Chapter 3
Construction of the cuspidal representations

3.1. Projective Representations and Central Extensions

Let G be a finite group and let H be a Hilbert space. Denote by U(H) the group of unitary automorphisms of H. Let U(1) denote the group {z ∈ C : z= 1} under multiplication.

Definition 3.1 (Projective representation). A projective representation of G on H is a function η : G U(H) such that for every g,h ∈ G, there exists a constant c(g,h) ∈ U(1) such that

η(gh) = c(g,h)η(g)η(h).

A projective representation where c(g,h) = 1 is a representation in the sense of Section 1.1 and, for emphasis, will be called an “ordinary representation”.

Exercise 3.2. Use the associative law on G to show that the function c : G × G U(1) defined above satisfies the cocycle condition:

c(g,h)c(gh,k) = c(g,hk)c(h,k).

It is natural to ask whether, given a projective representation η, is it possible to find suitable scalars s(g) ∈ U(1) for each g ∈ G such that η(g)s(g) is an ordinary representation. If such a set of scalars did exist, it would mean that

η(g)s(g)η(h)s(h) = η(gh)s(gh)
for all g,h ∈ G. Applying (3.1) gives the coboundary condition:
s(g)s(h) = c(g,h)s(gh).
This motivates the following definitions:

Definition 3.3.

  1. The abelian group of 2-cocycles of G in U(1) consists of functions c : G × G U(1) which satisfy (3.2). This group is denoted Z2(G,U(1)).
  2. Given a function s : G U(1), its coboundary is defined as the cocycle c(g,h) = s(g)-1s(h)-1s(gh). The subgroup of Z2(G,U(1)) consisting of all coboundaries is denoted B2(G,U(1)).
  3. The second cohomology group of G with coefficients in U(1) is the quotient H2(G,U(1)) = Z2(G,U(1))∕B2(G,U(1)).

Observe that

Proposition 3.4. For any projective representation η of G, there exists a function s : G U(1) such that η(g)s(g) is an ordinary representation if and only if the the cocycle defined by (3.1) is a coboundary.

Definition 3.5 (Central Extension). A central extension of G by U(1) is a group ~G, together with a short exact sequence

1 → U(1) → G → G →  1
such that U(1) is contained in the centre of ~G.

Given a central extension G~ of G by U(1), pick any function s : G G~ (which may not be a homomorphism) such that the image of s(g) in G is again g. Such a function is called a section. The failure of s to be a homomorphism is measured by

c(g,h) = s(gh)s(h)-1s(g)- 1 ∈ U(1).

Exercise 3.6. Show that c(g,h) defined in (3.3) satisfies the cocycle condition (3.2). Moreover, if s is replaced by another section s, and c is the resulting cocycle, then cc-1 is a coboundary.

Thus a central extension of G by U(1) determines a well-defined element of H2(G,U(1)).

Exercise 3.7. Given a cocycle c : G × G U(1) satisfying (3.2), show that G(c) = G × U(1) with multiplication defined by

(g,z)(g′,z′) = (gg′,zz′c(g,g′)-1),
is a central extension of G by U(1). Moreover if c: G × G U(1) is another cocycle, there is an isomorphism α : G(c) G(c) such that the diagram
 -----     -----     -----   -----
1     U (1)     G(c)      G|      1
        |        α|        |
        |                  |
1-----U (1)-----G(c′)-----G  -----1
commutes if and only if cc-1 is a coboundary.

In this way, H2(G,U(1)) classifies the central extensions of G by U(1). Thus, H2(G,U(1)) arises in two different contexts:

  1. It measures the obstruction to modifying a projective representation to an ordinary representation.
  2. It classifies the central extensions of G by U(1).

The two are related in the following way:

Exercise 3.8. If η is a projective representation and c is the cocycle associated to it by (3.1), then ~η : G(c) U(H) defined by ~η(g,z) = (g) defines an ordinary representation of G(c).

In other words, every projective representation can be resolved into an ordinary representation of the central extension corresponding to its cocycle.

3.2. The Heisenberg group

Assume that the finite group G is abelian. Let L2(G) denote the Hilbert space obtained when the space of complex valued functions on G is endowed with the Hermitian inner product xf(x)g(x). On L2(G), there are two natural families of unitary operators:

Translation operators:    (Txf)(y) = f(y - x),   x ∈ G,

Modulation operators: (Mχf)(y) = χ(y)f(y), χ ∈^G.

