that the dualizing sheaf on X is isomorphic to O_{X}. Let us fix such an isomorphism. Recall that a vector bundle V on X is called simple if End(V) = ℂ. The following result follows from Theorems 1 and 4 of [7].

Theorem 2.1. Let V_{1}, V_{2} be a pair of simple vector bundles on X such that Hom^{0}(V_{1}, V_{2}) = Ext^{1}(V_{1}, V_{2}) = 0. Let y_{1}, y_{2} be a pair of distinct smooth points of X. Consider the tensor

r_{y_{1},y_{2}}^{V_{1},V_{2}} ∈ Hom(V_{1,y_{1}}^{∗}, V_{2,y_{1}}^{∗}) ⊗ Hom(V_{2,y_{2}}^{∗}, V_{1,y_{2}}^{∗})

corresponding to the following composition

Hom(V_{1,y_{1}}, V_{2,y_{1}}) ^{^{Res_{y_{1}}^{−1}} \overline{}}Hom(V_{1}, V_{2}(y_{1})) ^{^{ev_{y_{2}}} \overline{}}Hom(V_{1,y_{2}}, V_{2,y_{2}}),

where V_{i,y} denotes the fiber of V_{i} at a point y ∈ X, the map

Res_{y} : Hom(V_{1}, V_{2}(y))˜→ Hom(V_{1,y}, V_{2,y})

is obtained by taking the residue at y, and the map ev_{y} is the evaluation at y. Then for a triple of simple bundles (V_{1}, V_{2}, V_{3}) such that each pair satisfies the above assumptions and for a triple of distinct points (y_{1}, y_{2}, y_{3}) one has

(r_{y_{1}y_{2}}^{V_{3}V_{2}})^{12}(r_{y_{1}y_{3}}^{V_{1}V_{3}})^{13} − (r_{y_{2}y_{3}}^{V_{1}V_{3}})^{23}(r_{y_{1}y_{2}}^{V_{1}V_{2}})^{12} + (r_{y_{1}y_{3}}^{V_{1}V_{2}})^{13}(r_{y_{2}y_{3}}^{V_{2}V_{3}})^{23} = 0  (19)

in Hom(V_{1,y_{1}}^{∗}, V_{2,y_{1}}^{∗}) ⊗ Hom(V_{2,y_{2}}^{∗}, V_{3,y_{2}}^{∗}) ⊗ Hom(V_{3,y_{3}}^{∗}, V_{1,y_{3}}^{∗}). In addition the following unitarity condition holds:

(r_{y_{1}y_{2}}^{V_{1}V_{2}})^{21} = −r_{y_{2}y_{1}}^{V_{2}V_{1}}.  (20)

Remark 2.2. The tensor r_{y_{1},y_{2}}^{V_{1},V_{2}} in the above theorem is a certain triple Massey product in the derived category of X, and the equation (19) follows from the appropriate A_{∞}-axiom (see [7]).

We are going to apply the above theorem for bundles V_{i} of the form V_{i} = V ⊗ L_{i}, where V is a fixed simple vector bundle of rank N on X and L_{i} are line bundles in Pic^{0}(X), the neutral component of Pic(X). Also, we let points y_{i} vary in a connected component X_{0} of X. Uniformizations of X_{0} ∩ X^{reg} and of Pic^{0}(X) allow to describe V_{i}’s and y_{i}’s by complex parameters. Thus, using trivializations of the bundles V_{i,y_{j}}^{∗} we can view the tensor r_{y_{1},y_{2}}^{V_{1},V_{2}} in the above theorem as a function of complex variables r(u_{1}, u_{2}; v_{1}, v_{2}) ∈ A ⊗ A, where A = Mat(N, ℂ), u_{i} describes V_{i}, v_{j} describes y_{j}. Note that equation (19) reduces to the AYBE in the case when r depends only on the differences of variables, i.e., r(u_{1}, u_{2}; v_{1}, v_{2}) = r(u_{1} − u_{2}, v_{1} − v_{2}).

A different choice of trivializations of V_{i,y_{i}}^{∗} would lead to the tensor r̃(u_{1}, u_{2}, v_{1}, v_{2}) given by

r̃(u_{1}, u_{2}; v_{1}, v_{2}) = (φ(u_{2}, v_{1})⊗φ(u_{1}, v_{2}))r(u_{1}, u_{2}; v_{1}, v_{2})(φ(u_{1}, v_{1})⊗φ(u_{2}, v_{2}))^{−1}

where φ(u, v) is a function with values in GL_{N}(ℂ). We say that tensor functions r̃ and r related in this way are equivalent. Note that the condition for functions to depend only on the differences u_{1} − u_{2} and v_{1} − v_{2} is not preserved under these equivalences in general. However, if (a, b) is a pair of commuting infinitesimal symmetries of r(u_{1} − u_{2}, v_{1} − v_{2}) then taking φ(u, v) = exp(ua + vb) we do get a tensor function r̃ that depends only on the differences, namely,

r̃(u, v) = exp[u(1 ⊗ a) + v(b ⊗ 1)]r(u, v) exp[−u(a ⊗ 1) − v(b ⊗ 1)]

(this kind of equivalence shows up in Theorem 0.2(ii)).

Since we are interested in trigonometric solutions, we will be using the multiplicative variables x_{i} = exp(u_{i}), y_{i} = exp(v_{i}). The solutions of (19) that we are going to construct in the next section will be equivalent to those depending only on the differences u_{1} − u_{2}, v_{1} − v_{2}. It will be convenient for us also to work with the intermediate form of the AYBE

\begin{matrix}r^{12}((x′)^{−1}; y_{1}, y_{2})r^{13}(xx′; y_{1}, y_{3}) − r^{23}(xx′; y_{2}, y_{3})r^{12}(x; y_{1}, y_{2})+ \\ r^{13}(x; y_{1}, y_{3})r^{23}(x′; y_{2}, y_{3}) = 0 \end{matrix}  (21)

for the tensor r(x; y_{1}, y_{2}) ∈ A ⊗ A, obtained from (19) in the case when r(x_{1}, x_{2}; y_{1}, y_{2}) = r(x_{1}/x_{2}; y_{1}, y_{2}). The corresponding unitarity condition has form

r^{21}(x; y_{1}, y_{2}) = −r(x^{−1}; y_{2}, y_{1}).  (22)

3 Simple vector bundles on cycles of projective lines

Let X = X_{0} ∪X_{1} ∪…∪X_{n−1} be the union of n copies of ℙ^{1}’s glued (transversally) in a configuration of type Ã_{n−1}, so that the point ∞ on X_{j} is identified with the point 0 on X_{j+1} for j = 0, …, n − 1 (where we identify indices with elements of ℤ/nℤ). A vector bundle V of rank N on X is given by a collection of vector bundles V_{j} of rank N on X_{j} along with isomorphisms (V_{j})_{∞} ≃ (V_{j+1})_{0}. Since every vector bundle on ℙ^{1} splits into a direct sum of line bundles, we can assume that

V_{j} = O_{ℙ^{1}}(m_{1}^{j}) ⊕ … ⊕ O_{ℙ^{1}}(m_{N}^{j})

for every j = 0, …, n − 1. Thus, the splitting types are described by the N × n-matrix of integers (m_{i}^{j}).

Let (z_{0} : z_{1}) denote the homogeneous coordinates on ℙ^{1}. We will use the standard trivialization of the fiber of O_{ℙ^{1}}(1) at 0 = (1 : 0) ∈ ℙ^{1} (resp., at ∞) given by the generating section z_{0} (resp., z_{1}). Note that a section s ∈ O_{ℙ^{1}}(1) is uniquely determined by its values s(0) and s(∞) (namely, s = s(0)z_{0} + s(∞)z_{1}).

Let us fix a splitting type matrix m = (m_{i}^{j}). For every λ ∈ ℂ^{∗} we define the rank-N bundle V^{λ} = V^{λ}(m) on X by using standard trivializations of V_{j} = ⊕_{i=1}^{n}O(m_{i}^{j}) at 0 and ∞ and setting the transition isomorphisms (V_{j})_{∞} ≃ (V_{j+1})_{0} to be identical for j = 0, …, n − 2, and the last transition map to be

λC^{−1} : (V_{0})_{0} → (V_{n−1})_{∞}

where C is the cyclic permutation matrix: Ce_{i} = e_{i−1}, where we identify the set of indices with ℤ/NZ. Note that in this definition only the cyclic order on the indices {1, …, N} is used. In particular, if we cyclically permute the rows of the matrix (m_{i}^{j}) (by replacing m_{i}^{j} with m_{i+1}^{j}) then we get the same vector bundle.

Lemma 3.1 below provides a criterion for simplicity of V^{λ}(m). This result is well-known (see [4], Theorem 5.3). For completeness we include the proof. It is also known that every simple vector bundle on X is isomorphic to some V^{λ}(m) (see loc. cit.). It will be convenient to extend the N × n-matrix (m_{i}^{j}) to the matrix with columns numbered by j ∈ ℤ using the rule m_{i}^{j+n} = m_{i−1}^{j}.

Lemma 3.1. The vector bundle V^{λ}(m) is simple iff the following two conditions are satisfied:

(a) the differences m_{i}^{j} − m_{i′}^{j} for i, i′ ∈ ℤ/Nℤ take values only {−1, 0, 1};

(b) for every i, i′ ∈ ℤ/Nℤ, i ≠ i′, the nN-periodic infinite sequence

(m_{i}^{j} − m_{i′}^{j}), j ∈ ℤ

is not identically 0, and the occurrences of 1 and −1 in it alternate. Furthermore, if (a) and (b) hold then V^{λ_{1}}(m) ≃ V^{λ_{2}}(m) iff (λ_{1}/λ_{2})^{N} = 1.

Proof. First, we observe that if m_{i}^{j} − m_{i′}^{j} = 2 then there exists a nonzero morphism O_{ℙ^{1}}(m_{i′}^{j}) → O_{ℙ^{1}}(m_{i}^{j}) vanishing at 0 and ∞. Viewing it as an endomorphism of V_{j} we obtain a non-scalar endomorphism of V^{λ}. Hence, the condition (a) is necessary. From now on let us assume that (a) is satisfied.

A morphism V^{λ_{1}} → V^{λ_{2}} is given by a collection of morphisms A_{j} : V_{j} → V_{j}, j = 0, …, n − 1, such that A_{j}(∞) = A_{j+1}(0) for j = 0, …, n − 2 and

A_{0}(0) = \frac{λ_{1}}{λ_{2}} CA_{n−1}(∞)C^{−1}.

We can write these maps as matrices A_{j} = (a_{ii′}^{j})_{1≤i,i′≤N}, where a_{ii′}^{j} ∈ H^{0}(ℙ^{1}, O(m_{i}^{j}−m_{i′}^{j})). Let us allow the index j to take all integer values by using the rule a_{ii′}^{j+n} = a_{i−1,i′−1}^{j}. Note that we still have a_{ii′}^{j} ∈ H^{0}(ℙ^{1}, O(m_{i}^{j} − m_{i′}^{j})) because of our convention on m_{i}^{j} for j ∈ ℤ. Then the equations on (A_{j}) can be rewritten as

a_{ii′}^{j}(0) = x^{δ(j)}a_{ii′}^{j−1}(∞)  (23)

for all i, i′ ∈ ℤ/Nℤ and j ∈ ℤ, where x = λ_{1}/λ_{2}, and δ(j) = 1 for j ≡ 0(n), δ(j) = 0 otherwise. Due to condition (a) we have the following possibilities for each a_{ii′}^{j}:

(i) if m_{i}^{j} < m_{i′}^{j} then a_{ii′}^{j} = 0; (ii) if m_{i}^{j} = m_{i′}^{j} then a_{ii′}^{j} is a constant, so a_{ii′}^{j}(0) = a_{ii′}^{j}(∞); (iii) if m_{i}^{j} > m_{i′}^{j} then a_{ii′}^{j} is a section of O(1), so it is uniquely determined by its values at 0 and ∞, and these values can be arbitrary.

From this we can immediately derive that (b) is necessary for V^{λ} to be simple. Indeed, if for some i ≠ i′ we have m_{i}^{j} = m_{i′}^{j} for all j ∈ ℤ then we can get a solution of (23) with x = 1 by setting a_{i+k,i′+k}^{j} = 1 for all j, k ∈ ℤ and letting the remaining entries to be zero. This would give a non-scalar endomorphism of V^{λ}. Similarly, if for some i ≠ i′ and some segment [j, k] ⊂ ℤ we have

(m_{i}^{j} − m_{i′}^{j}, m_{i}^{j+1} − m_{i′}^{j+1}, …, m_{i}^{k} − m_{i′}^{k}) = (1, 0, …, 0, 1)

then we get a solution of (23) with x = 1 by setting

a_{ii′}^{j} = z_{1}, a_{ii′}^{j+1} = 1, …, a_{ii′}^{k−1} = 1, a_{ii′}^{k} = z_{0}

and letting the remaining entries to be zero.

