Chapter 2
Representations constructed by
parabolic induction

2.1. Conjugacy classes in GL2(Fq)

Given a matrix (a b)
 c d in GL2(Fq) consider its characteristic polynomial

λ2 - (a + d)λ+ (ad- bc).

Exercise 2.1. If the roots (λ12) are distinct in Fq then the matrix is conjugate to (λ1 0 )
  0 λ2 and to (λ2 0 )
  0 λ1.

Exercise 2.2. If λ1 = λ2 then, either the matrix is (λ1 0)
 0 λ1 or it is conjugate to (     )
  λ1 1
  0 λ1 in GL2(Fq).

Exercise 2.3. If λ2 - (a + d)λ + (ad-bc) is irreducible in Fq[t], then the matrix is similar to (         )
 0 -(ad- bc)
 1   a+d.

To summarise, the conjugacy classes in GL2(Fq) are as follows:

  1. (q - 1) classes represented by (   )
  λ0 0 λ, with λ ∈ Fq*.
  2. (q - 1) classes represented by ( λ 1)
  0 λ, with λ ∈ Fq*.
  3. 1
2(q - 1)(q - 2) classes represented by ( λ1 0)
  0 λ2 with λ1⁄=λ2.
  4. 12(q2 - q) classes represented by (0 -a )
 1  a10, with λ2 - a1λ + a0 an irreducible polynomial in Fq[t].

In all, there are

                 q2 - q  (q- 1)(q- 2)
(q- 1)+ (q- 1)+  --2--+ ------2-----
(2.1)

conjugacy classes. Detailed information about the conjugacy classes is collected in Table 1.


Table 1: Conjugacy classes of GL2(Fq)





Name

element centraliser

no. of classes

size of class











Central

(a 0)
  0 a, a ∈ Fq*q(q - 1)2(q + 1) q - 1 1





Non-semisimple

(a 1)
  0 a, a ∈ Fq* q(q - 1) q - 1 (q - 1)(q + 1)





Split regular semisimple

(a 0)
  0 b
a⁄=b ∈ Fq*

(q - 1)2 (q--1)(2q--2) q(q + 1)





Anisotropic regular semisimple

Cp, p(t) ∈ Fq[t] quadratic, irreducible

(q - 1)(q + 1) q2-q
--2- q(q - 1)






2.2. Subgroup of upper-triangular matrices

Let B be the subgroup of GL2(Fq) consisting of invertible upper triangular matrices. Let N be the subgroup of upper triangular matrices with 1’s along the diagonal. Let T be the subgroup of invertible diagonal matrices.

Exercise 2.4. Show that

  1. Every element b ∈ B can be written in a unique way as b = tn, with t ∈ T and n ∈ N.
  2. N is a normal subgroup of B.
  3. B∕N~=T.

Let w = ( 0 1)
 -1 0.

Proposition 2.5 (Bruhat decomposition).

GL2(Fq) = B ∪ BwB,  a disjoint union.

Note that B is really a double coset B1B. So Proposition 2.5 really tells us that the double coset space B\GL2(Fq)∕B has two elements and that {1,w} is a complete set of representatives for these double cosets.

Proof. A matrix (  )
 ac b d lies in B if and only if c = 0. If c⁄=0, then

(   )  (     )  (         )
 ac bd =   1 a∕c w  -0cb--add∕c ∈ BwB.
         0 1

2.3. Parabolically induced representations for GL2(Fq)

Given characters χ1 and χ2 of Fq*, we get a character χ of T by

  (y  0)
χ  01y2  = χ1(y1)χ2(y2).
We extend χ to a character of B by letting N lie in the kernel. Thus
  (    )
χ  y01yx2  = χ1(y1)χ2(y2).
(2.2)

Let I(χ12) be the representation of GL2(Fq) induced from this character of B.

Proposition 2.6. Let χ1, χ2, μ1 and μ2 be characters of Fq*. Then

dimHomGL2(Fq)(I(χ1,χ2),I(μ1,μ2)) = e1 + ew,
where,
     ({
e  =  1    if χ1 = μ1 and χ2 = μ2,
 1   (0    otherwise,
and
     (
     { 1   if χ1 = μ2 and χ2 = μ1,
ew = ( 0   otherwise.

Proof. Let χ and μ be the characters of B obtained from the pairs χ1, χ2 and μ1, μ2 respectively as in (2.2). We regard χ and μ as one-dimensional representations of B acting on the space C. We may identify HomC(C,C) with C as well. Then, using Mackey’s description of intertwiners (Theorem 1.12), we see that we must compute the dimension of the space of functions Δ : GL2(Fq) C such that

Δ(b gb ) = μ(b )Δ(g)χ(b ), b ∈ B.
   2  1      2       1   i
(2.3)

It follows from the Bruhat decomposition that Δ is completely determined by its values at 1 and w.

