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ЧЕБЫШЕВСКИИ СБОРНИК
Том 21. Выпуск 4.
УДК 517.958:530.145
DOI 10.22405/2226-8383-2020-21-4-171-195
Локальные системы координат на квантовых флаговых многообразиях
Ф. Разавиниа
Фаррох Разавиниа — Московский физико-технический институт (г. Долгопрудный). e-mail: /.razavinia@phystech.edu
Этот документ состоит из 3 разделов. В первом разделе мы дадим краткое введение в «гомоморфизмы Фейгина» и увидим, как они помогут нам доказать наши основные и фундаментальные теоремы, связанные с квантовыми соотношениями Серра и операторами экранирования.
Во втором разделе мы введем локальный интеграл движений как пространство инвариантов нильпотента часть квантовых аффинных алгебр Ли и найдет двух- и трехточечные инварианты в случае ( Найв^), используя схему Волкова.
В третьем разделе мы введем решеточные алгебры Вирасоро как пространство инвариантов борелевской части ич(В+) в ич(д) для простой алгебры Ли д и найдем множество образующих Решеточная алгебра Вирасоро, соединенная с а12 и ич(в/2)
И как новый результат, мы нашли множество некоторых образующих решетки алгебры Вирасоро.
Ключевые слова: Решетки \¥ алгебры, квантовые группы, гомоморфизмы Фейгина.
Библиография: 14 названий.
Для цитирования:
Ф. Разавиниа Локальные системы координат на квантовых флаговых многообразиях // Че-бышевский сборник, 2020, т. 21, вып. 4, с. 171 195.
Аннотация
CHEBYSHEVSKII SBORNIK Vol. 21. No. 4.
UDC 517.958:530.145
DOI 10.22405/2226-8383-2020-21-4-171-195
Local coordinate systems on quantum flag manifolds1
F. Razavinia
Farrokh Razavinia — Moscow Institute of Physics and Technology (Dolgoprudnv). e-mail: f.razavinia@phystech.edu
_ Abstract
1the paper was supported by Russian Science foundation grant N 17-11-01377
This paper consists of 3 sections. In the first section, we will give a brief introduction to the "Feigin's homomorphisms" and will see how they will help us to prove our main and fundamental theorems related to quantum Serre relations and screening operators.
In the second section, we will introduce Local integral of motions as the space of invariants of nilpotent part of quantum affine Lie algebras and will find two and three-point invariants in the case of Uq(sl2) by using Volkov's scheme.
In the third section, we will introduce lattice Virasoro algebras as the space of invariants of Borel part Uq(B+) oi Uq(g) for simple Lie algebra g and will find the set of generators of Lattice Virasoro algebra connected to sl^d Uq(sl2)
And as a new result, we found the set of some generators of lattice Virasoro algebra.
Keywords: Lattice W algebras, quantum groups, Feigin's homomorphisms.
Bibliography: 14 titles.
For citation:
F. Razavinia, 2020, "Local coordinate systems on quantum flag manifolds" , Chebyshevskii sbornik, vol. 21, no. 4, pp. 171-195.
1. Introduction
In this section we will introduce Feigin's homomorphisms and we will see that how they will help us to prove our main and fundamental theorems on screening operators. "Feigin's homomorphisms "was born in his new formulation on quantum Gelfand-Kirillov conjecture, which came on a public view at RIMS in 1992 for the nilpotent part Uq(n), that are now known as "Feigin's Conjecture".
In that mentioned talk, Feigin proposed the existence of a family of homomorphisms from a quantized enveloping algebra to rings of skew-polynomials. These "homomorphisms"are became very useful tools for to study the fraction field of quantized enveloping algebra. [6]
Feigin's homomorphisms on Uq (n)
Here we will briefly try to show that what are Feigin's homomorphisms and how they will guide us to reach and to prove that the screening operators are satisfying in quantum Serre relations.
Set C as an arbitrary svmmetrizable Cartan matrix of rank r, and n = n+ the standard maximal nilpotent sub-algebra in the Kac-Moodv algebra associated with C (thus, n is generated by the elements E1,..., Er satisfying in the Serre relations). As always Uq(n) is the quantized enveloping algebra of n. And A = (Aij) = (diCij) is the symmetric matrix corresponding to C for non-zero relatively prime integers d1,...,dn such that didij = djaji for all i,j. And set g as a Kac-Moodv Lie algebra attached to A, on generators Ei,Fi,Hi, 1 < i < n .[11] Let us to mention some of the structures related to g that we will use them here: the triangular decomposition g = n- © h © n+; the dual space h*; elements of h* will be referred to as weights; the root space decomposition n± = ®«eA±ga,g<xi = CEf, the root lattice A e h*, iai, • • • , an} C A+ c h* being the set of simple roots; the invariant bilinear form A x A ^ Z defined by < ai, aj >= diaij. [11] Set Ai and A2 as a A — graded associative algebras and define a q— twisted tensor product as the algebra A1^A2 isomorphic with A1 0 A2 as a linear space with multiplication given by (a1 0 a2) • (ai 0 a'2) := q<№l'a2>a1ai 0 a2a'2, where o!x = deg(ai) and a2 = deg(a2). And by this definition A10>A2 become a A— graded algebra.
Proposition 1.1. Set g an arbitrary Kac-Moodv algebra, then the map
A : U±(g) ^ U±(g)0U±(g)
(1)
Such that _
A(1) := 1 ® 1
A(E-) := Ei ® 1 + 1 ® Ei
A(Fi) := Fi ® 1 + 1 ® Fi
for 1 ^ i ^ n, is a homomorphism of associative algebras. f9]f6]
Remark 1.2. there is no such map as U±(g) ^ U± (g) (g>U±(g) in the case that g is an associative algebra. [9]
And as always after defining a co-multiplication, A, then we can extend it by a iteration as a sequence of maps fl]
A : U~(g) ^ U~(grn,n = 2,3,... (2)
—2 — —^^ — —^_1
determined by A = A, A = (A ® id) o A
Now set C[Xi] as a ring of polynomials in one variable and by equipping it by grading structure degXi = a^ for any simple root a^ we can regard it as a A— graded. By this grading there will be a morphism of A— graded associative algebras
& : U-(g) ^ C[Xi]: xi (3)
By following this construction for any sequence of simple roots fii1, ■ ■ ■ ,fiik, there will be a morphism A—
_u _ _
(&i <g> ) o A : U~ (g) ^ n]® ■ ■ ■ ®C[Xktk] (4)
(the cause of double indexation here is the appearance of ijS more than once in the sequence). And finally, C[Xii1 ]® ■ ■ ■ ®C[Xfcik] is an ^^^^^ra of skew polynomials C[Xii1, ■ ■ ■ ,Xkik], with A— grading Xtit = q<'ait>XtitX.sls ^r s > t. But let us to simplify it as X-Xj = q<degxi,degxj>x.x.. the one that we will use it always.
So very briefly we constructed the already mentioned family of morphisms (Feigin's homo-morphisms) from U~ (g) (the maximal nilpotent sub-algebra of a quantum group associated to an arbitrary Kac-Moodv algebra) to the algebra of skew polynomials.
2. The contribution between Quantum Serre relations and screening operators
Theorem 2.1. Set Q = q2 and points x\, ■■■ ,xn such th at xi Xj = Qxixj for i < j. And set = x\ + ■ ■ ■ + xn. \i Qn = 1 and xf = 0 for some natural number N, then we claim that
(^x)N = o
proof. It's straightforward, just needs to use q-calculation. □ 2.1. sl(3) case As we know, M2 =
2 -1 -1 2
2 —1 q2 q 1
12
q 1 q2
is the generalized Cartan matrix for s/(3). Set Mq2 = and call it Cartan type matrix related to M2.
Theorem 2.2. Suppose we have two different types of points Xi, Namely, set (x2i-i)i, that we will call them of type 1 and (x2i)i, that we will call them of type 2 for i G I = {1, 2}, and the following q— commutative relations:
Xj Xj' = q2Xj' Xj if j < j ' andj,j ' G {1, 3}andj = j' XiXv = q2Xi'Xi if i < i'andi, i' G {2, 4}andi = i'
XiXj — Q X j Xi if i < j
Set Ef = Eje/x2i+i and Ef = Eje/œ2j. We will call them screening operators. Then we claim that Ef and Ef are satisfying on quantum Serre relations:
(Ef)2Ef - [2}q Ef Ef Ef + Ef (Ef)2 = 0 (5)
(Ef)2Ef - [2}q Ef Ef Ef + Ef (Ef)2 =0
proof. It's straightforward, just needs to use q-calculation. □
Theorem 2.3. Prove Theorem 2.2 in a general case, i.e. Set points Xi G [Xi, ■■■ ,Xn} and Yi G [Y, ■ ■ ■ ,Yn} with the following relations;
X, X, = q2X3 Xi ïîi<j YiYj = q% Yi ifi<j Xz Yj = q-1 Yj Xl iii<j
and the screening operators Ef = Ek= 1Xi and El = Ek-=-¡Yj.
We claim that Ef and El are satisfying in quantum Serre relations.
Proof. Proof bv induction on k.
As we see in theorem 2.2, it's true for k = 2.
Suppose that is true for k = n, we will prove that it's true for k = n + 1. As we set it out, n is a nilpotent Lie algebra, so the Cartan sub-algebra of n is equal to n with Chevallv generators Ef and El as they are satisfying in quantum Serre relations.
So we can define Uq(n) :=< Ef, El|(Ef)2E1 - (q + q- 1)EfE\Ef + El(Ef)2 = 0 > . Let Cq [X} be the quantum polynomial ring in one variable. We define:
Uq(n)®Cq[X} :=< Ef, El,X|(Ef)2E^ - (q + q-1)Ef E?Ef + El(Ef)2 = 0, EfX = q2XEf, ElX = q-1XE\ >.