Conversely, assume (a) and (b) hold. Then one can easily derive that V^{λ} is simple by analyzing the system (23) (with x = 1). Indeed, let us show first that a_{ii′}^{j} = 0 for i ≠ i′. It follows from (b) that in the case m_{i}^{j} = m_{i′}^{j} we can either find a segment [j_{1}, j] ⊂ ℤ such that m_{i}^{k} = m_{i′}^{k} for j_{1} < k < j and m_{j_{1}}^{k} < m_{j_{1}}^{k}, or a segment [j, j_{2}] ⊂ ℤ such that m_{i}^{k} = m_{i′}^{k} for j < k < j_{2} and m_{j_{2}}^{k} < m_{j_{2}}^{k}. In either case applying iteratively (23) we derive that a_{ii′}^{j} = 0 (recall that in this case a_{ii′}^{j} is a constant). In the case m_{i}^{j} > m_{i′}^{j} we can find both segments [j_{1}, j] and [j, j_{2}] as above, so that (23) implies that a_{ii′}^{j}(0) = a_{ii′}^{j}(∞) = 0. Hence, a_{ii′}^{j} = 0. The remaining part of the system (23) shows that all a_{ii}^{j} are equal to the same constant, i.e., V^{λ} has no non-scalar endomorphisms.

The above argument also shows that a morphism (a_{ii′}^{j}) : V^{λ_{1}}(m) → V^{λ_{2}}(m) has a_{ii′}^{j} = 0 (assuming conditions (a) and (b) hold), while the remaining components a_{ii}^{j} ∈ ℂ satisfy the equations

a_{ii}^{j} = x^{δ(j)}a_{ii}^{j−1}, a_{ii}^{j+n} = a_{i−1,i−1}^{j},

where x = λ_{1}/λ_{2}. This system has a nonzero solution iff x^{N} = 1, in which case the solution gives an isomorphism V^{λ_{1}}(m) ≃ V^{λ_{2}}(m). ⊓ ⊔

4 Computation of the associative r-matrix arising as a Massey product

Henceforward, we always assume that the matrix (m_{i}^{j}) satisfies the conditions of Lemma 3.1. Given a pair of parameters λ_{1}, λ_{2} ∈ ℂ^{∗} and a pair of points y, y′ ∈ X_{0} \ {0, ∞} we want to describe explicitly the maps

Res_{y} : Hom(V^{λ_{1}}, V^{λ_{2}}(y)) → Hom(V_{y}^{λ_{1}}, V_{y}^{λ_{2}}),
ev_{y′} : Hom(V^{λ_{1}}, V^{λ_{2}}(y)) → Hom(V_{y′}^{λ_{1}}, V_{y′}^{λ_{2}})

and especially the composition ev_{y′} ∘ Res_{y}^{−1} (for generic λ_{1}, λ_{2}). We will identify the target spaces of both maps with N × N-matrices using trivializations of the relevant line bundles over y induced by the appropriate power of z_{0} ∈ H^{0}(ℙ^{1}, O(1)). We also use the global 1-form trivializing ω_{X} that restricts to dz/z on each ℙ^{1} \ {∞} (where z = z_{1}/z_{0}).

A morphism V^{λ_{1}} → V^{λ_{2}}(y) is given by a collection of morphisms

A_{0} : V_{0} → V_{0}(y), A_{1} : V_{1} → V_{1}, …, A_{n−1} : V_{n−1} → V_{n−1}

with same equations as before. Writing these maps as matrices we can view Hom(V^{λ_{1}}, V^{λ_{2}}(y)) as the space of solutions of (23), where a_{ii′}^{j} ∈ H^{0}(ℙ^{1}, O(m_{i}^{j} − m_{i′}^{j})) for j ≢ 0(n) and a_{ii′}^{j} ∈ H^{0}(ℙ^{1}, O(m_{i}^{j} − m_{i′}^{j})(y)) for j ≡ 0(n).

Since the component X_{0} plays a special role, we will use a shorthand notation m_{i} := m_{i}^{0}, a_{ii′} := a_{ii′}^{0}. Let us also set b_{ii′} = Res_{y}(a_{ii′}). Recall that for every pair i, i′ ∈ ℤ/Nℤ we have the following three possibilities.

(i) If m_{i} < m_{i′} then we have a_{ii′} = \frac{yb_{ii′}}{z_{1}−yz_{0}} , so that

a_{ii′}(0) = −b_{ii′}, a_{ii′}(∞) = yb_{ii′}.  (24)

(ii) If m_{i} = m_{i′} then a_{ii′} = \frac{a_{ii′} (∞)z−a_{ii′} (0)y}{z−y} (where z = z_{1}/z_{0}), so we get the relation

a_{ii′}(∞) − a_{ii′}(0) = b_{ii′}.  (25)

(iii) If m_{i} > m_{i′} then a_{ii′} is uniquely determined by a_{ii′}(0), a_{ii′}(∞) and b_{ii′}. Namely, one can easily check that

a_{ii′} = \frac{z(b_{ii′} + a_{ii′}(0) − ya_{ii′}(∞)) − ya_{ii′}(0)}{z − y} · z_{0} + \frac{za_{ii′}(∞)}{z − y} · z_{1}.

Note that in the above three cases we also have the following expressions for a_{ii′}(y′):

a_{ii′}(y′) = \begin{cases}\frac{yb_{ii′}}{y′−y} , & m_{i} < m_{i′}, \\ \frac{y′b_{ii′}}{y′−y} + a_{ii′}(0) = \frac{yb_{ii′}}{y′−y} + a_{ii′}(∞), & m_{i} = m_{i′}, \\ \frac{y′b_{ii′}}{y′−y} + a_{ii′}(0) + y′a_{ii′}(∞), & m_{i} > m_{i′}. \end{cases}  (26)

To compute ev_{y′} ∘ Res_{y}^{−1} means to express all the entries a_{ii′}(y′) in terms of (b_{ii′}). The above formula gives such an expression in the case m_{i} < m_{i′}; in the case m_{i} = m_{i′} we need to know either a_{ii′}(0) or a_{ii′}(∞); and in the case m_{i} > m_{i′} we need to know both. Of course, in the latter two cases one has to use equations (23). Then condition (b) of Lemma 3.1 will guarantee that we get a closed formula for a_{ii′}(y′) in terms of all the entries b_{ii′}. To organize the computation it is convenient to use the complete order on the set of indices {1, …, N} given by

(⋆) i ≺ i′ if either m_{i} < m_{i′} or m_{i} = m_{i′} and the first nonzero term in the sequence (m_{i}^{j} − m_{i′}^{j}), j = 0, 1, …, is negative.

The fact that this is a complete order follows immediately from condition (b) of Lemma 3.1. We will write (ii′) > 0 if i ≺ i′ and (ii′) < 0 if i ≻ i′. We will also use the notation −(i, i′) = (i′, i).

Let us define a partially defined operation on pairs of distinct indices in ℤ/Nℤ by setting

τ(ii′) = (i − 1, i′ − 1) if (i − 1) ≺ (i′ − 1) and m_{i}^{j} = m_{i′}^{j} for 0 < j < n.

Note that τ is one-to-one. We denote by τ^{−1} the (partially defined) inverse and by τ^{k} the iterated maps. Condition (b) of Lemma 3.1 implies that for every pair of distinct indices (ii′) there exists k > 0 such that τ^{k} is not defined on (ii′). Case 1. Assume that i ≺ i′, i.e., (ii′) > 0. Then either m_{i} < m_{i′}, or there exists j > 0 such that m_{i}^{j′} = m_{i′}^{j′} for 0 ≤ j′ < j and m_{i}^{j} < m_{i′}^{j}. In the first case we can use formula (26). In the second case we have a_{ii′}^{j′} = const for 0 < j′ < j, j ≢ 0(n), while a_{ii′}^{j} = 0. Therefore, using (23) and (25) iteratively we get the following expression for a_{ii′}(∞):

−a_{ii′}(∞) = ∑_{k≥1} x^{−k}b_{τ^{k}(ii′)},  (27)

where the summation is only over a finite number of k’s for which τ^{k}(ii′) is defined. This gives the following formula

a_{ii′}(y′) = \frac{yb_{ii′}}{y′ − y} − ∑_{k≥1} x^{−k}b_{τ^{k}(ii′)} if (ii′) > 0,  (28)

that works also for the case m_{i} < m_{i′} (since in this case τ is not defined on (ii′)). Case 2. Assume that i ≻ i′, i.e., (ii′) < 0. Then either m_{i} > m_{i′}, or there exists j < 0 such that m_{i}^{j′} = m_{i′}^{j′} for j < j′ ≤ 0 and m_{i}^{j} < m_{i′}^{j}. Assume first that m_{i} > m_{i′}. Note that in this case there still exists j < 0 with the above property, and in addition there is k > 0 such that m_{i}^{j′} = m_{i′}^{j′} for 0 ≤ j′ < k and m_{i}^{k} < m_{i′}^{k} (by condition (b) of Lemma 3.1). Using equations (23) and (25) we derive that (27) still holds and also we have

a_{ii′}(0) = ∑_{k≥1} y^{ϵ(στ^{−k}σ(ii′))}x^{k}b_{στ^{−k}σ(ii′)},  (29)

where σ is the transposition: σ(i, i′) = (i′, i), the summation is only over those k for which τ^{−k}σ(ii′) is defined, ϵ(ii′) = 1 for (ii′) > 0 and ϵ(ii′) = 0 otherwise. This gives

a_{ii′}(y′) = \frac{y′b_{ii′}}{y′ − y} +∑_{k≥1} y^{ϵ(στ^{−k}σ(ii′))}x^{k}b_{στ^{−k}σ(ii′)}−y′ ∑_{k≥1} x^{−k}b_{τ^{k}(ii′)} if (ii′) < 0.

(30)

We observe that this formula still works in the case m_{i} = m_{i′} (the second summation becomes empty in this case). Case 3. Assume that i = i′. In this case we have relations

a_{ii}(0) = xa_{i+1,i+1}(0) + xb_{i+1,i+1}

for all i ∈ ℤ/Nℤ. Solving this linear system for a_{ii}(0) we get

a_{ii}(0) = (1 − x^{N})^{−1} ∑_{k=1}^{N} x^{k}b_{i+k,i+k}.

Finally, we derive

a_{ii}(y′) = \frac{y}{y′ − y} b_{ii} + (1 − x^{N})^{−1} ∑_{k=0}^{N−1} x^{k}b_{i+k,i+k}.  (31)

Formulas (28), (30) and (31) completely determine the map ev_{y′} ∘ Res_{y}^{−1}, so we can compute the associative r-matrix corresponding to the family of simple vector bundles V^{λ} on X: r(x; y, y′) = r_{const}(x, y/y′)+

∑_{α>0,k≥1}[−x^{−k}e_{−τ^{k}(α)} ⊗ e_{α} + y^{ϵ(−τ^{−k}(α))}x^{k}e_{τ^{−k}(α)} ⊗ e_{−α} − y′x^{−k}e_{−τ^{k}(−α)} ⊗ e_{−α}],

where

\begin{matrix}r_{const}(x, z) = \frac{z}{1−z} ∑_{α>0} e_{−α} ⊗ e_{α} + \frac{1}{1−z} ∑_{α>0} e_{α} ⊗ e_{−α}+ \\ \frac{z}{1−z} ∑_{i} e_{ii} ⊗ e_{ii} + (1 − x^{N})^{−1} ∑_{i} ∑_{k=0}^{N−1} x^{k}e_{i+k,i+k} ⊗ e_{ii}. \end{matrix}  (32)

In these formulas i is an element of ℤ/NZ, and α denotes a pair of distinct indices in ℤ/Nℤ. By a simple rearrangement of terms we can rewrite r(x; y, y′) in the following way:

\begin{matrix}r(x; y, y′) = r_{const}(x, y/y′)+ \\ ∑_{α>0,k≥1}[x^{k}e_{α} ⊗ e_{−τ^{k}(α)} − x^{−k}e_{−τ^{k}(α)} ⊗ e_{α} + yx^{k}e_{−α} ⊗ e_{−τ^{k}(−α)} − y′x^{−k}e_{−τ^{k}(−α)} ⊗ e_{−α}]. \end{matrix}

(33)

Recall that this is a solution of (21) with the unitarity condition (22).