The translation operators give a unitary representation of G on the Hilbert space L2(G). The modulation operators give a unitary representation of G^ on the same space. However, these operators do not commute:

Exercise 3.9. Show that

[Tx,M χ]f = χ(- x)f for each f ∈ L2(G).

The commutator is a scalar. Thus the map η : G×^G U(L2(G)) defined by

η(x,χ) = TxM χ
defines a projective representation of G ×^
G on L2(G).

Exercise 3.10. Show that the cocycle of G×^G with coefficients in U(1) associated to η in (3.1) is given by

c((x,χ),(x ′,χ′)) = χ(x′)-1.

Definition 3.11 (Heisenberg group). The Heisenberg group H(G) of G is the central extension of G × ^
G by U(1) corresponding to the cocycle (3.4) (see Exercise 3.7).

Explicitly, H(G) is the group whose underlying set of points is G ×^G × U(1) with multiplication given by

        ′ ′ ′        ′     ′   ′  ′
(x,χ,z)(xχ ,z ) = (x+ x ,χ + χ ,zzχ(x )).

The projective representation η of G × ^
G gives rise to an ordinary representation ~η of H(G) on L2(G), known as the Heisenberg representation (see Exercise 3.8). Explicitly, the Heisenberg representation is realized as

  ′  ′ ′        ′ ′    ′       ′
~η(x ,χ,z )f(x) = zχ (x- x )f(x - x ).

Remark 3.12. In the construction, and in all arguments relating to the Heisenberg group H(G), where G is a finite abelian group, U(1) can be replaced by an appropriate finite subgroup. Therefore, we may pretend that H(G) is a finite group.

Exercise 3.13. Verify that N := {0^
G×U(1) and ^
N := G×{0U(1) are normal subgroups of H(G). Z := {0}×{0U(1) is the centre of H(G). Here 0 denotes the identity element of either G or ^G.

Let θ : N C* be the character given by θ(0,χ,z) = z. Then the induced representation θH(G) is a representation of H(G) on the space

I := {f : H(G) → C ∣f(ng) = θ(n)f(g) for all n ∈ N,g ∈ H(G)}.

The action of H(G) on I is given by gf(g) = g(gg). For each f ∈ I, define ~f (x) = f(-x,0,1). Since the elements (-x,0,1), with x ∈ G form a complete set of representatives of the cosets in N\H(G), f↦→ ~
f is an isomorphism of I onto L2(G). Let g= (x,z) be an element of H(G)

^′          ′
g f(x)  =   gf(- x,0,1)
       =   f(x′ - x,χ′,z′)
       =   f((0,χ′,z′χ′(x ′ - x)- 1)(x′ - x,0,1)

       =   z′χ′(x′ - x)- 1f (x′ - x,0,1)
       =   z′χ′(x- x′)f~(x - x′).
Comparing with (3.6) shows that θH(G) is isomorphic the Heisenberg representation ~η .

Let ^
θ:  ^
N C* be the character given by ^
θ (x,0,z) = z. Then ^
θ H(G) is a representation of H(G) on the space

^                         ^
I := {f : H(G) → C ∣f (^ng) = θ(^n)f(g)}.
For each f ∈^I , define ~f (χ) = f(0,-χ,1). Since the elements (0,-χ,1), with χ ∈^G form a complete set of representatives of the cosets in N^\H(G), f↦→f~ defines an isomorphism of  ^
I onto L2(^

Exercise 3.14. Show that in this realization of θ^ H(G) on L2(G^), the action of H(G) is given by

((x′,χ′,z′)f)(χ) = z′χ(x′)-1f(χ - χ′)

Exercise 3.15. Show that the Fourier transform FT : L2(G) L2(^G) defined by

         ∑      ----         ^
FTf (χ) =   f (x)χ(x), for χ ∈G
is an isomorphism of H(G)-representations.

Theorem 3.16. The representation ~η is irreducible. Every irreducible representation of H(G) on which Z acts by the identity character of U(1) is isomorphic to ~η .

Proof. The irreducibility of ~η follows from the following exercise:

Exercise 3.17. Use Corollary 1.18 to show that θH(G) is irreducible.

