Uses of Quantum Spaces

HAL Id: cel-00374419

https://cel.archives-ouvertes.fr/cel-00374419 Submitted on 8 Apr 2009

HAL is a multi-disciplinary open access

archive for the deposit and dissemination of sci-entific research documents, whether they are pub-lished or not. The documents may come from teaching and research institutions in France or abroad, or from public or private research centers.

L’archive ouverte pluridisciplinaire HAL, est destinée au dépôt et à la diffusion de documents scientifiques de niveau recherche, publiés ou non, émanant des établissements d’enseignement et de recherche français ou étrangers, des laboratoires publics ou privés.

Oleg Ogievetsky

To cite this version:

Oleg Ogievetsky. Uses of Quantum Spaces. 3rd cycle. Bariloche (Argentine), 2000, pp.72. �cel-00374419�

Uses of Quantum Spaces

O. Ogievetsky Mame Contents 1. Introduction 2 2. Lie bialgebras 3

2.1. Deformation of the coproduct 3

2.1.1. Discrete groups 9

2.2. Lie algebras with an invariant scalar product 10 2.3. Belavin-Drinfeld triples 13

2.3.1. Maximal triples 15

3. Quantum spaces 20

3.1. GLtype 24

3.2. Technics of checking the Poincare series 25 3.2.1. Dierential calculus and Poincare series 26 3.3. Geometry of 3-dimensional quantum spaces 29

3.3.1. Gr obner base forE 32

3.4. slq(2) at roots of unity 33

3.4.1. Preliminaries 34

3.4.2. Formatted matrix algebras over graded rings 38

3.4.3. Matrix structure 39

3.4.4. Reduced function algebra 45

3.4.5. Centre 46

4. ^R-matrices 47

4.1. Skew-invertibility 47

4.1.1. Generalities on Hopf algebras 47

4.1.2. Matrix picture 50

4.2. Ice ^R-matrices 56

4.3. Construction of orthogonal and symplectic ^R-matrices 59

5. Real forms 67

5.1. General linear quantum groups 68 5.2. Orthogonal and symplectic quantum groups 70

References 71

1991Mathematics Subject Classication. 16W30, 16W35, 16W50, 81R50, 17B37, 20G42. .

1. Introduction

Quite often, a group appears as a set of symmetries of some object - a set equipped with geometrical, algebraic or combinatorial data. The theory of quan-tum groups enlarges the notion of symmetry a quanquan-tum group (often) describes \generalized symmetries" of an object. In the case of a linear (orthogonal, sym-plectic) quantum group, this object is a linear (orthogonal, symsym-plectic) quantum space - an algebra with certain quadratic relations. A study of these underlying objects, the quantum spaces, helps to understand the structure of the quantum groups. In these lectures I will illustrate the role of the quantum spaces on two examples: non-perturbative eects in the theory of Yang-Baxter operators and real forms of quantum groups.

To talk about non-perturbative eects, one should explain rst, what means \perturbative" or \deformational". This is the subject of the subsection 2.1. The initial data for a quantum deformation of a Lie algebraL is conveniently encoded

in terms of another Lie algebra D(L), the Drinfeld double ofL. The Lie algebra

D(L) has an invariant scalar product and I have included a subsection 2.2 on the

structure of Lie algebras with an invariant scalar product.

For a semi-simple Lie algebra L, the most important deformations are those

which are called quasitriangular. They are classied by Belavin-Drinfeld triples. The subsection 2.3 contains some information about the combinatorics of the Belavin-Drinfeld triples.

In section 3, after a geometrical interpretation of the quantum deformations of Lie groups, we introduce an algebra of functions on a quantum group a denition of GL-type quantum groups and quantum spaces is given in subsection 3.1. In subsection 3.2 we explain how to use a dierential calculus on a GL-type quantum space for calculating the Poincare series.

Subsection 3.3 is devoted to 3-dimensional quantum spaces. We exhibit an unexpected appearance of Yang-Baxter operators and give an example of a non-perturbative Yang-Baxter operator. We prove the Poincare-Birkho-Witt theorem for the quantum space dened by this Yang-Baxter operator.

Subsection 3.4 deals with eects specic to quantum groups at roots of unity. We introduce a terminology of formatted matrix algebras over local graded rings, which is useful in the study of non semi-simple algebras. We describe the matrix structure of the reduced quantum enveloping algebra and the reduced function algebra forslq(2).

Subsection 4.1 contains a summary of the theory of quasi-triangular Hopf alge-bras. In subsection 4.2 we classify Yang-Baxter matrices, which can have non-zero entries only at places where the pair of lower indices is a permutation of the pair of the upper ones. Subsection 4.3 gives a construction of the Yang-Baxter matrices for orthogonal and symplectic groups from the Yang-Baxter matrices forGL.

In section 5 we describe a method of classication of real forms of quantum groups. The method is based on the study of the corresponding quantum spaces.

Throughout the text, a sum over repeated indices is assumed. If X =fXijg

and Y = fYijg are two operators, the indices are summed as (XY)ij =XikYkj in

2. Lie bialgebras

A Hopf algebraH is a collection of datafHmSg, whereH is a vector

space over a ground eld k m :H H ! H a multiplication  : H !H H

a comultiplication  : H ! k is a counit and S : H ! H an antipode. For a

precise formulation of various relations between these maps see e.g. 

1

]. Let me just remind that for a Hopf algebraH one knows how to build tensor products of representations and it is given universally by  the counit gives rise to a trivial representation the antipode is needed to build contragredient representations.

The classical examples of Hopf algebras are group algebraskG] of nite groups

Gand universal enveloping algebrasU(L) of Lie algebrasL.

2.1. Deformation of the coproduct.

LetLbe a Lie algebra overC andUits

universal enveloping algebra. Denote byfXiga basis ofL. The classical coproduct

0:

U !UU is given on generatorsXiby 

0Xi=Xi

1+1Xi. The map  0

is a coassociative homomorphism (coassociativity means (0  I)

0= ( I 

0)0

for the mapsU!UUU here I is the identity map). In this subsection we shall

study deformations of the coproduct 0. A deformation of 0 is, by denition, a

coassociative homomorphism  :U !UU,

(a) = 0(a) +1(a) + 2

2(a) +::: :

(2.1.1)

The right hand side is a formal power series in the parameter, which is called a deformation parameter. The coecientsk(a) are elements ofUU.

Our task is to understand which deformations are \essential", in the sense that they cannot be removed by some redenition of generators. Here is the answer modulo2.

Theorem 1.

Any deformation of 0, after a change of generators, takes a form

(in the rst order in)

Xi = 0Xi+ jk

i XjXk:

(2.1.2)

The antisymmetric tensor jk_i (jk_i = ;kji) is a 1-cocycle with values in  2 L, 2Z 1( L 2 L) explicitly: Nab] ij] = ;sij ab s  (2.1.3)

whereN_ijab_{= ;ai}b_{j and ab] means antisymmetrization in indicesaand b,t}ab]=

tab;tba for a tensortab. Here ;kij are the structure constants of the Lie algebraL,

XiXj] = ;kijXk.

Proof. Assume that  is a deformation of the classical coproduct 0. On the

generatorsXi we have

(Xi) = 0(Xi) +i+:::

(2.1.4)

with somei 2UU, where dots denote higher powers in.

The coassociativity, in order1, is equivalent to a following equation on i in U 3 i1 + ( 0  I)i= 1i+ ( I 0)i  (2.1.5)

I is the identity operator. The algebraU

3is the enveloping algebra of

LLL.

respectively. Then the equation (2.1.5) can be rewritten as

i(XY) +i(X+YZ) =i(YZ) +i(XY +Z):

(2.1.6)

The statement that  is a homomorphism reads, in terms ofi, as

Xi+Yij];Xj+Yji] = ;kijk

(2.1.7)

(the algebraUU is the enveloping algebra ofLLXiandYi are the generators

of the rst and second copies ofL).

Let  : U U ! U U be the !ip, (xy) = yx. Decompose i into

symmetric and antisymmetric parts with respect to,

i=si+ai

(2.1.8)

with(si) =si and(ai) =;ai.

Proposition 2.

Ifisatises (2.1.5) and (2.1.7) then bothsiandaisatisfy (2.1.5)

and (2.1.7). Proof. We have 0(X i) = 0(Xi) + 0 i+:::, where0 i=(i) =si;ai. If  is a coproduct then 0 =  is a coproduct as well, so 0 isatises (2.1.7), Xi+Yi0 j] + Xj+Yj0 i] = ;kij0 k  (2.1.9) and (2.1.5), 0 i1 + ( 0  I) 0 i = 1 0 i+ ( I 0) 0 i : (2.1.10)

Take the sum and dierence of (2.1.5) and (2.1.10) (respectively, (2.1.7) and (2.1.9)) to nish the proof.