Example 4.1. Assume that n > N and the only nonzero entries of (m_{i}^{j}) are m_{1}^{N} = m_{2}^{N−1} = … = m_{N−1}^{1} = 1. Then the domain of definition of τ is empty, so in this case we have r(x; y, y′) = r_{const}(x, y/y′). Hence, r_{const}(exp(u), exp(v)) is a solution of the AYBE.

Later we will show that r(exp(u); exp(v_{1}), exp(v_{2})) is equivalent to an rmatrix depending only on the difference v_{1} − v_{2} (see Lemma 6.1), so that it gives a solution of the AYBE.

5 Associative Belavin-Drinfeld triples associated with simple vector bundles

The right-hand side of (33) depends only on the parameters x, y, y′ and on a certain combinatorial structure on the set S = {1, …, N}. We are going to show that this structure consists of an associative BD-structure as defined in the introduction together with a compatible complete order (see below). Later we will show that one can get rid of the dependence on a complete order by passing to an equivalent r-matrix (see Lemma 6.1). However, for purposes of studying splitting types of simple vector bundles on cycles of projective lines the full combinatorial structure described below may be useful.

Definition 5.1. We say that a complete order on a set S is compatible with the cyclic order given by a cyclic permutation C_{0} (or simply compatible with C_{0}) if C_{0} takes every non-maximal element to the next element in this order. In other words, if we identify S with the segment of integers [1, N] preserving the complete order then C_{0}(i) = i + 1 (where the indices are identified with ℤ/Nℤ). In this case we set α_{0} = (s_{max}, s_{min}) ∈ Γ_{C_{0}}, where s_{min} (resp., s_{max}) is the minimal (resp., maximal) element of S. A choice of a complete order on S compatible with C_{0} is equivalent to a choice of an element α_{0} ∈ Γ_{C_{0}}. By an associative BD-structure on a completely ordered set S we mean an associative BD-structure (C_{0}, C, Γ_{1}, Γ_{2}) on S such that the complete order is compatible with C_{0}.

Note that a choice of an associative BD-structure on the completely ordered set [1, N] such that α_{0} ∉ Γ_{1} and α_{0} ∉ Γ_{2}, is equivalent to a choice of a Belavin-Drinfeld triple in A_{N−1} equipped with an associative structure as defined in [8].

We will need the following characterization of associative BD-structures on completely ordered sets such that α_{0} ∉ Γ_{2}.

Lemma 5.2. Let (S, <) be a completely ordered finite set equipped with a transitive cyclic permutation C : S → S. Then to give an associative BDstructure on S with α_{0} ∉ Γ_{2} is equivalent to giving a pair of subsets P_{1} and P_{2} in the set of pairs of distinct elements of S, such that (C × C)(P_{1}) = P_{2} and the following properties are satisfied:

(a) For every (s, s′) ∈ P_{2} one has s < s′.

(b) Assume that s < s′ < s′′. If (s, s′′) ∈ P_{1} then (s, s′), (s′, s′′) ∈ P_{1}. The same property holds for P_{2}. Also, if (s′, s) ∈ P_{1} then (s′, s′′), (s′′, s) ∈ P_{1} (resp., if (s′′, s′) ∈ P_{1} then (s′′, s), (s, s′) ∈ P_{1}).

The proof is left for the reader. Let us observe only that property (b) assures that P_{ι} is determined by Γ_{ι} = P_{ι} ∩ Γ_{C_{0}}, where ι = 1, 2.

Now let us check that in the setting of section 4 we do get a completely ordered set with an associative BD-structure.

Lemma 5.3. Let (m_{i}^{j}) be a N ×n-matrix satisfying conditions of Lemma 3.1. Equip the set S = {1, …, N} with the complete order ≺ given by (⋆) and the cyclic permutation C(i) = i − 1. Also, let

P_{1} = {(ii′) | m_{i}^{j} = m_{i′}^{j} for 0 < j < n and C(i) ≺ C(i′)}.

Then these data define an associative BD-structure with α_{0} ∉ Γ_{2}.

Proof. We use Lemma 5.2. The only question is why property (b) holds. Let i ≺ i′ ≺ i′′.

Assume first that (i, i′′) ∈ P_{2}. Then m_{i+1}^{j} = m_{i′′+1}^{j} for j ∈ [1, n − 1]. Suppose there exists j ∈ [1, n − 1] such that m_{i+1}^{j} ≠ m_{i′+1}^{j}. Consider the maximal such j. We have either m_{i+1}^{j} < m_{i′+1}^{j} or m_{i′+1}^{j} < m_{i′′+1}^{j}. By condition (b) of Lemma 3.1, the former assumption contradicts to i ≺ i′, while the latter contradicts to i′ ≺ i′′. Hence, m_{i+1}^{j} = m_{i′+1}^{j} = m_{i′′+1}^{j} for all j ∈ [1, n − 1], so that (i, i′), (i′, i′′) ∈ P_{2}.

Assume that (i, i′′) ∈ P_{1}. Then m_{i}^{j} = m_{i′′}^{j} for j ∈ [1, n − 1]. Furthermore, since i ≺ i′′ and i − 1 ≺ i′′ − 1 we should have m_{i}^{0} = m_{i′′}^{0} (by condition (b) of Lemma 3.1). Suppose there exists j ∈ [0, n − 1] such that m_{i}^{j} ≠ m_{i′}^{j}. Consider the minimal such j. We have either m_{i}^{j} > m_{i′}^{j} or m_{i′}^{j} > m_{i′′}^{j}. But the former contradicts to i ≺ i′, and the latter contradicts to i′ ≺ i′′. Therefore, m_{i}^{j} = m_{i′}^{j} = m_{i′′}^{j} for all j ∈ [0, n − 1], so that (i, i′), (i′, i′′) ∈ P_{1}.

Finally, assume that (i′, i) ∈ P_{1} (resp., (i′′, i′) ∈ P_{1}). Then m_{i}^{j} = m_{i′}^{j} (resp., m_{i′}^{j} = m_{i′′}^{j}) for j ∈ [1, n − 1]. Also, since i′ ≻ i and i′ − 1 ≺ i − 1 (resp., i′′ ≻ i′ and i′′ − 1 ≺ i′ − 1), we necessarily have m_{i}^{0} < m_{i′}^{0} (resp., m_{i′}^{0} < m_{i′′}^{0}). Hence, m_{i′}^{0} = m_{i′′}^{0} (resp., m_{i}^{0} = m_{i′}^{0}). Suppose there exists j ∈ [1, n − 1] such that m_{i′}^{j} ≠ m_{i′′}^{j} (resp., m_{i}^{j} ≠ m_{i′}^{j}). Consider the minimal such j. Since i′ ≺ i′′ (resp., i ≺ i′), we have m_{i}^{j} = m_{i′}^{j} < m_{i′′}^{j} (resp., m_{i}^{j} < m_{i′}^{j} = m_{i′′}^{j}). But this contradicts to condition (b) of Lemma 3.1 (applied to i and i′′). Therefore, m_{i}^{j} = m_{i′}^{j} = m_{i′′}^{j} for all j ∈ [1, n − 1]. Since m_{i′}^{0} = m_{i′′}^{0} (resp., m_{i}^{0} = m_{i′}^{0}), we have i′ − 1 ≺ i′′ − 1 (resp., i − 1 ≺ i′ − 1), and hence (i′, i′′) ∈ P_{1} (resp., (i, i′) ∈ P_{1}). Also, i′′ − 1 ≺ i − 1 (by condition (b) of Lemma 3.1), so that (i′′, i) ∈ P_{1}. ⊓ ⊔

We will need below the following two operations on associative BDstructures.

Definition 5.4. For an associative BD-structure (C_{0}, C, Γ_{1}, Γ_{2}) on a finite set S we define

(i) the opposite associative BD-structure to be (C_{0}^{−1}, C, σ(Γ_{1}), σ(Γ_{2})), where σ is the permutation of factors in S × S (note that σ(Γ_{C_{0}}) = Γ_{C_{0}^{−1}}); (ii) the inverse associative BD-structure to be (C_{0}, C^{−1}, Γ_{2}, Γ_{1}).

Note that under passing to the opposite associative BD-structure each set P_{ι}, ι = 1, 2, gets replaced with σ(P_{ι}).

Theorem 5.5. An associative BD-structure on a completely ordered finite set S is obtained by the construction of Lemma 5.3 from some matrix (m_{i}^{j}) (satisfying conditions of Lemma 3.1) iff α_{0} ∉ Γ_{2} and C = C_{0}^{k} for some k ∈ ℤ (relatively prime to N = |S|).

Proof. “Only if”. Let us denote by t_{i} = ∑_{j=0}^{n−1} m_{i}^{j}, i = 1, …, N, the sums of entries in the rows of the matrix (m_{i}^{j}). Then we claim that for i ≺ i′ one has

t_{i} − t_{i′} = \begin{cases}−1, & if i − 1 ≻ i′ − 1, \\ 0, otherwise. \end{cases}  (34)

Indeed, assume first that m_{i}^{j} = m_{i′}^{j} for all j ∈ [0, n−1]. Then i−1 ≺ i′ −1 and t_{i} = t_{i′}, so the above equation holds. Next, assume that m_{i}^{j} ≠ m_{i′}^{j} for some j ∈ [0, n − 1]. Then the first nonzero term in the sequence (m_{i}^{j} − m_{i′}^{j})_{j∈[0,n−1]} is −1. Since −1’s and 1’s in this sequence alternate, we have t_{i} −t_{i′} = 0 (resp., t_{i} − t_{i′} = −1) iff the last nonzero term in the sequence (m_{i}^{j} − m_{i′}^{j})_{j∈[0,n−1]} is 1 (resp., −1). But this happens precisely when the first nonzero term in (m_{i−1}^{j} − m_{i′−1}^{j})_{j≥0} is −1 (resp., 1), so (34) follows.

Now assume that i ≺ i′ ≺ i′′. Then it follows from (34) that either C(i) ≺ C(i′) ≺ C(i′′) or C(i′) ≺ C(i′′) ≺ C(i) or C(i′′) ≺ C(i) ≺ C(i′). Since this holds for every triple (i, i′, i′′), it is easy to deduce that C = C_{0}^{k} for some k ∈ ℤ. “If”. First, note that the construction of the associative BD-structure on a completely ordered set S given in Lemma 5.3 can be rewritten as follows. Assume we are given a transitive cyclic permutation C of S and a matrix (m_{s}^{j}), where j ∈ [0, n], s ∈ S. Then we can extend the range of the index j to ℤ using the rule m_{s}^{j+n} = m_{C(s)}^{j}. Assuming that condition (b) of Lemma 3.1 holds for this extended matrix we can proceed to define the complete order by (⋆) and the set P_{1} as in Lemma 5.3. Of course, we can always identify S with {1, …, N} in such a way that C(i) = i − 1, so that we get to the setup of Lemma 5.3. The advantage of the new point of view is that we can also consider the set S = {1, …, N} with the cyclic permutation C(i) = i − k, where k ∈ ℤ/Nℤ is relatively prime to N. Then as was noted above we have to modify the definition of the extended matrix by using the rule m_{i}^{j+n} = m_{i−k}^{j}.

Note that changing (m_{i}^{j}) to (−m_{i}^{j}) changes the associative BD-structure on S to the opposite BD-structure, and the complete order on S gets reversed. Let us denote by w_{0} : S → S the permutation that reverses the order. Assume that we have C = C_{0}^{k}. Then conjugating by w_{0} the BD-structure associated with (−m_{i}^{j}) we get a BD-structure that is obtained from the original one by leaving the complete order the same, changing C = C_{0}^{k} to C_{0}^{−k}, and replacing P_{1} with (w_{0} × w_{0})σ(P_{1}). Therefore, it is enough to show that Lemma 5.3 produces all associative BD-structures with C = C_{0}^{−k}, where N/2 ≤ k < N.

Next, we describe a construction of a class of matrices (m_{i}^{j}) satisfying conditions of Lemma 3.1. Fix k, relatively prime to N, such that N/2 ≤ k < N. Start with a sequence (a_{1}, …, a_{N}) such that a_{1} = 1, a_{N} = n − 1 (where n > 1), and for every i ∈ [1, N − 1] one has either a_{i+1} = a_{i} or a_{i+1} = a_{i} + 1. Then set m_{i}^{0} = 1 for i ∈ [k + 1, N], m_{k+1−i}^{a_{i}} = 1 for i = 1, …, N, and let the remaining entries to be zero. We are going to check that this matrix satisfies conditions of Lemma 3.1 (with the modified definition of the extended matrix).

