MASTER PROJECT

Representations of SL2(K)

Supervised byProf. Steve Donkin, The University of York

Prof. Donna Testerman, Ecole Polytechnique Fédérale de Lausanne

Student:

Mikaël Cavallin

February 29, 2012

Acknowledgement

I would like to express my deepest thanks to Professor Stephen Donkin andProfessor Donna Testerman for giving me the opportunity to do my masterthesis under their supervision. I am very grateful to Professor Donna Testermanfor the time she spent organizing this exchange and correcting my dissertationand to Professor Donkin for the choice of the subject, his advice and the time hetook helping me throughout my stay in the University of York. I would also liketo thank everybody in the department of mathematics for involving me in theirseminars and group study, as well as the secretaries of the same department,who helped me with finding accommodation in York.

I would also like to thank the family I have been living with in York, whogave me a warm welcome and a very nice place to stay, as well as the friendsand acquaintances I have made in England, especially my officemate and friendHarry, with whom I had great times and a lot of fun.

Finally, I thank my parents for their support and their encouragement.

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Contents

1 Preliminaries 11.1 The special linear group . . . . . . . . . . . . . . . . . . . . . . . 11.2 Representations of algebras . . . . . . . . . . . . . . . . . . . . . 21.3 Duality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61.4 Symmetric and exterior powers . . . . . . . . . . . . . . . . . . . 7

2 Representations of SL2(q) 92.1 The theory of highest weights . . . . . . . . . . . . . . . . . . . . 92.2 Irreducible KSL2(p)-modules . . . . . . . . . . . . . . . . . . . . 132.3 Irreducible KSL2(q)-modules . . . . . . . . . . . . . . . . . . . . 15

2.3.1 The Frobenius twist of a KSL2(q)-module . . . . . . . . . 152.3.2 A complete set of irreducibles . . . . . . . . . . . . . . . . 152.3.3 A special family of KSL2(q)-modules . . . . . . . . . . . 17

3 Representations of sl2(K) 193.1 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

3.1.1 Representations of Lie algebras . . . . . . . . . . . . . . . 203.1.2 Weights and weight spaces . . . . . . . . . . . . . . . . . . 20

3.2 Irreducible sl2(K)-modules . . . . . . . . . . . . . . . . . . . . . 213.2.1 A complete set of irreducibles . . . . . . . . . . . . . . . . 213.2.2 Complete reducibility and the Clebsch-Gordan formula . . 22

4 Representations of SL2(K) 254.1 Some algebraic geometry . . . . . . . . . . . . . . . . . . . . . . . 25

4.1.1 Affine varieties . . . . . . . . . . . . . . . . . . . . . . . . 254.1.2 Algebraic groups . . . . . . . . . . . . . . . . . . . . . . . 274.1.3 The class of rational modules . . . . . . . . . . . . . . . . 284.1.4 Irreducibility and Connectedness . . . . . . . . . . . . . . 29

4.2 SL2(K) as a connected algebraic group . . . . . . . . . . . . . . . 314.2.1 Weights and weight spaces . . . . . . . . . . . . . . . . . . 314.2.2 The formal character . . . . . . . . . . . . . . . . . . . . . 334.2.3 The unipotent subgroups U and U− . . . . . . . . . . . . 34

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4.2.4 A dense subset of SL2(K) . . . . . . . . . . . . . . . . . . 354.3 Irreducible rational KSL2(K)-modules . . . . . . . . . . . . . . . 354.4 Additional results in characteristic zero . . . . . . . . . . . . . . 37

5 Decomposition in characteristic p > 0 395.1 The Schur algebra . . . . . . . . . . . . . . . . . . . . . . . . . . 40

5.1.1 The K-spaces AK(n) and AK(n, r) . . . . . . . . . . . . . 405.1.2 The category of polynomial KGLn(K)-modules . . . . . . 41

5.2 Irreducible SK(n, r)-modules . . . . . . . . . . . . . . . . . . . . 435.2.1 The formal character (revisited) . . . . . . . . . . . . . . 435.2.2 A complete set of irreducible SK(2, r)-modules . . . . . . 44

5.3 Indecomposable KSL2(K)-modules . . . . . . . . . . . . . . . . . 465.3.1 Projective indecomposable modules . . . . . . . . . . . . . 475.3.2 Tackling the problem . . . . . . . . . . . . . . . . . . . . . 48

5.4 Further results using ∇-filtrations . . . . . . . . . . . . . . . . . . 515.4.1 An example in characteristic two . . . . . . . . . . . . . . 52

Bibliography 53

iv

Introduction

The special linear group SLn(K) of degree n over a field K is defined as thekernel of the determinant morphism Det : GLn → K∗, ie.

SLn(K) = {A ∈ GLn(K) : Det(A) = 1}.

The aim of this project is to study the representation theory of SL2(K)over an algebraically closed extension L of K, where K could either denotea finite field or L itself, and then to study the decomposition of the tensorproduct of some particular LSL2(L)-modules (the symmetric powers) in termsof indecomposables.

In the first chapter, we recall some basic properties of the special lineargroup and of modules in general. We also define the symmetric and alternatingpowers of the natural LGLn(K)-module. Most of the proofs shall be omitted,but the reader can consult [Alp86], [ARO97] or [FH91] if interested.

In the second chapter, we consider q = pn, where p is a prime and n ∈ N∗a positive integer, and we give a complete set of non-isomorphic irreduciblerepresentations of SL2(q) over the algebraic closure of Fp. In order to do that,we start by finding a way to determine how many of them there are, following thetheory of highest weights and highest weight vectors presented in [Ste68] ratherthan studying conjugacy classes, which could have been another approach to theproblem. Finally, we construct this exact number of non-isomorphic irreduciblerepresentations.

In the third chapter, we let K be an algebraically closed field of character-istic zero, momentarily forget the special linear group and focus our attentionon the representation theory of the Lie algebra sl2(K). We first give a com-plete set of non-isomorphic irreducible finite-dimensional sl2(K)-modules, recallWeyl’s theorem of complete reducibility and finally, prove the Clebsh-Gordanformula, which tells us how to decompose the tensor product of any two irre-ducible sl2(K)-modules in terms of irreducibles. This chapter is mainly basedon [Hum78] and [Maz10].

In the fourth chapter, we let K be an algebraically closed field of any charac-teristic and study the representations of SL2(K) over K. In fact, we shall viewSL2(K) with its structure of connected algebraic group and consider its so-calledrational representations. In the case where K has characteristic zero, we shall

v

see that the representation theory of SL2(K) is similar to the representationtheory of the Lie algebra sl2(K). More precisely, the irreducibles are the same,every SL2(K)-module is completely reducible and the Clebsh-Gordan formulastill holds. On the other hand, in the case where K has positive characteristic,a complete set of non-isomorphic irreducible rational KSL2(K)-modules canbe given using the set obtained in the second chapter. The references for thischapter are [Bor91], [Hum75] and [Spr81].

In the last chapter, we let K be an algebraically closed field of positivecharacteristic and study the decomposition of the tensor product of any twosymmetric powers in terms of indecomposable rational KSL2(K)-modules. Ourapproach to the problem consists in finding such a decomposition as indecom-posable polynomial KGL2(K)-modules first and then in trying to deduce thedesired result. Our reason for considering these polynomial KGL2(K)-moduleslies in the fact that a very important tool is available in this situation, calledthe Schur algebra. This chapter takes up the ideas of [Gre81] and [Mar08],principally concerning the study of the Schur algebra and the classification ofthe irreducible polynomial KGLn(K)-modules in terms of partitions. We shallalso refer to [Don98].

vi

CHAPTER 1

Preliminaries

The aim of this chapter is to remind the reader of some basic knowledge, nec-essary to fully understand the project. As said in the introduction, we shallfix some notation and recall a few properties of the special linear group and ofmodules in general.

1.1 The special linear groupLet K be a field and let n ∈ N∗ be a positive integer. We first denote by U(n)Kand U(n)−K the upper and lower unipotent subgroups of SLn(K) respectively.In the same way, we let T (n)K be the subgroup of diagonal matrices. Finally,for every a ∈ K, we shall consider the elements u(a) ∈ U(2)K , u

−(a) ∈ U(2)−Kand h(a) ∈ T (2)K

1, respectively given as

u(a) =

(1 a0 1

), u−(a) =

(1 0a 1

)and h(a) =

(a 00 a−1

).

Lemma 1.1.1The special linear group SL2(K) is generated by its two unipotent subgroupsU(2)−K and U(2)K . In other words, we have

SL2(K) =⟨U(2)−K , U(2)K

⟩.

Proof. Let g = gij be an element of SL2(K). If g12 6= 0, then g can be writtenas g = u−

((g22 − 1)g−1

12

)u(g12)u−

((g11 − 1)g−1

12

). If g12 = 0 but g21 6= 0, then

g can be written as g = u((g11 − 1)g−121 )u−(g21)u((g−1

11 − 1)g−121 ). Finally, if

g12 = g21 = 0, then g can be written as g = u(−g11)u−(g−111 −1)u(1)u−(g11−1).

Hence any g ∈ SL2(K) can be written as a product of elements each belongingto U(2)−K or U(2)K and thus the proof is complete.

1The element h(a) is only defined for a ∈ K∗.

1

Remark 1.1.2Return now to the case where n ∈ N∗ is an arbitrary positive integer and for1 ≤ i, j ≤ n, let Eij ∈Mn×n be the matrix defined by

(Eij)µν =

{0 : (i, j) = (µ, ν),1 : otherwise.

A transvection of SLn(K) is a matrix of the form In + λEij , where λ ∈ Kand 1 ≤ i 6= j ≤ n. It is clear that such matrices are in SLn(K) and in fact,Lemma 1.1.1 is a special case of a more general result, which states that thespecial linear group SLn(K) is generated by its transvections.

Now consider the case where K is a finite field, that is, K = Fq for someq = pm, with p a prime and m ∈ N∗ a positive integer. Then the reader caneasily check that the order of the general linear group of rank n over K is givenas

|GLn(q)| = qn(n−1)

2

n∏i=1

(qi − 1).

As a direct consequence, since the special linear group SLn(q) is defined asthe kernel of the determinant morphism Det : GLn(q)→ F∗q , we obviously have

|SLn(q)| = qn(n−1)

2

n∏i=2

(qi − 1) (1.1)

and an easy computation yields the following result.

Lemma 1.1.3Both U(n)Fq and U−(n)Fq have order q

n(n−1)2 . In particular, they are p-Sylow

subgroups of SLn(q).

1.2 Representations of algebrasLet K be an algebraically closed field and let A be a finite-dimensional K-algebra. Throughout this report, we shall only consider finite-dimensional leftA-modules, unless otherwise indicated. Nevertheless, most of the following resultsstill hold in a more general setting. We refer to [Alp86] and [ARO97] for moredetails.

Lemma 1.2.1 (Schur’s Lemma)Let V, W be irreducible A-modules and let φ : V → W be a morphism of A-modules. Then φ is either trivial or an isomorphism. Furthermore,

HomA(V, V ) ∼= K.

Now recall that a composition series for an A-module V is a finite sequenceof submodules V = V0 ⊃ V1 ⊃ . . . ⊃ Vr−1 ⊃ Vr = 0 such that Vi/Vi+1 isirreducible, for every 0 ≤ i ≤ r − 1.

2

Theorem 1.2.2 (Jordan-Hï¿ 12 lder Theorem)

Assume that V = W0 ⊃ W1 ⊃ . . . ⊃ Ws−1 ⊃ Ws = 0 is another compositionseries for V. Then r = s and for any irreducible A-module S, we have

#{i : Vi/Vi+1∼= S} = #{i : Wi/Wi+1

∼= S}.

Proof. See [ARO97, Theorem 1.2, pp. 3].

Let V and S be two A-modules, with S irreducible, and consider any com-position series V = V0 ⊃ V1 ⊃ . . . ⊃ Vr−1 ⊃ Vr = 0 for V. Then the compositionmultiplicity of S in V (well-defined thanks to Theorem 1.2.2) is the non-negativeinteger

[V : S] = #{i : Vi/Vi+1∼= S}.

Theorem 1.2.3 (Krull-Schmidt Theorem)Any A-module is isomorphic to a direct sum of indecomposable A-modules.Moreover, the latter are uniquely determined, up to order and isomorphism.

Finally, recall that anA-module is said to be completely reducible (or semisim-ple) if it is isomorphic to a direct sum of irreducible A-modules. Again, thedecomposition has to be unique (up to order and isomorphism) by Theorem1.2.2 or Theorem 1.2.3. The following result gives a characterization of such amodule. Its proof can be found in [Alp86, Proposition 2, pp. 2-3].

Lemma 1.2.4An A-module is completely reducible if and only if every A-submodule is a directsummand.

Now the radical of A (denoted by Rad(A)) is the ideal consisting of all theelements of A which annihilate every irreducible A-module. The reader canconsult [Alp86, Proposition 4, pp. 3-5] for a proof of the following result.

Lemma 1.2.5The radical of A is equal to each of the following:

(a) The smallest submodule of A whose corresponding quotient is semisimple;

(b) The intersection of all maximal submodules of A;

(c) The largest nilpotent ideal of A.

We say that A is semisimple if Rad(A) = 0. By Lemma 1.2.5 (a), the K-algebra A is semisimple if and only if it is semisimple as an A-module. Moreover,we have that A is semisimple if and only if every A-module is semisimple sincewe only consider finite-dimensional A-modules.

3

Now let G be a finite group and let KG denote the group algebra of G.Then the reader can find a proof of the following result (in characteristic zero)in [FH91, Corollary 2.18, pp. 17] or [JL93, Theorem 11.9, pp. 100].

Lemma 1.2.6If the group algebra KG of G is semisimple, then every irreducible KG-moduleS appears exactly dim(S) times in any direct sum of irreducible KG-modules.

The trivial representation of G over K can be extended to a morphism ofK-algebras 1 : KG → K as follows: for every κ =

∑g∈G αgg in KG, set

1(κ) =∑g∈G αg. Its kernel is an ideal of KG, called the augmentation ideal of

KG and denoted by ∆(G). The following is a basic but very important resultin the representation theory of finite groups. We shall give some ideas of theproof and refer to [Alp86, Theorem 1, pp. 12-13] for the details.

Theorem 1.2.7 (Maschke’s Theorem)Let G be a finite group. Then the group algebra KG of G is semisimple if andonly if the characteristic of K does not divide the order of G.

Proof. The augmentation ideal ∆(G) of KG is a submodule of KG verifyingKG/∆(G) ∼= K. Now if char(K) divides |G|, the sum G of all elements in Gbelongs to ∆(G) and the reader can easily check that KG is a KG-submoduleof KG, also isomorphic to the trivial KG-module. Hence when the series

KG ⊃ ∆(G) ⊃ KG

is refined to a composition series, the trivial KG-module appears twice and thusKG is not semisimple by Lemma 1.2.6. Conversely, if char(K) does not divide|G|, then |G| is an invertible element of K. Hence if V is any KG-submodule andW is aKG-submodule of V, then one can show thatW is a direct summand of Vby considering a projection πW onto W and then defining a new KG-morphismπ′ : V → V by

π′(v) =1

|G|∑g∈G

gπW (g−1v), v ∈ V.

