Irreducible Representations of GL(2,F )
for non-Archimedian Local Fields
NAVA CHITRIK
Referenced heavily from Goldfeld and Hundley (2011), Automorphic Representations and L-Functions
for the General Linear Group,
Cambridge University Press and from notes on Jaquet-Langlands by
Michael Woodbury and Rob Rhodes and can be found at http://www.math.columbia.edu/~woodbury/research/jlnotes.pdf
Contents
Introduction
1
Preliminaries
2
Principal Series Representations
5
The Jacquet Module
6
Supercuspidal Representations
8
Matrix Coecients
9
The Jacquet Module for a principal Series
10
Classication of Principal Series Representations
11
Ramication
14
The Kirillov Model
14
The Whittaker Model
17
Local Zeta Integral and Functional Equation
18
The Global L-Function
20
Introduction
In these notes I will classify the admissible (to be dened) irreducible representations of GL(2,F ) where
F
is a non-archimedian local eld. It will turn out that
the only nite-dimensional ones are one-dimensional.
If we consider the innite
dimensional ones we can do as we did in the nite-eld case: we induce a repre-
principal series when
they are irreducible, and when they are not we get a special sub-representation or
sentation from the Borel subgroup.
These will be called
quotient. But these are not all of them, and there is a simple functor introduced by
1
Irreducible Representations of GL(2,F )
2
for non-Archimedian Local Fields
Jacquet to test if a representation is contained principal series. We group the ones
that aren't induced from the Borel subgroup into a class we call
supercuspidal.
Preliminaries
Let us rst of all dene a class of representations that we restrict ourselves to:
Denition 1.
(smooth) Let
(π, V )
be a representation of GL(2,F ). We say that
(π, V ) is smooth if for any g ∈ GL(2, F ), there exists a compact-open neighborhood
U
of
g
where
π(u) = π(g)∀u ∈ U.
Equivalently, for any vector
compact-open neighborhood of the identity
Remark 2. This does
constant.
not
U
such that
v∈V
there exists a
π(U ).v = v .
π : GL(2, F ) → End(V )
mean that the map
is locally
1
Before I continue, there are a couple of decompositions of
GL(2, F )
that will be
useful in the proofs ahead so I'll list them here without proof:
•
n
e1
( 1 u1 ) p
0
pe2
(Iwasawa)
u=
·k
o
where
e1 > e2 ∈ Z, k ∈ GL(2, Zp )
and
e1 −e
2 −1
X
ui pi
−N
•
G = P K or N AK
n
e1
o p
0
• (Cartan) k1 ·
· k2 = ( a a0 ) · k1 ·
pe 2
F
• (Bruhat) {P
P ωN }
(concise Iwasawa)
pn 0
1
· k2
Denition 3. (contragredient). Let (π, V ) be a representation of a group G. Let
V̂ = {smooth linear functionals on V }. I.e. linear functionals ṽ such that there
exists a neighborhood
Dene
U
of
I
such that if
u ∈ U ,ṽ(π(u).v) = ṽ(v)
for all
v ∈V.
π̂(g).v(v ∗ ) := v ∗ (π(g −1 ).v), or more neatly by < v, π̂(g).v ∗ >=< π(g −1 ).v, v ∗ >.
Note that this denition is dierent than the dual representation since we restrict
to
smooth linear functionals.
Denition 4.
that
We get that
(admissible) Let
(π, V )
V̂
is still of countable dimension.
be a representation of GL(2,F ).
We say
(π, V ) is admissible if it is smooth and for any compact-open subgroup K ,
subspace xed by
K
the
is nite-dimensional. In particular it is enough to check that
1If this happens we call the representation continuous.
Irreducible Representations of GL(2,F )
3
for non-Archimedian Local Fields
V Kn
is nite dimensional, where
neighborhood of
I
in
{Kn } are a fundamental system of open (compact)
GL(2, Zp )
Kn =
Fact 5.
(π, V
Lemma 6.
)
n
1+pn Zp pn Zp
pn Zp 1+pn Zp
⇐⇒ (π̂, V̂ )
is smooth (admissible)
Any smooth irreducible representation
o
is smooth (admissible).
(π, V ) of GL(2, F ) has countable
dimension.
Proof. We use the following decomposition
P
and
k
GL(2, F ) = P K
where
K = GL(2, Zp )
n1
=( ∗ ∗∗ ) and furthermore any matrix can be put uniquely in the form
u =
where
n1 −n
X2 −1
ui pi
.
Therefore
GL(2, F )/K
is
countable.