Taking g = 1 in (2.3), we see that for any t ∈ T,

μ(t)Δ(1) = Δ(t) = Δ(1)χ(t)
Therefore, if μ⁄=χ then Δ(1) = 0. On the other hand, if μ = χ, let Δ1 be the function such that
Δ1(b) = χ(b) for all b ∈ B,
and whose restriction to BsB is zero. If e1 = 0, we take Δ1 0.

Taking g = w in (2.3), we see that for any t ∈ T,

μ(t)Δ(w) = Δ(tw) = Δ(w(w - 1tw)) = Δ(w)χ(w- 1tw).

Exercise 2.7. w-1(t1 0)
  0 t2w = (t2 0)
  0 t1.

Therefore, if μ1⁄=χ2 or μ2⁄=χ1 then Δ(w) = 0. On the other hand, if μ1 = χ2 and μ2 = χ1, let Δw be the function such that

Δw(b2wb1) = χ(b1)μ(b2) for all b1,b2 ∈ B,
and whose restriction to B is 0. If ew = 0, we take Δw 0.

An arbitrary functions satisfying (2.3) can be expressed as a linear combination of Δ1 and Δw, so we see that the dimension of the space of such functions must be e1 + ew.

Theorem 2.8. Let χ1, χ2, μ1 and μ2 be characters of Fq*. Then I(χ12) is an irreducible representation of degree q + 1 of GL2(Fq) unless χ1 = χ2, in which case it is a direct sum of two irreducible representations having degrees 1 and q. We have

I(χ1,χ2) ~= I(μ1,μ2)
if and only if either
χ1 = μ1 and χ2 = μ2
(2.4)

or else

χ1 = μ2 and χ2 = μ1.
(2.5)

Proof. Apply Proposition 2.6 with χ1 = μ1 and χ2 = μ2. We see that

                         (
                         {
dim EndGLn(Fq)(I(χ1,χ2)) =  1   if χ1 ⁄= χ2,
                         ( 2   if χ1 = χ2.
Recall that if (π,V ) is a representation of a finite group G and V is a direct sum of distinct irreducible representations π1,⋅⋅⋅h with multiplicities m1,⋅⋅⋅,mh and with degrees d1,⋅⋅⋅,dh respectively, then the dimension of EndG(V ) is midi2. Hence I(χ12) is irreducible if χ1⁄=χ2, otherwise it is a direct sum of two irreducible representations because 2 = 12 + 12 is the only way of writing 2 as a sum of non-zero multiples of more than one non-zero squares.

Because the index of B in GL2(Fq) is q + 1, the dimension of I(χ12) is always q + 1. If χ1 = χ2, the representation of GL2(Fq) generated by the function f(g) = χ1(det(g)) clearly satisfies f(bg) = χ(b)f(g) for all b ∈ B and g ∈ G. Therefore f ∈ I(χ12). Moreover, (g f)(x) = χ1(det(g))f. Therefore the one-dimensional subspace spanned by f is invariant under the action of G, hence forms a one-dimensional representation of G. The other component is therefore q-dimensional.

If χ1⁄=χ2 then, I(χ12) is irreducible. By Proposition 2.6 there exists a non-zero element in Hom(I(χ12),I(μ12)) if and only if χ1 = μ1 and χ2 = μ2 or χ1 = μ2 and χ2 = μ1. By irreducibility, these homomorphisms must be isomorphisms. This proves the second part of the theorem.

Exercise 2.9. Find the isomorphism I(χ12) I(χ21) explicitly, when χ1⁄=χ2.

To summarise, in this section, we have constructed irreducible representations of GL2(Fq) corresponding to characters χ = (χ12) of T:

  1. When χ1⁄=χ2, there is a unique irreducible representation of GL2(Fq) of degree q + 1 corresponding to χ; the irreducible representation corresponding to (χ12) is isomorphic to the one corresponding to (χ21). We have 12(q-1)(q-2) irreducible representations of degree q + 1.
  2. When χ1 = χ2, there are two irreducible representations of GL2(Fq) corresponding to χ, one of degree 1 and the other of degree q. All these representations are pairwise non-isomorphic. Therefore we have q - 1 representations of degree 1 and q - 1 representations of degree q.

Recall from Schur theory, that the number of irreducible representations is the same as the number of conjugacy classes in a group. We have constructed

(q- 1)+ (q- 1)+ (q-- 1)(q--2)
                      2
irreducible representations so far. Comparing with (2.1), we see that there remain 1
2(q2 - q) representations left to construct.

Recall that for a group of order n whose irreducible representations are π1,r of degrees d1,,dr respectively,

    2        2
n = d1 + ⋅⋅⋅+ dr.

