3 Induced representations

3.6 Example: GL2(Fp) — nonexaminable

Let 𝔽p be the field of integers modulo p, with p a prime. We will construct some of the irreducible representation of G=GL2(𝔽p) by induction from the subgroup of upper triangular matrices. We will use Mackey theory to show that these representations are irreducible.

3.6.1 The group GL2(Fp)

Proposition 3.26.

The order of the group GL2(𝔽p) is

p(p+1)(p-1)2.

We let BGL2(𝔽p) be the subgroup of upper triangular invertible matrices:

B={(**0*)GL2(𝔽p)}.

The group B is sometimes called a Borel subgroup.

Proposition 3.27.

The index of B in G is p+1. A set of left coset representatives for G/B is

{(0110)}{(10i1):0ip-1}.

We write w=(0110). We will want to know the double cosets of B in G, and the result is

Proposition 3.28.

We have

B\G/B=BBwB.

For later use, we compute

w(xz0y)w-1=(y0zx)

so that

Bw=BBw={(x00y):x,y𝔽p×}.

3.6.2 Characters of B and their inductions.

Let χ1 and χ2 be two characters of 𝔽p×. Then there is a character

χ1χ2:B×

defined by

(χ1χ2)((xz0y))=χ1(x)χ2(y).

Define

π(χ1,χ2)=IndBG(χ1χ2).

If χ:𝔽p×× is a character, then we get a character

χdet:G×

taking g to χ(det(g)).

Lemma 3.29.

If χ1,χ2,χ:𝔽p×× are characters, then

π(χ1χ,χ2χ)(χdet)π(χ1,χ2).
Lemma 3.30.

On Bw, we have

(χ1χ2)w=χ2χ1.
Theorem 3.31.

If χ1,χ2:𝔽p×× are characters, then:

  1. 1.

    If χ1χ2, then

    π(χ1,χ2)

    is an irreducible representation of dimension p+1.

  2. 2.

    There is an irreducible representation σ of dimension p such that, for any χ:𝔽p××,

    π(χ,χ)(χdet)(𝟙σ).
  3. 3.

    There is an isomorphism

    π(χ1,χ2)π(χ1,χ2)

    if and only if (χ1,χ2)=(χ1,χ2) or (χ2,χ1).

  4. 4.

    For χχ, the representations (χdet)σ and (χdet)σ are not isomorphic.

One can prove the theorem using Mackey theory, but it is also possible to avoid that and directly compute the inner products of the induced characters.

The representations π(χ1,χ2), χdet, and χdetσ are not all of the irreducible representations of GL2(𝔽p). There is another family of p-1-dimensional representations called the cuspidal representations, which are more difficult to construct: see [1] Lecture 5, or [3] Chapter 28.

3.6.3 Computing the induced character

Finally, we can compute the character of π(χ1,χ2).

We will need to know the conjugacy classes of G and of B. The characteristic polynomial almost determines the conjugacy classes; the result is as follows. Recall that the size of the conjugacy class of gG is |G||CG(g)| where CG(g) is the centralizer of g in G.

Proposition 3.32.

Every element of G is conjugate to exactly one of the following elements:

  1. 1.

    (x00x),x𝔽p×

  2. 2.

    (x10x),x𝔽p×

  3. 3.

    (x00y),x,y𝔽p×,xy

  4. 4.

    (0-d1t),t𝔽p,d𝔽p× such that X2-tX+d=0 has no solutions in 𝔽p.

For each type of element G, the order of the centralizer subgroup is:

  1. 1.

    |G|=p(p+1)(p-1)2

  2. 2.

    p(p-1)

  3. 3.

    (p-1)2

  4. 4.

    (p-1)(p+1).

Proof.

We sketch the proof. Let X2-tX+d be the characteristic polynomial of an element gG. If it has distinct roots in 𝔽p, then g is diagonalizable and we are in case 3. If it has no roots in 𝔽p, then choosing a basis v,gv (for some V𝔽p2) puts us in case 4. If the roots are identical, both equal to x, there are two possibilities: either the matrix is a scalar matrix, giving case 1, or v,(g-xI)v is a basis for some v𝔽p2 and then we are in case 2.

In each case, the centralizer can be computed explicitly and checked to have the order claimed (the most difficult case is the fourth). ∎

Now let χ:𝔽p×× be a character. We compute the character of π(χ,𝟙); after twisting by a character of the form χdet this gives the general case. In the case χ=𝟙, we also record the character of σ.

Theorem 3.33.

The characters are as follows (where xy, and X2-tX+d has no solutions in 𝔽p):

(x00x)(x10x)(x00y)(0-d1t)1(p-1)(p+1)p(p+1)p(p-1)π(χ,𝟙)(p+1)χ(x)χ(x)χ(x)+χ(y)0σp01-1