−N
the representation π is irreducible we can take any non-zero
V = span{π(g).v|g ∈ G}
placing
g
by
pk
otherwise
v
v ∈ V
p
pn 2
Now since
and then
would generate an invariant subspace. Re-
as in the above we have that
V = span{π(pk).v}
K0 ⊂ K
Now we invoke smoothness to get a compact set
Now
K0
has nite index in
V = span(π(p).v)×{nite
K
so that
set}. Since
π(K).v
P
is a
such that
π(K 0 ).v = v .
nite set of vectors.
is countable,
V
Therefore,
is spanned by a countable
set.
Theorem 7.
Let
(π, V )
(Dixmier's Lemma, or Schur's Lemma for smooth representations)
be a smooth irreducible complex representation of
dim(EndG (V
GL(2, F ),
then
)) = 1
.
Proof. Let
any
L:V →V
c ∈ C.
be a intertwiner and
Then by irreducibility of
v ∈ V \{0}.
V , L − cI
L 6= cI
for
must be an isomorphism for allc.
Now consider the uncountable set
Suppose that
(L − cI)−1 v|c ∈ C ⊂ V
( 1 u1 )·
Irreducible Representations of GL(2,F )
4
for non-Archimedian Local Fields
And now, since
V
is spanned by a countable set, there must be a linear relation
among these elements. Suppose
N
X
ai (L − ci I)−1 v = 0
1
Then we apply each
polynomial of
L
(L − ci I) to the above to get a polynomial in L.
over
C
Factoring this
we get
N
Y
(L − di I).v = 0
1
This implies that
L − di I
dj .
isn't injective for some
Therefore
L − di I = 0
after
all.
Equality follows since constants are always intertwiners.
Proposition 8.
G = GL(2, F )
Any admissible irreducible, nite-dimensional representation of
is one-dimensional.
(π, V )
Proof. Let
be a nite-dimensional irreducible representation of
open set of the identity endomorphism
inverse image under
π
−1
( 1 x1 ) = b
( 1 x1 ) , y1 1 even
implies that since
so that
π
must be trivial on
g1 g2 g1−1 g2−1
of
V
Take an
that contains no subgroups. Then its
must be openwe can only do this by nite dimensionality
so it contains a matrices of the form
of the form
I
G.
∈ SL(2, F )
1
for
( 1 x1 ) ,
1
y 1
∈ G for a very small x and y .
This
b
( 1 bx
1 ) ( 1 )the inverse image contains all matrices
x, y
large. Now, such matrices generate
SL(2, F ).
Now for any matrices
−1 −1
therefore π(g1 g2 g1 g2 )
= I
SL(2, F )
g1 g2 ∈ GL(2, F ),
and the representation is
abelian. Therefore it must be one dimensional.
Here's an amazing fact whose proof is long and not included here
Proposition 9.
If
(π, V )
is a smooth and irreducible representation then it is also
admissible.
Proposition 10.
Let
(π, V )
be a smooth, irreducible representation. Then there
exists a so-called central character
ω : F× → C
such that
π ( a a ) .v = ω(a).v
.
Irreducible Representations of GL(2,F )
5
for non-Archimedian Local Fields
Proof. This follows directly from Schur's Lemma. The action
intertwiner since the scalar matrices are in the center of
G
π(a a) : V → V
therefore
π(
a
is an
a ) must
act by a (multiplicative) constant.
Principal Series Representations
GL(2, F )
Finally, we are able to introduce a specic class of representations of
called the principal series representations.
Denition 11.
χ1
Let
and
χ2
be characters of
B(χ1 , χ2 ) =
f : G → C|f
and let the action of
b ))
words, if π(( a c
G
(( a cb ) g)
on this vector space be given by
χ2
are and the
and
Proposition 13.
Proof. Let
f (g)
χ1
let
and
χ2
L-factors
B(χ1 , χ2 )
f (k · U ) = f (k).
For admissibility, let
that
c
f (kk 0 ) = f (k).
things. Therefore
locally constant
P
then
In other
B(χ1 , χ2 )
is the
B(χ1 , χ2 )
will be
will be unramied when
will match up with the
GL(1)
χ1
and
scenario. You'll see.
is an admissible representation.
From the denition and the decomposition
K
U1 , U2, ..., UN
and also
1
f
G = P K,
is locally constant. For every
f (k · Uk ) = f (k).
such that
suce to cover
K.
So if we let
Then by
U=
T
N
Ui
And so the representation is smooth.
K0
be a compact open subgroup in
Now since
only on what coset of
f is
One reason is that
B(χ1 , χ2 )
be a compact neighborhood of
compactness, nitely many
then
a 12
?
are. Also,
f (g) ∈ B(χ1 , χ2 ).
