CHARACTERS OF SL 2 (F q )

CHARACTERS OF SL 2 (F q ) 1. Recall and notations 1.1. Characters of finite groups. 1.1.1. Let Γ be a finite groups, K be a field of characteristic zero containing Γ -th roots of unity. We note by KΓ the group ring associated to the finite group Γ. As a result (Theorem 10.3 of [1]), the K-algebra KΓ is a splitting field for Γ, namely any irreducible K-representation of Γ is also K- irreducible. Moreover, any (finite dimensional) K-representation is afforded by a K-representation. Set Irr(Γ) the set of irreducible K-representations of Γ, then for any finite dimensional K-representation of Γ, we have the following decomposition V V (τ) eτ. τ Irr(Γ) 1.1.2. Let (V, ρ) be a representation of Γ, recall that the character of this representation is the application defined by χ ρ : Γ K, γ Tr(ρ(γ): V V ). We set deg(χ ρ ) = dim K V, and the character χ ρ is called linear if deg(χ ρ ) = 1. The basic properties of the characters are 1. If ρ = τ σ, then χ ρ = χ τ + χ σ. 2. The character χ ρ depends only on the conjugate class: for any γ Γ and any g Γ, we have χ ρ (g 1 γg) = χ ρ (γ). 3. Two representations with the same character are isomorphic. c 1997, Société MathéMatique de France

2 CHARACTERS OF SL 2 (F q) 4. For two representations V V of Γ, we define their scalar product by < V, V > Γ = dim K Hom KΓ (V, V ). By Schur s lemma, if < V, V >= 1, then V V are two isomorphic irreducible representations of Γ. 5. The cardinal Irr(Γ) is equal to the number of conjugacy classes of Γ, and we have moreover χ(1) 2 = Γ. 1.2. Notations. χ Irr(Γ) 1.2.1. Let Γ be a finite group, and KΓ be its group ring. Every representation considered in this report is of finite dimensional. We will denote by K 0 (KΓ) the Grothendieck groups of the category of finite dimensional Γ-representations. Which is then the same as the free abelian group generated by Irr(G). For V a KΓ-module, we denote by [V ] (or [V ] Γ when necessary) the element in the Grothendieck group K 0 (KΓ) corresponding to V. 1.2.2. In this report, we are interested in the characters of the group SL 2 (F q ). In the following, we will also use the following notations G = SL 2 (F q ), where q = p r for some odd prime p. U G the unipotent subgroup {( ) } 1 x U = : x F 0 1 q F q T G the multiplicative subgroup {( ) } a 0 T = 0 a 1 : a F q F q B G the solvable subgroup {( ) } a B = 0 a 1 : a F q. Recall also that in the group G = SL 2 (F q ), there are q + 4 conjugacy classes (Theorem 1.3.3 of [2]). In particular, the number of characters is equal to q+4.

CHARACTERS OF SL 2 (F q) 3 2. Harish-Chandra Induction The main reference of this is the Chapter 3 of [2]. 2.1. Bimodules. Let Γ and Γ be two finite groups, and let M be a (KΓ, KΓ )-bimodule of finite type. In particular, the dual M = Hom K (M, K) is naturally a (KΓ, KΓ)-bimodule. Define the following two functors: and We will denote by F M : KΓ mod KΓ mod, V M KΓ V. F M : KΓ mod KΓ mod, V M KΓ V. F M : K 0 (KΓ ) K 0 (KΓ), and F M : K 0 (KΓ) K 0 (KΓ ) the two induced morphisms between the Grothendieck groups. Lemma 2.1. With the notations as before, we have and Hom KΓ (V, F M V ) Hom KΓ ( F M V, V ) Hom KΓ ( F M V, V ) Hom KΓ (V, F M V ). Proposition 2.2. For any χ K 0 (KΓ ) and any γ Γ, we have F M (χ )(γ) = 1 Γ Tr M (γ, γ )χ (γ 1 ) γ Γ A similar equality holds for F M. Proof. We consider M K V, and define the action of Γ op Γ on M K V by the following formula (γ 1, γ 2) m x = mγ 1 1 γ 2x. Since the element e = 1 Γ (γ 1, γ ) γ is an idempotent, we get the following decomposition (here we identify e with the endomorphism of M K V defined by e) (1) M K V = ker(e) Im(e). Now we claim that (2) ker(e) = Vect K { m x mγ 1 γ x m M, γ Γ, x V }