Suppose that ρ is an irreducible representation of H(G) on which Z acts by the identity character of U(1). By Proposition 1.19,

Vρ =      Vχ,
where ^
N(ρ) consists of a single H(G)-orbit of characters of N. By hypothesis, the restriction of all these characters to Z is the identity character of U(1).

Exercise 3.18. Show that H(G) acts transitively on the set of characters of N1 whose restriction to Z is the identity character of U(1).

Exercise 3.19. Show that H(G)θ = N.

Therefore, θ ∈^N(ρ), and by Proposition 1.20, ρ~=θH(G).

Given an automorphism σ of H(G), let σ~η denote the representation of H(G) on the representation space V η of η given by σ~η (g) = ~η (σ-1 g). If σ fixes every element of Z, then σ~η is also an irreducible representation of H(G) on which Z acts by the identity character of U(1). By Theorem 3.16, ~η and σ~η are equivalent. Therefore, there exists ν(σ) : V η V η such that

ν(σ)∘ ~η(g) = σ~η(g) ∘ν(σ) for every g ∈ H(G).

Moreover, by Schur’s lemma, ν(σ) is uniquely determined modulo a scalar. Let B0(G) denote the group of all automorphisms of H(G) which fix the elements of Z.

Exercise 3.20. Show that

         ′        σ′σ              ′
ν(σ)∘ ν(σ)∘ ~η(g) =  η~(g)∘ ν(σ)∘ν(σ ).
Conclude that ν(σσ) and ν(σ) ν(σ) agree up to multiplication by a scalar.

It follows that the map σ↦→ρ(σ) = ν(σ-1) is a projective representation of B0(G) on L2(G). Projective representations of subgroups of B0(G) constructed in this way are known as Weil representations. In order to construct ν(σ) it is helpful to think of the realization of ~η as θH(G). The underlying vector space is the subspace I (see (3.7)) of C[H(G)]. Let r denote the representation of H(G) on C[H(g)], where H(G) acts by

r(g′)f(g) = f(gg′).
It is easy to come up with an isomorphism between r and σr, namely (νr(σ)f)(g) = f(σg). Unfortunately, νr(σ)f may no longer lie in I. This is rectified by modifying νr(σ) by an averaging operation to get ν(σ), as is seen in the following exercise:

Exercise 3.21. If f ∈ I, show that the function ν(σ)f defined by

(ν(σ)f)(g) =    f (σ((0,χ,1)g))

is also in I. The solution will use the fact that σ fixes every element of Z. Show that ν(σ) defined above satisfies (3.8).

Exercise 3.22. Let Q : G ×^
G U(1) denote the map

Q((x,χ),(x′,χ ′)) = χ(x′).
Let σ be any automorphism of G ×^
G such that
            ′  ′             ′ ′
Q( σ(x,χ),σ(x ,χ)) = Q((x,χ),(x,χ )).
Then the function ~σ : H(G) H(G) defined by
~σ(x,χ,z) = (σ(x,χ),z)
is an automorphism of H(G).

Exercise 3.23 (Symplectic form of the Heisenberg group). Assume that x↦→2x is an automorphism of G. Consider the bijection φ : H(G) G ×G^ × U(1) given by

φ(x,χ,z) = (x,χ,zχ(- x2)).
The multiplication map m : H(G)2 H(G) gives rise to a new multiplication map m: (G×G^×U(1))2 G×G^×U(1) determined by the commutativity of the diagram
     H(G)2 -----m-------H(G)
       |                  |
    φ×φ|                  |φ
                m′        |
(G × ^G × U(1))2 -----G × G^× U(1)
Show that
  ′         ′ ′  ′         ′     ′  ′  x′ ′  x
m ((x,χ, z),(x,χ ,z)) = (x + x,χ + χ,zz χ(2 )χ (-2)).

