edwin beggs- quasitriangular and differential structures on bicrossproduct hopf algebras

8/3/2019 Edwin Beggs- Quasitriangular and Differential Structures on Bicrossproduct Hopf Algebras

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arXiv:q

-alg/9701041v130Jan1997

Damtp/96-97

QUASITRIANGULAR AND DIFFERENTIAL STRUCTURES ON

BICROSSPRODUCT HOPF ALGEBRASEdwin Beggs

Department of MathematicsUniversity of Wales, Swansea

Singleton Park, Swansea SA2 8PP, UK

+

Shahn Majid1

Department of Mathematics, Harvard UniversityScience Center, Cambridge MA 02138, USA2

+Department of Applied Mathematics and Theoretical Physics

University of Cambridge, Cambridge CB3 9EW, UK

December 1996

ABSTRACT Let X = GM be a finite group factorisation. It is shown thatthe quantum double D(H) of the associated bicrossproduct Hopf algebra H =

kM

k(G) is itself a bicrossproduct kX

k(Y) associated to a group Y X, whereY = G Mop. This provides a class of bicrossproduct Hopf algebras which are qua-sitriangular. We also construct a subgroup YX associated to every order-reversingautomorphism of X. The corresponding Hopf algebra kXk(Y) has the samecoalgebra as H. Using related results, we classify the first order bicovariant differ-ential calculi on H in terms of orbits in a certain quotient space of X.

1 INTRODUCTION

The quantum double[1] of the bicrossproduct Hopf algebra H = kM k(G) associated to a finite

group factorisation X = GM has been studied recently in [2]. Here we continue this study withfurther results in the same topic, including a concrete application to the classification of the

bicovariant differential calculi on a bicrossproduct.

The bicrossproduct Hopf algebras have been introduced in [3] and [4], and extensively studied

since then. Factorisations of groups abound in mathematics, so these Hopf algebras, which are1Royal Society University Research Fellow and Fellow of Pembroke College, Cambridge2During 1995 & 1996

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tion of first order bicovariant differential calculi. By definition a first order differential calculus

over an algebra A means an A-bimodule 1 over A and a map d : A 1 obeying the Leibniz

rule (which makes sense using the bimodule structure). When A is a Hopf algebra the calculus is

said to be left-, right- or bi-covariant when d interwines the left regular coaction of A, the rightregular coaction, or both, with a coaction given on 1. It is known that first order bicovariant

calculi are related to the representation theory of the quantum double[6], allowing us to apply

our previous results[2] when A = k(M)kG. We find that the irreducible bicovariant calculi

correspond to the choice of an orbit in a certain quotient space of the group X along with an irre-

ducible subrepresentation of the isotropy subgroup associated to the orbit. We actually classify

the bicovariant quantum tangent spaces using the techniques developed in [7], and obtain the

corresponding 1-forms later by dualisation. The result is a constructive method which providesthe entire moduli space of bicovariant calculi on a bicrossproduct, as we demonstrate on some

nontrivial examples.

We note that in the physics literature an important bicrossproduct Hopf algebra is the -

deformed Poincare algebra[8], and for this Hopf algebra some examples of bicovariant calculi

have been obtained by other means[9]. Here the group G is the Lorentz group and M is R3>R.

Another important (and very similar) example is with G = SO(3) and M = R2>R in [4][10],

where the bicrossproduct is the algebra of observables (or quantum phase space) of a particularquantum mechanical system. It seems likely that a more geometrical version of the present

results should include such examples as well. Armed with a choice of differential calculus, one

may proceed to gauge theory (i.e. to bundles and connections) using the formalism of [ 11].

Preliminaries

We use the notation and conventions of [2], which are also the notations and conventions in the

text [10]. Briefly, let X = GM be a group which factorises into two subgroups. Then each group

acts on the other through left and right actions , defined by su = (su)(su) forall s M

and u G. Conversely, given two actions , obeying certain matching conditions

se = s, (su)v = s(uv); eu = e, (st)u = (s(tu)) (tu)

eu = u, s(tu) = (st)u; se = e, s(uv) = (su) ((su)v) .(1)

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we can build a double cross product group on G M with

(u, s)(v, t) = (u(sv), (sv)t), e = (e, e), (u, s)1 = (s1u1, s1u1). (2)

The associated bicrossproduct Hopf algebra H = kM k(G) has the smash product algebra

structure by the induced action of M and the smash coproduct coalgebra structure by the

induced coaction of G. Explicitly,

(s u)(t v) = u,tv(st v), (s u) =

xy=u s x sx y

1 =

u e u, (s u) = u,e, S(s u) = (su)1 (su)1 .

(3)

We work over a general ground field k. There is also a natural -algebra structure (s u) =

s1 su when the ground field has an involution. This happens over C, but can also be

supposed for any field with = for all k. The dual H

has a similar structure k(M)

kGon the dual basis,

(s u)(t v) = su,t(s uv), (s u) =

ab=s a bu b u

1 =

s s e, (s u) = s,e, S(s u) = (su)1 (su)1,

(4)

and (s u) = su u

1 when the ground field has an involution. The quantum double[1] is

a general construction D(H) = HopH built on H H with a double cross product algebra

structure and tensor product coalgebra structure. In our case the cross relations between H, Hop

are[2]

(1 t v)(s u 1) = ts(tvu1)1 (tvu1)u t (su)vu1 , (5)

where t = t(su)1.

2 More about D(H)

Here we extend results about the quantum double associated to a bicrossproduct in [2]. For our

first observation, it is known that to every factorisation X = GM there is a double factorisation

Y X where Y is also G M as a set and the action ofX is the adjoint action viewed as an action

on Y[10]. Here we give a similar but different double factorisation more suitable for our needs.

Proposition 2.1 Let Y = G Mop with group law (us).(vt) = uvts. Then there is a double

cross product group Y X (factorising into Y, X) defined by actions

usvt = ((sv)ts1u1)1(sv)

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usvt = us(vt)(us)1 = u(sv)((sv)ts1u1)((sv)ts1u1).

The second line is the adjoint action on X which we view as an action on the set Y.

Proof We show that these actions are matched in the required sense (see [10]). First note

that the following results are immediate from the definitions:

use = us , evt = e , use = e , evt = vt .

For the other results we must work rather harder. To derive the equation

us((vt).(wr) ) = (usvt).((usvt)wr) ,

begin by the formula given for as usvt = u(sv)yp where y = (sv)ts1u1 and p =

(sv)ts1

u1

. Then starting from usvt = y1

(sv) we calculate

(usvt)wr = y1 ((sv)w) ((svw)r(sv)1y) ((svw)r(sv)1y) ,

and on applying the rules for multiplication in Y we find the required formula,

(usvt).((usvt)wr) = u (sv) ((sv)w) ((svw)r(sv)1y) ((svw)r(sv)1y)p

= u (svw) ((svw)rts1u1) ((svw)rts1u1)

= us((vt).(wr)) .

Now we must prove that

((wr).(us))vt = (wr(usvt)).(usvt) .

Begin by calculating

wr(usvt) = wr(u(sv)yp) = ( (ru(sv)y)pr1w1)1 (ru(sv)y) ,

and then use the definition of multiplication in X to find

(wr(usvt)).(usvt) = ((ru(sv)y)pr1w1)1

(ru(sv)y) y1 (sv)

= ((ru(sv)y)pr1w1)1 ((ru(sv)y)y1) (ru(sv)) (sv)

= ((ru(sv)y)pr1w1)1

((ru(sv))y)1

((ru)sv)

= [((ru(sv))y) ((ru(sv)y)pr1w1)]1 ((ru)sv)

= [(ru(sv))y(pr1w1)]1 ((ru)sv) .

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Now consider a part of this equation

y(pr1w1) = ((sv)ts1u1)(((sv)ts1u1)(r1w1))

= (sv)ts1u1(r1w1) ,

and putting this into the previous equation we get

(wr(usvt)).(usvt) = [(ru(sv))(sv)ts1u1(r1w1)]1 ((ru)sv)

= [((ru)sv)ts1u1(r1w1)]1 ((ru)sv) .