4 CHARACTERS OF SL 2 (F q) Indeed, by definition, e(m x mγ 1 γ x) = 0. On the other hand, let i m i x i M K V an element in ker(e), we find hence i ( ) e m i x i i m i x i = 1 Γ = 1 Γ m i γ 1 γ x i = 0 i,γ ( mi x i m i γ 1 γ ) x i. In this way, we get the equality (2). By consequence, i,γ Im(e) M KΓ V. Now, for any γ Γ, which will be identified with the automorphism of M K V defined by γ, the automorphism γ preserves the decomposition (1). We have then γ = γ e + γ (id e) Hence Tr M KΓ V (γ) = Tr Im(e)(e) = Tr M K V (γ e). Therefore F M (V )(γ) = Tr(γ e: M K V M K V ) = 1 Γ Tr M (γ, γ 1 ) Tr V (γ ). γ In terms of characters, we get for any χ K 0 (KΓ ). F M (χ )(γ) = 1 Γ Tr M(γ, γ 1 )χ (γ ) 2.2. Harish-Chandra induction. We consider the (KG, KT )-bimodule K[G/U], where we define (g, t) hu := ghtu G/U. This element is well-defined since T normalizes U. Its dual K[G/U] = Hom K (K[G/U], K) can be identified with K[U \G]: indeed, let {e gu } be the canonical basis of K[G/U], and let {e gu } be the dual basis. For any element h G, and t T, we have h e gu = e h 1 gu, t e gu = e gt 1 U. Now, we define K[G/U] K[U \G], e gu e Ug 1,

CHARACTERS OF SL 2 (F q) 5 this establishes then an isomorphism of (KΓ, KΓ)-bimodules K[G/U] K[U \G]. On applying the general method in 2.1, we obtain the following two functors: and R K : KT modkg mod, V K[G/U] KT V, R K : KG modkt mod, W K[U \G] KG W, called Harish-Chandra induction and restriction respectively. As before, we denote by R K (or R) and R K (or R) the induced map between the Grothendieck groups. Definition 2.3 (Cuspidal character). An irreducible character χ of G is called cuspidal if there is no character α of T such that < χ, R(α) > G 0, i.e., R(χ) = 0. Remark 2.4. For any non zero virtual character α of T, its Harish- Chandra induction R(α) is not cuspidal. It is also possible to give an alternative definition of the Harish-Chandra induction (or restriction). For this, let V be a KT -module, we write V B the KB-module induced by the natural projection B T. Finally, we define Ind G B V B = KG KB V B. Moreover, for W a KG-module, if we look at the submodule W U, this gives a KT -module structure on W. Proposition 2.5. Let V ( resp. W ) be a KT -module ( resp. be a KGmodule). Then R K V Ind G B V B, 2.3. Mackey Formula. Proposition 2.6. Let α, β K 0 (KT ), then with s β(t) = β(s 1 ts) and R K W W U < R(α), R(β) > G =< α, β > T + < α, s β > T Remark 2.7. Note that s 1 ts = t 1 for any t T, hence s β = β is the dual character of β. Once β is a linear character, we have β = β 1. Now let α be any irreducible (linear) character of T µ q 1, since B G is of index q + 1, we find deg(r(α)) = q + 1. Moreover

