DESCRIPTION OF THE CENTER OF THE AFFINE TEMPERLEY-LIEB ALGEBRA OF TYPE Ã_{N}

MASHA VLASENKO

Abstract. Construction of the diagrammatic version of the affine Temperley-Lieb algebra of type Ã_{N} as a subring of matrices over the Laurent polynomials is given. We move towards geometrical understanding of cellular structure of the Temperley-Lieb algebra. We represent its center as a coordinate ring of the certain affine algebraic variety and describe this variety constructing its desingularization.

Introduction

In this paper we deal with a sequence of infinite-dimensional algebras T_{N}(δ), N = 1, 2, 3, …, introduced by Graham and Lehrer in [1]. These algebras are extended versions of the Temperley-Lieb quotients of the affine Hecke algebras of type Ã_{N}. Let us recall the naive definition of T_{N}(δ).

Consider a vertical cylinder with n marked points on the top circle component of its boundary and m marked points on the bottom circle component, with n ≡ m(mod 2). If we join these marked points pairwise by arcs on the (lateral) surface of the cylinder without intersection we get an affine diagram. One can also add to affine diagram a finite number of circles which circumnavigate the cylinder, if it is possible to do without intersections with arcs. Let us denote by D(m → n) the set of homotopy classes of such an affine diagrams. Take α ∈ D(m → n), β ∈ D(k → m). If we glue the bottom boundary circle of α to the top boundary circle of β so that corresponding marked points coincide, we get a cylinder again, and there can be some loops on its surface. Let us denote the number of loops by m(α, β), and the affine diagram obtained after removing all the loops by α ∘ β. Then α ∘ β ∈ D(k → n).

Definition 1. Let δ ∈ ℂ be a complex number. T_{N}(δ) is an associative ℂ-algebra spanned over ℂ by affine diagrams D(N → N) with multiplication

αβ = δ^{m(α,β)}α ∘ β

for α, β ∈ D(N → N).

One can find the proof that this multiplication really agrees with associativity and diagrams D(N → N) are linearly independent in [1]. Note that there is a unity 1 ∈ T_{N}(δ) given by diagram 1_{N} on the picture in section 1.

Let us enumerate the marked points on the top and bottom circles of cylinder by 1, …, N sequently, so that top 1 is over bottom 1, and so on. Take some point between N and 1 on the top and some point between N and 1 on the bottom circle, consider a line segment l on the surface connecting these points. For the diagram α ∈ D(N → N) its rank |α| is a minimal number of intersections with l among all the diagrams from the homotopy class α. A diagram is called monic if it has no arcs connecting bottom point to bottom point. It was proved in [2] that the Temperley-Lieb quotient of the affine Hecke algebra of type Ã_{N} is isomorphic to a subalgebra of T_{N}(δ), spanned by all non-monic diagrams of even rank together with the unity 1_{N}.

In section 1 of this paper we realize T_{N}(δ) as a subalgebra of matrices over the ring of Laurent polynomials in one variable ℂ[x, \frac{1}{x}] and construct a large simply arranged ideal inside of T_{N}(δ) (Theorem 1).

Denote T_{N} = {t ∈ ℤ | 0 ≤ t ≤ N, t ≡ N(mod 2)}. In section 2 we obtain the following result.

Corollary 1. The center of T_{N}(δ) is a coordinate ring of a certain affine algebraic variety. This variety has [\frac{n}{2}] + 1 irreducible components. Each irreducible component is a curve with finite number of singular points.

We also construct a certain desingularization of this variety (Theorem 2), and propose an algorithm to calculate the numbers of intersections (self-intersections) of its irreducible components. In section 3 we perform these calculations for T_{3}(δ).

Representation theory of algebras T_{N}(δ) is known: in [1] the set of all irreducible modules is described, and the composition factors of so called cell modules are determined. They use concept of cellular algebras ([3]). If some finite dimensional algebra has cellular structure its representation theory can be obtained in standard way. The generalization of the concept for infinite dimensional algebras is possible but requires some special technique in each concrete case ([4]). We also use cellular structure of T_{N}(δ) in our constructions (section 1), but we move towards its geometrical description. The problem of geometrical understanding of cellular structure in general was stated in [6].

1. Realization of T_{N}(δ) as a subalgebra of matrices over the ring of Laurent polynomials in one variable ℂ[x, \frac{1}{x}]

For k ∈ T_{N} let A_{k} be a linear span over ℂ of diagrams D(k → N). Then A_{k} is a left T_{N}(δ)-module with the action of diagrams on diagrams by multiplication:

αβ = δ^{m(α,β)}α ∘ β

for α ∈ D(N → N), β ∈ D(k → N). On the other hand, A_{k} is a right T_{k}(δ)- module. Multiplication law is associative ([1]), so these actions commute.

We call an arc a through arc if it connects bottom point to top point. Let B_{k} ⊂ A_{k} be a linear span of diagrams having strictly less then k through arcs. Note that the number of through arcs in α ∘ β is not greater then their number in each of the diagrams α, β. Thus B_{k} is preserved under both left T_{N}(δ)- and right T_{k}(δ)-action. So C_{k} = A_{k}/B_{k} is a T_{N}(δ) - T_{k}(δ)-bimodule as A_{k} and B_{k} are ones.

Note that C_{k} is spanned over ℂ by all monic diagrams from D(k → N). Now we consider left action of T_{k}(δ) on C_{k}.

Let k > 0. Denote by I_{k}^{k−2} a linear span of all non-monic diagrams in T_{k}(δ). I_{k}^{k−2} is an ideal, and all its elements have zero action on C_{k}. Consider the diagram τ_{k}, usually it is called twist:

[figure: •; •; •; •; •; •; •; •; 1; 2; k − 1; k; 1; 2; 3; k; τ_{k} : k → k (k > 0)]

[figure: τ_{0} : 0 → 0]

The quotient T_{K}(δ)/I_{k}^{k−2} is isomorphic to the group algebra of ℤ, or the algebra

of Laurent polynomials ℂ[x, \frac{1}{x}]. Indeed, this quotient is spanned by all monic diagrams D(k → k). The monic diagram τ_{k} is invertible: τ_{k}^{−1} is just the diagram obtained by reflection of τ_{k} in a horizontal line. And there is no monic diagrams in D(k → k) except powers of τ_{k}, powers of τ_{k}^{−1} and the unity 1_{k}:

[figure: •; •; •; •; •; •; •; •; 1; 2; k − 1; k; 1; 2; k − 1; k; 1_{k} : k → k]

Thus C_{k} becomes a right ℂ[x, \frac{1}{x}]-module with x acting by multiplication by τ_{k} from the right. It is a free right ℂ[x, \frac{1}{x}]-module of finite rank. Indeed, let us call a monic diagram standard if it has no circumnavigate circles and can be drawn without intersections of through arcs with the line segment l. Now k > 0 and every monic diagram α ∈ D(k → N) can be uniquely ”twisted” from the right to the standard one: there exist a unique r ∈ ℤ such that ατ_{k}^{r} is standard. Thus C_{k} has a basis of standard diagrams over ℂ[x, \frac{1}{x}].

