WEIL REPRESENTATION AND CENTRAL EXTENSION OF LOOP ... · preparation and writing style has been invaluable. I want to thank Erhard Neher for his never-ending source of wisdom that

WEIL REPRESENTATION AND CENTRAL EXTENSIONOF LOOP SYMPLECTIC GROUPS

By

Gabriel Bergeron-Legros

September 2014

A Thesis

submitted to the School of Graduate Studies and Research

in partial fulfillment of the requirements

for the degree of

Master of Science in Mathematics1

c© Copyright 2014

by Gabriel Bergeron-Legros, Ottawa, Canada

1The M.Sc. Program is a joint program with Carleton University, administered by the Ottawa-Carleton Institute of Mathematics and Statistics

Abstract

In this thesis, we present the Weil representation over loop symplectic groups. Then

we study the question of whether or not the Schrodinger representation and the Weil

representation are continuous. Finally, we define a cocycle of SL2(R((t))), adapt

Kubota’s theorem to this case and verify that this cocycle splits over SL2(R[[t]]).

ii

Acknowledgements

Above all else, I want to thank my supervisor Hadi Salmasian for the continuous

support and help he has provided, even years before the writing of this thesis. In

addition to the knowledge I have received through our meetings and four different

courses, his help towards scholarship and PhD applications, presentations, conference

preparation and writing style has been invaluable.

I want to thank Erhard Neher for his never-ending source of wisdom that he is

always willing to share, his organizing of the Kac-Moody seminar and the reading of

this thesis. Without Erhard, I would have missed much of the beauty of Lie Theory

and Algebra in general.

I would like to thank Paul Mezo for taking the time to read this thesis and the

knowledge he has shared in Galois Theory, two things I could not graduate without.

This thesis would not have been the same without the funding and support of

the National Sciences and Engineering Research Council of Canada. I have received

funding for two different Undergraduate Student Research Awards and a Canada

Graduate Scholarship, each of which helped immensely.

I would like to thank the Centre de Recherche Mathematiques and the Fields

Institute for the workshops and accommodation they have provided. I have learned

a lot from the time I have spent at these two places.

Lastly, I want to thank the University of Ottawa, the faculty of the department

of mathematics and statistics and more particularly professors Benoit Dionne, David

Handelman, Daniel Daigle, Barry Jessup and Monica Nevins, for their support and

teaching throughout my stay.

iii

Contents

Abstract ii

Acknowledgements iii

1 Introduction 1

2 Preliminaries 5

2.1 Local fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

2.2 The Schwartz space . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

2.3 Central extensions of groups . . . . . . . . . . . . . . . . . . . . . . . 10

3 The Weil Representation 18

3.1 The symplectic group . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

3.2 The Weil representation . . . . . . . . . . . . . . . . . . . . . . . . . 23

4 Continuity of the representations 43

4.1 Finite dimensional case . . . . . . . . . . . . . . . . . . . . . . . . . . 43

4.2 Infinite-dimensional case . . . . . . . . . . . . . . . . . . . . . . . . . 48

5 A specific central extension 56

5.1 Kubota’s theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56

5.2 Splitting of the cocycle over a subgroup . . . . . . . . . . . . . . . . . 69

iv

Chapter 1

Introduction

In 1959, Segal[9] defines and proves the existence of a certain projective repre-

sentation of infinite dimensional symplectic groups. Inspired by Shale’s[11] study

of this representation, Weil [8] introduces a generalization of this representation to

symplectic groups over an arbitrary local field. For this reason, this distinguished

representation of the symplectic group, or rather of the metaplectic group is called

the Segal-Shale-Weil representation, or simply Weil representation.

The usual starting point is to consider two different representations U and U s of

the Heisenberg group H = V ⊕ V ∗ ⊕ F on L2(V ) where V is a finite-dimensional

vector space over F, a local field of characteristic not 2, and s is a symplectic matrix.

Then by the Stone-Von Neumann theorem, these two representations will turn out to

be unitarily equivalent [7, Section 3.2]. In other words, there exists a unique up to

scalar unitary operator ξs such that

ξ−1s Uξs = U s.

Furthermore, by the uniqueness up to scalar of this intertwining operator, ξs is in

fact a projective representation of the symplectic group over the space of Schwartz

functions. An explicit formula for ξs can be found in [7, Lemma 3.2 (3)].

