Representations of Quantum Groups

1

Introduction

The term quantum group usually refers to two types of objects, both associatedwith a Lie group G and dependent on some parameter q in the underlyingfield. The quantised co-ordinate ring Oq(G) is a deformation of O(G), thering of polynomial functions on G, and the quantised enveloping algebra Uq(g)is a deformation of U(g), the universal enveloping algebra of the Lie algebracorresponding to g. These are dual as Hopf algebras, and we recover the classicalobjects O(G) and U(g) when we take the limit q −→ 1.

This essay will almost entirely deal with the simplest case of Uq(sl2), our mainresult being a description of all its irreducible representations over C. Uq(sl2)behaves very differently depending on whether or not q is a root of unity, so weneed to treat the two cases separately. The intuitive moral is that, when q isnot a root of unity, complex representations of Uq(sl2) exhibit behaviour strik-ingly similar to complex representations of sl2, so many techniques for analysingsl2-representations carry over almost verbatim. Everything works ”as well aspossible” in this non-root-of-unity setting - all finite-dimensional representationsare completely reducible, and a neat combinatorial algorithm using very littledata allows us to decompose any finite-dimensional representation into irre-ducibles. In contrast, when q is a root of unity, the complex representations ofUq(sl2) resemble modular representations of sl2 - for example, the dimension ofirreducible representations is bounded - and our results are far less pleasant. Wewill encounter representations which are not completely reducible, and findingthe irreducible composition factors becomes much more tricky. I will close byoutlining how this viewpoint extends to U(g) for arbitrary Lie algebras g.

Contents

1 The definition of Uq(sl2) 4

2 Finite-dimensional representations of Uq(sl2) when q is not aroot of unity 6

2.1 A baby example . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

2.2 The structure of irreducible representations . . . . . . . . . . . . 6

2.3 Complete Reducibility . . . . . . . . . . . . . . . . . . . . . . . . 10

2.4 Tensor representations, and decomposition into irreducibles . . . 15

3 Finite-dimensional representations of Uq(sl2) when q is a root ofunity 18

3.1 Some examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

2

3.2 Classification of irreducible representations . . . . . . . . . . . . 20

3.3 Composition factors and tensor representations . . . . . . . . . . 26

3

1 The definition of Uq(sl2)

Recall that is the Lie algebra with basis elements where ,

[H,E] = 2E, [H,F ] = −2F, [E,F ] = H (1.1)

Its universal enveloping algebra U(sl2) is an associative algebra containing sl2,with the Lie bracket given by the commutator operator. An explicit constructioninvolves taking the tensor algebra of sl2 and quotienting out by the two-sidedideal generated by the following Serre relations,

HE − EH = 2E, HF − FH = −2F, EF − FE = H (1.2)

By the Poincare-Birkoff-Witt theorem, the set of monomials ErHsF t : r, s, t ∈N form a basis for U(sl2). Another important corollary of this theorem is thatthe representations of U(sl2) are in bijection with those of sl2 - more precisely,any representation of U(sl2) is completely determined by the image on sl2, andany sl2-action on a vector space can be extended to a valid U(sl2)-action.

How might one derive some conditions for a quantised version of by ”deforming”the usual Serre relations (1.2)? In the spirit of quantum mechanics, set q = e~/2,and consider:

K = e~H/2 =∞∑i=0

1i!

(−~H

2

)i(1.3)

(Since we aim only for an intuitive argument, assume without proof that theusual exponential properties hold with this definition.) Differentiating expres-sions in K with respect to ~ and setting ~ = 0 would give expressions in H; wewill do this by naively taking first-order approximations in ~. For example, Kwill be approximated by 1− ~

2H.

First observe that as

K −K−1

q − q−1=

sinh(−~H/2)sinh(−~/2)

' (−~H/2)(−~/2)

= H (1.4)

as ~ → 0 so a plausible change to the third Serre relation reads EF − FE =K−K−1

q−q−1 .

We obtain a commutator when we differentiate a conjugate. Hence, using thefirst Serre relation, we see

KEK−1 '(

1− ~2H

)E

(1 +

~2H

)' 1− ~

2(HE − EH) = 1− ~

2(2E) ' q2E

(1.5)and we clearly can do the same with the second relation. So make the following

4

Definition. Uq(sl2) is the free algebra generated by the symbols E,F,K,K−1,quotiented out by the two-sided ideal generated by KK−1 = 1,K−1K = 1,together with the following quantised Serre relations:

KEK−1 = q2E (QSR1)

KFK−1 = q−2F (QSR2)

EF − FE =K −K−1

q − q−1(QSR3)

Without introducing Yang-Baxter equations, whose solution prompted the in-vention of quantum groups, it is difficult to justify setting K = qH rather thansome other function of q and H. One possible answer is that as defined aboveretains two desirable properties of U(sl2): the monomials ErHsF t : r, s, t ∈ Nprovide a PBW-type basis; and it admits the following Hopf algebra structure:

∆(E) = K ⊗ E + E ⊗ 1; ∆(F ) = 1⊗ F + F ⊗ 1K−1; ∆(K) = K ⊗K (1.6)

ε(E) = 0; ε(F ) = 0; ε(K) = 1

S(E) = −K−1E;S(F ) = −FK;S(K) = K−1

(Other Hopf algebra structures exist on Uq(sl2), but the above is most commonin the literature.) This enables us to define well-behaved tensor product anddual representations of Uq(sl2).

5

2 Finite-dimensional representations of Uq(sl2)when q is not a root of unity

2.1 A baby example

We defined Uq(sl2) as a free algebra quotiented out by an ideal of relations. So arepresentation of Uq(sl2) is precisely an algebra homomorphism ρ : A→ GL(V )for some vector space V with the relation ideal contained in ker ρ . Since ρis a homomorphism, it is completely specified by the image of the generators.Hence we aim to find linear maps E,F,K,K−1 which satisfy the quantised Serrerelations. Because we also require KK−1 = I and K−1K = I, K must be aninvertible linear map, and K−1 is the inverse map. Hence it suffices to defineE,F,K.

Let’s create our first representation by modifying the familiar action on sl2 givenby:

E =(

0 10 0

),F =

(0 10 0

),H =

(1 00 −1

)(2.1)

Naively set K = e~H/2 again. Since H is diagonal, the non-zero components ofpowers of H are simply powers of each component, so K has the form shownin (2.2) below. Direct computation shows that the quantised Serre relations aresatisfied; hence we have a representation defined by:

E =(

0 10 0

),F =

(0 10 0

),H =

(q 00 q−1

)(2.2)

The key feature here is that, analogous to the classical case, K is completelydiagonalisable, and E,F permute the K-eigenspaces in opposite directions. Thiscan be summarised pictorially, with each node representing a one-dimensionalK-eigenspace:

Figure 2.3 - L(1,+) representation of Uq(sl2)

It will follow from the upcoming theorem that every irreducible Uq(sl2)-modulehas a similar visual description.

2.2 The structure of irreducible representations

Our goal is to prove:

6

Theorem. The irreducible finite-dimensional representations of Uq(sl2), whenq is not a root of unity, are L(n,+) and L(n,−), for all n ≥ 0. L(n,+) andL(n,−) are both n+ 1-dimensional.

• For any v ∈ L(n,+) with Ev = 0,v,Fv,F2v, . . .Fnv

is a basis of

L(n,+), and Uq(sl2)-action on L(n,+) is given by:

EFrv =qr − q−r

(q − q−1)2(qn−r+1 − q−n+r−1

)Fr−1v ∀r > 0; Ev = 0

FFrv = Fr+1v ∀r < n; FFnv = 0

KFrv = qn−2rFrv

In other words:

E =

α1

α2

. . .αn

; F =

11

. . .1

; K =

qn

qn−2

. . .q2−n

q−n

where αr = qr−q−r

(q−q−1)2

(qn−r+1 − q−n+r−1

).

• For any v ∈ L(n,−) with Ev = 0,v,Fv,F2v, . . .Fnv

is a basis of

L(n,−), and Uq(sl2)-action on L(n,−) is given by:

EFrv = − qr − q−r

(q − q−1)2(qn−r+1 − q−n+r−1

)Fr−1v ∀r > 0; Ev = 0

FFrv = Fr+1v ∀r < n; FFnv = 0

KFrv = −qn−2rFrv

In other words:

E = −

−αα2

. . .αn

; F =

11

. . .1

; K = −

qn

qn−2

. . .q2−n

q−n

where αr = qr−q−r

(q−q−1)2

(qn−r+1 − q−n+r−1

).