Let k = 0. T_{0}(δ) itself is isomorphic to the algebra of polynomials in one variable ℂ[x] if we define a clean diagram (without arcs and circles) to be a unity 1_{0}. Indeed, D(0 → 0) consists of powers of τ_{0}. So, C_{0} is a free right module over ℂ[x] with basis of standard diagrams.

To avoid cumbersome notations it would be useful to consider not C_{0} but a similar module over ℂ[x, \frac{1}{x}]. Let C̃_{0} be a free right module over ℂ[x, \frac{1}{x}] with basis of all standard diagrams from D(0 → N). We can define left action of T_{N}(δ) on C̃_{0}. We prefer following symmetric version of this action to be consistent with computations made in [1]: let α ∈ D(N → N), let β ∈ D(0 → N) is standard, and α ∘ β = γτ_{0}^{r} where γ is also standard. Then put

αβ = δ^{m(α,β)}γ(x + \frac{1}{x})^{r}.

One can extend this to the action of T_{N}(δ) on C̃_{0} by linearity. Indeed, if ϱ, α ∈ T_{N}(δ) and β ∈ D(0 → N) is standard then in C_{0} one has

ϱ(αβ) = (ϱα)β = ∑_{s} γ_{s}Q_{s}(x)

for some standard γ_{s} and Q_{s} ∈ ℂ[x]. Then in C̃_{0}

ϱ(αβ) = (ϱα)β = ∑_{s} γ_{s}Q_{s}(x + \frac{1}{x}).

So, C̃_{0} is a T_{N}(δ) - ℂ[x, \frac{1}{x}]-bimodule, free over ℂ[x, \frac{1}{x}].

We denote by ℂ_{s}[x, \frac{1}{x}] ⊂ ℂ[x, \frac{1}{x}] the subset of polynomials symmetric in x and \frac{1}{x}. Then ℂ_{s}[x, \frac{1}{x}] ≅ ℂ[x].

Let us denote

M_{k} = \begin{cases}C_{k} & if k > 0, \\ C̃_{0} & if k = 0. \end{cases}

Consider

M = ∏_{k∈T_{N}} M_{k}

where the direct product is considered in the category of T_{N}(δ) - ℂ[x, \frac{1}{x}] - bimodules. Now we consider the action of T_{N}(δ) on M, we claim it is faithful.

Proposition 1. The kernel of the map T_{N}(δ) → End (M)_{ℂ[x, \frac{1}{x} ]} is zero.

Proof. Every element in b ∈ T_{N}(δ) is a unique linear combination of diagrams. Let us denote the set of this diagrams by supp(b). Also define deg b to be a maximal number of through arcs among supp(b), and a principal part b̂ to be a linear combination of diagrams with deg b through arcs such that deg(b − b̂) < deg b.

Suppose there is a nonzero element a ∈ T_{N}(δ) acting on M as zero. Then â ∈ T_{N}(δ) is nonzero. Let k = deg a. Since a − â has zero action on M_{k}, action of â on M_{k} coincides with action of a on M_{k} and is zero. We are going to deduce that â is zero then. So it will lead to a contradiction.

Let {β_{i}^{k}; i = 1, 2, …, d_{k}} be the set of standard diagrams from D(k → N). Then

â = ∑_{i,j=1}^{d_{k}} β_{i}^{k}P_{i,j}(τ_{k})(β_{j}^{k})^{∗}

for some Laurent polynomials P_{i,j}, where for the diagram α ∈ D(m → n) we define a^{∗} ∈ D(n → m) to be the reflection of α in a horizontal line. Moreover, the polynomials P_{i,j} are defined uniquely. Note that (β_{j}^{k})^{∗}β_{s}^{k} is a scalar multiple of a monic diagram from D(k → k). We have identified the linear span of monic diagrams from D(k → k) with ℂ[x, \frac{1}{x}] (with ℂ_{s}[x, \frac{1}{x}] for k = 0). Define the following d_{k} × d_{k} matrix over ℂ[x, \frac{1}{x}]

(1)  R_{k} = ((β_{j}^{k})^{∗}β_{s}^{k})_{j,s=1}^{d_{k}} ∈ M_{d_{k}} (ℂ[x, \frac{1}{x}])

in sense of this identification. Then the matrix of action of â in basis β_{s}^{k} is

(2)  \begin{cases}(P_{i,j}(x))_{i,j=1}^{d_{k}} R_{k} & for k > 0, \\ (P_{i,j}(x + \frac{1}{x}))_{i,j=1}^{d_{0}} R_{0} & for k = 0. \end{cases}

This matrix should be zero. The calculation in [1] (Corollary 4.7) shows that det R_{k} is a nonzero Laurent polynomial. Thus P_{i,j} = 0 for all pairs i, j, and â is zero. This is a contradiction. □

So we get an inclusion

T_{N}(δ) ⊂ ∏_{k∈T_{N}} M_{d_{k}} (ℂ[x, \frac{1}{x}]) ,

as End (M)_{ℂ[x, \frac{1}{x}]} ≅ ∏_{k∈T_{N}} M_{d_{k}} (ℂ[x, \frac{1}{x}]), where d_{k} is a number of standard diagrams in D(k → N). This number is obviously independent of choice of the line segment l of cut on the surface of the cylinder. It was calculated in [1] (Proposition 1.11):

d_{k} = ( N_{\frac{N−k}{2}} ) .

We proceed to know how large is the image of this inclusion.

Proposition 2. The commutant of T_{N}(δ) in ∏_{k∈T_{N}} M_{d_{k}} (ℂ[x, \frac{1}{x}]) is just ∏_{k∈T_{N}} ℂ[x, \frac{1}{x}], i.e. the set of matrices, scalar in each matrix component.