3.3. A special Weil representation

In this section SL2(Fq) will be realized as a subgroup of B0(G) for G = Fq2. The resulting Weil representation will turn out to be an ordinary representation (Proposition 3.26). All the cuspidal representations of GL2(Fq) and SL2(Fq) will be found inside this representation in Sections 3.5 and 3.6 respectively. Let G be the additive group of Fq2. The map x↦→(y↦→ψ(tr(xy))) defines an isomorphism of Fq2 onto ^Fq2 by Proposition B.11. Using this identification, the Heisenberg group H(Fq2) can be realized as Fq2 × Fq2 × U(1), with multiplication

m((x,y,z),(x′,y′,z′)) = (x+ x′,y+ y′,zz′ψ(tr(yx′))).
In the symplectic form (see Exercise 3.23), multiplication is given by
m ′((x,y,z),(x′,y′,z′)) = (x + x′,y + y′,zz′ψ(tr(1(yx′ - y′x)))).
To go from the Heisenberg group to its symplectic form, the transformation is given by φ(x,y,z) = (x,y,zψ(tr(1
2yx))). Suppose σ = (a b )
 c d∈ SL2(Fq). Then if Q((x,y),(x,y)) = tr(1
                     ′   ′   ′    ′            ′  ′
Q((ax+ by,cx+ dy),(ax + by ,cx + dy)) = Q((x,y),(x ,y)).
It follows that
(x,y,z) ↦→ (ax + by,cx + dy,z)
defines an automorphism of the symplectic form of the Heisenberg group. Using φ, we may associate to σ the automorphism
(x,y,z) ↦→ (ax+ by,cx+ dy,zψ(1tr(- yx+ (cx+-dy)(ax+ by)))).

of the Heisenberg group H(G) in its usual coordinates.

Exercise 3.24. Show that in the action defined by (3.11), t(a) = (a  0 )
 0 a-1, when a ∈ Fq*, acts by

(x,y,z) ↦→ (ax,a y,z),
w = (0 1 )
 -1 0 acts by
(x,y,z) ↦→ (y,- x,zψ(1tr(- xy - yx))),
and u(c) = (   )
 1c 0 1, when c ∈ Fq, acts by
(x,y,z) ↦→ (x,cx+ y,zψ(12tr(cxx))).

In the present context, (3.9) gives

                 1- ∑     σ            --
(ν(σ)f )(- x,0,1) = q2    f( (- x,y,ψ(tr(- yx)))).
σ(- x,y,ψ(tr(- yx)))
   =  (- ax+ by,- cx + dy,ψ(tr(- yx + 12(yx+ (- cx +dy)(- ax + by)))))
                     1    --   --------
      = (0,- cx+ dy,ψ(2tr(-yx - (- cx+ dy)(- ax +by))))(- ax + by,0,1).
  σ            --        1   --    --------
f( (- x,y,ψ(tr(- yx)))) = ψ( 2tr(-yx - (- cx+ dy)(- ax + by)))f(- ax + by,0,1).
(ν(σ)f)(- x,0,1)
           -1  ∑     1    --   --------
         = q2      ψ(2tr(- yx- (- cx + dy)(- ax+ by)))f(- ax + by,0,1).
Therefore, in the realization of ~η as L2(Fq2),
               ∑               --------
(ν(σ)~f)(x) =-12      ψ(12tr(- yx- (- cx +dy)(- ax + by)))~f(ax- by)
           q  y∈Fq2
for each ~f∈ L2(G), and therefore,
            1  ∑          --   -------
(ρ(σ)f~)(x) = q2     ψ(12tr(-yx - (cx +ay)(- dx - by)))~f(dx+ by).

Exercise 3.25. Show that, for any f~∈ L2(G),

           ({          ~
(ρ(σ)f~)(x) =  ψ(dc∑N (x))f(dx)     -               if b = 0,
           ( q12   y∈F 2 ψ(dN(x)--tr(ybx)+aN(y))~f(y)   otherwise.

We have already seen that ρ : SL2(Fq) GL(L2(Fq2)) is a projective representation. Let ~ρ be the modification of ρ by scalars given by

          { ψ(dcN (x))~f(dx)                      if b = 0,
ρ~(σ)f~(x) = (  1∑         dN(x)-tr(yx)+aN(y) ~
            - q  y∈Fq2 ψ(       b       )f(y)   otherwise.

Proposition 3.26. The function ~ρ : SL2(Fq) GL(L2(Fq)) defined by (3.13) is an ordinary representation.