Under this notation, L(1,+) is the example shown in Subsection 2.1, and L(0,+)is the trivial representation given by the counit map. Figure 2.4 illustrates twomore examples, L(3,+) and L(2,−) ; string diagrams for the other irreducible

7

representations are similar except in the number of nodes (since this is thedimension). The diagrams for L(n,+) and L(n,−) are identical up to labelling,since in both cases the K-eigenspaces are permuted by E and F in the samemanner - indeed, L(n,−) can be obtained L(n,+) from by composing E, Kwith −I.

–

Figure 2.4 - L(3,+) and L(2,−) representations of Uq(sl2), when qis not a root of unity

It is relatively straightforward to check that the matrices presented above definevalid representations - that is, they satisfy the quantised Serre relations. It iseasiest to check this separately for each basis vector (we have KEK−1(Frv) =q2E(Frv)∀r and so on). Now we show that L(n,+) and L(n,−) are irreducible.Any Uq(sl2)-submodule must contain a K-eigenvector; in all the cases listedabove, K has distinct eigenvalues, so this must be a multiple of one of the basisvectors. Given any pair of basis vectors, repeated application of E or F sendsone to a non-zero multiple of the other, as αr 6= 0. Hence this submodule mustcontain all basis vectors, and is therefore all of L(n,+) or L(n,−).

So the meat of the proof will be to show that no other irreducible representationscan exist. Before we start on this, we need some more terminology. Let Vbe an arbitrary finite-dimensional irreducible representation. Write Vµ for thesubspace of K-eigenvectors of eigenvalue µ; this is the µ-weight space of V , andany v ∈ Vµ is a µ-weight vector. If, in addition, Ev = 0 (so v ∈ Vµ∩ker E), thenv is a highest weight vector, and (provided v 6= 0 ) µ is a highest weight of therepresentation V (any µ with Vµ 6= 0 is a weight of V ). Hence highest weightvectors correspond to the end nodes of a string diagram. In the statement ofthe theorem, v is a highest weight vector, and the form of the matrices showthat, for each irreducible representation, the unique highest weight is preciselythe weight involving the highest power of q, hence the name. Explicitly, L(n,+)has highest weight qn and L(n,−) has highest weight q−n; since q is not a rootof unity, these are all distinct, so irreducible representations of Uq(sl2) can berecognised by their highest weights alone.

The proof of this classification theorem is only very slightly modified from thestandard procedure for analysing finite-dimensional irreducible sl2 representa-tions (found, for example, in Chapter 7 of [Humphreys]). In short, the argumentruns as follows: finite-dimensionality guarantees the existence of a highest weightvector; its image under repeated F-application is Uq(sl2)-invariant, and usingthe finite-dimensional hypothesis a second time gives a restriction on possiblevalues of the highest weight. Pay particular attention to the calculations (2.5),(2.6) and (2.7); we will use these over and over.

Step 1 V contains a highest weight vector v.

8

Since C is algebraically closed, some Vµ is non-zero. Take non-zero w ∈ Vµ .Recall that K is invertible, so µ 6= 0 and K−1w = µ−1w. Using (QSR1):

K(Ew) = KEK−1(µw) = q2E(µw) = q2µ(Ew) (2.5)

So Ew is another eigenvector of K, with eigenvalue q2µ. (2.5) holds for anyK-eigenvector, so by inductively applying it to Erw we find that, ∀r ≥ 0,Erw ∈ Vq2rµ. As q is not a root of unity, these eigenvalues are all distinct.Hence the non-zero elements of w,Ew,E2w, ... are linearly independent. V isfinite-dimensional, so, for at least one i, we must have Eiw = 0. By definitionof w, i ≥ 1, so we can define v = Ei−1w, then v ∈ Vq2(i−1)µ ∩ ker E as required.Call this highest weight λ instead of q2(i−1)µ.

Step 2 The non-zero vectors of the set v,Fv,F2v, ... form a basis for V .

Perform the same calculation as (2.5) above, with F in place of E:

K(Fw) = KFK−1(µw) = q−2F(µw) = q−2F(µw) (2.6)

for any w in any Vµ. Repeated application of (2.6) to Frv then shows that Frv ∈Vq−2rλ , so, by the previous argument, non-zero elements of v,Fv,F2v, ... arelinearly independent.

To show v,Fv,F2v, ... that span V , we show that E,F,K each send eachFrv into the subspace spanned by v,Fv,F2v, .... Since these three elementsgenerate Uq(sl2), all of Uq(sl2) must then fix this subspace, which must be allof V as V is irreducible. Our assertion is clear for F and K. As we saw in Step1, E maps Vq−2rλ to Vq−2(r−1)λ , so if v,Fv,F2v, ... indeed span V , EFrv mustbe some multiple αr of Fr−1v. We can verify this inductively, with the help of(QSR3) (in the base case of r = 1, use the fact that Ev = 0):

EFrv =K−K−1

q − q−1Fr−1v + FEFr−1v

=q−2r+2λ− q2r−2λ−1

q − q−1Fr−1v + Fαr−1Fr−2v

= αrFr−1v

(2.7) allows us to calculate αr explicitly, since α0 = 0:

αr =q−2r+2λ− q2r−2λ−1

q − q−1+ αr−1

=r∑i=1

q−2i+2λ− q2i−2λ−1

q − q−1

=qr − q−r

(q − q−1)2(q1−rλ− qr−1λ−1)

9

Step 3 For each integer n ≥ 0, there are precisely two irreducible representa-tions of dimension n+1, whose highest weights are qn and q−n respectively.

As V is finite-dimensional, the infinite set v,Fv,F2v, ... must contain someelements equal to zero. Let n be minimal with Fn+1v = 0; then for all r ≥ 0,and by minimality of n, v,Fv, ...,Fnv are all non-zero, so V has dimensionn+ 1. EFn+1v = 0 forces

αn+1 = 0⇒ q−nλ = qnλ−1 ⇒ λ = ±qn (2.9)

as claimed. Since q is not a root of unity, these two possibilities are distinct.

We now have enough information to write down E,F,K explicitly with respectto the basis v,Fv, ...,Fnv. The form of F is obvious; (2.6) in Step 2 statesthat Frv ∈ Vq−2rλ = V±qn−2r , so K is diagonal and has entries has shown in thetheorem for L(n,+); substituting λ = ±qn into (2.8) gives the correct form forE. QED

Finally, we isolate out Step 2 as a separate result, since this observation is usefulwhen decomposing representations into irreducibles. It concerns the structureof Uq(sl2)-modules which are generated by a single highest weight vector, calledhighest weight modules.

Proposition. Let v be a highest weight vector, and suppose q is not a root ofunity. Then the non-zero elements of v,Fv,F2v, ... is a basis of weight vectorsfor the Uq(sl2)-submodule generated by v.

2.3 Complete Reducibility

Recall that finite-dimensional representations sl2 of are completely reducible- in this subsection we prove the same for Uq(sl2), when q is not a root ofunity. We will again adapt an sl2 argument; this differs from the standardproof in the literature (found, for example in I.4.5 of [Brown, Goodearl] and2.9 of [Jantzen]). However, the core idea is the same - we locate some highestweight vectors, associate subspaces to them, then show that these are linearlyindependent, span V and are Uq(sl2)-invariant.

First, we define the Casimir element

Ω = FE +Kq +K−1q−1

(q − q−1)2((2.10))

Step 1 Ω is central in Uq(sl2)

10

It suffices to show that Ω commutes with the generators E,F,K. Taking theproduct of (QSR1) and (QSR2):

KFEK−1 = (KEK−1)(KFK−1) = q−2Fq2E = FE (2.11)

and K clearly commutes with the second term of Ω . Observe also

EΩ = EFE +EKq + EK−1q−1

(q − q−1)2

= FE2 +KE −K−1E

q − q−1+KEq−1 +K−1Eq

(q − q−1)2= ΩE

ΩF = FEF +KFq + EKFq−1

(q − q−1)2

= F 2E +FK − FK−1

q − q−1+FKq−1 + FK−1q

(q − q−1)2= FΩ

Now take any finite-dimensional representation of V and let 0 = V 0 ⊆ V 1 ⊆· · · ⊆ V d = V be a composition series for V - the V i are Uq(sl2)-submoduleswith each composition factor V i/V i−1 irreducible.

Step 2 The eigenvalues of Ω on V are of the form ε qn+1+q−n−1

(q−q−1)2 , where n ∈ N,ε ∈ +,−

Suppose v ∈ V satisfies Ωv = cv. Take i minimal with v ∈ V i, so v, theimage of v under quotienting by V i−1, is non-zero. Observe that Ωv = cv also.V i/V i−1 is an irreducible representation, so Step 1 and Schur’s lemma saysthat Ω acts on V i/V i−1 as scalar multiplication. Hence we must have Ωw = cw∀v ∈ V i/V i−1. By the classification Theorem 2.2, V i/V i−1 ' L(n, ε) for somen ∈ N, ε ∈ +,−. Then, taking w to be the highest weight vector shows thatc = ε q

n+1+q−n−1

(q−q−1)2 .