It is convenient to extend the range of the index i to ℤ by the rule m_{i}^{j} = m_{i+N}^{j}, so that we get a matrix (m_{i}^{j}) with rows and columns numbered by ℤ. Let us consider the subset Λ ⊂ [k + 1 − N, N] × [0, n − 1] defined by

Λ = ([k + 1, N] × {0}) ∪ {(k + 1 − i, a_{i}) | i = 1, …, N}.

Then we have

{(i, j) ∈ ℤ × ℤ | m_{i}^{j} ≠ 0} = ∪_{a∈ℤ}Λ_{a}, where
Λ_{0} = ∪_{b∈ℤ}(Λ + b(2N − k, n)), Λ_{a} = Λ_{0} + a(N, 0).

Note that each Λ_{a} intersects each row once, and if we denote by (i, j_{a}(i))) the intersection point of Λ_{a} with the ith row then either j_{a}(i − 1) = j_{a}(i) or j_{a}(i − 1) = j_{a}(i) + 1. In other words, as we go down one row the point of intersection either stays in the same column or moves one step to the right. It follows that the intersection of Λ_{a} with each column is a line segment. Moreover, it is easy to see that the number of elements in this intersection is at most N. Indeed, for columns corresponding to j ≡ 0(n) the intersection segment has N − k elements. On the other hand, for j ≢ 0(n) this number is equal to the number of i ∈ [1, N] such that j ≡ a_{i}(n), so it is at most N. This implies that Λ_{a} and Λ_{a′} are disjoint for a ≠ a′. Hence, j_{a}(i) < j_{a+1}(i) for all a ∈ ℤ and i ∈ ℤ.

Let us set E_{i} = {j_{a}(i) | a ∈ ℤ} for every i ∈ ℤ. We have to check that for every pair of rows, the i-th and the i′-th, where i < i′ < i + N, one has E_{i} ≠ E_{i′}, and the subsets E_{i} \ E_{i′} and E_{i′} \ E_{i} in ℤ alternate.

To prove that E_{i} ≠ E_{i′} we recall that by the construction, for every b ∈ ℤ the intersection of Λ_{0} with the bn-th column is the segment [k + 1 + b(2N − k), N +b(2N −k)]. The intersections of other sets Λ_{a} with the same column are obtained from the above segment by shifts in Nℤ. Since 2N − k is relatively prime to N, it follows that for appropriate b ∈ ℤ the intersection of ∪_{a}Λ_{a} with the bn-th column contains exactly one of the numbers i and i′. Hence, bn belongs to exactly one of the sets E_{i} and E_{i′}.

Finally, we have to prove that subsets E_{i} \ E_{i′} and E_{i′} \ E_{i} alternate. Note that for all a we have j_{a}(i′) ≤ j_{a}(i). Hence, our assertion would follow once we check that for every a ∈ ℤ one has j_{a}(i) ≤ j_{a+1}(i′). Suppose we have j_{a+1}(i′) < j_{a}(i). Then the intersection of Λ_{a+1} with the j_{a}(i)-th column is a segment [i_{1}, i_{2}], where i < i_{1} ≤ i_{2} < i′. Since Λ_{a+1} = Λ_{a} + (N, 0), the intersection of Λ_{a} with the j_{a}(i)-th column is [i_{1} − N, i_{2} − N]. Hence, i ≤ i_{2} − N < i′ − N, which contradicts our assumptions on i and i′.

Now given a BD-structure on a set S = {1, …, N} with the complete order 1 < 2 < … < N and the cyclic permutation C = C_{0}^{−k} (where N/2 ≤ k < N) we define the sequence (a_{1}, …, a_{N}) as follows. Set a_{1} = 1, and for i = 1, …, N − 1 set

a_{i+1} = \begin{cases}a_{i} & if α_{k−i} ∈ Γ_{1}, \\ a_{i} + 1 & otherwise \end{cases}

(this uniquely defines n). It is easy to check that the corresponding matrix (m_{i}^{j}) realizes our BD-structure. ⊓ ⊔

6 Solutions of the AYBE and associative BD-structures

Let (S, <, C, Γ_{1}, Γ_{2}) be a completely ordered finite set with an associative BD-structure such that α_{0} ∉ Γ_{2}. As in the introduction, for an element α = (i, j) ∈ S × S we set e_{α} = e_{ij} ∈ A_{S} ≃ Mat_{N}(ℂ) (where N = |S|, the rows and columns are numbered by S). We write (i, j) > 0 (resp., (i, j) < 0) if i < j (resp., i > j). Also, for α = (i, j) we set −α = (j, i). Mimicking formulas (32) and (33) we define

\begin{matrix}r_{const}(x, z) = \frac{z}{1−z} ∑_{α>0} e_{−α} ⊗ e_{α} + \frac{1}{1−z} ∑_{α>0} e_{α} ⊗ e_{−α}+ \\ \frac{z}{1−z} ∑_{i∈S} e_{ii} ⊗ e_{ii} + (1 − x^{N})^{−1} ∑_{i∈S} ∑_{k=0}^{N−1} x^{k}e_{i,i} ⊗ e_{C^{k}(i),C^{k}(i)}, \end{matrix} (35)

\begin{matrix}r(x; y, y′) = r_{const}(x, y/y′)+ \\ ∑_{α>0,k≥1}[x^{k}e_{α} ⊗ e_{−τ^{k}(α)} − x^{−k}e_{−τ^{k}(α)} ⊗ e_{α}]+ \\ ∑_{α<0,k≥1}[yx^{k}e_{α} ⊗ e_{−τ^{k}(α)} − y′x^{−k}e_{−τ^{k}(α)} ⊗ e_{α}]. \end{matrix}  (36)

In the last formula we use the operation τ defined on P_{1} ⊂ S ×S; the summation is extended only over those (k, α) for which τ^{k}(α) is defined. Below we will show that r(x; y, y′) is a solution of (21) (see Theorem 6.2). To deduce from this Theorem 0.2(i) we will use the following simple observation.

Lemma 6.1. In the above situation the A_{S} ⊗ A_{S}-valued function

−r(exp(\frac{u_{1} − u_{2}}{N}); exp(v_{1}), exp(v_{2}))

is equivalent to the one given in Theorem 0.2(i) for the inverse associative BD-structure (C_{0}, C^{−1}, Γ_{2}, Γ_{1}), where u = u_{1} − u_{2} and v = v_{1} − v_{2}.

Proof. We can assume that S = [1, N] (the segment of natural numbers) with the standard order. Let us set

φ(v)e_{j} = exp(−\frac{jv}{N})e_{j}.

Then the corresponding equivalent matrix r̃(u_{1}, u_{2}; v_{1}, v_{2}) is obtained from r(exp(\frac{u_{1}−u_{2}}{N}); exp(v_{1}), exp(v_{2})) by multiplying each term e_{ij} ⊗ e_{j′i′} with exp(\frac{(j−i)v_{1}−(j′−i′)v_{2}}{N}). Now we observe that r_{const}(x, y/y′) is a linear combination of e_{ij} ⊗ e_{j′i′}, where j − i = j′ − i′. Such a term gets multiplied by exp(\frac{(j−i)(v_{1}−v_{2})}{N}). The same is true about the terms in r(x; y, y′) not containing y or y′. Indeed, if i < j and τ^{k} is defined on (i, j) then C^{k}(j)−C^{k}(i) = j−i. On the other hand, the terms involving y = exp(v_{1}) and y′ = exp(v_{2}) are linear combinations of e_{i,j} ⊗e_{j′,i′}, where j′ −i′ = j −i+N. Indeed, this follows from the fact that if i > j and τ^{k} is defined on (i, j) then C^{k}(j)−C^{k}(i) = j −i+N (the proof reduces to the case (i, j) = (N, 1)). The only other observation we use to rewrite −r̃ in the form given in Theorem 0.2(i) (with C replaced by C^{−1} and Γ_{1} and Γ_{2} exchanged) is that for 0 < m < N and for i, j ∈ [1, N] we have j − i ≡ m(N) iff either i < j and j = i + m, or i > j and j = i + m − N. ⊓ ⊔

Since for every associative BD-structure on a finite set S we can choose a compatible complete order in such a way that α_{0} ∉ Γ_{2}, Theorem 0.2(i) will follow easily from the above lemma and the next result.

Theorem 6.2. Let (S, <, C, Γ_{1}, Γ_{2}) be a completely ordered finite set with an associative BD-structure such that α_{0} ∉ Γ_{2}. Then the function r(x; y, y′) given by (36) is a solution of (21) satisfying the unitarity condition (22).

Remark 6.3. By Theorem 5.5 we already know the statement to be true if C = C_{0}^{k}. Also, the work [8] deals with the case when in addition α_{0} ∉ Γ_{1} (this fact will be used below).

The rest of this section will be occupied with the proof of Theorem 6.2 (in the end we will also explain how to deduce Theorem 0.2(i)).

Let us denote by P = ∑_{i,j} e_{ij} ⊗ e_{ji} the permutation tensor. Then we can rewrite our r-matrix in the form

r(x; y, y′) = a(x) + yb(x) − y′c(x) + \frac{y}{y′ − y} P,

where

a(x) = (1 − x^{N})^{−1} ∑_{i∈S} ∑_{k=0}^{N−1} x^{k}e_{i,i} ⊗ e_{C^{k}(i),C^{k}(i)} + ∑_{α>0} e_{α} ⊗ e_{−α}+

∑_{α>0,k≥1}[x^{k}e_{α} ⊗ e_{−τ^{k}(α)} − x^{−k}e_{−τ^{k}(α)} ⊗ e_{α}],

b(x) = ∑_{α<0,k≥1} x^{k}e_{α} ⊗ e_{−τ^{k}(α)},

c(x) = b^{21}(x^{−1}) = ∑_{α<0,k≥1} x^{−k}e_{−τ^{k}(α)} ⊗ e_{α}.

Lemma 6.4. Assume that α_{0} ∉ Γ_{2}. Let us set Γ′_{1} = Γ_{1} \ {α_{0}}, Γ′_{2} = τ(Γ′_{1}). Then

a(x) + \frac{y}{y′ − y} P

is exactly the r-matrix corresponding to the associative BD-structure (S , < , C , Γ′_{1}).

Proof. It is easy to see that P′_{1} = {α ∈ P_{1} | α > 0}. Thus, the terms b(x) and c(x) in the r-matrix associated with the new associative BD-structure vanish. We claim that the term a(x) for the new associative BD-structure is the same as for the old one. Indeed, it is enough to check that α ∈ P′_{1} is in the domain of definition of τ^{k} iff it is in the domain of (τ′)^{k}, where τ′ : P′_{1} → P′_{2} is the bijection induced by τ. But this follows immediately from the fact that P_{2} consists only of α > 0 (due to the assumption that α_{0} ∉ Γ_{2}). ⊓ ⊔

Let us denote by AY BE[r](x, x′; y_{1}, y_{2}, y_{3}) the left-hand side of (21).

Lemma 6.5. Consider the r-matrix of the form

r(x; y_{1}, y_{2}) = a(x) + y_{1}b(x) − y_{2}c(x) + \frac{y_{1}}{y_{2} − y_{1}} P,  (37)

where a^{21}(x^{−1})+a(x) = P and b^{21}(x^{−1}) = c(x). Then r satisfies the unitarity condition (22). Also, AY BE[r] = 0 iff the following equations are satisfied:

(i) AY BE[a] = 0; (ii) b^{12}(x)b^{13}(x′) = 0; (iii) b^{13}(x)b^{23}(x′) = b^{21}(x′)b^{13}(xx′) + b^{23}(xx′)b^{12}(x); (iv) c^{13}(x)a^{23}(x′) + a^{12}((x′)^{−1})c^{13}(xx′) = c^{23}(xx′)a^{12}(x) − a^{13}(x)c^{23}(x′).

Proof. The unitarity condition follows immediately from our assumptions on a(x), b(x) and c(x). It is easy to check that

AY BE[a(x) + y_{1}b(x) − y_{2}c(x) + \frac{y_{1}}{y_{2} − y_{1}} P](x, x′; y_{1}, y_{2}, y_{3}) =

AY BE[a(x) + y_{1}b(x) − y_{2}c(x)](x, x′; y_{1}, y_{2}, y_{3})

− y_{1}c^{21}(x′)P^{13} − y_{2}c^{13}(x)P^{23} − y_{1}b^{23}(xx′)P^{12}.