On comparing this with

((wr).(us))vt = (w(ru)(ru)s))vt = [((ru)sv)ts1(ru)1(ru)1w1]1 ((ru)sv)

we get identical results since

(ru)1(ru)1w1 = ((ru)1(ru)1)(((ru)1(ru)1)w1) = u1(r1w1)

Theorem 2.2 D(H)=kXk(Y) as Hopf algebras, by

: D(H) kXk(Y), (s u t v) = (su)1t v(tv)1s1t .

Over a field with involution, the map preserves the star operation.

Proof The structure of D(H) in the basis used is in [2]. We check that the linear map is

an algebra isomorphism to the smash product induced by . Start with = x q t v, and

= s u r w in D(H), and multiply them together to get

= su,x v,rw (s uq tr w) ,

where

t = t(su)1 , v = (su)vu1 , s = ts(tvu1)1 , u = (tvu1)u .

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Now we can calculate

() = su,x v,rw ((suq)1tr w(trw)1s1tr)

= su,x v,rw (((su)q)1(su)1tr w(rw)1(t(rw))1s1tr)

= t(su)(tv)1,x (su)vu1,rw ((xq)1(tsu)1tr w(rw)1(tv)1s1tr)

= t(su)(tv)1,x (su)vu1,rw ((xq)1(t(su))(su)1r w(rw)1s1r)

= t(su)(tv)1,x (su)vu1,rw ((xq)1t(su)1r w(rw)1s1r) .

Here we have used

su = (t(su)1)s(tvu1)1((tvu1)u)

= ((t(su)1)su)((tvu1)1((tvu1)u))

= (t(su)1(su))(su)(tvu1u)1

= t(su)(tv)1 .

Conversely we can calculate the product in kXk(Y) as

()() = ((xq)1t v(tv)1x1t)((su)1r w(rw)1s1r)

= v(tv)1x1t , (su)1rw(rw)1s1r (xq)1t(su)1r w(rw)1s1r

= v(tv)1x1t , (su)1(rw)u(su)1 (xq)1

t(su)1

r w(rw)1s1r

= v , (su)1(rw)u (tv)1x1t , (su)1 (xq)1t(su)1r w(rw)1s1r .

It is now apparent that the expressions for () and ()() are the same.

To compare the coproducts, we use the coproduct of D(H), which is the tensor product one

D(H)(s u t v) =

ab=s , xy=v

(a bu t x) (b u tx y)

Applying to this, we find the following expression for ( )D(H)(s u t v) :

ab=s , xw=v

((su)1t x(tx)1a1t) ((bu)1(tx) w(tv)1b1(tx)) ,

Alternatively we can calculate

kXk(Y)(s u t v) = kX k(Y)((su)1t v(tv)1s1t)

=

yz=v(tv)1s1t

(su)1t y (su)1ty z ,

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where y, z Y, which is the smash coproduct for the coaction induced by the back-reaction .

We begin with the calculation

(x(tx)1a1t).(w(tv)1b1(tx)) = xw(tv)1b1a1t ,

which shows that if we replace y by x(tx)1a1t and z by w(tv)1b1(tx), that the conditions

of the summations are the same. It now remains to calculate

(su)1ty = (su)1tx(tx)1a1t = (a1su)1(tx) = (bu)1(tx) ,

as required.

If we apply to the unit of D(H), we get

s,v s e e v =

s,v e vs

1

,

which is the unit in kXk(Y).

Next we consider the counit,

kXk(Y)((s u t v)) = kXk(Y)((su)1t v(tv)1s1t) = v(tv)1s1t,e

The last -function splits into v,e(tv)1s1t,e, which is equal to v,es,e, which is in turn equal

to D(H)(s u t v ).

Finally we consider the antipode,

SkXk(Y)((s u t v)) = SkXk(Y)((su)1t v(tv)1s1t)

= ((su)1tv(tv)1s1t)1 ((su)1tv(tv)1s1t)

1

= (u1(tv))1 ((su)1(tv)s1(su))

1

= (tv)1u ((su)1(tv)u(su)1)

1

= (tv)1u u1(tv)1(su)(su) ,

where we remember in the last line to take the inverse for the Y group operation, and compare

this with

SD(H)(s u t v) = ((1 S(t v))(S1(s u) 1))

= ((1 (tv)1 (tv)1)((su)1 (su)1 1))

= (ts(tvu1)1 (tvu1)u t (su)vu1)

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where t = (tv)1, v = (tv)1, s = (su)1, u = (su)1 and t = t(su)1. Applying the

definition of , and using the fact that su = u1, we find

SD(H)(s u t v) = (tsu)1t (su)vu1(t(su)vu1)1(tvu1)s1t1t

= (tu1)1t u1vu1s1

= (tu1)u u1v(su)(su)

= (tv)1u u1(tv)1su ,

again as required. This concludes the proof of the Hopf algebra isomorphism. Now we show

that the star operation is preserved.

(s

u t v

) = ((I t1 tv

) (su

u1 I))

= (t(su)(t1(tv)u)1 (t1(tv)u)u1 t ((su)u1)(tv)u) ,

where t = t1(su). Applying the definition of , we get

(s u t v) = (t(su)u1)1t

(su)1(tv)u(t1(tv)u)1

(t1(tv)u)(su)1t1t

= ((t1su)u1)1t (su)1(tv)u(su)1

= t1(su) (su)1(tv)u(su)1 ,

(s u t v) = ((su)1t v(tv)1s1t)

= t1(su) (su)1tv(tv)1s1tt1(su)

= t1(su) (su)1(tv)u(su)1 ,

again as required.

This means that kXk(Y) inherits many of the nice properties of D(H). In particular,

it has a quasitriangular structure R and associated elements Q = R21R (the quantum inverse

Killing form) and u =

(SR(2))R(1) (the element which implements the square of the antipode)

in Drinfelds general theory of quasitriangular Hopf algebras[1].

Corollary 2.3 The bicrossproduct kXk(Y) is quasitriangular, with

R =

u,s,v,t

v1 us s1 (sv)t.

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The quantum inverse Killing form is

R21R =

u,vG s,pM

s1u1 u(sv)(pu)u1 v1p1 pusp1 .

and is non-degenerate as a bilinear functional k(X)kY. The elementu is the canonical element

u =

xX x x in kXk(Y), and is central.

Proof The computation is straightforward. Thus

( )(R) =

(s u e v) (t e s u)

=

(su)1 vs1 s u(su)1t1s

=

(s1v)1 us s1 vt

which yields the formula shown on a change of variables v to sv. We then compute R21R usingthe product in kXk(Y).

R21R =

u,s,v,t

s1 (sv)t v1 us

u,s,v,t

v1 us s1 (sv)t

=

u,s,v,t,u,s,v,t

s1 (sv)t

v1 us

v1 us

s1 (sv)t

=

u,s,v,t,u,s,v,t

(sv)t,v1 us us,s1(sv)t s1v1 us v

1s1 (sv)t

=

u,s,v,t,u

,s

,v

,t

(sv)t,v1u(sv)(sv) us,v(sv)1ts s1v1 us v

1s1 (sv)t

From the -functions here we can read off sv = v1u(sv), t = sv, u = v, and s =

(sv)1ts. If we rewrite these as v = u, s = tu1, u = u(sv)(tu1) and t = ts(tu1)1,

then on substituting p = tu1:

R21R =

u,vG s,pM

s1u1 u(sv)(pu)u1 v1p1 pusp1 .

Nondegeneracy ofR21R as a linear map D(H) D(H) is the so-called factorisability property

holding for any quantum double[10]. Hence it carries over in our case to a linear isomorphism

k(X)kY kXk(Y) or, equivalently, to a nondegenerate bilinear functional on k(X)kY.

Finally,

(u) =

(s u (su)1 (su)1)

=

(su)1(su)1 (su)1((su)1(su)1)1s1(su)1

=

(su)1(su)1 (su)1(su)1 =

us us

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on a change of summation variables in which su is replaced by s1 and su by u1. Finally,

the element u in any quasitriangular Hopf algebra implements the square of the antipode. But

for any bicrossproduct, the antipode is involutive, hence u here is central.