In 2008, Zhu [14] writes that the same construction can be done over loop sym-

plectic groups, namely by replacing the field F with a field of Laurent series with

coefficients in F. However, instead of using the Stone Von-Neumann theorem, one

carries over the explicit formula for the Weil representation to a specific subgroup

and checks that the expected properties are indeed satisfied[14, Proposition 2.8].

1

CHAPTER 1. INTRODUCTION 2

In a different, but related topic, Kubota [12] describes a cocycle of SL2(F), for Fagain a local field, which depends on so-called Hilbert symbols. He also shows that

the cocycle induces a non-trivial central extension.

The main contribution of this thesis is three-fold. Firstly, we rewrite Section 2

of [14] in detail in Chapter 3, and by doing so we fill some loose ends in the paper.

Secondly, we give in Chapter 4 explicit, elementary proofs of the continuity of some of

the representations defined in Chapter 3, including the Weil representation for loop

symplectic groups. Finally, we define a cocycle for SL2(F((t))) similar to Kubota’s

cocycle and prove that it splits over SL2(F[[t]]). The remainder of this introduction

serves to give an outline of each of the sections.

In Section 2.1, we define what a local field is without delving into the theory, and

we also describe the field F((t)) as a topological field in Proposition 2.1.5, or in other

words that the operations of addition, multiplication and inversion are continuous.

Furthermore, we give an explicit basis for this topology.

In Section 2.2, we define and discuss some properties of the Schwartz space S(V ).

We do so first over a finite dimensional vector space for both cases where F is an

archimedean local field or non archimedean. For an infinite dimensional space, we

will define a Schwartz function to be a function whose restriction to every finite

dimensional space is also Schwartz. If F = R, then by inducing a topology on the

Schwartz space with the use of seminorms, it will turn out that S(V ) inherits the

Frechet space structure of the finite case.

In Section 2.3, we introduce the study of central extensions through the use of

cocycles, and in fact prove that there is a bijection between isomorphism classes of

central extensions and equivalence classes of cocycles in Proposition 2.3.10.

These preliminary sections allow us to define the Weil representation over loop

symplectic groups in Chapter 3, Proposition 3.2.4. We will first consider a symplectic

form 〈., .〉 on an infinite dimensional vector space X = V ⊕V ∗, where V ∗ is the vector

space dual of V , and its isometry group Sp(X). However, Sp(X) will be too large

to allow a definition of the Weil representation which is compatible with its formula,

hence we will concern ourselves with a subgroup Sp(X, V ∗), which is the biggest

subgroup over which the formula for the Weil representation can be carried over. The

good thing is that Sp2n(F((t))) will be a subgroup of Sp(X, V ∗) (Lemma 3.2.19) for a


suitable choice of V , and this is a group which is well known and understood, hence

it is always possible to restrict the Weil representation to it.

The Stone Von-Neumann theorem in the finite dimensional case gives a certain

relationship between two representations of the Heisenberg group and the Weil rep-

resentation. We should expect the same relationship to appear in the infinite dimen-

sional case, and in fact, we will give a direct proof of this statement in Proposition

3.2.4.

To prove that the Weil representation we define is indeed a projective represen-

tation, we will have to reduce to the standard Weil representation. The trick will be

to write every element of Sp(X, V ∗) as a product g = pg′ where p and g′ lie in two

subgroups P and Spfin(X) which are much more manageable. Indeed, the Weil rep-

resentation for P will have a very simple formula where no integration is happening,

and the one for g′ will turn out to be very similar to the standard Weil represen-

tation, which we already know to be a projective representation. Lemmas 3.2.14,

3.2.15 and Corollary 3.2.9 will work out the technical details so that we obtain our

main result in Theorem 3.2.16. Finally we will end the chapter with a short study

of V = (t−1R[t−1])2n. We will prove that its vector space dual can be thought to be

R[[t]]2n in Lemma 3.2.17, and this will induce a symplectic form on R((t))2n.

