IRREDUCIBLE REPRESENTATIONS OF GL(2,F q ) A main tool that will be used is Mackey's Theorem. The specic intertwiner is given by (f) = f(x) = 1

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1 IRREDUCIBLE REPRESENTATIONS OF GL(2,F q ) NAVA CHITRIK Referenced heavily from Daniel Bump (99), Automorphic Representations, Section 4. In these notes I will give a complete description of the irreducible complex representations of the group GL(2,F q ) where F q is the eld with q = p n elements. This is accomplished by counting; we will describle several irreducible representations and show they are non-isomorphic and then, by some nite-group-representation-theory, show that these are all of them. A main tool that will be used is Mackey's Theorem Theorem. (Mackey's Theorem) Let H and H 2 be subgroups of a group G. Let (π, V ) and (π 2, V 2 ) be complex representations of H and H 2 respectively and let V G and V G 2 be the induced representations to G. Then Hom G (V G, V G 2 ) = { : G Hom C (V, V 2 ) (h 2 gh ) = π 2 (h 2 ) (g) π (h )} The specic intertwiner is given by (f) = f(x) = (xg )f(g). G g G Proof. (omitted for now) This theorem will make it a lot easier to tell if two representations are isomorphic since dealing with Hom G (V G, V G 2 ) is quite annoying given the quite complicated denition of an induced representation. Theorem 2. (Frobenius Reciprocity) Suppose that H is a subgroup of a group G. Let (π, V ) be a representation of H and let (π G, V G ) be its induction to G. Let (ρ, W ) be a representation of G and then, by restricting the action, it is also a representation of H. Then, Hom G (V G, W ) = Hom H (V, W )

2 IRREDUCIBLE REPRESENTATIONS OF GL(2,F q) 2 Let's talk about the maximal tori of GL(2,F q ). That is, maximal abelian subgroups. There are two: () T s = { } a b a, b F q (2) T a = { x y Dy x } = {( x+y D x y D the eld a bit) )} (the isomorphism is only if we extend We will show how all the representations are basically induced from these two subgroups. This is a nice fact since these subgroups are abelian so their representations take on particularly easy forms namely, characters. Lemma 3. Under multiplication by the Borel subgroup B={( )} GL(2,F q ) has double coset representative representatives ( 0 0 ) and ( ). That is, GL(2, F q ) = B\ ( 0 0 ) /B B\ ( ) /B Denition 4. Let χ and χ 2 be characters of F q. Then there's a representation of B on C by ( a b c ).z χ (a) χ 2 (c) z. Now we can induce this one dimensional representation of B to GL(2, F q ) by the usual process: B(χ, χ 2 ) = {f : GL(2, F q ) C f(( a b c ) g) = χ (a) χ 2 (c) f(g)} and we have GL(2, F q ) act by right translation, i.e. π(g).f(g ) = f(g g) which clearly preserves the space. We denote this induced representation B(χ, χ 2 ). This will be a large class of representations of GL(2, F q ) (about half). Proposition 5. Let B(χ, χ 2 ) and B(µ, µ 2 ) be the induced representations as described above, then 2 if χ = χ 2 = µ = µ 2 if χ χ 2 but {χ, χ 2 } = {µ, µ 2 } dim(hom(b(χ, χ 2 ), B(µ, µ 2 )) = (in some order) 0 otherwise

3 IRREDUCIBLE REPRESENTATIONS OF GL(2,F q) 3 In other words, B(χ, χ 2 ) is irreducible if and only if χ χ 2 and B(χ, χ 2 ) = B(µ, µ 2 ) if and only if, as sets {χ, χ 2 } = {µ, µ 2 }. Proof. We combine Mackey's theorem together with the double coset representatives for GL(2, F q ). By Mackey's theorem, we are looking for maps : GL(2, F q ) C (b gb 2 ) = χ χ 2 (b ) (g)µ µ 2 (b 2 ). By the double coset decomposition, there are at most two degrees of freedom for, namely, the values it takes on at the representatives = ( ) and ω ( ) 0. Consider () = (b b ) = χ χ 2 (b ) () µ µ 2 (b ). This implies that () 0 χ χ 2 (b ) = µ µ 2 (b ) χ = µ and χ 2 = µ 2 (you can see this by setting b = ( b ) to get the rst, or b = ( b ) to get the second equality ). Next, consider (ω 0 ( a b )) = (( b a ) ω 0) for an arbitrary a, b F q which implies that (ω 0 ) 0 χ 2 (a)χ (b) = µ (a)µ 2 (b) χ = µ 2 and χ 2 = µ So if χ = χ 2 = µ = µ 2 then there are two degrees of freedom for. If instead only one of the following is true : (χ, χ 2 ) = (µ, µ 2 ) or (χ, χ 2 ) = (µ 2, µ ) then there's one degree of freedom and if none of those are true then () = (ω 0 ) = 0, so there are no intertwiners Remark 6. It is a fact from representation theory that the dimension of an induced representation the index of the subgroup times the dimension of the induced representation. In this case, the Borel subgroup has index q +, so B(χ, χ 2 ) is a q + -dimensional representation What happens to the induced representations B(χ, χ)? By the proposition and the fact that complex representations of nite groups are unitarizable, we have that B(χ, χ) splits into two irreducible representations. It can easily be checked that there's is a one-dimensional invariant subspace coming from the function χ(det(g))