We also consider the -structure in the case where the ground field is equipped with an

involution.

Corollary 2.4 kXk(Y) is antireal-quasitriangular in the sense ( )(R) = R1.

Proof This is known for the quantum double D(H) of any Hopf -algebra[12], and hence

follows from Theorem 2.2; here we provide a direct proof for our particular case. Computing

using the bicrossproduct -structure, we have

( )(R) =

u,s,v,t

v v1 (us) s s1((sv)t) =

u,s,v,t

v v1u(sv)(sv) s v(sv)1ts

=

u,s,v,t

v u(sv) s vt

after a change of u, t variables in the last step. Meanwhile,

R1 = (S id)(R) =

u,s,v,t

(v1us)1 (v1 us)1 s1 (sv)t

=

u,s,v,t

sv (v1u(sv)(sv))1 s1 (sv)t =

u,s,v,t

sv (sv)1u1v(sv)1 s1 (sv)t

using the actions in Propsition 2.1. Note that the inversion in ( )1 is the inverse in Y =

G Mop. This gives the same as ( )(R) after changing variables to v = sv, s = s1,

u = (sv)1u1v. Here (sv)1 = s1(sv) = sv.

As an application, the finite-dimensional modules of any quasitriangular Hopf algebra have

a natural quantum dimension dim defined as the trace ofu in the representation. The modules

ofkXk(Y), as a cross product algebra, are just the Y-graded X-modules V such that |xv| =

x|v| for all v V homogeneous of degree | |.

Proposition 2.5 The quantum dimension of a general kXk(Y)-module V is

dim(V) =yY

traceVy (y)

where Vy is the subspace of degree y and (y) : Vy Vy is the restriction to Vy of the action of

y viewed as an element of X.

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Proof We write V = yVy for our Y-graded X-module. The action of f k(Y) is f(y) on

Vy. A general element x X acting on V sends Vy Vxy. Hence, in particular, y viewed in X

sends Vy Vy as yy = y from Proposition 2.1. This is the operator on Vy denoted (y). Let

{e

(y)

a } be a basis of Vy, with dual basis {fa(y)

}. Then

Tr (u) =

yY,xX

a

fa(y), (x x).e(y)a =

yY

a

fa(y),y.e(y)a =yY

Tr Vy(y).

For example, we may take the natural representation in k(Y) by left multiplication of k(Y)

and the left action of X induced by its action on Y. This is the so-called Schroedinger repre-

sentation of any cross product algebra. The spaces Vy are 1-dimensional with basis {y} and

(y)y = y(y1( )) = yy = y is the identity. So dim(k(Y)) = |Y| = dim k(Y), where |Y| is

the order of group Y. So for this representation the quantum dimension is the usual dimension.

Example 2.6 We consider the factorisation of the group S3 into a subgroup of order 3 and a

subgroup of order 2.

Proof Consider a factorisation of the group S3 of permutations of 3 objects, which we label 1,2

and 3. Let G be the subgroup consisting of the 3-cycles and the identity, and let the subgroup

M consist of the transposition (1, 2) and the identity. Then, in the notation of this section,

X = S3, and Y = GMop is a cyclic group of order 6. The left action of X on Y is the adjoint

action of the group S3 on the set S3, and the right action of Y on X is given by

uv = u , uv(1, 2) = u1 , u(1, 2)v = u(1, 2) , u(1, 2)v(1, 2) = u1(1, 2) ,

where u and v are any 3-cycles or the identity. This leads to a quasitriangular structure R on

kXk(Y), given by

R =

u,vG, tM

v1 u e vt +

u,vG, tM

v1 u(1,2) (1, 2) v1t .

What actually is the group Y X in this case? It is of order 36, and a short calculation will

show that it has no center. The possibilities, read off from a table of groups, are S3 S3 and

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of the group Y X shows that it has 15 elements of order 2, so it is isomorphic to S3 S3.

However the isomorphism does not seem to be obvious!

3 Subfactorisation from an order-reversing isomorphism

Let be an automorphism of X which reverses its factors GM (i.e. (G) = M and (M) =

G). It is shown in [2] that induces an semi-skew automorphism of D(H) (i.e. an algebra

antiautomorphism and coalgebra automorphism), which we denote :

(s u t v) = (tv) (tv) (su) (su). (6)

Via Theorem 2.2, we may view this as a semi-skew automorphism of kXk(Y). When the

ground field is equipped with an involution, we may follow by the star operation and obtain

an antilinear Hopf algebra automorphism .

Lemma 3.1 If is a factor-reversing automorphism of X then the induced antilinear automor-

phism of kXk(Y) is given by

(x y) = (x) (y)1

when y is viewed in X (and the inverse is also in X).

Proof We define via and (6). Thus,

((tv) (tv) (su) (su))

=

((tv)(tv))1(su) (su)((su)(su))

1

(tv)1(su)

=

(t1(su)) ((su)u1(tv)1(su))) = ((su)

1t) (t1s(tv)1v1)

= (s u t v) = ((su)1t v(tv)1s1t).

Comparing these expressions gives the result for after changing variables to general elements

of X, Y.

We observe that -invariant basis elements x y are characterised by the property that

x = x and (y) = y1 (computed in X).

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Proposition 3.2 There is a subgroup X ofX consisting of those elements x for whichx = x,

and a subset Y of Y consisting of those elements y for which y = y1 (inverse in X). The

actions , restrict to X, Y, forming a double cross product group YX factorising into

Y

, X

. The corresponding bicrossproduct kX

k(Y

) Hopf algebra has an isomorphic coalgebrato that of kM k(G).

Proof The proof that the action restricts is immediate. If we take x X and y Y, then

xy = xyx1 (adjoint action in the X multiplication). If we apply to this, then

(xy) = (xyx1) = (x)(y)(x1) = xy1x1 ,

which in the inverse (in X) of xy = xyx1, so xy Y.

The proof for the other action is rather more difficult. It will be most convenient to find

formulae for the elements of X and Y first.

If y = vt Y, then y = y1 (inverse in X), so we can substitute (y) = (v)(t) and

y1 = t1v1 and use uniqueness of factorisation to say that (v) = t1. Then we can write

y = Y(v) = v(v)1. Now we can write out a simple formula for the multiplication in Y as

y.y = Y(v).Y(v) = (v(v)1).(v(v)1) = vv(v)1(v)1 = vv(vv)1 = Y(vv) .

This shows that Y is actually isomorphic to G.

If x = us X, then x = x, so we can substitute

(x) = (u)(s) = x = us = (s1u1)1 (s1u1)1 ,

and use uniqueness of factorisation to say that (s) = (s1u1)1. Then u1 = s(s)1, so

we can write x = X(s) = (s(s)1)1s. This shows that X is bijective as a set with M.

Now consider the right action,

X(s)Y(v) = (s(s)1)1sv(v)1 = ((sv)(v)1s1(s(s)1))1 (sv)

= ((sv)(v)1(s)1)1 (sv)

= ((sv)(sv)1)1 (sv) = X(sv) .

In particular this shows that the result is in X.

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We therefore have a bicrossproduct Hopf algebra kXk(Y). Its coalgebra is determined

by the action of Y on X and the group structure of Y, hence we see that it is isomorphic via

the maps X, Y to the coalgebra of kM k(G).

Also, by construction, we see that if we equip k with the trivial involution then

kXk(Y) (kXk(Y))

as algebras. The right hand side denotes the fixed p oint subalgebra under the algebra auto-

morphism . The inclusion is clear from Lemma 5.1 (????) and the inclusion X X as

subgroups and k(Y) k(Y) (extension by zero) as X-module algebras. That is also a

coalgebra automorphism tells us further that the coproduct of kXk(Y) applied to elements

of the fixed subalgebra yields elements invariant under . It is natural to ask to what

extent the quasitriangular structure of kXk(Y) is likewise invariant.

Proposition 3.3 The quasitriangular structure of kXk(Y) obeys

( )(R) = R21

When the field has an involution, we have ( )(R) = R121 . Moreover, if 2 = id then

2

= id and ()2

= id.