Step 3 For each c ∈ε q

n+1+q−n−1

(q−q−1)2 : n ∈ N, ε ∈ +,−

, set V(c) =v ∈ V : (Ω− cI)dimV = 0

.

Then V =⊕

c V(c) where c ranges over the above set. For each c =

ε qn+1+q−n−1

(q−q−1)2 , V(c) is an Uq(sl2)-submodule, and each of its compositionfactors is L(n, ε).

The vector space decomposition V =⊕

c V(c) follows from considering the Jor-dan normal form of Ω. Each V(c) contains a Ω-eigenvector of eigenvalue c, so

by Step 2, only cs of the form ε qn+1+q−n−1

(q−q−1)2 occur in the sum. Since Ω,cI are

11

both central in Uq(sl2), (Ω − cI)dimV is central also, so V(c) , being its kernel,is Uq(sl2)-invariant as required.

From now on, fix c = ε qn+1+q−n−1

(q−q−1)2 and fix 0 = V 0(c) ⊆ V 1

(c) ⊆ · · · ⊆ V d(c) = V(c)

a composition series of V(c) . Suppose V i(c)/Vi−1(c) ' L(mi, δi) . For each i, take

vi ∈ V i(c)\Vi−1(c) and let vi ∈ V i(c)/V

i−1(c) be its image under natural projection.

As in the proof of Step 2, Ωvi = δiqmi+1+q−mi−1

(q−q−1)2 vi . But vi ∈ V(c) means(Ω− cI)dimV vi = 0 , so we must have

δiqmi+1 + q−mi−1

(q − q−1)2= c = ε

qn+1 + q−n−1

(q − q−1)2∀i (2.14)

⇒(εqn+1 − δiqmi+1

) (1− εδiq−n−1−mi−1

)= 0

⇒ εqn+1 = δiqmi+1 or εδi = q−n−1−mi−1

Since q is not a root of unity, the second alternative cannot hold, and the firstalternative is only true when mi = n, δi = ε. Hence all the composition factorsL(mi,δi) ∼= V i(c)/V

i−1(c) are L(n, ε) as desired.

So it suffices to show that each V(c) is completely reducible - to do this, werecycle many concepts from the proof of the classification Theorem 2.2.

Step 4 Any weight of V(c) must also be a weight of L(n, ε).

Take v ∈ V(c)with Kv = λv and choose i minimal with vi ∈ V i(c). As usual,write vi ∈ V i(c)/V

i−1(c) for its image under natural projection. Kv = λv , so λ is

a weight of V i(c)/Vi−1(c)∼= L(n, ε).

Step 5 For x ∈ V(c), x ∈ ker E ⇐⇒ (K− εqnI)mx = 0 for some m ∈ N

Suppose y ∈ ker E is a K-eigenvector. Then, as shown in Step 4, y ∈ V i(c)/Vi−1(c)

is a K-eigenvector with the same eigenvalue. The same logic says y ∈ ker E ,so from the structure of L(n, ε), we see that y must be a highest weight vector,with weight εqn. Hence y has weight εqn also.

From (QSR1), ker E is invariant under K. Consider the Jordan normal formof K-action on ker E. This is an upper triangular matrix all of whose diagonalentries are eigenvalues of the K-action on ker E - by the previous paragraph,these eigenvalues are all εqn. It follows that (K− εqnI)dim kerE must be the zeromap on ker E.

Conversely, if (K − εqnI)mx = 0 , then (K − εqnI)m(Ex) = 0 , since (QSR1)shows

(K − εqn+2)E = q2EK − εqn+2E = q2E(K − εqn) ((2.15))

12

repeated applications of which give

(K − εqn+2)mE = q2mE(K − εqn) (2.16)

From Step 4, we know εqn+2 is not an weight of V(c), so det(K− εqn+2I) 6= 0 .Hence (K− εqnI)m is non-singular, so we must have Ex = 0.

Step 6 Construct a basis x1, x2, .., xd of ker E such that Frxi : 1 ≤ i ≤d, 0 ≤ r ≤ n is a basis of V(c).

First let e1, e2, ..., ed(n+1) be a basis of V(c) such that, for each i, e1, e2, ..., ei(n+1)

span V i(c) and e(i−1)(n+1)+1, e(i−1)(n+1)+2, ..., ei(n+1) is a standard basis for V i(c)/Vi−1(c)∼=

L(n, ε). In other words, with respect to this basis, the matrices have this blockform:

E =

E′ ∗ · · · ∗

E′. . .

.... . . ∗

E′

; F =

F′ ∗ · · · ∗

F′. . .

.... . . ∗

F′

; K =

K′ ∗ · · · ∗

K′. . .

.... . . ∗

K′

(2.17)

where E′,F′,K′ are as displayed in the classification Theorem 2.2 and describeUq(sl2)-action on (* denotes unknown entries).

Take x1 = e1; then x1 ∈ ker E, and Frx1 = er−1 for 0 ≤ r ≤ n. By choice ofeis,

(K− εqn)e(n+1)+1 =n+1∑i−1

aiei (2.18)

for some ai ∈ C. Note that

(K− εqn)ei = (εqn−2(i−1) − εqn)ei 6= 0 (2.19)

for 2 ≤ i ≤ n = 1, so, by subtracting suitable multiples of these ei from e(n+1)+1,we find x2 with (K − εqn)x2 = a1e1. Ke1 = εqne1, so , (K − εqn)2x2 = 0 andx2 ∈ ker E” by Step 5.

Repeat this process with e2(n+1)+1: for each i with (n+ 1) + 2 ≤ i ≤ 2)n+ 1),

(K− εqn)ei = non-zero multiple of ei + element of V 1(c) (2.20)

so subtracting multiples of these ei from e2(n+1)+1 can ensure that(K− εqn)(resulting vector) ∈ span of e1, e2, ..., e(n+1)+2. Then we remove theei components, 2 ≤ i ≤ n+1, on the right hand side, in the same way as before,and produce x3 with (K− εqn)3x3 = 0.

Continuing this process gives x1, x2, .., xd ∈ ker E with xi = e(i−1)(n+1)+1.More precisely, xi = e(i−1)(n+1)+1 + vi−1 with vi−1 ∈ V i−1

(c) . From the form of

13

F, we see that Frxi = e(i−1)(n+1)+r+1 + element of V i−1(c) . Hence the matrix

expressing x1,Fx1,F2x1, ...,Fnx1, x2,Fx2, ...,Fnx2, x3, ...,Fnxd in terms ofe1, e2, ..., ed(n+1) is upper triangular with all diagonal entries equal to 1, andtherefore is invertible. So Frxi : 1 ≤ i ≤ d, 0 ≤ r ≤ n indeed form a basis ofV(c).

To see that x1, x2, .., xd span ker E, take any v ∈ ker E and set i minimal withvi ∈ V i(c). Then, since ker E ⊆ V i(c)/V

i−1(c) is one-dimensional, v ∈ V i(c)/V

i−1(c) is a

multiple of e(i−1)(n+1)+1. In other words, some linear combination of v and xilives in V i−1

(c) ∩ ker E. Applying induction on i allows v to be expressed in termsof the xi, so indeed x1, x2, .., xd span ker E.

Define Xi = span of Frxi : 0 ≤ r ≤ n. By Step 6, these span V(c) andare linearly independent. It remains to show that each Xi is Uq(sl2)-invariant.F-invariance will be clear once we figure out how F permutes the generalisedweight spaces, and deduce that Fn+1xi = 0; afterwards we prove that the Frxiare all weight vectors. E-invariance then follows from (QSR3) via (2.7), as inthe proof of the classification theorem.

Since complete reducibility means that all generalised weight spaces are in factweight spaces, the following should not come as a surprise:

Step 7 ∀x ∈ ker E, and all r ≥ 0, (K− εqn−2rI)mFrx = 0 for some m ∈ N

From (QSR2):

(K − εqn−2r)F = q−2FK − εqn−2rF = q−2F (K − εqn−2(r−1)) (2.21)

so repeated application shows

(K − εqn−2r)mF = q−2mF (K − εqn−2(r−1))m (2.22)

Now use induction on r:

(K− εqn−2rI)mFrx = q−2mF(K− εqn−2(r−1)I)mFr−1x (2.23)

the base case of r = 0 being the assertion of Step 5.