Here ® means quantum twisted tensor product.
We define the embedding Uq(n) ^ Uq(n)®Cq[X}: Ef ^ Ef + X ; El ^ E\. Claim 1:
(Ef + X)andE\ are satisfying on quantum Serre relations, proof of claim 1:
(Ef + X)2E1 - (q + q-1)(Ef + X)E?(Ef + X) + El(Ef + X)2 = (Ef )2E1 + EfXEl + XEf El + X 2E1 - (q + q-1)(EfE\Ef + X E^Ef + Ef E^X + X E\X ) + E^(Ef )2 + E^Ef X + E\X Ef + E\X2 = (Ef)2El - (q + q-1 ) EfE^Ef + E \(Ef)2 + ( q2XEfEf + XEfE\ + X2E1 - (q + q-1 )XE\Ef -(q 1 q-V)qXEfE\ - (q + q-v)q-1X21y1 + qXE\Ef + q- 1XE?Ef + q-2X2E\ = 0 + 0 = 0 So it's well defined. Now set X = Xn+1.
We will have the new operators Ef ' = Xi + ■■■ + Xn + Xn+1 and Ef = Y + ■■■ + Yn. Now define:
Uq (n) ^ Uq (n)®Cq [X} ^ Uq (n)®Cq [X}®CÇ [Y}
such that
f '
Ef i—> Ef + X i—> Ef
E1-^ E1-^ E1 + Y
Notice that Cq [X}®CÇ [Y} ^ C <X,Y|XF = q-1YX >.
And Define:
Uq(n)®Cq[X,Y] :=< Ef, E^,X,Y|Ef and E^ stisfying q-Serre relations and EfX = q2XEf, E^X = g- 1XE^, Ef Y = q-1YEf, E\Y = g2YE^, XY = g-1YX > . Claim 2:
Ef'' and(E\' + Y) are satisfying on quantum Serre relations, proof of claim 2:
(Ef )2(E^' + Y) — (q + q-1 )Ef'(E\' + Y)Ef + (E\' + Y)(Ef )2 = (Ef r)2E\' + (Ef ')2Y — (q + q-1) (Ef Ef Ef+EfY Ef )+Ef (Ef )2+Y (Ef )2 = (Ef )2Ef—(q+q-1)Ef Ef Ef+Ef (Ef )2+(Ef )2Y+ Ef Y = 0 + q-2Y(Ef )2 — (q + q-1)q-1Y(Ef)2 + Y(Ef)2 = 0 + 0 = 0. lets do some part of this computation that maybe make confusion:
(Ef)2Y = (Ef + Xn+x)2Y = (Ef )2Y+X2n+lY + EfXra+1Y+Xra+1EfY + q-2Y (Ef)2 + q-2YX2n+l + q-2Y EfXra+1 + q-2YXn+1Ei = q-2(Y ((Ef 2 + X^ + EfXra+1 + Xra+1Ef)) = q-2Y (Ef )2.
And Ef'Y = (Ef +Xn+1)Y = EfY + Xn+1Y = q-1YEf + q-1YXn+1 = q-1(Y(Ef + Xn+1)) = q-1YEf'. And by substituting these, we have the result. So our definition is well defined. Now set Y = Yn+1 and we are done. □
2.2. affinized Lie algebra sl(2)
As we know, M2 =
2 -2 22
is the generalized Cartan matrix for s ¿(2). Set Mq2 =
22 q2 q 2
22 q 2 q2
and call it Cartan type matrix related to M2.
sl(2) is satisfying in Theorems 2.2 and 2.3 as well; but what we need is just to change the quantum Serre relations in the following case:
(Ef)3El — (q2 + 1 + q-2)(ETl)2Ey1 Ef + (q2 + 1 + q-2)ETl E\ (Ef )2 — Ef (Ef)3 = 0 (6)
(El )3Ef — (q2 + 1 + q-2)(E\)2EfE^ + ( q2 + 1 + q~2)E\Ef(E? )2 — E\(E\ )3 = 0. —
X, X, = q2X3Xl Hi<j YiYj = q2Y3Y if i < j XiYj = q-2YJXl Hi <j
But lets try to prove it in the case of Laurent skew g—polvnomials C[X, X-1].
Theorem 2.4. Set points Xi g {Xi, ■■■ ,Xk} and X--1 g {X-1, ■■■ ,X-1} with the following relations;
f XtXj = q2XJXt Hi<j
\ Xa-1 = q~2X~1Xl Hi<j
and the screening operators Ef = E^=1Xi and Ef 1 = E^=1X-1.
Again we claim that Ef and Ef 1 are satisfying in quantum Serre relations (6).
Proof. Proof by induction on k.
For k = 2, Set Ef = xy + x2 and Ef = x-1 + x-1 and as we checked out, it's straightforward to
show that they are satisfying in quantum Serre relations (6).
Suppose that it's true for k = n components X\, ■ ■ ■ , xn. Again as before we define:
Ug(n) := {Ef, Ef1 |(Ef)3Ef1 — (q2 + 1 + (?"2)(Ef)2Ef1 Ef + (q2 + 1 + q~2)Ef1 Ef(Ef1 )2
—Sf-1 (Sf)3 = 0}
Define Uq(n) ^ Cq[X,X-1}; Sf ^ XX; Sf-1 ^ X-1 for A e C*
And define Uq(n) ^ Uq(n)®[/g(n) ^ Uq(n)®C[X, X-1}; Sf ^ Sf ® 1 + X®Sf ;Sf-1 ^ Sf-1 ® 1 + X-1®Sf-1.
And Uq(n) ^ Uq(n)®Uq(n)®-- ■®Uq(n) ^ C[X1,X-1}®C[X2,X2-1}®---®C[Xra,X-1} =
nterms
C[X1,X1 , ••• ,xn ,x- } □
3. Local integral of motions; Volkov's scheme
Set two screening operators
Sf-1 =S^-1, (7)
S = Sjxj ■
U 1-a.i q Si ) (Sj-
The project here is to find an analogue of R— matrix "R" such that
as we already saw, for these operators we have ( adqSf ) lJ (Sf) = 0.
(X1 + ••• + Xk)R(X1, ••• ,Xk) = R(X1, ••• ,Xk)(X1 + ••• + Xk) (8)
satisfy.
In this section we will try to find a solution for this equation as Volkov planed. We call these kind of solutions as "Local integral of motions".
In the sense of Feigin-Pugai [13], the main idea for to solve such kind of equations is to add "spectral parameter" ,0 to k points screening operator and to define an analogue of R— matrix:
(fix1 +X2 + ... +xk )R(X1,X2, ••• ,xk) = R(X1,X2, ••• ,xk )(X1 +X2 + ... + pxk), (9)
(0X-1 + x-1 +-----+ X-1)R(X1, X2,... , xk) = R(X1,X2, ... , xk)(x-1 + x-1 +-----+ fX-1)
Example Uq(s/2); two point invariants
X1 X2
In this case, our equations (9) will reduce to the following ones:
(fX1 + X2)R(X1 ,X2) = R(X1,X2)(X1 + 0X2), (10)
(fx-1 + x-1)R(x1, x2) = R(x1, x2)(x-1 + fx-1)
There is a solution for these equation in [2], but we are interested on re finding them again here. For to do this, let us change the equations (10) to the following one, for simplicity. Set a1 = x1x-1 and R(x1,x2) := R1,2,
(fx1 + x2)R(x1x-1) = R(x1x-1)(x1 + fx2), (11)
(fx-1 + x-1)R(x1x-1) = R(x1x-1)(x-1 + fx-1)
The solutions of these equations are identical to the previous one, we did this change just because to find the solutions in these ones are easier than the previous one and less confusing. Then both of (11) will reduce to this linear difference equation:
(f a1 + 1)R1,2(q-1v; f) = (u + f )ri,2(u] f), (12)
Let us do it for one of them (the first one) for to see the procedure (there is an identical approach for the second one):
(f3x\ + x2)R(x\x-1) = R(xix-1)(xi + fix2) (/3x1x-1 + 1)x2R(x1x-1) = R(x1 x-1)(x1x-1 + fi )x2
Set R(a1) = ECmoaY and then distribute x2 in it from the left and then bring it out to the right hand side, by using the g—commutation relations. The idea is to disappear x2 from the both sides by multiplying the equation by x-1 from the right side. So we have ((3x1x-1 + 1)R( q-1x1x-1)x2 = R(x1,x2 )(x1x-1 + fi)x2
(/3a,1 + 1)R1,2( q-1aV, /3) = RiMr; fi)(«1 + fi) (13)
Lets try to find R1,2(a1; fi):
(12) ^R1,2(a1;fi)'= f^RiM-1^; fi)
Q
_ fai+1 fig-1ai + 1 /3g-2ai+1 v — — i?
^I+f • q-1ai+i3 • q-2ai+f3 «Ь P) = ••• U=0 Ki,i+1
n<x fq %ai + 1 i=0 q-iai +f
But we need some thing more, so lets continue; For to find its recursive sequence we have to pass the following steps: (¡3x1 + X2)R(a1;fi) = R(a1;fi)(x1 + [3x2) (J3a1 + 1)R(q-1a1;fi) = R(cn;fi)(a1 + fi)
^ R(ai; I3)=z°=0cifia1 E<i=oCi,o Q-i + E°=0Cifl q-ia\ = E°=0Ci,oalf1 + E°=0Cifl[3a\
E=Ci-lflq-=+1f3a\ + E=Clflq-ia\ + Co,o = E=Ci-1,oa\ + E=Clflfial1 + pCo,a
E°=1(( q-i+1f3 — 1)Ci-1fl + (q-i — P)Cifl)a[ = 0 And then by comparing the coefficients in both side of the equation, we reach to the following key rule recursive sequence that we will use it for to find our final solution in the case of two points.