Proof. Suppose σ = (   )
 acb d, σ= (    )
 a′′ b′′
 c d, and σ′′ = (     )
 a′′′′ b′′′′
 c  d are elements of SL2(Fq) such that σ′′ = σσ. Let 10 ∈ L2(Fq2) denote the indicator function of {0}. In the case that b, band b′′ are all non-zero, we have

~ρ(σ ′′)10(0) = - q.
On the other hand,
(~ρ(σ′)1 )(x) = - 1ψ(d ′b′- 1N(x)).
      0       q
             ′            1  ∑         -1        1   ′′-1
      (~ρ(σ)~ρ(σ)10)(0)  =  - q      ψ(tr(ab  N (y)))(- qψ(d b  N (y)))
                         1- ∑         -1    ′′- 1
(3.14)                =   q2      ψ(tr(ab   + db   )N (y)))
                         1 (    ∑        ′′-1 ′-1     )
                     =   q2- 1+      ψ(tr(bb  b   N (y)) .
Now the norm map N : Fq2*Fq* is surjective, and takes each value q + 1 times (Exercise B.10). Therefore, as y ranges over Fq2*, b′′b-1b-1N(y) ranges over Fq* (q + 1) times. We get
 ∑                                  ∑
     ψ(tr(b′′b-1b′-1N (x)))  =  (q+ 1)    ψ(tr(u))
x∈F*2                              u∈F*q
   q                                ∑
                          =  (q+ 1)    ψ(tr(u)) - (q+ 1)
                          =  - (q +1).
        ′           1
(~ρ(σ)~ρ(σ )10)(0)  =  q2(1 - (q + 1))
                =  - -.
We already know that ~ρ (σ′′) and ~ρ (σ)~ρ (σ) differ by a scalar multiple. It follows from the above calculations that this scalar multiple is 1.

If b and bare non-zero, but b′′ = 0, then db-1 + ab-1 = 0, and the expression (3.14) equals 1, which is also the value of ~ρ (σ′′)10(0). Again, it follows that ρ~ (σ′′) = ~ρ (σ)~ρ (σ).

When exactly one of b and bis 0, then b′′⁄=0. In these cases, ~ρ(σ)~ρ (σ) = ~ρ (σσ) = -1

Exercise 3.27. For a ∈ Fq*, let t(a) = (     )
  a0 a0-1, let w = (    )
  0-11 0 and for c ∈ Fq, let u(c) = (1 0)
 c 1. Use (3.13) to show that for every ~f∈ L2(Fq2),

   (~ρ(t(a))f~)(x) = f~(a-1x),
   (ρ~(w)f~)(x) = q-FTf~(x),
(~ρ(u(c))~f)(x) = ψ(cN(x))~f(x).


Here, the Fourier transform of ~f∈ L2(Fq2) is once again thought of as a function of Fq2, since Fq2 is identified with its Pontryagin dual. Explicitly,
    ~     ∑    ~        --
FT f(x) =     f(y)ψ(tr(- yx)).

Exercise 3.28. Any element of SL2(Fq) can be written as a product of elements of the above types. Consider the matrix (   )
 ac bd ∈ SL2(Fq). If b = 0, then d = a-1 and (a 0)
 c d = t(a)u(ac). On the other hand, if b⁄=0, then (a b)
 c d = u(d∕b)wu(ab)t(b-1).

3.4. The degrees of cuspidal representations

In Chapter 2 we constructed all the representations (π,V ) of GL2(Fq) for which

HomGL2(Fq)( π,I(χ1,χ2)) ⁄= 0 for some characters χ1,χ2 ∈ ^F*q.
Thus for the representations that remain,
HomGL2(Fq)(π,I(χ1,χ2)) = 0 for all characters χ1,χ2 ∈ ^F*q.

Representations (π,V ) satisfying (3.18) are known as the cuspidal representations of GL2(Fq). By Frobenius reciprocity (Section 1.3), we have

HomB( πB,χ) = 0 for all characters χ : B → C such that χ∣N ≡ 1.

Given a representation (π,V ) of any group G, let V * be the dual space HomC(V,C) of V . Let π* be the representation of G on V * given by

(π*(g)ξ)(v) = ξ(π(g-1)v).
The representation (π*,V *) is called the contragredient of (π,V ).

Proposition 3.29. A representation (π,V ) of GL2(Fq) is cuspidal if and only if there exists no non-zero vector ξ ∈ V * such that

π*(n)ξ = ξ for all n ∈ N.