Step 8 K acts on ker E as a scalar map

From Step 7, ∀i, (K − εq−n−2I)mFn+1xi = 0 for some m, but εq−n−2 is not aweight of L(n, ε), so (K−εq−n−2I)m cannot be the zero map, forcing Fn+1xi = 0(this is the same argument as the last paragraph of Step 5). Hence EFn+1xi = 0.Later (see (3.13)) we will show that

14

EFn+1 =qn+1 − q−n−1

(q − q−1)2Fn(q−nK + qnK−1) + Fn+1E (2.24)

Since q is not a root of unity, we must have Fn(q−nK + qnK−1)xi = 0∀i. ker Eis K-invariant, so (q−nK + qnK−1)xi =

∑dj=1 aijxj for some aij ∈ C. Then

Fn(q−nK + qnK−1)xi =d∑j=1

aijFnxj (2.25)

Fnxj form part of a basis set, so they are linearly independent. Consequently, wemust have aij = 0, so q−nK+qnK−1 is the zero map on ker E. Rearranging, wesee K2 = q2nI on ker E. Write K in Jordan normal form, and recall from Step 5that all the diagonal entries are εqn. Explicitly calculating the square of such amatrix shows that all off-diagonal entries must be 0, so Kx = εqnx ∀x ∈ ker E.

So the xi are weight vectors as claimed, and iteratively applying (QSR2) (as in(2.6)) shows that all Frxi must also be weight vectors. By the discussion beforeStep 7, this completes the proof. QED

2.4 Tensor representations, and decomposition into irre-ducibles

In this subsection, we address how to decompose any arbitrary finite-dimensionalUq(sl2)-representation into a direct sum of irreducibles. The proof of completereducibility suggests examining ker E or Casimir action. However, if we are sim-ply interested in the isomorphism types of these submodules rather than theirparticular location within V , then there is a far easier combinatorial methodwhich only requires knowing the weights of V and their multiplicities (whichis the dimension of the corresponding weight space). We illustrate such analgorithm on the example of tensor products, and deduce an analogue of theClebsch-Gordan formula, which we then rephrase in terms of character polyno-mials.

First, consider the simple case of L(1,+) ⊗ L(1,+), which is four-dimensional.Denote by v and w the highest weight vectors of the two factors. So Ev =Ew = 0, Kv = qv, Kw = qw. (We abuse notation and write K for the K-action on both factors of L(1,+)⊗L(1,+), and similarly for E,F.) Let E, F, Kbe matrices describing Uq(sl2)-action on the tensor product. From the Hopfalgebra structure of (1.6):

K(v ⊗ w) = Kv ⊗Kw = qv ⊗ qw = q2v ⊗ wK(v ⊗ Fw) = Kv ⊗KFw = qv ⊗ q−1Fw = q2v ⊗ Fw

K(Fv ⊗ w) = KFv ⊗Kw = q−1Fv ⊗ qw = q2Fv ⊗ wKF(v ⊗ Fw) = KFv ⊗KFw = q−1Fv ⊗ q−1Fw = q−2Fv ⊗ Fw

15

so v ⊗ w, v ⊗ Fw,Fv ⊗ w,Fv ⊗ Fw is a basis of weight vectors (that this setis a basis follows from the definition of tensor product). Hence the weights ofL(1,+)⊗L(1,+) are precisely q2, 1 with multiplicity two (ie dim [L(1,+)⊗ L(1,+)]1 =2), and q−2. For n > 2, qn,−qn are not weights of L(1,+) ⊗ L(1,+), socannot contain L(n,+) or L(n,−). Having ruled these out, the only remain-ing irreducible representation containing q2 as a weight is L(2,+). HenceL(1,+) ⊗ L(1,+) must contain copies of L(2,+) - in fact, exactly one copy,since q2 occurs as a weight of L(1,+)⊗ L(1,+) with multiplicity one. Indeed,

E(v ⊗ w) = Kv ⊗Ew + Ev ⊗ w = 0 (2.27)

so v ⊗ w is a highest weight vector of weight q2. By Proposition 2.2 on thestructure of highest weight modules, v ⊗ w generates a copy of L(2,+).

The weights of L(2,+) are q2,1 and q−2, each with multiplicity one, so the onlyweight unaccounted for in L(1,+)⊗ L(1,+) is 1. The only irreducible Uq(sl2)-module with 1 as its sole weight is L(0,+). Hence we deduce L(1,+)⊗L(1,+) ∼=L(2,+)⊕ L(0,+).

Now turn to the general case - take two irreducible representations L(n, ε) andL(m, δ) (where δ, ε ∈ +,−) with highest weight vectors v, w respectively.Recall that Frv : 0 ≤ r ≤ n and Fsw : 0 ≤ s ≤ m are bases of weightvectors for the two factors. Now, generalising (2.26):

K(Frv ⊗ Fsw) = KFrv ⊗KFsw

= εqn−2rFrv ⊗ δqm−2sFsw

= εδqn+m−2(r+s)Frv ⊗ Fsw

So Frv ⊗ Fsw : 0 ≤ r ≤ n, 0 ≤ s ≤ m is a basis of weight vectors forL(n, ε) ⊗ L(m, δ), and the weight of any such basis vector is the product ofthe weights of its factors. We deduce that the weights of L(n, ε) ⊗ L(m, δ) areεδqn+m, εδqn+m−2, ..., εδq−n−m where the εδqn+m−2t-weight space has dimen-sion |(r, s) : 0 ≤ r ≤ n, 0 ≤ s ≤ m, r + s = t|. If t ≤ minm,n (equivalently,n+m− 2t ≥ |n−m|), this expression simplifies to |r : 0 ≤ r ≤ t| = t+ 1.

The highest power of q, up to sign, which occurs here is εδqn+m, with multiplicityone. By the same reasoning as in the simple example above, L(n, ε) ⊗ L(m, δ)must contain exactly one copy of L(n + m, εδ) and no copies of L(i,±) fori > n+m.

We calculated above that the εδqn+m−2-weight space has dimension two, andL(n + m, εδ) contains only a linear subspace of this. Hence a single copy ofL(n+m− 2, εδ) is responsible for the second dimension.

Continuing like this, we see the irreducible summands of L(n, ε)⊗L(m, δ) includeL(n+m, εδ), L(n+m−2, εδ), L(n+m−4, εδ),..., L(|n−m|, εδ), and our simplifiedexpression for the weight multiplicities does not allow us to go further. However,the total dimension of these summands is (n+m+ 1) + (n+m−1) + · · ·+ (|n−m|+ 1) = (n+ 1)(m+ 1) = dimL(n, ε)⊗L(m, δ), so this list is in fact complete.We have proved the quantised Clebsch-Gordan formula:

16

Proposition. L(n, ε)⊗L(m, δ) = L(n+m, εδ)⊕L(n+m, εδ)⊕· · ·⊕L(|n−m|, εδ)

Figure 2.29 illustrates a few examples as string diagrams.

–

Figure 2.29 - L(2,+)⊗ L(2,−) and L(3,−)⊗ L(1,−)representations of when q is not a root of unity

This algorithm can be applied to any finite-dimensional representation, to givethe decomposition

V ∼=n⊕i=0

(dimVqi − dimVqi+1 − · · · − dimVqn

)L(i,+)

⊕m⊕j=0

(dimV−qj − dimV−qj+1 − · · · − dimV−qm

)L(i,−)

where n = maxi : Vqi 6= 0, m = maxi : V−qi 6= 0. Since this formula onlyinvolves the dimensions of weight spaces, the following is true:

Theorem. Finite-dimensional representations of Uq(sl2) are determined up toisomorphism by two polynomials in q:

χ+(V ) =∞∑n=0

dimVqnqn

χ−(V ) =∞∑m=0

dimV−qmqm

(The notation is my own; there seems to be no standard definition in the liter-ature)

From the classification Theorem 2.2:

χ+(L(n,+)) = qn + qn−2 + · · ·+ q−n = qn+1−q−n−1

q−q−1 = χ−(L(n,−))χ−(L(n,−)) = 0 = χ+(L(n,+))

(2.31)completely analogous to the classical case. Generalising (2.28) then gives thisalternative formulation of the quantised Clebsch-Gordan rule:

χ+(V ⊗W ) = χ+(V )χ+(W ) = χ−(V )χ−(W ) (2.32)

χ−(V ⊗W ) = χ+(V )χ−(W ) = χ−(V )χ+(W )

Under this framework, identifying the irreducible components of a represen-tation equates to expressing Laurent polynomials as linear combinations ofqn+1−q−n−1

q−q−1 : n ≥ 0

.

17

3 Finite-dimensional representations of Uq(sl2)when q is a root of unity

3.1 Some examples

Before we develop the general theory, let’s work through a couple of examplesto get a feeling for some of the differences that arise when q is a root of unity.Throughout this subsection, we consider the simplest setting where q3 = 1 ieq = e2πi/3.