6 CHARACTERS OF SL 2 (F q) (1) If α 2 1, < R(α), R(α) > G =< R(α), R(α 1 ) > G =< α, α > T = 1. As a result, R(α) = R(α 1 ) are irreducible of degree q + 1. (2) If α = α 0 is the unique linear character of order 2 of T F q µ q 1, then < R(α 0 ), R(α 0 ) > G =< α 0, α 0 > T + < α 0, α 0 > T = 2. As a result, R(α 0 ) has two non isomorphic irreducible factors: R(α 0 ) = R + (α 0 ) R (α 0 ). In particular, R(α 0 ) = R + (α 0 ) + R (α 0 ), with R + (α 0 ) R (α 0 ). (3) If α = 1 T, we have equally < R(1 T ), R(1 T ) >= 2. By Frobenius reciprocity, we have the following natural isomorphism Hom KG (1 G, Ind G B 1 B ) Hom KB (Res G B 1 G, 1 B ) = Hom KB (1 B, 1 B ) As a result, R K (1 T ) contains a copy of 1 G of multiplicity 1. We will denote by St G the other irreducible factor of R K (1 T ) (called the Steinberg character). Hence R(1 T ) = 1 G + St G Since deg(1 G ) = 1, and deg(r(1 T )) = q + 1. Hence deg(st G ) = q. (4) If α / {β, β 1 }, then < R(α), R(β) > G = 0. In this way, we obtain 4 + q 3 G = SL 2 (F q ). 2 = q+5 2 nonisomorphic irreducible characters of Proposition 2.8. deg(r + (α 0 )) = deg(r (α 0 )) = q+1 2. 3. Drinfeld curve 3.1. Definition of Drinfeld curve. In this, F is a fixed algebraic closure of F q (recall that q = p r with p an odd prime), and let Y A 2 F = Spec(F[X, Y ]) be the affine curve defined by the following equation: XY q Y X q = 1. It is easy to verify that this curve is smooth, irreducible. Moreover, G = SL 2 (F q ) acts on A 2 F via g (x, y) = (ax + by, cx + dy) with ( ) a b g = G. c d This action stabilizes Y. µ q+1 acts on A 2 F by homotheties, which induces an action on Y.

CHARACTERS OF SL 2 (F q) 7 the Frobenius endomorphism stabilizes equally Y. F : A 2 F A2 F, (x, y) (xq, y q ) These three actions satisfy the following relations: for any g G = SL 2 (F q ) and any ξ µ q+1, g ξ = ξ g g F = F g. F ξ = ξ 1 F Now consider the monoid G (µ q+1 <F> mon ) where the action of F on µ q+1 is given by ξ ξ 1. The previous relations give then an action of G (µ q+1 <F> mon ) on Y. Proposition 3.1. With the notations as before. 1. The group G acts freely on Y. 2. The group µ q+1 acts freely on A 2 F, and therefore also on Y. 3.2. Interesting quotients. 3.2.1. The following proposition is very useful for the determination of quotients. Recall that for a quasi-projective variety V/K, which is endowed with an action of finite group Γ. Then the fppf-quotient V/Γ is representable. When V = Spec(A) is affine, then V/Γ is representable by Spec(A G ) (SGA1). Proposition 3.2. Let V and W be two smooth and quasi-projective irreducible varieties over a field k, α : V W be a morphism of k-varieties, and Γ a finite group acting on V. Suppose that the following properties are satisfied: (1) α: X Y is surjective; (2) For any v, v V ( k), α(v) = α(v ) if and only v and v are in the same Γ-orbit; (3) There exists x 0 V ( k) such that the induced map of α between the tangent spaces v 0 is surjective. Then α: V/Γ W induced by α is an isomorphism of varieties. Proof. By definition, V/Γ is the cokernel of the following double morphism Γ V pr 2 Φ V α W p ᾱ? V/Γ