In particular, each part (symmetric or antisymmetric) of i alone denes a

coproduct in order1.

Clearly, a redenition of generators can change only the symmetric part ofi.

We start by analyzing this case (the case of symmetrici).

Proposition 3.

Assume that  is symmetric in order1, 0

i =i. Then the1

terms can be removed by a redenition of generators.

Proof. U is the algebra of polynomials in the generators Xi. It is ltered by the

degree of polynomials,FkU are polynomials of degreek. The associated graded

termFkU=Fk ;1

U is isomorphic to SkL, the symmetric power of L. Any element

u2Uhas a well-dened \highest symbol": ifu2FkUnFk ;1

U(nis the set-theoretic

complement) then its highest symbol is the image of uin SkL. Denote by xi the

basis of commuting variables corresponding to generatorsXi. The highest symbol

is a homogeneous polynomial in a set of commuting variablesxi.

The algebraUU is the enveloping algebra ofLL, the highest symbols are

homogeneous polynomials in two sets of variables,xi andyi.

Letfi be the symbol of i. Then fi is a polynomial in two sets of variables, fi=fi(xy). The symmetry condition implies thatfi(xy) =fi(yx).

The coassociativity implies, in order1, an equation

f(x+yz) +f(xy) =f(xy+z) +f(yz) (2.1.11)

Lemma 4.

Letf(xy) be a homogeneous polynomial, symmetric with respect to the !ip x $ y. The polynomial f satises (2.1.11) if and only if there exists a

homogeneous polynomial g(x) (a polynomial in only one set of variablesxi) such

that

f(xy) =g(x+y);g(x);g(y):

(2.1.12)

Proof. It is straightforward to see that f(xy) = g(x+y);g(x);g(y) satises

(2.1.11).

Assume now thatf satises (2.1.11). Let M be a total degree off. If M= 0 thenf =cis a constant and it is enough to takeg=;c. Assume that M >0.

Applying _@x@_i to (2.1.11) and evaluating atx= 0, we obtain an equation (after replacingy!x andz!y) @i 1f jxy=@i 1f j 0x+y ;@i 1f j 0x  (2.1.13) where@i

1 are the partial derivatives in the rst set of variables.

Applying _@z@_i to (2.1.11) and evaluating atz= 0, we obtain an equation

@i 2f jxy=@i 2f jx +y0 ;@i 2f jy 0  (2.1.14) where@i

2 are the partial derivatives in the second set of variables.

Since f is homogeneous of degree M, we have (xi@1i +yi@ i

2)f = Mf, which,

together with (2.1.13) and (2.1.14), gives

Mf=xi@1if j 0x+y+yi@ i 2f jx +y0 ;xi@i 1f j 0x ;yi@i 2f jy 0 : (2.1.15)

The symmetry off,f(xy) =f(yx) implies that@1

ifjxy=@ 2

ifjyx. Therefore we

can rewrite (2.1.15) in the form (2.1.11) withg(x) = 1 Mxi@1

ifj

0x. The proof of the

Lemma 4 is nished.

We proved that for eachithere existsgi(x) such that fi(xy) =gi(x+y);gi(x);gi(y):

(2.1.16) Letg_

i(X) be an element whose highest symbol isgi(x). The combinationg_ i(X+ Y);g

_ i(X);g

_

i(Y) satises the equation (2.1.6). Therefore, an elementi(XY);

g_

i(X+Y)+g_

i(X)+g_

i (Y), which has the ltration degree smaller than the degree

ofi(XY), satises (2.1.6) as well, and we can apply the Lemma 4 again.

Repeating this process a needed number of times, we shall nally build a set of elementsi2U such that

i(XY) =i(X+Y);i(X);i(Y):

(2.1.17)

LetXi =Xi;i(X). It is straightforward to see that in the order  1 the

coproduct for the generatorsXi is classical, (Xi) =Xi1 + 1Xi.

It is left to show that one can choosei in such a way that the generatorsXi

satisfy the same Lie algebraic relations as the original generatorXi. It will be so if

and only if Xij];Xji];;kijk = 0 for alliandj.

Since  is a homomorphism, it follows immediately that elementsij= Xij];

Xji];;kijk satisfy relations

ij(X+Y) =ij(X) +ij(Y):

(2.1.18)

We shall need a short digression into the general theory of universal enveloping algebras (see, e.g. 

2

]).

An elementu2U can be uniquely decomposed into a sum

u= symb0

(u) + symb1

(u) ++ symbd(u)

(2.1.19)

where d is the ltration degree ofu and symbA(u) =ci1:::iAX

i1:::XiA for some

completely symmetric tensorci1:::iA. Elements of the formci1:::iAX

i1:::XiAwith a

completely symmetricci1:::iAform a subspace

UAUand the above decomposition

ofuimplies that U is a direct sum of UA, U = 1 A=0

UA. Each UA is aL-module

(that is, commutators of generatorsXi with symbA(u) are again in UA) in other

words,

symbA(Xiu]) = Xi symbA(u)]:

(2.1.20)

The moduleUA is isomorphic to the symmetric powerSAL.

Letj =P

symbA(j) be a decomposition of the form (2.1.19) of the element j. We have seen that ij is in U

1 for each iand j. It follows then from (2.1.20)

thatij = Xi symb1( j)];Xj symb 1( i)];;kij symb 1( k). Therefore, Xi~j];

Xj~i];;kij~k = 0 for alli and j, where ~i = i; symb 1(

i). Therefore, the

elements ~Xi =Xi;~i(X) satisfy the same Lie algebraic relations as the original

generatorsXi,  ~XiX~j] = ;kijX~k.

Moreover, since for an elementg1 2U

1, the combinationg1(X+Y) ;g

1(X) ;

g1(Y) vanishes, the elements ~

i =i; symb 1(

i) still verify (2.1.16). Therefore,

as before, the coproduct for the elements ~Xi is classical.

Thus, the elements ~Xi provide the needed redenition of the generatorsXi.

The proof of the Proposition 3 is nished.

Using, if necessary, the redenition of the Proposition 3, we get rid of the symmetric part ofi. Assume therefore thati is antisymmetric. Again, letfi be

the highest symbol ofi. The symmetry condition is nowfi(xy) =;fi(yx). As

before, the coassociativity implies, in order1, the equation (2.1.11) for eachi.

Proposition 5.

Let f(xy) be a homogeneous polynomial, antisymmetric with respect to the !ipx$y. Assume that the polynomialf satises (2.1.11). Then

f(xy) = jk_xjyk

(2.1.21)

for some antisymmetric tensor , jk₌; kj.

Proof. The derivatives of f satisfy equations (2.1.13) and (2.1.14). There is one more equation which we didn't need for the Lemma 4. It is obtained by applying_@

@yi to (2.1.11) and evaluating at y= 0 (we change variables,z!y)

@i 1f jxy+@i 2f jx 0=@ i 2f jxy+@i 1f j 0y : (2.1.22)

The antisymmetry of f, f(xy) = ;f(yx), implies @ 1 ifjxy =;@ 2 ifjyx. Substi-tuting @i 1f jxy and @i 2f

jxy from (2.1.13) and (2.1.14) into (2.1.22) and using the

antisymmetry, we nd @i 1f j 0x+y=@ i 1f j 0x+@ i 1f j 0y: (2.1.23)

Thus, @i 1f

j

0x is a linear function. Substituting (2.1.23) into (2.1.13), we nd

@i 1f j 0x+y =@ i 1f j 0y. Thus,@ i 1f

jxyis a linear function which depends on the second

set of variables only. In other words,@i 1f jxy= ij 1yj. Similarly,@i 2f

jxyis a linear function which depends on the rst set of variables

only,@i 2f jxy= ij 2 xj. The antisymmetry,@1 ifjxy=;@ 2

ifjyx, implies that ij 1 = ; ij 2. Let ij ₌ ij 1 . Then @1if jxy= ijyj and @i 2f jxy=; ijxj : (2.1.24)

Since the derivatives of f are homogeneous of degree 1, the function f itself is homogeneous of degree 2. So 2f = (xi@i1+yi@ i 2)f. Substituting expressions (2.1.24), we nd 2f =xi ijyj;yi ijxj= ij]x iyj  (2.1.25)

where the square brackets mean antisymmetrization, ij] = ij

; ji and the

assertion of the Proposition 5 follows.

After the Propositions 2, 3 and 5 it is only left to check a condition that  is a homomorphism in the rst order in . A straightforward calculation gives the cocycle condition (2.1.3). The proof of the Theorem 1 is nished.