Exercise 2.10. Show that the order of GL2(Fq) is (q2 - 1)(q2 - q).

The sum of squares of degrees of the representations that we have constructed so far is

1(q- 1)(q- 2)(q + 1)2 + (q - 1)(q2 + 1).
2
The difference between the above numbers is
1  2         2
2(q - q)(q - 1) .

We will see in Section 3.4 that there are 12(q2 - q) irreducible representations of degree q - 1 remaining. These will be constructed in Section 3.5.

2.4. Conjugacy classes in SL2(Fq)

Let Aut(SL2(Fq)) denote the group of all automorphisms of SL2(Fq). GL2(Fq) acts on SL2(Fq) by conjugation. This gives rise to a homomorphism GL2(Fq) Aut(SL2(Fq)). The kernel of this automorphism consists of scalar matrices in GL2(Fq), and is therefore isomorphic to Fq*. The image is therefore isomorphic to the group PGL2(Fq), which is the quotient of GL2(Fq) by the subgroup of invertible scalar matrices. The orbits of PGL2(Fq) on SL2(Fq) are precisely the conjugacy classes of GL2(Fq) which are contained in SL2(Fq) (note that SL2(Fq) is a union of conjugacy classes of GL2(Fq)).

On the other hand, the image of SL2(Fq) in Aut(SL2(Fq)) is PSL2(Fq), the quotient of SL2(Fq) by the subgroup {1}. The conjugacy classes of SL2(Fq) are precisely the PSL2(Fq) orbits.

Now, PSL2(Fq) is a subgroup of PGL2(Fq) (when both groups are viewed as subgroups of Aut(SL2(Fq))) of index two. Therefore, each conjugacy class of GL2(Fq) whose elements lie in SL2(Fq) is either a single conjugacy class in SL2(Fq) or a union of two conjugacy classes in SL2(Fq). If ε is an element of Fq* which is not a square (since q is assumed to be odd, there are q-1
 2 such elements), then the image of (ε 0)
 0 1 in Aut(SL2(Fq)) does not lie in the image of SL2(Fq).

Let σ ∈ SL2(Fq). Whether or not the conjugacy class of σ in GL2(Fq) splits or not can be determined by counting. The basic principle here is that the number of elements in an orbit for a group action is the index of the stabiliser of a point in the orbit.

With the above observations in mind, it is not difficult to prove that

Theorem 2.11. Let σ ∈ SL2(Fq). Let Z denote the centraliser of σ in GL2(Fq). Then [Z : Z SL2(Fq)] is either q - 1 or q-1
 2. In the former case, the conjugacy class of σ in GL2(Fq) is a single conjugacy class in SL2(Fq). In the latter case, the conjugacy class of σ in GL2(Fq) is a union of two conjugacy classes in SL2(Fq), represented by σ and ( ε 0)
 0 1σ(ε 0)
 0 1-1 respectively.

Exercise 2.12. Prove Theorem 2.11.

Exercise 2.13. Show that the conjugacy classes in SL2(Fq) are as follows:

  1. 2 central classes, represented by (     )
 1  0
  0 1.
  2. 4 non-semisimple classes, represented by (1 1 )
  0 1 and (     )
 1  ε
  0 1.
  3. 1
2(q-3) split regular semisimple classes, represented by (a  0 )
  0 a-1, a ∈ Fq*.
  4. 1
2(q-1)2 anisotropic semisimple classes, represented by (0 -1)
  1 a, where λ2 - + 1 is an irreducible polynomial in Fq[t].

In all, there are

       q--3-  q--1-
2+ 4 +  2   +  2
(2.6)

conjugacy classes.


Table 2: Conjugacy classes of SL2(Fq)





Name

representative centraliser

no. of classes

size of class










Central

( 1 0 )
  0 1 q(q - 1)(q + 1) 2 1





Non-

(      )
  0111 2q 2 (q-1)(2q+1)




semisimple

( 1 ε )
  0 1 2q 2 (q-1)(q+1)
    2





Split regular semisimple

(a  0 )
 0 a-1
a ∈ Fq*\{1}

q - 1 q-3
 2 q(q + 1)





Anisotropic regular semisimple

Cp, p[t] ∈ Fq[t] irreducible, p(0) = 1

q + 1 q-1
 2 q(q - 1)






2.5. Parabolically induced representations for SL2(Fq)

Let B now consist of the upper triangular matrices in SL2(Fq), N the upper triangular matrices with 1’s along the diagonal, and T the matrices in SL2(Fq) which are diagonal. Note that the results of Exercise 2.4 are still valid, as is the Bruhat decomposition:

SL2(Fq) = B ∪ BwB,  a disjoint union.
Given a character χ of Fq*, we may think of it as a character of T by
  (y  0 )
χ  0 y-1  = χ(y).
It can be extended to a character of B which is trivial on N by setting
  (y  x )
χ  0 y-1  = χ(y).
Let I(χ) be the representation of SL2(Fq) induced from this character of B. There is an analogue of Proposition 2.6 for SL2(Fq).