Proof It is easier to do the first computations in D(H). There, we have

( )(R) =

u,v,s,t

(s u e v) (t e s u)

=

u,v,s,t

((ev) (ev) (su) (su)) ((su) (su) (te) (te))

=u,s

(1 (su) (su)) ((su) (su) 1)

where the sum over v, t are replaced by sums over t = (v), v = (t) and give the unit elements

of k(M) and k(G) respectively. Then we change variables from u, s to s = (su), u = (su),

to recognise R21 in D(H). Hence the same result applies for kXk(Y). This combines with

Corollary 2.4 to obtain the corresponding property for . Also, the automorphism in (6)

clearly obeys

2(s u t v) = 2(s) 2(u) 2(t) 2(v)

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and hence 2 = id when 2 = id. The same feature for is immediate from Lemma 3.1.

Hence R is not invariant in the usual sense (unless G, M are trivial), due to the nonde-

generacy of R21R in Corollary 2.3. Rather, one should note that for any quasitriangular Hopf

algebra, R121 defines a second conjugate quasitriangular structure. It corresponds in topo-

logical applications to reversing braid crossings; we see that our quasitriangular structure is

invariant up to conjugation in this sense. Although kXk(Y) does not in general inherit a

quasitriangular structure from R, its inclusion as fixed point subalgebra in a quasitriangular

Hopf algebra equipped with such an automorphism might b e a useful substitute. Of course, it

may still happen that kXk(Y) is quasitriangular for some other reason, which is the case

in the first example below.

Another natural question, in view of Proposition 3.2, is whether kXk(Y) is in fact

isomorphic as a Hopf algebra to our original bicrossproduct kM k(G). The following example

also demonstrates that it is not necessarily isomorphic to it.

Example 3.4 We consider the example in [BGM] of the doublecrossproduct of two cyclic groups

of order 6 (C6) which gives the product of two symmetric groups S3S3. In this case kXk(Y)

is isomorphic to kS3k(S3) in Example 2.6 and hence quasitriangular.

Proof Consider the group X = S3 S3 as the permutations of 6 objects labelled 1 to 6,

where the first factor leaves the last 3 objects unchanged, and the second factor leaves the first

3 objects unchanged. We take G to be the cyclic group of order 6 generated by the permutation

1G = (123)(45), and M to be the cyclic group of order 6 generated by the permutation 1M =

(12)(456). Our convention is that permutations act on objects on their right, for example

1G applied to 1 gives 2. The intersection of G and M is just the identity permutation, and

counting elements shows that GM = M G = S3 S3. We write each cyclic group additively,

for example G = {0G, 1G, 2G, 3G, 4G, 5G}. The action of the element 1M on G is seen to be

given by the permutation (1G, 5G)(2G, 4G), and that of 1G on M is given by the permutation

(1M, 5M)(2M, 4M).

The factor reversing automorphism can be taken to be conjugation by the permutation

(1, 4)(2, 5)(3, 6). Then if we split the elements ofX into S3 S3, we see that the elements of X

are of the form , for S3, and that the elements of Y are of the form 1. Then

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X is isomorphic to S3, and the action of X on Y is the adjoint action of the group S3 on the

set S3.

This is enough information to show that, despite having the same dimension, that kXk(Y)

is not isomorphic to kG

k(M) or k(G)

kM. All these can, as algebras overC

, be decomposedinto a direct sum of matrix algebras. According to [2], the algebras kGk(M) or k(G)kM

have at most 2 2 matrices in their decompositions. However the existence of an orbit of size 3

in the adjoint action of the group S3 on the set S3 shows that at least one 3 3 matrix occurs

in the decomposition of kXk(Y).

So what is kXk(Y)? In fact we have already met it, it is the bicrossproduct given by the

product ofS3 and a cyclic group of order 6 in example 2.6. The explicit correspondence is given

by deleting the points 4, 5 and 6 from the example here. Then we have the maps forX, and 1 for Y, corresponding to the subsets of S3 used in in example 2.6. But

from the previous example we see that kXk(Y) is actually the double of a bicrossproduct

obtained from a group of order 2 and a group of order 3. We deduce that as a result kXk(Y)

is actually quasitriangular. However there is no obvious relation between this quasitriangular

structure and the standard structure on kXk(Y).

Example 3.5 The upper-lower triangular example, in which kXk(Y) is isomorphic to

kM k(G).

Proof Let G be the group T+ of upper triangular n n matrices with ones on the diagonal,

with the usual matrix multiplication. Also let M be the group T of lower triangular n n

matrices with ones on the diagonal. As in [10] we define actions, using (a) = (aT)1,

su = 1 + (s)(u 1) , su = 1 + (s 1)(u) .

Using these actions we find

X(t).X(s) = X((t(s(s)1)1)s) = X(2ts s t + 1) .

Hence there is a group isomorphism X=T defined by X(t) 2t I. We call the group

isomorphism : T X, where (s) = X((s + 1)/2). Then we calculate

(s)Y(v) = X((s+1)/2)Y(v) = X(s + 1

2v) = X(1+(

s + 1

21)(v)) = X(1+

s 1

2(v)) = (sv) .

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As the map v Y(v) gives a group isomorphism from T+ to Y we might think that we are

on our way to showing that the XY doublecross product is isomorphic to the original T+T

doublecross product. To check this, we must perform the calculation of the left action:

X(s)Y(v) = Y(1 + (2 (s))1(s)(v 1)) = Y(1 + (2(s)1 1)1(v 1))

Now we use the fact that (s)1 = sT, and calculate

(t)Y(v) = X((t + 1)/2)Y(v) = Y(1 + (tT)1(v 1)) = Y(tv) ,

as required to prove the isomorphism of the doublecross products.

4 A *-representation ofD(H) on a Hilbert space

In this section we provide a Hilbert space representation of D(H) which is one of the motivations

behind Theorem 2.2. Recall that it was shown in [2] that representations of D(H) are G M-

bigraded bicrossed G M-bimodules. We shall use |w| for the G-grading and w for the

M-grading of a homogeneous element w of the representation.

Proposition 4.1 There is a representation of D(H) on a vector space E with basis {s,u|s

M, u G}, with gradings

|s,u| = u , s,u = s

and the group actions

ts,u = ts(tu)1,tu , s,uv = sv,(sv)1uv .

The corresponding action of D(H) is

(s u t v)r,w = v,w t1s(tv),r su,(su)1(tw)u .

Proof The definition of the group actions is made precisely so that the matching conditions

in [2] are true. The corresponding actions of the Hopf algebras H and H are

(t v)s,u = v,u ts,u , s,u(t v) = s,t s,uv ,

and the formula (a h)w = (hw)a gives the action of D(H) as

(I t v)s,u = v,u ts(tu)1,tu , (t v I)s,u = t,s sv,(sv)1uv ,

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which gives the formula stated.

As far as the original group doublecross product is concerned, the E representation is more

symmetric than the standard[10] Schroedinger representation ofD(H) in H, as we do not have

to decide to take the group algebra of one factor subgroup, and the function algebra of the other.

The E representation motivates the isomorphism in Theorem 2.2 in the following manner: If we

rewrite wr1 = r,w, then the action above gives

(s u t v)wr1 = v(tv)1s1t,wr1 (su)1t(wr1)t1(su) .

Compare this to the action of a single bicrossproduct kXk(Y), with Y-grading on homo-

geneous elements:

(x y)( ) = y,w x( ).

We see that the formulae agree if we set x = (su)1t, y = v(tv)1s1t, wr1 = wr1, and

use the adjoint action of X, xwr1 = x(wr1)x1 . This suggests trying a formula of the type

used for in Section 2.

Proposition 4.2 Over C, there is an inner product (, ) on E (conjugate-linear in the first

variable and linear on the second), defined by

(s,u, r,w) = s,r u,w .

With respect to this, D(H) with its natural -structure is represented as a -algebra.

Proof The inner product given is non-degenerate (in fact the s,u form an orthonormal basis).

We can then check the condition that

(,) = (, )

for any D(H). We shall only prove this in the case = I t v, and leave the other case

to the reader. First we calculate

(su, su) = v,u ts(tu)1,s tu,u .