Now the conditions (i)-(iv) are obtained by equating to zero coefficients with various monomials in y_{1}, y_{2} and y_{3} (of degree ≤ 2). Namely, (i) is obtained by looking at the constant term (i.e., by substituting y_{i} = 0). Conditions (ii), (iii) and (iv) are obtained by looking at the coefficients with y_{1}^{2}, y_{1}y_{2} and y_{3}, respectively. To see that these conditions imply AY BE[r] = 0 we can use the identity

AY BE[r](x, x′; y_{2}, y_{3}, y_{1})^{231} = AY BE[r]((xx′)^{−1}, x; y_{1}, y_{2}, y_{3})

that holds for any r satisfying the unitarity condition (22). ⊓ ⊔

Let us introduce the following notation. For every k ≥ 1 we denote by P (k) ⊂ P_{1} the domain of definition of τ^{k} and by P (k)^{+} ⊂ P (k) (resp., P (k)^{−} ⊂ P (k)) the set of all α > 0 (resp., α < 0) contained in P (k). Note that P (1) = P_{1}. The assumption α_{0} ∉ Γ_{2} implies that τ(P (k)) ⊂ P (k − 1)^{+}. Using this notation we can rewrite our formulas for a(x), b(x) and c(x) as follows:

a(x) = (1 − x^{r})^{−1} ∑_{0≤k<r,i} x^{k}e_{i,i} ⊗ e_{C^{k}(i),C^{k}(i)} + ∑_{i<j} e_{i,j} ⊗ e_{j,i}+

∑_{(i,j)∈P (k)^{+}}[x^{k}e_{i,j} ⊗ e_{C^{k}(j),C^{k}(i)} − x^{−k}e_{C^{k}(j),C^{k}(i)} ⊗ e_{i,j}],

b(x) = ∑_{k≥1,(i,j)∈P (k)^{−}} x^{k}e_{i,j} ⊗ e_{C^{k}(j),C^{k}(i)}.

c(x) = ∑_{k≥1,(i,j)∈P (k)^{−}} x^{−k}e_{C^{k}(j),C^{k}(i)} ⊗ e_{i,j}.

The following two combinatorial observations are also going to be useful in the proof.

Lemma 6.6. Let (i_{1}, i_{2}, i_{3}) be a triple of elements of S and let k ≥ 1. Then the following two conditions are equivalent:

(a) (i_{1}, i_{3}) ∈ P (k)^{−} and i_{1} < i_{2} (resp., i_{2} < i_{3});

(b) (i_{1}, i_{2}) ∈ P (k)^{+} and (i_{2}, i_{3}) ∈ P (k)^{−} (resp., (i_{1}, i_{2}) ∈ P (k)^{−} and (i_{2}, i_{3}) ∈ P (k)^{+}).

The proof is straightforward and is left to the reader.

Lemma 6.7. Let k ≥ 1. Then for every (i_{1}, i_{2}) ∈ P (k)^{−} one has a decomposition S = S_{1} ⊔ S_{2}, where

S_{1} = {i | i < i_{1}, C^{k}(i) > C^{k}(i_{1})}, S_{2} = {i | i > i_{2}, C^{k}(i) < C^{k}(i_{2})}.

Proof. We can assume that S = [1, N] with the standard order. Note that the map C^{k} restricts to a bijection

[i_{1}, N] ⊔ [1, i_{2}]˜→[C^{k}(i_{1}), C^{k}(i_{2})].

Passing to the complements we derive that the open segment (i_{2}, i_{1}) is the disjoint union of its intersections with S_{1} and S_{2}. Next, if i ≤ i_{2} then (i_{1}, i) ∈ P (k)^{−} (by Lemma 6.6), so that C^{k}(i_{1}) < C^{k}(i). Hence, [1, i_{2}] ⊂ S_{1} \ S_{2}. Similarly, [i_{1}, N] ⊂ S_{2} \ S_{1}. ⊓ ⊔

Proof of Theorem 6.2. Let us check that equations (i)-(iv) of Lemma 6.5 hold in our case. Equation (i) follows from Lemma 6.4 and Theorem 3.4 of [8]. More precisely, one can easily check that in the case when α_{0} ∉ Γ_{1} and α_{0} ∉ Γ_{2} our r-matrix coincides with the associative r-matrix constructed in [8] for the opposite associative BD-structure on S. Equation (ii) follows from the fact that for any (i, j), (i′, j′) ∈ P (1)^{−} one has i′ > j and i > j′ (otherwise we would have Γ_{1} = Γ_{S}). To check equation (iii) we write

b^{13}(x)b^{23}(x′) =

∑_{k≥1,m≥1;(i,j)∈P (k)^{−},(i′,j′)∈P (m)^{−};C^{k}(i)=C^{m}(j′)} x^{k}(x′)^{m}e_{i,j} ⊗ e_{i′,j′} ⊗ e_{C^{k}(j),C^{m}(i′)}.

Note that we cannot have k = m since this would imply that i = j′ contradicting the assumption that (i, j) ∈ P (k)^{−} ⊂ P (1)^{−} and (i′, j′) ∈ P (m)^{−} ⊂ P (1)^{−}. Hence, we can split the summation into two parts: one with k > m and one with k < m. Denoting k − m (resp., m − k) by l in the first (resp., second) case, we can rewrite these sums as

Σ_{1} = ∑_{l≥1,m≥1;(i,j)∈P (m+l)^{−},(i′,C^{l}(i))∈P (m)^{−}} x^{l}(xx′)^{m}e_{i,j}⊗e_{i′,C^{l}(i)}⊗e_{C^{m+l}(j),C^{m}(i′)},

Σ_{2} = ∑_{l≥1,m≥1;(i′,j′)∈P (m+l)^{−},(C^{l}(j′),j)∈P (m)^{−}}(xx′)^{m}(x′)^{l}e_{C^{l}(j′),j}⊗e_{i′,j′}⊗e_{C^{m}(j),C^{m+l}(i′)}.

On the other hand, we have b^{23}(xx′)b^{12}(x) =

∑_{l≥1,m≥1;(i,j)∈P (l)^{−},(i′,C^{l}(j))∈P (m)^{−}} x^{l}(xx′)^{m}e_{i,j} ⊗ e_{i′,C^{l}(i)} ⊗ e_{C^{m+l}(j),C^{m}(i′)}.

We claim that this is equal to Σ_{1}. Indeed, the condition (i, j) ∈ P (m + l)^{−} is equivalent to the conjuction of (i, j) ∈ P (l)^{−} and (C^{l}(i), C^{l}(j)) ∈ P (m)^{+}. Now our claim follows from Lemma 6.6 applied to the triple (i′, C^{l}(i), C^{l}(j)) (recall that C^{l}(i) < C^{l}(j) since (i, j) ∈ P (l)). Similarly, we check that b^{21}(x′)b^{13}(xx′) = Σ_{2}, which finishes the proof of equation (iii).

Finally, let us verify equation (iv). We can split both terms in the left-hand side of this equation into four sums according to the four pieces comprising a(x):

c^{13}(x)a^{23}(x′) = L_{1} +L_{2} +L_{3} −L_{4}, a^{12}((x′)^{−1})c^{13}(xx′) = −L_{5} +L_{6} +L_{7} −L_{8},

where

L_{1} = (1 − (x′)^{N})^{−1}×

∑_{0≤m<N,k≥1;(i,j)∈P (k)^{−}} x^{−k}(x′)^{m}e_{C^{k}(j),C^{k}(i)} ⊗ e_{C^{N−m}(j),C^{N−m}(j)} ⊗ e_{i,j},

L_{2} = ∑_{m≥1;i<j,(i′,j)∈P (m)^{−}} x^{−m}e_{C^{m}(j),C^{m}(i′)} ⊗ e_{i,j} ⊗ e_{i′,i},

L_{3} = ∑_{k≥1,m≥1;(i,j)∈P (k)^{+},(i′,C^{k}(j))∈P (m)^{−}} x^{−m}(x′)^{k}e_{C^{k+m}(j),C^{m}(i′)} ⊗ e_{i,j} ⊗ e_{i′,C^{k}(i)},

L_{4} = ∑_{k≥1,m≥1;(i,i′)∈P (m)^{−},(i′,j)∈P (k)^{+}} x^{−m}(x′)^{−k}e_{C^{m}(i′),C^{m}(i)} ⊗ e_{C^{k}(j),C^{k}(i′)} ⊗ e_{i,j},

L_{5} = (1 − (x′)^{N})^{−1}×

∑_{0≤m<N,k≥1;(i,j)∈P (k)^{−}}(x′)^{N−m}(xx′)^{−k}e_{C^{k}(j),C^{k}(i)} ⊗ e_{C^{k+m}(j),C^{k+m}(j)} ⊗ e_{i,j},

L_{6} = ∑_{k≥1,(i,j)∈P (k)^{−},i′<C^{k}(j)}(xx′)^{−k}e_{i′,C^{k}(i)} ⊗ e_{C^{k}(j),i′} ⊗ e_{i,j},

L_{7} =

∑_{k≥1,m≥1;(i,j)∈P (k)^{−},(i′,C^{k}(j))∈P (m)^{+}}(x′)^{−m}(xx′)^{−k}e_{i′,C^{k}(i)} ⊗ e_{C^{k+m}(j),C^{m}(i′)} ⊗ e_{i,j},

L_{8} =

∑_{k≥1,m≥1;(i,j)∈P (k)^{−},(i′,j′)∈P (m)^{+},C^{m}(i′)=C^{k}(j)}(x′)^{m}(xx′)^{−k}e_{C^{m}(j′),C^{k}(i)} ⊗ e_{i′,j′} ⊗ e_{i,j}.

We split each of the sums L_{4} and L_{8} into 3 parts according to the ranges of summation k = m, k > m, and k < m (in the last two cases we make substitutions k ↦ k + m and m ↦ k + m, respectively):

596  A. Polishchuk
L_{4} = L_{4,1} + L_{4,2} + L_{4,3}, L_{8} = L_{8,1} + L_{8,2} + L_{8,3},

where

L_{4,1} = ∑_{k≥1,(i,j)∈P (k)^{−},i′<j}(xx′)^{−k}e_{C^{k}(i′),C^{k}(i)} ⊗ e_{C^{k}(j),C^{k}(i′)} ⊗ e_{i,j},

L_{4,2} = ∑_{k≥1,m≥1;(i,i′)∈P (m)^{−},(i′,j)∈P (k+m)^{+}} x^{−m}(x′)^{−k−m}e_{C^{m}(i′),C^{m}(i)} ⊗ e_{C^{k+m}(j),C^{k+m}(i′)} ⊗ e_{i,j},

L_{4,3} = ∑_{k≥1,m≥1;(i,i′)∈P (k+m)^{−},(i′,j)∈P (k)^{+}} x^{−k−m}(x′)^{−k}e_{C^{k+m}(i′),C^{k+m}(i)} ⊗ e_{C^{k}(j),C^{k}(i′)} ⊗ e_{i,j},

L_{8,1} = ∑_{k≥1,(i,j′)∈P (k)^{−},j<j′} x^{−k}e_{C^{k}(j′),C^{k}(i)} ⊗ e_{j,j′} ⊗ e_{i,j},

L_{8,2} = ∑_{k≥1,m≥1;(i,j)∈P (k+m)^{−},(C^{k}(j),j′)∈P (m)^{+}} x^{−k−m}(x′)^{−k}e_{C^{m}(j′),C^{k+m}(i)} ⊗ e_{C^{k}(j),j′} ⊗ e_{i,j},

L_{8,3} = ∑_{k≥1,m≥1;(i,C^{m}(i′))∈P (k)^{−},(i′,j′)∈P (k+m)^{+}} x^{−k}(x′)^{m}e_{C^{k+m}(j′),C^{k}(i)} ⊗ e_{i′,j′} ⊗ e_{i,C^{m}(i′)}.

Making appropriate substitutions of the summation variables and using Lemma 6.6 one can easily check that

L_{2} = L_{8,1}, L_{3} = L_{8,3}.

It follows that the left-hand side of (iv) is equal to

(L_{1} − L_{5}) + (L_{6} − L_{4,1}) + (L_{7} − L_{4,2}) − (L_{4,3} + L_{8,2}).

Next, making the substitution m ↦ N − k − m in the sum L_{5} we find

L_{1} − L_{5} = − ∑_{0<m<k,(i,j)∈P (k)^{−}} x^{−k}(x′)^{−m}e_{C^{k}(j),C^{k}(i)} ⊗ e_{C^{m}(j),C^{m}(j)} ⊗ e_{i,j}.

Also, substituting i′ by C^{k}(i′) in L_{6}, switching k and m in L_{4,2}, and using Lemma 6.6 we find that

L_{6}−L_{4,1} = ∑_{k≥1,(i,j)∈P (k)^{−},i′>j,C^{k}(i′)<C^{k}(j)}(xx′)^{−k}e_{C^{k}(i′),C^{k}(i)}⊗e_{C^{k}(j),C^{k}(i′)}⊗e_{i,j},

L_{7} − L_{4,2} =

∑_{k≥1,m≥1,(i,j)∈P (k)^{−},(C^{k}(i′),C^{k}(j))∈P (m)^{+},i′>j} x^{−k}(x′)^{−k−m}e_{C^{k}(i′),C^{k}(i)} ⊗ e_{C^{k+m}(j),C^{k+m}(i′)} ⊗ e_{i,j}.