Proof. Suppose (π,V ) is not cuspidal. Then there exists a non-zero element ξ ∈ HomB(V,χ) for some χ : B C* such that χN 1. Such a ξ can be regarded as an element of V *. We have, for any n ∈ N and v ∈ V ,

  *                 -1
(π (n)ξ)(v)  =  ξ(π(n   )v)
            =  ξ(χ(n)v)

            =  ξ(v),
so that ξ satisfies (3.19).

Conversely, look at the space V *N of all vectors in V * satisfying (3.19). This space is preserved under the action of T (since tNt-1 = N for all t ∈ T). Therefore, one can write

V*N =     V*χN ,
where V χ*N is the space of vectors v ∈ V *N which transform under T by χ. If V *N⁄=0, then there exists χ such that V χ*N⁄=0. Therefore, HomB(V,χ)⁄=0, from which it follows that (π,V ) is not cuspidal.

Exercise 3.30. Show that (π,V ) is cuspidal if and only if (π*,V *) is cuspidal.

Corollary 3.31. The degree of every cuspidal representation of GL2(Fq) is always a multiple of (q - 1).

Proof. Suppose that (π,V ) is a cuspidal representation. For each a ∈ Fq, let V a* be the space of all ξ ∈ V * such that

  (   )
π* 10 x1 ξ = ψ(ax)ξ.
Then the map
     *(t 0)
ξ ↦→ π   0 1 ξ
is an isomorphism of V *(a) with V *(ta) for all t ∈ Fq*. Hence for a⁄=0, the q - 1 spaces V *(ta), with t ∈ Fq* have the same dimension. The space V *(0) is just V *(N), hence is trivial. Therefore the dimension of V *, hence the degree of V must be a multiple of q - 1.

From Corollary 3.31 and the discussion at the end of Section 2.3 it follows that besides the representations constructed in that section, there are exactly 12(q2 - q) irreducible cuspidal representations, each of degree q - 1. These representations are constructed in Section 3.5.

A cuspidal representation of SL2(Fq) can be defined in a similar manner. A representation (π,V ) of SL2(Fq) is said to be cuspidal if

HomSL2(Fq)(π,I(χ)) = 0 for all characters χ ∈ ∈ Fq.

Exercise 3.32. Verify that Proposition 3.29 continues to hold when GL2(Fq) is replaced by SL2(Fq).

However, Corollary 3.31 does not hold as stated

Exercise 3.33. Show that the degree of a cuspidal representation of SL2(Fq) is always a multiple of q--1

3.5. Construction of cuspidal representations of GL2(Fq)

Let ω be a character of Fq2* such that ω⁄=χ N for any character χ of Fq* (here N denotes the norm map Fq2 Fq). Such a character is called primitive.

Exercise 3.34. Show that there are q2 - q such characters.


  *           *
(Fq2)1 = {y ∈ F q2 ∣N (y) = 1}.

Exercise 3.35. Show that a character ω : Fq2* C* is primitive if and only if its restriction to (Fq2*)1 is non-trivial.


           2                 -1                 *
W ω = {~f ∈ L (Fq2) ∣ ~f(yx) = ω(y) f~(x) for all y ∈ (F q2)1}.

Exercise 3.36. Show that Wω is preserved by the action of ~ρ (σ) for every σ ∈ SL2(Fq). [Hint: note that if N(x) = 1, then x = x-1.]

Therefore, ρ~ gives a representation (πω,Wω) for each such ω. For any x ∈ Fq2, the set of elements xsuch that N(x) = N(x) coincides with the set of elements of the form x′′x, where x′′∈ (Fq2*)1. Hence, if f ∈ Wω, then the value of ~f at x determines the value of f~ at any element xwith N(x) = N(x). However, if x = 0, there is an additional constraint, namely that f~ (0) = ω(y)-1f~ (0) for every y ∈ (Fq2*)1. By Exercise 3.35, if ω is primitive, then it is forced that  ~
f (0) = 0. Since there are q - 1 non-zero values for the norm, we have

Lemma 3.37. When ω is primitive Wω has dimension q - 1. For every f~∈ Wω, f~ (0) = 0.

Each matrix σ in GL2(Fq) can be written in a unique way as a product of (1  0  )
 0 det(σ) and a matrix in SL2(Fq). Define

(        )
 ~ρ (1 0)~f (x) = ω(~a)~f(~ax),
    0 a

where ∈ Fq2* is chosen so that N() = a.