Now = I t1 tv, and

(su, su) = tv,u s,t1s(t1u)1 u,t1u ,

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which is the same condition.

Proposition 4.3 E has a coalgebra structure

E(s,u) =

ab=s, wz=u

a,w b,z =

(1) (2) , (s,u) = s,e u,e ,

and becomes a left module coalgebra under the action of D(H)

Proof To show that E is a coalgebra, we first show that it is coassociative, which is

(I E)E(s,u) =

abc=s, wzv=u

a,w b,z c,v = (E I)E(s,u) .

Next we show that is a counit;

(I )E(s,u) =

ab=s, wz=u

a,w e,zb,e = s,u = ( I)E(s,u) .

For the module coalgebra condition, first we have to prove

E() =

(1)(1) (2)(2) = ()E() .

Again we shall only prove this in the case = I t v, and leave the other case to the reader.

We set = s,u

. The coproduct of D(H) is

() =

(1) (2) =

xy=v

(I t x) (I tx y) ,

and so we obtain

(1)(1) (2)(2) = v,u

xy=u , ab=s

ta(tx)1,tx (tx)b(tu)1,(tx)y .

Now we use = v,u ts(tu)1,tu and calculate

E() = v,u

xy=tu , ab=ts(tu)1

a,x b,y .

If we now use the correspondences a = ta(tx)1, b = (tx)b(tu)1, x = tx and y = (tx)y

we see that the formulae for (1)(1) (2)(2) and ()(1) ()(2) agree.

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Lastly we must prove that () = ()(). Using = s u t v and = r,w, we

have

() = v,w t1s(tv),r (su,(su)1(tw)u) = v,w t1s(tv),r su,e (su)1(tw)u,e

= v,w t1s(tv),r s,e (su)1(tw)u,e = v,w t1(tv),r s,e u1(tw)u,e

= v,w t1(tv),r s,e (tw),e = v,w t1(tv),r s,e w,e

= v,e t1(tv),r s,e w,e = v,e e,r s,e w,e

= (s u t v) (r,w) .

Now suppose that is an order reversing isomorphism of the doublecross product group

X = GM. As previously mentioned, from [2] we have an antilinear Hopf algebra automorphism

: D(H) D(H).

Proposition 4.4 There is an antilinear map : E E defined by

(s,u) = (u),(s) ,

which obeys

() = ()

() D(H).

Proof As usual we shall only prove this in the case = I t v, and leave the other case

to the reader. We begin with

(I t v) = ((tv) (tv)) I = (v) (tv)1 I ,

where we have used the equation (tv)(tv) = (v). Now

((v) (tv)1 I)(u),(s) = (v),(u)

(u)(tv)1 , ((u)(tv)1)1

(s)(tv)1.

Since is 1-1, we can replace (v),(u) by v,u. Also we calculate

(u)(tv)1 = ((tv)u1)1

and (u)(tv)1 = ((tv)u1)1

,

so the equation above becomes

((v) (tv)1 I)(u),(s) = v,u

((tv)u1)1

, ((tv)u1)(s)(tv)1.

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Now using the v,u to put v = u in the equation;

((v) (tv)1 I)(u),(s) = v,u (tu) , (t)(s)(tu)1 = v,u (tu) , (ts(tu)1) ,

which is the formula for (s,u) as required.

Proposition 4.5 The coproduct on E and the inner product are related by the formula

(E , E) = |X| ( , ) ,

where we use the tensor product inner product on E E.

Proof It is sufficient to prove this for basis elements,

(Es,u , Es,u) =

ab=s, ab=s, wz=u, wz=u

(a,w , a,w) (b,z , b,z)

=

ab=s, ab=s, wz=u, wz=u

a,a w,w b,b z,z

=

ab=s=s, wz=u=u

1 = s,s u,u |G| |M| = |X| (s,u , s,u) .

5 First order bicovariant differential calculi on H

In this section, we regard the Hopf algebra A = H = k(M)kG associated to a group factori-

sation GM as a coordinate ring of some non-commutative geometric phase space. This is the

point of view introduced in [4], where H is an algebraic model for the quantization of a particle

on M moving along orbits under the action of G. Here we develop some of the noncommutative

geometry associated to this point of view.

First of all, on any algebra A, one may define a first-order differential calculus or cotangent

space 1 in a standard way cf[13]

1. 1 is an A-bimodule.

2. d : A 1 is a linear map obeying the Leibniz rule d(ab) = (da)b + adb.

3. A A 1 given by a b adb is surjective.

When A is a Hopf algebra, it is natural to add to this left and right covariance (bicovariance)

under A. Thus[6]

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4. 1 is an A-bicomodule and the the bimodule structures and d are bicomodule maps.

Here 1 A and A 1 have the induced tensor product bicomodule structures, where A is a

bicomodule under its coproduct.

Cf. [14] one knows that compatible bimodules and bicomodules (Hopf bimodules) are of theform (say) 1 = V A for some left crossed A-module V. The latter in our finite-dimensional

setting means nothing more than left modules of the Drinfeld quantum double D(A). A par-

ticular module is ker A, a restriction of the canonical Schroedinger representation whereby

D(A) acts on A (by left multiplication and the coadjoint action, see[10]). As observed in [6],

the further conditions for 1, d amount to requiring V to be a quotient of ker as a quantum

double module. Then

da = ( id)(a(1) a(2) 1 a), a A, (7)

where : ker V is the quotient map. The right (co)module structure on V A is by

right (co)multiplication in A, the left module structure is the tensor product of V and left

(co)multiplication in A.

In the finite-dimensional setting which concerns us, one may equally well work in the dual

picture in terms of H = kM k(G) and L = V. By definition cf[6], a bicovariant quantum

tangent space L for H is a submodule of ker H under the quantum double D(H). Here

D(H) acts on ker by

hg = h(1)gSh(2), ag = a, g(1)g(2) a, g1, h H, g ker H, a H

as a projection to ker of the Schroedinger representation. We say that a quantum tangent

space is irreducible ifL is irreducible as a quantum double module. It corresponds to 1 having

no bicovariant quotients. This dual point of view has been used recently in [7], where the ir-

reducible bicovariant quantum tangent spaces over a general quasitriangular Hopf algebra have

been classified under the assumptions of R21R non-degenerate and a Peter-Weyl decomposition

for the left regular representation. In the same manner, we now classify the irreducible bicovari-

ant quantum tangent spaces L when H is a bicrossproduct. The corresponding 1 will be given

as well.

The Schroedinger representation ofD(H) in Hhas already been computed for H = kM k(G)

in [2]. Using the description of D(H) modules as M G bicrossed bimodules (as recalled in

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Section 4), it has[2]

|t v| = (tv)v1, t v = t

s(t v) = sts1 sv, (t v)u = tu u1v,

(8)

where s = s(tv)v1. We now use the isomorphism D(H)=D(X) (the quantum double of the

group algebra of X) also in [2] to transfer to an action of D(X). Since D(X) = k(X)>kX as

an algebra, this should make it easier to decompose representations into irreducibles. One can

also use our new isomorphism in Theorem 2.2 to transfer to an action of kXk(Y), but this

appears to be less natural for the present purpose.

Lemma 5.1 The action (8) defines an action of k(X)>kX on H given by

us(t v) = (sts1)u1 u(sv), us(t v) = u,v((tv)1v1)sv,(tv)1 t v,

where s = sv(tv)1.

Proof According to the general results in [2], the corresponding X-grading || || and X-action

can be computed from the M G-bicrossed bimodule structure as ||w|| = w1|w| and usw =

((s|w|1)w)u1 for all w in the module. Hence,

||t v|| = t1(tv)v1 = v((t1(tv))v1)(t1((tv)v1))

= v((tv)1v1)((tv)1v1) = v(tv)1v1

and

us(t v) = (s(t v))u

1 = (sts1 sv)u1 = (sts1)u1 u(sv).

Motivated by the form of ||t v|| in the proof of the preceding lemma, we chose new bases

for the vector space on which D(X) acts.