The matrices describing L(n,±), as given in the classification Theorem 2.2, obeythe quantised Serre relations whatever the value of q, so, when q3 = 1, thesematrices still define valid representations. For example, L(4,+) is given by

E =

α1

α2

α3

α4

; F =

11

11

; K =

q4

q2

1q−2

q−4

(3.1)

where

αr =qr − q−r

(q − q−1)2(q5−r − qr−5

)=

qr − q−r

(q − q−1)2(q2−r − qr−2

)(3.2)

q3 = 1 means that α2 = α3 = 0, so the string diagram for L(4,+) has two”missing links”, as shown in Figure 3.3 below:

–

Figure 3.3 - The representation L(4,+), when q3 = 1, and itsinvariant submodules

Now L(4,+) is not irreducible: the two left-most nodes in Figure 3.3 representa Uq(sl2)-invariant submodule (since no arrows leave this pair), and the same istrue of the three left-most nodes. To be specific, α2 = 0 and α3 = 0 causes thethird and fourth basis vectors respectively to become additional highest weightvectors, which each generate a proper submodules, as we saw in the proof of theclassification Theorem 2.2.

More interestingly, neither submodule identified above has a Uq(sl2)-invariantcomplement - from the form of the matrix F, or from the string diagram, it isclear that, given any vector, some power of F sends this to a multiple of thelast basis vector (the left-most node). Hence every Uq(sl2)-invariant submodulecontains this basis vector, so pairs of complementary submodules cannot exist.Contrary to the non-root-of-unity case, then, we have here a representationwhich is not completely reducible.

18

When we classified the irreducible representations in Section 2, a crucial con-sequence of q not being a root of unity is that repeated applications of E orF cannot send a weight space back to itself. This restriction disappears whenq3 = 1: given v ∈ Vλ a highest weight vector, F3v ∈ Vq−6λ = Vλ, using (QSR2)as in (2.6). So F3v could well be a multiple of v. Indeed, assume F3v = v,and continue our analysis in the usual way by working out the matrix E using(QSR3), as in (2.7). As usual, this calculation shows that E maps the span ofv,Fv,F2v, ... to itself, so the linearly independent members of v,Fv,F2v, ...form a basis. In this example, the basis is v,Fv,F2v, with respect to which(after some manipulation with the αis,):

E = 1q−q−1

λ− λ−1

qλ−1 − q−1λ

; F =

1

11

; K =

λ

q−2λq−4λ

(3.4)

In the non-root-of-unity case, our next step would be to set EFn+1v = 0 andobtain a condition on λ. This is unnecessary here as α3 contains q3 − q−3 = 0as a factor. Indeed, (3.4) satisfies the quantised Serre relations, for any λ ∈ C,and it turns out to always be irreducible.

–

Figure 3.5 - The representation Z1(λ), with q3 = 1

In the pictorial description of Figure 3.5, we see that E,F are ”rotating” theweight spaces, compared to the ”translation” we saw previously. We shall seethat this ”rotation” is typical of irreducible representations in the root-of-unitycase.

Both phenomena exhibited above have analogues in the representations of clas-sical sl2 over F3, the field of 3 elements. Reducing modulo 3 the usual five-dimensional complex representation of gives sl2 :

E =

46

64

=

10

01

; F =

11

11

; H =

42

0−2

−4

(3.6)

As in (3.1), E has more zero entries than expected, which creates sl2-invariantproper submodules. Again, from the form of F we see that this representationis not completely reducible.

We also have the following three-dimensional representation where F ”rotates”the weight spaces (check that this does satisfy the Serre relations):

E =

22

; F =

11

1

; H =

λλ− 2

λ− 4

(3.7)

19

Our naive viewpoint K = qH gives some explanation for the similarities betweenrepresentations of Uq(sl2) when q is a pth root of unity and representations ofsl2 over fields of characteristic p, where p is a prime - namely, Kp = qpH is theidentity under both these circumstances.

3.2 Classification of irreducible representations

As in Section 2, we now proceed to give a complete description of all irreducibleUq(sl2) representations, following 2.13 of [Jantzen]. Throughout this subsection,q2 will be a primitive lth root of unity - if q is an even root of unity, then l ishalf of the multiplicative order of q; otherwise l is the order of q. The result wework towards is:

Theorem. Let q2 be a primitive lth root of unity. Then the irreducible finite-dimensional representations of Uq(sl2) are:

• L(n,+) for n ≤ l − 2. These are n+ 1-dimensional:

E =

α1

α2

. . .αn

; F =

11

. . .1

; K =

qn

qn−2

. . .q2−n

q−n

where αr = qr−q−r

(q−q−1)2

(qn−r+1 − q−n+r−1

)• L(n,−) for n ≤ l − 2. These are n+ 1-dimensional:

E = −

α1

α2

. . .αn

; F =

11

. . .1

; K = −

qn

qn−2

. . .q2−n

q−n

where αr = qr−q−r

(q−q−1)2

(qn−r+1 − q−n+r−1

)• Z0(λ) for λ ∈ C×\±1,±q,±q2, ...,±ql−2, which are l-dimensional:

E = −

α1

α2

. . .αl−1

; F =

11

. . .1

; K = −

λq−2λ

. . .. . .

q−2(l−1)λ

where αr = qr−q−r

(q−q−1)2

(q1−rλ− qr−1λ−1

)20

• Zb(λ) for λ ∈ C×, b ∈ C×, with the extra identifications Zb(qr−1) ∼=Zb(q−r−1), Zb(−qr−1) ∼= Zb(−q−r−1) for 1 ≤ r ≤ l − 1. These are l-dimensional:

E = −

α1

α2

. . .αl−1

; F =

b1

1. . .

1

; K = −

λq−2λ

. . .. . .

q−2(l−1)λ

where αr = qr−q−r

(q−q−1)2

(q1−rλ− qr−1λ−1

)• ωZb(λ) for λ ∈ C×, b ∈ C×, with the extra identifications ωZb(q1−r) ∼=ω

Zb(q1+r), ωZb(−q1−r) ∼=ω Zb(−q1+r) for 1 ≤ r ≤ l − 1. These are l-dimensional:

E =

b1

1. . .

1

; F = −

ωα1ωα2

. . .ωαl−1

; K = −

λq2λ

. . .. . .

q2(l−1)λ

where ωαr = qr−q−r

(q−q−1)2

(q1−rλ−1 − qr−1λ

)• Wa,b(λ), for λ, b, a ∈ C× with αr 6= −ab∀r ≥ 1, up to the relation

(λ, a, b) ∼ (q−2rλ, αr/b+ a, b). These are l-dimensional:

E = −

α1+ab

.... . .

αl−1+ab

a

; F =

b1

1. . .

1

; K = −

λq−2λ

. . .. . .

q−2(l−1)λ

where αr = qr−q−r

(q−q−1)2

(q1−rλ− qr−1λ−1

)(Wa,b(λ) is non-standard notation; these representations do not seem to have aname in the literature.)

–

Figure 3.8 - Some irreducible representations of Uq(sl2), whenq5 = 1

21

Observe that, topologically, the parametrising set ofWa,b(λ) is a three-dimensionalcomplex space whilst the other parametrising sets have lower dimension. Hence,a generic irreducible representation has the form Wa,b(λ), and we should thinkof the other five scenarios as degeneracies of Wa,b(λ). Figure 3.8 displays theassociated string diagrams for the case l = 5.

This looks considerably more complicated than the non-root-of-unity case: notonly do we have more types of irreducible representations, most types haveuncountably many members. Another major contrast is that the dimension ofany one of these irreducible representations is at most l, whereas previously theirreducible representations had arbitrarily large dimension. Indeed, when q isnot a root of unity, infinite-dimensional irreducible representations exist (forexample, analogues of the Verma modules of sl2, see 2.4 of [Jantzen]), but theproof which follows does not invoke the finite-dimensional hypothesis, whichmeans there are no infinite-dimensional irreducible representations in the root-of-unity setting. These differences arise because the centre of Uq(sl2) is muchlarger when q is a root of unity, as we shall see two paragraphs below.

As with the non-root-of-unity case, we start the proof by checking that thesematrices satisfy the quantised Serre relations, so they define valid representa-tions; this is a straightforward, if slightly long, calculation. Next we show thatall these representations are irreducible. In every case, K is diagonal with dis-tinct eigenvalues, so, by the same reasoning as in Subsection 2.2 (when q wasnot a root of unity) we need only show ”transitive permutation” of the weightspaces. (Strictly speaking, this is not a permutation action as weight spacesmay be sent to 0, but what I mean is that it is possible to reach any weightspace from any other via Uq(sl2) action.) In Zb(λ) and Wa,b(λ), this is achievedby powers of F alone (as b 6= 0); powers of E play this role for ωZb(λ). In theother three families, sending one weight space to another may require E or F(depending on the ”direction”); these function as desired since the restrictionson n and λ ensure that all αr are non-zero.