Finally, we can rewrite the sum of the other remaining terms as follows: L_{1} − L_{5} − L_{4,3} − L_{8,2} =

− ∑_{k≥1,m≥1,(i,i′,j)∈Π(k,m)} x^{−k−m}(x′)^{−k}e_{C^{k+m}(i′),C^{k+m}(i)} ⊗ e_{C^{k}(j),C^{k}(i′)} ⊗ e_{i,j},

where Π(k, m) is the subset of {(i, i′, j) | (i, j) ∈ P (k)^{−}, (C^{k}(i), C^{k}(i′)) ∈ P (m)^{+}} consisting of (i, i′, j) such that either i′ ≤ j or C^{k}(j) < C^{k}(i′). It follows from Lemma 6.7 that

Π(k, m) = {(i, i′, j) | (i, j) ∈ P (k)^{−}, (C^{k}(i), C^{k}(i′)) ∈ P (m)^{+}, i′ < i}.

We deal similarly with the right-hand side of equation (iv). Namely, we write

c^{23}(xx′)a^{12}(x) = R_{1} + R_{2} + R_{3} − R_{4}, a^{13}(x)c^{23}(x′) = R_{5} + R_{6} + R_{7} − R_{8},

where the parts correspond to the summands in a(x). We also have a decomposition R_{3} = R_{3,1} + R_{3,2} + R_{3,3} (resp., R_{8} = R_{8,1} + R_{8,2} + R_{8,3}) obtained by collecting terms with x^{k}(xx′)^{−m} (resp., x^{−k}(x′)^{−m}) with k = m, k > m and k < m. Now one can easily check that

R_{6} = R_{3,1}, R_{7} = R_{3,2}.

Also, we have

R_{1} − R_{5} = ∑_{m≥1,0<k≤m;(i,j)∈P (m)^{−}} x^{−k}(x′)^{−m}e_{C^{k}(i),C^{k}(i)} ⊗ e_{C^{m}(j),C^{m}(i)} ⊗ e_{i,j}.

We denote by (R_{1} − R_{5})_{k=m} and by (R_{1} − R_{5})_{k<m} parts of this sum corresponding to the ranges k = m and k < m. Then we have

(R_{1} − R_{5})_{k=m} + R_{2} + R_{8,1} =

∑_{k≥1,(i,j)∈P (k)^{−},i≤i′ or C^{k}(i′)<C^{k}(i)}(xx′)^{−k}e_{C^{k}(i′),C^{k}(i)} ⊗ e_{C^{k}(j),C^{k}(i′)} ⊗ e_{i,j}.

Using Lemma 6.7 it is easy to see that the condition on (i, j, i′) in this summation can be replaced by the conjuction of (i, j) ∈ P (k)^{−}, j < i′ and C^{k}(i′) < C^{k}(j) (same as in the formula for L_{6} − L_{4,1}). Finally, we have

R_{8,2} − R_{4} =

− ∑_{k≥1,m≥1,(i,j)∈P (m)^{−},j′<i,(C^{m}(i),C^{m}(j′))∈P (k)^{+}} x^{−k−m}(x′)^{−m}e_{C^{k+m}(j′),C^{k+m}(i)} ⊗ e_{C^{m}(j),C^{m}(j′)} ⊗ e_{i,j},

(R_{1} − R_{5})_{k<m} + R_{3,3} + R_{8,3} =

∑_{k≥1,m≥1,(i,j′,j)∈Π′(k,m)} x^{−k}(x′)^{−k−m}e_{C^{k}(j′),C^{k}(i)} ⊗ e_{C^{k+m}(j),C^{k+m}(j′)} ⊗ e_{i,j},

where Π′(k, m) is the subset of {(i, j′, j) | (i, j) ∈ P (k)^{−}, (C^{k}(j′), C^{k}(j)) ∈ P (m)^{+}} consisting of (i, j′, j) such that either i ≤ j′ or C^{k}(j′) < C^{k}(i). By Lemma 6.7, we get

Π′(k, m) = {(i, j′, j) | (i, j) ∈ P (k)^{−}, (C^{k}(j′), C^{k}(j)) ∈ P (m)^{+}, j < j′}.

Now it is easy to see that parts of the left-hand side and the right-hand side of equation (iv) match as follows:

L_{6} − L_{4,1} = (R_{1} − R_{5})_{k=m} + R_{2} + R_{8,1},
L_{7} − L_{4,2} = (R_{1} − R_{5})_{k<m} + R_{3,3} + R_{8,3},
L_{1} − L_{5} − L_{4,3} − L_{8,2} = R_{8,2} − R_{4}.
⊓ ⊔

Proof of Theorem 0.2(i). As was already observed, the fact that r(u, v) is a unitary solution of the AYBE follows from Lemma 6.1 and Theorem 6.2. It follows from Theorem 1.5 that r(u, v) also satisfies the QYBE for fixed u. It remains to check the unitarity condition for the quantum R-matrix given by (6). In view of the unitarity of r(u, v) this boils down to proving the identity

s(u, v) = ([exp(\frac{v}{2} ) − exp(−\frac{v}{2} )]^{−2} − [exp(\frac{u}{2}) − exp(−\frac{u}{2})]^{−2}) · 1 ⊗ 1. (38)

To this end we observe that from Theorem 1.5 and Lemma 1.9 we know that

s(u, v) = (f(u) + g(v)) · 1 ⊗ 1,

where f(u) = \frac{1}{N} tr μ(\frac{∂r(u,0)}{∂u}) and g(v) = −\frac{1}{N} (tr ⊗ tr)(\frac{∂r(0,v)}{∂v}). Now (38) follows immediately from the equalities

f(u) = \frac{d}{du} (\frac{1}{exp(u) − 1}) = −[exp(\frac{u}{2}) − exp(−\frac{u}{2})]^{−2},

g(v) = \frac{d}{dv} (\frac{1}{exp(−v) − 1}) = [exp(\frac{v}{2} ) − exp(−\frac{v}{2} )]^{−2}.

⊓ ⊔

Remark 6.8. The following interesting observation is due to T. Schedler. Assume that Γ_{1} does not contain two consecutive elements of Γ_{C_{0}}, say, (C_{0}^{−1}(i_{0}), i_{0}) and (i_{0}, C_{0}(i_{0})). Then the function r(u, v) given by Theorem 0.2(i) is equivalent to the one of the form \frac{1⊗1}{exp(u)−1} + r(v). Indeed, let us denote by O(i_{0}, i) the minimal k ≥ 0 such that C^{k}(i_{0}) = i. Then one can easily check that

a = ∑_{i} \frac{O(i_{0}, i)}{N}e_{ii}

is an infinitesimal symmetry of r(u, v) and

exp[u(1 ⊗ a)]r(u, v) exp[−u(a ⊗ 1)] = \frac{1 ⊗ 1}{exp(u) − 1} + r(v),

where r(v) depends only on v. Note that the fact that r(u, v) is a unitary solution of the AYBE is equivalent to the following equations on r(v):

AY BE[r](v, v′) = r^{13}(v + v′), r^{21}(−v) + r(v) = 1 ⊗ 1.

7 Meromorphic continuation

As was shown in the proof of Theorem 6 of [7] (see also Lemma 4.14 of [8]), a unitary solution of the AYBE with the Laurent expansion (5) at u = 0 is uniquely determined by r_{0}(v). Therefore, it is not surprising that some of the results from [3] about solutions of the CYBE (such as meromorphic continuation) can be extended to solutions of the AYBE.

First, we apply the above uniqueness principle to infinitesimal symmetries.

Lemma 7.1. Let r(u, v) be a nondegenerate unitary solution of the AYBE with the Laurent expansion (5) at u = 0. Then the algebras of infinitesimal symmetries of r(u, v) and of r_{0}(v) are the same (and are contained in the algebra of infinitesimal symmetries of \overline{r}_{0}). If in addition \overline{r}_{0} has a period then these coincide with the commutative algebra of infinitesimal symmetries of \overline{r}_{0}.

Proof. Let a ∈ A be an infinitesimal symmetry of r_{0}(v). Then for any t ∈ ℂ the function

exp[t(a ⊗ 1 + 1 ⊗ a)]r(u, v) exp[−t(a ⊗ 1 + 1 ⊗ a)]

is a solution of the AYBE with the same r_{0}-term in the Laurent expansion at u = 0. By the uniqueness mentioned above this implies that exp[t(a⊗1+1⊗a)] commutes with r(u, v), so a is an infinitesimal symmetry of r(u, v). Recall that by Theorem 1.5(i), \overline{r}_{0}(v) is nondegenerate. It is easy to see that if \overline{r}_{0} is either elliptic or trigonometric then the algebra of infinitesimal symmetries of \overline{r}_{0} is commutative. Indeed, in the elliptic case this algebra is trivial (see Lemma 5.1 of [7]). In the trigonometric case this follows from the fact proven in [3] that there exists a pole γ of \overline{r}_{0} such that

\overline{r}_{0}(v + γ) = (ϕ ⊗ id)(\overline{r}_{0}(v)),

where ϕ is a Coxeter automorphism of sl_{N}. Thus, any infinitesimal symmetry is contained in the commutative algebra of ϕ-invariant elements. ⊓ ⊔

Proposition 7.2. Let r(u, v) be a nondegenerate unitary solution of the AYBE with the Laurent expansion (5) at u = 0, such that the equivalent conditions of Theorem 1.5(ii) hold. Then r(u, v) admits a meromorphic continuation to D × ℂ, where D is a neighborhood of 0 in ℂ. If r(u, v) has a pole at v = γ then this pole is simple and \overline{r}_{0}(v) also has a pole at v = γ.

Proof. Note that \overline{r}_{0}(v) has a meromorphic continuation to ℂ with at most simple poles by Theorem 1.1 of [3]. First, we want to deduce a meromorphic continuation for r_{0}(v). From the condition (c) in Theorem 1.5(ii) we know that

r_{0}(v) = \overline{r}_{0}(v) + b ⊗ 1 − 1 ⊗ b + h(v) · 1 ⊗ 1,  (39)

where b is an infinitesimal symmetry of \overline{r}_{0}(v) (by Lemma 1.10). Note that b is also an infinitesimal symmetry of r_{0}(v). Hence, by Lemma 7.1, b is an infinitesimal symmetry of r(u, v). Applying the equivalence transformation

r(u, v) ↦ exp[u(1 ⊗ b)]r(u, v) exp[−u(b ⊗ 1)]

we can assume that b = 0. In this case we have

AY BE[r_{0}](v_{12}, v_{23}) − AY BE[\overline{r}_{0}](v_{12}, v_{23}) ≡
[h(v_{13}) − h(v_{23})]\overline{r}_{0}^{12}(v_{12}) + c.p.(1, 2, 3) mod (ℂ · 1 ⊗ 1 ⊗ 1)

where we use the notation from the proof of Theorem 1.5 (the omitted terms are obtained by cyclically permuting (1, 2, 3); we denote v_{ij} = v_{i} − v_{j}). Applying pr ⊗ pr ⊗ id and using (18) we obtain

[h(v + v′) − h(v′)]\overline{r}_{0}^{12}(v) = [(pr ⊗ pr)r_{1}(v)]^{12} − (pr ⊗ pr ⊗ id)AY BE[\overline{r}_{0}](v, v′).

(40)

Note that AY BE[\overline{r}_{0}](v, v′) is meromorphic on the entire ℂ × ℂ and has at most simple poles at v = γ, v′ = γ and v + v′ = γ, where γ is a pole of \overline{r}_{0}(v). Also, by Lemma 1.7, r_{1}(v) is holomorphic near v = 0. Choose a small disk D around zero such that r_{1}(v) is holomorphic in D and \overline{r}_{0}(v) has no poles or zeros in D \ {0}. Assume that we already have a meromorphic continuation of h(z) to some open subset U ⊂ ℂ containing zero. Then the above formula gives a meromorphic continuation of h(z) to U + D. Iterating this process we continue h(z) meromorphically to the entire complex plane. Furthermore, it is clear from (40) that h(v) has only simple poles and is holomorphic outside the set of poles of \overline{r}_{0}(v). Therefore, the same is true for r_{0}(v).

Next, considering the constant terms of the Laurent expansions of the AYBE in u′ (keeping u fixed) we get

r_{0}^{12}(v_{12})r^{13}(u, v_{13})+r^{13}(u, v_{13})r_{0}^{23}(v_{23})−r^{23}(u, v_{23})r^{12}(u, v_{12}) = \frac{∂r^{13}}{∂u} (u, v_{13}).