Lemma 5.2 Let vt = t1v1 v. Here {vt} is a basis of H labelled by vt X. Then the

action in Lemma 5.1 is

usvt = usvt(sv)1 , usvt = us,vtv1vt

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Proof Here ||vt|| = vtv1 = v(tv1)(tv1) gives the action of us by evaluation against

the degree. This can be written more explicitly as usvt = u,v(tv1)s,tv1vt and is thereby

equivalent to the action in Lemma 5.1. Moreover,

s(t1v1)s1 = (sv((t1v1)v)1)(t1v1)s1 = (sv(t1v1))(t1v1)s1

= (((sv)t1)v1)s1 = ((sv)t1(sv)1)(sv)1 = ((sv)t(sv)1)1(sv)1.

Note that (sv)1(sv)1 = v1 and (sv)1(sv)1 = s1 for any matched pair of groups.

Then

usvt = us(t1v1 v) = (s

(t1v1)s1)u1 u(sv)

= ((sv)t(sv)1)1(u(sv))1 u(sv) = u(sv)(sv)t(sv)1 = usvt(sv)1 .

Our task is to decompose ker H into irreducibles under this action of k(X)>kX. We

begin by decomposing the action in Lemma 5.2 into irreducibles and afterwards projecting to

ker . Note that Lemma 5.2 tells us that when we identify H=kX as linear spaces by the above

basis, the action of X is the linear extension of a certain group action of X on itself.

Proposition 5.3 LetX act on itself by the action usvt = usvt(sv)

1

as in Lemma 5.2. Let||vt|| ||vt|| = vtv

1 as an X-valued function on X.

(i) Let denote the equivalence relation onX defined by vt us if and only if ||us|| = ||vt||.

Then descends to an action of X on the quotient space X/.

(ii) Let [vt] X denote the isotropy subgroup of an equivalence class [vt] X/. Then

[vt] = {us X| us||vt|| = ||vt||us},

the centraliser of ||vt|| in X.

Proof (i) may be verified directly. However, it follows from Lemma 5.2 since an action

of k(X)>kX (where X acts on X by the adjoint action in the semidirect product) requires

||usvt|| = us||vt||(us)1. In terms of the group X, this is ||usvt|| = us||vt||(us)1. This

also implies (ii) since the group [vt] consists of us X such that usvt vt, i.e. such that

||usvt|| = ||vt||.

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We denote by O[vt] the orbit containing the point [vt] in X/.

Example 5.4 We may restrict attention to orbits of the form O[s], where s M. Then the

elements of the equivalence class [s] may be identified with the subset of M fixed under the action

of s, [s] = {sv| sv = v}. The stabiliser [s] consists of all elements of X which commute with

s. The action of[s] on elements of the equivalence class [s] is given by utsv = su(tv). In the

particular case where s = e, we get [e] = G and [e] = X.

Proof This may seem to be a rather specialised example, but in fact any orbit O[us] in X/

contains a point of the given form, since [s] O [us]. We compute,

[s] = {vt| vtv1 = s} = {vt| v(tv1) = e , tv1 = s} = {v(sv)| sv = v} .

This can be simplified if we note that if sv = v, then v(sv) = (sv)(sv) = sv, giving the

result stated. The action of [s] is computed as

utsv = utv(sv) = utv(sv)(tv)1 = utsv(tv)1 = sutv(tv)1 = su(tv)

as stated.

For X/, define S = kp1() kX, where p is the canonical projection to X/ . Here

S is the linear span of the elements of viewed as a leaf in X, and is a linear representation

since, by definition, the action of sends p1() to itself.

Proposition 5.5 LetO be an orbit inX/ under the action ofX. ThenMO = OS kX

is a subrepresentation under the action of k(X)>kX in Lemma 5.2. Morever, kX = OMO

is a decomposition of kX into subrepresentations.

Proof The action of usu1 k(X) on kX in Lemma 5.2 is the projection operator

usu1vt = usu1,vtv1vt.

This is evident since the action of k(X) is evaluation against ||vt|| (or, explicitly, put usu1 =

u(su1)(su1) into Lemma 5.2.) We see that [us] usu1 projects kX onto the subspace

S[us] and

kX =

S = O

O

S = O

MO (9)

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as vector spaces. Then the operator

O =

O

commutes with the action of k(X)>kX, and is a projection to MO. To see that O doescommute with the action of the algebra, we can calculate

.(x y) = (x y).x1 ,

and note that the operation x1 is a 1-1 correspondence on the set O.

In what follows, we fix an orbit O X/ and a base point 0 on it. We denote by the

isotropy subgroup at 0, and S = S0.

Proposition 5.6 Let S = S1 S n be a decomposition into irreducibles under the action

of . Let .Si = usSi when = us0 (this is independent of the choice of us). Then Mi =

O.Si MO are irreducible subrepresentations under k(X)>kX. Moreover, MO =

iMi is a decomposition of MO into irreducibles.

Proof Let x X be a choice of us such that us0 = . Define .Si = xSi. Now if we

take any x so that x0 = , then x1 x , so

xSi = x(x1 xSi) = xSi = .Si ,

as Si is a representation of . Moreover, .Si .Si = {0} for , distinct, so Mi spanned as

shown is a direct sum.

Next we show that k(X)>kX acts on Mi. Clearly, xMi Mi for all x X since

x.Si = xxSi = .Si, where = x = xx0 is another point in O. Meanwhile, The

action of the element us in k(X) is either either zero or one of the projections associated to

(9), and these are all zero or the identity on each S. Therefore the whole action of k(X)>kX

preserves the decompositions of the S, with the result that Mi are subrepresentations and

MO = iMi.

We now prove irreducibility. Let m Mi be nonzero. Then there is some such that the

projection m to .Si is nonzero, and then x1 m is a nonzero element of Si. Since Si is

irreducible under X, we know that vectors of the form x1 m, for , span all of Si.

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Since the projection is itself the action of an element of k(X)>kX, we see that Si is contained

in the space spanned by the action of this algebra on m. Moreover by using xSi = .Si we see

that every .Si is contained in the image of m under k(X)>kX. Hence Mi = (k(X)>kX).m,

i.e. Mi is irreducible.

These two propositions give a total decomposition of kX into irreducibles. In particular,

we obtain irreducible subrepresentations for every choice of orbit and every irreducible subrep-

resentation of the associated isotropy group. The converse also holds by similar arguments to

those in the preceding proposition.

Proposition 5.7 LetM kX be an irreducible subrepresentation underk(X)>kX in Lemma 5.2.

Then as a vector space, M is of the form

M = O

.M0

for some orbitO (with base point 0) and some irreducible subrepresentation M0 S under .

(Here .M0 = usM0 when = us0.)

Proof Consider M kX and let M = (M) for any X/. Choose a 0 so that M0 is

nonzero, write M0 = M0, and let be the stabiliser of 0. Now M0 must be a representation

of . IfS1 is an irreducible subrepresentation of M0 under , then the previous proposition

shows that

O.S1 M

is an (irreducible) representation of the algebra, where O is the orbit containing 0. Since M

is irreducible, this representation must be equal to M, and in particular M0 is an irreducible

representation of .

It is now a short step to obtain the classification for subrepresentations of ker H. Note

that every Hopf algebra is a direct sum k1 ker as vector spaces, with associated projection

(h) = h 1(h). In our case, remembering that

(vt) = (t1v1 v) = v,e ,

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we see that ker is spanned by the projected basis elements

vt (vt) = vt v,euG

ue.

Lemma 5.8 The action in Lemma 5.2 and the projected action

usvt = usvt, usvt = us,vtv1 vt

on ker are intertwined by : kX ker .

Proof First we check the kX actions, using and then acting in ker by us, to get

usvt = (usvt) v,e(1) = (usvt) ,

using the facts that us1 = 1 and (1) = 0. Now we check the k(X) actions, using and then

acting in ker by us, to get

usvt = (usvt) v,e(us1) = (usvt) v,e(1)us,e = (usvt) .

Theorem 5.9 The irreducible quantum tangent spaces L ker are all given by the following

two cases:

(a) For an orbit O = {[e]} in X/ , choose a base point 0 O. For each irreducible

subrepresentation M0 S of we have an irreducible quantum double subrepresentation M =

O.M0 kX, and an isomorphic quantum double subrepresentation L = (M) ker .