C0,0 = 1
Ci,o = fori = 1,...,
Co,o = f3Co,o ^ fi = 1
And now let us to set an general agreement for to simplify writing:
Set ((3)n := (1 — fi)(1 — q/3)(1 — q2/3) ■ ■ ■ (1 — qn-1(3) and let our summation be finite, i.e. set i £ {0, ■ ■ ■ ,n} and R(a1; fi) = Ef=0Ci,oa\ and in the next step we can extend our radius of convergence.
Now lets try to find it:
C = \-c-i+1|C
Ci,0 = q-i-I Ci-1,0
n _ 1-q-i+1f 1-q-i+2fjn
= q-i-f ■ q-i+1-I Ci-2,0
C = 1-q-i+1f 1-q-i+2f 1-q-i+3fC
= q-i-f ^ q-i+1-fI ^ q-i+2-fI
C = 1-q-i+1f 1-q-i+2f 1-q-i+3f 1-q-3f 1-q-2f 1-q-1f Ci'° = q-i-f ^ q-i+1-f ^ q-i+2-f q2-f ■ q-f ' 1-f
C = (1-q-i+1fi)(1-q-i+2fi)(1-q-i+3fi)-(1-q-3f)(1-q-2f)(1-q-1f) C0 = (q-i-f)( q-i+1-f)( q-i+2-f)( q-i+3-f)-( q2-f)(q -f)(1-f)
C = _(1-q-i+1f)(1-q-i+2f)(1-q-i+3f)...(1-q-3 f)(1-q-2 f)(1-q-1 f)_
Ci'° q-i(1-qifi) q-i+1(1-qi-1fi) q-i+2(1-qi-2fi) q-i+3(1-qi-3f)■■■ q-2(1-q2fi) q-1(1-qfi)(1-fi)
C = _(1-q-i+1f)(1-q-i+2f)(1-q-i+3py„(1-q-3 f)(1-q-2 f)(1-q-1 f)_
Ci'° q-i q-i+1 q-i+2 q-i+3-q-2 q-1(1-qif)(1-qi-1f)(1-qi-2f)(1-qi-3P)-(1-q2 f)(1-qf)(1-f)
C = _(1-q-i+1f)(1-q-i+2f)(1-q-i+3f)_
= q(-i+0) + (-i+1) + (-i+2) + (-i+3) + --- + (-i+(i-2)) + (-i+(i-1)) + (-i+(i+0))
(1-g-3f){1-g-2f){1-g-1ff)
(1-qHf)(1-qi-1f)(1-qi-2f)(1-qi-3ff)-(1-q2f)(1-q f)(1-f)
In infinity when i ^ we have q 1 ^ 1; So we have:
C = (1-q-i+1 i3)(1-q-i+2i3)(1-q-i+3i3)
= g(0) + (1) + (2) + (3) + - + ((i-2)) + ((i-1)) + ((i+0))
(1-q-3l3)(1-q-2l3)(1-q-1l3)
(1-qi/3)(1-qi-1 P)(1-qi-2P)(1-qi-3P)-(1-q2P)(1-qP)(1-P)
rt,0 — n(n-1)
q 2 (qP)n
r — (1-g-i q/3)(1—q-i+1g^)(1-g-i+2q^)...(l-q-4q^)(l-q-3q^)(l-q-2q^)
rl,0 — n(n-1)
_ _ q^^ (,qP)n _ _ _
r — 1---i-i )№)(. -g-i+1)(^)( -g-*+2)-(^)( -q-i)(q[i)( ^-g-3)(gl3)( ^-g-2)(1-1 )
ri,0 — n(n-1)
q^^ (ql3)n (-q/3)-1
7 — 1 7 — 2 7 — 3 3 2 1
r — (q/3)"-1(q-iq-i+1q-'+2 ■■■q-2q-1)( y-1)( y-1)( -1)-( fr-1)( fr-1)( fr-1)(1-1 )
ri,0 — n(n-1)
q^^ (q/3)n(-ql3)-1 r — (-qfi)n(q ^^ )(1-)(1-)(1-><1-1 )(1-1 )(1-1 )(1-1 )
ri,0 — n(n-1)
(qp)r.
ri,o — : ^(i4)
(— P )n( i )n
(9 P)r
Example Uq(sl2); three point invariants
As what we had for previous example in two points; we will proceed the same steps for to find the solution of the equation (8) for three points x0,x\,x2.
Set (Xi = xiX-^; such that a^aj = q(Xj(Xi for i,jel as usual, and R(a0, a) = Sn,mCn,ma'^a^.
R
(fxo +xi +X2)R(ao,a ;f) = R(ao,a\;f)(xo +fx2), (15)
The process is exactly same as the previous one, so we will skip writing them here.
For this equation We got the following recursive sequences, that will guide us to reach to our main
solution for n,m = 1, ■■■, .
Co,o = 1
C = q-m + 1-q-n-m+2l3 c I 1-q-n-m+1 c
Cn,m — q-m-p Cn-1,m-1 + q-m-p Cn,m-1
n _ 1-q-rn+1 -C0,m = p-q-m C0,m-1
And bv considering the second sequence as our main key, and following it; we arrived to a nice and important sequence:
'in i in —11
Cn,m = (f)m(Qf )n-m 1 j
m = 0
calculation.
4. Lattice Virasoro algebra
In this section we are interested on solutions S1x of system of difference equation
{XiXj — qXjXi deg&u ) — 0 [E+£X,, E^j, — 0
that will be a generator of lattice Virasoro algebra.
If we be able to find such kind of solution; then we can extend it to an another generator by a shift operator:
S2X = S^ [X1 ^ X2, X2 ^ X3,X3 ^ X4, ■ ■ ■ (17)
S3^ = Sf[x2 ^ X3,X3 ^ X4,X4 ^ X5, ■ ■ ■
where S^ = S^ (X1,X2, ■■■ ,xk).
Lattice Virasoro algebra connected to sl2
Here as always, we have the g—commutation relation XiXj = qXjXi, i < j between the points in sI2.
Let us try to find three points invariants; this means to try to solve the following system of difference equation:
(XiXj — qXjXi deg(Elx)
0
[ (X1 +X2 + X3)SU(X1, X2,X3) = S^ (X1,X2, X3)(X1 +X2 + X3) One can find easily the trivial solutions of the second equation as follows:
|S11^(Xl, X2, X3) = X1 + X2 + X3S12x(X1 ,X2,X3) = X1X2 1X3
but as we see, non of them have zero grading. So we should find another solution.
By just keeping to look at them for a while, we can see that by multiplying these kind of solutions, one can find a zero grading expression, but it's not satisfying for these two ones. Again we note that for a solution; it's inverse is again a solution, so by this remark, we have two options here. We can inverse S11;c or S12x and then multiply it with the other one. In both case we will have same set of generators except that in the first case (inverse of S11^), lattice Virasoro algebra is generated by elements of form S^ = XiX~+1Xi+2(Xi +Xi+1 + Xi+2)-1 and in the second case (inverse of S12.J, lattice Virasoro algebra is generated by elements of form S^ = (Xi + Xi+1 + Xi+2)X-1 Xi+1X~+2. And by a simple fact that our space of working is closed under multiplication, so these new recently found objects can be a trivial solution for our system of difference equation. And then by shift operators (17), we will have the set of generators for our lattice Virasoro algebra connected to s12.
Lattice Virasoro algebra connected to Uq(sl2)
Set A = the algebra of polynomials in variables q, Xi over C for i e /(our ordered
yx^x j qx j Xj/)
index set), such that
{qxi = Xiq fori e I
XiXj — C^X j Xi <
Our first project is to extend the usual binomial expansion to this algebra, for example we can see the shape of such expansion in a lower exponent 3:
(xi + Xj) - (xi + xj )(xi + xj ^(xi + xj)
- ry . ry . ry . I ry . ry . ry . I ry . ry . ry . I ry . ry . ry . I ry . ry . ry . I ry . ry . ry . I ry . ry . ry . I ry . ry . ry .
- JbiJbiJbi | JUiJUjJUi | JUjJUiJUi | JbjJbjJbi\ JUiJUiJUj | JbiJbjJbj\ JbjJbiJbj\ <Ajjd.jjd.jj
22
- Jb i Jb i Jb i | y J I J I J J 1 ' y J 1 1 ' y JbjJbjJbi\ K^JbjJbjJbi |
ry.ry.rY».
- ry3 I ff ry ■ ry2 I ry ■ ry2 I ry2 ry ■ I f~f2 ry ■ ry»2 I ff2 ry2 ry» ■ I ffy2 - I Of3
- Jj ^ \ K^JjjJJ,^ \ JJ jJJ ^ \ JJ \ y JJ jJJ ^ \ y JJ \ y^^^ T Js j
= x\ + (1 + q + q2)xjxi + (1 + q + q2)xjxf + x'3
= vk=0\k) qxj xi
But for to prove it in a general case, we will use some techniques from combinatorics:
Suppose xj and xi as above and set w as a word formed bv xj and xi. Then it's easy to see that any such kind of words can be permuted to xjxk along with a factor power of q, by using the g—commutation rule. For example, xixjxjxixixjxjxixixjxi = q^x^x6, as the first xj, should pass
xj xi xj xi xj
sixth will be stable.