Proposition 2.14. Let χ and μ be characters of Fq*. Then,

dim HomSL  (F )(I(χ),I(μ)) = e1 + ew,
         2  q
where,
    (                            (
    { 1   if μ = χ,               { 1   if χ = μ-1,
e1 = (                and   ew = (
      0   otherwise,               0   otherwise.

Taking μ = χ in Proposition 2.14 gives that

                     ({           - 1
dim EndSL  (F )(I(χ)) =  2   if χ = χ
         2  q        (1   otherwise.
Take ε ∈ Fq* to be a generator (this is cyclic of even order q - 1 by Theorem B.5). Note that χ is completely determined by χ(ε), which can be any (q - 1)st root of unity in C*. Furthermore, χ = χ-1 if and only if χ(ε) = χ(ε)-1, i.e., if and only if χ(ε) = 1. Therefore, there are q - 3 characters χ for which I(χ) is irreducible. For each of these, I(χ)~=I(χ-1), and there are no other isomorphic pairs. We get q--3
 2 such irreducible representations, each of degree q + 1. There remain the characters χ for which χ(ε) = 1. Each of these give rise to two irreducible non-isomorphic representations. We consider the two cases separately:

Case χ(ε) = 1

In this case, I(χ) contains the invariant one dimensional subspace of constant functions on G. Therefore I(χ) splits into a direct sum of two irreducible representations, the trivial representation and a representation of dimension q, which is called the Steinberg representation.

Case χ(ε) = -1

In this case it is necessary to make a closer analysis of EndSL2(Fq)(I(χ)). Let Δ1 denote the unique function SL2(Fq) C for which Δ1(1) = 1, Δ1(w) = 0, and Δ(b1gb2) = χ(b1)Δ(g)χ(b2). Also let Δw denote the unique function SL2(Fq) C for which Δw(1) = 0, Δw(w) = 1, and Δ(b1gb2) = χ(b1)Δ(g)χ(b2). These two functions form a basis of EndSL2(Fq)(I(χ)). Write I(χ) = ρ+ ρ-, where ρ+ and ρ- are the two irreducible summands of of I(χ). The identity endomorphism in I(χ) can be written as a sum of two idempotents, coming from the identity endomorphisms of ρ+ and ρ-.

Exercise 2.15. Show that the identity endomorphism of I(χ) is given by f↦→q-1(q - 1)-1Δ1 * f.

Exercise 2.16. Show that

Δ1 * Δ1 = q(q - 1)Δ1, Δ1 * Δw = q(q - 1)Δw,
Δw * Δ1 = q(q - 1)Δw, Δw * Δw = q2(q - 1)χ(-1)Δ1.

Exercise 2.17. Besides q-1(q - 1)-1Δ1 and 0, show that the only idempotents in EndSL2(Fq)(I(χ)) are

                  √---
12q-1(q - 1)- 1(Δ1  ( - 1)κq-1Δw).
Here κ = 0 if χ(-1) = 1 and κ = 1 if χ(-1) = -1.

Let εI(χ) be the representation of SL2(Fq) on the representation space of I(χ), but where the action of SL2(Fq) is given by

(a b)        ( ( ε 0)-1 (a b)( ε 0))
 c d f(x) = f x  0 1    c d  0 1  .

Exercise 2.18. Show that f↦→~f , where ~f (x) = f(x(ε0 01)) is an isomorphism I(χ) εI(χ).

Exercise 2.19. Show that ~Δ1 = Δ1 and ~Δw = -Δw. Conclude that ερ+ = ρ- and ερ- = ρ+.

Therefore, the two representations ρ+ and ρ- must have equal degrees. It follows that I(χ) is a sum of two irreducible representations, each of degree q+1
 2.

In this section, we have constructed

2+ 2+ q---3
        2
irreducible representations of SL2(Fq). Comparing with (2.6) we see that there remain 2 + q-1
 2 irreducible representations to construct. The sums of squares of the degrees of the representations that we have constructed so far is:
              (     )
q--3-     2     q+-2- 2      2
 2  (q+ 1) + 2    2    + 1+ q .
The order of SL2(Fq) is q3 - q. The sums of the degrees of the irreducible representations that remain is therefore,
  (     )2
2  q-- 1  + (q- 1)2q--1.
     2               2
We will see in Section 3.6 that, among the representations that remain to be constructed, there are two of degree q--1
 2, and q--1
 2 of degree q - 1.