Exercise 3.38. Check that the right hand side of (3.20) does not depend on the choice of such that N() = a, and that it preserves Wω for each primitive ω.

Extend πω to GL2(Fq) by ~ρ ((1 0 )σ)
  0 a = ~ρ (1 0)
 0 a~ρ (σ). For this extended function to be a homomorphism of groups, it is necessary that, for all a,a∈ Fq* and all σ,σ∈ SL2(Fq),

  ((   )  (   )  )    ((  )  ) ((   )  )
~ρ   10 0 a σ 10 a0′ σ ′= ρ~ 10 0 a σ ~ρ  10 0a′ σ′ .


                       [(     )         ]
(1 0)σ (1 0′)σ′ = (1 0′)  1 ′0-1  σ(1 0′) σ′ ,
 0 a    0 a       0 aa   0 a      0 a
and (     )
 1  0′-1
 0 aσ(1 0′)
  0 aσ∈ SL2(Fq).

Exercise 3.39. Using this to expand both sides of (3.21) in terms of (3.20), show that it is sufficient to check that for each a ∈ Fq*, f ∈ L2(Fq2) and each element σ of SL2(Fq),

 (1 0)     (1 0)-1    ((1 0) (1 0)-1)
~ρ 0 a ~ρ(σ)~ρ 0 a    = ~ρ  0 a σ  0 a    .

Exercise 3.40. Verify (3.22) for σ of the form t(a), w and u(c) (see Exercise 3.28). Conclude that it holds for all σ ∈ SL2(Fq).

We will denote again by (πω,Wω) the restriction of ~ρ to the subspace Wω.

Proposition 3.41. For every primitive character ω, the representation (πω,Wω) is cuspidal.

Proof. We will show that Wω contains no non-zero vectors fixed by N, the subgroup consisting of matrices of the form (1 0)
  c 1, c ∈ Fq. This suffices, for ~f is fixed by N if and only if πω(w)~f is fixed by N. Suppose that f~0 is a vector fixed by N. By Lemma 3.37, f~ 0(0) = 0. On the other hand, if x ∈ Fq2*, then choose c ∈ Fq so that ψ(cN(x))⁄=1. Then, by (3.17)

       (         )
f~0(x) = ρ~(1 0)~f0 (x) = ψ(cN (x))~f0(x),
           c 1
we have ~f 0(x) = 0.

Clearly, any sub-representation of a cuspidal representation is also cuspidal. Therefore, by Corollary 3.31 (πω,Wω) is simple for each ω of the type considered above.

Lemma 3.42. Let ω and η be two characters of Fq2* as above. If the representations (πω,Wω) and (πη,Wη) are isomorphic, then either ω = η or ω = η F, where F is the Frobenius automorphism Fq2* Fq2* (see Section B.3).

Proof. For each u ∈ Fq*, fix an element ũ ∈ Fq2 such that N(ũ) = u. Let 1u ∈ Wω be the unique function such that 1u(ũ) = 2 and 1u(x) = 0 if N(x)⁄=u. The set {1u u ∈ Fq*} is a basis of Wω. Therefore, for any σ ∈ GL2(Fq), tr(πω(σ)) = u∈Fq*(πω(σ)1u)(ũ).

For any a ∈ Fq2*, (   )
 a1 0a = (     )
  a1 a0-1(    )
  10 a02. From (3.13) and (3.20), we have that

   (   )
(πω  a1 0 a 1u)(~u) = ω(a)ψ(a-1u)1u(~u).
tr(π  (a 0)) =   ∑   ω(a)ψ(a- 1u)
   ω  1 a      u∈F *
            =  ω(a)     ψ(u)
                   u∈F *q
            =  - ω(a).

Exercise 3.43. Show that if ω and η are two characters of Fq2*, then their restrictions to Fq* are equal if and only if either ω = η or ω = η F.

If (πω,Wω) and (πη,Wη) were isomorphic, then we would have

  (  (a 1))    (  (a 1))
tr πω  0 a  = tr πη 0 a  ,
which by Exercise 3.43 would mean that either ω = η or ω = η F.