(b) For the orbitO = {[e]} inX/, choose the base point 0 = [e] O . For any irreducible

subrepresentation M0 S of other than the trivial one (multiples of 1), we have an irreducible

quantum double subrepresentation M = O.M0 kX, and an isomorphic quantum double

subrepresentation L = (M) ker . Here S = kG and = X, as in Example 5.4.

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Proof In the cases above, M = O.M0 is an irreducible representation of the un-

projected action. By the previous lemma : M L is a map of representations, where

L ker uses the projected representation. The map is onto, and if it is 1-1 then the two

representations are isomorphic, and hence L is also irreducible.The only case where the map is not 1-1 is where 1 M. Since k1 is a representation

and M is assumed irreducible, this is just the nontriviality exclusion stated in the theorem. We

have shown that the cases described lead to irreducible representations of ker .

Now suppose that we have L, an irreducible representation of ker . Then its inverse image

1L H is a representation of k(X)>kX, and it contains the subrepresentation k1. If

L = {0}, then there is at least one other irreducible subrepresentation M H so that k1M

1

L. But now M must be of the form described earlier, and by irreducibility M = L.

We are now left with the problem of finding the irreducible subrepresentations of S under

the group . We can decompose the representation S into a sum of irreducibles S = S1

. . . Sr, so in this case there would be at least r irreducible subrepresentations. However if

there are any equivalent representations in this list, there are many more possible irreducible

subrepresentations. This is a standard situation in representation theory and we briefly recall

its resolution, as follows.

Lemma 5.10 LetN be an irreducible representation of the group , and letM N U be an

irreducible representation of, for another representation U. Suppose that there is a vector inM

with a nonzero N component. Then there is a-map T : N U so thatM = {wT w| w N }.

Proof First we show that any w N occurs as the first coordinate of a vector in M. The

projection : N U N is a -map, so M N is a subrepresentation of N. As N is

irreducible, we see that M = N.

Now we show that for any w N there is at most one u U for which w u M. If we had

w u M and w u M, then on subtraction we would also have 0 (u u) M. But now

using 0 (u u) as a cyclic vector we could construct a representation contained in M which

consisted purely of vectors with zero N component. This would contradict the irreducibility of

M unless u = u.

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Now simply label this unique u as T(w). As (w T w) = w T w M for , we

must have T(w) = T(w).

Proposition 5.11 Suppose that M, an irreducible representation of the group , is containedin Nn11 . . . N

nkk , where the Ni are inequivalent irreducible representations, and ni gives

the corresponding multiplicities in the sum. Then there is an index j, and (1, . . . , nj) knj

(not all zero) so that

M =

0 . . . 0 (1w . . . njw) 0 . . . 0 Nn11 . . . N

nkk | w Nj

.

Proof A nonzero vector in M must have a nonzero component in at least one of the irreducible

summands. To reduce confusion, we shall assume that the first N1 summand has a such a

nonzero component. Then we can write M N1 U, where U is the sum of the rest of the

Nis. By the previous lemma, there is a -map T : N1 U so that M = {w T w| w N1}.

Now we can follow the map T with a projection to one of theNi summands in U and examine

the resulting map. If we project to another of the N1 summands, the result is a -map from an

irreducible representation to itself, and by Schurs lemma this must be scalar multiplication. If

we project to a Ni summand for i = 1, the result is a -map from an irreducible representation

to an inequivalent irreducible representation, and by Schurs lemma this must be zero.

Thus the irreducible subrepresentations of S in Theorem 5.9 are themselves classified as

follows. For a given orbit O with base point 0, decompose S into irreducible representations

under the action of . Write this as S = Mn11 . . . Mnkk , where the Mi are inequivalent

irreducible representations, and ni gives the corresponding multiplicities in the sum. The data is

then (O, M0, ), where M0 is an irreducible representation of occuring in S with multiplicity

n > 0, and P(kn). (Note that we take the projective space as only the ratios of the scalars

in the last proposition are required to specify the subspace.)

Bicrossproducts interpolate between group algebras and group function algebras. As a check,

we recover the seemingly quite disparate results known separately for these two cases.

Example 5.12 Suppose that G is trivial. Then H = kM. In this case is the same as equality

and X/= X. In this case the equivalence classes are singletons corresponding to points in M

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and st = sts1 is conjugation. Hence the orbits O are conjugacy classes in M. The action of

the isotropy group is trivial and hence this is the only data. We recover the result [15][7 ] that

the irreducible quantum tangent spaces are the projected spans of the conjugacy classes.

Example 5.13 Suppose that M is trivial. Then H = k(G). In this case X is one entire

equivalence class and X/ = {[e]}. Then there is only one orbit O = [e] and Theorem 5.9

reduces to the classification of quantum tangent spaces in ker k(G) (so O = kG) which

have been classified by the second author in [7 ] as irreducible subspaces under the left regular

representation induced by uv = uv.

We can also keep M, G non trivial but let or be trivial in the matched pair. In these

cases X is a semidirect product. Note that the two cases are quite different. In one, H is a

tensor product algebra. In the other it is a cross product kM


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The orbit decomposition First we find the allowed orbits:

Orbit 1 : O = {[e]}, choosing base point [e]. Then [e] = {e, (123), (321)} and = S3. The

action of on [e] is given by the formula utv = utvt1. We have the eigenspaces of the G action

M0 = e + (123) + (321), M1 = e + (123) + 2

(321), and M2 = e + 2

(123) + (321).Now M0 forms a 1-dimensional representation of , but this is annihilated by . The action of

(12) is to swap M1 and M2, so we get a 2 dimensional irreducible representation M1 M2

of , giving a 2 dimensional irreducible subrepresentation in ker .

Orbit 2 : O = {[(12)], [(13)], [(23)]}, choosing base point [(12)]. Then [(12)] = {(12)}

and = M. The action of on [(12)] is the trivial 1 dimensional representation, giving a 3

dimensional irreducible representation in ker .

The direct approach The space ker is spanned by the vectors {x| x S3}, where

x = x, and there is the linear relation e + (123) + (321) = 0. We use the relation to rewrite

(321) = e (123), giving a basis consisting of 5 elements. This is then split into two parts

by the action of X:

(1) The space spanned by the elements {e, (123)}. This has the X action use = u

and us(123) = us(123)s1 , where we remember to rewrite (321) = e (123). This gives a

2 dimensional irreducible representation.

(2) The space spanned by the elements {(12),

(13),

(23)}. This has the Xaction us

v(12) =

usvs1(12) for all v G. This gives a 3 dimensional irreducible representation.

Example 5.17 We apply the theory above to Example 3.4 where X = C6C6 and spell out the

final result. This is rather complicated to do directly (hence justifying our methods). It is one

of the simplest examples[2 ] of a true bicrossproduct Hopf algebra.

First we identify the possible values of ., and the orbits of these values under X = S3 S3.

Since vt = vtv1, the possible values and the orbits are simply given by looking at the

conjugacy classes of the elements t M in X. These are:

(Orbit 0) t = 0M gives the conjugacy class consisting only of the identity. We choose

[0M] as the base point for this orbit. Then [0] = X, and the equivalence class is [0M] = G (as

previously noted). The action of [0] = X on [0M] = G is given by the formula utv = u(tv).

Let us now look at the decomposition of S = kG into irreducibles under the action of

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[0] = X. The element 1G acts on any irreducible M S, and its action diagonalises, that

is M is a sum of Mr (r Z6), where each Mr is zero or frk, where

fr = 0G + r1G +

2r2G + 3r3G +

4r4G + 5r5G ,

for a primitive 6th root of unity. The action of 1M is to send fr to fr, so we get the 4

irreducible representations S1 = f0, S2 = f1, f5, S3 = f2, f4, and S4 = f3. The two

1-dimensional representations S1 and S4 are not equivalent, as S1 is the trivial representation

and S4 is not trivial. To show that S2 and S3 are not equivalent we use the trace of 1G on the

representations, which is + 1 on S2 and 2 + 2 on S3.