We now establish a property of irreducible representations of abelian groupsover an algebraically closed field K of any characteristic.

Lemma 1.2.8Let G be an abelian group and let V be an irreducible KG-module. Then V isone-dimensional.

Proof. Let g ∈ G. Since G is abelian, the endomorphism v 7→ gv is a morphismof KG-modules and thus by Lemma 1.2.1, there exists λ(g) ∈ K∗ such thatgv = λ(g)v for every v ∈ V. Therefore, every K-subspace of V is stable underthe action of G and thus the result follows.

4

In the next chapter, we shall work with the unipotent subgroups of SL2(Fq),where m ∈ N∗ and q = pm, which are p-groups by Lemma 1.1.3. Therefore, itwill be interesting to give a few properties of representations of p-groups overalgebraically closed fields of characteristic p. Let then K denote such a field andlet G be a finite p-group.

Lemma 1.2.9The following assertions hold:

(a) Every KG-module 0 6= V contains non-zero vectors fixed by G,

(b) Every irreducible KG-module is trivial,

(c) The ideals ∆(G) and Rad(KG) of KG are equal.

Proof.

(a) We proceed using induction on |G|. In the case where |G| = 1, the resultis obvious. Also if G = Cp is cyclic of order p, then any irreducibleKG-submodule W of V is one-dimensional by Lemma 1.2.8 and eventrivial. Indeed, let λ : G→ K∗ be the representation afforded byW. Thenλp = λ = 1 and thus the result for |G| = p. Assume then that |G| = pn

for some positive integer n > 1. Since G is a p-group, it contains a normalsubgroup N of index p and by the induction hypothesis, the subspaceV N ⊆ V of N -fixed vectors in V is non-zero. Also G/N acts on V N andusing the induction hypothesis one more time, we can affirm that thereexists a non-zero vector v ∈ V N fixed by G/N.

(b) Let V be an irreducible KG-module. By (a), V contains a non-zero vectorv fixed by G and thus contains the trivial KG-module. Now since V wasassumed irreducible, we get the desired result.

(c) The ideal ∆(G) is clearly maximal since its codimension in KG is one andhence Rad(KG) ⊂ ∆(G) by condition (b) of Lemma 1.2.5. Now considerany composition series for KG. Then G acts trivially on every compositionfactor by (b) and thus ∆(G) is nilpotent. Therefore, ∆(G) is containedin Rad(KG) by condition (c) of Lemma 1.2.5 and the result follows.

Finally, we recall a very important result in the representation theory offinite groups, which provides a way of finding a complete set of non-isomorphicirreducible KSL2(q)-modules, where K denotes the algebraic closure of Fp. Itsproof can be found in [Alp86, pp. 14-20].

Theorem 1.2.10The number of irreducible KG-modules equals the number of conjugacy classesof G of elements of order not divisible by the characteristic of K.

5

1.3 DualityLet A be a finite-dimensional K-algebra and let V be a left A-module. Thenits linear dual V ∗ = HomK(V,K) can be provided with the structure of a rightA-module. Indeed, if a ∈ A and θ ∈ V ∗, then define the element θ ·a of V ∗ suchthat

(θ · a)(v) = θ(av), v ∈ V . (1.2)

However, it will be interesting to provide V ∗ with the structure of a leftA-module. Suppose that σ : A → A is an involutive anti-automorphism of A,ie. a K-linear map such that σ(1) = 1, σ2(x) = x and σ(xy) = σ(y)σ(x), thisfor every x, y ∈ A. Then V ∗ is a left A-module, with corresponding action

(a · θ)(v) = θ(σ(a)v), a ∈ A, θ ∈ V ∗ and v ∈ V,

called the contravariant dual of V modulo σ and often denoted by V σ.

Example:

Let G be a group and let V be a left KG-module. Then we usually considerthe anti-automorphism −1 : KG → KG sending any element g ∈ G to its in-verse g−1 (extended linearly to KG) in order to give V ∗ the structure of a leftKG-module.

Now the following results hold for both kinds of duals presented before,but we only show them for contravariant ones and so we drop the left or rightassumptions. First let σ : A → A be an anti-automorphism of A. Then iff : V → W is a morphism of A-modules, its transpose fσ : Wσ → V σ, definedby fσ(θ)(v) = θ(f(v)) for any v ∈ V , is a morphism of A-modules as well. Moregenerally, this construction leads to the linear isomorphism

HomA(V,W ) ∼= HomA(W ∗, V ∗). (1.3)

In fact, all the usual properties of duality carry over to A-modules. Forexample, if V is an A-module, then the natural isomorphism V σσ ∼= V of K-spaces is a morphism of A-modules. Similarly, if V and W are two A-modules,the isomorphisms (V ⊕ W )σ ∼= V σ ⊕ Wσ and (V ⊗ W )σ ∼= V σ ⊗ Wσ areisomorphisms of A-modules. Finally, consider a short exact sequence of A-modules 0→W → V → V/W → 0.

Lemma 1.3.1The induced sequence 0→

(V/W

)σ → V σ →Wσ → 0 is a short exact sequenceof A-modules. In particular, V is irreducible if and only if V σ is irreducible.

Proof. First notice that the A-submodule Z = {θ ∈ V σ : θ|W = 0} of V σ canbe identified with (V/W )σ using the map φ : Z →

(V/W

)σ, defined such that

φ(θ)(v + W ) = θ(v) for every θ ∈ V σ and every v ∈ V. Now the A-morphismψ : V σ → Wσ sending θ ∈ V σ to θ|W is surjective and has kernel Z, whencethe result.

6

1.4 Symmetric and exterior powersFirst recall that ifG is an infinite group, then the group algebraKG ofG consistsof all finite linear combinations of elements each belonging to G, with additionand multiplication as in the finite case. Let then n ∈ N∗ be a positive integerand let L be a field extension of K. Denote by E = EL(n) the n-dimensionalvector space of column vectors over L, ie. E = 〈e1, . . . , en〉L , where (ei)j = δijfor 1 ≤ i, j ≤ n. Now the general linear group GLn(K) acts naturally on E,which becomes an LGLn(K)-module with action as follows: for 0 ≤ k ≤ n andg = gij ∈ GLn(K), we have

g · ek =

n∑r=1

grker.

Furthermore, recall that if G denotes a group and V and W are two LG-modules, then V ⊗ W is an LG-module with corresponding action given ongenerators as

g · (v ⊗ w) = (gv)⊗ (gw). (1.4)

Consequently, the tensor algebra T (E) =⊕

r∈NE⊗r of E admits the struc-

ture of an LGLn(K)-module and it is easy to see that the ideals I generatedby all differences of products x ⊗ y − y ⊗ x (x, y ∈ E) and J generated by allelements of the form x ⊗ x (x ∈ E) are LGLn(K)-submodules of T (E). Thisimmediately leads to the definitions of the symmetric and exterior algebras onE, respectively given as

S(E) = T (E)/I and Λ(E) = T (E)/J.

Now it is possible to view the symmetric algebra S(E) on E as the polynomialalgebra S(E) = L[e1, . . . , en], on which GLn(K) acts as algebra automorphisms,extending the action on E according to the rule

g

( n∏k=1

eikk

)=

n∏k=1

g(ek)ik , (1.5)

for any g ∈ G and any i1, . . . , in ∈ N. With this vision, for r ∈ N, we define therth symmetric power SrE of E to be the rth homogeneous subspace of S(E),ie. the L-subalgebra of S(E) spanned by elements of the form ei1 · · · eir , where1 ≤ i1 ≤ . . . ≤ ir ≤ n.We then have an obvious grading of S(E) as an L-algebra

S(E) ∼=⊕r∈N

SrE. (1.6)

Lemma 1.4.1For every r ∈ N, the L-space SrE is an LGLn(K)-submodule of S(E). Thus thegrading (1.6) is a decomposition as LGLn(K)-modules.

Proof. Consider 0 ≤ i1, . . . , ir ≤ n and g ∈ GLn(K). Then since geik ∈ E forany 1 ≤ k ≤ n and since the action of G on S(E) is given by Equation (1.5),we clearly have g(ei1 · · · eir ) ∈ SrE and the L-algebra SrE is stable under theaction of G.

7

In a similar way, for a given r ∈ N, it is possible to define the rth exteriorpower ΛrE of E as the L-subalgebra of Λ(E) spanned by the elements of theform ei1 ∧ · · · ∧ eir , where 1 ≤ i1 < i2 < . . . < ir−1 ≤ n and x ∧ y denotes theimage of x ⊗ y under the quotient map π : T (E) → Λ(E). Again, we have thefollowing grading of Λ(E) as an L-algebra

Λ(E) ∼=⊕r∈N

ΛrE (1.7)

and the following assertion holds.

Lemma 1.4.2For every r ∈ N, the L-space ΛrE is an LGLn(K)-submodule of Λ(E). Thus thegrading (1.7) is a decomposition as LGLn(K)-modules.

8

CHAPTER 2

Representations of SL2(q)

Throughout this chapter, we let p be a prime and q = pn, where n is a positiveinteger. We then let K be an infinite field containing Fp and denote by G thespecial linear group of degree 2 over Fq, which has order |G| = (q − 1)q(q + 1)by Equation (1.1). We also let U, U− and H respectively denote the threesubgroups U(2)Fq , U(2)−Fq and T (n)Fq introduced in the previous chapter anddefine the elements

U =∑a∈Fq

u(a) ∈ KU, U− =∑a∈Fq

u−(a) ∈ KU−.

Finally, we denote by w the element of G given as

w =

(0 1−1 0

).

The aim of this chapter consists in giving a complete set of non-isomorphicirreducible KG-modules. In order to do this, we choose to view G with itsChevalley group structure and come up with a classification in terms of highestweights. Another approach would have been to use Theorem 1.2.10 togetherwith the well-known table of conjugacy classes of G, which can be found in[FH91, ï¿ 1

2 5.2, pp.71], for example.

2.1 The theory of highest weightsWe now introduce the definition of highest weight and highest weight vector andstudy their relationship with the irreducible KG-modules. All the argumentscan be translated into the language of Chevalley groups and so are still valid ina more general context. Indeed, this section is basically a simplified version of[Ste68, pp. 230-236], in which the case of any Chevalley group of rank one istreated.

9

Definition 2.1.1A non-zero vector v in a KG-module V is called a highest weight vector if itsatisfies the following properties:

(a) xv = v for every x ∈ U,

(b) hv = λ(h)v for every h ∈ H and λ some character on H,

(c) Uwv = µv for some µ ∈ K.

Remark 2.1.1In the situation above, the couple (λ, µ) is called the weight of v. At this point,the reason why condition (c) was introduced is quite unclear. However, weshall see later that it allows us to distinguish the ‘smallest’ KG-module fromthe ‘largest’ and replaces a density argument used in the case whereK is infinite.

We might now ask about existence and uniqueness of highest weight vectorsin an arbitrary KG-module, but first observe that for every a ∈ F∗q , we havewu(a)w = u(−a−1)h(−a−1)wu(−a−1) and since H normalizes U, we get

(Uw)2 = −U +∑a∈F∗q

h(a)Uwu(a). (2.1)

Theorem 2.1.2 (Existence)Every non-zero KG-module V contains a highest weight vector v. Furthermore,for such a v, we have KGv = KU−v.

Proof. By Lemma 1.1.3, U is a p-group and hence by Lemma 1.2.9 (a), thereexists a non-zero vector v ∈ V fixed by U. Now since H normalizes U, for anyh ∈ H and x ∈ U, one has xhv = hh−1xhv = hv, ie. V U is a KH-module. ButH is abelian and so by Lemma 1.2.8, we can find v ∈ V U satisfying conditions(a) and (b) of Definition 2.1.1. Now if Uwv = 0, then v is a highest weightvector of weight (λ, 0). If on the contrary Uwv 6= 0, set v1 = Uwv. Then forany x ∈ U, xv1 = xUwv = Uwv = v1 and thus v1 verifies condition (a) ofDefinition 2.1.1. Morever, for any h ∈ H, hv1 = hUwv = Uwh−1v = λ(h−1)v1

and thus condition (b) of Definition 2.1.1 holds replacing λ by wλ, where wλis the character on H defined by wλ(h) = λ(h−1), for every h ∈ H. Finally,Equation (2.1) yields

Uwv1 = −Uv︸ ︷︷ ︸=0

+∑a∈F∗q

h(a)Uw u(a)︸︷︷︸∈U

v

=

( ∑a∈F∗q

wλ(h(a))

)︸ ︷︷ ︸

=µ

v1.

Consequently, v1 satisfies the third and last condition of Definition 2.1.1 and istherefore a highest weight vector of weight (wλ, µ).

In order to prove the second statement, let v be a highest weight vector ofweight (λ, µ). It is easy to verify that G can be written as G = U−B ∪ wB,

10

where B = UH denotes the subgroup of upper triangular matrices in G (thisdecomposition still holds for any Chevalley group of rank one). Therefore, wehave

KGv = KU−Bv ∪KwBv = KU−v ∪Kwv,

where for any subset X of V, we let KX denote the K-subspace of V generatedby all finite linear combinations of elements each belonging to X. It only remainsto show that wv ∈ KU−v. Now for every a ∈ F∗q , we have

u(a)wv = w2u(a)w−1v

= wwu(a)w−1v

= wu(−a−1)h(−a−1)w−1u(−a−1)v

= wu(−a−1)h(−a−1)w−1v

= wu(−a−1)w−1h(−a)v

= λ(h(−a))wu(−a−1)w−1v.

and thus using the equality wu(a)w = u(−a−1)h(−a−1)wu(−a−1) one moretime yields

µv = wv +∑a∈F∗q

λ(h(a))wu(a−1)w−1︸ ︷︷ ︸∈U−

v. (2.2)

Theorem 2.1.3 (Uniqueness)Let V be an irreducible KG-module and let v ∈ V be a highest weight vector.Then the line Kv is the unique line in V fixed by U.

Proof. Set V ′ = Kv and V ′′ = ∆(U−)v, which are K-subspaces of V. As weknow, KU−/∆(U−) ∼= K as a KU−-module and thus KU− = ∆(U−) ⊕K asa K-vector space. Now KU−v = V by Theorem 2.1.2, so

V = V ′ + V ′′.

Moreover, if x ∈ V ′ ∩ V ′′, then x = αv for some α ∈ K and x = zv for somez ∈ ∆(U−). Then applying z recursively to x and using Lemma 1.2.9 (c), onegets x = 0. Consequently, V can be written (as a K-vector space) as

V = V ′ ⊕ V ′′.

Assume now that there exists some U -fixed vector v1 ∈ V \V ′. Without loss ofgenerality, one can suppose that v1 ∈ V ′′ and since H is abelian, that v1 isan eigenvector for H. Moreover, we have Uwv1 = wU−v1 = wU−yv for somey ∈ ∆(U−) and hence v1 is a highest weight vector of weight (λ, 0), where λ isa character on H. By Theorem 2.1.2, one gets

V = KU−v1 ⊂ KU−V ′′ = V ′′,

which is a contradiction.

11

By Theorem 2.1.3, any irreducibleKG-module V has a unique highest weightvector (modulo a scalar multiple) and hence V determines uniquely its highestweight (λ, µ). We can then talk about the highest weight of an irreducible KG-module without ambiguity. Now it is obvious that two isomorphic KG-moduleshave same highest weight. In order to show that the converse is true, we use adensity argument which also appears in the case where K is infinite.