Now according to this fact , each word w consist of k x^s and n — k = I x/s in (xi + xj)n. That corresponds to a partition which lies inside an (n — k) x k rectangle. On the other hand each such partition corresponds to a unique word w. Lets look at our example again; we have w = xixjxjxixixjxjxixixjxi, and the partition is 533110. If w = qmxfn-kxk, then as we see m is the sum of the parts of the partition. And the generating function for all partitions lie inside an (n — k) x k rectangle is the definition of the q — binomial coefficient (nnk)q = (n)q■ So for a positive n
(xi + xj)n = ^ kx>k
But what will happen for the negative exponents?
nk n
- n
-
(n\ = (l-gr
k =
(l-gk ){l-g k-l)...(l-gl)
n - n
( k )q
(l-qn-1)(l-qn-2)---(l-q r-k+1)
(l-gk )(l-g k-1)...(l-g1)
l-q-1n)(l-q-1n — 1)(l—q-1n-2)-(l-q-1n+k-1)
(l-q-1-k)(l-g-1-k+1)...(l-q-1-1)
l- q-1"+k-1)(l-g-1n + k — 2)...(l-g-1r + k-(k-2))(l-g-1r+k-(k-1))(l-q-1n + k-(k))
(q-1)-k ((g-1)k-l))( g-1)-k+1({ g-1)k-1-l))-( g-1)-1(( Q-1)1-l))
l- g-1r+k-1)(l-q-1r+k-2)...(l-q-1r+k-(k-2))(l-q-1r+k-(k-1))(l-q-1"+k-(k))
((q-1)-k( q-1)-k+1-(q-1)1)(-l)k (l-( Q-1)k )(l-( g-1)k-1)~(l-( Q-1)1)
l- q-1"+k-1)(l-q-1"+k-2)...(l-q-1r+k-(k-2))(l-q-1r+k-(k-1))(l- q-1r+k-(k))
k(-k+1)
(q )(-l)k ((l-( q-1)k)(l-(q-1)k-1 )-(l-( q-1)1))
-l)-kq-k(-2k+1) (^q-1r+k-1)(l-q-1n+k-2)...(l-q-1r+k-(k-2))(l-q-1r+k-(k-1))(l-q-1r+k-(k))
(l-(g-1)k)(l-( q-1)k-1)...(l-(q-1)1)
(-l)k g(k)(l-g-1r + k-1 )...(l-g-1r)
(l-( q-1)k )(l-(g-1)k-1)...(l-( q-1)1) = ( — 1)kq(k) (n+kk-l)q-1.
So as what we had for a positive exponent, we will have the result for negative exponent as follows:
(xi + x3)-n = E£o ( n) x-n-kxk = sr=c(—1)k<?(k) (n + k ^ fin-kx
Remark 4.1. And it's identical to write the summation (18) from to 0 as follows:
(xi + xj)-n = ( k) _y+kx-k = Vl=-^(—1)kq^ (n —k 1) x-n+kx-k (19)
Formulation for to extend to four and more invariant points
Set = U- + Elk=lXi + U+, where U+ = X+~+lXi and U- = E0=-^Xi Set ( F* k)(0) = f(xl, ■■■ ,xk) = ECpX1?1 ■ ■ ■ Xlk such that [ U+, (F*k)(0)] = U+(Ffik)(0) — (F*k)(0)U+
= U+(Ffik )(0) — (F*k )(0)Xk+l — (F^k )(0)Xk+2
,)(0) (
= U+{Flk)(0) - ^^)(0) (Xfc+1 + Xk+2 + ■ ■ ■ )(Ff>fc)(0) = U+(Ffk)(0) - ^^)(0)U+ (Ffk)(0)
= (1 — F?, k)(0 )U+( Fftk )(0) (20)
U-
[ U-, (F*k)(0)] = (1 — q^(F1,k)(0))U-(Fi,k)(0) (21)
If we suppose deg(Ffk)(0) = 0, then both of [U+, (Ffk)(0)^d [U-, (Ffk)(0)] will be zero and
we will have to check the correctness of [Sk=l, (F*k)(0)] = 0. If it was true? then we will have a generator for lattice Virasoro algebra and by the shift operator, we will have all set of generators for it.
Let us define a Poisson bracket as Drinfeld defined and then compute some results by using
r
{Xj ,Xin} := Limq^l—i-f- and then we have: adxiXn = 0, in both classical and quantum case.
adxiX'j = (1 — qn)XiX'n and adXjX'i = (1 — Qn)XjX!n for i < j in quantum case. adXiXjn = ^d adXjXin = 0 for i < j in classical case.
X Xin
{X, Xin} = Limq^l(1 — gn)XiXr = —nX.X'ndxJori < 1 (22)
1—
{X, X™} = Limq^l(1 — \ -)XlX" = nX^dxJori > 1 (23)
1—
For example we have {Xi,X2} = XiX2dX2
By using this operators in the case of the equation (21) we see that this part when q ^ 1 will be zero and so after this time we just will deal with SXi = Vkk=lXi + U+.
Xi
follows:
And let us consider E^ Fj,, Hj, as the generators for Uq(sl2), then Ej, and H will produce the Borel part B+; One of the ways that we can act B+ on the C[Xi,Xi-lj is as follows
k : B+ x C[Xi,X"lj ^ C[Xi,X"lj : (Ej, P) ^ *(Ej)P := ad^xP = [XX, P]g (24) k : B+ x C[Xj, X~l] ^ C[Xj, X-1] : (Hi, P) ^ k(H)P :=< aj, degP > P (25)
where ai is a simple root related to Hi and P an arbitrary homogeneous elenient of C[Xi,X~l]
the functional basis of space Invjjq(b+ (C[Xi, X~l\). And for to find these generators we need to solve the following functional equations;[13]
*, ] =0 andHiEix = 0(**)
Xi
for equations (**)?
dim(B+) + 1. [13] So in the case of sl2 it will be 3, the number of variables.[13] Now let us to go back to our example; Xlf (X0) = f(q-lX0)Xl Set q = exp(H);
= f(e-HX0)Xl When q^ 1 then e-H ^ (1 — H) and eH ^ (1 + H); = /((1 — H)X0)Xl = f(X0 — HX0)Xl = ( f(X0) — f(HX0))Xl = (f(X0) — Hf (X0))Xl = ( f(X0) — XlHf (X0)) = (/(X0) — XlX0&Xo f(X0))
^ {Xl, f (X0)} = —XlX0dXof (X0) So in general if we repeat the process for any Xj<]_, we will have
{Xj, f(X3<j)} = — XjXjdX, f(Xj<i) (26)
(**)
form
E&ix = (Xl(Xl +X2 + X3 + U+)dX1 +X2(X2 +X3 + U+)dX2 +X3U+dX3 + U\dj+ = 0 (27) HjEiX = (XldX1 + X2dX2 + X3dX3 + U+dj+ = 0
Xl , X2 X3
As well as there is a same process with a minor differ for when we have j > 1 ;
{Xj, f(X3>j)} = XjXjdXj f(Xj>i) (28)
Xi ( Xi) = ( Xi) Xi = ( H Xi) Xi
= /((1 + H)Xi)Xj = f(Xj + HXj)Xj = f(Xj )Xj + XjHf (Xj) = f(Xj)Xj + Xj2dXi f(Xj)
{Xj, f(Xj)} =X?dXz f(Xj) (29)
Now set f = f(Xl,X2) = (F*2)(-(here (—l) means that our polynomial is of the degree — 2 ) then by previous definition of dj+, we have dj+f = 0.
Set (Ff2)(2) = [XX, (Ff2)(-1 )]g = adsx(Ff2)(-1), then by using the previous discussion we have ( Ff2)(2) = ( Ff2)(2)( U+,XUX2).
( 2)
F = dj+
H = U+dj+ + XdX1 + X2dX2 (30)
E = U\dj+ + (x2 + XX + XU+)dX1 + (X22 + X2U+)dX2
We need the highest weight vector of this representation that is the solution of equations (30). So we should have E(Ff2)(- 2) = adsx [XX, (Ff2)(-1)] = 0. There is a solution for these equations in [13]. Here we use the same solution and procedure.
The Idea is to suppose existing of the local fields [13] ( Ffk)(0), (F2k+i)(0"1, ■ ■ ■ (here (0) means According to [13] we try to find the exchange algebra relations
( Fifc)« = [E*, [E*, [••• , [E*, (F^)(0)] ••• ]]]( z - Umes) (31)
And then again will use the shift operator and will shift it once for to find another module as follows (F^,k+1 )U) = [Ex, [Ex, [... , [Ex, (F^k+i)(0)] ■ •• ]]](j - Umes)
Let us to proceed as what we planned: Set k e {2} and i = -\
(FfJ-2) =X?X- 2 (Xi + X2)-2 (32)
,2) 2 = X1
as [13]. Then
[Ex, ( F^)(-2)] = [ U- + (Xi +X2) + U+,X?X2 2 (Xi +X2)-2]
= (1 - q-2)U+X2X22 (Xi +X2)-2 [13]. Where U+ = E+^. Let us to call it (Ff2)( 1 because it's degree is 2 and to find another module from it by using shift X1 ^ X3 and X2 ^ X4 as follows:
(F0-2) = X^X- 2 (X3 + X4)-1 (33)
And again in a same process we will have
1 1
(F02) = [Ex,F0-2] = (1 - q-1 )(U+ -X3 -X4)X32X4-2(X3 +X4)-2 [13] (34)
Now let us multiply the equations (33) and (34) (because we need zero degree) with each other for to see what will happen?