Proof. Take an arbitrary matrix A = ∏_{k∈T_{N}} A_{k} commuting with T_{N}(δ). For an arbitrary 1 ≤ i, j ≤ d_{k} consider the element b^{i,j} = β_{i}^{k}(β_{j}^{k})^{∗} ∈ T_{N}(δ), let B^{i,j} ∈ M_{d_{k}} (ℂ[x, \frac{1}{x}]) be the matrix of action of b^{i,j} on M_{k}. Then A_{k}B^{i,j} = B^{i,j}A_{k}. But ImB^{i,j} = β_{i}^{k}ℂ[x, \frac{1}{x}], so A_{k}β_{i}^{k} = Q_{i}(x)β_{i}^{k} for some Q_{i} ∈ ℂ[x, \frac{1}{x}]. Recall nonzero monomials (R_{k})_{i,i} ∈ ℂ[x, \frac{1}{x}] defined by (1). Then

Q_{i}(x)(R_{k})_{j,j}(x)β_{i}^{k} = Q_{i}(x)β_{i}^{k}(β_{j}^{k})^{∗}β_{j}^{k} = A_{k}B^{i,j}β_{j}^{k} = B^{i,j}A_{k}β_{j}^{k}
= Q_{j}(x)B^{i,j}β_{j}^{k} = Q_{j}(x)(R_{k})_{j,j}(x)β_{i}^{k},

hence Q_{i} = Q_{j}. So, A_{k} is a scalar matrix. □

We often identify elements of T_{N}(δ) with corresponding matrices over ℂ[x, \frac{1}{x}] in basis of standard diagrams.

Now we are going to construct a large ideal in T_{N}(δ). For these purposes we will define a set of Laurent polynomials \{P_{k}(x); k ∈ T_{N}\} ⊂ ℂ[x, \frac{1}{x}] such that for k > 0 if A ∈ M_{d_{k}} (ℂ[x, \frac{1}{x}]) is divisible by P_{k}(x^{k}) then the matrix equal to A in k-th component and zero in all other components belongs to T_{N}(δ). The 0-th matrix component for each element of T_{N}(δ) is a matrix with symmetric polynomials in its entries, so P_{0} will be a polynomial of x + \frac{1}{x} such that any d_{0} × d_{0} matrix A of polynomials of x + \frac{1}{x} divisible by P_{0} will belong to T_{N}(δ) when we put zeros in all other matrix components.

We denote ℂ^{×} = ℂ\\{0\}. Consider q ∈ ℂ^{×} such that δ = −q − \frac{1}{q}. Of coarse every symmetric Laurent polynomial in q is a polynomial in δ. Put

G_{k}(x) = ∏_{r∈T_{N} ,r>k}(x^{2} + \frac{1}{x^{2}} − q^{r} − \frac{1}{q^{r}})^{d_{r}}

for k < N and G_{N}(x) = 1. For each k ∈ T_{N} we defined a matrix R_{k} ∈ M_{d_{k}} (ℂ[x, \frac{1}{x}]) by (1). It was calculated in [1](Corollary 4.7) that either

(3)  det R_{k} = G_{k}(x) or det R_{k} = −G_{k}(x).

For k > 0 take H_{k} ∈ ℂ[x, \frac{1}{x}] to be the Laurent polynomial of smallest possible degree such that H_{k}(x^{k}) is divisible by G_{k}(x), put H_{0}(x) = G_{0}(x). Put

P_{k}(x) = ∏_{r∈T_{N} ,r≤k} H_{r}(x).

Theorem 1. There are inclusions: for odd N

∏_{k∈T_{N}} P_{k}(x^{k})M_{d_{k}} (ℂ[x, \frac{1}{x}]) ⊂ T_{N}(δ) ⊂ ∏_{k∈T_{N}} M_{d_{k}} (ℂ[x, \frac{1}{x}]) ,

and for even N

P_{0}(x)M_{d_{0}} (ℂ_{s}[x, \frac{1}{x}]) × ∏_{k∈T_{N} ,k>0} P_{k}(x^{k})M_{d_{k}} (ℂ[x, \frac{1}{x}]) ⊂ T_{N}(δ)

⊂ M_{d_{0}} (ℂ_{s}[x, \frac{1}{x}]) × ∏_{k∈T_{N} ,k>0} M_{d_{k}} (ℂ[x, \frac{1}{x}]) .

Proof. Consider a filtration of T_{N}(δ) by the set of ideals

T_{N}(δ) = I_{N}^{N} ⊃ I_{N}^{N−2} ⊃ I_{N}^{N−4} ⊃ …

where for k ∈ T_{N}

I_{N}^{k} = {a ∈ T_{N}(δ)| deg a ≤ k} .

Consider a homomorphism π_{k} : T_{N}(δ) → M_{d_{k}} (ℂ[x, \frac{1}{x}]) where π_{k}(a) is the matrix of action of a ∈ T_{N}(δ) on M_{k} written in basis of standard diagrams.

Note that if k < s then π_{s}(I_{N}^{k}) = 0. Recall the matrix R_{k} ∈ M_{d_{k}} (ℂ[x, \frac{1}{x}]) defined by (1). Then

π_{k}(I_{N}^{k}) = \begin{cases}M_{d_{k}} (ℂ[x, \frac{1}{x}]) R_{k} & for k > 0, \\ M_{d_{0}} (ℂ_{s}[x, \frac{1}{x}]) R_{0} & for k = 0 \end{cases}

due to (2).

The theorem statement is a consequence of the following more general one: if A ∈ M_{d_{k}} (ℂ[x, \frac{1}{x}]) (A ∈ M_{d_{0}} (ℂ_{s}[x, \frac{1}{x}])) is devisible by P_{k}(x^{k}) (by P_{0}(x)) then there exist an element a ∈ I_{N}^{k} with π_{k}(a) = A and π_{m}(a) = 0 for m ∈ T_{N}, m ≠ k. We are going to prove it by induction moving up in k ∈ T_{N}.