In Chapter 4, we will study whether or not the Heisenberg group representation

and the Weil representation are continuous in the case F = R. In fact, we will start

with an explicit proof in the finite dimensional case for the Heisenberg group, which

will turn out to be much more technical than one would initially expect. The main

strategy will be to consider the action of three subgroups of the Heisenberg group

and how they interact with each other, which will be done in the numerous technical

Lemmas 4.1.1, 4.1.2, 4.1.34.1.5, 4.1.6, 4.1.7 and 4.1.8 to obtain the main result in

Proposition 4.1.9. The proofs for the infinite dimensional case V = t−1R[t−1] will be

in appearance easier, because the topologies involved are much more easier to work

with than the Euclidean topology. Thus, we will not need any technical Lemmas and

we will directly proceed to the proofs in Proposition 4.2.2 and Theorem 4.2.4.

In Chapter 5, we define a cocycle very similar to Kubota’s cocycle, but for

SL2(R((t))). Then we carry over Kubota’s results[5] to SL2(R((t))) in Chapter 5

in full detail. The definition of the cocycle is very difficult to work with, and a proof


by brute force would involve over a hundred of cases. Therefore, we spend time

obtaining Lemmas 5.1.3, 5.1.4, 5.1.5, 5.1.6, 5.1.7, 5.1.8 and 5.1.10 to significantly

reduce the number of cases needed for Theorem 5.1.11. Finally, we also prove that

the cocycle splits over the subgroup SL2(R[[t]]) in Theorem 5.2.3 with a case by case

argument, or more precisely that it is equal to a coboundary.

Chapter 2

Preliminaries

2.1 Local fields

Suppose F is a field. We say that F is a local field if (F,+) is a locally compact

topological group with a non-discrete topology. Denote by F((t)) the vector space of

all Laurent series over t, i.e. F((t)) = F[[t]]⊕ t−1F[t−1]. Then F((t)) with the obvious

operations of addition and multiplication, is a field. [1, p. 238, Exercise 4]

The following theorem is a combination of Theorem 5 and Theorem 8 in [12].

Theorem 2.1.1. Suppose F is a local field. Then F is either R, C, a finite extension

of Qp, or F ((t)) where F is a finite field.

We will refer to R and C as archimedean, and to the finite extensions of Qp or

F ((t)) for F a finite field as non-archimedean. We will often be interested in F((t))

where F is a local field. Suppose x =∞∑

i=−nait

i. Then we set v(x) = mini : ai 6= 0

for x 6= 0, and v(0) = +∞. Let

Un = x ∈ F((t)) : v(x) ≥ n.

We equip F((t)) with the topology generated by all sets of the form x + Un where

x ∈ F((t)). Remark that if m ≥ n, then Um ⊆ Un. Furthermore, Un + Un ⊆ Un. The

following proposition will be very useful :

Proposition 2.1.2. The collection of open sets of the form x+ Un is a basis for the

topology on F((t)).

5

CHAPTER 2. PRELIMINARIES 6

Proof. Suppose U is an open set. Then it can be written as an infinite union of finite

intersections of sets of the form x+Un. We need to show that for all y ∈ U , there exist

a set of the form y′+Un containing y and contained in U . Without loss of generality,

we can assume U = (x1+Un1)∩...∩(xm+Unm) is a non-empty intersection of elements

of the form x + Un. Let N = maxn1, ..., nm and y ∈ U . We claim that y + UN is

contained in xi +Uni for all i. Indeed, we know that xi− y ∈ Uni . If y+ vN ∈ y+UN ,

then xi−(y+vN) ∈ Uni−UN ⊆ Uni . Therefore, y+vN = xi−(xi−(y+vN)) ∈ xi+Uni ,hence y + UN ⊆ xi + Uni for all i and then y + UN ⊆ U .

The main application of this proposition is the following :

Lemma 2.1.3. Suppose U ⊆ F((t)) is open and x ∈ U . Then there exists m ∈ Zsuch that x+ Um ⊆ U .

Proof. Choose m and x′ ∈ F((t)) such that x ∈ x′+Um ⊆ U . Then x−x′ ∈ Um. But

then x+ Um = x′ + Um ⊆ U .

Remark 2.1.4. Often, we will choose m to be greater than or equal to some other

arbitrary number. This is not a problem because m ≥ n implies that Um ⊆ Un.

Proposition 2.1.5. The operations of addition, multiplication and inversion of F((t))

are continuous.

Proof. Let us write x =∑

i≥−Mait

i and y =∑i≥−N

biti for M,N ≥ 0 such that a−M 6= 0,

b−N 6= 0. Then we have x+y =∑

i≥−maxM,N(ai+bi)t

i and xy =∑

i≥−(M+N)

i+N∑k=−M

akbi−kti.