Theorem 2.1.4Any two irreducible KG-modules having the same highest weight are isomorphic.

Proof. Consider two irreducibleKG-modules V1 and V2 having the same highestweight (λ, µ) and let v1 and v2 be highest weight vectors of V1 and V2 respec-tively. If v = v1 + v2 ∈ V1 ⊕ V2, Theorem 2.1.2 yields

KGv = KU−v.

Now v2 /∈ KU−v, since otherwise we could write v2 = αv + v′′, with α ∈ Kand v′′ ∈ ∆(U−)v and then projecting on V1 and V2 would yield 0 = α = 1.Therefore, the projection π1 : KGv → V1 is injective. Furthermore, Theorem2.1.2 yields KGv1 = V1 and consequently π1 is surjective as well. The sameargument works for the projection π2 : KGv → V2 and so

V1

∼=←− KGv∼=−→ V2.

We now establish the main result of this section, which gives us an upperbound for the number of irreducible KG-modules. Moreover, it tells us infor-mation on the possible values of the highest weights of these modules.

Theorem 2.1.5Let (λ, µ) be the highest weight of an irreducible KG-module. If λ 6= 1, then itcompletely determines µ. If on the contrary λ = 1, then µ = 0 or −1. Therefore,there are at most |H|+ 1 = q non-isomorphic irreducible KG-modules.

Proof. First consider a one-dimensional irreducible KG-module V. Then V isirreducible as a KU -module (and as a KU−-module as well) and by Lemma1.2.9 (b), V is trivial as a KU -module (and as a KU−-module as well) . UsingLemma 1.1.1, one concludes that G acts trivially on V and hence V has highestweight (λ, µ) = (1, 0). Now if dim(V ) > 1, write V = V ′ ⊕ V ′′ as in the proofof Theorem 2.1.3 (clearly, V ′′ 6= 0 since dim(V ) > 1). Then since U− fixes V ′′,it must fix a line in it and by Theorem 2.1.3 applied to U− instead of U, thisline is uniquely determined in V. Now since U− fixes wv, we get wv ∈ V ′′ andprojecting Equation (2.2) onto V ′ yields

µ =∑h∈H

λ(wh−1w). (2.3)

Hence if λ = 1 and dim(V ) = 1, then µ = 0. If λ = 1 and dim(V ) > 1, thenEquation (2.3) yields µ = −1. Finally, if λ 6= 1, then dim(V ) > 1 and thus byEquation (2.3), µ is determined by λ.

12

2.2 Irreducible KSL2(p)-modulesWe first consider the special case where q = p is a prime and start by provinga result of linear algebra, which we shall use several times in the remainder ofthe project.

Lemma 2.2.1Let K be a field, V a K-vector space and U a subspace of V. In addition supposethat v0, . . . , vm ∈ V are such that v0 + tv1 + · · ·+ tmvm ∈ U for m+ 1 values oft ∈ K, then vi ∈ U for 0 ≤ i ≤ m.

Proof. Ab absurdo, assume that there exists some 0 ≤ i ≤ m such that vi /∈ U,ie. such that vi + U ∈ V/U is non-zero. Then one can choose φ ∈ (V/U)∗ suchthat φ(vi + U) 6= 0. Lifting φ to V gives us a new linear functional φ̃ ∈ V ∗

verifying φ̃(v) = φ(v+U) for every v ∈ V. Clearly, φ̃(vi) 6= 0. But then applyingφ̃ to v0 + tv1 + · · ·+ tmvm ∈ U yields

φ̃(v0) + tφ̃(v1) + · · ·+ tmφ̃(vm) = 0, (2.4)

this for every t ∈ K. But the polynomial in Equation (2.4) is of degree m inthe indeterminate t and has m+ 1 roots, whence φ̃(vi) = 0, a contradiction.

Return to the case where K is the algebraic closure of Fp and let E = E(2)Kdenote the two-dimensional vector space over K having K-basis {e1, e2}. Forany 0 ≤ r ≤ p−1, the rth symmetric power SrE on E is a KGL2(q)-module byLemma 1.4.1 and so is a KSL2(q)-module as well. We write V (r) = SrE andrecall that a K-basis of V (r) is given as {er1, er−1

1 e2, . . . , e1er−12 , er2}.

Finally, let µ, ν ∈ N be two non-negative integers with ν ≤ µ. Then we shallwrite Cµν for the binomial coefficient

Cµν =

(µν

)=

µ!

ν!(µ− ν)!.

Lemma 2.2.2The vector er1 generates V (r) as a KU−-module.

Proof. For any a ∈ Fp, we have

u−(a)er1 = (u−(a)e1)r

= (e1 + ae2)r

=

r∑i=0

Cri ar−iei1e

r−i2 .

Now since 0 ≤ r ≤ p − 1, the binomial coefficients Cri are non-zero and thenLemma 2.2.1 yields ei1e

r−i2 ∈ KU−er1, for every 0 ≤ i ≤ r.

13

Lemma 2.2.3The line Ker1 is the only U -invariant line in V (r).

Proof. Let v ∈ V (r) \Ker1, ie. v =∑ri=0 αie

i1er−i2 , where αi 6= 0 for at least one

i different from r. Then for any a ∈ Fp, we have

u(a)v =

r∑i=0

αiu(a)(ei1er−i2 )

=

r∑i=0

αiei1(ae1 + e2)r−i

=

r∑i=0

r−i∑j=0

αiajCr−ij ei+j1 e

r−(i+j)2

=

r∑i=0

ai( r−i∑j=0

αjCr−ji ei+j1 e

r−(i+j)2

)︸ ︷︷ ︸

vi

and it only remains to check that at least two vi’s are non-zero. Indeed, byLemma 2.2.1, this would imply that KUv is not one-dimensional. Let thenj0 be the smallest integer satisfying αj0 6= 0. Since Cµν is non-zero for every0 ≤ ν ≤ µ ≤ r, the vector vi is non-zero for every 0 ≤ i ≤ r − j0. We thenconclude using the fact that j0 < r.

We are now able to prove the main result of this section, which provides acomplete description of the irreducible KSL2(p)-modules.

Theorem 2.2.4For every 0 ≤ r ≤ p−1, the KSL2(p)-module V (r) is irreducible. Moreover, thelist {V (r)}r=0,...,p−1 forms a complete set of non-isomorphic KSL2(p)-modules.

Proof. Let 0 ≤ r ≤ p− 1 be fixed and consider a non-zero KSL2(p)-submoduleV of V (r). Clearly, V is a KU -module and therefore contains an irreducibleKU -submodule W. However, by Lemma 1.1.3, U is a p-group and so it followsfrom Lemma 1.2.9 (b) that W is trivial as a KU -module, ie. V contains aU -invariant line Kv. This line is obviously a U -invariant line in V (r) and henceby Lemma 2.2.3, one can set v = er1. Finally, by Lemma 2.2.2, we have

V ⊃ KU−er1 = V (r),

which concludes the proof of the first affirmation. Now by Theorem 2.1.5, thereare at most p non-isomorphic irreducibleKSL2(p)-modules. But since the V (r)’shave different dimensions, they clearly are distinct from each other and so thesecond statement directly follows from the first.

14

2.3 Irreducible KSL2(q)-modulesWe now return to the general case G = SL2(q) and give a classification of theirreducible KG-modules. First observe that since SL2(p) ⊂ G, the KG-modulesK,E, S2E, . . . , Sp−1E are irreducible, since they were irreducible as KSL2(p)-modules by Theorem 2.2.4. However, Theorem 2.1.5 tells us that we still haveq − p irreducible KG-modules to determine. Now if r ≥ p, the module SrE isnot irreducible in general. Indeed, consider SpE as a first example. Then forany g = gij ∈ G, we have

gep1 = (g11e1 + g21e2)p = gp11ep1 + gp21e

p2 ∈ Ke

p1 +Kep2,

gep2 = (g12e1 + g22e2)p = gp12ep1 + gp22e

p2 ∈ Ke

p1 +Kep2.

Hence Kep1 +Kep2 is a non-zero proper KG-submodule of SpE and so the latteris not irreducible. Consequently, it seems that we need to look somewhere elsein order to find the remaining irreducible KG-modules.

2.3.1 The Frobenius twist of a KSL2(q)-moduleThe Frobenius morphism on G is the application F : G→ G defined by

F

(a bc d

)=

(ap bp

cp dp

).

The reader can check that F is a well-defined morphism of groups. The rea-son for introducing this particular morphism of groups is that it allows us toconstruct new KG-modules from the irreducible ones we already know at thispoint, ie. the SrE’s, where 0 ≤ r ≤ p − 1. Indeed, let V be a KG-module onwhich G acts according to the representation ρ : G→ GL(V ).

Definition 2.3.1The Frobenius twist of V, denoted V F , is the KG-module consisting in the sameunderlying K-vector space but on which G acts according to the representationρF = ρ ◦ F.

As an example, consider G = SL2(9). By Theorem 2.1.5, there are at mostnine non-isomorphic irreducible KG-modules and according to Theorem 2.2.4,K, E and S2E are three of them. Moreover, since the Frobenius morphism issurjective, the Frobenius twist of any irreducibleKG-module is again irreducibleand so one can show directly that EF and

(S2E

)F are two new irreducible KG-modules. However, four irreducible KG-modules may be missing.

2.3.2 A complete set of irreduciblesAs we saw in the previous subsection, we need a few more tools if we want to givea complete description of the irreducible KG-modules. Let then 0 ≤ r ≤ q − 1be a non-negative integer and observe that r can be written in a unique way asr = a0 + a1p + a2p

2 + . . . + an−1pn−1, where 0 ≤ a0, . . . , an−1 ≤ p − 1 (such a

writing is called the p-adic expansion of r). We then define the new KG-module

L(r) = Sa0E ⊗(Sa1E

)F ⊗ · · · ⊗ (San−1E)Fn−1

. (2.5)

15

Remark 2.3.1For every 0 ≤ r ≤ q − 1, the KG-module L(r) defined in Equation (2.5) canbe viewed as a submodule of SrE using the injective morphism of KG-modulesφ : L(r)→ SrE defined on generators by

φ(u0 ⊗ u1 ⊗ · · · ⊗ un−1) = u0up1 · · ·u

pn−1

n−1 .

In order to check that the map φ defined in Remark 2.3.1 is an injectivemorphism of KG-modules, the following statement is necessary.

Lemma 2.3.2 (Lucas Formula)Let s, r ∈ N be two non-negative integers with s ≤ r having p-adic expansionsr = a0 + a1p+ · · ·+ amp

m and s = b0 + b1p+ · · ·+ bmpm. Then we have

Crs =

Ca0b0 C

a1b1· · ·Cambm : if bi ≤ ai for every 0 ≤ i ≤ m,

0 : otherwise.

Proof. Since the field K has characteristic p, we have the polynomial equality

(x+ 1)r = (x+ 1)a0(xp + 1)a1 · · · (xpm

+ 1)am

and then applying the well-known Binomial theorem to both sides of the equa-tion yields

r∑i=0

Cri xi =

( a0∑j0=0

Ca0j0 xj0

)( a1∑j1=0

Ca1j1 xj1p

)· · ·( am∑jm=0

Camjm xjmp

m

).

We now proceed exactly as we did in the previous subsection, ie. we firstprove that for 0 ≤ r ≤ q− 1 fixed, the KG-module L(r) is generated by er1 as aKU−-module and then that Ker1 is the unique U -invariant line in L(r).

Lemma 2.3.3The KG-module L(r) is generated by er1 as a KU−-module.

Proof. Let r = a0 + a1p+ . . .+ an−1pn−1 be the p-adic expansion of r. We can

then see that a K-basis of L(r) is given by the vectors

ei0+i1p+...+in−1p

n−1

1 er−(i0+i1p+...+in−1p

n−1)2 ,

where 0 ≤ iµ ≤ aµ for every 0 ≤ µ ≤ n − 1. By Lemma 2.3.2, we can thenaffirm that the vector ei1e

r−i2 (0 ≤ i ≤ r) belongs to this basis if and only if the

binomial coefficient Cri is non-zero. Now for every a ∈ Fq, we have

u−(a)er1 =

r∑i=0

Cri ar−iei1e

r−i2

and thus applying Lemma 2.2.1 yields the desired result.

16

Lemma 2.3.4The line Ker1 is the unique U -invariant line in L(r).

Proof. Similar to the proof of Lemma 2.2.3.

We are now able to prove the main result of this chapter, which provides acomplete description of the irreducible KG-modules.

Theorem 2.3.5For every 0 ≤ r ≤ q − 1, the KG-module L(r) is irreducible. Moreover, thelist {L(r)}r=0,...,q−1 forms a complete set of non-isomorphic irreducible KG-modules.

Proof. The first statement can be shown following exactly the same method aswe did in the proof of Theorem 2.2.4, using Lemmas 2.3.3 and 2.3.4 instead ofLemmas 2.2.2 and 2.2.3 respectively.

Now we cannot use a simple dimension argument to see that L(r) and L(r′)are non-isomorphic whenever r 6= r′ and so we shall use an argument on high-est weights and highest weight vectors instead. By Lemma 2.1.3, the highestweight vector of an irreducible KG-module V is uniquely defined (up to scalarmultiplication) to be the generator (as a K-space) of the unique U -invariantline in V. Hence by Lemma 2.3.4, the element v = er1 is a highest weight vectorof L(r) and its corresponding weight (λ, µ) is given as (λ : h(t)→ tr, µ), where

µ =

{0 : if r ∈ {0, . . . , q − 2},−1 : if r = q − 1.

To see this, use the well-known fact that the multiplicative group F∗q is cyclic oforder q − 1 together with the formula (1 − xn) = (1 + x + . . . + xn−1)(1 − x).Therefore by Lemma 2.1.4, the KG-modules L(r) and L(r′) are non-isomorphicwhenever r 6= r′ and thus we conclude using Theorem 2.1.5.

2.3.3 A special family of KSL2(q)-modulesIn the previous subsection, Theorem 2.3.5 gave us a classification of the irre-ducible KG-modules in terms of highest weights and we saw that in general,the KG-module SrE is not irreducible if r ≥ p. Indeed, it contains the properirreducible KG-module L(r). Nevertheless, consider the special case r = pm−1,for some 1 ≤ m ≤ n. Then r = p−1+(p−1)p+. . .+(p−1)pm−1 is the p-adic ex-pansion of r and the injective morphism ofKG-modules φ : L(r)→ SrE definedin Remark 2.3.1 becomes an isomorphism, since dim(L(r)) = pm = dim(SrE).

Definition 2.3.2A KG-module of the form SrE with r as above is called a Steinberg module.

As an example, considerG = SL2(9). The Steinberg modules are L(2) = S2Eand L(8) = S2E ⊗ S2EF ∼= S8E.

17

CHAPTER 3

Representations of sl2(K)

Let K be an algebraically closed field of characteristic zero and let n ∈ N∗be a positive integer. Then the special linear Lie algebra sln(K), defined bysln(K) = {g ∈Mn×n : Tr(g) = 0}, has dimension n2 − 1 and standard basis

{eij : 1 ≤ i 6= j ≤ n} ∪ {ekk − enn : 1 ≤ k ≤ n− 1},

where (eij)µν = δµiδνj . Now in the case n = 2, we let x, y and h respectivelydenote the elements e12, e21 and e11 − e22, ie.

x =

(0 10 0

), y =

(0 01 0

), h =

(1 00 −1

).