The base of induction (either k = 0 or k = 1) follows from the description of π_{s}(I^{k}) for s ≥ k given above. Indeed, for these cases P_{k}(x) is devisible by det R_{k} due to (3). Consider the matrix

(4)  N_{k} = (N_{i,j} = (−1)^{i+j}R̂_{k}^{j,i})_{i,j=1}^{d_{k}} ,

where R̂_{k}^{i,j} is a minor of matrix R_{k} obtained by deleting of i-th row and j-th column. Then N_{0} ∈ M_{d_{0}} (ℂ_{s}[x, \frac{1}{x}]) and N_{k} ∈ M_{d_{k}} (ℂ[x, \frac{1}{x}]) for k > 0, and N_{k}R_{k} = det R_{k}(x). So,

π_{1}(I_{N}^{1}) = M_{d_{1}} (ℂ[x, \frac{1}{x}]) R_{1} ⊃ M_{d_{1}} (ℂ[x, \frac{1}{x}]) N_{1}R_{1} ⊃ P_{1}(x)M_{d_{1}} (ℂ[x, \frac{1}{x}]) ,

π_{0}(I_{N}^{0}) = M_{d_{0}} (ℂ_{s}[x, \frac{1}{x}]) R_{0} ⊃ M_{d_{0}} (ℂ_{s}[x, \frac{1}{x}]) N_{0}R_{0} ⊃ P_{0}(x)M_{d_{0}} (ℂ_{s}[x, \frac{1}{x}]) .

Suppose we have proved the statement for all k ∈ T_{N}, k < r. Take an arbitrary matrix B ∈ M_{d_{r}} (ℂ[x, \frac{1}{x}]) and consider A = P_{r}(x^{r})B. Take Q(x) = \frac{H_{r}(x^{r})}{det R_{r}}. It is a Laurent polynomial due to (3). Then A = Q(x)P_{r−2}(x^{r})BN_{r}R_{r}, so A ∈ π_{r}(I_{N}^{r}). Let C = Q(x)BN_{r}. Recall that {β_{i}^{r}; i = 1, 2, …, d_{r}} is a basis of standard diagrams in M_{r}. Consider element

a_{r} = ∑_{i,j=1}^{d_{r}} β_{i}^{r}P_{r−2}(τ_{r}^{r})C_{i,j}(τ_{r})(β_{j}^{r})^{∗}.

Then a_{r} ∈ I_{N}^{r} and π_{r}(a_{r}) = P_{r−2}(x^{r})CR_{r} = A due to (2). So, for s > r we have π_{s}(a_{r}) = 0 as a_{r} ∈ I_{N}^{r}. We claim that π_{k}(a_{r}) is devisible by P_{k}(x^{k}) for r > k > 0 and π_{0}(a_{r}) = 0 if N is even. This will finish the proof. Indeed, by inductional assumption we can take a_{k} ∈ I_{N}^{k} for each k ∈ T_{N},r > k > 0 such that π_{s}(a_{k}) = 0 for all s ≠ k and π_{k}(a_{k}) = −π_{k}(a_{r}). Then the element ∑_{0<k≤r} a_{k} will satisfy our conditions.

To prove the claim note that for arbitrary monic diagrams μ, ν ∈ D(r → N) we have for k ∈ T_{r}

π_{k} (μτ_{r}^{r}ν^{∗}) = x^{k}π_{k} (μν^{∗}) .

This statement is equivalent to μτ_{r}^{r}ν^{∗}β_{t}^{k} = μν^{∗}β_{t}^{k}τ_{k}^{k} for 1 ≤ t ≤ d_{k}. For k = 0 one has μτ_{r}^{r}ν^{∗}β_{t}^{k} = μν^{∗}β_{t}^{k} obviously. For k > 0 the substitution of τ_{r}^{r} by τ_{k}^{k} is shown on the following picture:

[figure: 1_{k}; 1; •; •; k; •; •; β_{t}^{k}; •; •; 1; •; •; •; k; •; ν^{∗}; 1; •; 2; •; 3; N−1; •; N; •; τ_{r}^{r}; •; 1; •; 2; •; r−1; •; r; 1; 2; r−1; r; μ; •; 1; 2; •; •; 3; •; •; N−1; •; N; =; τ_{k}^{k}; 1; •; •; k; •; •; β_{t}^{k}; •; •; 1; •; •; •; k; •; ν^{∗}; 1; •; 2; •; 3; N−1; •; N; •; 1_{r}; •; 1; •; 2; •; r−1; •; r; 1; 2; r−1; r; μ; •; 1; 2; •; •; 3; •; •; N−1; •; N]

When we apply either τ_{r}^{r} between μ and ν^{∗} or τ_{k}^{k} below β_{t}^{k} we get the same result: every through line of the resulting diagram makes one more turn around the cylinder.

Then for k ∈ T_{r} we have

π_{k}(a_{r}) = ∑_{i,j=1}^{d_{r}} π_{k} (β_{i}^{r}P_{r−2}(τ_{r}^{r})C_{i,j}(τ_{r})(β_{j}^{r})^{∗}) = ∑_{i,j=1}^{d_{r}} π_{k} (β_{i}^{r}C_{i,j}(τ_{r})(β_{j}^{r})^{∗}) P_{r−2}(x^{k})

= P_{r−2}(x^{k})π_{k} ( ∑_{i,j=1}^{d_{r}} β_{i}^{r}C_{i,j}(τ_{r})(β_{j}^{r})^{∗}) .

So if 0 < k < r then π_{k}(a_{k}) is divisible by P_{k}(x^{k}) because P_{r−2}(x^{k}) is divisible by P_{k}(x^{k}). For k = 0 one has π_{0}(a_{r}) = 0 because P_{r−2}(1) = 0. The theorem is proved. □

Note that Proposition 2 can be deduced independently from Theorem 1.

Proof. Consider the matrix A = ∏_{k∈T_{N}} A_{k} commuting with T_{N}(δ). Let k > 0. Then A_{k} commutes with M_{d_{k}} (ℂ[x, {1 \atop x}]) P_{k}(x^{k}). Consider z ∈ ℂ^{×}\ roots of P_{k}(x^{k}). For every matrix B ∈ M_{d_{k}} (ℂ) there exist a matrix C ∈ M_{d_{k}} (ℂ[x, {1 \atop x}]) P_{k}(x^{k}) such that C(z) = B. Thus A_{k}(z) commutes with M_{d_{k}} (ℂ), hence it is diagonal. All nondiagonal entries of A_{k} equal zero in infinitely many points of ℂ, so they are zero everywhere on ℂ. All diagonal entries of A_{k} coincide in infinitely many points, so they coincide as elements of ℂ[x, {1 \atop x}]. We proceed analogously for k = 0. □

2. Geometry of the center

In this section we consider the center Z (T_{N}(δ)) and prove Corollary 1. We identify T_{N}(δ) with the subalgebra of ∏_{k∈T_{N}} M_{d_{k}} (ℂ[x, \frac{1}{x}]).