4 IRREDUCIBLE REPRESENTATIONS OF GL(2,F q) 4 amd what remains is a q-dimensional subrepresentation which is the tensoring of χwith B(, ) which is sometimes called the Steinberg representation. Let's count how many representations we have so far and their dimensions. For each character χ, B(χ, χ) gives us a -dimensional and a q-dimensional irreducible representation... and there are (q ) of each For each unequal, unordered pair, (χ, χ 2 ) we get a q + -dimensional irreducible representation... and there are (q )(q 2)/2 of these How many representations do we expect? Well, we use the fact that the regular representation decomposes as G reg = irrep V dimvi i Then, taking the dimensions of each side we get G = (dimv ) 2 irrep And in our case, GL(2, F q ) = (q 2 )(q 2 q) (because the rst row is any element of F 2 q\{0} and the second row is anything in F 2 q that isn't one of the q multiples of the rst row.) In any case, you can count it, and we're missing some. It turns out that the rest of them are induced from characters of the other torus. There's a tidy way to state all these representations under one denition, but rst we'll need some generators of GL(2, F q ). Proposition 7. (Generators) SL(2, F q ) has generators ( y ) y, ( z ) and ( ) varying over y F q and z F q. Together with the matrices ( r ) these also generate GL(2, F q ). There are also some nite number of generators... Now, let ψ be an additive character on F and let χ be a character of E where F F or E = F( D) the unique quadratic extension

5 IRREDUCIBLE REPRESENTATIONS OF GL(2,F q) 5 We call the rst case, the split case. E ismorphic to the isotropic maximal torus T 0. The second is called the anisotropic, or, the non-split case, then E is isomorphic to the non-split torus. There's an involution on each of these modules. In the rst case its (a, b) = (b, a) and in the second it's the galois conjugate. This allows us to dene norm and trace. We require that χ doesn't factor through the norm. Otherwise the representation we are about to construct will not be irreducible. Notice, that in the case E = F F this is the same as the requirement above that χ χ 2. Denition 8. We can dene a fourier transform on functions on E with respect to ψ ˆΦ(x) = ±q E Φ(y)ψ(tr( xy)) Where the ± depends on what case E is. (+ in the rst case, - in the nonsplit case) Denition 9. We dene a representation on GL(2, F q ) called the Weil Representation, denoted (W(χ), ω) as follows W(χ) = { f : E C f(yx) = χ (y)f(x) whenever y has norm } (( y )) ω y.φ(x) = Φ(yx) ω (( z )).Φ(x) = ψ(n(x) z)φ(x) ω (( )).Φ(x) = ˆΦ(x) ω (( a )).Φ(x) = χ(b)φ(bx) where b is such that N(b) = a Proposition 0. The Weil Representation is a group representation of GL(2, F q ). Proof. You basically need to check the relations. Most of this is easy. Which I haven't even stated. Remark. By the way, in the last bullet point, why doesn't it matter which b we choose? Well, if we chose a dierent one, it would be of the form ub where u has norm. Then, by the denition of W(χ), the choice of b is irrelevant. Now for the amazing fact:

6 IRREDUCIBLE REPRESENTATIONS OF GL(2,F q) 6 Proposition 2. If E = F F then χ is of the form (χ, χ 2 ). (W(χ), ω) = B(χ, χ 2 ) We have that Proof. The dimension of W(χ) is equal to q + since the norm one elements of F F are of the form (a, a ) so that any function in W(χ) is specied by what it does on (a, ) for a F and on (, 0). So there's q + degrees of freedom. Recall now that the dimension of B(χ, χ 2 ) is also q + and it is irreducible. Therefore, if we nd an intertwiner (W(χ), ω) B(χ, χ 2 ) then it must be an isomorphism. Indeed there's an easy one. Let Then, dene L : (W, ω) (B, π) by τ : W(χ) C by τ(φ) = Φ(, 0) LΦ(g) = τ(ω(g).φ) LΦ(g) is indeed an element of B(χ, χ 2 ) since LΦ (( a c b ) g) = τ (ω (( a c b ) g).φ) = τ ( ω ( ( ac ) ( ) ( ) ) ) c bc a c g.φ = χ (a)χ 2 (c)τ(φ). We can also check easily that this is an intertwiner: L(ω(h).Φ) = τ(ω(g)ω(h).φ) = τ(ω(gh)φ) = π(h).τ(ω(g)φ) = π(h).l(φ) It remains only to show that in the non-split case, under the condition that χ E (χ doesn't factor through the norm), the Weil representation is irreducible. For this we must introduce one more notion. Denition 3. A representation is called cuspidal if there does not exist a linear functional l V such thatl (π ( x ).v) = l(v) for all x F for all v 0. Lemma 4. In the nite case, a representation is cuspidal there does not exist an element v in V such that π ( x ).v = v for all x F. Proof. For a nite group, the multiplicity of the trivial representation is the same as the multiplicity of the trivial representation of its contragredient. Apply this to the group {( x )} and π. Lemma 5. A cuspidal representation has dimension m(q )