(Ffo)(-1 )(Fl4)( 2
(35)
= Xi2X-2(Xi + X2)-2((1 - q-2)(U+ - X3 - X4)X32X4-2 (X3 + X4)-2) = x\X—* (Xi + X2)-2U+ X2X4-i2 (X3 + X4)-2 -X2X— 2 (Xi + X2)-2X3X32X4- 2 (X3 + X4)-2 -X2X-2 (Xi + X2)-2X4X3^X4- 2 (X3 + X4)-2 -q - 2X2X2- 2 (Xi +X2)-2U+ X^X— 2 (X3 + X4)-2 -q - 2XfX2- 2 (Xi +X2)-^2X3X32X:- 2 (X3 +X4)-2 -q - 2X 2 X— 2 (X! +X2)-2X4X3^X4- 2 (X3 +X4)-2 And again let us proceed as [13] and to find:
-q2 (Ff 2)(2 )(F?4)(-2)
= q2 ((1 - q-2)U+X^X2 2 (Xi +X2)-2)X32 X4 2 (X3 + X4)-2 = -q 2u+X2X-2 (X! +X2)-2X1X4- 2 (X3 + X4)-2
+U+X2 x2
(Xi + X2)-2Xf X- 2(X3 + x-)-1
= -x2 X2 1 (Xi + X2)-2 u+xl x41 (X3 + X-)-2
+q'
■tX2 x21
(Xi + X2)-1U+ X2X-2(X3 + X-)-1
And again by following [13], by adding the equations (35) and (36) for to find an object with zero grading that can be a four point invariant generator of lattice Virasoro algebra
Pi,4 = (Ff)(-1) №)( 1) - Q 2 №)( 2 H^-2)
(37)
= -(1 + q-2 )X2 X2 2 № + X2)-2 (X3 + X-)X|X- 2 (X3 + X-)-2 But now let us follow precisely the notation from [13] for to not be confused
Ai,3 = X2X2 1 (Xi + X2)-2(X3 + X-)X2X- 2 (X3 + X-)-1
(38)
that is a four point generator of lattice Virasoro algebra, but we are not looking for such kind of solutions, because to extend it to a general form is somehow difficult. So we are looking for a simple solution.
Now let us to define another such kind of solutions as we experienced.
(F|3)(-1) := XfX3-2(X2 + X3)-2
(39)
then define
(^3)( 1) := (1 - 2)(0+ - *3)X22X3 2 (X2 + X3)-2
pi,3 = (^f)(-2) (^ 1) - Q1 №)( 1 )(F2^^3)(-2)
(40)
(41)
Let us calculate it;
(^2)(-1) (^2,3 )(1)
(42)
= X2X2 1 (Xi + X2)-1 (1 - g-1 )(tf+ - X3)X22X3 1
= x? X21
1
(Xi + X2)-2 - X3)X22X3 2
2 (X2 + X3)'
(X2 + X3)- 2
= g- 2X2 X2 2 (X: + X2)-2 - X3)X22X3 2 (X2 + X3)
= x2 x2-1 (x: + X2)-2 ^+x2 x3-1 (X2 + X3)-2
1 _1 1 1 _1 1
-X2 X2 2
1 1 ' 2
(X! + X2)- 2 X3X22 X3 2 (X2 + X3)- 2 -g-1X2 x2- 2 (X: + X2)-2 ^+x2 x3-1 (X2 + X3)-
+9
1 „-1
1
"2XIX2 2
(X: + X2)-2X3X21 X3 2(X2 + X3)-
Now let us calculate the other part
-q1 (Ff)( 2) (^2,3)(-1)
= -q2(1 - g-2)^+X2X2 2 (Xi + X2)-1X2x3 2 = -q1 X2 X-1 (Xi + X2)-1X2 x3- 2 (X2 + X3)
(X2 + X3)- 2
= -xf X-1 (Xi + x2)- 2 ^+x2 x3 2 11-1 1 1 _1 +g-1 Xi2X- J- Y„\- 1TT. YtY 2
+^+X2 X- 2 (Xi + X2)-2 X2 x3- 2 (X2 + X3)-1
(X2 + X3)-2 (Xi + X2)-2 ^+X2 x3- 2 (x2 + X3)-1
And again by following [13], by adding the equations (42) and (43), we have another generator of lattice Virasoro algebra, but of three point invariant:
Pi,3 = ( q-2 - 2№ + X2)-2X3X^X3 2(X2 + X3)-2
(44)
But now let us follow precisely the notation from [13] for to not be confused Ai,2 = Xi2X— 2 (Xi +X2)-2 X3XJ2 X— 2 (X2 +X3)-2
(Ai;3)-i = (X3 + X4)2X^X3 2(X3 +X4)-i(Xi + X2)2X|Xj is again a solution.
Now let us multiply the equations (45) and (46) for to see what will happen?
(45)
(46)
(Ai;a)-1 Ai,2 =
(X3+X4) 2 x42 x3- 2 (X3+X4)-1 (Xi+X2)1 xj x-1 x2 x2- 2 (Xi+X2)- 2 x3xj x-1 (X2+X3)- 2 = (X3 + X4) 2xjx-2 (X3 + x4)-ix3xlx3-1 (X2 + X3)-2
by using the equations (45) and (46) we have the following result which has been mentioned in
[13]:
^ 1 1
£ = (X3 + X4)-2x2x2 (X2 + X3)
(47)
Our next goal is to prove that S is a generator for lattice Virasoro algebra. Now let us to set some q—commutation relations such that any other relation comes from them by using the inverse and multiplication operators.
xx = i
qX3Xz
XX-i =
-iX7iXi
XX- 2
j
= q- 1X- 1X, 11
XtXf = 1
x2x =
_1 X-1x[
x{xr2
xjx2 -
xhj 2
x" 1xj2
1
XX2 x!x~i i ХгХ12
q2X3 2 Хг = q2x~
1 1 1
q-2ХгХ2
-, 1 i = q- 1X-i X2
1 _1 1
= Q22X} 2Xi
-- q^XjXf
1 -1 1 = q- 4X, 2Хг2
11
ОТ 4 X,2 Хг
2
-q-2X?X3
= T 2x~ixl
= q2x; 2Хг
2
Set Sx = Sj^+^Xj. We want to show that Sx will commute with
£ = (X3 +X4) 2X42X32 (X2 +X3) 2
by using the usual commutator[x, y] = xy — yx, i.e. to show that the equation [Sx, S] = 0 is correct.
For to show it, one can easily check that the contributions of many entries vanishes. Namely, the elements Xj with indexes j from minus infinity to 1 and from 5 to plus infinity definitely commute
due to the rules mentioned above. And after that we can concentrate on the sum + £3 + ^ So lets do it,
_1 1 1 _1
[®2 + ^3 + (®3 + 2 (®2 + ^3) 2 ]
? _1 1 1 _1 _1 1 1 = («2 + ^3 + ®-)((®3 + 2 (®2 + ^3) 2 ) - ((®3 + 2 ^3 (^2 +
X3)-1 )(®2 + ^3 +
for to do this job, let us divide the project to the following small projects. We must demonstrate that the following equations are satisfying.
_1 1 1 _1 ? _1 1 1 _1
(®2)((®3 + 2 (®2 + ^3) 2 ) = ((«3 + 2 (®2 + ^3) 2 )(®2) (48)
(®3)((®3 + 2 (®2 + £3)
1 1 U2i
1 ? 1 1 1 1 "2 ) = ((®3 + 2(®2 + ^3)-2)(®3)
(«-)((®3 + 1 ^3 (®2 + £3) 1 ) = ((®3 + ^^
1 1
2 2 L- X3
(«2 + ^3) 1 X^)
(49)
(50)
Let start with equation (48):
_1 1 1 _1 ? _1 1 1 _1
(®2)((®3 + 2 ^3 (®2 + £3) 2 ) = ((®3 + 2 ^3 (®2 + £3) 2 )(®2) multiply the equation from the left side with (®3 + 1.
1 ? 11
(«3 + ^) 2^2(^3 + 2^3 (®2 + £3)-2 = x^ («2 + ^3)-2£2 Now we have two options for to move, it can go to the left direction and act on the (#3 + 1 or to the right direction and act on the (#3 + 2.
As we see, in (®3 + x^-1, our summation will take part from k = 0 to k = We can use of this fact for to skip some factors in the powers of q that will appear in our calculation when that our partners in action are not two.
In the case of in the left hand side we have no problem in our action; because £2 will act on £3 and ^ and we have two different partners in our action. So we can move £2 to the left hand side and act it on (®3 + 1:
1 1 ____1 _ _ 1 r _ _ 1
(Sfc=0(I)a^ ^Xfc)^2(^3 + 1 ®3 (®2 + ^3)-2 = ®3 (®2 + ^3)-1 ^2
1 -fc
^k) q
1 , 1s
_1 1 1 \ o ry 2 ry 2 I -1 ^ A ^ Q
1 ? 11.
ry 2 /Y> 2 I -JLj A Jb O
— - 1 1— fc 111 1 ? 11 1
^2(£fc=0 (2) 2-fc9-fc)(®3 + 2(®2 + Z3)-1 = (®2 + ^3)-2^2
1 ,1, ' ' ' ' ' ' 2
2 (Sfc=0( fc) fc^3)(^3 + 2 ®3 (®2 + £3) 1 = ®3 (®2 + ^3) 1 ^2
1?
q 2 ^2^-2 ^3 (®2 + ^3) 2 = a- (®2 + ^3) 2 ^2
Now multiplv both side of the equation from the right hand side bv («2 + ®3) 1; we will have:
_1 1 1 ? 11 _1 1 "
q 2 £2^2 ^3 = ^3 (®2 + ^3) 2 ^2 (®2 + ^3) 2
In the right hand side, we have just one partner in action, i.e. we just have the action of £2 on £3; So there will be a factor of the power of q. And as I mentioned it already for to skip this factor we will use of limit in infinity. So should act on (®2 + £3)-1. Lets see what will happen; (here we have to use the equation (18) for to expand (#2 + £3)-2):
1 11 ? 1 -q-1 £2^2 ^3 = a- (-1)fc9
""+"( ^fc')
„ 2+fc™—fc
)®2(®2 + £3) 2
_1 11 ? 11 q 2 ^3 = '
(sfc=-^(-1)fc ? ^^ (-2 r)
^(^+1) /-2-fcA 1 ^-1-fc
J
3
)^2(^2 + ^3) 2
Bv using the fact that when k ^ then q 2 fc ^ q 2 we have:
21 11 ? 111, ,1
Q
<T
11 ? 1 1 1 1
"2 x^ == ^1 + ®3)1
1 11 ? 1 i 1 1
" 2 x^ = q 2 (®2 + £3) 2
1 11 ? 1 1111 1
"2 ^3 = q 2 2 £2 g-2 (®2 + ) 2
-1 31 1 -1 3
2 ^32 = q 2 ^32
g
So it's correct in the case of x2.