The rest of this subsection will be spent proving that the above list indeedexhausts all the irreducible representations, and is irredundant. Compared tothe non-root-of-unity case, the only extra idea in this proof is the examinationof central elements:

Lemma. Let q2 be a primitive lth root of unity (so q2l = 1). Then El, F l arecentral elements of Uq(sl2).

It suffices to show that El, F l commute with the generators E,F,K, and we doso simply by applying the quantised Serre relations repeatedly. By (QSR1) and(QSR2) respectively:

KErK−1 = (KEK−1)r = (q2E)r = q2rEr

⇒ KEr = q2rErK, K−1F r = q−2rErK−1

KF rK−1 = (KFK−1)r = (q−2F )r = q−2rF r

⇒ KF r = q−2rF rK, K−1F r = q2rF rK−1

22

In particular, taking r = l shows that El, F l commute with K. Next, use(QSR3) to show by induction that

EF r =K −K−1

q − q−1F r−1+F

K −K−1

q − q−1F r−2+· · ·+F r−1K −K−1

q − q−1+F rE (3.11)

FEr = ErF−Er−1K −K−1

q − q−1−Er−2K −K−1

q − q−1E+· · ·−K −K

−1

q − q−1Er−1 (3.12)

Using (3.9) and (3.10) to substitute for each term in the equations above:

EF r =[q−2(r−1) + · · ·+ q−2 + 1

] F r−1K

q − q−1−[q2(r−1) + · · ·+ q2 + 1

] F r−1K−1

q − q−1+F rE

(3.13)

FEr = ErF−[1 + q2 + · · ·+ q2(r−1)

] Er−1K

q − q−1+[1 + q−2 + · · ·+ q−2(r−1)

] Er−1K−1

q − q−13.14

Again, set r = l. Each square bracket above is the sum of all the distinct lthroots of unity, which is zero, so El commutes with F , F l commutes with E, asrequired. QED

It’s worth remarking that, throughout the above proof, we applied identicalarguments to E and to F - this symmetry between E and F is something wewill exploit in case 3 below. Also notice that the commutation formulas (3.13)and (3.14) show that no power of E or F can be central if q is not a root ofunity.

The point of the above lemma is that, given any irreducible representation V ,Schur’s lemma says that El, F l must act as scalars. So four scenarios can occur:

Case 1 El,Fl are both the zero matrix : L(n,±) and Z0(λ)

Here, ker E is non-trivial, and (QSR1) shows that K(ker E) ⊆ ker E. Henceker E contains some K-eigenvector v, with some corresponding eigenvalue λ.

As we did in (2.6), for the non-root-of-unity case, repeated applications of(QSR2) give Frv ∈ Vq−2rλ. Take n minimal with Fn+1v = 0; since Fl is thezero map, such an n exists, and n+ 1 ≤ l. As q−2rλ are distinct for 0 ≤ r < l,and hence for 0 ≤ r < n, v,Fv, ...,Fnv is a set of linearly independent weightvectors. K-action on this set is clear from the fact that Frv ∈ Vq−2rλ.. To findthe E-action, reuse (2.7) and (2.8) without change - that is, use (QSR3) to showinductively that EFrv = αrFr−1v where

αr =qr − q−r

(q − q−1)2(q1−rλ− qr−1λ−1

)(3.15)

23

As in the non-root-of-unity setting, EFn+1v = 0 implies that

αn+1 =qn+1 − q−n−1

(q − q−1)2(qnλ− q−nλ−1

)= 0 (3.16)

Since ql = q−l = ±1 (depending on whether q has odd or even order), (3.16) istrue for n = l − 1 irrespective of λ, and in this case we find no restrictions onour highest weight ; this is the representation Z0(λ). Otherwise we obtain thefamiliar condition λ = ±qn, which gives the previously seen modules L(n,±).(Observe that L(l− 1,±) ∼= Z0(±ql−1), which is why n runs up to l− 2 only inthe statement of the theorem.)

Note that, if αr = 0, then EFrv = 0, so, as we saw in the first example of thissection, Frv,Fr+1v, ...,Fnv spans an Uq(sl2)-invariant submodule. Hence wemust require αr 6= 0∀r, 1 ≤ r < n + 1, or equivalently, λ = ±qr−1, by (3.15).This accounts for the removal of ±1,±q, ...,±ql−2 from the range of λ for therepresentations Z0(λ).

Case 2 El = 0,Fl = bI 6= 0: Zb(λ)

As above, ker E is non-trivial, so we can find a highest weight vector v ∈ Vqfor some λ ∈ C×. Flv = bv 6= 0, so Frv 6= 0∀r ≥ 0. By (QSR2), as in(2.6), Frv ∈ Vq−2rλ, and these weight spaces are distinct for 0 ≤ r < l, sov,Fv, ...,Fl−1v is a linearly independent set. Since (2.7), the formula forEFrv, still applies, Uq(sl2)-action with respect to the basis v,Fv, ...,Fl−1vagrees with the matrices given in the statement of the theorem for Zb(λ).

Case 3 El = bI 6= 0,Fl = 0: ωZb(λ)

Recalling what we said earlier about symmetry, we tackle this by essentiallyexchanging the roles of E and F. So take v ∈ ker F with Kv = λv ((QSR2)ensures that K preserves ker F). By (QSR1), as in (2.5), Erv ∈ Vq2rλ, andthese weight spaces are distinct for 0 ≤ r < l, so v,Ev, ...,El−1v is a linearlyindependent set. The inductive computation in (2.7), with E and F exchanged,gives

FErv =qr − q−r

(q − q−1)2(q1−rλ−1 − qr−1λ

)Er−1v∀r ≥ 1 (3.17)

so, with respect to the basis v,Ev, ...,El−1v, E,F,K have the form shown inthe theorem for ωZb(λ). These representations are so named because they are”twisted” versions of Zb(λ), where E and F are exchanged and q−1 replaces q.

Case 4 El non-zero, Fl = bI 6= 0 : Wa,b(λ)

Here we have no highest weight vectors, but there is no reason why we cannotstill examine the F-span of some K-eigenvector v ∈ Vλ - in fact, the usual

24

argument shows that v,Fv, ...,Fl−1v are linearly independent with Frv ∈Vq−2rλ. What remains is to find out how E acts on this set.

Recall that the Casimir element

Ω = FE +Kq −K−1q−1

(q − q−1)2(3.18)

is central in Uq(sl2); hence, by Schur’s lemma, it must act on V as scalarmultiplication by some c ∈ C. So

FEv = cv − Kq −K−1q−1

(q − q−1)2v = some multiple of v

Fl = bI 6= 0 allows us to invert F: F−1 = b−1Fl−1. So

Ev = F−1(FEv) = b−1Fl−1(some multiple of v) = aFl−1v (3.20)

for some a ∈ C×. Now we apply the commutation relation (3.13) to find that

EFrv =r−1∑i=0

q−2iλ− q2iλ−1

q − q−1Fr−1v + FrEv

=[qr − q−r

(q − q−1)2(q1−rλ− qr−1λ−1) + ab

]Fr−1v

Since El 6= 0, these coefficients must all be non-zero. So we require αr 6=−ab∀r ≥ 1 ( αr as defined in the theorem) - this is Wa,b(λ).

This completes the list of irreducible representations. As a result, the follow-ing algorithm allows us to determine the type and parameters of an unknownirreducible representation of Uq(sl2) (when q2l = 1):

—

Observe that Wa,b(λ) contains no distinguished weight spaces - its string dia-gram (Figure 3.8) is entirely symmetric. However, according to this algorithm,finding λ, a depends on a choice of weight. Hence these two parameters are notuniquely determined by the representation (unlike b, whose definition requires nochoices). Indeed, assume we picked some other weight q−2rλ(from the statementof the theorem we know all weights of Wa,b(λ) have this form). Then our basiswould be some multiple of the set Frv,Fr+1v, ...,Fl−1v, bv, bFv, ..., bFr−1v,and we would name this representation is Wαr/b+a,b(q

−2rλ). Hence we need toquotient out the parametrising set by the relation

(λ, a, b) ∼ (q−2rλ, αr/b+ a, b) (3.22)

A similar complication can occur with Zb(λ) and ωZb(λ) . In the generic case,the highest weight (or equivalent in ωZb(λ) ) is unique, but this fails whenαr = 0 (in Zb(λ)) or (ωαr = 0 in ωZb(λ)) for some r, 1 ≤ r ≤ l − 1. Take

25

the first case, which happens exactly when λ = ±qr−1. As noted previously,this can hold for at most one value of r in this range. Then q−2rλ = ±q−r−1 isanother highest weight, so Zb(qr−1) ∼= Zb(q−r−1) and Zb(−qr−1) ∼= Zb(−q−r−1),as stated in the theorem. Figure 3.23 depicts the r = 3 case, with q5 = 1. Theanalogous argument for shows that ωZb(q1−r) ∼=ω Zb(q1+r), and ωZb(−q1−r) ∼=ω

Zb(−q1+r) .