Furthermore, the reader can check that the latter satisfy the relations

[h, x] = 2x, [h, y] = −2y and [x, y] = h.

The aim of this chapter is to study the representations of sl2(K) overK.Moreprecisely, we shall give a complete set of irreducible sl2(K)-modules, mentionWeyl’s Theorem of complete reducibility and study how to decompose the tensorproduct of two irreducible sl2(K)-modules. We principally follow the ideas of[Hum78] and [Maz10].

3.1 PreliminariesWe start by recalling some elementary background on representation theory ofLie algebras and explore some basic properties of sl2(K) in particular. Most ofthe results are still valid in the more general context of semisimple Lie algebrasand we refer the reader to [Hum78] if interested in seeing more details.

19

3.1.1 Representations of Lie algebrasLet L be a Lie algebra over a field K. Then an L-module is a K-vector space Vtogether with a bilinear action ( , ) : L× V −→ V satisfying

([x, y], v) = (x, (y, v))− (y, (x, v)),

for every x, y ∈ L and v ∈ V. Throughout this chapter, we shall only work withfinite-dimensional L-modules, unless otherwise indicated. As in group represen-tation theory, one can then define a representation of L to be a morphism ofLie algebras ρ : L→ gl(V ), where V is a K-vector space, and see that there is aone-to-one correspondence between L-modules and representations of L. Now anL-module is said to be irreducible if it has no non-zero proper submodules andanalogously a representation of L is said to be irreducible if the correspondingL-module is irreducible. Finally, an L-module is said to be completely reducibleif it is isomorphic to a direct sum of irreducible L-modules and a representationof L is called completely reducible if the corresponding L-module is completelyreducible. The following result corresponds to Lemma 1.2.4 in the representa-tion theory of groups. Its proof is exactly the same and so we refer the interestedreader to the previously given reference.

Lemma 3.1.1A finite-dimensional L-module is completely reducible if and only if every L-submodule is a direct summand.

3.1.2 Weights and weight spacesLet K be an algebraically closed field, L be a nilpotent Lie algebra over K andlet ρ : L→ gl(V ) and λ : L→ K be two representations. Then λ is said to be aweight of ρ if there exists 0 6= v ∈ V such that (ρ(x)− λ(x)IdV )v = 0 for everyx ∈ L. The K-space Vλ = {v ∈ V : ∀x ∈ L,∃Nx s.t. (ρ(x)− λ(x)IdV )Nxv = 0}is called the weight space of weight λ.

Theorem 3.1.2Let L be a nilpotent Lie algebra over K and V an L-module. Then V admits adecomposition (as an L-module) into a direct sum of weight spaces.

By Theorem 3.1.2, the Cartan subalgebra h = Kh of sl2(K) acts diagonallyon any L-module V, that is,

V =⊕λ∈K

Vλ,

where Vλ = {v ∈ V : h · v = λv for every h ∈ h} is the eigenspace associatedto the eigenvalue λ ∈ K. Since V is a finite-dimensional sl2(K)-module, thereare finitely many λ for which Vλ is different from zero, that is, a finite numberof weights and weight spaces. Furthermore, the reader can easily check that ifv ∈ Vλ, then xv ∈ Vλ+2 and yv ∈ Vλ−2. Hence, there exists at least one weightλ ∈ K such that Vλ+2 = 0 and any non-zero vector v ∈ Vλ is called a maximalvector of weight λ. For such a fixed maximal vector v0 ∈ Vλ, we set v−1 = 0and vk = 1

k!ykv0, for k > 0.

20

Lemma 3.1.3The following assertions hold:

(a) hvk = (λ− 2k)vk;

(b) yvk = (k + 1)vk+1;

(c) xvk = (λ− k + 1)vk−1.

Moreover, if r ∈ N denotes the smallest integer for which vr 6= 0, vr+1 = 0, thenthe K-subspace V (r) of V spanned by the vectors v0, . . . , vr is an irreduciblesl2(K)-submodule of V.

Proof. The fact that formulas (a), (b) and (c) hold follows from straigthforwardcalculations and thus its proof is left to the reader, who can consult [Hum78,Lemma 7.2 pp. 32] for details. Also, it is very easy to see that the K-vectorspace V (r) is an sl2(K)-submodule of V. Finally, let W be a non-zero sl2(K)-submodule of V (r) and pick 0 6= w ∈W, which can be written as

w =

r∑k=0

αkvk,

where α0, . . . , αr ∈ K. Then by applying x and y recursively, one can show thatvk ∈ W for every 0 ≤ k ≤ r and hence W = V (r), which leads to the desiredresult.

3.2 Irreducible sl2(K)-modulesIn this section, we first give a description of the irreducible sl2(K)-modules.Then we recall Weyl’s Theorem, which states that every sl2(K)-module is com-pletely reducible. Finally, we study the decomposition of the tensor product ofany two irreducible sl2(K)-modules, via the Clebsch-Gordan formula.

3.2.1 A complete set of irreduciblesLet E = E(2)K denote the two-dimensional vector space over K having K-basis{e1, e2} (see Section 1.4) and extend the natural action of sl2(K) on E to thesymmetric algebra S(E) as follows: for x ∈ sl2(K) and f, g ∈ S(E), let

x · (fg) = (xf)g + f(xg).

Then provided with this action, S(E) becomes an sl2(K)-module. Moreover,the reader can check that for every r ∈ N, the rth symmetric power SrE of Eis an sl2(K)-submodule of S(E).

Theorem 3.2.1For every r ∈ N, the sl2(K)-module SrE is irreducible. Moreover, the list{SrE}r∈N forms a complete set of non-isomorphic irreducible sl2(K)-modules.

21

Proof. Let r ∈ N be fixed. Then there is an isomorphism of sl2(K)-modulesφ : V (r) → SrE, which sends vk to er−k1 ek2 for every 0 ≤ k ≤ r. Therefore,SrE is an irreducible sl2(K)-module. On the other hand, by Lemma 3.1.3, anyirreducible sl2(K)-module is isomorphic to V (r) for some r ∈ N, which concludesthe proof.

Remark 3.2.2Let V be an irreducible sl2(K)-module and consider any maximal vector v0 ∈ Vof weight λ ∈ K. Setting v−1 = 0 and vk = 1

k!ykv0, for k > 0 as before, Lemma

3.1.3 yields V ∼= V (r), where r denotes the smallest integer for which vr 6= 0,vr+1 = 0. Now applying formula (c) of the same Lemma to k = r + 1, we get0 = (λ − r)vr and hence λ = r ∈ N. In other words, the weight of a maximalvector is a non-negative integer, called the highest weight of V. Consequently,Theorem 3.2.1 gave us a classification of the irreducibles in terms of highestweights.

3.2.2 Complete reducibility and the Clebsch-Gordan for-mula

Let r ∈ N be a non-negative integer and suppose that char(K) = 0. From nowon, we denote by V (r) the irreducible sl2(K)-module of highest weight r. Alsorecall the following result, which in fact holds for any semisimple Lie algebraover an algebraically closed field. Its proof can be found in [Hum78, ï¿ 1

26.3, pp.28-29].

Theorem 3.2.3 (Weyl’s Theorem)Any sl2(K)-module is completely reducible.

Now let V andW be two finite-dimensional L-modules with basis {v1, . . . , vn}and {w1, . . . , wm} respectively. Then their tensor product V ⊗W can be pro-vided with the structure of an L-module, defining the action on its basis byx · (vi ⊗ wj) = (xvi) ⊗ wj + vi ⊗ (xwj), for any x ∈ L, any 1 ≤ i ≤ n and any1 ≤ j ≤ m.

Theorem 3.2.4 (Clebsch-Gordan formula)Let s, r ∈ N be two non-negative integers, with s ≤ r. Then we have

V (r)⊗ V (s) ∼=s⊕i=0

V (r + s− 2i).

Proof. We proceed using induction on s. For s = 0, the result is obviously true.Now for s = 1, recall that the action of sl2(K) on V (1) = E is given as xe1 = 0,xe2 = e1, ye1 = e2, ye2 = 0, he1 = e1, and he2 = −e2.

Now let v be a maximal vector of V (r) and consider v′ = v ⊗ e1. Then thereader can verify that xv′ = 0, hv′ = (r + 1)v and so V (r + 1) is an irreduciblesubmodule of V (r) ⊗ V (1). Finally, Theorem 3.2.3 makes V (r + 1) a directsummand.

22

The same argument applied to v′′ = (yv) ⊗ e1 − rv ⊗ e2 gives us anotherdirect summand V (r − 1) and a simple dimension argument yields

V (r)⊗ V (1) ∼= V (r + 1)⊕ V (r − 1).

Finally, assume the result true for 0 ≤ s ≤ k − 1. Then

(V (r)⊗ V (k − 1)

)⊗ V (1) =

( r+k−1⊕i=0

V (r + k − 1− 2i)

)⊗ V (1)

=

r+k−1⊕i=0

(V (r + k − 1− 2i)⊗ V (1)

)

=

r+k−1⊕i=0

(V (r + k − 2i)⊕ V (r + k − 2− 2i)

).

On the other hand,

V (r)⊗(V (k − 1)⊗ V (1)

)= V (r)⊗ V (k)⊕ V (k − 2)⊗ V (r)

= V (r)⊗ V (k)⊕( r+k−2⊕

i=0

V (r + k − 2− 2i)

)and thus we conclude using the associativity of the tensor product.

23

24

CHAPTER 4

Representations of SL2(K)

In this chapter, we let K be an algebraically closed field and study the rationalrepresentations of SL2(K) over K.We start by introducing some elementary no-tions from algebraic geometry and then explore some basic properties of SL2(K)as an algebraic group. We then give a complete description of the irreducibleKSL2(K)-modules, depending on the characteristic of K and observe in addi-tion that if the latter is zero, everyKSL2(K)-module is completely reducible andwe can easily decompose the tensor product of any two irreducible KSL2(K)-modules. In fact, the representation theory of SL2(K) over K is similar to therepresentation theory of the Lie algebra sl2(K). On the other hand, if the char-acteristic of K is positive, not every KSL2(K)-modules is completely reducibleand decomposing tensor products becomes far more complicated.

4.1 Some algebraic geometryThe aim of this first section is to familiarize the reader with basic conceptsin algebraic geometry such as affine varieties, the Zariski topology, algebraicgroups, rational representations and connectedness. We shall not give everydetail in our construction, but the reader is welcome to consult [Bor91], [Hum75]or [Spr81], which are the classical references.

4.1.1 Affine varietiesLet V be a set. Then the set KV of functions from V to K is obviously a com-mutative unitalK-algebra, having addition and multiplication defined pointwiseand unit 1 : g 7→ 1K . Given a subalgebra A ⊂ KV and v ∈ V , we define theevaluation map εv ∈ HomK-alg(A,K) at v by

εv : A −→ Kf 7−→ f(v).

25

Definition 4.1.1An affine variety over K is a pair (V,A) consisting of a set V and a finitelygenerated K-subalgebra A of KV such that the map

ε : V −→ HomK−alg(A,K)v 7−→ εv

is a bijection. We write K[V ] for A and call it the coordinate algebra of V.

As an example, consider V = Kn and let A = K[X1, . . . , Xn] denote theK-algebra generated by the coordinate functions Xi ∈ KV , which are given asXi(v) = vi, for v = (v1, . . . , vn) ∈ V. By definition, the K-algebra A is finitelygenerated. Furthermore,

εv = εw ⇐⇒ εv(f) = εw(f) for every f ∈ A⇐⇒ f(v) = f(w) for every f ∈ A⇐⇒ Xi(v) = Xi(w) for every 0 ≤ i ≤ n⇐⇒ vi = wi for every 0 ≤ i ≤ n

and hence the map ε : V → HomK-alg(A,K) is injective. Finally, if θ is amorphism of K-algebras from A to K, then θ is easily seen to be the image ofv = (θ(X1), . . . , θ(Xn)) under ε. Consequently, the latter is surjective as welland (V,A) becomes an affine variety over K, which is usually written An andcalled the affine n-space over K.

Definition 4.1.2A map φ : V → W between two affine varieties is called a morphism of affinevarieties if g ◦ φ ∈ K[V ] for every g ∈ K[W ]. This gives rise to a new mapφ# : K[W ] → K[V ] such that φ#(g) = g ◦ φ, for g ∈ K[W ], which we call thecomorphism corresponding to φ.

Let (V,A) be an affine variety over K, let S ⊂ A be a subset of A and denoteby V(S) = {v ∈ V : f(v) = 0 ∀f ∈ S} the set constitued of those v ∈ V whichare anihilated by every f ∈ S. Then one can show that the sets V(S), S ⊂ Aform the closed sets of a topology on V, known as the Zariski topology.

Lemma 4.1.1Any closed subset of V is an affine variety in its own rights. In particular, anysubset of affine n-space specified by polynomial equations is an affine variety.

Proof. Let W = V(S) be a closed subset of V and consider the K-algebraB = {f |W : f ∈ A}. Clearly B is finitely generated and if ε′w : B → Kdenotes the restriction of the evaluation map εw at w ∈ W , then obviouslyw1 = w2 whenever ε′w1

= ε′w2. Moreover, if θ ∈ HomK-alg(B,K), then the map

θ̄ : A → K sending f ∈ A to θ(f |W ) is a morphism of K-algebras from A toK. Hence there exists some v ∈ V such that εv = θ̄. Now for every f ∈ S, wehave f(v) = εv(f) = θ̄(f) = θ(f |W ) = 0, thus v ∈ W. Finally, for any g ∈ B,we have θ(g) = θ̄(g|W ) = εv(g|W ) = g|W (v) = g(v) = ε′v(g) and so the mapε′ : W → HomK-alg(B,K) is a bijection as desired.

26

We now show a similar result concerning a special class of open sets. Let(V,A) be an affine variety and let 0 6= f ∈ A. The set Vf = {v ∈ V : f(v) 6= 0}is the complement of V(f) in V and thus is open in V. Such sets are said to beprincipal open sets. Also define the K-algebra

Af =

{a

fr: a ∈ A, r ≥ 0

}.

Lemma 4.1.2The pair (Vf , Af ) is an affine variety. Hence any principal open set is an affinevariety in its own rights.

Proof. Left to the reader.

Examples:

(a) Consider the set V = Mn×n(K) of n× n matrices with entries in K. ThenV can be viewed as the affine n-space over K with coordinate algebraK[V ] = K[X11, X12, . . . , Xnn], where Xij(A) = Aij for any A ∈ V.

(b) Let V be as in (a) and let d : V → K denote the determinant map.Then SLn(K) = V(d − 1) and thus is a closed subset of V and so isan affine variety by Lemma 4.1.1, with corresponding coordinate algebraK[SLn(K)] = K[X ′11, . . . , X

′nn], where X ′ij denotes the restriction of Xij

to SLn(K).

(c) Let V and d be as in (b). Then GLn(K) = Vd and thus is a principal openset of V and so is a affine variety, by Lemma 4.1.2, with correspondingcoordinate algebra K[Vd] = K[X11, X12, . . . , Xnn]d.