3.6. The cuspidal representations of SL2(Fq)

Let ω be a non-trivial character if (Fq*)1, the subgroup of Fq* consisting of elements of norm one (there are exactly q such characters). As in section 3.5 define

W ω = {~f ∈ L2(Fq2) ∣f~(yx) = ω(y)-1f~(x) for all x ∈ Fq2}.
Each such character ω can be extended to a primitive character of Fq2*, and therefore, the Wω’s are the same as the spaces defined in Section 3.5, and are invariant under the representation ~ρ of SL2(Fq) on L2(Fq2). Each such representation is of dimension q - 1. Let πω denote the representation of SL2(Fq) on Wω. These are just the restrictions of the representations of GL2(Fq) constructed in Section 3.5 to SL2(Fq). It follows that they are cuspidal. However, it no longer follows that these representations are irreducible, as the degree of a cuspidal representation of SL2(Fq) is only known to be a multiple of q-1
 2 by Exercise 3.33.

We shall analyze the representations πω through their characters. We already know that tr(πω((1 0)
  c 1)) = -1 from the proof of Lemma 3.43.

Exercise 3.44. Show that tr(   (     ))
 πω  a0 a0-1 = 0 if a⁄= 1.

Lemma 3.45. For every character ω of (Fq2*)1 and d ∈ Fq such that λ2 - + 1 is irreducible with roots z and z-1 in Fq2,

     (    )
tr(πω( 01 -d1 )) = - ω(z)- ω(z- 1),

Proof. By (3.13), we have

  (    )          ∑
ρ~ 01 -d1 ~f(x) = - 1   ψ(tr(yx)- dN(x))~f(y).
Using the notation of Lemma 3.43, we have
~ρ(0 -1)1  (~u) = - 1 ∑   ψ(tr(y~u)- du)1 (y).
   1 d   u      q y∈F               u
Now, 1u(y) = 0 unless y = zũ for some z ∈ (Fq2*)1. We have
  (0 - 1)          1   ∑         -1--          -1
ρ~ 1 d  1u(~u) =   -q        ψ(tr(z  ~u~u) - du)ω(z)
                     z∈(F *q2)1
                   1   ∑
              =   -q        ψ(u(z + z-1)- du)ω(z)-1
                     z∈(F *q2)1
                   1   ∑
              =   --        ψ(u(z + z-1 - d))ω(z)-1.
                   q z∈(F *q2)1
    (0 -1)        1 ∑     ∑            -1         -1
tr(πω  1 d )  =  - q            ψ(u(z + z  - d))ω(z)
                    u∈F *qz∈(F *q2)1
                  1  ∑        - 1 ∑          -1
             =  - q        ω(z)      ψ(u(z + z  - d)).
                    z∈(F*q2)1       u∈F*q
If d⁄=z + z-1, then
 ∑                     ∑
     ψ(u(z + z-1 - d)) =   ψ(u) = - 1.
u∈F*q                  u∈F*q
On the other hand, if d = z + z-1, then
    ψ(u(z + z- 1 - d)) = q- 1.
     (0 -1)        1  ∑        -1        1   ∑        -1
tr(π ω 1 d  )  =  - q       ω(z)  (q+ 1)- q        ω(z)
                    z[+z-1=d                z+z-1⁄=]d
                   1  ∑       -1    ∑         -1
              =  - q    * ω(z)  +         qω(z)
                     z∈Fq2        z+z-1=d
              =  - ω(z) - ω(z)-1.

Exercise 3.46. Suppose that ω is the unique non-trivial character of (Fq2*)1 taking only the values 1. Show that σ∈SL2(Fq)tr(πω(σ)) = 2(q3-q). Conclude that πω(σ) is a sum of two non-isomorphic irreducible representations of SL2(Fq).

These representations must be irreducible of degree q-1
-2- by Exercise 3.33. Using the book-keeping at the end of Section 2.5, we see that there remain q-2-1 irreducible representations of SL2(Fq).

Exercise 3.47. Define an equivalence relation on the set of non-trivial characters of (Fq2*)2 by ω ~ ω, where ω= ωF. Here F is the Frobenius automorphism (Section B.3). Observe that tr(πω) = tr(πω). Show that the characters of the representations πω, where ω runs over the equivalence classes of non-trivial characters of (Fq2*)1 are pairwise orthogonal.

It follows that πω, ω non-trivial and different from the character considered in Exercise 3.46 give the remaining q-1
 2 irreducible representations of SL2(Fq).