–Figure 3.23 - The irreducible representation Zb(q2) ∼= Zb(q4), with

q5 = 1

It follows from the algorithm that, up to these relations, all the representationslisted in the theorem are distinct. Observe that the set of weights of any of thel-dimensional irreducible representations is precisely all the lth roots of somecomplex number λl. QED

3.3 Composition factors and tensor representations

In the absence of complete reducibility, there is no one-size-fits-all algorithmfor identifying irreducible composition factors. We demonstrate a few usefultechniques on the examples of Z0(±qn), L(n,±) and Wa,b(λ)⊗Wc,d(µ).

Firstly, recall from Case 1 of the previous subsection that Z0(±qn) is not ir-reducible for n < l − 1, since αn+1 = 0. Let be a generating highest weightvector as usual, and write V j for the span of the last l − j basis vectors, ieFjv,Fj+!v, ...,Fl−1v is a basis for V j . Under this notation, V n+1 is Uq(sl2)-invariant; to show that this is the only proper invariant submodule, we first needa

Lemma. Suppose v ∈ V j, v /∈ V j+1. Then any Uq(sl2)-invariant submodulecontaining v must contain V j.

Apply induction on l − j. The assertion is clear if j = l − 1, since V l−1 is aone-dimensional subspace. If j < l − 1, then Fv ∈ V j+1, Fv /∈ V j+2. AnyUq(sl2)-invariant submodule containing v also contains Fv, hence, by inductivehypothesis, it contains V j+1. So this submodule contains the span of v andV j+1, which is precisely V j . QED.

Now if j 6= n + 1, E(Fjv) ∈ V j−1\V j , so a Uq(sl2)-invariant submodule con-taining V j must also contain V j−1. Applying this argument iteratively showsthat V n+1 is indeed the unique proper invariant submodule, therefore it must beirreducible. By dimension considerations, V n+1 ∼= L(l−n− 2,±), and the quo-tient Z0(±qn)/V n+1 ∼= L(n,±). To determine the signs, the flow diagram asksus to find the highest weights. The highest weight of Z0(±qn)/V n+1 is that ofZ0(±qn), which is ±qn, and V n+1 has highest weight ±qn−2(n+1) = ±qlql−n−2.Recalling that ql = 1 if q is an odd root of unity, and ql = −1 otherwise, wehave proved

26

Proposition. Suppose q2 is a primitive lth root of unity. Then:

• If q is an odd root of unity,

– Z0(qn) has unique proper invariant submodule V n+1 ∼= L(l−n−2,+)and the quotient Z0(qn)/V n+1 is L(n,+);

– Z0(−qn) has unique proper invariant submodule V n+1 ∼= L(l − n −2,−) and the quotient Z0(qn)/V n+1 is L(n,−);

• If q is an even root of unity,

– Z0(qn) has unique proper invariant submodule V n+1 ∼= L(l−n−2,−)and the quotient Z0(qn)/V n+1 is L(n,+);

– Z0(−qn) has unique proper invariant submodule V n+1 ∼= L(l − n −2,+) and the quotient Z0(qn)/V n+1 is L(n,−);

Next, we generalise our example of L(4,+)from Subsection 3.1. Take L(n,±)and write n+ 1 as ml + s with 0 ≤ s < l. So

αr =qr − q−r

(q − q−1)2(qml+s−r − q−ml−s+r

)(3.24)

which means αl = α2l = · · · = αml = 0, αl+s = α2l+s = · · · = αml+s = 0,and all other αis are non-zero. Then, by the reasoning in the previous exam-ple, V l, V 2l, ..., V ml, V l+s, V 2l+s, ...V (m−1)l+s are precisely the Uq(sl2)-invariantsubmodules. So, if s 6= 0, the composition series for L(n,±) reads

0 ⊆ V ml ⊆ V (m−1)l+s ⊆ V (m−1)l ⊆ · · · ⊆ V l ⊆ V s ⊆ L(n,±) (3.25)

Each composition factor has dimension s < l or l−s < l, so the flow algorithm ofthe last subsection identifies them as L(s−1,±) and L(l−s−1,±). For example,when q3 = 1, L(4,+) has composition series 0 ⊆ V 3 ⊆ V 2 ⊆ L(4,+), withalternating two- and one-dimensional quotients (l = 3,m = 1, s = 2), as inFigure 3.3. The signs of these representations again depend on the order ofq. The highest weight of V rl/V rl+s is ±qml+s−1−2rl = ±qs−1(ql)m; similarly,V (r−1)l+s/V rl has highest weight ±ql−s−1(ql)m−1. This proves the followingresult, as illustrated in Figure 3.26:

Proposition. Suppose q2 is a primitive lth root of unity, and n + 1 = ml + swith 0 < s < l. Then:

• If q is an odd root of unity, then

– the composition factors of L(n,+) are m copies of L(s− 1,+) alter-nated with m− 1 copies of L(l − s− 1,+);

27

– the composition factors of L(n,−) are m copies of L(s− 1,−) alter-nated with m− 1 copies of L(l − s− 1,−);

• If q is an even root of unity, then

– the composition factors of L(n,+) are m copies of L(s − 1, ε) alter-nated with m− 1 copies of L(l − s− 1,−ε);

– the composition factors of L(n,−) are m copies of L(s−1,−ε) alter-nated with m− 1 copies of L(l − s− 1, ε)

where ε is the sign of (−1)m.

–

Figure 3.26 - The decomposition of L(ml + s− 1,±) when q2l = 1and s 6= 0

Returning to L(4,+) with q3 = 1, its submodules L(4,+), V 3 and V 2 havehighest weights q4 = q, 1, q−2 = q respectively, so the composition factors areL(1,+), L(0,+) and L(1,+). If q6 = 1, then L(4,+) still has V 3 and V 2 as itsproper invariant submodules (since l,m, s are unchanged), but now the highestweights are q4 = −q, 1, q−2 = −q, so the quotients become L(1,−), L(0,+) andL(1,−) as guaranteed by the proposition.

Our final example investigates the structure of Wa,b(λ) ⊗Wc,d(µ), the tensorproduct of two generic irreducible representations. We will give an argumentfor complete reducibility in the generic case, then study in detail one of the”singular” cases.

For simplicity, take q4 = 1, so l = 2. Let v be any weight vector in Wa,b(λ), wany weight vector in Wc,d(µ), and write wt(v) for the weight of v. Then, as in(2.28):

K(v ⊗ w) = Kv ⊗Kw = wt(v)wt(w)v ⊗ w (3.27)

so v ⊗w form a basis of weight vectors. (As before, E, F, K refer to Uq(sl2)-action on Wa,b(λ) ⊗Wc,d(µ).) Hence the weights of Wa,b(λ) ⊗Wc,d(µ) are λµand −λµ, each with multiplicity 2.

Finding F is a similarly straightforward calculation. Since comultiplication ∆is an algebra homomorphism,

∆(F 2) = (1⊗ F + F ⊗K−1)2

= 1⊗ F 2 + F ⊗ (FK−1 +K−1F ) + F 2 ⊗K−2

= 1⊗ F 2 + F 2 ⊗K−2

using (QSR2) in the last equality. Therefore

F2 = v ⊗ dw + bv ⊗ wt(w)−2 = (d+ bµ−2)v ⊗ w (3.29)

28

since all weights of Wc,d(µ) square to µ2. As v ⊗ w is a basis for Wa,b(λ) ⊗Wc,d(µ), this shows that F2 = bI where b = d+bµ−2 .To generalise this to largerl, we can calculate ∆(F r) inductively and find

∆(F l) = 1⊗ F l + F l ⊗K−1 (3.30)

so F = (d+ bµ−l)I.