The case of x3 and x4 are almost identical to the case of x2 with just some minor differs.
x
_i 1 1 _2 ? _i 2 2 _2 (X3 )((x3 +X4) 2X42x| (X2 +x3) 2 ) = ((x3 +X4) 2 x42x32 (X2 + X3) 2 )(x3)
multiply the equation with (x3 + x4)2 from the left hand side; * 2 _2 2 2 _i ? 2 2 _i (X3 + X4) 2X3(x3 +X4) 2X42X2 (X2 + X3) 2 = X42X2 (X2 + X3) 2 X3
2 3 _2_fc 2 2 2 ? 2 2 2
(x3 + X4)2X3(E+=°0(-1)kQ(2) ("fc2)q-lX4 2 X3)x42x2 (X2 + X3)-2 = X_2 Xg (X2 + X3)-2X3
3 1 2 k 2 2 i^!!
" o— fc ,„ ,„N o o / . x_I ? O o
fc ^ g-!
-,-2-fc
1 3 1 _1_к 2 2 i ? 2 2 i
(x3 + X4)2(£+=0(-1)kg(2)(T2) _2q-2-2x4 2 qkxk)x3x|x3 (x2 +x3)-2 — x4x3 (x2 + x3)-2x3
By using the fact that when k ^ then q 2 fc ^ q 2 we have:
22 2 2 _x ? 2 2 _x
q 2X3X_X2 (X2 + X3) 2 = X_X2 (X2 + x3) 2X3
multiply the equation with (x2 + x3)2 from the right hand side;
_i '22 ? 2 2 _x x
q 2 ^x42 X2 = X42X2 (X2 + X3) 2X3 (X2 + x3)2
2 2 2 ? 2 2 f 3 _2_fc 2
<T2x3x42x32 = x_x32 (E+=^(-1)fc<?(2)Cfc2) 2 xfc)x3(X2 +x3)2
here again we have to change our equation to (18);
_2 22 ? 22 0 , fe(fc+i) 3_fc\ "2 +fc _fc 2
q 2 x3x42 x32 = x42 x32 (Efc=_^(-1)fc<7 2 ( -fc ) 2 x2 )x3(X2 + x3)2
— 2 2 2 ? 2 2 0 fc fe(fe + 2) , 3_fc\ _2 +fc , _fc 2
q 2x3x|x3 = x_x32x3(Efc=-TO(-1)fc<? 2 ( -fc ) 2 q-fcx2 )(x2 + x3)2
Bv using the fact that when k ^ then q-fc ^ 1 we have:
\ 22 ? 22 q 2 ^x42 x2 = x42x2 x3
_2 11 _ 2 2 2 q 2 ^x42 x2 =q 2x3x|x3
x4
_2 2 2 ? _2 2 2
(X4)((X3 +X4) 2 X42X32 (X2 + X3) 2 ) = ((X3 +X4) 2X42X32 (X2 + X3) 2 )(X4)
multiply the equation with (x3 + x4)2 from the left hand side;
"2 2 2 ? 2 2
(x3 + X4) 2x4(x3 +X4) 2X42X2 (x2 + X3) 2 = x42x2 (x2 + X3) 2 x4
2 + h (k) i-3\ -1-fc 22 ? 22
(x3 + X4) 2X4(E+=°0(-1)fcq(2) { fc2) q_1X4 2 Xfc )x_ X 2 (X2 + X3) 2 =X42 X32 (X2 + X3) 2 X4 2 3 _2_fc 2 2 2 ? 2 2 2
(x3 + X4)2 (E+=^(-1)fcg(2) (_fc2)q-2X4 2 (?-fcXfc)X_X^X2 (X2 + x3)_2 = x42x2 (X2 + X3)_2X4
By using the fact that when k ^ then q-fc ^ 1 we have:
" 2 2 ? 2 2 x4x4!x2 (x2 + X3) 2 = x42x2 (x2 + X3) 2 x4
multiply the equation with (x2 + x3)2 from the left hand side;
2 2 ? 2 2 2
X4X4 Xg — X4 Xg (x2 + X3) 2X4(x2 + X3) 2
22 ? 2 2 1 22 2 _fc
x_xlx2 = X4X2 (x2 + x3) 2x4(Efc=0(fc)gx2 x2)
22 ? 22 i2i 1 2—fc
x4x4x2 = X42x2 (x2 + x3)_2(Efc=0(2) q_2 +fcx3 q-fcx2)x4
2 2? 22 1 1 2 1 2_fc
2™2 ™ 2™2/™ 1 ™ n — 2 ^ - 2^2 (2\ m 2 fc™fc
x4x42x32 — x4 x2 (x2 +xz ) 2q 2 (E2=o{k) qx 2 xk )x4
2 2 ? 2 2 _2
rf> . rf> 2 rf> 2 - ry> 2 ry 2 ri 2 ^У t
x4x4 xo — x4 xo 2 x4
22 ? _ 2 222
ry . ry2 ry2 - rt 2 ry.ry2yi2^f2
x4x4 xo — L[ 2 x4x4 L[ 2 xo
2 2 2 2 2 2 2 2
x4 x4 x — x4 x4 x
And then again by using the shift operators (17), we will have the set of generators Eix for our lattice Virasoro algebra connected to Uq(sl2).
Generators of lattice Virasoro algebra coming from 2-dimensional representation
of sl2
Claim 4.3. , (®3 + + £3)-i] = 0
Proof. The proof is identical to the proof for fractional exponent 2,above. □ Claim 4.4. [E^-^j, («2 + £3 + ^)-i(^3 + ^^(^i + ^2)-i] = 0
Proof. The proof is identical to the proof for fractional exponent 2- Just what you need is to set £3 + ^ = The rest is exactly identical. □
Claim 4.5. [E^-^, («2 +-----+ Xfc)-i(^3 +-----+ Xfc+ ^2)-i] = 0
Proof. Proof by induction on k. It is true for k = 3.
Suppose that it is true for k - 1 component. Then for k component we have:
, ((«2 +-----+ ®fc-i) + Xfc)-i((^3 +-----+ ®fc-i) + Xfc)«2(®i + ^2)-i]
Set («2 +-----+ ®fc-i) = ®'fc-i;
= [Ej=+~^, («'fc-i + Xfc)-i(^fc-i + ^fc)«2(®i + ^2)-i]; So the rest will come from k = 3 and we are done. □
And then by using the shift operators (17), we will have the set of all generators.
Results; Generators of lattice Virasoro algebra coming from 3 and 4-dimensional representation of sl2
Let us suppose the following 3-dimensional representation of s/2. The process of defining this
(30)
Define:
F = d(u+ _X3)
H = U+du+ + Xi dXl + X2%2 + ^3^X3 (51)
E = - X3)29(u+-X3) + №2 + + XiX3 + Xi(^+ - X3))9X1 +(X22 + X2X3 + X2(^+ - X3))9x2 + (*! + - ^3))9X3
As before set
(^,fc+i,fc+2)W = [EX, [XX, [■ ■ ■ , [EX, (F^+i)«]] ■ ■ ■ ]] (^fc+i,fc+2,fc+3)(j) = [EX, [EX, [■ ■ ■ , [EX, (F^fc,fc+i)(^)]] ■ ■ ■ ]
Set k = 1 and i = -i and set (^i,2,3)(-2) = X12X2 2 (X2 + X3)-^ because it's satisfying in the
relation [EX, [EX, (Ff,2,3)(-2)]] = 0 and is our highest vector in this representation. Then
[EX, (^2,3)(-1)] = [tf- + № + X2 + X3) + (0+ - X3), X2 X2-1 (X2 + X3)-1 ]
= (1 - 2)(tf+ - X3)X2X2-1 (X2 + X3)-1 where as usual = E+f^Xj. We call it as usual (^f,2,3)( 1), because it has degree
Set Xi ^ Х3 and Х2 ^ Х4 and Х3 ^ Х5 in (Ff2 3)( 2 \ then we will have,
(Я
3^4,5)(-2 ) = 4 *4-1 № + Х5Г1
(52)
and again in a same process we have,
№f,4,5)(-1) = [EX, (Ff;4;5)(-1)] ^ ^
= (1 - 1 )(tf+ - X3 - X4 - X5)X3^X4-1 (X4 + X5)-2 Now let us multiply (Fj
7>x
i,2,3) 2' — v4 ^2
(1 - 2 )(^+ - X3 - X4 - X5)^32 x-2 - - 5)ф
(^1,2,3)(—2
)(-1) — x2X2 2(X2 + X3)—Irnth (F3X,4,5)( 1) — 4 (X4 + X5)—1 for to have
3,4,5)
q 2
q 2
(X2 + X3)— 2 (1 - 2)(tf+ - x3 - X4 - X5)X|X4 2 (X4 + X5) — 2
(X4 + X5) — 2 -(X4 + X5)—2
— X\ X2— Г , ,
— X2X2—1 (X2 + X3)-1 ((tf+ - X3 - X4 - X5)^2X4— 2 (X4 + Х5Г2) -
- x3 - X4 - X5)X2 X4—1 (X4 + X5)—2 )
1 — X^ X2—1 (X2 + X3 )—1 - x3 - X4 - X5)X2 X4— —-1
x2 x2— 1 (X2 + x3)-2 - x3 - X4 - X5)X2 x4— 1
— Xi2 x—1 (X2 + x3 )—1 ^+x2 x—2 (X4 + X5)—2
-X2 X—1 (X2 + X3 ) — 1X3X2 X4— 2 (X4 + X5)— 1 -Xi2 x—1 (X2 + x3 )—2 X4X2 X4— 2 (X4 + X5)—2
-x2 x—1 (X2 + x3 )—2 X5X2 X4—1 (X4 + X5)—2
2 x2 x—2 (X2 + x3 )—2 x2 x—1 (X4 + X5)—1 2 XiJ x— J (X2 + x3 )—2 x3x| x— 2 № + X5)—1 -g-2 x2 x—2 (X2 + x3 )-2 X4X31 x—1 (X4 + X5)—1
-Г1 X2 X—1 (X2 + X3 )- 1X5X2 X— 2 (X4 + X5)— 1
— (1 - g-1 )X2 X2—1 (X2 + X3)- 1 ^+X2 X—1 (X4 + X5)-
+(1 - 2 )x * x— 1 (X2 + Х3Г 2 ХзХ| x— 2 (X4 + X5)—
+(1 - Г 2 )Xf X2— 2 (X2 + X3)-1X4X3^X— 2 (X4 + X5) — +(1 - 1 )x2 X2—1 (X2 + Х3Г1X5X31 x—1 (X4 + X5)—
And again let us calculate the multiplication:
-Ç-2 (Fj
x 1,2,3
x 3,4,5
)(-2 )
(53)
— -g-2(1 - 2)(tf+ - X3)X2X 2
i i 2 № + X3)-2 X32 X42 (X4 + X5)-2
— 1 (1 - Г2)^+X2X2-2(X2 + X3)-2X32X4-1 (X4 + X5)-1 +g-1 (1 - Г2)ХзХ 1X—1 (X2 + X3)-2X|X4-1 (X4 + X5)-1
— -(1 - Г 2)X2X2-2 (X2 + X3)-2ВД2X—1 (X4 + X5)-1
-(1 - g-2)X2X2-1 (X2 + X3)-2Х3ХЗ1 X4-1 (X4 + X5)-2 i i i
^ №Х2,з)(- 1 )(^3x,4,5)( 2 ) - Г2 (FX,2,3)( 2 ) (^3X4,5 )(-2 ) — (1 - Г2 )4 X2- 2 (X2 + Хз)- 1 X|
x4 2(X4 + X5)-1
Lets call it pi,5. But the coefficients here are not so important for us, so let us skip it and write it as follows:
Pi,5 = X?X2 2(X2 + X3)-2X|X_ 2(X_ +X5)-2
(54)
that is a five points generator of lattice Virasoro algebra, but again we are not interested on it and still looking for a simplest one of type ABCD.