8 CHARACTERS OF SL 2 (F q) The second condition implies then the existence of ᾱ, which is surjective and purely inseparable by (2). The last condition implies then α is separable, as a result, so is ᾱ. Therefore, ᾱ is proper, quasifinite, hence is finite. Moreover, ᾱ is birational. The conclusion follows then from the fact that the scheme Y is normal. 3.2.2. Consider the map γ : Y A 1 F, (x, y) xyq2 yx q2 which is µ q+1 <F>-equivariant. Proposition 3.3. The map γ : Y which is an isomorphism. A 1 F induces a map γ : Y/G A1 F Proof. For the first assertion, since the varieties are separated over F, we only need to verify the corresponding statement on the F-points. Let then (x, y) Y (F) any element, and ( ) a b g = G, c d then g (x, y) = (ax + by, cx + dy). Further more (ax + by) (cx + dy) q2 (cx + dy) (ax + by) q2 = (ax + by) (cx q2 + dy q2 ) (cx + dy)(ax q2 + by q2 ) = xy q2 yx q2. That is, γ(g (x, y)) = γ(x, y). Whence the existence of the morphism γ. To verify that γ is an isomorphism, we apply then the criteria (proposition 3.2). 3.2.3. The morphism v : Y A 1 F {0}, (x, y) y is µ q+1 <F> mon -equivariant, where the action of µ q+1 on A 1 F {0} is given by ξ z = ξz. This map is constant on the U-orbits, and induces by passing to quotient a map v : Y/U A 1 F {0} which is again µ q+1 <F> mon -equivariant. Proposition 3.4. The map v is an isomorphism.

CHARACTERS OF SL 2 (F q) 9 3.2.4. The morphism π : Y P 1 F P1 F (F q), (x, y) [x : y] is well-defined and is G < F > mon -equivariant, which induces clearly an isomorphism π : Y/µ q+1 P 1 F P1 F (F q) 3.3. Fixed points under certain Frobenius endomorphism. In order to apply the Lefschetz fixed point theorem, we will need the following results. Proposition 3.5. Let ξ µ q+1, we have Y ξf =. Proof. Since π : Y P 1 F P1 F (F q) is µ q+1 <F> mon -equivariant. Hence we only need to show that ( P 1 F P 1 F (F q) ) F =, which is clear. Proposition 3.6. Let ξ µ q+1. Then { Y ξf 2 0 if ξ 1 = q 3 q if ξ = 1. 3.4. Compactification. If we identify A 2 F as the open subset {[x : y : z] P 2 (F) z 0} P 2 F = Proj(F[X, Y, Z]) then the Zariski closure Y of Y in P 2 F is given by the equation The complement XY q Y X q = Z q+1 Y Y = {[0 : 1 : 0]} {[1 : a : 0] a F q } P 1 F (F q). The action of G (µ q+1 <F> mon ) on Y extends naturally to Y : ( ) a b g [x : y : z] = [ax + by : cx + dy : z] with g = SL c d 2 (F q ); ξ [x : y : z] = [ξx : ξy : z]; F [x : y : z] = [x q : y q : z q ]. These actions stabilize Y. Moreover, one shows that Y /F is smooth. Finally, consider the morphism π 0 : Y P 1 F, [x : y : z] [x : y]

10 CHARACTERS OF SL 2 (F q) By the definition of π 0, this map π 0 is well-defined, G-equivariant and surjective. Moreover, it is constant on µ q+1 -orbits, and therefore induces, after passing to the quotient, a morphism of varieties π 0 : Y /µ q+1 P 1 F Proposition 3.7. The morphism π 0 : Y /µ q+1 P 1 F is a G <F > monequivariant isomorphism, which extends naturally the isomorphism π considered in 3.2.4. 4. Deligne-Lusztig Induction 4.1. Notations and Definitions. We will use the following notations: p = an odd prime, and q = p r. Y/F q the Drinfeld curve as defined in the previous ; l a prime different from p; K/Q l a large enough finite extension of Q l containing G -th roots of unity; We note H c(y, K) the compact support l-adic cohomology with coefficient in K. We will briefly denote it by H c(y ). These groups inherit hence a structure of G (µ q+1 <F> mon )-module. Definition 4.1. If θ is a character of µ q+1, we set R (θ) = ( 1) i [ H i ] c(y ) Kµq+1 V θ i 0 with V θ the Kµ q+1 -module admitting the character θ. In this way, we get a morphism of groups R : K 0 (Kµ q+1 ) K 0 (KG) which will be called Deligne-Lusztig induction. Remark 4.2. Let V be a representation of G µ q+1, and for any linear character θ of µ q+1, let V (θ) be the component of V such that µ q+1 acts via θ. We have then the following decomposition of G-representations V = V (θ) θ µ q+1 Let K θ0 be the irreducible representation associated to the linear character θ 0. Then we find { 0 if θ0 θ; V (θ) Kµq+1 K θ0 V (θ) if θ 0 = θ. G