Remark. It is not necessary to assume that the ground eld isC. The Theorem 1

holds for an arbitrary eld of characteristic 0 (and it is not true if the characteristic is dierent from 0).

Repeating the proof of the Proposition 3 consecutively in powers of , one obtains a version of the Milnor-Moore theorem (for its general formulation see, e. g. 

3

]):

Corollary 6.

A formal (i.e. given by a formal power series in) cocommutative deformation of the coproduct on a universal enveloping algebra is always trivial, that is, it can be removed by a formal redenition of generators.

In the rest of this subsection we explain what happens in the next order in, in the2-terms.

It turns out that the consistency in the 2 terms imposes new conditions on

jk_i .

Assume that we can extend the deformation (2.1.2) to2-terms,

Xi= 0Xi+ jk i XjXk+ 2 i  (2.1.26)

and  is coassociative up to3. Coassociativity implies

i1 + ( 0  id)i+abci XaXbXc = 1i+ ( id 0)i+ cba i XaXbXc  (2.1.27)

with a notation: abc_{i =jci}ab_{j for a tensor}jk_{i .}

To cancel the abc_{i -terms, one has (in order to get trilinear in X expressions)}

to choosei in the form

i=Aabci XaXbXc+Babci XcXaXb :

with tensors Aabc_{i andBi}abc _{symmetric in indices}fabg. In the next formulas, the

lower index is omitted.

Coassociativity (2.1.27) gives

2(Bbca;Aabc) =abc+bca:

(2.1.29)

Exchangeb and c in (2.1.29) and subtract from (2.1.29), taking into account the symmetry ofBabc _infabg:

2(Aacb;Aabc) =bca+Jabc 

(2.1.30) where

Jabc_=abc_+bca_+cab _:

(2.1.31)

The tensorJabc _{is totally antisymmetric. Under the exchangea}$b, eqn. (2.1.30)

becomes

2(Abca;Abac) =acb+Jbac:

(2.1.32)

Under the exchangea$c, eqn. (2.1.30) becomes

2(Acab;Acba) =bac+Jcba:

(2.1.33)

The combination (2.1.30) - (2.1.32) - (2.1.33) gives (due to the symmetry of Aabc

infabg)

0 = 4Jabc _:

(2.1.34)

This is the Jacobi identity (2.1.40) for.

Now from 6 permutations offabcgone gets only two equations

2(Aacb;Aabc) =bca 

(2.1.35)

2(Abca;Abac) =acb :

(2.1.36)

The tensor Aabc_{, being already symmetric in the rst two indices, can have two}

types of symmetry, corresponding to Young diagrams and . The totally symmetric part (the diagram ) ofAcannot be dened by (2.1.35)-(2.1.36), it is arbitrary. The part, corresponding to the diagram

satises

Aabc_+Abca_+Acab_{= 0:}

(2.1.37)

Together, eqs. (2.1.35), (2.1.36) and (2.1.37) can be easily solved and we conclude (taking into account the totally symmetric part) that the general solution forAis

Aabc_{= 16(}cba_+cab_{) +}abc

(2.1.38)

with totally symmetricabc_.

From (2.1.29) it follows then that

Bbca_{= 16(}abc_+acb_{) +}abc_:

(2.1.39)

In particular,Aabc_=Babc_.

The totally symmetric part abc _{can be removed by a redenition (solve forg}

We conclude that the coassociative extension of  to the -terms is possible only if the Jacobi identity foris satised,

abc_{i + cyclein(abc) = 0:}

(2.1.40)

This extension has the form Xi = 0Xi+ jk i XjXk+ 2 6 (cbai +cabi )(XaXbXc+XcXaXb): (2.1.41)

In general,  does not preserve the original commutation relations XiXj] =

;kijXk, the multiplication structure ofU has also to be deformed in the order 2.

Exercise. The map  preserves the following relations 

4

] XiXj] = ;kijXk+

2

18sai tbj;cstX(aXbXc)

(2.1.42)

with round brackets in X(aXbXc) denoting the symmetrization, X(i 1Xi2Xi

3) = P

Xi(1)Xi(2)Xi(3), the sum is over all permutations 2S

3.

Given the multiplication (2.1.42) and the comultiplication (2.1.41), one needs to know the counit and the antipode to complete the Hopf algebra structure. Exercise. The counit stays undeformed,

(Xi)0 mod 3

(2.1.43)

for the antipode mod3one has

S(Xi) =;Xi+ 12MaiXa+ 14 2bliMal;cabX c  (2.1.44) whereMva =;vca ;kck; kj

k ;vjasta;vst. (Note thatS stays linear in generators.)

It is known today that in higher orders inno further restriction onappears in other words, if jk_i satises the Jacobi and cocycle conditions, there exist, as formal power series in , the multiplication, which begins as (2.1.42), and the comultiplication, which begins as (2.1.41) (and the counit and antipode). Moreover, there exists such deformation that each term in the formal power series (for the multiplication, comultiplication and the antipode) is expressible in terms of the tensorsjk_i and ;kij only.

2.1.1. Discrete groups. The situation with discrete groups is dierent. It is an easy exercise to analyze the formal deformations of the coproduct for the group algebras of discrete groups. In contrast to the case of universal enveloping algebras, the result is trivial.

LetGbe a discrete group, U =CG] its group algebra over complex numbers.

Theorem 7

. Udoes not admit a nontrivial deformation of the standard coproduct. Proof. Assume that there is a rst order deformation of a coproduct,

g=gg+ X kl C kl g kl  (2.1.45) where2= 0.

(i) The coassociativity condition in the rst order ingives X kl C kl g (klg+kkl;gkl;kll) = 0: (2.1.46)

Collecting termsabc with xedaand c,a6=g andc6=g, one nds

C_gac_a=Cgac_{c :}

(2.1.47)

This holds for all a andc dierent from g. Therefore, only C_ggg_{, Cg}kg_{, Cg}gk _and C_gkk _{might dier from 0.}

Now the condition (2.1.46) becomes

X k6=g

fCggk(gkg;gkk) +Cgkg(kkg;gkg)

+C_gkk_(kkg;gkk)g= 0:

(2.1.48)

This implies that there is a set of constantsBkg fork6=g, s. t.

Cggk=Cgkg=Bkg  Cgkk=;Bkg  k6=g :

(2.1.49)

This solves the coassociativity condition. Thus, we have g= (1 +cg)gg+

X k6=g

B_kg(gk+kg;kk):

(2.1.50)

(ii) The condition that  is a homomorphism implies (in the rst order in):

Bg;1k h +Bkh;1 g =Bkgh and cgh=cg+ch: (2.1.51) Let g0= (1 +c g)g+X k6=g B_kgk : (2.1.52)

A direct calculation shows that (2.1.51) is exactly the condition saying thatg!g 0

is an algebra homomorphism,g0h0 = (gh)0.

Again a straightforward calculation shows that g0 =g0

g 0 :

(2.1.53)

Therefore, given a deformation of the standard coproduct, we can explicitly con-struct an isomorphism with the original bialgebra. The proof is nished.

In the same way as the Corollary 6 followed from the Proposition 3, we obtain the information about the formal deformations in this case.

Corollary 8.

At formal level, all deformations of the coproduct for the group algebras of discrete groups are trivial.

2.2. Lie algebras with an invariant scalar product.

We have seen in the previous subsection that the essential role in the theory of deformations of the coproduct on universal enveloping algebras is played by a tensor jk_i . All the conditions on the tensorare expressed in terms of the Lie algebra itself, without any reference to the deformation theory. The relevant classical notion is a \Lie bialgebra".

Denition.

A Lie bialgebra is a Lie algebraLequipped with a map:L! L,

Xi=jki XjXk, where the tensor(antisymmetric in the upper indices) satises

the Jacobi identity and belongs toZ1( L

2 L).

Both ; and  satisfy the Jacobi identity. The condition  2 Z

1, written

explicitly as (2.1.3), is symmetric in  $ ;. So the notion of the Lie bialgebra

is self-dual (like the notion of the Hopf algebra). In other words, if L is a Lie

bialgebra then there is a Lie bialgebra structure on the dual space L

, the roles

of ; and  being interchanged. There is an object which explicitly realizes this symmetry betweenand ;. It turns out (Exercise: verify it) that all the data for a Lie bialgebra can be conveniently expressed as the Jacobi identity for a larger Lie algebra with generatorsXi andXi, satisfying

XiXj] = ;kijXk  for generators of L

(2.2.1) Xi_Xj_{] =}ij_kXk _{ for generators of} L   (2.2.2) XiXj] =;;ikj Xk+jki Xk : (2.2.3)

This Lie algebra is called a Drinfeld double of the Lie bialgebraLand denotedDL.

