IRREDUCIBLE REPRESENTATIONS OF GL2(Fq)MICHAEL A. HILL1. PreliminariesThe group G = GL2(Fq) is not only the slightly overweight cousin of GL2(C)(it is a “Flabby” S2) but also a tremendously interesting finite group. It is boththe automorphism group of the vector space of dimension 2 over the field Fqand afinite group of order q(q − 1)(q2− 1), so naturally, representation theorists wouldlike very much to understand its representations. We follow the exposition in [1].Because the representations of GL2(Fq) of dimension higher than 1 are difficultto describe, we need to better describe the structure of G. Probably the mostimportant subgroup of G is B, the Borel subgroup. B, together with the non-trivialpermutation matrix w allow G to be decomposed via the Bruhat decomposition.In order to find the representations of G, we will induce representations from B.B contains the group P of “shears,” matrices of the form [ab01], in addition to thegroup D of diagonal matrices and Z = Z(G), the scalar multiples of the identity.Finally, B contains a copy of F+q, the group U of matrices of the form [1 a01].We are now in a position to begin describing the representation of G, but howmany are there? This question is equivalent to the question of the number of con-jugacy classes of G, so we begin by computing this. Linear algebra tells us that twomatrices in GL2are conjugate iff they satisfy the same minimal polynomial. Thus,we can determine the conjugacy classes by looking at the minimal polynomial of anarbitrary matrix. Since G is the group of 2 × 2 matrices, the characteristic polyno-mial is the minimal polynomial whenever the matrix is not a scalar multiple of theidentity, our job is actually quite easy. We classify the characteristic polynomial ofa matrix A basedonitsroots:If the roots αiare equal (and the minimal polynomial is not the characteristicpolynomial), then A is αi· IIf the roots αiare equal and the minimal polynomial is the characteristic poly-nomial, then A is conjugate toα110 α1 If the roots αiare unequal, then A is conjugate toα100 α2 If the roots are not in the field Fq, then they are in the unique quadratic extensionFq2.If¯· denotes the conjugate of an element of Fq2over Fq,thenN(α)=α¯α ∈ Fq,and Tr(α)=α +¯α ∈ Fq, and clearly, A is conjugate toh0 − N(α)1Tr(α)i, α ∈ Fq2− Fq.So, how many conjugacy classes are there? Clearly, any given matrix in G fallsin one of the above conjugacy classes, so we need only find the number of conjugacyclasses of each type. There are q − 1 conjugacy classes of the first two types (1 foreach possible α ∈ F×q),12(q − 1)(q − 2) conjugacy classes of the third type, and12(q2− q) of the fourth type. Thus, there are exactly q2− 1 conjugacy classes.Now that we know the number of representations, we can begin finding them.The easiest place to start is with the representations of B. Since B = Z × P ,we12 MICHAEL A. HILLcan find the irreducible representations of B by finding those of Z and P . Z isisomorphic to F×q, the cyclic group of order q − 1, and so it has q − 1 irreduciblerepresentations ρi: F×q→ C×. P clearly also has q − 1 1-dimensional irreduciblerepresentations ρj= ρj◦ det. Thus, B has (q − 1)21-dimensional charactersρji= ρi⊗ ρj.P has one more irreducible representation. Fix a non-trivial representationψ : F+q→ C×,andforeacha ∈ F×q, define a representation of U byψa(u)=ψ(αuα−1),α=a 001The representation ψ induces a representation on P which satisfies the interestingproperty thatResPUIndPU(ψ)=Ma∈F×qψaFrobenius reciprocity shows then that IndPU(ψ) is irreducible, and the direct sumdecomposition of the restriction shows that it is of dimension q − 1. Thus,φ =IndPU(ψ)is the final irreducible representation of P (the sums of the squares of the dimensionsof the irreducible representations is |P|).We lift this representation to B to find the final q − 1 representations of B:φi= ρi⊗ φ2. Inducing Representations of G from BFirst, we observe that [G : B]=q +1,sodim(IndGB(ρji)) = q +1We begin by associating to each 1-dimensional representation σ of B an orderedpair of 1-dimensional representations of F×q(σ1,σ2), such thatσa ∗0 b= σ1(a)σ2(b)That this can be done is immediate from the fact that B/U = D and D∼=F×q×F×q.For each σ =(σ1,σ2), we also define σwbyσw=(σ2,σ1)Thus,σw(d)=σ(wdw−1) ∀d ∈ DUsing this characterization of a 1-dimensional representation, we analyze the in-duced representations. Since it provides a very nice classification tool, we constructthe Jacquet Module of a representation ρ of G:J(Vρ)={v ∈ Vρ|ρ(u)(v)=v, ∀u ∈ U }Clearly,J mMn=1Vρn!=mMn=1J(Vρn)Theorem 1. If ˆρ is induced from a 1-dimensional representation ρ of B,thendim(J(Vˆρ)) = 2.IRREDUCIBLE REPRESENTATIONS OF GL2(Fq)3The proof of this is actually surprisingly simple, and since the line of reasoningis fairly indicative of the reasoning used for most theorems pertaining to represen-tations of GL2(Fq), we present it here:Proof. Vˆρis the set of all functions f : G → C satisfyingf(bg)=ρ(b)f (g), ∀b ∈ B,g ∈ G.If a vector f is in J(Vˆρ), then f also satisfies the relationf(bu)=f (b)Thus,f(b)=ρ(b)f(1) and f(bwu)=ρ(b)f(w)The Bruhat decomposition of G = B`(BwU) shows that any f is determineduniquely by its value on 1 and w.Thus,J(Vˆρ)=hf1,fwi,wheref1(1) = 1 f1(w)=0, fw(1) = 0 fw(w)=1. This theorem, in conjunction with the following, shows that IndGB(ρji) has atmost two irreducible components.Theorem 2. J(Vˆρ) 6=0iff ρ is a direct summand of IndGB(ρji) for some i and j.A simple dimension argument shows that IndGB(ρji) has at most 2 irreduciblecomponents:Proof. Assume ρ =IndGB(ρji)=ρ1⊕ρ2⊕···⊕ρn. Since each ρiis a direct summandof ρ,weknowthatJ(Vρi) 6=0. Thus,J(Vρ)=J(Vρ1) ⊕···⊕J(Vρn)has dimension greater than or equal to n. However, by the previous theorem,dim (J (Vρ)) = 2, so n ≤ 2 The introduction of the description of a representation of B by an ordered pair ofrepresentations of F×qcan now be used to determine exactly when a representationhas 2 irreducible components:Theorem 3. If ρ is a 1-dimensional representation of B, then the q+1 dimensionalrepresentation IndGB(ρ) is reducible iff ρ = ρw. If it is reducible, then it has a 1-dimensional component and an irreducible q-dimensional component.The proof is lengthy and therefore will be omitted.This