Encouraged by Subsection 2.3, the proof of complete reducibility in the non-root-of-unity case, we look for eigenspaces of the Casimir operator

Ω = FE +qK− q−1K−1

(q − q−1)2

Since Ω commutes with K, Ω preserves each weight space, so we can study itsaction on each weight space separately. Returning to l = 2, take weight vectorsv0 ∈ [Wa,b(λ)]λ, w0 ∈ [Wc,d(µ)]µ and set v1 = Fv0, w1 = Fw0. We have

∆(FE) = (1⊗ F + F ⊗K−1)(K ⊗ E + E ⊗ 1)= K ⊗ FE + E ⊗ F + FK ⊗K−1E + FE ⊗K−1

so, with respect to the basis v0⊗w0, v1⊗w1 of the weight space [Wa,b(λ)⊗Wc,d(µ)]λµ,Ω is represented byλcd+ µ−1ab− iλµ−iλ−1µ−1

4

(λ−λ−1

2i + ab)d− λµ−1b

(µ−µ−1

2i + cd)

a− λcµ−1 −λ(µ−µ−1

2i + cd)− µ−1

(λ−λ−1

2i + ab)

+ iλµ−iλ−1µ−1

4

(3.33)

Generically, the characteristic polynomial of this matrix will have two distinctroots ν1 and ν2. Then the corresponding eigenvectors x1, x2 give a basis of[Wa,b(λ)⊗Wc,d(µ)]λµ. Make the additional assumption that F2 = (d+bµ−2)I 6=0, so F is an isomorphism. Then Fx1, Fx2 inherit linear independence fromx1, x2 , so is a basis of [Wa,b(λ)⊗Wc,d(µ)]−λµ.

We see from the definition of Ω,

Ex1 = b−1F2Ex1 = b−1F

(ν1x1 −

qK + q−1K−1

q − q−1x1

)= a1Fx1 (3.34)

for some a1 ∈ C, and

E(Fx1) = FEx1 +K + K−1

q − q−1x1 = a1bx1 +

λµ+ λ−1µ−1

q − q−1x1 (3.35)

using (3.33) in the second equality. So x1, Fx1 form the basis of a Uq(sl2)-module, and the same is true for x2, Fx2, as we can find a2 ∈ C with Ex2 =

29

a2Fx2. Since x1, Fx1, x2, Fx2 is a basis for Wa,b(λ)⊗Wc,d(µ), these two sub-modules are disjoint. Hence, in the generic case where a1, a2, a1b+ λµ+λ−1µ−1

q−q−1 ,

a2b+ λµ+λ−1µ−1

q−q−1 are all non-zero,

Wa,b(λ)⊗Wc,d(µ) = Wa1 ,b(λµ)⊕Wa2 ,b

(λµ) (3.36)

where Wai ,b(λµ) has basis xi, Fxi. (If some of the quantities above are zero,

then we still have two irreducible two-dimensional summands, but they may beof type Zb(λ).)

Applying the ideas above to general gives the following

Theorem A. If Ω has linearly independent eigenvectors x1, x2, ..., xl ∈ [Wa,b(λ)⊗Wc,d(µ)]λµthen Wa,b(λ) ⊗ Wc,d(µ) =

⊕li=1X

i where Xi has basis xi, Fxi, ..., Fl−1xi.Furthermore, if b := d+ bµ−l is non-zero, then every Xi is irreducible.

A generalisation of (3.34) will compute Exi and thus identify the type andparameters of Xi. This theorem has a partial converse

Theorem B. Suppose b := d+bµ−l 6= 0 and Wa,b(λ)⊗Wc,d(µ) is completely re-ducible. Then Ω has l linearly independent eigenvectors in [Wa,b(λ)⊗Wc,d(µ)]λµ.

This is because Ω acts as a scalar on each irreducible summand; since b =6= 0 ,these all have l weight spaces of distinct weights and hence must each containa linear subspace of [Wa,b(λ)⊗Wc,d(µ)]λµ.

Theorem B states that, if Ω is not diagonalisable, then this tensor product repre-sentation is not completely reducible. We now construct one such example whereq4 = 1, λ = 1, µ = −1 - this will greatly simplify our calculations, since the αisof both factors, which are λ−λ−1

q−q−1 and µ−µ−1

q−q−1 , vanish, as does iλµ − iλ−1µ−1.

This last condition means that [Wa,b(1)⊗Wc,d(−1)]−1 ⊆ ker qK−q−1K−1

(q−q−1)2 , so Ω

and FE agree on this subspace, and is given by(cd− ab abd+ bcda+ c −cd+ ab

)(3.37)

This has characteristic polynomial ξ2 + (b+ d)(a2b+ c2d), which gives repeatedroots precisely when b+ d = 0 or a2b+ c2d = 0. Assume that the latter holds,and that b + d 6= 0 , so F is again an isomorphism. This forces cd − ab 6= 0,a+ c 6= 0. Thus x1 := (cd− ab)v0⊗w0 + (a+ c)v1⊗w1, x2 := v0⊗w0 is a basisfor [Wa,b(1)⊗Wc,d(−1)]−1.

By direct computation, FE = Ω has the form(

0 10 0

)with respect to the

basis x1, x2. Since F is injective,FEx1 = 0 implies Ex1 = 0. By (QSR3)

E(Fx1) = FEx1 +K + K−1

q − q−1x1 = 0 + 0 (3.38)

30

So x1, Fx1spans a Uq(sl2)-invariant submodule where E-action is zero. Sincex1 belongs to the −1-weight space, this submodule is Zb(−1) ∼= Zb(1).

As we saw earlier, linear independence of x1, x2 passes through to Fx1, Fx2,so Fx1, Fx2 is a basis for the 1-weight space of Wa,b(1) ⊗ Wc,d(−1). So,to complete our analysis, we need to find how E acts on x2 and Fx2. SinceFEx2 = x1 ,

E(x2) = b−1F2Ex2 + b−1Fx1 (3.39)

and, using (QSR3) as in (3.37):

E(Fx2) = FEx2 +K + K−1

q − q−1x2 = x1 (3.40)

Hence, with respect to the basis x1, Fx1, x2, Fx2, the representation Wa,b(1)⊗Wc,d(−1) is given by the matrices:

E =

0 10 b−1

00

; F =

b1

b1

; K =

−11−1

1

(3.41)

From either the matrices (3.41) or the string diagram Figure 3.42, we deducethat the two composition factors of Wa,b(1)⊗Wc,d(−1) are both Zb(−1) ∼= Zb(1).

–

Figure 3.42 - The representation Wa,b(1)⊗Wc,d(−1), whena2b+ c2d = 0 and b+ d 6= 0, which is not completely reducible

31

Conclusion

Despite restricting our attention to Uq(sl2), we have actually glimpsed manyof the important properties of general quantised enveloping algebras. Suchan algebra is created by taking a simple Lie algebra and suitably ”bending”its Serre relations - I will not reproduce the eyesore of these formulae here,but the spirit is no different from our Section 1. So Uq(g) is generated byE1, E2, ..., En, F1, F2, ...Fn,K1,K2, ...Kn, where n is the number of simple rootsin g; this is again a Hopf algebra.

We can generalise the concept of weight vector to denote vectors which are simul-taneously eigenvectors of all the Ki; a highest weight vector is then one whichis annihilated by all Ei. Then, in the non-root-of-unity setting, the Uq(sl2)-theory we developed shows (via a modified classical argument) that any finite-dimensional irreducible representation of Uq(g) has a basis of weight vectors. Inparticular, as we saw with Uq(sl2), each such representation is uniquely spec-ified by its highest weight. In Section 2, we labelled these highest weights bya positive integer and a sign ε ∈ +,− which relates to K-action. In Uq(g),there are n analogues of K, so n signs are needed to describe the irreducibleUq(g)-modules, in addition to a dominant weight of g which plays the role ofthe positive integers. The usual proof of this is directly adapted from that of theclassical version (that dominant weights index the finite-dimensional irreduciblerepresentations of g). The fact that this combinatorial data specifies finite-dimensional irreducible representations uniquely means we can define analogouscharacter formulae and use them to decompose arbitrary finite-dimensional rep-resentations, since these are again completely reducible.

As for the root-of-unity case, things are again complicated by the larger centre,which contains lth powers of all the Eis and Fis. As we saw with Uq(sl2),the dimension of irreducible representations is bounded (by lN , where N is thenumber of positive roots in g) and the great majority of them have the maximaldimension.

References

[Brown, Goodearl] K.A. Brown and K.R. Goodearl, Lectures on Algebraic Quan-tum Groups, Advanced Courses in Math. CRM Barcelona, Birkhauser, Basel,2002

[DPRR] C. DeConcini, C. Procesi, N.Reshetikhin and M.Rosso, Hopf algebraswith trace and representations, found at http://arxiv.org/abs/math/0304313v1

[Humphreys] J.E. Humphreys, Introduction to Lie Algebras and RepresentationTheory, Grad. Studies in Math. 9, Springer-Verlag, New York, 1972

[Jantzen] J.C. Jantzen, Lectures on Quantum Groups, Grad. Studies in Math.6, Amer. Math. Soc., Providence, 1996

32

[Sweedler] M.E. Sweedler, Hopf Algebras, Benjamin, New York, 1969

33