(41)

Since we already know that r_{0}(v) is meromorphic on the entire ℂ, we can use this equation to get a meromorphic continuation of r(u, v). Indeed, assume that r(u, v) is meromorphic in D × D for some open disk around zero D ⊂ ℂ. For fixed v_{13} ∈ D the above equation gives a meromorphic extension of

r^{23}(u, v_{21} + v_{13})r^{21}(−u, v_{21}) = −r^{23}(u, v_{23})r^{12}(u, v_{12})

to D × ℂ. By the nondegeneracy of r(u, v) this allows to extend meromorphically r(u, v) from D × U to D × (U + D). Iterating this process we get the required meromorphic extension. The assertion about poles follows easily from (41) by fixing v_{13} such that r(u, v) has no pole at v = v_{13} and r(u, v_{13} − γ) is nondegenerate, and considering the polar parts at v_{12} = γ. ⊓ ⊔

The argument in the following Lemma is parallel to that in Proposition 4.3 of [3].

Lemma 7.3. With the same assumptions as in Proposition 7.2 for every pole γ of \overline{r}_{0}(v) there exists an algebra automorphism ϕ_{γ} of A and a constant λ ∈ ℂ such that

r(u, v + γ) = exp(λu)(ϕ_{γ} ⊗ id)(r(u, v)).

Proof. From Proposition 7.2 we know that the pole of r(u, v) at v = γ is simple. Set τ(u) = lim_{v→γ}(v − γ)r(u, v). Recall that lim_{v→0} vr(u, v) = cP for c ∈ ℂ^{∗}. Let us define an operator ϕ(u) ∈ End(A) by the equality

τ(u) = (ϕ(u) ⊗ id)(cP).

Considering polar parts near v = γ in (1) we get

τ^{12}(−u′)r^{13}(u + u′, v′ + γ) = r^{23}(u + u′, v′)τ^{12}(u).

The right-hand side can be rewritten as follows:

r^{23}(u + u′, v′)τ^{12}(u) = c(ϕ(u) ⊗ id ⊗ id)(r^{23}(u + u′, v′)P^{12}) =
c(ϕ(u) ⊗ id ⊗ id)(P^{12}r^{13}(u + u′, v′)),

Hence, we have

τ^{12}(−u′)r^{13}(u + u′, v′ + γ) = c(ϕ(u) ⊗ id ⊗ id)(P^{12}r^{13}(u + u′, v′)). (42)

Taking the residues at v′ = 0 we find

τ^{12}(−u′)τ^{13}(u + u′) = c^{2}(ϕ(u) ⊗ id ⊗ id)(P^{12}P^{13}).

This means that ϕ(u) satisfies the identity

ϕ(u_{1} + u_{2})(XY) = ϕ(u_{1})(X)ϕ(u_{2})(Y),

where X, Y ∈ A. Let D be a small disk around zero in ℂ such that ϕ(u) is holomorphic on D \ {0}. For every u ∈ D \ {0} we denote by I(u) ⊂ A the kernel of ϕ(u). Then from the above identity we derive that I(u)A ⊂ I(u+u′) and AI(u) ⊂ I(u+u′) whenever u, u′, u+u′ ∈ D\{0}. In particular, we deduce that I(u) ⊂ I(u + u′), so I(u) = I ⊂ A does not depend on u ∈ D \ {0}. It follows that I is a two-sided ideal in A. Since ϕ(u) is not identically zero, we derive that I = 0. Therefore, ϕ(u) is invertible for every u ∈ D \ {0}. Now as in the proof of Lemma 1.3 we derive that

ϕ(u) = exp(λu)ϕ_{γ}

for some λ ∈ ℂ, where ϕ_{γ} is an algebra automorphism of A. Applying ϕ_{γ}^{−1} ⊗ id ⊗ id to (42) we derive

exp(−λu′)P^{12}(ϕ_{γ}^{−1} ⊗ id ⊗ id)r^{13}(u + u′, v′ + γ) = exp(λu)P^{12}r^{13}(u + u′, v′).

This implies the required identity. ⊓ ⊔ Lemma 7.4. Keep the same assumptions as in Proposition 7.2. Assume that \overline{r}_{0}(v +p) = \overline{r}_{0}(v) for some p ∈ ℂ^{∗}. Then r(u, v +p) = exp(λu)r(u, v) for some constant λ ∈ ℂ.

Proof. Consider the decomposition (39) again. The identity (40) implies that h(v +v′)−h(v′) is periodic in v′ with the period p. Hence, h(v +p) = h(v)+λ for some λ ∈ ℂ. It follows that r_{0}(v + p) = r_{0}(v) + λ · 1 ⊗ 1. Applying the rescaling r(u, v) ↦ exp(−λuv)r(u, v) we can assume that r_{0}(v + p) = r_{0}(v). Now Lemma 7.3 implies that r(u, v + p) = (ϕ_{p} ⊗ id)r(u, v), where ϕ_{p} is an automorphism of A. Since r_{0}(v) is nondegenerate (as follows from Lemma 1.7), we derive that ϕ_{p} = id. ⊓ ⊔

We will use the following result in the proof of Theorem 0.3.

Proposition 7.5. Let r(u, v) be a nondegenerate unitary solution of the AYBE with the Laurent expansion at u = 0 of the form (5) such that the equivalent conditions of Theorem 1.5(ii) hold. Then one has

r_{0}(v) = \overline{r}_{0}(v) + b ⊗ 1 + 1 ⊗ b + h(v) · 1 ⊗ 1,
h(v) = λv + ch_{0}(c′v),

where b ∈ sl_{N} is an infinitesimal symmetry of \overline{r}_{0}(v), λ ∈ ℂ, c, c′ ∈ ℂ^{∗}, and h_{0}(v) is one of the following three functions: Weierstrass zeta function ζ(v) associated with a lattice in ℂ; \frac{1}{2} coth(\frac{v}{2} ); or \frac{1}{v}. Furthermore, if \overline{r}_{0}(v) is equivalent to a rational solution of the CYBE then h_{0}(v) = \frac{1}{v}.

Proof. The equation (18) implies that

[r_{0}^{12}(v_{12}) + r_{0}^{23}(v_{23}) + r_{0}^{31}(v_{31})]^{2} = x^{12}(v_{12}) + x^{23}(v_{23}) + x^{31}(v_{31}), (43)

where x(v) = r_{0}(v)^{2} − 2r_{1}(v) (and v_{ij} = v_{i} − v_{j}). On the other hand, it is easy to see that x(v) is the constant term of the Laurent expansion of s(u, v) = r(u, v)r(−u, v) at u = 0. Rescaling r(u, v) we can assume that its residue at v = 0 is equal to P. Then we have

s(u, v) = [f(u) + g(v)] · 1 ⊗ 1,

where

f(u) = \frac{1}{N} tr μ(\frac{∂r(u, 0)}{∂u}), g(v) = −\frac{1}{N} (tr ⊗ tr)(\frac{dr_{0}(v)}{dv})

(see Lemma 1.9). If we change r(u, v) to exp(λuv)r(u, v) for some λ ∈ ℂ then f(u) changes to f(u)+Nλ (this operation also changes r_{0}(v) to r_{0}(v)+λv·1⊗ 1). Therefore, we can assume that f(u) has no constant term in the Laurent expansion at u = 0. In this case we obtain x(v) = g(v) · 1 ⊗ 1. Hence, denoting

T (v_{1}, v_{2}, v_{3}) = r_{0}^{12}(v_{12}) + r_{0}^{23}(v_{23}) + r_{0}^{31}(v_{31})

we can rewrite (43) as follows:

T (v_{1}, v_{2}, v_{3})^{2} = [g(v_{12}) + g(v_{23}) + g(v_{31})] · 1 ⊗ 1.  (44)

Viewing T (v_{1}, v_{2}, v_{3}) ∈ A ⊗ A ⊗ A as an endomorphism of V ⊗ V ⊗ V, where A = End(V), we obtain

T (v_{1}, v_{2}, v_{3}) = T_{0}(v_{1}, v_{2}, v_{3}) + [h(v_{12}) + h(v_{23}) + h(v_{31})] · id_{V ⊗V ⊗V} , (45)

where T_{0} is a traceless endomorphism and h(v) is defined from the decomposition (39). Note also that for fixed (generic) v_{2} and v_{3} we have

lim_{v_{1}→v_{2}}(v_{1} − v_{2})T (v_{1}, v_{2}, v_{3}) = P^{12}.

The latter operator has S^{2}V ⊗ V and ⋀^{2} V ⊗ V as eigenspaces. Therefore, for v_{1} close to v_{2} we have a decomposition

V ⊗ V ⊗ V = W_{1} ⊕ W_{2},

where dim W_{1} = N^{2}(N + 1)/2, dim W_{2} = N^{2}(N − 1)/2, and

(T (v_{1}, v_{2}, v_{3}) − λ id)(W_{1}) = 0, (T (v_{1}, v_{2}, v_{3}) + λ id)(W_{2}) = 0, where
λ^{2} = g(v_{12}) + g(v_{23}) + g(v_{31}).

Comparing the traces of both sides of (45) we derive

λ = N[h(v_{12}) + h(v_{23}) + h(v_{31})].

Since g(v) = −Nh′(v), we obtain

N[h(v_{12}) + h(v_{23}) + h(v_{31})]^{2} + h′(v_{12}) + h′(v_{23}) + h′(v_{31}) = 0.

Replacing h(v) by h(Nv) we get

[h(v_{12}) + h(v_{23}) + h(v_{31})]^{2} + h′(v_{12}) + h′(v_{23}) + h′(v_{31}) = 0. (46)

We are interested in solutions of this equation for an odd meromorphic function h(v) in the neighborhood of zero having a simple pole at v = 0. It is easy to see that the Laurent expansion of h(v) at v = 0 should have form h(v) = 1/v + c_{3}v^{3} + . . .. As shown in the proof of Theorem 5 of [7], all such solutions of (46) have form c · h_{0}(cv), where h_{0} is one of the three functions described in the formulation.^{1}

Finally, if \overline{r}_{0}(v) is rational then its only pole is v = 0 (see [3]). Therefore, by Proposition 7.2, r_{0}(v) also cannot have poles outside zero, which implies that h_{0}(v) = \frac{1}{v}. ⊓ ⊔

Remark 7.6. In the case when \overline{r}_{0}(v) is either elliptic or trigonometric the assertion of the above proposition can also be deduced from the explicit formulas for r(u, v) (the elliptic case is discussed in [7], sec.2; the trigonometric case is considered in Theorem 0.2).

1Solutions of (46) were first described by L. Carlitz in [5].

8 Classification of trigonometric solutions of the AYBE

Recall (see [3]) that to every nondegenerate trigonometric solution \overline{r}_{0}(v) of the CYBE for sl_{N} with poles exactly at 2πiℤ Belavin and Drinfeld associate an automorphism of the Dynkin diagram A_{N−1} by considering the class of the automorphism ϕ of sl_{N} defined by

\overline{r}_{0}(v + 2πi) = (ϕ ⊗ id)(\overline{r}_{0}(v)).  (47)

They also show that ϕ is a Coxeter automorphism. The next lemma shows that in the case of trigonometric solutions coming from a solution of the AYBE the automorphism of the Dynkin diagram is always trivial.

Lemma 8.1. Let r(u, v) be a nondegenerate unitary solution of the AYBE with the Laurent expansion (5) at u = 0. Assume that \overline{r}_{0}(v) is a trigonometric solution of the CYBE with poles exactly at 2πiℤ. Then the automorphism ϕ in (47) is inner.

Proof. This follows immediately from Lemma 7.3, since every algebra automorphism of A is inner. ⊓ ⊔

Now let us recall the Belavin-Drinfeld classification of trigonometric solutions of the CYBE for sl_{N} corresponding to the trivial automorphism of A_{N−1}. Let us denote by 𝔥_{0} ⊂ sl_{N} the subalgebra of traceless diagonal matrices. For every Belavin-Drinfeld triple (Γ_{1}, Γ_{2}, τ) for Ã_{N−1} we have the corresponding series of solutions

\overline{r}_{0}(v) = t + \frac{1}{exp(v)−1}(pr ⊗ pr) ∑_{0≤m<N,j−i≡m(N)} exp(\frac{mv}{N})e_{ij} ⊗ e_{ji}+ ∑_{0<m<N,k≥1;j−i≡m(N),τ^{k}(i,j)=(i′,j′)}[exp(−\frac{mv}{N})e_{ji} ⊗ e_{i′j′} − exp(\frac{mv}{N})e_{i′j′} ⊗ e_{ji}],

(48)

where t ∈ 𝔥_{0} ⊗ 𝔥_{0} satisfies

t^{12} + t^{21} = (pr ⊗ pr)P^{0},  (49)
[τ(α) ⊗ id + id ⊗α]t = 0, α ∈ Γ_{1},  (50)

where P^{0} = ∑_{i} e_{ii} ⊗e_{ii}. The result of Belavin and Drinfeld in [3] is that every nondegenerate unitary trigonometric solution of the CYBE for sl_{N} that has poles exactly at 2πiZ and the residue (pr ⊗ pr)P at 0, is conjugate to

exp[v(b ⊗ 1)]\overline{r}_{0}(v) exp[−v(b ⊗ 1)],

where \overline{r}_{0}(v) is one of the solutions of the form (48) and b ∈ sl_{N} is an infinitesimal symmetry of \overline{r}_{0}.