CHARACTERS OF SL 2 (F q) 11 In the following, we will use the following results in l-adic cohomology. Let X/F be a quasi-projective variety with an action of a monoid Γ via endomorphisms. Proposition 4.3. Let Γ be a finite subgroup normalized by Γ, then we have an isomorphism of Γ/ -modules H i c(x/ ) H i c(x) Proposition 4.4. Let s and u be two invertible elements of Γ such that su = us, s is of order prime to p, and u is of order a power of p, then Tr X(su) = Tr X s(u). Proposition 4.5. Suppose Γ is a finite group, and let T be a torus acting on X which commutes with the action of Γ, then H c(x) H c(x T ) 4.2. First properties. As the curve Y/F is affine and irreducible of dimension 1, we have H i c(y ) = 0, for i / {1, 2}. Hence R (θ) = [ H 1 ] c(y ) Kµq+1 V θ G [ H 2 c(y ) Kµq+1 V θ ]G Moreover, by Poincaré duality, we know that Hence H 2 c(y, K) H 0 (Y, K) 1 G µq+1. [ H 2 c (Y ) ] G µ q+1 = 1 G µ q+1 and we get Proposition 4.6. If θ is a non-trivial linear character of µ q+1, then R (θ) = [ H 1 c(y ) Kµq+1 V θ ] G In particular, R (θ) is a character of G. Proposition 4.7. Let θ be a virtual character of Kµ q+1. Then R (θ) = R (θ ) = R (θ). Proof. Denote by φ: µ q+1 µ q+1 the morphism ξ ξ 1, then θ = φ θ. Let φ H i c(y ) the (KG, Kµ q+1 )-bimodule on which the action of µ q+1 is twisted by φ. To show the first equality, we only need to prove that (3) H i c(y ) φ H i c(y ) as (KG, Kµ q+1 )-modules. Since the Frobenius F induces an automorphism F : H i c(y ) H i c(y )

12 CHARACTERS OF SL 2 (F q) such that F g = g F, and F ξ = ξ 1 F = ξ 1 F, we get in this way the isomorphism (3) as (KG, Kµ q+1 )-bimodule. To show the second equality, for any g G, we have R (θ 1 )(g) = Tr Y (g, ξ 1 )θ (ξ) q + 1 ξ µ q+1 Similarly, R (θ) (g) = = = 1 q + 1 1 q + 1 1 q + 1 ξ µ q+1 Tr Y (g, ξ)θ(ξ). Tr Y (g 1, ξ 1 )θ (ξ 1 ) ξ µ q+1 Tr Y (g 1, ξ 1 )θ(ξ). ξ µ q+1 Hence, we are reduced to show that Tr Y (g, ξ) = Tr Y (g 1, ξ 1 ), which follows from the fact that (g, ξ) is of finite order and the fact that Tr Y (g, ξ) Z. This finishes then the proof. 4.3. The character R (1). By definition, we have Since R (1) = [ H 1 c(y ) µ q+1 ] G [ H 2 c(y ) µ q+1 ] = [ H 1 c(y/µ q+1 ) ] G 1 G. Y = Y P 1 (F q ) with µ q+1 acts trivially on Y Y, we obtain [H c(y/µ q+1 )] G = [H c(p 1 F )] G [H c(p 1 (F q ))] G. Lemma 4.8. We have [H c(p 1 (F q ))] G = K[G/B] G = 1 G + St G, and [H c(p 1 F )] = 2 1 G. Proof. Since H 1 c(p 1, K) = 0, we get immediately the equality concerning P 1 F. For the finite set P1 (F q ), recall that the action of G on P 1 (F q ) is given by ( ) a b g [x : y] = [ax + by : cx + dy], g = G = SL c d 2 (F q ). This action is transitive, and the for x 0 = [1 : 0] P 1 (F q ), we have the isotropic subgroup is B. Hence we find P 1 (F q ) G/B, and thus H c(p 1 (F q )) = H 0 c(p 1 (F q )) K[G/B]. It remains to compute K[G/B], which is Ind G B1 B = R K (1 T ).