Uk
and
g.f (x) = f (xg).
is a representation of
is determined by what it does on
k∈K
2
Why do we use the normalization
unitarizable when
Let
a 12
= χ1 (a)χ2 (c) f (g)
c
1
= χ1 (a)χ2 (b) ac 2
induced representation.
F ×.
K
0
K0 ⊂ K
it is in, hence
B(χ1 , χ2 )K
0
K.
Let
f ∈B(χ1 , χ2 )K
has nite index, the values of
f (k)
0
so
depends
f (k) is determined by what it does on many
is a nite dimensional vector space.
2
Denition 12.
With perhaps the additional constraint that the functions be uniformly locally
constant, that is, smooth. (it happens to be automatic in this case). According to Wikipedia,
when you induce from a subgroup you place the constraint that the functions are in
L2 (G).
Irreducible Representations of GL(2,F )
6
for non-Archimedian Local Fields
We would like to prove some things about these principal series representations:
their composition series length, when they are irreducible, which are isomorphic and
so on. But rst we will introduce a powerful tool for dealing with representations
of
GL(2, F ).
The Jacquet Module
For any representation, we will dene a certain module called the Jacquet Module
and a functor, the Jacquet Functor which simply takes a representation to its
Jacquet module. The Jacquet module will provide us with many things, but mainly
it will allow us to determine if a representation is isomorphic to a principal series.
Namely, if the Jacquet module is trivial then it will not be a sub-representation of
a Principal-Series representation.
Denition 14.
The Jacquet module of a representation
(π, V )
is dened as
J(V ) = V span {π ( 1 x1 ) .v − v}
and the Jacquet functor
V 7→ J(V )
Remark 15. You can easily check that the diagonal matrices
span {π ( 1 x1 ) .v − v}
. This makes the Jacquet module a
A = ( a1
A-module,
a2
)
x
a fact which
will be useful later on.
There is a nice criterion to determine if a representation is has trivial Jacquet
module give by the following.
Lemma 16.
w ∈ span {π ( 1 x1 ) .v − v} ⇐⇒
´
π(u).wdu = 0
for some compact
U
open subgroup
U ⊂ F.
Proof. Suppose
w ∈ span {π ( 1 x1 ) .v − v},
pick some neighborhood
U
w=
1x
so that
containing all of the
i
1
N
X
π
1
. Then,
1 xi
1
.vi − vi .
Then
Irreducible Representations of GL(2,F )
7
for non-Archimedian Local Fields
ˆ
ˆ
π(u).wdu =
U
π(u).
N ˆ
X
1
=
π
.vi − vi
du
ˆ
1 xi
1
π(u)π
du
−
U
N ˆ
X
1
!
1 xi
1
1
U
=
N
X
π(u)du
U
ˆ
π(u0 )du0 −
U
π(u)du
u0 = u
(take
1 xi
1
)
U
=0
Conversely, suppose
´
π(u).wdu = 0.
Let
U0
U 0 .w = w
be such that
(smoothness).
U
Then, intersecting if necessary ensure that
U0 ⊂ U
so that
U=
N
[
ui U 0
and
u1 = 1.
i=1
Then
ˆ
0=
Xˆ
π(u).wdu =
i
U
X
=
π(u).wdu
ui U
µ(ui U ) · π(ui ).w
i
X
=
ai · π(ui ).w
i
by adding
w=
X
=
X
wto
both sides,
ai · π(ui )w − (−
i
1
)w
N
a0i · π(ui )w0 − w0
i
Proposition 17.
The Jacquet functor
V 7→ J(V )
is exact.
Proof. Suppose
α
β
0 → V 0 → V → V 00 → 0
was an exact sequence of representations. Then, we need to show that
[α]
[β]
0 → J(V 0 ) → J(V ) → J(V 00 ) → 0
Irreducible Representations of GL(2,F )
8
for non-Archimedian Local Fields
is exact.
Now, taking quotients is right exact so we only need to check that
is still injective. For this we will use the lemma. Let
span {π ( 1 x1 ) .v − v} i.e. v 0
U ⊂G
is in the kernel of
[α].
v0 ∈ V 0
such that
[α]
α(v 0 ) ∈
Then there's some neighborhood
with
ˆ
π(u).αv 0 du
0=
U
ˆ
π(u).v 0 du
=α
α
is an intertwiner +linearity
U
ˆ
π(u).v 0 du
0=
since
αis
injective
U
So that
v0 = 0
in
J(V 0 ).
Remark 18. For convenience many sources use the notation
VN = J(V ).
So it may
appear in these notes.
Remark 19. We will also use the notation
V (N ) = span {π ( 1 x1 ) .v − v},
so that
J(V ) = V V (N ).
Supercuspidal Representations
Denition 20.
An admissible representation is called supercuspidal if
J(V ) = 0.
Now we're ready for the big theorem which will essentially classify the admissible
irreducible representations of
Lemma 21.
Let
(π, V )
sense that there exists
N
X
π(gi )vi .
Then
GL(2, F )
. We just need a small lemma
be a nitely generated representation of a group G in the
v1 , v2, , ..., vN ∈ V
(π, V )
such that for any element
v ∈ V, v =
has an irreducible quotient.
i=1
Proof. If we take a chain of quotients
V /V0 ⊃ V /V1 ⊃ V /V2 ⊃ ... ⊃ V /Vk = {0}.
then this corresponds to a chain of subspaces
V0 ⊂ V1 ⊂ V2 ⊂ ... ⊂ Vk = V
. By
nite generation the rst chain has an upper bound always, and so the rst chain
has a lower bound - that, is, we can't keep taking smaller and smaller quotientseventually one will be irreducible.
Irreducible Representations of GL(2,F )
Proposition 22.
(π, V )
9
for non-Archimedian Local Fields
Let
(π, V )
be an irreducible admissible representation.
⇐⇒
is not supercuspidal
Then
it is a sub-representation of a principal series
B(χ1 , χ2 ).
J(V ) = VN 6= 0.
Proof. Suppose that
VN
is a nitely generated
A is abelian,
quotient. Since
θ
A-module,
A = {( a a )},
Let
then we can show that
and so, by the lemma, it has an irreducible
we conclude that this quotient is isomorphic to
C. Let
denote the map
θ : V VN C
Since
P
must act on
C
by some character, say
we can dene the intertwiner
21
χ1 (a1 )χ2 (a2 ) aa21 ,
L : V → B(χ1 , χ2 )
χ1 , χ2 ,
for some
by
L(v) = θ(π(g).v)
Conversely, dene the following functional on
L : B(χ1 , χ2 ) → C,
B(χ1 , χ2 )
f 7→ f (1)
Since
L(π ( 1 x1 ) .f (y)
21
1
− f (y)) = L(χ1 (1)χ2 (1) f (y) − f (y)) = L(0) = 0
1
L descends to a linear functional on B(χ1 , χ2 )N .
we can say that
L is not identically zero on any subspace of B(χ1 , χ2 ):
f
and an element
g∈G
L(π(g)f ) = f (g) 6= 0.
no subspace of
such that
f (g) 6= 0
we can always nd a function
(otherwise its the zero-subspace) then
So there's a non-zero linear functional on
B(χ1 , χ2 )
On the other hand,
has trivial Jacquet module.
B(χ1 , χ2 )N
so that
Matrix Coefficients
Denition 23.
to
C
A matrix coecient of a representation
of the following form. Let
v∈V
and
ṽ ∈ V ∗ ,
(π, V ) is a function from G
then we construct the function
fv,ṽ (g) =< π(g).v, ṽ >
Irreducible Representations of GL(2,F )
Theorem 24.
10
for non-Archimedian Local Fields
Let
(π, V )
be supercuspidal. Then matrix coecients are compactly
(π, V )
supported modulo the center. If
is special then the matrix coecients are in
L2 (Z\G).
Corollary 25.
pairing on
V
If the central character is unitary, we can dene a G-invariant
by xing
ṽ
and letting
ˆ
< v1, v2 >=
fv1 ,ṽ (g)fv2 ,ṽ (g)dg
Z\G
The Jacquet Module for a principal Series
Theorem 26.
dimB(χ1 , χ2 )N
Proof. The dual of
where
n ∈ N =
=2
B(χ1 , χ2 )N
{( 1 ∗1 )}.
L(π(n)f ) = L(f )
are linear functionals satisfying
Therefore it is enough to show that the space of such
linear functionals is two dimensional. We recall the Bruhat decomposition
n G
o
G= P
P ωN
Let's rst consider functions in
with support on
functions
3
P ωN
S(Qp )
into
B(χ1 , χ2 )N
that vanish at 1. These are the functions
- the big cell. Then we map the set of Bruhat -Schwartz
B(χ1 , χ2 )
h(x) 7→ fh (g) =
by

1

χ1 (a1 )χ2 (a2 ) a1 2 h(x) g = ( a1
a2

0
g
∗
a2
) ω ( 1 x1 )
in the little cell
This is clearly an isomorphism with the set of functions such that
f (1) = 0.
Now, if
L is a linear functional on S(Qp ) which is translation invariant (that's the condition
that
L(π(n)f ) = L(f )) then we are in a position to invoke the Reese representation
theorem which tells us that
ˆ
L(f ) − c f (ω ( 1 x1 )) dx = L − c1 L1 = 0
Qp
3compactly supported and locally constant
Irreducible Representations of GL(2,F )
11
for non-Archimedian Local Fields
on functions that vanish at 1. Then, since
L(f ) − c1 L1
takes on the value zero on functions such that
functions vanishing at
1.
This condition makes
a linear functional that vanishes on
linear multiple of the functional
L
W
f (1) = 0.
W
Let
W
be the set of
a co-dimension 1 set. If we have
then it must be a constant, i.e. it must be a
L2 (f ) = f (1).
Therefore
must be an element in the vector space spanned by
Corollary 27.
is a linear functional that
L(f ) − c1 L1 = c2 L2 .
L1 and L2 .
So
A composition series i.e. a chain
V0 ( V1 ( ... ( V = B(χ1 , χ2 )
such that
Vi /Vi−1
is irreducible has length at most two for
B(χ1 , χ2 ).
Proof. We can take the Jacquet module of this chain and it will still be injective
and
VN
has dimension two. So the chain can have been at most 2 long.
Corollary 28.
N
−1
−1 ∼
−1
\
∼
B(χ
B(χ1 , χ2 )
1 , χ2 ) = B(χ1 , χ2 ) = (χ1 , χ2 )
Proof. We can dene an inner product on
−1
−1
\
B(χ
1 , χ2 ) × B(χ1 , χ2 )
by
´
< v1, v2 >= v1 (k)v2 (k)dk
K
this will give an injection
by symmetry
−1
−1
\
B(χ
1 , χ2 ) ,→ B(χ1 , χ2 ).
We can repeat the process and
−1
−1
−1
B(χ−1
1 , χ2 ) ,→ B(χ1 , χ2 ).
Classification of Principal Series Representations
We can use the results above to classify the principal series representations. We
want to know which of these are isomorphic, and which are irreducible.
Theorem 29.
B(χ1 , χ2 )
is irreducible
if
1
(1)
−2
χ1 χ−1
2 (a) = |a|
(2)
2
χ1 χ−1
2 (a) = |a| .
1
Proof. In the rst case, let
1
1
χ(a) = χ1 (a)|a| 2 = χ2 (a)|a|− 2 .
Then
χ(det(g))
invariant subspace
a 21
χ(det(( a xb ) g) = χ(ab)χ(det(g)) = χ1 (a)χ2 (b) det(g)
b
is an
Irreducible Representations of GL(2,F )
12
for non-Archimedian Local Fields
If we quotient out by this we get an irreducible representation called the steinberg
representation.
In the second case we have the opposite,
1
1
−1
−2
2
χ(a) = χ−1
.
1 (a)|a| = χ2 (a)|a|
But
here
a 12
−1
χ(det(( a xb ) g) = χ(ab)χ(det(g)) = χ−1
1 (a)χ2 (b) det(g)
b
which is an element of
−1
B(χ−1
1 , χ2 ), the contragredient representation.
we look at the space of functions in
we get a
G
Therefore, if
B(χ1 , χ2 ) that are xed by this linear functional
xed sub-representation.
Theorem 30.
Suppose
χ1 6= χ2
Hom(B(χ1 , χ2 ), B(η1 , η2 ))
=


C
(χ1 , χ2 ) = (η1 , η2 )

0
otherwise
or
(η2 , η1 )
V ⊂ B(χ1 ,χ2 ) be the set of functions
which are supported only on the
that is, V =
f ∈ B(χ1 , χ2 )f (1) = 0 . We look at the exact sequence
Proof. Let
big cell,
ev(1)
0 → V → B(χ1 , χ2 ) → C → 0
and we take the Jacquet functor
ev(1)
0 → VN → B(χ1 , χ2 )N → C → 0
V = Cc∞ (F ), so you can show that the action of the diagonal matrices A
21
a on VN is given by χ2 (a1 )χ1 (a2 ) 1 by showing that the linear functional (integraa2
12
a tion) on V acts this way. We get that the action of A is given by χ1 (a1 )χ2 (a2 ) 1 a2
Recall that
on the right. Then if you had an intertwiner
have a map on the Jacquet modules
B(χ1 , χ2 ) → B(η1 , η2 ),
we would also
Irreducible Representations of GL(2,F )
13
for non-Archimedian Local Fields
0→
ev(1)
VN →
B(χ1 , χ2 )N →
C→
0
C→
0
↓
0→
The Jacquet modules are
as
C
L
C
ev(1)
VN →
B(η1 , η2 )N →
A−modules
(Where
as in the above exact sequence.
B(χ1 , χ2 )N → B(η1 , η2 )N
A = {( a1
4
a2
)})
and they decompose
Now any intertwiner gives a map
each of which is isomorphic to
C
L
C
so the intertwiner
can either swap the two components, or preserve them. In the rst case, by comparing the action of
and
χ2 = η1 .
A
it must be that
χi = η i
and in the second case that
χ1 = η 2
In either case, there's at most 1-dimension of homomorphisms be-
tween the Jacquet modules. This assumes that
χ1 6= χ2
(otherwise both cases can
happen) so we deal with that case below.
Before that, though, we will show that
dim End(B(χ1 , χ2 )) ≤ 1
when
χ1 6= χ2 .
We
use the same process as above. Suppose we had an intertwiner, then it would descent
to a map
M
B(χ1 , χ2 )N → B(χ1 , χ2 )N .
Now there's a bijection between intertwiners
and linear functionals
(
L : B(χ1 , χ2 ) → CL(π ( a1
x
a2
)
21
a1 ) f = χ1 (a1 )χ2 (a2 ) L(f )
a2
(you just do the obvious thing to go back and forth).
Now it's clear that if we
had such a linear functional and it was the zero functional on the Jacquet module,
then it had to have been trivial to begin with. Therefore there's an injection from
intertwiners to maps between the Jacquet modules. Therefore there's at most one
dimension of intertwiners of
Theorem 31.
B(1, 1)
B(χ1 , χ2 )
is irreducible.
4
B(χ1 , χ2 )N
f1 and f2 were basis vectors
1
a1 2
then π
a2 (f1 + V (N )) = χ1 (a1 )χ2 (a2 ) a2 f1 + V (N ) because A xes V (N ). So f1 and f2
are mapped to their own subspaces under the action of A. We can always take the rst function
f1 to vanish at 1.
Proof. The reason that
a1
decomposes so nicely is because if
Irreducible Representations of GL(2,F )
14
for non-Archimedian Local Fields
B(1, 1)
Proof. (Sketch) We dene an injection of
c
|x|
which look like
·
into the space of
C∞
functions
χ1 (−1)χ1 (x)χ−1
2 (x). Then we show that any non-zero vector
generates this space under the action of
This will proves the case for all
G.
B(χ, χ) = χ
N
B(1, 1).
By Schur's Lemma
dim End(B(χ, χ)) =
1.
One big question I have not answered is if there are any
other
principal series
representations which are reducible, besides for the ones mentioned above.
The
answer is that there are no others. I haven't found a nice proof of this yet. There's
one in Bump on pages 475-476.
Ramification
Denition 32.
GL(2, Zp )-xed
We say a representation is unramied if there exists a
K =
vector. Otherwise we say it is ramied.
Claim 33. The only possible unramied representations are the principal series
where each character is unramied. In that case, we have the
spherical vector
a 21
f (( a xb ) k) = χ1 (a) · χ2 (b) b
We need the characters to be unramied in order for this to be well-dened.
Claim 34. For an irreducible representation there is at most a 1-dimensional space
of
K−xed
vectors.
This follows since the
K−xed
vectors are an irreducible
Hecke-module and the Hecke-functions are commutative. Ignore this.
The Kirillov Model
Denition 35.
(Twisted Jacquet Module)
X = V /span {π ( 1 x1 ) .v − ψ(x)v}
X
is sometimes denoted
Lemma 36.
Jψ (V ) to emphasis the similarity with the Jacquet module.
w ∈ span {π ( 1 x1 ) .v − ψ(x)v} ⇐⇒
compact open subgroup
U.
´
U
ψ(−u)π(u).v = 0
for some
Irreducible Representations of GL(2,F )
15
for non-Archimedian Local Fields
Proof. We do the same thing as when we proved this about the Jacquet module.
Denition 37.
A Kirillov model
(π 0 , K)
of a representation
(π, V )
is one where
π 0 ( a 1b ) f (y) = ψ(by)f (ay)
(0.1)
We construct this by creating the vector space of functions
K = f : F × → X|fv (y) := π ( y 1 ) .v
So the isomorphism
of
V →K
is simply given by
v 7→ fv (y) = π ( y 1 ) .v
as an element
X.
The action is given by
π 0 (g).fv (y) = fπ(g).v (y) = π ( y 1 ) π(g).v
Claim 38. The above construction actually satises 0.1.
Proof. We need to show that
We let
U = p−n Zp
π 0 ( a 1b ) fv (y)−ψ(by)fv (ay) is in span {π ( 1 x1 ) .v − ψ(x)v}.
(we will choose
n
later) and use the lemma to compute:
π 0 ( a 1b ) fv (y) − ψ(by)fv (ay) = π ( y 1 ) π ( a 1b ) .v − ψ(by)π ( ya 1 ) .v
and so
ˆ
ˆ
ψ(−u)π ( 1 u1 ) π ( y 1 ) π ( a 1b ) .v
p−n Z
ψ(−u)π ( 1 u1 ) π
du =
ya yb
1
.v du
p−n Zp
p
now choose
n
such that
|by| < pn
and make the change of variables
u 7→ u − by
so
we get that
ˆ
ˆ
ψ(−u)ψ(by)π ( ya u1 ) .v du
ψ(−u)π ( 1 u1 ) π ( y 1 ) π ( a 1b ) .v du =
p−n Zp
which exactly cancels out the second piece.
p−n Zp
Irreducible Representations of GL(2,F )
16
for non-Archimedian Local Fields
Claim 39. The functions
fv (y)
are locally constant and compactly supported.
Proof. Since the representation is smooth, there exists a small range of
π(
a
1 ) .v
=v
π ( 1 1b ) .v = v
so that with
when
b
so
fv
is locally constant. Similarly for any
but the Kirillov denition gives us that
in a small range,
fv (y) = ψ(by)fv (y)
b
a
there's a range where
π 0 ( b 1 ) .fv (y) = ψ(by)fv (y)
therefore
fv (y)
is non-zero only
ψ(by) = 0.
Claim 40.
where
×
K = SX (Q×
p ) + π(ω0 )SX (Qp )
Proof. We know from the denition of a Kirillov model how a function acts on the
groupP (plus we need the central character). The Bruhat decomposition then tells
us that these functions are determined by what they do to
1
and
ω0 .
(These two
values might not be independent though)
Theorem 41.
The Kirillov Model is unique
Proof. I'll just sketch this. What we are actually going to prove is that the space
X, or what we will now call
Jψ (V ) = V
span {π
( 1 u1 ) .v − ψ(u)v}
is actually a one dimensional space. We do this by counting the number of linear
functionals on this space. We dene:
J(t, ω) : X → X
by
J(t, ω) = π
0
−1
1
i h −1 0
−1 0
. ω(t y ) · 1Z×
(t y ) · x p
Then one can show that any linear operator on
X
that commutes with all
is a constant endomorphism. Finally you show that the
other. Therefore all endomorphisms of
Theorem 42.
Let
n =the
X
(x ∈ X)
y 0 =1
J(t, ω)
S(Q×
p)
in
K.
commute with each
are constant.
co-dimension of
J(t, ω)
Then
Irreducible Representations of GL(2,F )
17
for non-Archimedian Local Fields
Proof. Let
(π, K)
We will show that
n=0
⇐⇒
V
is supercuspidal
n=1
⇐⇒
V
is a special representation
n=2
⇐⇒
V
is a principal series
be the Kirillov model of an innite-dimensional representation.
V (N ) = span {π ( 1 u1 ) .v − v} = SX (Q×
p ).
theorem, because we already proved that
V (N )
has co-dimension 2 for principal
series, 1 for special and 0 for supercuspidal. Suppose that
then then there exists a (very positive)
n
such that
ˆ
p−n Z
f
V (N )
pn Zp .
Then
Conversely, suppose that
then
ˆ
π ( 1 u1 ) .f (y)du
=
ψ(uy)f (y)du = 0
p−n Zp
U
for some neighborhood
U = pn Zp ⊂ F ×
again by 16. Also
ˆ
ψ(uy)f (y)du = f (y) ·
0=
p−n Zp
f
f ∈SX (Q×
p)
ψ(uy)f (y)du = 0
by the criterion in 16.
ˆ
so that
vanishes on
and
p−n Zp
p
is in fact in
f ∈ V (N )
f
f ∈K
ˆ
π ( 1 u1 ) .f (y)du =
so that
That will give us the
must be zero on
pn Zp .
So


pn
y ∈ pn Zp

0
otherwise
f ∈ S(Q×
p)
The Whittaker Model
Denition 43.
A Whittaker function
W : G → C is one that satises W (( 1 x1 ) g) =
ψ(x)W (g )
Remark 44. We can make a representation out of a Whittaker function by acting
on it by the right regular action. This is called the Whittaker Model
Irreducible Representations of GL(2,F )
18
for non-Archimedian Local Fields
We could construct a Whittaker functional by
ˆ
ψ(−u)π ( 1 u1 ) .vdu
L(v) =
F
Proposition 45.