Put

I_{N} = \begin{cases}∏_{k∈T_{N}} P_{k}(x^{k})ℂ[x, \frac{1}{x}] & if N is odd, \\ P_{0}(x)ℂ_{s}[x, \frac{1}{x}] × ∏_{k∈T_{N} ,k>0} P_{k}(x^{k})ℂ[x, \frac{1}{x}] & if N is even, \end{cases}

K_{N} = \begin{cases}∏_{k∈T_{N}} ℂ[x, \frac{1}{x}] & if N is odd, \\ ℂ_{s}[x, \frac{1}{x}] × ∏_{k∈T_{N} ,k>0} ℂ[x, \frac{1}{x}] & if N is even. \end{cases}

Then

(5)  I_{N} ⊂ Z (T_{N}(δ)) ⊂ K_{N}

by Theorem 1 and Proposition 2.

Proposition 3. The center Z (T_{N}(δ)) is a finitely generated algebra without nilpotent elements.

To prove the proposition we need the following facts.

Lemma 1. Let P (x) ∈ ℂ[x]. The ideal I_{P} = P (x)ℂ[x] of ℂ[x] is finitely generated as algebra.

Proof. Let n = deg P. If n = 0 then I_{P} = ℂ[x] and elements 1, x are the generators. If n > 0 one can take Y_{0} = P (x), …, Y_{n−1} = x^{n−1}P (x) as a set of generators. □

Lemma 2. Let P (x) ∈ ℂ[x, \frac{1}{x}]. The ideal I_{P} = P (x)ℂ[x, \frac{1}{x}] of ℂ[x, \frac{1}{x}] is finitely generated as algebra.

Proof. If P (x) is invertible in ℂ[x, \frac{1}{x}] then I_{P} = ℂ[x, \frac{1}{x}] and generators are x and \frac{1}{x}. Recall that the only invertible elements in ℂ[x, \frac{1}{x}] are αx^{m} for some α ∈ ℂ^{×}, m ∈ ℤ. If P (x) is equal to x+α for some α ∈ ℂ^{×} up to multiplication by invertible element of ℂ[x, \frac{1}{x}], then the generators are x + α and \frac{1}{α} + \frac{1}{x}. Otherwise we can suppose that P (x) = α_{n}x^{n} + ⋯ + α_{−m} \frac{1}{x^{m}} for some n, m ≥ 1 and α_{n}, α_{−m} ∈ ℂ^{×}. Indeed, one can make P (x) to be of this form multiplying P by some invertible element of ℂ[x, \frac{1}{x}]. Then Y_{−m} = x^{−m}P (x), …, Y_{n−2} = x^{n−2}P (x), Y_{n−1} = x^{n−1}P (x) ∈ I_{P} is the set of generators. □

Proof of Proposition 3. There are no nilpotent elements in Z (T_{N}(δ)) as it is included into algebra without nilpotent elements by (5).

I_{N} is finitely generated algebra as it is a direct product of finitely generated algebras (Lemma 1 and Lemma 2).

Z (T_{N}(δ))/I_{N} is finite dimensional. Indeed, Z (T_{N}(δ))/I_{N} ⊂ K_{N}/I_{N} and

K_{N}/I_{N} ≅ \begin{cases}∏_{k∈T_{N}} (ℂ[x, \frac{1}{x}]/I_{P_{k}}) & if N is odd, \\ (ℂ[x]/IP̃_{0}) × ∏_{k∈T_{N} ,k>0} (ℂ[x, \frac{1}{x}]/I_{P_{k}}) & if N is even, \end{cases}

where P̃_{0} ∈ ℂ[x] is such that P_{0}(x) = P̃_{0}(x + \frac{1}{x}). The algebra on the right is of finite dimension.

Let s^{1}, …, s^{L} ∈ Z (T_{N}(δ)) are such that their images under factorization modulo I_{N} form a linear basis. Then Z (T_{N}(δ)) is generated by generators of I_{N} together with s^{1}, …, s^{L}. □

Any finitely generated commutative algebra with 1 over an algebraically closed field and without nilpotent elements is a coordinate ring of a certain affine algebraic variety (it is an exercise in [5]). Let us denote this variety for Z (T_{N}(δ)) by V_{N}.

We are going to describe V_{N} in this section. We will find what are its irreducible components and construct a desingularization of V_{N}.

Consider the sets S_{k} = ℂ^{×} for k > 0 and S_{0} = ℂ. Consider the map

(6)  Ψ : ⨆_{k∈T_{N}} S_{k} → V_{N}

defined by the following. Any element f ∈ Z (T_{N}(δ)) is a set of Laurent polynomials {f_{k}; k ∈ T_{N}}. For z ∈ S_{k} consider a homomorphism ψ_{k,z} : Z (T_{N}(δ)) → ℂ defined by

ψ_{k,z}(f) = \begin{cases}f_{k}(z) & if k > 0, \\ f_{0}(t) where t + \frac{1}{t} = z & if k = 0. \end{cases}

Then Kerψ_{k,z} is a maximal ideal in Z (T_{N}(δ)), so it defines a unique point Ψ(z) ∈ V_{N}.

Theorem 2. ImΨ = V_{N} and Ψ is one-to-one on the set

\begin{cases}⨆_{k∈T_{N}} S_{k}\ \{roots of P_{k}(x^{k})\} & if N is odd \\ S_{0}\ \{t + \frac{1}{t}|P_{0}(t) = 0\} ⨆ ⨆_{k∈T_{N} ,k>0} S_{k}\ \{roots of P_{k}(x^{k})\} & if N is even \end{cases}

To prove this statement we need the following fact.

Lemma 3. Let P_{1}, …, P_{L} ∈ ℂ[x]. Consider the map ϕ : ℂ → ℂ^{L} given by ϕ(x) = (P_{1}(x), …, P_{L}(x)). Then Imϕ is algebraically closed.