As a vector space,DL is isomorphic toLL .

Denition.

A scalar product hxyi on a Lie algebra L (i.e. a nondegenerate

symmetric pairingLL ! k) is called invariant if hxy]zi = hxyz]ifor all

xyz2L.

Example: The Killing form on a semi-simple Lie algebra is invariant. The natural pairing betweenL andL

, given by

hXiXji= 0 hXiXji= 0 hXiXji=ji

(2.2.4)

is an invariant scalar product onDL. Moreover, the commutation relations between

XiandXjcan be reconstructed (with relations (2.2.1) and (2.2.2) being given) from

the demand of invariance of the natural pairing. Indeed, let XiXj] = Aj_ikXk+ B_ijkXk. Then

;Ajia=hXjXi]Xai=hXjXiXa]i=hXj;biaXbi= ;jia

and similarly forB_ijk.

Denition.

A set of data fgL 1

L 2

gwhere gis a Lie algebra with an invariant

scalar product,L 1and

L

2are isotropic Lie subalgebras of dimension = dim L

2 and

g=L 1

L

2 is called a Manin triple.

A Lie bialgebra L denes a Manin triple fDLLL  g. Conversely, a Manin triple fgL 1 L 2

g denes a Lie bialgebraL

1. From this perspective, the study of

Lie bialgebras splits into two parts: Lie algebras with an invariant scalar product their maximal isotropic subalgebras. I will shortly comment on the rst part.

Denote byfgga Lie algebragwith an invariant scalar product((xy) = hxyi). The pairfggis called indecomposable if it cannot be represented as a

direct sumfg 11

gfg 22

Example of fgg. Let M be a Lie algebra with generators Xi. Let g =

MnM

(the semi-direct product with respect to the coadjoint action). Then the

scalar product

hXiXji= 0 hXiXji=ij  hXiXji=ji 

(2.2.5)

where(XiXj) =ij is an arbitrary bilinear symmetric form, is an invariant scalar

product.

Generalization of this example. LetfWgbe a Lie algebra with an invariant

scalar product. Suppose that a Lie algebragacts onW by derivations, Taxy] =

Taxy] + xTay], whereT is the action,T :gW !W,aw7!Ta(w) suppose

that the operatorsTa,a2g, are antisymmetric with respect to the scalar product

onW,(Taxy) =;(xTay) for allxy 2W anda2g.

Exercise. Show that the map : 2W ! g

 dened by

ha(xy)i =(Taxy),

where hi is the natural pairing betweengand g

, is a 2-cocycle, 2Z 2(W g ) (g

 is considered here as a trivialW-module).

As a 2-cocycle, denes a central extension ofW by g

. In other words, the

bracket

xy] = xy]W +(xy)

(2.2.6)

where xy]W is the commutator of x and y in the Lie algebraW, denes a Lie

algebra structure onW g

. Denote this Lie algebra by ~W.

Exercise. Fora2gx2W andf 2g  let ~ Ta(x+f) =Tax+ ad af  (2.2.7) where ad

is the coadjoint action. Show that the formula (2.2.7) denes an action ofgon ~W.

We have therefore a Lie algebra structure on the space A = gW g : a

semi-direct productgnW~ with respect to the action (2.2.7).

Dene a scalar productAonA: the pairings between the generators ofgand g

are given by (2.2.5) the restriction of

AonW is all the other pairings are 0.

Exercise. The scalar productA is invariant.

The Lie algebraAwith the scalar productAis called the double extension of fWgbyS (and the action ofS onW).

Theorem (



5

]

).

If a Lie algebra with an invariant scalar product is not simple or 1-dimensional then it is either decomposable or a double extension. Moreover, one can always choosegto be simple or 1-dimensional.

This theorem gives a way to construct higher-dimensional Lie algebras with an invariant scalar product from lower-dimensional ones. However, this is not a classication.

Example of a nontrivial double extension: g=so(n),W is the n-dimensional

fundamental representation ofg considerW as an abelian Lie algebra. The cocycle,

giving a bracket on W g

 is given by the natural map  : W

^W ! g , and

A=gn(V g ).

Exercises.

In dimension 2 there is only one non abelian Lie algebra choose a basisfxyg

in such a way that the commutation relation is xy] =y. Denote this Lie algebra byL2.

1. Show that any bialgebra structure on L2 can be written (after possible

redenitions) in one of two forms:

x= 0 y=x^y

(2.2.8) or

x=x^y  y= 0:

(2.2.9)

2. Show that for the bialgebra structure (2.2.8) the double isgl

2 for (2.2.9)

the double is a semi-direct product C nN of a one-dimensional Lie algebra (with

a generatorW) and the three-dimensional Heisenberg algebraN (with generators

XYZ and relations XY] = Z, ZX] = ZY] = 0) the action of W on N is

given by WX] = 2X, WY] =;2Y and WZ] = 0.

3. Show that operations

x=x1 + 1x  y=y1 +exy (2.2.10) and x=x(1;2y) ;1+ 1 x  y=y1 + (1;2y)y (2.2.11)

provide Hopf algebra structures on corresponding completions ofUL 2.

4. Show that the Hopf algebra structure, dened by (2.2.10) (respectively, (2.2.11)) is a quantization of the Lie bialgebra structure (2.2.8) (respectively, (2.2.9)). Note that the terms of order 1 in the deformation parameter are not antisym-metric, but, as you remember, the symmetric part can be removed by redenitions.

5. LetL=sl 2

sl

2. Show that any invariant scalar product onLhas a form

c , where is the Killing form onsl

2 andc is a constant.

6. Let c = ;1. Show that the diagonalg 1 = sl 2 is isotropic. A subalgebra g 2 with a basis f(e +0)(0e;)(h

;h)gis a complementary isotropic subalgebra

(fhe +e; g is a standard basis in sl 2, he] = 2e , e+e;] =h). Thus, this

Manin triple provides a Lie bialgebra structure onsl 2.

7. Classify all Manin triples onL.

8. Classify 3-, 4- and 5-dimensional Lie algebras with an invariant scalar prod-uct.

2.3. Belavin-Drinfeld triples.

LetLbe a simple Lie algebra overC. In this

case, every 2-cocycle is a coboundary (see any textbook on Lie algebras, e.g., 

6

]) so one can solve the cocycle condition for : =@or, explicitly, jk_i XjXk =

0Xi], with  = 

ab_Xa Xb an element of the wedge square of L (that is,

ab₌;ba).

Now the Jacobi identity forcan be rewritten as a non-linear equation for the element.

Notation: for an elementA2UU,A= P xyletA 12= P xy1, A13= P x1y andA 23= P 1xy the elementsA 12,A13andA23 are fromUU U:

Exercise. Show that the Jacobi identity for can be rewritten in terms ofas 1213] + 1223] + 1323]Xi

11 + 1Xi1 + 11Xi] = 0

(2.3.1) for alli.

The element 1213] + 1223] + 1323] belongs to the third wedge power

of L, i.e., it has a form AijkXiXjXk with totally antisymmetric Aijk. The

space of invariant elements in 3

L, for the simple L, is known (see, e.g., 

6

]) to

be one-dimensional it is generated by an element  = ;ijk_XiXjXk, where

;ijk = ;kabBai_Bbj_{, Bij =}hXiXjifor the Killing form hi andBij is inverse to

Bij,BijBjk=ik (ikis the Kronecker delta). We conclude that the Jacobi identity

for=@is satised i 1213] + 1223] + 1323] is proportional to .

LetC=Bij_XiXj.

Exercise. Show that C12C13] + C12C23] + C13C23] is proportional to .

Therefore we can nd a combinationr=+ constC for which

r12r13] + r12r23] + r13r23] = 0:

(2.3.2)

Note that we still have jk_i XjXk = Xir] sinceB commutes with  0Xi for

alli. The equation (2.3.2) is called the classical Yang-Baxter equation (cYBe). We explained that for a simple Lie algebraLthe problem of nding the Lie bialgebra

structures onL reduces to cYBe for rwhich satises: r+r

0 is proportional toC, r+r0=xC withx 2C (r 0 is the !ip ofr,r0=rijX jXi forr 0=rijX iXj). If

x6= 0 one can setx= 1 by rescalingr.

The Yang-Baxter equation (which reduces to the cYBe in the classical limit) is

R 12 R 13 R 23= R 23 R 13 R 12: (2.3.3)

Solutions of the cYBe for whichx6= 0 are the most interesting - their quantizations

nd lots of applications in statistical models, knot theory, representation theory etc.

