Character theory

Definition 2.4.1 LetAbe ak-algebra andV anA-module of finitek-dimension. After choosing a basis, every elementa∈Aacts onV via a matrix, so it makes sense to speak of its trace.

The character ofV is defined as the map

A−→k, a7→Tr_V(a).

IfK/kis a field extension andW anA_K-module, then we will often also consider the character ofW as anA-module, i.e.

A→A_K −→^χ K.

We use the same notation.

Remark 2.4.2 (i) IfB ⊂Ais a spanning set ofAas aK-vector space (e.g. a basis), any charac-terχis uniquely determined by the valuesχ(b)forb∈B, asχis a vector space homomorphism.

The standard application of this concerns group algebrasA=k[G]. Any characterχis deter-mined byχ(g)forg∈G. The usual use of the word ’character’ is in this sense.

Another important application is to the situation of a field extensionK/kand anA_K-moduleW, affording a characterχ:A_K →K. It is uniquely determined by its values onA(via the natural embeddingA ֒→ A_K). This will be very important in the sequel, in particular, for applications to the definability of Galois representations.

(ii) In certain cases, it makes sense and it is very useful not to consider the natural matrix action from the definition. For instance, letDbe a division quaternionk-algebra, which is split overK.

Then thek-dimension of the simpleD-moduleDis4, but, splitting the algebraD_K = Mat₂(K) has the consequence thatDKhas a2-dimensional simpleK-module. Its trace and determinant are called the reduced trace/determinant. They are half (resp. the square root) of the other trace (resp. determinant).

Proposition 2.4.3 Letkbe a field of characteristic0,Aak-algebra andV,V^′ two semi-simple A-modules of finitek-dimension. If the characters ofV andV^′ are the same (i.e. ifTr_V(a) = Tr_V^′(a) for alla∈A), thenV andV^′are isomorphic asA-modules.

Proof. (See [Bourbaki], p. 136.) Since the statement is about semi-simple modules, we may assume that A is a semi-simple algebra. Letting it act on V ⊕V^′ me may further assume, using Corollary 2.1.27, that we have

A∼= Yr

i=1

Matri Endk(Si) with the simpleA-modulesSioccuring in one ofV orV^′.

The only question is whether the multiplicitiesn, n^′ with which a given simple moduleS, say S =S1, appears inV andV^′are equal. For that we just take the elementc∈Awhich is the identity

2.4. CHARACTER THEORY 35

onS and0elsewhere. We haveTrV(c) = n·dimS and similarly forV^′, from which the result is clear, using thatkhas characteristic0, asn·dimS =n^′·dimSfollows. 2 Theorem 2.4.4 (Burnside-Frobenius-Schur) Let A be a finite dimensionalk-algebra. The char-acters of the simple A-modules arek-linearly independent, ifk is a splitting field of A or kis of characteristic0.

Proof. (See [CurtisReiner], 27.8.) Since all the modules in question are simple and thusJac(A) acts trivially, we may assume thatAis semi-simple. By Corollary 2.1.27 we have

A∼=

Assume first thatkis a splitting field. Then the decomposition becomes A∼=

Yr

i=1

Mat_ri(k).

The characters of the simple modulesS_i ∼= k^ri are now obviously linearly independent, since in A we can choose elements that are zero in all matrix algebras except one, where we put a single1on the diagonal.

Corollary 2.4.5 Absolutely irreducible modules are uniquely characterized by their characters.

Proof. (See [CurtisReiner], 30.15.) LetS1 andS2be absolutely irreducible modules. Replacing Aby its image inEnd_k(S₁⊕S₂)if necessary, we may and do assume thatkis a splitting field ofA.

By Theorem 2.4.4, the corresponding characters arek-linearly independent and hence different. 2

36 CHAPTER 2. GENERAL REPRESENTATION THEORY

This corollary, for example, tells us that simple modules are uniquely determined by their charac-ters, when the field is algebraically closed. The most complete result about characters is the following theorem by Brauer and Nesbitt, where the assumption of the simplicity is dropped. Moreover, it also works when the field over which the algebra is defined is not the splitting field (in the proof we will see that it is no loss of generality to pass to the splitting field).

The Brauer-Nesbitt theorem says that the composition factors of a module are uniquely character-ized by the character of the module. This is, of course, best possible: The characteristic polynomial does not see the difference between a diagonal matrix and a matrix with the same diagonal and some non-zero entries above the diagonal.

Theorem 2.4.6 (Brauer-Nesbitt) Letkbe any field andAak-algebra. LetM, N be twoA-modules which are finite-dimensional ask-vector spaces. If for alla ∈ A, the characteristic polynomials on M andN are equal, thenM andN have the same composition factors, i.e. define the same element in the Grothendieck group.

Proof. (See [CurtisReiner], 30.16.) For this question we may and do assume thatA is a semi-simple finite dimensionalk-algebra: if necessary, replaceM andN by the direct sum of its compo-sition factors; this does not change the characteristic polynomials; and, if necessary, replaceAby its image inEnd_k(M⊕N).

Assume that the action ofAonM andNhas the same characteristic polynomials, but thatMand N are not isomorphic. By splitting off common composition factors, we also assume thatM andN do not have any composition factor in common. We want to show thatM andN are zero.

LetK be a Galois splitting field ofA of finite degree overk. By Proposition 2.2.3, alsoAK is semi-simple. The characteristic polynomials of the action of anya ∈ A_K on M_K andN_K are the same. We again split off all common composition factors, in order to assume thatM_K andN_K do not have any factor in common. We have decompositions into simpleA_K-modules: M_K ∼= L

iS_i^ei andN_K ∼=L

jT_j^fj. Letσ_i andτ_j be the characters corresponding to the simple modulesS_i andT_j, respectively. As the characteristic polynomials are the same, we have the equality

X

i

e_iσ_i =X

j

f_jτ_j.

By Theorem 2.4.4, it follows thate_iandf_j are all0inK. If the characteristic ofKis zero, the proof is finished. Assume now that the characteristic ofkisp, then we obtainp|ei andp|fj for alli, j.

We now crucially use that we know that the characteristic polynomials are the same and not only the traces. The above yields the existence ofA_K-modules M₁ and N₁ such that M_K ∼= M₁^p and N_K ∼= N₁^p. It follows that the characteristic polynomials of theA_K-action onM₁ andN₁ are the same, since takingp-th roots is unique. By Theorem 2.4.4 we again conclude that M₁ ∼= M₂^p and N1∼=N₂^p. Since the degree of the polynomials is divided bypin each step, we obtain a contradiction.

2

Remark 2.4.7 There are different formulations of Theorem 2.4.6: