IRREDUCIBLE REPRESENTATIONS OF GL(2,Fq )
NAVA CHITRIK
Referenced heavily from Daniel Bump (1991), Automorphic Representations, Section 4.1
In these notes I will give a complete description of the irreducible complex representations of the group GL(2,Fq ) where
Fq
is the eld with
q = pn
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 1.
(Mackey's Theorem)
Let H1 and H2 be subgroups of a group G.
Let (π1 , V1 ) and (π2 , V2 ) be complex representations of H1 and H2 respectively and
let V1G and V2G be the induced representations to G. Then
HomG (V1G , V2G ) ∼
= {∆ : G → HomC (V1 , V2 )|∆(h2 gh1 ) = π2 (h2 ) ◦ ∆(g) ◦ π1 (h1 )}
The specic intertwiner is given by ∆(f ) = ∆ ∗ f (x) =
1
|G|
X
∆(xg −1 )f (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
HomG (V1G , V2G )
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,
HomG (V G , W ) ∼
= HomH (V, W )
1
IRREDUCIBLE REPRESENTATIONS OF GL(2,Fq )
2
Let's talk about the maximal tori of GL(2,Fq ). That is, maximal abelian subgroups.
There are two:
(1)
Ts =
(2)
Ta =
a
×
b |a, b ∈ Fq
n
o
√
x y
x+y D
∼
√
(the isomorphism is only if we extend
Dy x =
x−y D
the eld a bit)
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,Fq ) has
double coset representative representatives ( 10 01 ) and
−1
1
.
That is,
GL(2, Fq ) = B\ ( 10 01 ) /B
Denition 4.
of
B
on
C
by
Let
χ1
and
χ2
be characters of
( a cb ) .z 7→ χ1 (a) · χ2 (c) · z
representation of
B
to
G
GL(2, Fq )
B\
F×
q .
−1
1
/B
Then there's a representation
. Now we can induce this one dimensional
by the usual process:
B(χ1 , χ2 ) = {f : GL(2, Fq ) → C|f (( a cb ) g) = χ1 (a) · χ2 (c) · f (g)}
and we have
GL(2, Fq )
act by right translation, i.e.
π(g).f (g 0 ) = f (g 0 g)
clearly preserves the space. We denote this induced representation
will be a large class of representations of
Proposition 5. Let
B(χ1 , χ2 )
and
GL(2, Fq )
B(µ1 , µ2 )
which
B(χ1 , χ2 ).
(about half ).
be the induced representations as
described above, then
dim(Hom(B(χ1 , χ2 ), B(µ1 , µ2 )) =
This



2






1
if χ1 = χ2 = µ1 = µ2








0
(in some order)
if χ1 6= χ2 but {χ1 , χ2 } = {µ1 , µ2 }
otherwise
IRREDUCIBLE REPRESENTATIONS OF GL(2,Fq )
In other words,
B(χ1 , χ2 )
B(χ1 , χ2 ) ∼
= B(µ1 , µ2 )
Proof.
is irreducible if and only if
if and only if, as sets
χ1 6= χ2
3
and
{χ1 , χ2 } = {µ, µ2 }.
We combine Mackey's theorem together with the double coset representa-
tives for
GL(2, Fq ).
By Mackey's theorem, we are looking for maps
C|∆(b1 gb2 ) = χ1 χ2 (b1 )∆(g)µ1 µ2 (b2 ).
at most two degrees of freedom for
sentatives
Consider
1=
(1
1 ) and
ω0
−1
1
By the double coset decomposition, there are
∆,
namely, the values it takes on at the repre-
.
−1 −1
∆(1) = ∆(b1 · 1 · b−1
1 ) = χ1 χ2 (b1 ) · ∆(1) · µ1 µ2 (b1 ).
∆(1) 6= 0 ⇐⇒ χ1 χ2 (b1 ) = µ1 µ2 (b1 ) ⇐⇒ χ1 = µ1
(you can see this by setting
∆ : GL(2, Fq ) →
b1 = ( b 1 )
to get the rst, or
This implies that
and
χ2 = µ2
b1 = ( 1 b )
to get the
second equality ).
Next, consider
∆ (ω0 ( a b )) = ∆ (( b a ) ω0 )
for an arbitrary
a, b ∈ Fq
which implies
that
∆(ω0 ) 6= 0 ⇐⇒ χ2 (a)χ1 (b) = µ1 (a)µ2 (b) ⇐⇒ χ1 = µ2
So if
χ1 = χ2 = µ1 = µ2
and
χ2 = µ1
then there are two degrees of freedom for
only one of the following is true :
(χ1 , χ2 ) = (µ1 , µ2 )
or
∆.
If instead
(χ1 , χ2 ) = (µ2 , µ1 )
there's one degree of freedom and if none of those are true then
∆(1) = ∆(ω0 ) = 0,
so there are no intertwiners
Remark 6.
then
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 + 1-dimensional
q + 1,
so
B(χ1 , χ2 )
is a
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))
IRREDUCIBLE REPRESENTATIONS OF GL(2,Fq )
amd what remains is a
χwith B(1, 1)
q -dimensional
4
subrepresentation which is the tensoring of
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 1-dimensional and a
(q − 1)
ducible representation..... and there are
For each unequal, unordered pair,
representation..... and there are
(χ1 , χ2 )
q -dimensional
irre-
of each
we get a
(q − 1)(q − 2)/2
q+1
-dimensional irreducible
of these
How many representations do we expect? Well, we use the fact that the regular
representation decomposes as
Greg =
M
VidimVi
irrep
Then, taking the dimensions of each side we get
|G| =
X
(dimV )2
irrep
And in our case,
of
F2q \{0}
|GL(2, Fq )| = (q 2 − 1)(q 2 − q) (because the rst row is any element
and the second row is anything in
F2q
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, Fq ).
Proposition 7. (Generators) SL(2, Fq ) has generators
y
y
−1
, ( 1 z1 ) and
r
varying over y ∈ F×
q and z ∈ Fq . Together with the matrices ( 1 )
generate
GL(2, Fq ).
Now, let
ψ
−1
E=
F
and let
χ
be a character of


F L F
or
√

F( D)
the unique quadratic extension
E
these also
There are also some nite number of generators...
be an additive character on
1
where
IRREDUCIBLE REPRESENTATIONS OF GL(2,Fq )
E×
We call the rst case, the split case.
T0
.
The second is called the
5
ismorphic to the isotropic maximal torus
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
and in the second it's the galois conjugate.
We require that
trace.
χ
This allows us to dene norm and
doesn't factor through the norm.
Otherwise the
representation we are about to construct will not be irreducible.
the case
E=F
Denition 8.
to
L
F
(a, b) = (b, a)
Notice, that in
this is the same as the requirement above that
We can dene a fourier transform on functions on
χ1 6= χ2 .
E
with respect
ψ
Φ̂(x) = ±q −1
X
Φ(y)ψ(tr(x̄y))
E
± depends on what case E
Where the
Denition 9.
is. (+ in the rst case, - in the nonsplit case)
We dene a representation on
tation, denoted
(W(χ), ω)
GL(2, Fq )
as follows
W(χ) = f : E → C|f (yx) = χ−1 (y)f (x)
• ω
y
y −1
• ω
−1
1
whenever
y
has norm
= ψ(N (x) · z)Φ(x)
.Φ(x) = Φ̂(x)
• ω (( a 1 )) .Φ(x) = χ(b)Φ(bx)
where
b
is such that
N (b) = a
Proposition 10. The Weil Representation is a group representation of
Proof.
1
.Φ(x) = Φ(yx)
ω (( 1 z1 )) .Φ(x)
•
called the Weil Represen-
You basically need to check the relations.
GL(2, Fq ).
Which I haven't even stated.
Most of this is easy.
Remark
.
11
By the way, in the last bullet point, why doesn't it matter which
choose? Well, if we chose a dierent one, it would be of the form
norm 1. Then, by the denition of
amazing fact:
W(χ),
ub
where
b
u
we
has
the choice of b is irrelevant. Now for the
IRREDUCIBLE REPRESENTATIONS OF GL(2,Fq )
Proposition 12. If
E = F
L
6
F then χ is of the form (χ1 , χ2 ). We have that
(W(χ), ω) ∼
= B(χ1 , χ2 )
Proof.
F
L
F
W(χ)
The dimension of
are of the form
does on
(a, 1)
−1
(a, a
a∈F
for
)
q+1
(1, 0).
B(χ1 , χ2 )
is also
W(χ)
q+1
So there's
q+1
(W(χ), ω) → B(χ1 , χ2 )
if we nd an intertwiner
since the norm one elements of
so that any function in
and on
now that the dimension of
is equal to
is specied by what it
degrees of freedom. Recall
and it is irreducible. Therefore,
then it must be an isomorphism.
Indeed there's an easy one. Let
τ : W(χ) → C
Then, dene
L : (W, ω) → (B, π)
by
τ (Φ) = Φ(1, 0)
by
LΦ(g) = τ (ω(g).Φ)
LΦ(g)
B(χ1 , χ2 ) since LΦ (( a cb ) g) = τ (ω (( a cb ) g) .Φ) =
1 bc−1 a−1 g .Φ = χ (a)χ (c)τ (Φ).
1
2
1
is indeed an element of
τ ω ( ac 1 )
c−1
c
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 × 6≡ 1
1
(χ doesn't factor through the norm), the Weil representation is irreducible. For this
we must introduce one more notion.
Denition 13.
functional
`∈V
A representation is called cuspidal if there
∗
such that` (π ( 1 x
1 ).v)
= `(v)
for all
x∈F
does not exist a linear
for all
Lemma 14. In the nite case, a representation is cuspidal
exist an element v in V such that
Proof.
π ( 1 x1 ).v
=v
for all
v 6= 0.
⇐⇒ there
does not
x ∈ F.
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
{( 1 x1 )}
and
π.
Lemma 15. A cuspidal representation has dimension m(q − 1)
IRREDUCIBLE REPRESENTATIONS OF GL(2,Fq )
Proof.
We decompose
M
V∗ =
7
V ∗ (a) where V ∗ (a) = {` ∈ V ∗ |π̂ ( 1 x1 ) .` = ψ(ax).`}
a∈Fq
ψ
and where
is some xed non-trivial character. Now the requirement that
V (0) = 0.
cuspidal is simply that
×
Furthermore, the action of Fq by
V ∗ must
mutes these spaces transitively. Therefore the dimension of
of
(t
V
is
1 ) per-
be a multiple
q − 1.
Proposition 16. In the nonsplit case
√
E = Fq ( D), W(χ) is cuspidal and irre-
ducible.
Proof.
x∈
Suppose that there existed a
Φ∈W
such that
F. We simply apply the denition, π (( 1 x1 )) Φ(s)
s
for any
we can certainly nd an
x
such that
π (( 1 x1 )) Φ(s) = Φ(s)
for all
= ψ(N (s)x)Φ(s)
ψ(N (s)x) 6= 1
. This proves that
W
is cuspidal.
Next, the dimension of
is
q−1
because if we look at functions on
2
degrees of freedom. On the other hand, there are
E × so
there are
q −1
1
W
of an function in
q−1
W
q+1
degrees of freedom for functions on
E×
there are
norm 1 elements in
E×.
Finally the value
at 0 must be 0. Together with 15 this implies that
W(χ)
irreducible.
Remark
17
.
is
None of the
W(χ)
are isomorphic to any other unless
¯,
χ1 (x) = χ2 (x)
i.e. they are Galois conjugates. You can see this by restriction to matrices
Now how many such equivalence classes of characters
looking at characters of
E
×
χ
( r 1 ).
are there? Again, we are
/E1× by the restriction that it can't factor through the
norm and only the galois pairs.
Theorem 18. The representations B(χ1 , χ2 ) with χ1 6= χ2 together with the Weil
representations W(χ) coming from the non-split torus comprise all the irreducible
representations of GL(2,Fq )
Proof.
1You
We can just count them. As I noted in 6 we know that
|E/ker(N )| = |Im(N )|. We consider
n
N : E × → F×
q . If the image isn't trivial (it isn't) then it has to be a factor of p − 1 we
n
n
n
n
n
have that (p − 1)(p + 1)/|kerN | divides p − 1. So |kerN | = m(p + 1) where m divides p -1.
×
2
On the other hand, all function on E have dimension q − 1....
can see this from the isomorphism theorems of groups
the norm
IRREDUCIBLE REPRESENTATIONS OF GL(2,Fq )
X
|GL(2, Fq )| = (q 2 − 1)(q 2 − q) =
8
dim(Vi )2
irrep
let's list o what we have
dimTotal
Type:Number of them
B(χ1 , χ2 ) χ1 6= χ2
(q − 1)(q − 2)
2
Contribution
q+1
B(χ, χ) 1d
portion(q
− 1)
1
B(χ, χ) qd
portion(q
− 1)
q
W(χ)???
q−1
Anyhow, the sum of the total contributions should be
(q 2 − 1)(q 2 − q) = |G|.
Whittaker Models
There is a certain naturally-arising class of representations of GL(2,Fq ) that contains every irreducible representation except for the one dimensional ones and with
multiplicity one. Such a representation is the following: Let
the set of matrices of the form
dene the representation
from
ψ
G
{( 1 x1 )}.
Let
ψ(x)
N (F ) ⊂ GL(2, Fq )
be a character of
N (F )
.
be
We
to be the representation of GL(2,Fq ) that is induced
.
Theorem 19.
G contains all the representations of GL(2,Fq ) except the one di-
mensional ones. It also contains each with multiplicity one.
By a Whittaker Model of an irreducible representation
ization of the representation inside of
G.
(π, V )
we mean the real-
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, Fq ) → C|W (( 1 x1 ) g) = ψ(x)W (g)}
with action by right translation. These are called Whittaker functions.
IRREDUCIBLE REPRESENTATIONS OF GL(2,Fq )
Now recall that by Frobenius Reciprocity, a map
N (F )−)
homomorphism
V → C.
V → W(π)
9
is the same as a (an
This functional is called a Whittaker functional.
To be more explicit, a Whittaker functional is a linear functional
L
on
V
that
satises:
L (( 1 x1 ) .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
space
W(π)
given by
V ∼
= W(π).
W 7→ W (1).
Then we have a functional on this
This certainly satises the condition to be a
Whittaker functional. On the other hand, if we had a linear functional
could construct functions
to
W(π)
L(π(g).v)
and with the isomorphism
L
then we
which form a vector space of functions
V → W(π)
by
v 7→ L(π(g).v).
equal