Exercise. In the situation of the exercise 6 from the previous subsection, show that the corresponding coproduct onsl

2arises from anr-matrix,r= 1 4h h+e ; e +.

Verify the cYBe for thisr.

We shall now explain how the solutions of cYBe withr+r0 =Care classied

in terms of so called Belavin-Drinfeld triples. A procedure of quantizing these solutions is known today 

7, 8

]. It is however interesting to enumerate the Belavin-Drinfeld triples, which is a combinatorial question in the end of this subsection we shall discuss and partly answer it.

Classication of solutions.

Fix a Cartan subalgebrah. LetRbe the set of roots,R=R +

R

;, and ;the

set of simple positive roots.

Denition.

A Belavin-Drinfeld triple (;1;2) consists of the following data: ;1

and ;2 are subsets in ; and : ;1 ! ;

2 is a one-to-one mapping which satises

(i) preserves the scalar product, that is,h()()i=hifor alland

from ;1.

(ii)  is \nilpotent". It means the following. Assume that (), which is an element from ;2 is still in ;1. Then 

2() is dened. If again2() 2 ;

1 then

there is 3(). Nilpotency means that the sequence must terminate, that is, for

somek2N, an elementk() is not any more in ;

1for any 2;

1.

Given a Belavin-Drinfeld triple, consider a system of equations for a tensor

r0 2hh, r0+r 0 0=t 0 (() id + id)(r 0) = 0 for all  2; 1 : (2.3.4)

Here t0 is the \Cartan part" of t: for a basis H of

hlet Bo = hHHi then

t0=B

o_{HH whereB}o _{is the inverse to B, B}o_{B = .}

The system (2.3.4) is compatible 

9

]. Recall thatg=h

L 2R

g, where hx] =(h)xforx2g, dimg= 1.

LetAi be a Lie subalgebra generated bye with 2;i,i = 12. Then Ais

the direct sum of those g for which the expansion of in terms of simple roots

contains simple roots from ;i only.

The map : ;1 !; 2extends to an isomorphism : A 1 !A 2(denoted also by ), by the formulae7!e

(). It is an isomorphism because the only relations in Aiare Serre relations which depend on the scalar producth::ionly and respects

the scalar product.

For each2R choosee in such a way that

(i)he ;e i= 1, (ii)e()=(e) whenever(e) is dened (e 2A 1).

Dene a partial order:  <  for2R means that there exists a naturalk

such thatk_{() =.}

Theorem



9

]

.

Let r=r0+ X 2R + e; e+ X 2R< (e; e;ee ;) (2.3.5) wherer0 is a solution of (2.3.4). Then

(i) the tensorr satises cYBE(r) = 0 andr+r0=t

(ii) any solution of equations cYBE(r) = 0 andr+r0 =t, after a suitable change

of the basis, is of the form (2.3.5).

Thercorresponding to the trivial Belavin-Drinfeld triple (;1and ;2are empty

sets), withr0= 1 2t

0, is called the \standard"r.

2.3.1. Maximal triples. I shall say several words about the combinatorics of Belavin-Drinfeld triples. The whole information about scalar products is contained in the Dynkin diagram for the algebra g. We shall consider the most interesting

case of the Lie algebras of the type A(that is, Lie algebras sl(n)), for which the Dynkin diagram is

Given a Belavin-Drinfeld triple, it is useful to draw a diagram, corresponding to it, like: x x x x x x x x x x @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ Fig. 2

The upper and lower rows are two copies of the Dynkin diagram A5, the lines

between the rows carry an information about the triple the lines should be thought as going from the upper low to the lower one the roots from ;1are the roots at the

upper row from which the lines start the roots from the lower row are those at which the lines end they are from ;2. The angles between the roots are determined by the

number of edges connecting the corresponding vertices of the Dynkin diagram it is therefore easy to understand, looking at the picture, whether the map preserves scalar products. To check the nilpotency one needs to draw more than two rows -depicting the powers of. For example, for the diagram on Fig. 2 one draws:

x x x x x x x x x x x x x x x x x x x x x x x x x @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ Fig. 3

The meaning of Fig. 3 is clear: the lines going from the rst row to the third one represent2, from the rst row to the fourth one represent3, etc.

There are two types of natural equivalences for triples: (i) Tl : (;

1;2) 7! (;

2;1

;1) this corresponds to a re!ection of the

picture of the triple in the horizontal mirror,T2 l = id

(ii) if a Dynkin diagram has a symmetry  then ((G1)(G2)

;1) is a

triple and the equivalence isT : (G1G2)

7!((G

1)(G2) ;1).

For An-diagram there is a symmetry: a re!ection of Fig. 1 in the vertical

mirror letT$ be the corresponding equivalence we haveT 2 $= id.

If ~;1 is any subset of ;1 then (~;1(~;1) j~

;1) is clearly a triple. So it is

interesting to look only for \maximal" triples, i. e. those to which one cannot add any more vertices.

For example, the only nontrivial triple forA2 is x x x x @ @ @ @ Fig. 4

Exercises. 1. Show that for A3 there are, up to equivalences, two maximal

triples: x x x x x x x x x x x x @ @ @ @ @ @ @ @ H H H H H H H Fig. 5 Fig. 6

2. Show that forA4 there are, up to equivalences, four maximal triples:

x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x @ @ @ @ @ @ @ @ @ @ @ @ H H H H H H H H H H H H H H            P P P P P P P P P P P @ @ @ @ P P P P P P P P P P P ; ; ; ; Fig. 7 Fig. 8 Fig. 9 Fig. 10

With the growth of rank it becomes more and more dicult to decide if a given triple is maximal. For example, the triple on Fig. 11

x x x x x x x x x x X X X X X X X X X X X X X X H H H H H H H            Fig. 11

is maximal, but one has to draw several rows (like on Fig. 3) to see that loops appear when one adds one more vertex.

If ;n;

1 consists of only one vertex (or #;1 = #;

;1) then the triple is

certainly maximal. We shall enumerate triples with #;1= #; ;1.

Proposition 9.

For the Dynkin diagramAl, the number of triples with #;1=l ;1

is 1

2%(l+ 1), % is the Euler function, %(n) = #

fj2f1:::ngjj is coprime tong.

Proof.

(i) %(l+ 1) is the number of primitive roots of unity of order (l+ 1).

We shall rst associate a Belavin-Drinfeld triple to any primitive root of unity of order (l+ 1). Let  = exp(2i

l+1). Label the vertices of the Dynkin diagramAl as

shown on Fig. 12:

x x x x

Fig. 12 . . .

2 3 l

Ifaandb are labels of two vertices thenais connected by an edge tobif and only if a=b.

Fix a primitive root q. Let ;1 = fqq 2:::ql;1 g and ; 2 = fq 2q3:::ql g

(more precisely, ;i,i= 12, are the sets of vertices ofAllabeled by the

correspond-ing roots of unity). Sinceqis primitive, each of the sets ;1 and ;2 contain (l ;1)

distinct elements. Let  : ;1

! ;

2 be the multiplication by q. Multiplying a label q

i _{byq, we}

obtain a sequenceqi!qi +1

!!ql, and the sequence terminates sinceql +1= 1

is not a label of any vertex. Thus, the map  is nilpotent.

The condition of being neighbors,qi_=qj is stable under the multiplication

byq, therefore preserves scalar products.

Thus, (;1;2) is a Belavin-Drinfeld triple. Call it Tq.

Consider an arbitrary Belavin-Drinfeld tripleT = (;

1;2) with #;1=l ;1.

We shall prove that it coincides with one ofTq's.

(ii) Denote the vertex omitted from ;1byq

;1. It divides the row of the diagram

Al (as on Fig. 1) into two segmentsI1 andI2:

w w k w w w w w Fig. 13 . . . . I1 z }| { I2 z }| { q ;1

We have q;1 =a for some a. Making, if necessary, a vertical re!ection, we

can, without loss of generality, assume that #I1 #I

2.

Let q0 be a label of a vertex omitted from ;

2. The lower row of the picture

corresponding to T is also divided by q

0 into two segments J

1 and J2 (J1 to the

left ofq0,J

2 to the right of it). The map preserves neighbors and it follows that

either  :I1 !J 1, I2 !J 2 or  :I1 !J 2,I2 !J

1. The former case is excluded

since otherwiseI1 =J1, I2 =J2 and restrictions of on the sets Ii, i = 12, are

permutation of these sets and therefore cannot be nilpotent. Thus, :I1 !J 2,I2 !J 1 andq 0=q.

We cannot have #I1 = #I2 { then restrictions of on Ii, i= 12, would be

permutations of these sets. Therefore, #I1<#I2.