CHARACTERS OF SL 2 (F q) 13 The second equality concerning H 0 c(p 1 (F q )) follows from Remark 2.7 (3). As a corollary, we get and therefore R (1) = St G 1 G, [ H i c (Y/µ q+1 ) ] 1 G if i = 2, = St G if i = 1, 0 otherwise 4.4. Dimensions. Let ξ be a non-trivial element of µ q+1. Then Tr Y (ξ) = Tr Y ξ (1) since Y ξ =, we find Tr Y (ξ) = 0. Hence as a character of µ q+1, [H c(y )] µq+1 is a multiple of the character of the regular representation. So for any θ µ q+1, we have deg(r (θ)) = deg(r (1)) = q 1. 4.5. Cuspidality. Proposition 4.9. Let α and θ be characters of T F q µ q 1 and µ q+1 respectively, then < R(α), R (θ) > G = 0. Proof. We may assume that θ is a linear character. Since R (1) = St G 1 G, and R(1) = St G + 1 G, we have < R(1), R (1) >= 0. Moreover, if α does not contain 1 G, we have equally < R(α), R (1) >= 0 (see Remark 2.7). Therefore, we may assume that θ 1. In this case, we know that R (θ) = [ H 1 c(y ) KT K θ ]G is a character of G (i.e., not only a virtual character), therefore, we only need to prove this result when α = reg T is the character of the regular representation. Since B/U T we get Hence reg T B = K[T ] = K[B/U] = Ind B U (1 U ), R(reg T ) = Ind G B(reg T B ) = Ind G B(Ind B U 1 U ) = Ind G U (1 U ). < R (θ), R(reg T ) >=< R (θ) U, 1 U >= dim K (H 1 c(y ) U Kµq+1 K θ ). Since H 1 c(y ) U = H 1 c(y/u) = H 1 c(a 1 F {0}) = 1 µ q+1 Therefore, for θ a linear character different from 1 µq+1, the θ-isotropic part of H 1 c(y ) U is trivial. This finishes the proof.

14 CHARACTERS OF SL 2 (F q) 4.6. Mackey formula. Proposition 4.10. Let θ and η be two virtual characters of µ q+1. We have < R (θ), R (η) > G =< θ, η > µq+1 + < θ, η > µq+1 We set Z = Y Y with the diagonal action of G, and denote by µ (1) q+1 (resp. µ (2) q+1 ) the µ q+1 µ q+1 -set with underlying set µ q+1 and such that, if ζ µ q+1 and (ξ, ξ ) µ q+1 µ q+1, then (ξ, ξ ) ζ = ξξ ζ (resp. (ξ, ξ ) ζ = ξ 1 ξ ζ). This defines two K[µ q+1 µ q+1 ]-modules. Lemma 4.11. With the notations as before. Then [ [ [H c (Z/G)] µq+1 µ q+1 K[µ (1) q+1 ]µ ] + K[µ (2) q+1 ]. q+1 µ q+1 ]µ q+1 µ q+1 Proof. Define and Z 0 = {(x, y, z, t) Z xt yz = 0} Z 0 = {(x, y, z, t) Z xt yz 0} Then Z 0 and Z 0 are (G µ q+1 µ q+1 )-stable subvarieties of Z. As a result, [H c(z/g)] µq+1 µ q+1 = [H c(z 0 /G)] µq+1 µ q+1 + [H c(z 0 /G)] µq+1 µ q+1 To compute the first term, consider the following isomorphism of F q -varieties µ q+1 Y Z 0, (ξ, x, y) (x, y, ξx, ξy) which induces then an action of G µ q+1 µ q+1 on µ q+1 Y given by Hence Z 0 /G µ q+1 A 1, and (g, ξ, ξ ) (ζ, x, y) = (ξ 1 ξ ζ, g (x, y)) [H c(z 0 /G)] µq+1 µ q+1 = [ K[µ (2) q+1 ] ] µ q+1 µ q+1 For the second term, consider the variety V = { (u, a, b) (A {0}) A 2 u q+1 ab = 1 } and the morphism ν : Z 0 V, (x, y, z, t) (xt yz, xt q yz q, x q t y q z) One can prove that ν induces an isomorphism of F q -varieties (see [2] 4.2): ν : Z 0 /G V. Under this isomorphism, the action of µ q+1 µ q+1 on V is given by (ξ, ξ ) (u, a, b) = (ξξ u, ξξ 1 a, ξ 1 ξ b).