We see from these formulas that v(x+ y) ≥ minv(x), v(y) and v(xy) = v(x) + v(y)

for x and y 6= 0.

Now suppose U ⊆ F((t)) is open and (x, y) is in the pre-image of U under the

map (x, y) 7→ x + y. Then there exists m ∈ Z such that x + y + Um ⊆ U . But now,

(x+Um)× (y +Um) is an open set of F((t))× F((t)) containing (x, y). Furthermore,

if x′ ∈ x + Um and y′ ∈ y + Um, we have x′ + y′ ∈ x + y + Um. We conclude that

(x+ Um)× (y + Um) is contained in the pre-image of U and therefore the pre-image

of U is open.

If U ⊆ F((t)) is open and (x, y) is in the pre-image of U under the map (x, y) 7→ xy,

then once again there exists m ≥ 0 such that xy + Um ⊆ U . Let

R = maxm− v(x),m− v(y),m.


Clearly, (x + UR) × (y + UR) is an open set of F((t)) containing (x, y). But now, if

x′ ∈ x+ UR and y′ ∈ y + UR, we have

x′y′ ∈ xy + xUR + yUR + URUR ⊆ xy + Um + Um + U2m ⊆ xy + Um.

We conclude that (x+UR)×(y+UR) is contained in the pre-image of U and therefore

the pre-image of U is open.

Finally, suppose U ⊆ F((t)) and that x 6= 0 is in the pre-image of U under the

map x 7→ x−1. Because 0 = v(xx−1) = v(x) + v(x−1), we know that v(x−1) = M .

Write x−1 =∑i≥M

citi. Then

xx−1 =∑i≥0

i−M∑k=−M

akci−kti = 1.

Therefore, we obtain that cM = a−1−M and then recursively that

ci+M = −a−1−M

i−M∑k=−M+1

akci−k.

In particular, ci+M depends on ak for −M ≤ k ≤ −M + i. Now, choose m ≥ 1

such that x−1 + Um ⊆ U . Then ci for M ≤ i < m is determined uniquely by a−M , ...

a−M+m−1. We claim that x+U−M+m ⊆ (x−1 +Um)−1. Indeed, if y ∈ x+U−M+m, then

y−1 (which exists because a−M 6= 0 and m ≥ 1, hence y 6= 0) will share the same first

m coefficients with x−1, because they are uniquely determined by a−M ,...,a−M+m−1,

which are the same for x and y. Therefore, y−1 ∈ x−1 + Um ⊆ U , and thus the

pre-image of U is open.

2.2 The Schwartz space

Suppose F is a non-archimedean local field. If V is a finite-dimensional vector space

over F, we say that f is a Schwartz function if f is locally constant with compact

support. If V is not finite-dimensional, we say that f is a Schwartz function if its

restriction to every finite-dimensional subspace of V is Schwartz. We will denote the

space of Schwartz functions by S(V ). For the rest of this section, we deal with F = R.


Remark 2.2.1. For F = C, we simply think of a vector space over F as a vector

space over R with twice the dimension and the Schwartz space is then defined in the

same way as in the real case.

Suppose V ∼= Rn, f ∈ C∞(V ) and let α = (α1, ..., αm) and β = (β1, ...βm) ∈ Nm.

Write

pα,β(f) = supx∈V

∣∣xα∂βf(x)∣∣ ,

where x = (x1, ..., xn), xα = xα11 x

α22 ...x

αnn , ∂β = ∂β11 ...∂

βnn . Then let

S(V ) = f ∈ C∞(V ) : pα,β(f) <∞ for all multi-indices α, β.

Proposition 2.2.2. Each pα,β is a semi-norm.

Proof. This proof is an easy exercise.

Consider the sets Uα,β,ε = f ∈ S(V ) : pα,β(f) < ε. We equip S(V ) with the

topology generated by the collection of all sets of the form f + Uα,β,ε.

Definition 2.2.3. Suppose X is a Hausdorff topological vector space. Then we say

that X is a Frechet space if

1. Its topology is induced by a countable family of seminorms ‖.‖k in the following

way : we say that U ⊆ X is open if and only if for all v ∈ U , there exists finitely

many seminorms ‖.‖k1 , ..., ‖.‖kn and ε > 0 such that

n⋂i=1

w ∈ X : ‖w − v‖ki < ε ⊆ U

2. It is complete with respect to the seminorms, i.e., every sequence which is

Cauchy with respect to every seminorm ‖.‖k has a limit in X.