Proof. Let m = max_{1≤i≤L} deg P_{i}. Consider the map of projective spaces ϕ̂: ℙ^{1} → ℙ^{L} given in homogenous coordinates as

ϕ̂([X_{0}, X_{1}]) = [X_{0}^{m}, P̂_{1}(X_{0}, X_{1}), …, P̂_{L}(X_{0}, X_{1})]

where for the polynomial Q(x) = ∑_{k} a_{k}x^{k} we denote Q̂(X_{0}, X_{1}) = ∑_{k} a_{k}X_{1}^{k}X_{0}^{m−k}. This map is correctly defined: for X_{0} ≠ 0 one has nonzero first coordinate X_{0}^{m} ≠ 0; otherwise X_{1} ≠ 0 and there is a number k with deg P_{k} = m, so P̂_{k}(0, X_{1}) ≠ 0. So ϕ̂ is regular map of projective variety to projective variety, thus Imϕ̂ is closed. There is a canonically defined open set U_{0} = {Y_{0} ≠ 0} ⊂ ℙ^{L} isomorphic to ℂ^{L}, and the intersection of any closed projective variety with U_{0} is isomorphic to a closed affine variety. Isomorphism is given by

[Y_{0}, Y_{1}, …, Y_{L}] → (\frac{Y_{1}}{Y_{0}} , …, \frac{Y_{L}}{Y_{0}} ).

Then Imϕ̂ ∩ U_{0} is isomorphic to Imϕ. Lemma is proved. □

Similar statement is true for Laurent polynomials but under additional assumptions. Let P (x) = ∑_{i=m}^{n} a_{i}x^{i} where a_{m}, a_{n} ≠ 0 be Laurent polynomial. We denote mindegP = m and maxdegP = n.

Lemma 4. Let P_{1}, …, P_{L} ∈ ℂ[x, \frac{1}{x}]. Consider the map ϕ : ℂ^{×} → ℂ^{L} given by ϕ(x) = (P_{1}(x), …, P_{L}(x)). Suppose there exist m > 0 such that

min_{1≤i≤L}mindegP_{i} = −m, max_{1≤i≤L}maxdegP_{i} = m

and there exist i ≠ j such that mindegP_{i} = −m, maxdegP_{i} < m and mindegP_{j} > −m, maxdegP_{j} = m. Then Imϕ is algebraically closed.

Proof. Consider the map ϕ̂: ℙ^{2} → ℙ^{L} given in homogenous coordinates as

ϕ̂([X_{0}, X_{1}, X_{2}]) = [X_{0}^{m}, P̂_{1}(X_{0}, X_{1}, X_{2}), …, P̂_{L}(X_{0}, X_{1}, X_{2})]

where for Q(x) = c + ∑_{i>0} a_{i}x^{i} + ∑_{j>0} b_{i} \frac{1}{x^{j}} we define

Q̂[X_{0}, X_{1}, X_{2}] = cX_{0}^{m} + ∑_{i>0} a_{i}X_{1}^{i}X_{0}^{m−i} + ∑_{j>0} b_{i}X_{2}^{j}X_{0}^{m−j}.

It is again correctly defined because either P̂_{i}(0, X_{1}, X_{2}) ≠ 0 or P̂_{j}(0, X_{1}, X_{2}) ≠ 0. Then Imϕ is isomorphic to U_{0} ∩ Imϕ̂ ({X_{0}^{2} = X_{1}X_{2}}), so it is closed. □

Consider an algebra A_{N} = ℂ1 + I_{N}. A_{N} is obviously finitely generated (see arguments in proof of Proposition 3) and has no nilpotent elements. So A_{N} is a coordinate ring of some affine algebraic variety, let us denote this variety by W_{N}. Then we can construct the map

Φ : ⨆_{k∈T_{N}} S_{k} → W_{N}

in the same way as Ψ for V_{N}.

Proof of Theorem 2. First, we choose useful coordinates on variety V_{N}. Let s^{1}, …, s^{L} ∈ Z (T_{N}(δ)) be such that form a linear basis in Z (T_{N}(δ))/I_{N} after factorization modulo I_{N}. Recall that for f ∈ Z (T_{N}(δ)) we denote by {f_{k}; k ∈ T_{N}} its components which are Laurent polynomials.

Due to Lemma 1 and Lemma 2 one can choose generators Y_{1}^{k} = x^{m_{k}}P_{k}(x^{k}), Y_{2}^{k} = x^{m_{k}+1}P_{k}(x^{k}), …, Y_{L_{k}}^{k} = x^{n_{k}}P_{k}(x^{k}) in P_{k}(x^{k})ℂ[x, \frac{1}{x}] for k > 0 and Y_{1}^{0} = P_{0}(x), Y_{2}^{0} = (x + \frac{1}{x})P_{0}(x), …, Y_{L_{0}}^{0} = (x + \frac{1}{x})^{L_{0}−1}P_{0}(x) in P_{0}(x)ℂ_{s}[x, \frac{1}{x}]. Then for each k > 0 we can make m_{k} smaller and n_{k} large for the k^{th} set of generators so that Laurent polynomials

Y_{1}^{k}, …, Y_{L_{k}}^{k}, s_{k}^{1}, …, s_{k}^{L}

satisfy conditions of Lemma 4. So, we suppose this condition to be satisfied. If N is even, let us consider such polynomials Ỹ_{1}^{0}, …, Ỹ_{L_{0}}^{0}, s̃_{0}^{1}, …, s̃_{0}^{L} ∈ ℂ[x] that

Y_{i}^{0}(x) = Ỹ_{i}^{0}(x + \frac{1}{x}) and s_{0}^{i}(x) = s̃_{0}^{i}(x + \frac{1}{x}).

So {Y_{i}^{k}, s^{j}|k ∈ T_{N}, 1 ≤ i ≤ L_{k}, 1 ≤ j ≤ L} can be considered as a coordinates on V_{N} because they are generators of Z (T_{N}(δ)). Then

Ψ(z) = \begin{cases}(0, …, 0, Y_{1}^{k}(z), …, Y_{L_{k}}^{k}(z), 0, …, 0, s_{k}^{1}(z), …, s_{k}^{L}(z)) & if z ∈ S_{k}, k > 0, \\ (Ỹ_{1}^{0}(z), …, Ỹ_{L_{0}}^{0}(z), 0, …, 0, 0, …, 0, s̃_{0}^{1}(z), …, s̃_{0}^{L}(z)) & if z ∈ S_{0}. \end{cases}

Then ImΨ is dense in V_{N} because there is no nonzero element f ∈ Z (T_{N}(δ)) which is zero on ImΨ, i.e. f_{k}(z) = 0 for each k ∈ T_{N}, z ∈ ℂ^{×}. But ImΨ = ⋃_{k∈T_{N}} Ψ(S_{k}), and each Ψ(S_{k}) is closed by Lemma 4 for k > 0 and by Lemma 3 for k = 0. So, V_{N} = ImΨ.