CHARACTERS OF SL 2 (F q) 15 On the other hand, we can also define an action of G m,f on V as follows: λ (u, a, b) = (u, λa, λ 1 b) and this action commutes with that of µ q+1 µ q+1. As a consequence, we get H c(z 0 /G) µq+1 µ q+1 H c(v ) µq+1 µ q+1 H c(v G m,f ) µq+1 µ q+1. Now V G m,f = µ q+1 {0} {0}. Therefore [H c(z 0 /G)] µq+1 µ q+1 = This gives the formula of the Lemma. [ K [ ]] µ (1) q+1. µ q+1 µ q+1 Proof of Proposition 4.10. By Proposition 4.7, we have < R (θ), R (η) > G =< 1 G, R (θ) R (η) > G =< 1 G, R (θ ) R (η) > G On the other hand, we have ( R (θ ) R (η) ) G = (( H c (Y ) Kµq+1 K θ ) K ( H c (Y ) Kµq+1 K η )) G where = H c ((Y Y )/G) K[µq+1 µ q+1 ] K θ η θ η : µ q+1 µ q+1 K, (ξ, ξ ) θ (ξ)η(ξ ) According to the key formula (Lemma 4.11), we have [ [ [H c ((Y Y )/G)] µq+1 µ q+1 K[µ (1) q+1 ]µ ] + K[µ (2) q+1 ]. q+1 µ q+1 ]µ q+1 µ q+1 Let µ (1) = {(ξ, ξ 1 ) ξ µ q+1 } and µ (2) = {(ξ, ξ) ξ µ q+1 }. Then [ ] K µ (i) q+1 = Ind µ q+1 µ q+1 1 µ (i) µ (i). Hence ( ) < R (θ), R (η) > G = dim K K[µ (1) q+1 ] K[µ q+1 µ q+1 ] K θ η ( ) +dim K K[µ (2) q+1 ] K[µ q+1 µ q+1 ] K θ η This finishes the proof. = < K θ η, Ind µ q+1 µ q+1 µ (1) 1 µ (1) > µq+1 µ q+1 + < K θ η, Ind µ q+1 µ q+1 µ (2) 1 µ (1) > µq+1 µ q+1 = < θ η, 1 µq > µq+1 + < θ η, 1 µq > µq = < θ, η > µq+1 + < θ, η > µq+1.

16 CHARACTERS OF SL 2 (F q) 4.7. Parametrization of Irr(G). By Mackey formula (Proposition 4.10) and 4.3, we have 1. R (1) = 1 G + St G ; 2. R (θ) = R (θ 1 ) Irr(G) if θ µ q+1 such that θ2 1. 3. R (θ 0 ) = R +(θ 0 ) + R (θ 0 ), where R ±(θ 0 ) Irr(G) and R +(θ 0 ) R (θ 0 ). 4. If θ 2 1, η 2 1 and θ / {η, η 1 }, then R (θ) R (η). Remark 4.12. In this way, we obtain (q 1)/2 + 2 = (q + 3)/2 cuspidal irreducible characters. Indeed, by Proposition 4.6, for any linear character θ 1, R (θ) is a character. Moreover, by definition, for any linear character α of T µ q 1, R(α) is a character. As a result, by the orthogonality property (Proposition 4.9), we have < R(α), R (θ) > G = 0. Hence, each irreducible component of R (θ) is different from those obtained in Remark 2.7. Hence by combining with Remark 2.7, we obtain q + 5 2 + q + 3 2 = q + 4. On the other hand, as Irr(G) = q + 4, we conclude that these give all the characters of G = SL 2 (F q ). Proposition 4.13. We have deg(r ±(θ 0 )) = q 1 2. Proof. Let d + = deg(r +(θ 0 )), and d = deg(r (θ 0 )). Then G = 1 2 + q 2 + q 3 ( ) q + 1 2 2 (q + 1)2 + 2 + q 1 2 2 (q 1)2 + d 2 + + d 2. Moreover, d + + d = deg(r (θ 0 )) = q 1. In this way, we get d + = d = q 1 2. 4.8. Action of the Frobenius Endomorphism. Recall that the Frobenius F acts by automorphism on H i c(y ), and is given by multiplication by q when i = 2. Let now θ be a linear character of µ q+1, and e θ = 1 θ(ξ 1 )ξ Kµ q+1. q + 1 ξ µ q+1