Finally, consider two affine varieties V, W over K. We have a natural mor-phism of K-algebras i : K[V ]⊗K[W ]→ KV×W defined on generators by

i(f ⊗ g)(v, w) = f(v)g(w).

Since i is injective, we may view the tensor product K[V ] ⊗ K[W ] as a K-subalgebra of KV×W . Now the reader can check that (V ×W,K[V ]⊗K[W ]) isan affine variety (see [Spr81, ï¿1

2 1.5 pp. 13-14] for details) and so from now on,we write K[V ×W ] for K[V ]⊗K[W ].

4.1.2 Algebraic groupsAn algebraic group is a group G which is also an affine variety and such that themultiplication map m : G×G→ G and inversion map i : G→ G are morphismsof affine varieties. A map φ : G → H between two such groups is a morphismof algebraic groups if it is both a morphism of groups and of affine varieties.

Examples :

(a) Denote by Ga the additive group over K. Then Ga is the affine 1-spaceover K and so has coordinate algebra K[Ga] = K[X], where X(α) = αfor every α ∈ K.Moreover, it is easy to see that X ◦m = 1⊗X+X⊗1 andso belongs to K[Ga]⊗K[Ga] = K[Ga×Ga]. Similary, X ◦i = −X ∈ K[Ga]and hence Ga is an algebraic group.

27

(b) Denote by Gm the multiplicative group over K∗. Then Gm is the principalopen set A1

X and so has coordinate algebra K[Gm] = K[X,X−1], whereX(α) = α as before. Moreover, it is easy to see that X ◦ m = X ⊗ X,X−1 ◦m = X−1 ⊗X−1 and X ◦ i = X−1. Therefore, Gm is an algebraicgroup.

(c) Let G = GLn(K). Recall that G is an affine variety overK with correspond-ing coordinate algebra K[G] = K[X11, . . . , Xnn]d, where Xij(g) = gij forevery g ∈ G and d : G → K∗ is the determinant map. Now for any0 ≤ i, j ≤ n, one can easily show that

Xij ◦m =

n∑k=1

Xik ⊗Xkj ∈ K[G]⊗K[G] = K[G×G]

and similarly,

d−1 ◦m = d−1 ⊗ d−1 ∈ K[G]⊗K[G] = K[G×G].

Therefore, since K[G] is generated as a K-algebra by the coordinate func-tions together with d−1 and since the map f 7→ f ◦m is a morphism ofK-algebras, the multiplication map m is a morphism of affine varieties.A similar argument shows that the inversion map i also is a morphism ofaffine varieties and thus G is an algebraic group.

(d) Any closed subgroup of GLn(K) is an algebraic group. In particular,SLn(K) is an algebraic group, since it is the subgroup of GLn(K) con-stitued of those elements anihilated by the map d− 1 ∈ K[GLn(K)].

4.1.3 The class of rational modulesLet G be a group, V be a KG-module and ρ : G→ GL(V ) be the representationafforded by V. If {v1, . . . , vn} denotes a K-basis of V, we have

ρ(g)(vi) =

n∑j=1

ρji(g)vj ,

this for any g ∈ G, 1 ≤ i ≤ n. The functions ρij ∈ KG are called coefficientfunctions of the KG-module V. Also, the K-space

cf(V ) = 〈ρij : 1 ≤ i, j ≤ n〉Kgenerated by the coefficient functions is called the coefficient space of V. Thereader can check that cf(V ) is independent of the choice of a K-basis of V.

Definition 4.1.3Let G be an algebraic group and let V be a KG-module. Then V is said to berational if its coefficient space lies in the coordinate algebra K[G] of G.

The reader can verify that the class of rational KG-modules is closed underduality, tensor products, sums and subquotients. As a first example, considerG = GLn(K) and let E = E(n)K denote the n-dimensional space of columnvectors over K. Then E is clearly a rational KG-module and thus the symmetricand exterior algebras S(E) and Λ(E) are rational KG-modules.

28

Lemma 4.1.3Let V be a rational KG-module, W a K-subspace of V and v ∈ V. Then the setS = {g ∈ G : gv ∈W} is closed with respect to the Zariski topology on G.

Proof. Let ρ : G→ GL(V ) be the representation afforded by V and fix a basis{v1, . . . , vm} of W, which we extend to a basis {v1, . . . , vm, vm+1, . . . , vn} of V.Then writing v in this basis as v =

∑ni=1 αivi, we see that gv belongs to W if

and only if there exist λ1, . . . , λm ∈ K such thatn∑i=1

n∑j=1

αi ρji︸︷︷︸∈K[G]

(g)vj −m∑k=1

λkvk = 0.

Therefore the set S = {g ∈ G : gv ∈ W} corresponds to the set of zeros offunctions lying in some subset of K[G] and hence is closed in G.

4.1.4 Irreducibility and ConnectednessLetX be a topological space. ThenX is said to be Noetherian if every ascendingchain of open sets is stationary, or equivalently, if every descending chain ofclosed sets is stationary. Notice that any affine variety V is Noetherian. Indeed,if Z1 ⊃ Z2 ⊃ . . . is a descending chain of closed subsets in V, then we get anascending chain of ideals N (Z1) ⊂ N (Z2) ⊂ . . . in K[V ], where for Z ⊂ V, weset N (Z) = {f ∈ K[V ] : f(z) = 0 for every z ∈ Z}. Therefore, since K[V ] isa Noetherian ring by Hilbert’s basis theorem, we get the desired result. AlsoX is said to be irreducible if it is impossible to write it as a union of twoproper closed subsets, or equivalently, if any two non-empty open subsets neverintersect trivially.

Lemma 4.1.4An affine variety V is irreducible if and only if its coordinate algebra K[V ] isan integral domain.

Proof. Let f, g ∈ K[V ] with fg = 0. Then V = V(fg) = V(f) ∪ V(g). Conse-quently, if V is irreducible, we must have V(f) = V or V(g) = V. Therefore,either f or g is identically zero and thus K[V ] is an integral domain. Conversely,if K[V ] is an integral domain, then fg = 0 if and only if f = 0 or g = 0 and soeither V(f) = V or V(g) = V .

Let X be a set and let X1, . . . , Xn be subsets whose union is X. Then theexpression X = X1∪ . . .∪Xn is said to be irredundant if there are no inclusionsamong the Xi’s and we refer to [Spr81, Proposition 1.2.4 pp. 4-5] for a proof ofthe following statement.

Theorem 4.1.5Let X be a Noetherian topological space. Then X can be written as an irredun-dant union of finitely many irreducible closed subsets X = X1 ∪ . . . ∪ Xn in aunique way (up to order).

29

Now let G be an algebraic group. By Theorem 4.1.5, G = X1 ∪ . . . ∪Xn forsome irreducible closed subsets X1, . . . , Xn of G and without loss of generality,we can assume 1 ∈ X1. For g ∈ G, we have G = gG = gX1 ∪ . . . ∪ gXn andeach gXi is a closed irreducible subset of G. Hence by Theorem 4.1.5, for every1 ≤ i ≤ n, there exists 1 ≤ j ≤ n such that gXi = Xj .

Lemma 4.1.6The irreducible components of G are disjoint. Hence any connected algebraicgroup is irreducible.

Proof. Assume for a contradiction that x ∈ Xi ∩ Xj for some x ∈ G andi 6= j. Then for any g ∈ G, we have g ∈ gx−1Xi ∩ gx−1Xj , so every g ∈ Gbelongs to two irreducible components. Consequently, X1 ⊂ X2 ∪ . . . ∪ Xn

and X1 = (X1 ∩ X2) ∪ . . . ∪ (X1 ∩ Xn) which by irreducibility of X1, givesX1 ∩Xi = X1 for some i 6= 1 and thus X1 ⊂ Xi, contradicting Theorem 4.1.5.

By Lemma 4.1.6, X1 is the unique irreducible component of G containing 1.Hence for any g ∈ X1 we have g = g · 1 ∈ X1 ∩ gX1 and thus gX1 = X1. Wealso have g−1X1 = X1 and hence g−1 ∈ X1. Therefore, X1 is a closed subgroupof G, denoted G0 and the cosets of G0 are the irreducible components of G, ie.

G = G0 t g1G0 t . . . t gn−1G

0

for some g1, . . . , gn−1 ∈ G. Since there are finitely many of them, G0 has finiteindex in G and since G0 is irreducible, it is in particular connected.

Lemma 4.1.7Let H and K be two closed connected subgroups of G and assume in addition thatK has finite index in G. Then H ⊂ K. In particular, the irreducible componentG0 of G is the unique closed connected subgroup of G of finite index.

Proof. Since K has finite index in G, there exist g1, . . . , gn ∈ G such thatG = K t g1K t . . . t gnK. Therefore, we can write H as

H = (H ∩K) t (H ∩ g1K) t . . . t (H ∩ gnK),

which is a union of finitely many closed subsets of H and thus by connectednessof the latter, we have H ∩K = H and H ∩ giK = ∅ for every 1 ≤ i ≤ n.

We now use Lemma 4.1.7 in order to show that the special linear groupSL2(K) over K is connected. In fact, the exact same argument works forSLn(K), n ∈ N arbitrary, by the well-known fact that it is generated by transvec-tions (see Remark 1.1.2), and even for all Chevalley groups, since they also aregenerated by groups which are isomorphic to Ga.

Corollary 4.1.8The group SL2(K) is a connected algebraic group. In particular, its coordinatealgebra K[SL2(K)] is an integral domain.

30

Proof. The unipotent subgroups U and U−1 of SL2(K) are closed and connected,since they are isomorphic to Ga. Hence by Lemma 4.1.7, U,U−1 ⊂ G0 andso Lemma 1.1.1 yields SL2(K) ⊂ G0. Finally, the second statement followsimmediately from Lemma 4.1.4.

4.2 SL2(K) as a connected algebraic groupLet K be an algebraically closed field and set G = SL2(K). The aim of thissection is to introduce the prerequisites necessary to classify the irreduciblerational KG-modules.

4.2.1 Weights and weight spacesLet V be a rational KG-module. The diagonal subgroup T of G obviously actson V and we will show that V is completely reducible as a KT -module. Thisstatement still holds if G denotes any algebraic group and where we take T tobe a torus, ie. a subgroup of G isomorphic to some Dn. We refer the reader to[Spr81, pp. 52-53] for a proof of the general result.

Lemma 4.2.1Let V be a one-dimensional rational KT -module. Then there exists r ∈ Z suchthat the action of T on V is given as

h(t)v = trv, for every t ∈ K∗, v ∈ V .

Proof. Let λ : T → K∗ denote the representation afforded by V and recall thatthe coordinate algebra K[T ] of T is given as K[T ] = K[X11, X

−111 ]. Hence we

haveλ(h(t)) = P (t), t ∈ K∗,

where P ∈ K[x, x−1] is a Laurent polynomial in the indeterminate x. Now usingthe fact that λ is a representation, one can show that P has to be of the formP (x) = xr for some r ∈ Z and so the statement holds.

Theorem 4.2.2Every rational KT -module is completely reducible.

Proof. Let V be a rational KT -module and let ρ : T → GL(V ) denote therepresentation afforded by V. We proceed by induction on dim(V ) = n. If Vis one-dimensional, then V is irreducible. If n = 2, V is either irreducible orcontains a one-dimensional KT -submodule W. In the latter case, Lemma 4.2.1yields

ρ(h(t)) =

(tr P (t)0 ts

), t ∈ K∗,

where P ∈ K[x, x−1] is a Laurent polynomial in the indeterminate x. Again, us-ing the fact that ρ is a representation, one can show that P has to be identicallyzero and so we get the desired result for n = 2.

31

Now assume the Theorem true for n ≥ 2, let V be an (n + 1)-dimensionalKT -module and consider a maximal KT -submodule W of V. By Lemma 1.2.8,W has codimension one in V and is completely reducible by our inductionhypothesis. Therefore there exist two proper KT -submodules W1, W2 of Wsuch that W = W1 ⊕W2 and the morphism of KT -modules

φ : V → V/W1 ⊕ V/W2

has Kernel Ker(φ) = W1 ∩W2 = 0. Consequently, V can be viewed as a KT -submodule of the direct sum of two completely reducible KT -modules and sothe result follows.

Let V be a rational KG-module. By Theorem 4.2.2, V is isomorphic to a di-rect sum of irreducibleKT -modules, which by Lemma 1.2.8 are one-dimensional.Thus using Lemma 4.2.1, we conclude that V has the grading as a KT -module

V =⊕r∈Z

Vr,

where Vr = {v ∈ V : h(t)v = trv for every t ∈ K∗} . In this situation, r ∈ Z iscalled a weight of V whenever Vr is non-zero. The latter is called the weightspace associated to the weight r. Finally, the greatest integer r such that Vr isnon-zero is called the highest weight of V. Again, this can be done in a moregeneral context (see [Spr81, pp.187]).

Lemma 4.2.3For every r ∈ Z, the weight spaces Vr and V−r are isomorphic. Hence the highestweight of a rational KG-module is a non-negative integer.

Proof. Let g = h(t) ∈ T. Then a simple calculation yields w−1gw = g−1 andhence for v ∈ Vr, we get g(wv) = (gw)v = (ww−1gw)v = (wg−1)v = w(g−1v) =t−rwv. This gives rise to a morphism of KT -modules θ : Vr → V−r sendingv ∈ Vr to wv ∈ V−r and the reader can check that θ is an isomorphism.

Now let 0 → X → V → Y → 0 be an exact sequence of rational KG-modules. Then the proof of the following result is left to the reader.

Lemma 4.2.4The naturally induced sequence of K-spaces 0 → Xr → Vr → Yr → 0 is exact.In particular, for any two rational KG-modules V, W and any r ∈ Z, we have

(V ⊕W )r = Vr ⊕Wr.

Finally, let V,W be two rational KG-modules. Then the reader can checkthat the following assertion holds.

Lemma 4.2.5For any r ∈ Z, we have (V ⊗ W )r =

⊕Vi ⊗ Wj , where the sum is over all

i, j ∈ Z satisfying i+ j = r.

32

4.2.2 The formal characterLet V be a rational KG-module. Then the formal character of V is the Laurentpolynomial ΦV ∈ Z[x, x−1] defined by

ΦV =∑r∈Z

dim(Vr)xr.

By Lemma 4.2.3, ΦV is a symmetric Laurent polynomial. Moreover, it doesnot depend on the characteristic of the field K, which is an obvious but veryimportant property.

Examples:

(a) The formal character of the trivial KG-module K is ΦK = 1.

(b) The weights of the KG-module E = E2(K) = 〈e1, e2〉K are −1, 1 and thenthe formal character of E is ΦE = x+ x−1.

(c) More generally, let r ∈ N be a non-negative integer and consider the rthsymmetric power SrE with its KG-module structure. The weights of SrEare r, r − 2, . . . ,−r + 2,−r and then the formal character of SrE is

ΦSrE =

r∑i=0

xr−2i.

In particular, we observe that the list {ΦSrE}r∈N is linearly independantand hence forms a basis of the space of symmetric Laurent polynomials in theindeterminates x, x−1.

Lemma 4.2.6Let W be a submodule of a rational KG-module V . Then ΦV = ΦW +ΦV/W . Inparticular, the formal character of the direct sum of finitely many KG-modulesequals the sum of the formal character of each direct summand.