L(v)
Proof.
L
is nonzero if the representation is innite-dimensional.
is only zero on the elements in
Vψ (N )
denition. And
Denition 46.
Vψ (N ) = span {π ( 1 u1 ) .v − ψ(u)v}
V
is codimension 1 in
if
V
is innite dimensional.
A Whittaker functional is a map
L:V →C
by
that satises
L (π ( 1 x1 ) v) = ψ(x)L(v)
We can make a Whittaker function (and hence a Whittaker model) out of this by
Wv (g) = L(π(g).v)
Conversely, if we have a Whittaker model we can put a whittaker functional on it,
namely
Wv 7→ Wv (1).
Local Zeta Integral and Functional Equation
Denition 47.
Let
s ∈ C, φ
Schwartz) function on
tation
(π, V )
be a locally constant, compactly supported (Bruhat-
GL(2, Qp )
and
β(g)
a matrix coecient for some represen-
We dene the local-zeta integral
ˆ
1
φ(g)β(g)|det(g)|s+ 2 d× g
Zp (s, φ, β) :=
GL(2,Qp )
Theorem 48.
•
(local functional equation).
The local zeta integral
• Zp
Zp
converges in some half-plan.
is a rational function of
a common divisor
Lp (s, π).
p−s
and as we range over all
Furthermore, there exist
φ
Zp (s, φ, β) = Lp (s, π)
• Zp (s, φ, β) = γ(s, π)Zp (1 − s, φ̂, β̌)
•
Zp (s,φ,β)
L(s,π)
= (s, π)
Zp (1−s,φ̂,β̌)
L(1−s,π̃)
where
β̌(g) = β(g −1 )
φ, β
they have
β
such that
and
Irreducible Representations of GL(2,F )
19
for non-Archimedian Local Fields
The proof of this theorem is done individually for each type of representation and
relies heavily on the behavior of matrix coecients. There's also a lot of detail and
a lot of computing... I will omit most of this and just give a brief discussion of the
main results.
(1) Unramied principal series.
B(χ1 , χ2 )
where both characters have no ram-
ication we get the L-function
Lp (s, π) = (1 − χ1 (p)p−s )(1 − χ2 (p)p−s )
and the epsilon factor
(s, π) = (s, χ1 )(s, χ2 )
where the factors are in the
GL(1)
case (they're going to be
recall). All this is achieved using the function
coecient given by the unramied vectors
1,
if you
1M at(2,Zp ) (g) and the matrix
< π(g).f, f˜ >where f (( a1
12
k) = χ1 (a1 )χ2 (a2 ) aa21 x
a2
)·
(2) Ramied principal series
Lp (s, π) = Lp (s, χ1 )Lp (s, χ2 )
as in
GL(1)
and
(s, π) = (s, χ1 )(s, χ2 )
as in
GL(1)
(3) Special Representation (unitary)
1
Lp (s, π) = (1 − χ(p)p−s− 2 )−1
and
1
p (s, π) = −χ(p)−1 (s, χ1 )(s, χ2 )ps− 2
this follows from the previous case if we just set
χ1 χ2 = |a|±1 .
(4) Supercuspidal Representations.
Lp (s, π) = 1
we can achieve this as follows. Take any matrix coecient
is compactly supported modulo the center so take
β.
We know it
φ(g) = β(g)
supported
Irreducible Representations of GL(2,F )
for non-Archimedian Local Fields
20
only on the elements with determinant one. Then
ˆ
|β(g)|2 dg
Zp (s, φ, β) :=
Z\G
is an element of
C.
So its holomorphic.
The epsilon factor is a bit more
complicated to show, here. But we can show that it is some holomorphic
monomial
γ(s).
The Global L-Function
Where does all this lead us? The aim of classifying these representations is to obtain
the functional equation of the Jacquet-Godement L-function of an automorphic
representation.
That is, if
(π, V )
were an automorphic representation then it is
know that
π=
0
O
πv
and that all but nitely many of the
πv
will be unramied.
We can dene the
L-function of such a representation as
L(s, π) =
Y
Lv (s, π)
In a manner similar to Tate's thesis, we dene a global Zeta-integral and show that
it satises a global functional equation. Finally using the above data we can nd
(s, π)
such that
L(s, π) = (s, π)L(1 − s, π̂)
which gives us the functional equation.
Question: Why do we care about this
Answer: Special values!
L−function?