Proposition 2.2.4. S(V ) with the topology defined above, is a Frechet space.

Proof. This is Proposition 8.2 in [4].

Next, suppose V is an arbitrary vector space. Then we define S(V ) to be the space

of all functions f : V → C such that for any finite-dimensional subspace W ⊆ V , we

have that f |W ∈ S(W ).


We will be particularly interested in the case where V = (t−1R[t−1])2n. We can

write an element v of V as

v =2n∑i=1

t−1pi(t−1)ei

for polynomials pi ∈ R[t], and the ei’s is the standard basis of V as a t−1R[t−1]-module.

Next, suppose m ∈ N. Then let α = (αi,j)1≤i≤m,1≤j≤2n and β = (βi,j)1≤i≤m,1≤j≤2n and

Vm = spant−jei : 1 ≤ j ≤ m, 1 ≤ i ≤ 2n.

If f ∈ S(V ), we know that f |Vm is a Schwartz function in the ordinary sense, hence

we can define the maps

pα,β,m(f) = supx∈Vm

∥∥∥∥∥xα0,0

0,0 ...xαm,2nm,2n

∂β0,0

∂xβ0,00,0

...∂βm,2n

∂xβm,2nm,2n

f(x)

∥∥∥∥∥for x =

∑i≤m,j≤2n

xi,jt−ij ∈ Vm.

Proposition 2.2.5. Each pα,β,m is a seminorm.

Proof. Follows immediately from Proposition 2.2.2.

We equip S(V ) with a topology in the same way that we did in the finite-

dimensional case with the seminorms pα,β, but Uα,β,ε is replaced by Uα,β,m,ε. Namely,

Uα,β,m,ε = f ∈ S(V ) : pα,β,m(f) < ε

We also obtain

Proposition 2.2.6. The space S(V ) with the topology induced by the seminorms

pα,β,m is a Frechet space.

Proof. Suppose f1, f2 ∈ S(V ) and that f1 6= f2. Then there exists v ∈ V such that

f1(v) 6= f2(v). In particular, v is contained in Vm for some m. Write c = f1(v)−f2(v).

Then the open sets U0,0,m,c/2 + f1 and U0,0,m,c/2 + f2 are disjoint open sets separating

f1 from f2. We conclude that S(V ) is Hausdorff.

Now suppose fn is a sequence of functions in S(V ) which is Cauchy with respect

to every seminorm pα,β,m. Suppose v ∈ V . Then v ∈ Vm for some m. In particular,

fn|Vm is a Cauchy sequence with respect to p0,0,m and therefore fn(v) is also a Cauchy


sequence, hence has a point-wise limit. We let f(v) be the point-wise limit of fn(v).

Now suppose W ⊆ V is finite-dimensional. Then fn|W are Schwartz in the ordinary

sense. Because W ⊆ Vm′ for some m′, we see that fn|W form a Cauchy sequence

of Schwartz functions on W and thus by Proposition 2.2.4 converges to a Schwartz

function g. However, it is not hard to see that g agrees with f |W . Therefore, f is a

Schwartz function.

Notation 2.2.7. We will sometimes replace pα,β(f) or pα,β,m(f) by the notation

‖f‖α,β or ‖f‖α,β,m. Furthermore, we will also sometime replace p0,0(f) by ‖f‖∞.

2.3 Central extensions of groups

In Chapter 5, we will be interested in studying central extensions of loop symplectic

groups. We recall some of the standard facts about central extensions in this section.

Definition 2.3.1. Let K,G,H be groups. An extension is a short exact sequence

1 // H i // Gj // K // 1 .

The extension is said to be central if i(H) is contained in the center of G.

Example 2.3.2. Suppose G = K n H. Define i′ : H → G by i′(h) = (1, h) and

j′ : G→ K by j′(k, h) = k. Then,

1 // H i′ // Gj′ // K // 1 (1)

is an extension. If G = K ×H and H is abelian, then (1) is a central extension.

Proof. The proof is an easy, straightforward computation.