Proof. This follows directly from Lemma 4.2.4.

Lemma 4.2.7Let V and W be two rational KG-modules. Then ΦV⊗W = ΦV ΦW .

Proof. This follows directly from Lemma 4.2.5.

Now let V be a rational KG-module and assume that {L(λ) : λ ∈ Λ} formsa complete set of non-isomorphic irreducible rational KG-modules.

Theorem 4.2.8The formal character ΦV of V is given as

ΦV =∑λ∈Λ

[V : L(λ)]ΦL(λ).

Proof. Consider a composition series V ⊃ V1 ⊃ . . . ⊃ Vr−1 ⊃ Vr = 0 for V.Then using recursively Lemma 4.2.6 yields the desired result.

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4.2.3 The unipotent subgroups U and U−

We now explore some very important properties verified by the two unipotentsubgroups U and U− of G. Let then V be a rational KG-module, r ∈ Z be aweight of V and consider 0 6= v ∈ Vr.

Lemma 4.2.9For every i, j ∈ Z, there exist vi ∈ Vr+2i and wj ∈ Vr−2j such that for everya ∈ K, we have

(a) u(a)v = v0 + av1 + . . .+ akvk,

(b) u−(a)w = w0 + aw1 + . . .+ alwl.

Proof. The arguments are the same in the two cases and so we only give a proofof (a). Let {x1, · · · , xn} be a K-basis of V, with x1 = v and let ρ : G→ GL(V )be the representation afforded by V. Then for any a ∈ K, we have

u(a)v = u(a)x1

=

n∑k=1

ρk1(u(a))xk

=

n∑k=1

Pk(a)xk,

where for every 1 ≤ k ≤ n, Pk ∈ K[x] denotes a polynomial. Thus there existv0, . . . , vk ∈ V such that for every a ∈ K, we have

u(a)v = v0 + av1 + . . .+ akvk.

Now for t ∈ K∗ fixed, an easy calculation yields h(t)u(a)h(t−1) = u(at2) andhence

h(t)u(a)h(t−1)v = v0 + at2v1 + . . .+ akt2kvk.

On the other hand,

h(t)u(a)h(t−1)v = h(t)u(a)t−rv

= t−rh(t)u(a)v

= t−rh(t)(v0 + av1 + . . .+ akvk)

= t−rh(t)v0 + at−rh(t)v1 + . . .+ akt−rh(t)vk

and thus for every t ∈ K∗, a ∈ K, we have

(t−rh(t)− 1)v0 + (t−rh(t)− t2)av1 + . . .+ (t−rh(t)− t2k)akvk = 0.

Therefore, for every 0 ≤ i ≤ k, we have (t−rh(t)− t2i)vi = 0, which is the sameas saying that vi ∈ Vr+2i.

34

Remark 4.2.10Observe that if we replace a by 0 in Lemma 4.2.9, we get v0 = v and w0 = w.In particular, if v is a highest weight vector of a KG-module V, then U actstrivially on Kv.

Lemma 4.2.11Every irreducible rational KU -module is trivial (the statement also holds forU− instead of U). Hence the only one-dimensional rational KG-module is thetrivial one.

Proof. Since U is abelian, every irreducible KU -module V is one-dimensional,by Lemma 1.2.8. Now for any a ∈ K and any v ∈ V, we have u(a)v = P (a)v,where P ∈ K[x] is a polynomial in the indeterminate x, and since V is a KU -module, we must have P (a + b) = P (a)P (b) for every a, b ∈ K, which forcesP = 1. Finally, the second affirmation follows immediately from Lemma 1.1.1.

4.2.4 A dense subset of SL2(K)

Consider the subset U−TU of G. Then the reader can easily check that anelement g ∈ G belongs to U−TU if and only if X11(g) is non-zero. In otherwords, we have U−TU = GX11

= {g ∈ G : X11(g) 6= 0}, ie. U−TU is aprincipal open set. Hence G = GX11

∪ V(X11), a union of closed subsets of G,the second of which is proper; thus Lemma 4.1.8 yields the following statement.

Lemma 4.2.12The principal open set U−TU is dense in G.

Now let V be a rational KG-module having highest weight r ∈ N and let0 6= v ∈ Vr be a vector of weight r.

Theorem 4.2.13The K-vector space KU−TUv is a KG-submodule of V.

Proof. Consider the subset S = {g ∈ G : gv ∈ KU−TUv} of G. Then Sobviously contains U−TU, which is dense in G by Lemma 4.2.12. On the otherhand, S is closed by Lemma 4.1.3 and so S = S ⊃ U−TU = G, ie. KU−TUvis a KG-submodule of V.

4.3 Irreducible rational KSL2(K)-modulesWe are finally ready to give a complete set of irreducibleKG-modules, wheneverK has characteristic zero or p > 0. The idea is to proceed in the same way as wedid for SL2(q) (and sl2(K)), that is, to give a classification in terms of highestweights. We first need a replacement for Lemma 2.1.4.

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Lemma 4.3.1Any two irreducible KG-modules having the same highest weight are isomorphic.

Proof. Consider two irreducible KG-modules V1 and V2 having the same high-est weight r ∈ N and let v1 and v2 be highest weight vectors of V1 and V2

respectively. If v = v1 + v2 ∈ V1 ⊕ V2, Lemmas 4.2.9 and 4.2.13 yield

KGv = KU−TUv

= KU−Tv

= KU−v

= Kv ⊕( r⊕i=1

Kxi

),

where xi ∈ Vr−2i for every 1 ≤ i ≤ r. Therefore, v2 does not belong to KGv andso the projection map π1 : KGv → V1 is an injective morphism of KG-modules.Moreover, π1(v) = v1 ∈ V1 and thus π1 is surjective as well. The same argumentworks for the projection map π2 : KGv → V2 and so

V1

∼=←− KGv∼=−→ V2.

Let r ∈ N be fixed. By Lemma 4.3.1, there is at most one irreduciblerational KG-module of highest weight r. Consequently, the formal characters ofthe irreducible rational KG-modules are linearly independant and then Lemma4.2.8 gives rise to the following result.

Theorem 4.3.2Two KG-modules have the same set of composition factors (up to isomorphismand including multiplicities) if and only if they have the same formal character.

Let V be an irreducible rational KG-module and let σ : G → G be theinvolutive anti-automorphism of G sending any g ∈ G to its inverse g−1.

Corollary 4.3.3The irreducible rational KG-module V is self-dual, ie. V σ ∼= V as rationalKG-modules.

Proof. By Lemma 1.3.1, the KG-module V σ is irreducible. Moreover, it is easyto see that V σ has same highest weight as V and thus Lemma 4.3.1 yields thedesired result.

Now assume that K has characteristic zero and denote by E = E(2)K thetwo-dimensional K-vector space having K-basis {e1, e2} (see Section 1.4). Forany r ∈ N, recall that the K-space SrE is a rational KGL2(K)-module and sois obviously a rational KG-module as well, which we denote by V (r).

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Theorem 4.3.4For every r ∈ N, the KG-module V (r) is irreducible. Furthermore, the list{V (r)}r∈N forms a complete set of non-isomorphic irreducible rational KG-modules in characteristic zero.

Proof. Let r ∈ N be fixed. In order to prove the first statement, we proceedin the exact same way as in the proof of Theorem 2.2.4, ie. we first show thatthere is a unique U -invariant line Kvr in V (r) and that KU−vr = V (r). Thisargument works since Lemma 4.2.11 plays the same role as Lemma 1.2.9 (b)did in the proof of Theorem 2.2.4. Finally, it is easy to check that V (r) hashighest weight r and then Lemma 4.3.1 yields the desired result.

Now assume that K has characteristic p > 0 and for r ∈ N, let L(r) denotethe KG-module defined as in (2.5). Notice that L(r) is a rational KG-module,since it is a tensor product of rational KG-modules.

Theorem 4.3.5For every r ∈ N, the KG-module L(r) is irreducible. Furthermore, the list{L(r)}r∈N forms a complete set of non-isomorphic irreducible rational KG-modules in positive characteristic.

Proof. The KG-module L(r) is irreducible as a KSL2(p)-module by Theorem2.3.5 and thus is irreducible as a KG-module as well, since Fp ⊂ K. Finally, itis easy to check that L(r) has highest weight r and then Lemma 4.3.1 yields thedesired result.

4.4 Additional results in characteristic zeroIn this section, we consider an algebraically closed field K of characteristic zeroand study how rational KG-modules decompose as direct sums of indecompos-able (even irreducible by the next result) rational KG-modules.

Theorem 4.4.1Every rational KG-module is completely reducible.

Proof. Let V be a rational KG-module and first assume that it contains aKG-submodule W of codimension one. As in the proof of Theorem 3.2.3 (see[Hum78, ï¿ 1

26.3, pp. 28-29]), we can assume W irreducible and so by Theorem4.3.4, W = V (r) for some r ∈ N. Now by Lemma 4.2.11, the sequence 0 →V (r) → V → K → 0 is exact. Dualizing (modulo −1 : KG → KG) and usingLemma 4.3.3, we get the short exact of rational KG-modules

0 −→M −→ V ∗ −→ V (r) −→ 0, (4.1)

where M is a KG-submodule of V ∗ which is isomorphic to the trivial KG-module K.

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If r is an odd integer, let V ′ be the sum of those weight spaces in V ∗ havingeven weight and V ′′ the complement of V ′ in V ∗ as defined below:

V ′ =⊕i∈Z

V ∗2i and V′′ =

⊕i∈Z

V ∗2i+1.

It is clear that V ∗ is isomorphic to the direct sum of V ′ = V (r) and V ′′ = Kas a K-vector space (even as a KT -module) and by Lemma 4.2.9, V ′ and V ′′are stable under the action of U and U− and hence by Lemma 1.1.1, under theaction of G. Therefore V ∗ ∼= V ′ ⊕ V ′′ = V (r)⊕K as KG-modules.

On the other hand, if r is an even integer, take 0 6= v ∈ V ∗r . Then as in theproof of Lemma 4.3.1, we have

KGv = Kv ⊕( r⊕i=0

Kxi

),

where xi ∈ V ∗r−2i for every 0 ≤ i ≤ r. Therefore, KGv has a one-dimensionalzero weight space N. By exactness of (4.1), M ∩N = 0 and hence

V ∗ ∼= V (r)⊕M.

Now let W be an arbitrary rational KG-submodule of V and recall thatthe K-space HomK(V,W ) has the structure of a KG-module, where for everyx ∈ G, v ∈ V and f ∈ HomK(V,W ), we put (x · f)(v) = xf(x−1v). Nowdefine the sets V = {f ∈ HomK(V,W ) : f |W = αIdW for some α ∈ K} andW = {f ∈ V : f |W = 0}, which are two KG-submodules of HomK(V,W ) fittingin the short exact sequence 0 → W → V → K → 0 by Lemma 4.2.11. Indeed,codim(W,V) = 1.

According to the special case considered above, V has a one-dimensionalKG-submodule complementary to W, which is spanned by some f ∈ HomK(V,W ).Without loss of generality, we can assume f |W = IdW . Now f is a morphismof KG-modules and thus Ker(f) is a KG-submodule of V. One can then easilyconclude that V = W ⊕Ker(f) and thus the proof is complete by Lemma 1.2.4.

We are now ready to study the decomposition of the tensor product of anytwo symmetric powers, which are, sinceK is a field of characteristic zero, exactlythe irreducible rational KG-modules by Theorem 4.3.4.

Theorem 4.4.2 (Clebsch-Gordan Formula)Let s, r ∈ N be two non-negative integers with s ≤ r. Then we have

V (r)⊗ V (s) ∼=s⊕i=0

V (r + s− 2i).

Proof. We proceed in a completely different way than in the case of the Lie al-gebra. In fact, the introduction of the formal character earlier makes everythingeasy. Indeed, using Lemmas 4.2.6 and 4.2.7, one can show that the KG-modulesV (r) ⊗ V (s) and

⊕si=0 V (r + s − 2i) have same formal character and thus by

Theorem 4.3.2, they have same composition factors including multiplicities (upto isomorphism). Finally, Theorem 4.4.1 concludes the proof.

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CHAPTER 5

Decomposition in characteristic p > 0

In the previous chapter, Theorem 4.4.2 gave us a way to decompose the tensorproduct of any two symmetric powers into a direct sum of irreducible KSL2(K)-modules, where K denoted an algebraically closed field of characteristic zero.Now in this situation, the symmetric powers formed a complete set of non-isomorphic irreducible rational KSL2(K)-modules by Theorem 4.3.4. Let thenK be an algebraically closed field of characteristic p > 0. We could wonderhow to decompose the tensor product of two irreducible KSL2(K)-modules aswell as how to decompose the tensor product of two symmetric powers. Thefirst problem is studied in [DH05] and [DD09]. The second is unsolved even incharacteristic 2, but we shall see a way to tackle it.

Lemma 5.0.3Let V be an indecomposable GLn(K)-module and let W be a K-subspace ofV. Then W is a KGLn(K)-submodule of V if and only if it is a KSLn(K)-submodule of V .

Proof. Let Z denote the center of GLn(K) and let z ∈ Z be fixed. Then as aK-space, V can be written as

V =⊕λ∈K

V λ, (5.1)

where V λ = {v ∈ V : zv = λv}. Now let λ ∈ K, v ∈ V λ and g ∈ GLn(K). Thenwe have z(gv) = (zg)v = (gz)v = g(zv) = λgv and thus (5.1) is a decompositionof V as a KGLn(K)-module. Since we assumed V indecomposable, V = V λ

for some λ ∈ K, ie. z acts by scalar multiplication on V. Finally, using the factthat GLn(K) = ZSLn(K) yields the desired result.

39

Remark 5.0.4A direct consequence of Lemma 5.0.3 is that a decomposition of V in termsof indecomposable KSLn(K)-submodules can be immediately deduced from adecomposition in terms of indecomposable KGLn(K)-submodules. Now thereis a real advantage in working with KGLn(K)-modules instead of KSLn(K)-modules and we shall clarify this point in the next pages.

5.1 The Schur algebraFrom now on, we let K be an algebraically closed field of characteristic p > 0and we write G = GLn(K).We also denote by AK(n) the K-subalgebra of K[G]generated by the coordinate functions Xij , ie.

AK(n) = K[X11, . . . , Xnn].

Since the coordinate functions Xij are algebraically independent over K,AK(n) can be regarded as the algebra of all polynomials over K in n2 indeter-minates Xij . Finally, for r ∈ N, we let AK(n, r) be the K-subspace of AK(n)consisting of the elements expressible as homogeneous polynomials of degree rin the Xij ’s. Clearly we have a grading (as K-spaces)

AK(n) =⊕r∈N

AK(n, r). (5.2)

5.1.1 The K-spaces AK(n) and AK(n, r)

First recall that aK-coalgebra A = (A,∆, ε) is aK-space together withK-linearmaps ∆ : A→ A⊗A and ε : A→ K satisfying

(IdA ⊗∆) ◦∆ = (∆⊗ IdA) ◦∆,

(IdA ⊗ ε) ◦∆ = (ε⊗ IdA) ◦∆ = IdA.

The maps ∆ and ε are respectively called comultiplication and counit maps.Also a K-bialgebra is a K-space A which is both a unital associative K-algebraand a K-coalgebra. In addition, we require the counit and muliplication mapsto be morphisms of K-algebras.