We consider {Y_{i}^{k}|k ∈ T_{N}, 1 ≤ i ≤ L_{k}} as a coordinates on W_{N} because they are generators of I_{N}. Then Φ can be written in this coordinates in the similar way as Ψ, and one has ImΦ = W_{N} again. Now V_{N} ⊂ ℂ^{∑_{k} L_{k}+L}, W_{N} ⊂ ℂ^{∑^{k} L_{k}}, and

Φ = π ∘ Ψ,

where π is a projection in ℂ^{∑_{k} L_{k}+L} on first ∑_{k} L_{k} coordinates. Consider z ∈ S_{k}. If P_{k}(z^{k}) = 0 (P_{0}(t) = 0 with t + \frac{1}{t} = z) then Φ(z) = (0, …, 0). Otherwise Φ(z) ≠ (0, …, 0) and z and k both can be calculated from Φ(z) in obvious way. So, Φ is one-to-one on the set ⨆_{k∈T_{N}} S_{k}\Φ^{−1} ((0, …, 0)). Then Ψ is also one-to-one on this set. □

DESCRIPTION OF THE CENTER OF THE AFFINE TEMPERLEY-LIEB ALGEBRA 11

So, W_{N} is a bouquet of [\frac{n}{2}] + 1 irreducible curves, it has exactly one singular point.

Proof of Corollary 1. Denote ν_{k} = Ψ(S_{k}). This is an irreducible curve with finite number of singular points (points of self-intersection) by Theorem 2. Then V_{N} = ∪_{k∈T_{N}} ν_{k} and ν_{k} ∩ ν_{l} is a finite set due to Theorem 2 again. □

The desingularization Ψ constructed above gives us a way to give a proper description of V_{N}. For those one should calculate points on sheets S_{k} correspondent to the points of intersections of the curves ν_{k} = Ψ(S_{k}). We deduce from Theorem 2 that all intersection (self-intersection) points are among Φ^{−1}(0, …, 0). If one finds some successive family of central elements in T_{N}(δ) it will separate some of those points one from each other. Following proposition gives us one more necessary condition for the intersections.

Proposition 4. Let x ∈ S_{k}, y ∈ S_{m}. Then Ψ(x) = Ψ(y) only if (i) P_{k}(x^{k}) = 0 (or x = z + \frac{1}{z} with P_{0}(z) = 0 if k = 0) (ii) P_{m}(y^{m}) = 0 (or y = z + \frac{1}{z} with P_{0}(z) = 0 if m = 0) (iii) x^{k} = y^{m}

Proof. Two first statements are due to Theorem 2. Let us prove the third one.

Obviously \{F (τ_{n}^{n}); F ∈ ℂ[x, \frac{1}{x}]\} ⊂ Z (T_{N}(δ)). Then Ψ(x) = Ψ(y) only if F (τ_{n}^{n})_{k}(x) = F (τ_{n}^{n})_{m}(y) for each F. But F (τ_{n}^{n})_{k}(x) = F (x^{k}) and F (τ_{n}^{n})_{m}(y) = F (y^{m}). So, x^{k} = y^{m}. □

3. Example for T_{3}(δ)

Let us consider T_{3}(δ) now. T_{3} = {1, 3}, d_{3} = 1 (d_{N} = 1 because the only standard diagram in D(N → N) is 1_{N}). Following diagrams are the standard ones in D(1 → 3)

[figure: •; •; •; •; 1; 2; 3; 1; β_{1}; •; •; •; •; 1; 2; 3; 1; β_{2}; •; •; •; •; 1; 2; 3; 1; β_{3}]

where we omit upper indices (β_{i} instead of β_{i}^{1}). So, d_{1} = 3 and

T_{3}(δ) = \{S(τ_{3}) + ∑_{i,j=1}^{3} β_{i}F_{i,j}(τ_{1})β_{j}^{∗}|S, F_{i,j} ∈ ℂ[x, \frac{1}{x}]\} ⊂ M_{1} (ℂ[x, \frac{1}{x}])×M_{3} (ℂ[x, \frac{1}{x}]) .

Proposition 5. Z (T_{3}(δ)) ≅ \{S × T ∈ ℂ[x, \frac{1}{x}]^{2}|S(x) − T (x^{3}) ⋮ x^{2} + \frac{1}{x^{2}} + δ\}

Proof. For P ∈ ℂ[x, \frac{1}{x}] denote

P̂ = (\begin{matrix}P_{0}(x) & \frac{1}{x}P_{2}(x) & P_{1}(x) \\ xP_{1}(x) & P_{0}(x) & P_{2}(x) \\ P_{2}(x) & P_{1}(x) & P_{0}(x) \end{matrix}) ,

where P (x) = P_{0}(x^{3}) + xP_{1}(x^{3}) + \frac{1}{x}P_{2}(x^{3}). Then

P̂ + Q̂ = { \atop P + Q}, P̂Q̂ = P̂ Q

and if P̂A = AP̂ is a scalar matrix then A = Q̂ for some Q.

Let a = S(τ_{3})+∑_{i,j=1}^{3} β_{i}F_{i,j}(τ_{1})β_{j}^{∗} ∈ T_{3}(δ). Then a acts on M_{3} ×M_{1} by matrix

S(x) × (Ŝ + F R_{1})

where F = (F_{i,j}) and

R_{1} = ((β_{i})^{∗}β_{j})_{i,j=1}^{3} = (\begin{matrix}δ & 1 & \frac{1}{x} \\ 1 & δ & x \\ x & \frac{1}{x} & δ \end{matrix})

was defined by (1). Indeed, consider the action of τ_{3} on M_{1}:

τ_{3}β_{1} = β_{2}τ_{1} = β_{2}x, τ_{3}β_{2} = β_{3}, τ_{3}β_{3} = β_{1}.