By an isomorphism of extensions (or of central extensions), we mean a commuta-

tive diagram

1 // H

1H

i // G

φ

j // K //

1K

1

1 // Hi′// G′

j′// K // 1

(2)

where φ is an isomorphism between G and G′ and each row is an extension (or a

central extension).


Proposition 2.3.3. Suppose

1 // H i // Gj // K // 1 (3)

is an extension. Then the following conditions are equivalent :

1. There exists a group homomorphism q : K → G such that jq = 1K.

2. (3) is isomorphic to (1) where G = K nH for some semidirect product.

Proof. Suppose that there exist a group homomorphism q : K → G such that jq =

1K . Note that the image of H in G is normal, because i(H) = ker j. Define a

map α : K → Aut(H) by α(k)h = khk−1. For each k, α(k) is a well-defined group

homomorphism because H is normal, and is clearly injective and surjective, hence

an automorphism. Consider the subgroups q(K) ⊆ G and i(H) ⊆ G. Suppose

x ∈ q(K) ∩ i(H). Then x ∈ ker j by exactness, hence j(x) = 1. But then x = q(y)

for some y ∈ K, hence j(x) = jq(y) = y = 1. Therefore, x = q(1) = 1. We conclude

that q(K) ∩ i(H) = 1.

Now suppose x ∈ G. Then qj(x−1)x ∈ ker j. Therefore, qj(x−1)x = i(h) for some

h ∈ H. But now, x = qj(x)i(h) ∈ q(K)i(H). Therefore, G = q(K)i(H). We conclude

from [1, p. 180, Theorem 12] that G ∼= q(K) n i(H) ∼= K nH with the isomorphism

φ : q(k)i(h) 7→ (k, h), and it is easy to check from the definition of φ that Diagram

(2) commutes.

Next suppose G ∼= K nH as in Example 2.3.2. Then in K nH, we have i′(H) =

1 n H and j′(k, h) = j(k, 1). Define q : K → K n H by q′(k) = (k, 1). Clearly

j′q′ = 1K and q′ is a group homomorphism. But then, by the commutativity of

the diagram it is easy to check that q := φ q′ is a group homomorphism K → G

satisfying jq = 1K .

In the previous proof, the assumption that H is contained in the center of G was

not needed. For central extensions, we obtain the following :

Corollary 2.3.4. Suppose that

1 // H i // Gj // K // 1 (4)

is a central extension. Then the following conditions are equivalent :


1. There exists a group homomorphism q : K → G such that jq = 1K.

2. (4) is isomorphic to (1) where G = K ×H.

Proof. Suppose that there exists a group homomorphism q : K → G such that

jq = 1K . Then by Proposition 2.3.3, (4) is isomorphic to (1). Therefore, there exists

φ : K → Aut(H) a homomorphism such that G ∼= K nφ H with operation

(k1, h1)(k2, h2) = (k1k2, h1φk1(h2))

However,

(1, h) = (k, 1)(1, h)(k−1, 1) = (k, φk(h))(k−1, 1) = (1, φk(h)φk(1)) = (1, φk(h))

and therefore φk(h) = h for all k and the semidirect product is in fact a direct product.

The converse follows immediately from the previous proposition because a direct

product is also a semidirect product.

A central extension satisfying the conditions of the previous corollary is also called

split. Our next objective is to establish a bijection between isomorphism classes of

central extensions and equivalence classes of maps which we call cocycle.

Definition 2.3.5. Suppose H and K are groups where H is abelian. Then a map

α : K ×K → H is called a cocycle if

1. α(k, 1) = α(1, k) = 1 for all k ∈ K

2. α(k1k2, k3)α(k1, k2) = α(k1, k2k3)α(k2, k3) for all k1, k2, k3 ∈ K

Proposition 2.3.6. Suppose H,K and α are as in definition 2.3.5. Then the set

G = K ×H with operation

(k1, h1)(k2, h2) = (k1k2, α(k1, k2)h1h2)

is a group. Furthermore, we obtain a central extension

1 // H i // Gj // K // 1

with i(h) = (1, h) and j(k, h) = k.


Proof. First, note that

α(kk−1, k)α(k, k−1) = α(k, k−1k)α(k−1, k)

hence

α(k, k−1) = α(k−1, k).

Second, we verify the axioms of a group.