(iii) Consider the restriction of  on the set I2,  : I2 ! J

1. There are two

possibilities:  preserves the order or reverses it. We shall prove that  cannot reverse the order. Indeed, if reverses the order then mapsqtoq;1(it is useful

to draw a picture here). Then induces a permutation on the set ;n(qq

;1) and

cannot be nilpotent.

Let us collect obtained information about the tripleT in a picture:

x x k x x x x x x x x x x k x x .                                                                       . . . . . . . . Fig. 14

(iv) For the restriction of  on the set I1 we have again two possibilities, the

order is reversed or preserved. We shall prove that it is preserved. If the cardinality ofI1 is 0 or 1, there is nothing to prove, so, without a loss of generality we assume

that #I1>1, in other words,a

3. Thus we havel6 since #I

2>#I1.

Extend to a map ~ : ;!; by ~ :q ;1

7!q (and ~ = on ;

1). The map ~

is a permutation of ;. Decompose ~ into a product of cycles. Since q;1 maps to

q, the decomposition contains a cycle c= (:::q;1q :::). If there are other cycles,

~

 =cc 1

c

2::: then the productc1 c

2::: is a permutation of some set S. This

permutation is the restriction of onS, thus cannot be nilpotent. We conclude that ~ is a cycle.

Explicitly, the action of is

8 < : a+i 7!i  i= 1:::n;a  i7! ;i  i= 1:::a : (2.3.6)

We shall follow a sequence ~n_(q;1). First, q;1 = a maps to q;1 = l+1;a.

Then it goes back, l+1;a 7!l

+1;2a

7!7!l

+1;ka, wherel+ 1

;kaa but

l+ 1;(k;1)a > a orl+ 1(k+ 1)abut l+ 1> ka. This requiresksteps (i.e.

this is the result of the action of ~k _onq;1). At the next step,l+1;ka maps to

ka_{and then again goes back,}ka7! (k;1)a

7!7!a. This takeskmore steps.

Thus, ~2k(q;1) =q;1.

But 3kak < l+ 1. Therefore, 2k < 2l+2

3 < lbecause lis at least 6.

Therefore, the permutation ~2k has a xed point and 2k < l. Thus ~ cannot

be a cycle.

We are left with only one possibility: the restriction of  onI1 preserves the

order.

(v) The map  preserves the order on I2 by (iii) and on I1 by (iv). Written

explicitly, it means that the map is the multiplication byq. It will not be nilpotent ifqis not primitive, therefore, the tripleT coincides with one ofTq's.

(vi) The group, generated by !ipsTlandT$, is Z 2 Z 2. But on triples Tq the

operationsTl andT$coincide: each of them is

Tq 7!Tq

;1 (in general, it is not so,

even for maximal triples: consider the triple corresponding to Fig. 11). Therefore, onlyZ

2acts on the space of triples

Tq. This action does not have xed points and

we conclude that the number of triples, under equivalences, is 1

2%(l+ 1) as stated.

The proof of the Theorem 9 is nished.

3. Quantum spaces

We shall brie!y, without going into details, give geometrical motivations which lead to the notion of quantum spaces.

LetGbe a Lie group,gits Lie algebra. In the section 2, Lie bialgebras appeared

in the study of deformations of the coproduct on the universal enveloping algebra

Ug. Geometrically, Lie algebra g is the Lie algebra of left-invariant vector elds

on G. The universal enveloping algebra of the Lie algebra g can therefore be

realized as the algebra of left invariant dierential operators onG. Up to topological and functional analytic considerations (convergence, etc.), a function on Rn can

be reconstructed, as a Taylor series, from the knowledge of its derivatives at the origin. For a Lie group G, the knowledge of derivatives of a function f at the origin is replaced by the knowledge of values on f of all left invariant dierential operators at the unity of G. The elements of Ug are linear functionals on the

space FG of functions on G, so, up to topological considerations, the spaces Ug

and FG are dual to each other, the pairing between X 2 Ug and f 2 FG is

given by hXfi = X(f)je, where e 2 G is the unity element. It follows then

that the coproduct onUgcorresponds to the product onFG- the usual product of

functions. Thus, deformations of the coproduct onUgcorrespond to deformations of

the commutative algebraFGof functions. Innitesimal deformations from Section

1 correspond to particular Poisson brackets on G { Poisson brackets which are compatible with the group structure. One says that Poisson brackets are compatible with the group structure if the multiplication m :GG!G is a Poisson map.

In other words: dene, for a given function f on G, a function f~on GG by

the rule f~(xy) =f(xy) the compatibility of the Poisson bracketsf::gmeans ffgg~=ff~g~g, where the Poisson brackets onGGare Poisson brackets of the

direct product of two Poisson manifolds. Groups with compatible Poisson brackets are called Poisson-Lie groups.

In the other direction, it is not dicult to check that if Gcarries compatible Poisson brackets then its Lie algebraggets a Lie bialgebra structure.

Compatible Poisson brackets are of a very special form. We shall illustrate it on an example of a matrix groupG(a subgroup of the group of invertible matrices). Leta_ijbe matrix elements. Assume thatfaikamsg= %imks(a) are compatible Poisson

brackets. Then faijb j

kamnbnsg = %imks(ab) (this is the equality ff~g~g = ffgg~

for f = a_ik and g = a_ms). On the other hand, faijb j kamnbnsg = faijamngb j kbns+ fb j kbnsgaijamn = %imjn(a)bikbns + % jn

ks(b)aijamn because GG is equipped with the

Poisson structure of the direct product. Therefore % must be homogeneous of degree 2 (this re!ects the fact that i in the deformation of the coproduct on Ug belongs to gg). Thus, the Poisson brackets are quadratic. To quantize

constant or linear Poisson brackets, one simply replaces the Poisson brackets by the commutator. However, it is not obvious how to quantize quadratic Poisson

brackets - we cannot replace Poisson brackets by the commutator because we don't know how to order consistently the quadratic right hand side.

Exercises. 1. Show that formulas

fabg=ab  facg=ac  fbdg=bd  fcdg=cd  fbcg= 0 fadg= 2bc

(3.0.1)

are Poisson brackets for four variablesa, b,c andd.

2. Show that if a,b,c and dare matrix elements of a matrix A=



a b c d



then the Poisson brackets (3.0.1) provideGL(2) with a Poisson-Lie structure. 3. Show that the Poisson brackets (3.0.1) of the determinant ofA, detAwith all matrix elements vanish.

The main interest about the most commonly appearing groups, likeGL, SO,

Sp, ... - is that they arise as groups of symmetry (of a vector space, of a vector space with a bilinear form, ... ). The Poisson-Lie groups one can interpret in this way too. One says that a Poisson-Lie groupGacts on a Poisson manifoldMin a

Poisson way if (i)Gacts onM

(ii) the actionGM!Mis a Poisson map, whereGMis equipped with

a Poisson structure of the direct product.

Again, for a matrix groupG acting on a vector space V2, xi

7!aijxj, a

sim-ilar calculation shows that the Poisson brackets of coordinates, fxixjgmust be

quadratic inxi_.

Exercise. For a two dimensional vector space with coordinatesx1andx2 let fx

1x2 g=x

1x2 :

(3.0.2)

Show thatGL(2) with the Poisson structure given by (3.0.1) acts in a Poisson way on this Poisson vector space.

It turns out that from the point of view of the theory of quantum groups, the appropriate way to quantize the Poisson brackets (3.0.2) is provided by the following commutation relations:

x1x2=qx2x1:

(3.0.3)

This is the rst example of a quantum vector space. Denote this quantum vector space (that is, the algebra of polynomials in x1 and x2 subjected to the relation

(3.0.3)) byV2 q.

The linear group of transformations preserving the relation (3.0.3) is poor, it consists only of rescalings xi 7! cixi with some constantsci. It is the quantum

group which is the right analogue of the symmetry group of the quantum vector space.

General picture.

LetU be a quasitriangular Hopf algebra with a universalR-matrixR. LetU 

be a dual Hopf algebra. The pairing betweenU andU

satises

haxyi=haxyi

We shall think of U as of analogue of the universal enveloping algebra of a

semi-simple Lie algebra. Ideologically (in the spirit of the Peter-Weyl theorem) the dual Hopf algebra is generated by matrix elements of representations ofU.

Letbe a representation ofU on a vector spaceV,maps an elementx2U

to a matrixTij(x).