Then

τ_{3} = x × (\begin{matrix}0 & 0 & 1 \\ x & 0 & 0 \\ 0 & 1 & 0 \end{matrix}) , τ_{3}^{2} = x^{2} × (\begin{matrix}0 & 1 & 0 \\ 0 & 0 & x \\ x & 0 & 0 \end{matrix}) , τ_{3}^{3} = x^{3} × (\begin{matrix}x & 0 & 0 \\ 0 & x & 0 \\ 0 & 0 & x \end{matrix}) ,

thus S(τ_{3}) = S(x) × Ŝ. Also ∑_{i,j=1}^{3} β_{i}F_{i,j}(τ_{1})β_{j}^{∗} = 0 × F R_{1} due to (2).

a ∈ Z (T_{3}(δ)) if and only if a = S(x) × T (x)1_{3} for some T ∈ ℂ[x, \frac{1}{x}] due to Proposition 2. Hence, Ŝ + F R_{1} = T 1_{3}. Then F = V̂ for some V ∈ ℂ[x, \frac{1}{x}]. Indeed, F det R_{1} = { \atop (T (x^{3}) − S(x))}N_{1} where N_{1} was defined by (4) such that R_{1}N_{1} = (det R_{1})1_{3}. Note that R_{1} = { \atop x^{2} + \frac{1}{x^{2}} + δ}. Then N_{1} = Q̂ for some Q ∈ ℂ[x, \frac{1}{x}], and

F det R_{1} = { \atop (T (x^{3}) − S(x)) Q(x)}.

Now, S(x) + V (x)(x^{2} + \frac{1}{x^{2}} + δ) = T (x^{3}). Obviously, one can take an arbitrary V ∈ ℂ[x, \frac{1}{x}] and put F = V̂. □

We have an explicit description of Z (T_{3}(δ)). Recall the desingularization

Ψ : ℂ^{×} ⊔ ℂ^{×} → V_{3}

of corresponding affine algebraic variety introduced by (6). We take z as a coordinate on S_{3} = ℂ^{×} and w as a coordinate on S_{1} = ℂ^{×}.

Proposition 6. Consider the points z_{1} = \sqrt{q}, z_{2} = −\sqrt{q}, z_{3} = \frac{1}{\sqrt{q}} , z_{4} = −\frac{1}{\sqrt{q}}, put w_{i} = z_{i}^{3} for i = 1, 2, 3, 4.

Then Ψ(z_{i}) = Ψ(w_{i}) for i = 1, 2, 3, 4 and Ψ maps ℂ^{×} \ {z_{i}|i = 1, 2, 3, 4} ⊔ ℂ^{×} \ {w_{i}|i = 1, 2, 3, 4} onto V_{3} \ {Ψ(z_{i})|i = 1, 2, 3, 4} bijectively.

Proof. Ψ(w_{1}) = Ψ(w_{2}) only if w_{1} = w_{2} due to Proposition 4 (iii).

Put P (x) = x^{2} + \frac{1}{x^{2}} − q^{3} − \frac{1}{q^{3}} where δ = −q − \frac{1}{q}. Now P_{3}(x) = P_{1}(x) = P (x), and for w such that P (w) = 0 the map Ψ may glue some cubic roots of w either with w or with each other. We will consider these two possibilities.

(I) Let Ψ(z) = Ψ(w). This is equivalent to

{S × T ∈ Z (T_{3}(δ)) |S(z) = 0} = {S × T ∈ Z (T_{3}(δ)) |T (w) = 0}

by definition of the map Ψ. Recalling Proposition 5 the last is equivalent to

\{(V, T) ∈ ℂ[x, \frac{1}{x}]^{2}|T (z^{3}) + V (z)(z^{2} + \frac{1}{z^{2}} + δ) = 0\}

= \{(V, T) ∈ ℂ[x, \frac{1}{x}]^{2}|T (w) = 0\} ,

or

(7)  z^{2} + \frac{1}{z^{2}} + δ = 0

since z^{3} = w.

Conversely, if (7) holds then Ψ(z) = Ψ(w) where w = z^{3}.

(II) Let Ψ(z) = Ψ(x) for some unequal z, x ∈ S_{3}. Then z^{3} = x^{3} by Proposition 4 (iii), denote w = z^{3}. If z and x are roots of (7) both then Ψ(z) = Ψ(x) = Ψ(w) by part (I). Suppose z is not a root of (7), put a = z^{2}+ \frac{1}{z^{2}} +δ ≠ 0. Then Ψ(z) = Ψ(x) implies

\{(V, T) ∈ ℂ[x, \frac{1}{x}]^{2}|V (z) = −\frac{T (w)}{a}\}

= \{(V, T) ∈ ℂ[x, \frac{1}{x}]^{2}|V (x)(x^{2} + \frac{1}{x^{2}} + δ) = −T (w)\}

what is obviously a contradiction. □

So, for generic q ∈ ℂ^{×} variety V_{3} looks like two copies of ℂ^{×} glued in four points (z_{i} = w_{i}). If q^{6} = 1 some of those points coincide, so the number of singular points will be less than 4. Especially:

q = 1. Then we have two points z = −1, 1 glued to w = −1, 1 correspondingly. q = \frac{1±3\sqrt{−1}}{2}. Then we have two points z = α, −\frac{1}{α} glued to w = i and two points

z = −α, \frac{1}{α} glued to w = −i correspondingly where α = cos \frac{π}{6} + \sqrt{−1} sin \frac{π}{6} . q = \frac{−1±3\sqrt{−1}}{2}. Then we have two points z = α, \frac{1}{α} glued to w = −1 and two

points z = −α, −\frac{1}{α} glued to w = 1 correspondingly where α = \frac{1+3\sqrt{−1}}{2}.

q = −1. Then we have two points z = −\sqrt{−1}, \sqrt{−1} glued to w = \sqrt{−1}, −\sqrt{−1} correspondingly.

Case q = −1 (δ = 2) is when the corresponding Hecke algebra is a group algebra of the affine Coxeter group.

References

[1] J. J. Graham and G. I. Lehrer, The representation theory of affine Temperley-Lieb algebras // Enseign. Math. (2) 44 (1998), no. 3-4, 173–218

[2] C. K. Fan and R. M. Green, On the affine Temperley-Lieb algebras // J. London Math. Soc. (2) 60 (1999), no. 2, 366–380

[3] J. J. Graham and G. I. Lehrer, Cellular algebras// Inventiones Math. 123 (1996), 1–34 [4] J. J. Graham and G. I. Lehrer, Cellular algebras and diagram algebras in representations

theory // to appear in Advansed Studies in Pure Mathematics

[5] R. Hartshorne, Algebraic geometry //Springer, New York, 1977

[6] G. I. Lehrer, Geometric themes in representation theory //Beijing Normal University Press, to appear

   E-mail address: mariyka@imath.kiev.ua

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