1. (1, 1)(k, h) = (k, α(1, k)h) = (k, h) = (k, α(k, 1)h) = (k, h)(1, 1)

2. For all k1, k2, k3 ∈ K and h1, h2, h3 ∈ H,

((k1, h1)(k2, h2))(k3, h3) = (k1k2, α(k1, k2)h1h2)(k3, h3)

= (k1k2k3, α(k1k2, k3)α(k1, k2)h1h2h3)

= (k1k2k3, α(k1, k2k3)α(k2, k3)h1h2h3)

= (k1, h1)(k2k3, α(k2, k3)h2h3)

= (k1, h1)((k2, h2)(k3, h3))

3. (k, h)(k−1, α(k, k−1)−1h−1) = (kk−1, α(k, k−1)α(k, k−1)−1hh−1) = (1, 1). Fur-

thermore, we have (k−1, α(k, k−1)−1h−1) = (k−1, α(k−1, k)−1h−1). Therefore,

(k−1, α(k, k−1)−1h−1)(k, h) = (1, α(k−1, k)α(k−1, k)−1h−1h) = (1, 1)

We conclude that G is a group. It is easy to see that i injective group homomorphism.

Similarly, it is easy to see that j is a surjective group homomorphism. Furthermore,

(k, h) ∈ ker j if and only if k = 1 and it is clear that (1, h) ∈ Im i. Finally, for all

h, h′ ∈ H and k ∈ K,

(1, h′)(k, h) = (k, α(1, k)h′h) = (k, h′h) = (k, α(k, 1)hh′) = (k, h)(1, h′).

Therefore, i(H) ⊆ Z(G) and the extension is a central extension. This completes the

proof.

Definition 2.3.7. Suppose K,H are as above. Then a coboundary α is a map

α : K ×K → H satisfying

α(x, y) = s(xy)s(x)−1s(y)−1

for some map s : K → H.


Remark 2.3.8. It is easy to see that every coboundary is a cocycle.

The next two results allow us to classify all central extensions in terms of cocycles.

Lemma 2.3.9. For two cocycles α and β, define

α ∼ β if and only if αβ−1 is a coboundary.

Then ∼ is an equivalence relation.

Proof. 1. Clearly, αα−1 = 1, and 1 is obtained from s(x) = 1.

2. Suppose α ∼ β. Then αβ−1(x, y) = s(xy)s(x)−1s(y)−1. Therefore, by tak-

ing the inverse, we have βα−1(x, y) = (s(xy))−1s(x)s(y). Defining s−1(x) =

(s(x))−1, we obtain that β ∼ α.

3. Suppose α ∼ β and β ∼ γ. Then αβ−1(x, y) = s(xy)s(x)−1s(y)−1 and βγ−1(x, y) =

r(xy)r(x)−1r(y)−1. But then,

αγ−1(x, y) = s(xy)r(xy)s(x)−1r(x)−1s(y)−1r(y)−1

= sr(xy)(sr)−1(x)(sr)−1(y).

The following justifies why we study central extensions using cocycles and cobound-

aries.

Proposition 2.3.10. There is a bijection between isomorphism classes of central

extensions and equivalence classes of cocycles.

Proof. Firstly, suppose that

1 // Hi // G

j // K // 1

is a central extension. We prove that it is obtained from a cocycle.

The map j is surjective, hence using the axiom of choice we can choose a set-

theoretic map s : K → G such that js = 1K . Setting

s′(x) =

1 if x = 1

s(x) if x 6= 1


we can assume that s(1) = 1. We claim that G = i(H)s(K). Indeed, suppose g ∈ G.

Then j(gs(j(g)−1)) = j(g)j(g)−1 = 1, hence gs(j(g)−1) ∈ ker j = Im i. Therefore,

gs(j(g)−1) = i(h) for some h. Then g = i(h)s(j(g)), as claimed.

Now, consider the product s(k1)s(k2). Applying j, we obtain k1k2. However, we

also obtain k1k2 by applying j to s(k1k2). Therefore,

s(k1)s(k2)s(k1k2)−1 ∈ ker j = Im i.