So let us to define another such kind of solutions as we proposed and experienced already;
t U 2-2
)(-2) :=X22X32 (X3 + X_)-
(55)
/ 1 \ 1 2 ( F2,3;4)(2) :=(1 - 0T2)( U+ -X3 -X_)X22X3 2 (X3 + X_)-2
and then define;
Pi,4 = №.3)(-2)(F23,_)(2) - Q-2(F
?i,2,3,
Let us calculate it;
(Ff,2,3)(-2 )(F23,4)( 2)
22.3,4)( 2)
(56)
(57)
(58)
= Xi2X22 (X2 + X3)-
2 ((1 - q-2)( U+ - X3 - X_)X2X3 2 (X3 + X_)-2
)X2X2 2(X2 + X3)-
)X2X;
= (1 -= (1 --(1 - --(1 - q-2)X2X2-2 (X2 +X3)
)X2X2
2 ( U+ -X3 -X_)X22X3
r(X2 + X3)-(X2 +X3)-
~2X-
2
X3 X2 X3
2 X4X22 X3 2 (X3 + X4)
Now let us calculate the other part as well;
(X3 +X_)-2
2U+X22 X3 2 (X3 +X_)-2 2 (X3 +X_)-2
-q - 2 (Fi
\X
i,2,3
2,3,4 )(-2)
(59)
= -q-2 (1 - q-2)( U+ -X3)Xi2X2 2 (X2 + X3)-2X22X3 2 (X3 + X_) = -q-2(1 - q-2)U+X2X2-2(X2 +X3)-2X22X3-2(X3 +X_)-2
+q-2 (1 - q-2 )X3X2X2 2 (X2 +X3)-^^ 2(X3 + X_)
= -(1 - q-2)X2X2 2(X2 +X3)-2U+X22X3 2(X3 + X_)-2
1 2 _2 1 2
~2Y 2(Y~ _L Y„\- 1Y,,Y2X 2/
+(1 - q-2)X2X2 2 (X2 + X3)'
X3X2 X3
!(X3 +X_)"
So we have
Pi,4 = -(1 - q-2)Xi2X2 2 (X2 + X3)-2X_X22X3 2 (X3 + X_)-2
(60)
that has degree zero, so it can be one of the generators for lattice Virasoro algebra, but still again is not interested for us, so we should look for a simplest one. And again the coefficient is not important for us, so lets skip it. So we have
and
Pi,4 = X2X2 2 (X2 + X3)-2X_X22 X3 2 (X3 + X_)-2
P-5 = (X_ + X5)2x_2X32(X_ + X5)-i(X2 + X3)2X22X1
(61) (62)
Let us calculate the multiplication p- 5pi,4 for to see what will happen?
Claim 4.6. The claim is that p- 5pi,4 should give us the answer?
2
2
PROOF. p_5pM = (X4 + X5)1 X2X_1 (X4 + X5)-1 X4X2 X_ 2(X3 + X4)-2
= (X4 + X5) 2 X42 q 2 (X4 + X5)-1X-1X4XIX33 2 (X3 + X4)-1 = (X4 + X5) 2 g-2g 2 (X4 + x-1X4X2 x-2 (x3 + X4)-1
= (X4 + X5)-1X2 X- 2 X4X2 X-1 (X3 + X4)-1
= (X4 + X5)-1 x2 2 X4X-1 x22 x-1 (x3 + X4)-1
= (X4 + X5)-1X2 g-1X4X21 q1X-1 (X3 + X4 )-2
1 1 1 3 1
= (X4 + X5)-1X42 X22 g-4 X4X-1(^ + X4)-2 3 1 1 1 1 = g - 4 (X4 + X5)-1X42 X22 (X3 + X4)-1
And as always the coefficients are not important for us, so let us the set the final solution as follows:
P-5P1,4 = № + X5)-1X41 X22 (X3 + X4)-2 (63)
as we were looking for. □
Now suppose the following representation of s/2. Define:
F =
H = - X3 - X4)9(^+_X3_X4) + X^ + X20X2 + X,^ (64)
E = - X3 - X4)29(^+_X3_X4) + (X2 + X1X2 + X1X3 + X^+ - X3 - X4))9X1 +(X22 + X2X3 + X2(0+ - X3 - X4))9x2 + (X2 + X3- X3 - X4))%3 As before set
№+3)W = (^fc;fc+1;fc+2;fc+3)(t) := P*, , [■ ■ ■ , , (^+3)«]] ■ ■ ■ ]] (^+1,fc+4)œ = (^fc+1;fc+2;fc+3;fc+4)(j) = P*, [E*, [■ ■ ■ , [E*, (^ë+1,fc+4)(^)]l ■ ■ ■ ]
Set к = 1 Mid г = - 1 and s et (Ff4)(-2 ) = X2 X2 2 (X2 + X3 + X4)-2.
2
and as what we had already, set
2) = (1 - 2))(^+ - X3 - X4)X2X2 2(X2 + X3 + X4)-2 (65)
where as usual = Xj.
Now as before, set Xi ^ J^d X2 ^ X4 and X3 ^ J^d X4 ^ X6 in (Ff4)(-2\ then we will have,
(^6)(-2) = №M,5;6)(-2) = 42 № + X5 + X6)-2 (66)
2) = (1 - 2))(^+ - X3 - X4 - X5 - X6)X32X4-2(X4 + X5 + X6)-2 (67) and then we will proceed as before again
(^4)(-2)(^6)(2) (68)
2-2 1 / In 2 _2 1
= X-2 X2 2 (X2 + X3 + X4)-2 (1 - 4(-2))(^+ - X3 - X4 - X5 - X6)X32 X4 2 (X4 + X5 + X6)-2
/ U 2 _2 2 2 _ 2 2
= (1 - 4(-2))X2 X2 2 (X2 + X3 + X4)-2 - X3 - X4)X32 X4 2 (X4 + X5 + X6)-2
t U 2 _2 2 2 _2 2
-(1 - 4(-2))X12 X2 2 (X2 + X3 + X4)-2X5X32 X4 2 (X4 + X5 + X6)-2
, ^ 1 -1 1 1 _1 1
-(1 - q(-1 ))Xi2X2 2 (X2 + X3 + X_)-2X5X32X_ 2 (X_ + X5 + Xo)-2 and on the other hand, we have
-q(- 1)(Ff,_)( 2)( Ff,6)(-1) (69)
= -q(-2)(1 - q(-2))( U+ -X3 -X_)X2X2-1 (X2 + X3 + X_)- ^X^X^1 (X_ + X5 + Xo)-1
t 1\ 1 -1 1 1-1 1
= -(1 - q(-1 ))Xi2X2 2 (X2 + X3 + X_)-2 (U+ - X3 - X_)X32X4 2 (X_ + X5 + Xo)-2
and so ( Ff_)(-1)( Ff,6)( 1) - q(-1)( Ff_)(2)(Ff,6)(-1)
^ ^ 1 -1 , 1 , 1-1 1
= -(1 - q(-1 ))Xi2X2 2 (X2 + X + X_)-2 (X5 + X6)X32X_ 2 (X_ + X5 + X-)-2 as always let us call it
pi,6 = X2X2-1 (X2 + X + X_)-2 (X_ + X5 + X6)X2X_- 2 (X_ + X5 + X-)-1 (70)
-I 2 1 -1 ! 2 1 -1
P-,6 = (X_ + X5 + X6) 2X42 X- 2 (X_ + X5 + X6)-i(X2 + X + X4) 2X22 X- 2 (71)
Now again let us define another such kind of solutions:
( F2,5)(-1) = (F2,3,4,5)(-2) := X2^X3-1 (X +X_ +X5)-1 (72)
( F2,5)( 1) = ( F2,3,4,5)( 1) := (1 - Q(-2))( U+ - X3 - X_ - X5)X22X3-2(X3 + X_ + X5)-1 (73) and we are looking for the value of the following objects;
(Fi,_)(-2)(F2,5)(2) (74)
1-1 2 < u 1-1 2
= Xi2X2 2 (X2 + X3 + X_)-2 ((1 - q(-1 ))(U+ - X3 - X_ - X5)X22 X3 2 (X3 + X_ + X5)-2
I 2\ 1 -1 2 1-1 2
= ((1 - q(-2))Xi2X2 2 (X2 + X3 + X_)-2(U+ - X3 - X_)X22 X3 2 (X3 + X_ + X5)-2 / 2\ 1 -1 2 1 -1 2 -((1 - q(-1 ))Xi2X2 2 (X2 +X3 +X_)-2X5X22 X3 2 (X3 +X_ +X5)-2
and on the other side we have:
-q(-2)(Ff,_)( 2)( F2,5>(-1) (75)
= -q(-2)(1 - q(-2))(U+ - X3 - X_^fX-1 (X + X3 + X_)-^^1 (X3 + X_ + X5)-1 = -(1 - q(-2))X2-2 (X2 + X3 + X_)-1 (U+ - X3 - X_)X22X3-2 (X3 + X_ + X5)-1 so we have: Pi,5 = (F0-2)(Ff^2) - q(-1 )(F?_)( 1 )(F2,5)(-2)
Pi,5 = X2X2-1 (X2 + X3 + X_)-1X5X2 X-2 (X3 + X_ + X5)-2 (76)
Claim 4.7. p-^p^ has degree zero.