It easy to see that the decomposition (5.2) is not a K-algebra grading. In-deed, the K-spaces AK(n, r)’s are not subalgebras of AK(n). Nevertheless, con-sider the maps ∆ : AK(n) → AK(n) ⊗ AK(n) and ε : AK(n) → K given ongenerators as

∆(Xij) =

n∑k=1

Xik ⊗Xkj and ε(Xij) = δij .

Lemma 5.1.1The maps ∆ and ε defined above provide the K-algebra AK(n) with the structureof a K-bialgebra. Moreover, the K-subspaces AK(n, r)’s are subcoalgebras ofAK(n) and thus (5.2) is a grading as K-coalgebras.

Proof. The proof only consists of straightforward calculations and thus is leftto the reader.

40

Let A = (A,∆, ε) be aK-coalgebra. Then the linear dual A∗ = HomK(A,K)of A can be provided with the structure of a unital associative K-algebra, wherethe multiplication of any two α, β ∈ A∗ is given as (α · β)(v) = (α ⊗ β)∆(v),this for every v ∈ A, and the unit 1A∗ = ε.

Definition 5.1.1The Schur algebra SK(n, r) is the linear dual of AK(n, r), provided with multi-plication and unit as above.

Finally, G acts as algebra automorphisms on AK(n) by (g · f)(h) = f(hg)and by (f · g)(h) = f(gh), g ∈ G, f, h ∈ AK(n), making it a KG-bimodule.Now the reader can check that (5.2) is a grading as KG-bimodules and that theaction of any element g ∈ G on the coordinate functions is given as

g ·Xij =

n∑k=1

gkjXik and Xij · g =

n∑k=1

gikXkj . (5.3)

5.1.2 The category of polynomial KGLn(K)-modulesLet MK(n) (resp. MK(n, r)) denote the category of KG-modules V whosecoefficient space cf(V ) lies in AK(n) (resp. AK(n, r)). It is possible to prove (see[Mar08] p. 2-3) that any indecomposable module V ∈ MK(n) is homogeneous,ie. V ∈ MK(n, r) for some r ∈ N. Therefore, from now on, we shall considersuch an r to be fixed.

Definition 5.1.2The objects in MK(n, r) are called polynomial KG-modules (of degree r) andthe afforded representations are called polynomial representations (of degree r)of G over K.

Given f ∈ KG, there is a unique linear extension f ∈ KKG of f , which sendsany κ =

∑αgg ∈ KG to

f(κ) =∑

αgf(g), (5.4)

called the evaluation of f at κ. Now for g ∈ G, let eg denote the element ofSK(n, r) such that eg(f) = f(g) for every f ∈ AK(n, r). This gives rise to a mape : G → SK(n, r) and an easy calculation yields egeh = egh for every g, h ∈ Gand e1G = ε. Hence if we extend linearly the map e, we get a morphism ofK-algebras e : KG −→ SK(n, r), which for any κ ∈ KG corresponds to theevaluation at κ seen in (5.4).

Lemma 5.1.2The morphism of K-algebras e defined above is surjective.

Proof. Suppose for a contradiction that Im(e) is a proper subspace of SK(n, r).Hence there exists 0 6= f ∈ AK(n, r) such that for every g ∈ G, eg(f) = f(g) = 0,a contradiction.

41

Lemma 5.1.3An element f ∈ KG belongs to AK(n, r) if and only if f(Ker(e)) = 0.

Proof. Let f ∈ AK(n, r) and consider κ ∈ Ker(e). Then f(κ) = e(κ)(f) = 0and thus the first direction is proved. Conversely, let f ∈ KG be such thatf(Ker(e)) = 0. By applying Lemma 1.3.1 to Lemma 5.1.2, the sequence

0 −→ SK(n, r)∗ −→ (KG)∗ −→ Ker(e)∗ −→ 0

is a short exact sequence. Therefore, there exists some y ∈ SK(n, r)∗ such thaty(e(κ)) = f(κ), for all κ ∈ KG. Consequently, there exists some c ∈ AK(n, r)satisfying y(ξ) = ξ(c), this for all ξ ∈ SK(n, r). In particular, choosing ξ = e(κ)leads to the desired result.

Lemma 5.1.4Let V be a KG-module. Then V belongs toMK(n, r) if and only if Ker(e)V = 0.

Proof. Let {v1, . . . , vn} be a K-basis of V and denote by ρij the matrix affordedby the action of KG on this basis. Then

Ker(e)V = 0 ⇐⇒ ρij(Ker(e)) = 0 ∀ 1 ≤ i, j ≤ n⇐⇒ ρij ∈ AK(n, r) ∀ 1 ≤ i, j ≤ n (by Lemma 5.1.3)

⇐⇒ cf(V ) ⊂ AK(n, r)

⇐⇒ V ∈MK(n, r).

We now prove that there is an equivalence of categories between the categoryMK(n, r) of polynomial KG-modules and the category of SK(n, r)-modules.This is of a capital interest, since contrary to the group algebra KG of G,the Schur algebra SK(n, r) is a finite-dimensional algebra over K. Hence thestudy of SK(n, r)-modules is far more pleasant than the study of polynomialKG-modules in MK(n, r).

Theorem 5.1.5The category of SK(n, r)-modules and the category MK(n, r) of polynomial KG-modules are equivalent.

Proof. Let V ∈MK(n, r) be a polynomialKG-module. Then V can be providedwith the structure of an SK(n, r)-module as follows: for any ξ ∈ SK(n, r), chooseκξ ∈ KG such that e(κξ) = ξ and let ξ act on v ∈ V via

ξ · v = κξv.

By Lemma 5.1.2, there always exists such an element κξ, this for any ξin SK(n, r). Moreoever, if κ′ξ is another element in KG satisfying e(κ′ξ) = ξ,then κξ − κ′ξ ∈ Ker(e) and hence by Lemma 5.1.4, κξv = κ′ξv, so that V is awell-defined SK(n, r)-module.

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Conversely, let M be an SK(n, r)-module. Then M can be provided withthe structure of a KG-module as follows: for any κ ∈ KG, let κ act on v ∈ Vvia

κv = e(κ) · v.

By Lemma 5.1.4, M ∈MK(n, r) and thus the proof is complete.

Finally, let V ∈ MK(n, r) be a polynomial left KG-module and considerthe anti-automorphism −1 : KG → KG defined by sending any g ∈ G to itsinverse g−1 and then extended linearly to the wholeKG. Then the contravariantdual V −1 of V modulo −1 is a left KG-module, but not a polynomial one.However, it is possible to make the dual V ∗ of V into a left polynomial KG-module if we consider another anti-automorphism tr : KG → KG, defined bysending any g ∈ G to its transpose gtr, extended linearly toKG. Using Theorem5.1.5, one can show that this anti-automorphism of KG corresponds to the anti-automorphism σ of SK(n, r), defined on generators ξij , 1 ≤ i, j ≤ n, such thatit sends ξij to ξji (see [Gre81] or [Mar08] for details on the K-basis of SK(n, r)).

5.2 Irreducible SK(n, r)-modulesIn this section, we first give a description of the irreducible polynomial KG-modules in terms of partitions. Then we give a complete set in the case wheren = 2, using the complete set of rational irreducible KSL2(K)-modules givenin Theorem 4.3.5. Since this was not main part of this project, we shall not doeverything in details and thus refer the reader to [Gre81] or [Mar08] if interested.

5.2.1 The formal character (revisited)Let n, r ∈ N be two non-negative integers. An n-composition of r is a sequenceλ = (λ1, . . . , λn) of non-negative integers such that

∑ni=1 λi = r. We denote by

Λ(n, r) the set of all n-compositions of r. Now given λ = (λ1, . . . , λn) ∈ Λ(n, r),the multiplicative character of the diagonal subgroup T of G associated to λ isthe map χλ : T → K∗ sending d(t) = diag(t1, . . . , tn) to tλ1

1 · · · tλnn . Moreover,for V ∈MK(n, r), we define the λ-weight space V λ of V to be the set

V λ = {v ∈ V : d(t)v = χλ(d(t))v for every d(t) ∈ T}

and call it the weight space of weight λ whenever it is non-zero. Now since Kis an algebraically closed field, one can show that any polynomial KG-moduleV ∈MK(n, r) admits a decomposition (as a K-space)

V =⊕

λ∈Λ(n,r)

V λ.

Observe that the weight spaces verify similar properties to those verifiedby the weight spaces introduced in Section 5.2.1 for KSL2(K)-modules. Forexample, the weight spaces V (λ1,...,λn) and V (λσ(1),...,λσ(n)) are isomorphic forany σ ∈ Perm(n), which is a more general case of Lemma 4.2.3. In a similarway, Lemmas 4.2.4 and 4.2.5 can be generalized as well.

43

Definition 5.2.1The formal character of a polynomial KG-module V ∈MK(n, r) is the polyno-mial Φ ∈ Z[x1, . . . , xn] defined by

ΦV =∑

λ∈Λ(n,r)

dim(V λ)xλ11 · · ·xλnn .

Notice that ΦV is homogeneous of degree r. Moreover, ΦV is a symmetricpolynomial by the generalization of Lemma 4.2.3. As an example, considerthe n-dimensional K-vector space E = E(n)K having K-basis {e1, . . . , en} andconsider the rth exterior power ΛrE of E, which we recall has K-basis

{ei1 ∧ . . . ∧ eir : 1 ≤ i1 < . . . < ir ≤ n}.

Lemma 5.2.1The formal character ΦΛrE of ΛrE is the rth elementary symmetric function

er =∑

1≤j1<...<jr≤n

xj1 · · ·xjr .

Proof. This follows immediately from straigthforward calculations, which weleave to the reader.

5.2.2 A complete set of irreducible SK(2, r)-modulesAn n-composition λ = (λ1, . . . , λn) ∈ Λ(n, r) of r is called an n-partition of rif λ1 ≥ λ2 ≥ . . . ≥ λn. The conjugate partition λ′ = (λ′1, . . . , λ

′m) is defined

by λ′i = |{j : λj ≥ i}| (observe that m = λ1). The set of all n-partitions ofr is written Λ+(n, r). Now if λ = (λ1, . . . , λn) ∈ Λ+(n, r) is given, the Youngdiagram for λ is the subset

[λ] = {(i, j) ∈ N∗ × N∗ : 1 ≤ i ≤ n, 1 ≤ j ≤ λi}

of Z×Z. Usually, such a diagram is represented by λi boxes in the ith row, therows of the boxes lined up on the left. Notice that the conjugate partition λ′

of λ is the partition corresponding to the Young diagramm in which rows andcolumns were interchanged.

As an example, Let n = 4, r = 9 and λ = (3, 2, 1, 1). Then the conjugatepartition λ′ of λ is given as λ′ = (4, 2, 1) and the associated Young diagrams are

[λ] = , [λ′] = .

Lemma 5.2.2For each λ ∈ Λ+(n, r), there exists an irreducible SK(n, r)-module Fλ whosecharacter ΦFλ has leading term xλ1

1 · · ·xλnn .

44

Proof. Choose λ ∈ Λ+(n, r) and let λ′ denotes its conjugate partition. Bythe generalization of Lemma 4.2.7 and by Lemma 5.2.1, the SK(n, r)-moduleV = Λλ

′1E ⊗ . . .⊗ Λλ

′sE has formal character

ΦV = eλ′1 · · · eλ′s

and so has leading term xλ11 · · ·xλnn . Hence by Lemma 4.2.4, there exists at least

one composition factor S of V whose character has leading term xλ11 · · ·xλnn .

Setting Fλ = S concludes the proof.

In fact, it turns out that the SK(n, r)-module Fλ described in Lemma 5.2.2is the unique irreducible SK(n, r)-module whose formal character has leadingterm xλ1

1 · · ·xλnn . We shall not give the details of the proof, but the idea is asfollows: first for a polynomial KG-module V ∈MK(n, r), we give a relation be-tween the formal character ΦV of V and the usual character χV of V, which foreach g ∈ G corresponds to the trace of the endomorphism g· : V → V. We thendeduce that the list {ΦVλ : λ ∈ Λ} is linearly independant, where {Vλ : λ ∈ Λ}forms a complete set of non-isomorphic irreducible polynomial KG-modules.Finally, using some facts on symmetric functions, we show that the formal char-acters ΦFλ(λ ∈ Λ+(n, r)) form a Z-basis of the space Sym(n, r) of homogeneoussymmetric polynomials of degree r in the indeterminates x1, . . . , xn. Therefore,we get the following result.

Theorem 5.2.3For every λ ∈ Λ+(n, r), the irreducible SK(n, r)-module Fλ described in Lemma5.2.2 is the only irreducible SK(n, r)-module whose character has leading termxλ1

1 · · ·xλnn . Thus there is a one-to-one correspondance between the set of irre-ducible polynomial KG-modules in MK(n, r) and the set of n-partitions Λ+(n, r).

Now setG = GL2(K) and let L(s, 0) denote the rational irreducibleKSL2(K)-module L(s) considered as a KG-module. Then for a 2-partition λ = (λ1, λ2)in Λ+(2, r), we set

L(λ) =(Λ2E

)⊗λ2 ⊗ L(λ1 − λ2, 0).

Theorem 5.2.4For every λ ∈ Λ+(2, r), the SK(2, r)-module L(λ) is irreducible. Moreover, thelist {L(λ)}λ∈Λ+(2,r) forms a complete set of non-isomorphic irreducible SK(2, r)-modules.

Proof. As a KSL2(K)-module, Λ2E is trivial and hence L(λ) is isomorphic toL(λ1 − λ2) as a KSL2(K)-module. Therefore, L(λ) is irreducible as a KG-module by Theorem 4.3.5 and since SL2(K) ⊂ G. Moreover, it is easy to checkthat ΦL(λ) has leading term xλ1

1 xλ22 and hence Theorem 5.2.3 allows us to con-

clude.

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As an example, let p = 3. Then Λ+(2, 6) = {(6, 0), (5, 1), (4, 2), (3, 3)} and sothe irreducible SK(2, 6)-modules are L(6, 0) = (S2E)F , L(5, 1) = Λ2E⊗E⊗EF ,L(4, 2) = Λ2E ⊗ Λ2E ⊗ S2E and L(3, 3) = Λ2E ⊗ Λ2E ⊗ Λ2E.

5.3 Indecomposable KSL2(K)-modulesLet n, r ∈ N be two non-negative integers and set G = GLn(K). For an n-composition λ = (λ1, . . . , λn) ∈ Λ(n, r), we let SλE denote the polynomialKG-module Sλ1E ⊗ · · · ⊗ SλnE. Moreover, we let Zλ be the K-subspace ofAK(n, r) consisting of all finite linear combinations of elements of the formz1 · · · zn with zi ∈ Zλii , where Zνi = 〈Xij1 · · ·Xijν : 1 ≤ j1, . . . , jν ≤ n〉K . Thereader can easily check that for every λ ∈ Λ(n, r), the K-vector space Zλ is apolynomial KG-submodule of AK(n, r). Thus the latter admits the grading

AK(n, r) =⊕

λ∈Λ(n,r)

Zλ (5.5)

as a polynomial KG-module. Also let ν ∈ N be a non-negative integer and let1 ≤ k ≤ n be arbitrary.