CHARACTERS OF SL 2 (F q) 17 Then for the Kµ q+1 -module H i c(y ), we have the following isomorphism H i c(y ) Kµq+1 K θ H i c(y )e θ H i c(y ) Since F ξ = ξ 1 F, and F is an automorphism of H i, as a result, we find F (H i c(y )e θ ) = H i c(y ) eθ 1. In particular, F stabilizes H 1 c(y )e 1 and H 1 c(y )e θ0 4.9. Action of F on H 1 c(y )e 1. The KG-module H 1 c(y )e 1 = H 1 c(y ) µ q+1 = H 1 c(y/µ q+1 ) St G is irreducible, and since F commutes with the action of G, F acts H 1 c(y )e 1 as multiplication by a scalar ρ 1. By the Lefschetez fixed point theorem (ref...), we have 0 = (P 1 (F) P 1 (F q )) F = q ρ1 dim K (H 1 c(y/µ q+1 )) = q(1 ρ 1 ) As a result, ρ 1 = 1. 4.10. Action of F on H 1 c(y )e θ0. Set H 1 c(y )e θ0 = V + θ 0 V θ 0 with V ± θ 0 the irreducible factors of H 1 c(y )e θ0 with characters R (θ 0 ) ±. As a result, F acts on V ± θ 0 by multiplication of scalars ρ ±. Moreover, since V ± θ 0 are two non isomorphic representations, we find in this way F = ρ + ρ. By Lefschetz fixed-point theorem, we have 0 = Y (F) F = q qρ1 (q 1)(ρ + + ρ 1 ) Tr ( F, 2 θ:θ2 1H 1 ) c(y )e θ As a result, we find ρ + = ρ. To explicitly calculate ρ ±, we will study the action of F 2. As F 2 stabilizes H 1 c(y )e θ, and it acts by multiplication by a scalar λ θ. Hence λ θ0 = ρ 2 ±. Theorem 4.14. Let θ be a linear character of µ q+1. then { 1 if θ = 1, λ θ = θ( 1)q if θ 1. Proof. The equality λ 1 = 1 has been verified. By the Lefechetz fixed-point theorem, we have Y ξf 2 = q 2 qλ 1 1)θ(ξ)λ θ θ 1(q for all ξ µ q+1. Hence Y ξf 2 = (q 2 1) (q 1) (q 1)θ(ξ)λ θ θ µ q+1

18 CHARACTERS OF SL 2 (F q) As a result, θ µ q+1 θ(ξ)λ θ = (q + 1) Y ξf { 2 1 q q 1 = 2 if ξ = 1, q + 1 if ξ 1. These give then a system equations, which admits a unique solution. One can verify that the numbers λ θ give the unique solution. This finishes the proof. Corollary 4.15. We have ρ ± = ± θ 0 ( 1)q 4.11. Action on H 1 c(y ) eθ H 1 c(y ). eθ 1 Suppose θ µ q+1 θ 2 1. Since F stabilizes such that and sends H 1 c(y ) eθ each of which has multiplicity q 1. H 1 c(y ) eθ H 1 c(y ) eθ 1 onto H 1 c(y ) eθ 1, therefore F has two eigenvalues θ( 1)q, θ( 1)q References [1] M. Isaacs. Character theory of finite groups. Dover Publications, INC. New York. 1976. [2] C. Bonnafé. Reprentations of SL 2 (F q ). Algebra and Applications, 13. Springer.