The coproduct  on matrix elementsT_ij looks especially simple:

hTijxyi=hTijxyi=Tij(xy) =Tik(x)Tkj(y) =hTikxihTkjyi=hTikTkjxyi (3.0.4) and therefore T_ij=T_ikTkj : (3.0.5)

The commutation relations between the matrix elementsTijare expressed in terms of a numericalR-matrixR, the image of the universalR-matrixR=

P iaibiin the representation,R=P i(ai)(bi). Let x= P uv. hT 1T2x i=hT 1 T 2 X uvi= X T1(u)T2(v): (3.0.6)

By the denition ofRwe have P uv=R ;1 P vuR. Therefore hT 1 T 2 P uvi=hT 1 T 2 R ;1 P vuRi =T1 T 2( R ;1 P vuR) =T1 T 2( R ;1) T 1 T 2( P vu)T 1 T 2( R) =R;1 P T1(v)T2(u)R=R ;1 P T2(u)T1(v)R : (3.0.7)

The \" means matrix multiplication.

Arguments u and v are now in the same order as in (3.0.6). Therefore, T1T2=R

;1T

2T1R or

RT1T2=T2T1R :

(3.0.8)

Because of the form of this relation, this algebra is often called the \RTT"-algebra. The algebraU

 was rst written in the form (3.0.8) and (3.0.5) in 

10

].

We shall write the relations (3.0.8) in a dierent way. LetP be a permutation of factors inV V,P(xy) =yx. Let ^R=PR. Then (3.0.8) is equivalent to

^

RT1T2=T1T2R :^

(3.0.9)

A motivation to use ^R instead of R: the eigenvalues of ^R have a representation-theoretic meaning. A theorem, due to Drinfeld 

11

], says that there exists an element F such that R = F

21q tF

;1, where q = exp() ( is the deformation

parameter), t is the invariant tensor Bij_XiXj. Let C = BijXiXj. Now let

V V =

P

W be a decomposition of the tensor square of the space V into

irreducible representations. We have 0C=C

1 + 1C+ 2t, where  0 is the

classical coproduct. Therefore,tjW = 1 2C

jW ;CjV. Denote this quantity byt.

We haveR=PFPq (t)F;1, therefore, ^R=FPq (t)F;1.

First of all, V V decomposes into the symmetric and antisymmetric parts,

V V =S 2V



2V. The operatorP takes the value (+1) on S2V and ( ;1) on

2V. EveryW

 is either is either inS2V or in 2V so the \sign" ofW

, depending

on whetherW isS2V or 2V, sign(W

Sinceq t jW =qt we have ^ RjFW =FPq (t) jW=qt sign(W): (3.0.10)

Thus, the projector decomposition of ^R re!ects the decomposition of the tensor product of representations.

Polynomials in the matrix elementsT_ijare \quantized" functions on the group. We also had the Poisson brackets on the coordinates xi _{and x}i 7! Tijxj was a

Poisson map. On the quantum level we have the ^RTT relations for Tij. Which relations can we impose onx's in such a way that the mapxi7!Tijxj is an algebra

homomorphism? These relations should also quantize the Poisson brackets forx's. Since the quantization of the Poisson brackets forT_ij produced quadratic relations, we expect to have quadratic relations for the algebra of x's as well. Impose a set of quadratic relations forx's,E_ijxi_xj_{= 0,labels relations. Then forTxwe have} EijTiaxa_Tbj_xb_=EijTiaTbj_xa_xb_{, so we have to understand which tensors we can move}

throughT_iaT_bj.

The dening relation (3.0.9) shows that we can move ^R and therefore any function of ^R. As we have seen, ^R =P

(, where  =qt sign(W) and (

is the projector on the space FW. So essentially, all functions of ^R are linear

combinations of projectors.

Conclusion: covariant algebras are given by relations (()ijklxkxl= 0 for some (one or several). Denote byAl (lfor \left") the quadratic algebra dened by one projector (. Equally well, there is a covariant \right" algebraAr (the algebra of

covectorsxi), dened byxixj(()ijkl= 0, the covariance isxi7!xaTai.

An important fact is that the RTT-relations can be reconstructed from the requirement that all the algebrasA_l (or all the algebrasA_r) are covariant. Indeed,

A_l is covariant means

(()ijklTkaTlbxaxb = 0

(3.0.11)

therefore, (()ijklTkaTlbmust be proportional to ( whose lower indices areab:

(()ijklTkaTlb= )ijuv(()uvab :

(3.0.12)

Multiplying (3.0.12) by (()abcd with 6=, we nd

(T1T2( = 0

(3.0.13)

for all pairsfj 6=g.

Lemma 10.

The system (3.0.13) of relations for the matrix elementsT_ij is equiv-alent to the RTT-relations ^RT1T2=T1T2R^.

Proof.

(i) (3.0.9) implies (3.0.13). We have ^R = P

( moreover, P

(g = 1 is a decomposition of unity.

Multiplying ^RT1T2 =T1T2R^ by ( from the left and ( from the right, we nd

(T1T2( = (T1T2(. If

6

= then  6= and it follows that the relation

(T1T2( = 0 is satised.

We have ^RT1T2= ^RT1T2

1 =  (T

1T2(. The last expression, due to

the relations (3.0.13), can be rewritten as

X  (T1T2( : (3.0.14) Similarly, T1T2R^ = 1 T 1T2R^ = P

 (T1T2(. Again, due to (3.0.13), this

equalsP

 (T1T2(, which coincides with (3.0.14). Thus, (3.0.9) holds.

If ^R appears in the process of a deformation then there is a candidate for an especially nice quantum space. Again, V V =

L

W classically denote the set

of fg by J, J = J +

J

; where J =

fj signW = 1g. Then projectors

( = P 2J ( have ranks rk(  = N(N1) 2 . Therefore, a set (( ;) ij klxkxl = 0 containsN(N;1)

2 relations - exactly the number of relations which we have classically

for commuting variables. The quantum space dened by relations ((;) ij

klxkxl= 0 is

the only reasonable candidate for the quantization ofCN on which a groupGacted

in a Poisson way. Similarly, the quantum space dened by relations ((+) ij

klxkxl= 0

is the quantization of the algebra of odd (grassmanian) variables. ForGL(N), in the decompositionV V =S

2V 

2V, the summands S2V

and 2V are irreducible. It is natural therefore to call an ^R, which contains only

two projectors, an ^R-matrix of GL type. One usually rescales ^R to have ^R =

q(+ ;q

;1(

;, where (+ and (; are projectors, which are called, due to their

origin, the q-symmetrizer and the q-antisymmetrizer respectively (and we shall often denote (+ byS and (; byA).

To conclude:

 Geometrically and physically meaningful ^R-matrices decompose into

projec-tors 1( + 1 +:::k( + k | {z } S +1( ; 1 +:::l( ; l | {z } A (3.0.15)

and we know which projectors constitute the q-symmetrizer S and the q -antisymmetrizerA, respectively (as shown by underbracing in (3.0.15)).

The ranks of the projectors S = (+ = ( + 1 +:::( + k and A = (; = (; 1 + + ( ; l are classical, rk( = N(N1) 2 .

 Covariant algebras are quadratic algebras of the form (() ij

klxkxl = 0 or xixj(()ij_kl = 0 where ( is one of the projectors in the decomposition

(3.0.15).

 The algebra of functions on a quantum group is given by the relations (3.0.9)

and these relations are equivalent to the condition that all the algebrasA_l

are covariant in other words, the RTT-algebra can be reconstructed from theA_l algebras. The same holds if one replaces left algebras by the right ones.

3.1.

GL

type.

ForGL-type algebras we have only two projectors1, ^R=qS ;

q;1AIn this case, to reconstruct the RTT-algebra, it is enough to require covariance

of two algebras, Al + and A l ; dened by relations S ij klxkxl = 0 and Aijklxkxl = 0 respectively 

12

].

1_{When ^}Rhas two eigenvalues, one says that ^Ris of Hecke type. We require additionally that

the ranks of the projectors are xed rkS= N(N+1)

2 , rkA= N(N;1)

As we have seen, covariance of the algebraAlimplies the condition (T1T2(=

0 for all dierent from. Thus, in theGLcase, the covariance of the algebraAl +

implies

ST1T2A= 0

(3.1.1)

while the covariance ofAl

; implies

AT1T2S= 0:

(3.1.2)

On the other hand, the covariance of the algebra Ar

+ implies the same relation

(3.1.2). Together, relations (3.1.1) and (3.1.2), are equivalent to the RTT-relations. This shows that in theGLcase, one can interpret the RTT-relations in two ways: either as the condition of the covariance of the algebras Al

+ and A l

; or as the

condition of the covariance of the algebras Al

+ and A r

+. We shall use the latter

interpretation in the sequel.