7 IRREDUCIBLE REPRESENTATIONS OF GL(2,F q) 7 Proof. We decompose V = V (a) where V (a) = {l V ˆπ ( x ).l = ψ(ax).l} a F q and where ψ is some xed non-trivial character. Now the requirement that V is cuspidal is simply that V (0) = 0. Furthermore, the action of F q by ( t ) permutes these spaces transitively. Therefore the dimension of V must be a multiple of q. Proposition 6. In the nonsplit case E = F q ( D), W(χ) is cuspidal and irreducible. Proof. Suppose that there existed a Φ W such that π (( x )) Φ(s) = Φ(s) for all x F. We simply apply the denition, π (( x )) Φ(s) = ψ(n(s)x)φ(s) for any s we can certainly nd an x such that ψ(n(s)x). This proves that W is cuspidal. Next, the dimension of W is q because if we look at functions on E there are q 2 degrees of freedom. On the other hand, there are q + norm elements in E so there are q degrees of freedom for functions on E. Finally the value of an function in W at 0 must be 0. Together with 5 this implies that W(χ) is irreducible. Remark 7. None of the W(χ) are isomorphic to any other unless χ (x) = χ 2 ( x), i.e. they are Galois conjugates. You can see this by restriction to matrices ( r ). Now how many such equivalence classes of characters χ are there? Again, we are looking at characters of E /E norm and only the galois pairs. by the restriction that it can't factor through the Theorem 8. The representations B(χ, χ 2 ) with χ χ 2 together with the Weil representations W(χ) coming from the non-split torus comprise all the irreducible representations of GL(2,F q ) Proof. We can just count them. As I noted in 6 we know that You can see this from the isomorphism theorems of groups E/ker(N) = Im(N). We consider the norm N : E F q. If the image isn't trivial (it isn't) then it has to be a factor of p n we have that (p n )(p n + )/ kern divides p n. So kern = m(p n + ) where m divides p n -. On the other hand, all function on E have dimension q 2...

8 IRREDUCIBLE REPRESENTATIONS OF GL(2,F q) 8 let's list o what we have GL(2, F q ) = (q 2 )(q 2 q) = dim(v i ) 2 irrep Type:Number of them dimtotal Contribution B(χ, χ 2 ) χ χ 2 (q )(q 2) 2 q + B(χ, χ) d portion(q ) B(χ, χ) qd portion(q ) q W(χ)??? q Anyhow, the sum of the total contributions should be (q 2 )(q 2 q) = G. Whittaker Models There is a certain naturally-arising class of representations of GL(2,F q ) that contains every irreducible representation except for the one dimensional ones and with multiplicity one. Such a representation is the following: Let N(F ) GL(2, F q ) be the set of matrices of the form {( x )}. Let ψ(x) be a character of N(F ). We dene the representation G to be the representation of GL(2,F q ) that is induced from ψ. Theorem 9. G contains all the representations of GL(2,F q ) except the one dimensional ones. It also contains each with multiplicity one. By a Whittaker Model of an irreducible representation (π, V ) we mean the realization of the representation inside of G. The theorem then is basically that the Whittaker model is unique. More explicitly, a Whittaker model of a representation is a space of functions W(π) = {W : GL(2, F q ) C W (( x ) g) = ψ(x)w (g)} with action by right translation. These are called Whittaker functions.

9 IRREDUCIBLE REPRESENTATIONS OF GL(2,F q) 9 Now recall that by Frobenius Reciprocity, a map V W(π) is the same as a (an N(F ) ) homomorphism V C. This functional is called a Whittaker functional. To be more explicit, a Whittaker functional is a linear functional L on V that satises: L (( x ).v) = ψ(x)l(v) and so, again by the theorem, this space of linear functionals is one-dimensional. Whittaker functionals and Whittaker models can be obtained from eachother. Suppose we had a Whittaker model V = W(π). Then we have a functional on this space W(π) given by W W (). This certainly satises the condition to be a Whittaker functional. On the other hand, if we had a linear functional L then we could construct functions L(π(g).v) which form a vector space of functions equal to W(π) and with the isomorphism V W(π) by v L(π(g).v).