Proof. p-,6Pi,5 = (X_ + X5 + X6)^^2(X_ + X5 + X6)-i(X2 + X3 + X_) ^X-2X2
X2- 2 (X2 + X3 + X_)- 2X5X2^X3-1 (X3 + X_ + X5)- 2
= (X_ + X5 + X6) 1X_^X3- 2 (X_ + X5 + X6)-iX5X2^X3- 2 (X3 + X_ + X5)-2
= q(-3 )(X_ + X5 + X6)-21xjxl (X3 + X_ + X5)-1 □ So we have:
P->i,5 = (X_ + X5 + X)-244 (X3 + X_ + X5)-2 (77)
5. Conclusion
Four point invariant, that's coming from the 3-dimensional representation of s/2;
[Xl^X,, (X4 + X5)-1X4X2(^ + X4)-1 ], = 0 Five point invariant, that's coming from the 4-dimensional representation of s/2;
(X4 + X5 + X6)-1X4X2(X3 + X4 + X5)-1], = 0
Claim 5.1. We have the following n-point invariant that's coming from the n-dimensional representation.
(X4 + ■ ■ ■ + Xra) 1X4X2 (X3 + ■ ■ ■ + Xra-1) 1
And then by using the shift operators (17), we will have the space of all nontrivial generators of lattice Virasoro algebra.
We call these kind of generators that are the only nontrivial ones:
Generatorso/type"
These (new lattice) algebras are so important and may in principle lead to a new integrable chain equations which people can hardly provide.
Now let us check the satisfactory of our generators in quantum Serre relations.
We need to show the correctness of (X2+X3 + X4 + X5)((X4+X5)-1X42 X| X-1 (X3 + X4)1) —
((X4 + X5)-2 X42 xj X3-1(X3 + X4)-1 )(X2 + x3 + X4 + X5) So let us proceed it as before on each component:
.31 1 ? 1 2 1 1
X2(X4 + X5)-1X42 X| x-1(^ + X4) 1 = (X4 + X5)-2 X42 x22 x-1(^ + X4)-1X2 1 .21 ? 21 1 1 (X4 + X5) 1 X2(X4 + X5)-2 X42X22X-1 = X42X22X-1(^ + X4)-2 X2(^ + X4) 1
3 1 ? x 3 1
V V 2 V 2 v -1 „1 Y 2 V 2 Y -1y A2A4 A2 A3 — q 2 A4 A2 A3 A2
3 1 ? 1 3 1
X2X2 X22 X.-1 — q 1gX| X2X22 X.-1
3 1 ? 1 3 1
X2X2 X22 X3-1 — q 19Xf X2XJ X3-1
3 1 1 3 3 1
X2X2 X22 X.-1 — q 1 3 X| X2X22 X.-1
3 1 3 1
X2X42 X22 X3 — X2X42 X22 X3
Lets do it for X3;
X3(X4 + X5)-1X42 X2 X-1 (X3 + X4) 2 — (X4 + X5)-1 xj X.2 X3-1(X3 + X4)-1X3
1 3 1 ? 1 3 1
2 X3X4J X| X3-1 — 1 XJ X| X-1X3 g-1X3XJ X| X-1 — 1 q1 xj X3XJ X-1
1 3 1 ? 1 3 1
^ 2 X3X42 X22 X-1 — 1X3X42 X| X-1
And after repeating a similar trend for J^d X5 we will get the desired result. Acknowledgments
It is a great pleasure to thank Professor Boris Feigin for to suggesting this interesting problem to me and enlightening discussions during the preparation for this paper and also I wish to thank Professor Yaroslav Pugai for his wonderful article and his very useful advices and comments.
СПИСОК ЦИТИРОВАННОЙ ЛИТЕРАТУРЫ
1. Alekseev, A., Faddeev, L. and Semenov-Tian-Shanskv, M., 1992. Hidden quantum groups inside KaoMoody algebra. //Communications in mathematical physics, 149(2), pp.335-345.
2. Brown, K. and Goodearl, K.R., 2012. Lectures on algebraic quantum groups. //Birkhâuser.
3. Frenkel, E., 1995. Free field realizations in representation theory and conformai field theory. //In Proceedings of the International Congress of Mathematicians (pp. 1256-1269). Birkhâuser, Basel.
4. Feigin, B.L. :// talk at RIMS 1992.
5. Gelfand, I. and Dickey, L. A Lie algebra structure in a formal variations calculus, //Functional Anal. Appi., 10 (1976), 16-22.
6. Iohara, K. and Malikov, F., 1994. Rings of skew polynomials and Gel'fand-Kirillov conjecture for quantum groups. //Communications in mathematical physics, 164(2), pp.217-237.
7. Majid, S., 2002. A quantum groups primer (Vol. 292).// Cambridge University Press.
8. Kac, V.G., 1990. Infinite-dimensional Lie algebras. //Cambridge university press.
9. Kassel, C., 2012. Quantum groups (Vol. 155). //Springer Science k, Business Media.
10. Kupershmidt, B.A. and Wilson, G., 1981. Conservation laws and symmetries of generalized sine-Gordon equations.// Communications in Mathematical Physics, 81(2), pp.189-202.
11. Klimvk, A. and Schmiidgen, K., 2012. Quantum groups and their representations. //Springer Science k, Business Media.
12. Laugwitz, R., Quantum Groups and Their Representations.
13. Pugav, Y.P., 1994. Lattice W algebras and quantum groups.// Theoretical and Mathematical Physics, 100(1), pp.900-911.
14. Sklvanin, E.K., 1985. On an algebra generated by quadratic relations. //Usp. Mat. Nauk, 40, p.214.
15. Stanley, R.P., 1986. Enumerative combinatorics, //Wadsworth Publ. Co., Belmont, CA. REFERENCES
1. Alekseev, A., Faddeev, L. and Semenov-Tian-Shanskv, M., 1992. "Hidden quantum groups inside Kac-Moodv algebra". Communications in mathematical physics, 149(2), pp.335-345.
2. Brown, K. and Goodearl, K.R., 2012. "Lectures on algebraic quantum groups". Birkhâuser.
3. Frenkel, E., 1995. "Free field realizations in representation theory and conformai field theory". In Proceedings of the International Congress of Mathematicians (pp. 1256-1269). Birkhâuser, Basel.
4. Feigin, B.L. talk at RIMS 1992.
5. Gelfand, I. and Dickey, L. "A Lie algebra structure in a formal variations calculus", Functional Anal. Appi., 10 (1976), 16-22.
6. Iohara, К. and Malikov, F., 1994. "Rings of skew polynomials and Gel'fand-Kirillov conjecture for quantum groups". Communications in mathematical physics, 164(2), pp.217-237.
7. Majid, S., 2002. "A quantum groups primer" (Vol. 292). Cambridge University Press.
8. Kac, V.G., 1990. "Infinite-dimensional Lie algebras". Cambridge university press.
9. Kassel, C., 2012. "Quantum groups" (Vol. 155). Springer Science к Business Media.
10. Kupershmidt, B.A. and Wilson, G., 1981. "Conservation laws and symmetries of generalized sine-Gordon equations". Communications in Mathematical Physics, 81(2), pp.189-202.
11. Klimvk, A. and Schmiidgen, K., 2012. "Quantum groups and their representations". Springer Science к Business Media.
12. Laugwitz, R., "Quantum Groups and Their Representations".
13. Pugav, Y.P., 1994. "Lattice W algebras and quantum groups". Theoretical and Mathematical Physics, 100(1), pp.900-911.
14. Sklvanin, E.K., 1985. "On an algebra generated by quadratic relations". Usp. Mat. Nauk, 40, p.214.
15. Stanley, R.P., 1986. "Enumerative combinatorics", Wadsworth Publ. Co., Belmont, CA.
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