The algebras Alr+ are the left and right quantum spaces. If they are good

deformations then the dimension d(Nk) of the space of polynomials of degree k

coincides with the dimension of the space of polynomials inNcommuting variables,

d(Nk) =  N+k;1 k  : (3.1.3)

So, quantum spaces are quadratic algebras with correct Poincare series. As we shall see below, the behavior of Poincare series is intimately related to the theory of quantum groups.

Denition.

Given a set of tensors E = fE = Eijg, ij = 1:::n, dene an

algebraAE with generatorsx

i _{and relations}

E_ijxi_xj_{= 0 for all  :}

(3.1.4)

Letd(Nk) be the dimension of the space of polynomials of degreekin xi_{. We say}

that AE has a Poincare-Birkho-Witt property (or that AE is a PBW-algebra) if

(3.1.3) is satised. In particular the range of the indexisf1::: N(N;1)

2 g.

Relation (ij_klxk_xl _{= 0 (with ( a projector) is an example of (3.1.4) but in}

general the tensorsE_{are not organized in a projector.}

Requirement thatxi_{are covariant, that is, that the relations (3.1.4) are satised}

byTijxj _{implies some relations betweenTij's.}

Assume that we are given two quantum spaces, Al

E with generators x i _and

relationsE_ijxi_xj _{= 0 andA}r

F with generatorsxi and relationsxixjF ij  = 0.

Denition.

We say that the quantum spacesAl

E and A r

F are compatible if the

covariance algebra ofTijhas the PBW property (that is, its Poincare series coincide with the Poincare series forN2variables).

Next subsection is a digression on the Poincare series then we shall continue with a discussion of compatible quantum spaces in dimension 3.

3.2. Technics of checking the Poincare series.

Consider an algebra with generators xi _{and relations (3.1.4). Sometimes it is useful to try to apply the}

diamond lemma 

13, 14

]. In its easiest form it says: assume that there is a basis

fx

1:::xN

witha < band all the monomials in the sum are lexicographically smaller or equal

xj_xi_{. Take these relations as instructions: replace x}j_xi _{by the right hand side.}

Apply these instructions to xk_xj_xi_{, wherek > j > i, in two ways, starting from} xk_xj _{or fromx}j_xi_{. If both ways will eventually produce the same result, to which}

no more instructions can be applied then the ordered monomialsx1 1 :::x

N

N form

a linear basis in the algebra, which implies that the algebra has the PBW property. Note that whether one can apply this procedure depends on a choice of a basis in the algebra. Such a basis might not exist.

We shall now describe another way of checking the Poincareseries which one can apply in theGLcase. It uses a dierential calculus on quantum planes, developed in 

15

].

3.2.1. Dierential calculus and Poincare series. Assume that the quadratic re-lations are given by

Aij_klxk_xl_{= 0 }

(3.2.1)

whereAis theq-antisymmetrizer in the Hecke ^R-matrix, ^R=qS;q ;1A.

Leti _{be generators of the odd quantum space for ^R, that is, the relations for} i _are

S_klijk_l_{= 0 :}

(3.2.2)

One can unify generatorsxi_andi_{into one quadratic algebra by requiring that} xi_j_{= ^R}ij_klk_xl _:

(3.2.3)

Exercise. Verify that relations (3.2.3) are compatible with (3.2.1) and (3.2.2). The compatibility here means the following. Let be a combination of quadratic ex-pressionsS_klijk_l_{. Then= 0 in the algebra with generators}i_{. Take an element} xi_{with someiand use (3.2.3) to movexto the right. We obtain an element of the}

form P

jjxj, with some quadratic (in 's) elements j. Since was 0, we must

havej = 0. In other words, for eachj, the elementj must be a combination of

expressionsS_klijk_l_{. In the same manner, there is a compatibility check when one}

movesi _{to the left through relations (3.2.1) forx's.}

At the next step, one adds \q-derivatives"@i in the generatorsxi. An algebra

of the derivatives@i is the algebra with generators@i and relations @i@jAjikl = 0

(3.2.4)

(note the order ofij).

To add @i, one needs cross-commutation relations with the already existing

generatorsxi _andi_{. These relations are:}

@ixj = 1 +qR^jsitxt@s

(3.2.5)

@ij = ( ^R;1)jsitt@ s:

(3.2.6)

Exercise. Verify that relations (3.2.5) and (3.2.6) are compatible with (3.2.1), (3.2.2) and (3.2.3) (in the above sense of compatibility).

Finally, one adds derivatives i in i. An algebra of the derivatives i is the

algebra with generatorsi and relations ijSklji= 0 :

(3.2.7)

The cross-commutation relations betweeni and the generatorsxi,i and@i are: ixj= ^Rjsitxts (3.2.8) ij= 1;qR^jsitts  (3.2.9) i@j = ( ^R;1)klji@ lk : (3.2.10) .

Exercise. Verify that relations (3.2.8), (3.2.9) and (3.2.10) are compatible with (3.2.1)- (3.2.7).

We shall need the following singlets, the Euler operators Ee and Eo and the

dierentialsdand:

Ee=xi@i Eo=ii  d=i_{@i  =x}i_{i :}

(3.2.11)

Their relations with the generators of the algebra are (Exercise: verify the relations): Eexi =xi(1 +q2E e) @iEe= (1 +q2E e)@i  Eei=iEe  iEo=Eoi  (3.2.12) E0x i_=xi_{Eo @iEo=Eo@i } Eoi=i(1 +q2E o) iEo= (1 +q2E o)i  (3.2.13) dxi_=i_+qxi_{d  @id=qd@i } di₌;qid  id= (1 + (q 2 ;1)Eo)@i;qdi  (3.2.14) xi_=qxi_{  @i= (1 + (q}2 ;1)Ee)i+q@i  i_=xi;qi  i=;q ;1 i : (3.2.15) Using operators Ne= 1 + (q2 ;1)Ee and No = 1 + (q 2 ;1)Eo, appearing in the

right hand sides of (3.2.14) and (3.2.15), one can rewrite (3.2.12) and (3.2.13) in a form Nexh1j=q 2x h1jNe Ne@j1i=q ;2@ j1iNe  Neh1j=h1jNe Nej1i=j1iNe (3.2.16) Noxh1j=xh1jNo No@j1i=@j1iNo N0h1j=q 2 h1jNo Noj1i =q ;2 j1iNo : (3.2.17) Exercise. Verify:

1. The Euler operators commute,

EeEo=EoEe :

(3.2.18)

2. Commutation relations between the Euler operators and the dierentials are

dEe= (1 +q2E e)d or Ned=q;2dN ed  Eod=d(1 +q2E o) or Nod=q2dN o  Ee=(1 +q2E e) or Ne=q2N e  Eo= (1 +q2E o) or No=q;2N o : (3.2.19)

3. The dierentials square to zero,

d2= 0  2= 0 :

(3.2.20)

4. For the anticommutator ofdandwe have  :=d+d=Ee+Eo+ (q2 ;1)EoEe= 1 q2 ;1( NoNe;1): (3.2.21)

The last exercise implies that

xi _=xi_(q2 + 1) i=i(q2 + 1):

(3.2.22)

LetMab be a space of polynomials inxand , of degreeainx and of degreeb in . For2Mab one nds, by induction,

=((a+b)q2 +q

2(a+b))

(3.2.23)

where theq-number (n)q is dened by (n)q = 1;qn

1;q = 1 +q+q 2+ +qn ;1. LetMn=a +b=nMabandM=  1 n=0Mn. The spaceMis a Z 2-graded vector

space, the grading is given by the degree of a monomial in's.

One can consider @i and i as operators acting on the spaceM. To this end,

one introduces a vacuum Vac, which satises @i Vac = 0 andi Vac = 0. LetX

be an expression inxi_,i_{,@i andi. To evaluate it on an element}2M, take an

element X. Using the commutation relations, we move all @i andi to the right

and evaluate on the vacuum. This gives an element ofM which we denoteX(). The consistency requires only that the vacuum is a representation of the algebra of

@i andi which is clearly true.

For instance, we have () = (a+b)q2for2Mab.

For each n, the space Mn = a

+b=nMab is nite dimensional. We have

dimMab = de(a)do(b), where de(a) and do(b) are dimensions of the spaces of

polynomials inxof degreeaand in of degreeb, respectively. The grading ofMab

is (;1)b.

The spaceMn is closed under the action ofdand. Therefore, the supertrace

of their anticommutator (of the operator ) vanishes, Str = 0, which implies

X a+b=n de(a)do(b)(;1)b(a+b)q 2 = 0 (3.2.24) for eachn.

One can write the set (labeled byn) of identities (3.2.24) in a compact form. Lett be an indeterminate. Denote byPe andPo the Poincare series for even and