There are four inequivalent irreducible representations for this orbit, but one is annihilated

by , leaving three irreducible representations of ker on application of .(Orbit 1) t = 1M and t = 5M give the conjugacy class consisting of elements of the form

(any 2-cycle in 1,2,3)(any 3-cycle in 4,5,6). We choose [1M] as the base point for this orbit.

Then [1] = M, and the equivalence class is [1M] = {1Mv| 1Mv = v}. Since 1M acts on G by

the permutation (1G, 5G)(2G, 4G), we find that [1M] = {1M0G, 1M3G}. The action of [1] = M

on [1M] is given by the formula t1Mv = 1M(tv), which is the trivial action since both 0G and

3G are fixed by the left action of M.

The decomposition of S = k{1M0G, 1M3G} into irreducibles under the action of [1] = Mgives two copies of the trivial one-dimensional representation.

(Orbit 2) t = 2M and t = 4M give the conjugacy class consisting of elements of the form

(any 3-cycle in 4,5,6). We choose [2M] as the base point for this orbit. Then [2] = S3 C3,

where C3 is the group of permutations of{4, 5, 6} consisting of the identity and the two 3-cycles.

In terms of the factorisation, [2] consists of all elements of the form ut for u {0G, 2G, 4G}

and t M. The equivalence class is [2M] = {2Mv| 2Mv = v}. Since 2M has the trivial action

on G, we find that [2M] = {2M0G, 2M1G, 2M2G, 2M3G, 2M4G, 2M5G}. Under the [2] actionutsv = su(tv), [2M] splits into two orbits, {2M0G, 2M2G, 2M4G} and {2M1G, 2M3G, 2M5G}.

First we decompose k{2M0G, 2M2G, 2M4G} into irreducibles under the action of [2]. The

action of 2G on this vector space diagonalises, that is k{2M0G, 2M2G, 2M4G} = M0 M1 M2,

where M0 = 2M0G + 2M2G + 2M4Gk, M1 = 2M0G + 2M2G + 22M4Gk, and M2 =

2M0G + 22M2G + 2M4Gk ( being a primitive 3rd root of unity). The effect of 1M on these

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eigenspaces is to swap M1 and M2. The decomposition into irreducibles gives S1 = M0 (trivial

1-dimensional) and S2 = M1 M2.

Next we decompose k{2M1G, 2M3G, 2M5G} into irreducibles under the action of [2]. The

action of 2G on this vector space diagonalises, that is k{2M1G, 2M3G, 2M5G} = N0 N1 N2,where N0 = 2M1G + 2M3G + 2M5Gk, N1 = 2M1G + 2M3G +

22M5Gk, and N2 = 2M1G +

22M3G + 2M5Gk. The effect of 1M on these eigenspaces is to swap N1 and N2. The

decomposition into irreducibles gives S3 = N0 (trivial 1-dimensional) and S4 = N1 N2.

In fact the two 2-dimensional representations S2 and S4 are isomorphic, using the map

2M0G + 2M2G + 22M4G 2M1G + 2M3G +

22M5G and 2M0G + 22M2G + 2M4G

2(2M1G + 22M3G + 2M5G).

(Orbit 3) t = 3M gives the conjugacy class consisting of elements of the form (any 2-cycle in 1,2,3). We choose [3M] as the base point for this orbit. Then [3] = C2 S3, where

C2 is the group of permutations of {1, 2, 3} consisting of the identity and (1, 2). In terms of

the factorisation, [3] consists of all elements of the form ut for u {0G, 3G} and t M.

The equivalence class is [3M] = {3Mv| 3Mv = v}. Since 3M acts on G by the permutation

(1G, 5G)(2G, 4G), we find that [3M] = {3M0G, 3M3G}. The [3] action on [3M] is given by

utsv = su(tv) = suv.

Now decompose S = k{3M0G, 3M3G} into irreducibles under the action of [3]. The actionof 3G on this vector space diagonalises, that is S = M1 M2, where M1 = 3M0G + 3M3Gk

and M1 = 3M0G 3M3Gk. The decomposition into irreducibles gives S1 = M1 and S2 = M2,

which are not equivalent.

This completes our classification of the bicovariant quantum tangent spaces on bicrossprod-

ucts. It remains to dualise the results to obtain the corresponding first order differential calculi,

as outlined at the start of the section. Explicitly, the dual of the inclusion iL : L ker H is a sur-

jection iL : (ker H)

V, where V = L

. On the other hand, the inclusion j : ker H H du-alises to a map j : A (ker H)

where A = H. Since kerj = (imagej) = (ker H) = k1A,

the restriction j|ker A : ker A (ker H) is an isomorphism. Putting these together, we get a

quotient map V : ker A V. Recall also that we can describe a representation L of D(H) as

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a left H and right H-module, with actions obeying the compatibility condition

h(xa) =

((h(1)a(1))x)(h(2)a(2)) ,

for all x L, which can be further computed in terms of the mutual coadjoint actions. (We freelyidentify a left Hop module as a right H-module.) For a given a H, the action a : L L

dualises to (a) : V V where V = L. Similarly for the operators h. We obtain in this way

a left action of A = H and a right action of H on V, by av = (a)(v) and vh = (h)(v),

which make V into a D(A) representation. This is a general observation about the dualisation

of quantum double modules. Combined with the projection V and d in (7), we obtain the

corresponding first order differential calculus 1 = V A.

To fully specify the resulting exterior differential d it is equivalent and more convenient togive its evaluations v = (v, id) d against all v L. These operators v : A A are the

braided vector fields associated to elements v L. The term is justified because they obey a

braided version of the Leibniz rule[7]. Since we obtain L from the map : M L, we specify

equivalently the operators

m = (m, id)( id) d, m M.

Example 5.18 The operators in a few cases from Example 5.17.

We take orbit 0 in Example 5.17, and write

fr = 0M 0G + r0M 1G +

2r0M 2G + 3r0M 3G +

4r0M 4G + 5r0M 5G ,

where is a primitive 6th root of unity. From its definition,

fr(t v) =

ab=t

a bv,fr b v

a

a 0G, fr t v ,

After a little calculation this reduces to

fr(t v) = (r(tv) 1) t v .

We use the allowed spaces for M for this orbit, which are f3k, f1, f5k, and f2, f4k.

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Next we take a case from orbit 2, the irreducible representation given by M0 = g1, g2k,

where

gr = 2M0G + r2M2G +

2r2M4G = 4M 0G + r4M 2G +

2r4M 4G ,

and is a primitive 3rd root of unity. In this case the orbit consists of more than one point, in

fact O = {[2M], [4M]}. We choose x[4M] X so that x[4M][2M] = [4M], for example x[4M] = 1G.

Now we can add 1Gg1 and 1Gg2 to g1 and g2 to get a basis of the 4-dimensional representation

specified by O and M0, that is M = g1, g1, 1Gg1, 1Gg2k, where

1Ggr = 1G2M + r3G2M +

2r5G2M = 2M 1G + r2M 3G +

2r2M 5G .

Then we may calculate the evaluations

gr(t v) =

(t4)v,0 + r(t4)v,2 +

2r(t4)v,4

t4 v t v ,

1Ggr(t v) =

(t2)v,1 + r(t2)v,3 +

2r(t2)v,5

t2 v .

References

[1] V.G. Drinfeld. Quantum groups. In A. Gleason, editor, Proceedings of the ICM, pages

798820, Rhode Island, 1987. AMS.

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181:112151, 1996. To appear.

[3] M. Takeuchi. Matched pairs of groups and bismash products of Hopf algebras. Commun.

Alg., 9:841, 1981.

[4] S. Majid. Physics for algebraists: Non-commutative and non-cocommutative Hopf algebras

by a bicrossproduct construction. J. Algebra, 130:1764, 1990. From PhD Thesis, Harvard,

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-Poincare group and on the -Minkowski space. Czech J. Phys., 46:201208, 1996.

[10] S. Majid. Foundations of Quantum Group Theory. Cambridge Univeristy Press, 1995.

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Math. Phys., 157:591638, 1993. Erratum 167:235, 1995.

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edwin beggs- quasitriangular and differential structures on bicrossproduct hopf algebras

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