It is easy to see that the solution of the CYBE for sl_{N} obtained from the associative r-matrix in Theorem 0.2(i) for S = [1, N] and C_{0}(i) = i + 1 is given by the above formula with

t = \frac{1}{2}(pr ⊗ pr)P^{0} + s_{C},  (51)

where

s_{C} = ∑_{0<k<N,i}(\frac{1}{2} − \frac{k}{N})e_{ii} ⊗ e_{C^{k}(i),C^{k}(i)} ∈ 𝔥_{0} ∧ 𝔥_{0}.

The proof of the next result is almost identical to that of Lemmas 4.19 and 4.20 in [8]. Let us denote by e_{i} : 𝔥 → ℂ the functional on diagonal matrices given by e_{i}(e_{jj}) = δ_{ij}.

Lemma 8.2. Let r(u, v) be a nondegenerate unitary solution of the AYBE with the Laurent expansion (5), such that \overline{r}_{0}(v) is given by (48). Then there exists a unique transitive cyclic permutation C of [1, N] such that (51) holds. Furthermore, for any (i, i + 1) ∈ Γ_{1} with τ(i, i + 1) = (i′, i′ + 1) one has C(i) = i′ and C(i + 1) = i′ + 1 (i.e., τ is induced by C × C).

Proof. We will make use of the identity

(pr ⊗ pr ⊗ pr)(AY BE[\overline{r}_{0}]) = 0  (52)

that follows from Theorem 1.5. First, considering the projection of AY BE[\overline{r}_{0}] to 𝔥 ⊗ 𝔥 ⊗ 𝔥 we get

(pr ⊗ pr ⊗ pr)(AY BE[\frac{1}{exp(v) − 1}P^{0} + t]) = 0.

Using the fact that t^{12} + t^{21} ≡ P^{0} mod (ℂ · 1 ⊗ 1) this can be rewritten as

(pr ⊗ pr ⊗ pr)(AY BE[t]) = 0.

Therefore, we have

[(e_{i} − e_{1}) ⊗ (e_{j} − e_{1}) ⊗ (e_{k} − e_{1})](AY BE[t]) = 0  (53)

for all i, j, k. Set t = ∑_{i,j} t_{ij}e_{ii}⊗e_{jj}. Note that t_{ij}+t_{ji} = 0 for i ≠ j and t_{ii} = \frac{1}{2} for all i. Let us set t′_{ij} = t_{ij} − t_{1j} − t_{i1}. Then substituting t_{ij} = t′_{ij} + t_{1j} − t_{1i} into t and then into (53) we deduce that

t′_{ij}t′_{ik} − t′_{jk}t′_{ij} + t′_{ik}t′_{jk} = \frac{1}{4}, 1 < i, j, k ≤ N.  (54)

As shown in the proof of Lemma 4.20 in [8], the above equation implies that t′_{ij} = ±\frac{1}{2} for 1 < i, j ≤ N, i ≠ j, and there is a unique complete order ≺ on [2, N] such that t′_{ij} = \frac{1}{2} iff i ≺ j (for i, j ∈ [2, N], i ≠ j). We define the cyclic permutation C of [1, N] by the condition that it sends each element of [2, N] to the next element with respect to this complete order. As in the proof of Lemma 4.20 in [8] this easily implies that t − \frac{1}{2}P^{0} ≡ s_{C} mod (ℂ · 1 ⊗ 1).

Next, we want to check that τ is induced by C × C. Assume that τ(i, i + 1) = (j, j + 1) and consider the coefficient A_{ijk} with e_{i+1,i} ⊗ e_{j,j+1} ⊗ e_{kk} in AY BE[\overline{r}_{0}]. Let us denote by ⟨e_{lm} ⊗ e_{np}, r(v)⟩ the coefficient with e_{lm} ⊗ e_{np} in r(v). Then we have

\begin{matrix}A_{ijk} = ⟨e_{i+1,i} ⊗ e_{j,j+1}, r(v_{12})⟩⟨e_{ii} ⊗ e_{kk}, r(v_{13})⟩ \\ −⟨e_{jj} ⊗ e_{kk}, r(v_{23})⟩⟨e_{i+1,i} ⊗ e_{j,j+1}, r(v_{12})⟩+ \\ ⟨e_{i+1,i} ⊗ e_{k,k+1}, r(v_{13})⟩⟨e_{j,j+1} ⊗ e_{k+1,k}, r(v_{23})⟩. \end{matrix}  (55)

Note that we cannot have τ^{n}(j + 1, j) = (i + 1, i) since this would imply that Γ_{1} (resp., Γ_{2}) is the complement to (j, j + 1) (resp., (i, i + 1)), N is even, j − i ≡ N/2(N), and τ(l, l + 1) = (l + N/2, l + 1 + N/2), in which case the nilpotency condition is not satisfied. Therefore,

⟨e_{i+1,i} ⊗ e_{j,j+1}, r(v)⟩ = exp(−\frac{v}{N}),

⟨e_{j,j+1} ⊗ e_{i+1,i}, r(v)⟩ = − exp(\frac{v}{N}).

Next, we claim that the third summand in the right-hand side of (55) is zero unless k = i or k = j. Indeed, τ (resp., τ^{−1}) cannot be defined on both (k, k+1) and (k + 1, k). This leaves only two possibilities with k ≠ i and k ≠ j: either τ^{n_{1}}(i, i+1) = (k, k+1) and τ^{n_{2}}(k, k+1) = (j, j+1), or τ^{n_{1}}(j+1, j) = (k+1, k) and τ^{n_{2}}(k + 1, k) = (i, i + 1) (where n_{1}, n_{2} > 0). The latter case is impossible since j ≠ k. In the former case we derive that τ^{n_{1}+n_{2}}(i, i+1) = (j, j +1) which contradicts to our assumption that τ(i, i + 1) = (j, j + 1) (since n_{1} + n_{2} ≥ 2). Thus, recalling that

⟨e_{i+1,i} ⊗ e_{i,i+1}, r(v)⟩ = \frac{exp(\frac{(N−1)v}{N})}{exp(v) − 1},

⟨e_{j,j+1} ⊗ e_{j+1,j}, r(v)⟩ = \frac{exp(\frac{v}{N})}{exp(v) − 1},

we can rewrite (55) as follows:

A_{ijk} = exp(−\frac{v_{12}}{N})[t_{ik} − t_{jk} + \frac{δ_{ik}}{exp(v_{13}) − 1} − \frac{δ_{jk}}{exp(v_{23}) − 1}]

− δ_{ik} exp(\frac{v_{23}}{N})\frac{exp(\frac{(N−1)v_{13}}{N})}{exp(v_{13}) − 1} + δ_{jk} exp(−\frac{v_{13}}{N})\frac{exp(\frac{v_{23}}{N})}{exp(v_{23}) − 1}.

Hence,

exp(\frac{v_{12}}{N})A_{ijk} = t_{ik} − t_{jk} − δ_{ik}.

Since pr ⊗ pr ⊗ pr(AY BE[\overline{r}_{0}]) = 0, it follows that A_{ijk} does not depend on k. Therefore,

t_{ik} − t_{jk} − δ_{ik} = [(e_{i} − e_{j}) ⊗ e_{k}]s_{C} − \frac{1}{2}(e_{i} + e_{j}, e_{k})

does not depend on k (note that (e_{i}, e_{j}) = δ_{ij}), i.e.,

[(e_{i} − e_{j}) ⊗ α]s_{C} = \frac{1}{2}(e_{i} + e_{j}, α)  (56)

for all roots α ∈ Γ. Repeating the above argument for the coefficient with e_{j,j+1} ⊗ e_{i+1,i} ⊗ e_{kk} in AY BE[\overline{r}_{0}] we derive that

[(e_{i+1} − e_{j+1}) ⊗ α]s_{C} = \frac{1}{2}(e_{i+1} + e_{j+1}, α)  (57)

for all α ∈ Γ. As shown in the proof of Lemma 4.20 in [8], (56) and (57) imply that C(i) = j and C(i + 1) = j + 1. ⊓ ⊔

Lemma 8.3. Assume that N > 1. Then a nondegenerate unitary solution r(u, v) of the AYBE with the Laurent expansion at u = 0 of the form (5) such that r_{0}(v) ≡ \overline{r}_{0}(v) mod (ℂ · 1 ⊗ 1), is uniquely determined by \overline{r}_{0}, up to rescaling r(u, v) ↦ exp(λuv)r(u, v).

Proof. This follows from the proof of Theorem 6 in [7]: one only has to observe that \overline{r}_{0}(v) is nondegenerate by Theorem 1.5(i), so it has rank > 2 generically.

⊓ ⊔

Proof of Theorem 0.2(ii). Let r(u, v) be a nondegenerate unitary solution of the AYBE with the Laurent expansion at u = 0 of the form (5) such that \overline{r}_{0}(v) is trigonometric. Changing r(u, v) to cr(cu, c′v) we can assume that \overline{r}_{0}(v) has poles exactly at 2πiℤ and lim_{v→0} v\overline{r}_{0}(v) = (pr ⊗ pr)P. Recall that we are allowed to change r(u, v) to an equivalent solution

r̃(u, v) = exp[u(1 ⊗ a) + v(b ⊗ 1)]r(u, v) exp[−u(a ⊗ 1) − v(b ⊗ 1)],

where a and b are infinitesimal symmetries of r(u, v) (note that a and b always commute by Lemma 7.1). This operation changes \overline{r}_{0}(v) to an equivalent solution in the sense of Belavin-Drinfeld [3] and also changes r_{0}(v) to r_{0}(v) − a ⊗ 1 + 1 ⊗ a. Therefore, in view of Lemma 8.1 and of (39), changing r(u, v) to an equivalent solution we can achieve that r_{0}(v) ≡ \overline{r}_{0}(v) mod (ℂ · 1 ⊗ 1) and \overline{r}_{0}(v) has the form (48). Note that in this case any infinitesimal symmetry of \overline{r}_{0}(v) is diagonal (since it has to commute with the corresponding Coxeter automorphism ϕ from (47)). It remains to use Lemmas 8.2 and 8.3. ⊓ ⊔

Proof of Theorem 0.3. Let r(v) be a nondegenerate unitary solution of the AYBE, not depending on u. Then one can easily check that

r(u, v) = \frac{1 ⊗ 1}{u} + r(v)

is also a nondegenerate unitary solution of the AYBE. By Lemma 1.4, r(u, v) (and hence, r(v)) has a simple pole at v = 0 with the residue cP, where c ≠ 0. Now applying Lemma 1.9 we obtain

s(u, v) = −\frac{1 ⊗ 1}{u^{2}} + g(v) · 1 ⊗ 1,

where g(v) = −\frac{c}{N} (tr ⊗ tr)(\frac{dr(v)}{dv}). Hence, by Theorem 1.5, r(u, v) is a solution of the QYBE and \overline{r}(v) is a nondegenerate solution of the CYBE. It is easy to see that \overline{r}(v) cannot be equivalent to an elliptic or a trigonometric solution. Indeed, if this were the case then r(u, v) would have a pole of the form u = u_{0} with u_{0} ≠ 0 (in the elliptic case this follows from the explicit formulas for elliptic solutions in [7], sec.2; in the trigonometric case this follows from Theorem 0.2(ii)). Now Proposition 7.5 gives the required decomposition of r(v). Therefore, g(v) = −c^{2}/v^{2}, which shows that R(u, v) = (1/u + c/v)^{−1}r(u, v) satisfies unitarity condition (7). ⊓ ⊔

Remark 8.4. The function of the form r(v) = \frac{P}{v} + r, where r ∈ A ⊗ A does not depend on v, is a unitary solution of the AYBE iff r is a skew-symmetric constant solution of the AYBE for A = Mat(N, ℂ). Some information about such solutions can be found in [2], sec.2 (including the classification for N = 2, see Ex. 2.8 of loc. cit.).

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