Lemma 5.3.1The KG-modules SνE and Zνk are isomorphic.

Proof. Consider the map θ : E⊗ν → Zνk defined on generators by θ(ej1 ⊗ . . . ⊗ejν ) = Xkj1 · · ·Xkjν . The reader can easily check that θ is a surjective morphismof KG-modules. On the other hand, consider the morphism of KG-modulesψ : E⊗ν → SνE given on generators by ψ(ej1 ⊗ . . .⊗ ejν ) = ej1 · · · ejν . Clearly,Kerψ = Kerθ and so the result.

We are now able to give a very useful decomposition of AK(n, r) as a poly-nomial KG-module, which was the main goal of this subsection.

Theorem 5.3.2The polynomial KG-module AK(n, r) admits the following grading as a KG-module:

AK(n, r) ∼=⊕

λ∈Λ(n,r)

SλE.

Proof. Let m : Zλ11 ⊗ . . .⊗Zλnn → Zλ1

1 · · ·Zλnn denote the multiplication isomor-phism, which sends a generator y1 ⊗ . . .⊗ yn to y1 · · · yn. Then we have

SλE = Sλ1E ⊗ . . .⊗ SαnE∼= Zα1

1 ⊗ . . .⊗ Zλnn (by Lemma 5.3.1 )∼= Zλ1

1 · · ·Zλn (via m )= Zλ

and Equation (5.5) yields the desired result.

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5.3.1 Projective indecomposable modulesLet K be an algebraically closed field, A be a finite-dimensional K-algebraand let {L(λ) : λ ∈ Λ} form a complete set of non-isomorphic irreducible A-modules. We first give a description of the projective indecomposable A-modulesin terms of the irreducible A-modules, following the approach of [Alp86]. Noticethough that the same result can be proven in the more general context of finitelength modules over Artin algebras, but this theory needs the notions of rightminimality, essential epimorphisms and projective covers to be introduced. Ifinterested, the reader can consult [ARO97, I: Artin Rings].

Theorem 5.3.3For each λ ∈ Λ, there exists an indecomposable projective A-module P (λ) con-taining a unique maximal A-submodule M(λ) such that P (λ)/M(λ) ∼= L(λ). Infact, P (λ) is uniquely determined by these properties and the set {P (λ) : λ ∈ Λ}forms a complete set of non-isomorphic indecomposable projective A-modules.

Proof. See [Alp86, Theorem 3, pp. 31-33] for a proof of this statement.

Remark 5.3.4For each λ ∈ Λ, the A-module P (λ) in Theorem 5.3.3 is in fact the projectivecover of the irreducible A-module L(λ). We refer the reader to [ARO97, ï¿ 1

21.4,pp. 12-14] if interested in the details.

Now by Theorem 1.2.3, any A-module P is isomorphic to a direct sum of in-decomposable A-modules in a unique way (up to order and up to isomorphism).If in addition P is projective, any direct summand is projective as well and thusTheorem 5.3.3 yields

P ∼=⊕λ∈Λ

P (λ)(mλ), (5.6)

where for λ ∈ Λ, the coefficient mλ denotes the number of times that P (λ)appears in the decomposition (we call mλ the multiplicity of P (λ) in P ).

Lemma 5.3.5Let λ1, λ2 be in Λ. Then the following assertion holds :

HomA(P (λ1), L(λ2)) ∼={K : if λ1 = λ2,0 : otherwise.

Proof. Let φ ∈ HomA(P (λ1), L(λ2)) be a non-trivial morphism of A-modules.Since L(λ2) is irreducible, φ is surjective and hence P (λ1)/Ker(φ) ∼= L(λ2),which implies the maximality of Ker(φ) in P (λ1) and hence Theorem 5.3.3yields

L(λ2) ∼= P (λ1)/Ker(φ) = P (λ1)/M(λ1) ∼= L(λ1).

Therefore, since we assumed that {L(λ) : λ ∈ Λ} forms a complete set of non-isomorphic irreducible A-modules, HomA(P (λ1), L(λ2)) is non-trivial if and onlyif λ1 = λ2, in which case Lemma 1.2.1 allows us to conclude.

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Corollary 5.3.6The multiplicities mλ in (5.6) are given as mλ = dim

(HomA(P,L(λ))

).

Proof. Let µ ∈ Λ. Then an easy computation yields

HomA(P,L(µ)) =⊕λ∈Λ

HomA(P (λ), L(µ))(mλ)

and applying Lemma 5.3.5 yields the desired result.

Now if M is a projective A-module, one can easily show that M∗ is aninjective A-module and thus the following result holds.

Theorem 5.3.7For each λ ∈ Λ, there exists an indecomposable injective A-module I(λ) con-taining a unique irreducible A-submodule L(λ)∗. In fact, I(λ) is uniquely deter-mined by these properties and the set {I(λ) : λ ∈ Λ} forms a complete set ofnon-isomorphic indecomposable injective A-modules.

Proof. For λ ∈ Λ, let I(λ) = P (λ)∗ denote the linear dual of the indecompos-able projective A-module P (λ) given in Theorem 5.3.3 and then conclude usingLemma 1.3.1.

Finally, let I denote an injective A-module. Then using elementary prop-erties of the linear dual introduced in Section 1.3, we see that Equation (5.6)becomes

I ∼=⊕λ∈Λ

I(λ)(mλ), (5.7)

where mλ = dim(HomA(L(λ)∗, I)

), applying Equation (1.3) to Corollary 5.3.6.

5.3.2 Tackling the problemLet r ∈ N be a non-negative integer. By Theorem 5.3.2, we have

SK(n, r)∗ ∼=⊕

α∈Λ(n,r)

SαE

as SK(n, r)-modules, or equivalently, as polynomial KG-modules. Observe thatSK(n, r) is projective as an SK(n, r)-module and hence SK(n, r)∗ is injective asan SK(n, r)-module. Consequently, for α ∈ Λ+(n, r), the SK(n, r)-module SαEis injective and thus Equation (5.7) yields

SαE ∼=⊕

λ∈Λ+(n,r)

I(λ)(mλ), (5.8)

where mλ = dim(HomG(L(λ)∗, SαE)

).

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Lemma 5.3.8Let A be a finite-dimensional algebra over K, 0 6= e = e2 ∈ A and let V be anA-module. Then the K-vector spaces HomA(Ae, V ) and eV are isomorphic.

Proof. Consider the linear map φ : HomA(Ae, V )→ eV sending θ to θ(e). Firstφ is well-defined, since θ(e) = θ(e2) = eθ(e) ∈ eV. Moreover, φ(θ) = 0 if andonly if θ(Ae) = 0 and thus it is injective. Finally, consider ev ∈ eV. Then themap θ : se 7→ sev is a well-defined morphism of A-modules and thus the proofis complete.

Let α ∈ Λ(n, r) be an n-composition for r and let V be an SK(n, r)-module.Then there exists an idempotent ξα ∈ SK(n, r) such that V α ∼= ξαV (see [Gre81]or [Mar08] for details).

Lemma 5.3.9The K-spaces HomSK(n,r)(V

∗, SαE) and V α are isomorphic. In particular, themultiplicities in (5.8) are given as mλ = L(λ)α, λ ∈ Λ+(n, r).

Proof. By Lemma 5.3.8 applied to A = SK(n, r) and e = ξα, the K-spacesHomSK(n,r)(SK(n, r)ξα, V ) and V α are isomorphic. Then using equation (1.3)yields

HomSK(n,r)(V∗, AK(n, r)ξα) ∼= V α

and the reader can check that AK(n, r)ξα = Zα and thus Lemma 5.3.1 concludesthe proof of the first statement.

Now let V ∈MK(2, r) be a polynomial KGL2(K)-module, which affords therepresentation ρ : GL2(K)→ GL(V ) and denote by V ′ the rational KSL2(K)-module whose associated representation is the restriction ρ|SL2(K). The proofof the following result is straightforward and so is left to the reader.

Lemma 5.3.10Let λ ∈ Λ(2, r) be a 2-decomposition of r. Then the K-spaces V λ and (V ′)λ1−λ2

are equal.

Now let α = (α1, α2) ∈ Λ(2, r) be a 2-partition of r. Then Equation (5.8)becomes

Sα1E ⊗ Sα2E ∼=⊕s

I(r − s, s)(ms), (5.9)

where s runs through those non-negative integers satisfying (r − s, s) ∈ Λ(2, r)andms denotes the multiplicitymr−s,s of I(r−s, s) in SαE.Moreover, applyingLemmas 5.3.9 and 5.3.10 respectively yields

ms = dim(L(r − s, s)α) = dim(L(r − 2s)α1−α2). (5.10)

We are now ready to give a way to compute the multiplicities ms whichappear in the decomposition (5.9).

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Theorem 5.3.11The multiplicities ms in Equation (5.9) are given as

ms =

1 : r − 2α2 = ξ0 + . . .+ pmξm for some ξi ∈ {ai, ai − 2, . . . ,−ai},where r − 2s = a0 + a1p+ . . .+ amp

m,0 : otherwise.

Proof. Let such an s be fixed. Then by Equation (5.10), we have

ms =

{1 : r − 2α2 is a weight of L(r − 2s),0 : otherwise.

Now by definition, L(r − 2s) = Sa0E ⊗ (Sa1E)F ⊗ . . . ⊗ (SamE)Fm

, wherer − 2s = a0 + a1p + . . . + amp

m is the p-adic expansion of r − 2s and whencethe result.

At this point, we know how the tensor product of any two symmetric powersdecomposes in terms of indecomposable rationalKSL2(K)-modules but we haveno idea of what they look like. The first step in trying to solve this problemconsists in finding a way to compute their formal character. Let then r ∈ N bea fixed odd (resp. even) integer, α = (α1, α2) ∈ Λ(2, r) be a 2-partition of r,0 ≤ s ≤ α2 be an integer and denote by mα2

(s) the multiplicity of I(r − s, s)in Sα1E ⊗ Sα2E. By Theorem 5.3.11, the matrix M ∈ M(

⌈r2

⌉,⌈r2

⌉) (resp.

M( r+22 , r+2

2 )) whose coefficients satisfy Mij = mi−1(j − 1) is lower unipotent.We then have the following result.

Theorem 5.3.12The formal characters of the indecomposable SK(n, r)-modules can be computedusing the recursive formula

ΦI(r−α2,α2) = ΦSr−α2EΦSα2E −α2−1∑s=0

mα2(s)ΦI(r−s,s)

Let us give a concrete application of Theorem 5.3.11 and Theorem 5.3.12, byconsidering an algebraically closed field K of characteristic 2 and r = 7. Firstusing Theorem 5.3.11, the reader can verify that the multiplicities are given as

m0(s) =

{1 : s = 0,0 : otherwise.

m1(s) =

{1 : s = 0, 1,0 : otherwise.

m2(s) =

{1 : s = 0, 1, 2,0 : otherwise.

m3(s) =

{1 : s = 0, 2, 3,0 : otherwise.

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In other words, Theorem 5.3.11 yields the decompositions S7E ∼= I(7, 0),S6E ⊗E ∼= I(7, 0)⊕ I(6, 1), S5E ⊗ S2E ∼= I(7, 0)⊕ I(6, 1)⊕ I(5, 2) and finallyS4E ⊗ S3E ∼= I(7, 0)⊕ I(5, 2)⊕ I(4, 3). Now in terms of formal characters, wehave the linear system

1 0 0 01 1 0 01 1 1 01 0 1 1

ΦI(7,0)

ΦI(6,1)

ΦI(5,2)

ΦI(4,3)

=

ΦS7E

ΦS6EΦEΦS5EΦS2E

ΦS4EΦS3E

,

which is a special case of the result stated in Theorem 5.3.12 and the readercan check that we get ΦI(7,0) = ΦS7E , ΦI(6,1) = ΦS5E , ΦI(5,2) = ΦS3E andΦI(4,3) = ΦS5E + ΦE . In order to be efficient while doing these calculations, onecan use the formal character argument used in the proof of the Clebsch-GordanFormula 4.4.2. Notice that the formal characters ΦI(λ) are expressible in termsof positive sums of formal characters of symmetric powers. We shall see in thenext section that this is always the case, for every λ ∈ Λ+(2, r) and every r ∈ N.

Let us give an interesting consequence of Theorem 5.3.11 in characteristic two.

Corollary 5.3.13Let ν ∈ N be a non-negative integer. Then we have

SνE ⊗ SνE ∼= I(ν, ν).

Hence the KG-module SνE ⊗ SνE is always indecomposable if char(K) = 2.

Proof. In this case, α1 = α2 = ν, r = α1 + α2 = 2ν and thus r − 2α2 = 0. Onthe other hand, for any s 6= ν, r − 2s is even and different from zero. We thenconclude using Theorem 5.3.11.

This is where we would stop if there were no alternative to computing thecharacters of the I(λ)’s. However, using the theory of good filtrations, we cango further in our research.

5.4 Further results using ∇-filtrationsLet G = SL2(K) and let r ∈ N be a non-negative integer. We denote by ∇(r)the rational KG-module SrE. A ∇-filtration for a rational KG-module V is afinite sequence V = V0 ⊃ V1 ⊃ . . . ⊃ Vk−1 ⊃ Vk = 0 such that for 0 ≤ i ≤ k− 1,the quotient Vi/Vi+1 is either 0 or isomorphic to some ∇(ri), ri ∈ N.

Lemma 5.4.1Let V be a rational KG-module which admits a ∇-filtration. Then any directsummand of V admits a ∇-filtration as well. Furthermore, if W is anotherrational KG-module having ∇-filtration, then their tensor product V ⊗W alsoadmits a ∇-filtration.

51

Hence since the symmetric powers admit good filtrations (they are them-selves good filtrations), then Lemma 5.4.1 applied to Equation (5.8) tells usthat I(λ) admits a good filtration as well, this for every λ ∈ Λ(2, r), r ∈ N.More precisely, the following assertion holds.

Lemma 5.4.2Let r ∈ N and let λ, µ ∈ Λ(2, r). Then (I(λ) : ∇(µ)) = [∇(µ), L(λ)]. Hence wehave

ΦI(λ) =∑

µ∈Λ+(2,r)

[∇(µ) : L(λ)]Φ∇(µ).

5.4.1 An example in characteristic twoLet K be an algebraically closed field of characteristic two. First an easy com-putation yields

Φ∇(1+2r) = Φ∇(1)Φ∇(r)F , (5.11)

Φ∇(2r) = Φ∇(r)F + Φ∇(r−1)F . (5.12)

Therefore, we have an alternative way to compute the characters of theindecomposables recursively. To illustrate this, let us consider r = 7. ThenEquations (5.11) and (5.12) together with Theorem 4.3.2 yield Φ∇(7) = ΦL(7)

(indeed, L(7) is a Steinberg module for G), Φ∇(5) = ΦL(5)+ΦL(1), Φ∇(3) = ΦL(3)

(again a Steinberg KG-module) and Φ∇(1) = ΦL(1). Hence by Lemma 5.4.2, thereader can check that ΦI(7,0) = Φ∇(7), ΦI(6,1) = Φ∇(5), ΦI(5,2) = Φ∇(3) andΦI(4,3) = Φ∇(5)+Φ∇(1)

, as in the previous section.

52

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