We denote by ck1,k2 elements of H such that i(ck1,k2) = s(k1)s(k2)s(k1k2)−1 and define

α(k1, k2) = ck1,k2 . It is easy to see that α(1, k) = α(k, 1) = s(1) = 1. Suppose that

k1, k2, k3 ∈ K. Then we have

i(ck1,k2ck1k2,k3) = i(ck1,k2)i(ck1k2,k3)

= i(ck1,k2)s(k1k2)s(k3)s(k1k2k3)−1

= (s(k1)s(k2))s(k3)s(k1k2k3)−1

= s(k1)(s(k2)s(k3))s(k1k2k3)−1

= s(k1)i(ck2,k3)s(k2k3)s(k1k2k3)−1

= i(ck2,k3)i(ck1,k2k3)

= i(ck2,k3ck1,k2k3)

or equivalently from the fact that i is injective,

α(k1k2, k3)α(k1, k2) = α(k1, k2k3)α(k2, k3)

which is exactly the second cocycle condition.

Now consider G = K×H with operation given by the cocycle α as in Proposition

2.3.6. Then we claim that the map φ : G → G, (k, h) 7→ i(h)s(k) is a group isomor-

phism. Indeed, (1, 1) goes to 1. Furthermore φ is surjective because G = s(K)i(H),

as proven earlier. φ is also a group homomorphism. Indeed,

φ((k1, h1)(k2, h2)) = φ(k1k2, ck1k2h1h2) = i(ck1,k2h1h2)s(k1k2)

= i(h1)i(h2)s(k1)s(k2) = φ(k1, h1)φ(k2, h2).

The map φ is injective because if φ(k, h) = i(h)s(k) = 1, we have jφ(k, h) = k = 1

and then φ(1, h) = i(h) = 1 implies that h = 1 by the injectivity of i. We conclude


that G ∼= G and it is not difficult to check directly that φ is in fact an isomorphism

of central extensions. Indeed, denoting the maps in Proposition 2.3.6 by i′ and j′, we

have

φi′(h) = φ(1, h) = i1H(h)

and

jφ(k, h) = j(i(h)s(k)) = j(s(k)) = k = 1Kj′(k, h).

Secondly, suppose we have two isomorphic central extensions G,G′. Then we can

assume that G = K × H with product given by some cocycle α and G′ = K × Hwith product given by some cocycle β as in Proposition 2.3.6. We wish to show that

these cocycles differ by a coboundary. First, note that from the commutativity of the

diagram,

j′φ(k, 1) = 1Kj(k, 1) = k.

Therefore, φ(k, 1) = (k, φ′(k)) for some map φ′ : K → H. Furthermore, again by the

commutativity of the diagram,

φ(1, h) = φ(i(h)) = i′1H(h) = (1, h).

Therefore, φ(1, h) = (1, h) and then

φ(k, h) = φ(k, 1)φ(1, h) = (k, φ′(k))(1, h) = (k, φ′(k)h).

However,

φ((k1, h1)(k2, h2)) = φ(k1k2, α(k1, k2)h1h2) = (k1k2, φ′(k1k2)α(k1, k2)h1h2)

and

φ(k1, h1)φ(k2, h2) = (k1, φ′(k1)h1)(k2, φ

′(k2)h2) = (k1k2, φ′(k1)φ′(k2)β(k1, k2)h1h2).

Equating the two equations we obtain

α(k1, k2)β(k1, k2)−1 = φ′(k1)φ′(k2)φ′(k1k2)−1

and therefore αβ−1 is a coboundary.

Finally, we must show that two equivalent cocycles induce an isomorphism be-

tween their central extensions. Suppose α, β are cocycles such that αβ−1(k1, k2) =


s−1(k1k2)s(k1)s(k2) for some s : K → H. Let G1 = K ×H with the group structure

induced by α(k1, k2) and G2 = K ×H with the group structure induced by β(k1, k2).

Then it is easy to check that

φ(k, h) = (k, s(k)h)

is an isomorphism of central extensions.

Corollary 2.3.11. Suppose α is a coboundary. Then the central extension defined

by Proposition 2.3.6 is split.

Proof. If α is a coboundary, then it is equivalent to the trivial cocycle β(x, y) = 1.

Therefore, by Proposition 2.3.10, G is isomorphic (as central extension) to a direct

product, hence is split.

Chapter 3

The Weil Representation

For this chapter, we let F be a local field of odd or zero characteristic, and ψ : F→ C×

be a non-trivial continuous additive character of F. By an additive character, we mean

a group homomorphism (F,+) → C×. The main reference for this chapter will be

[14], from which we rewrite many of the proofs in full detail.

F is a local field, hence a locally compact group under addition, so there exists a

Haar measure on F [3, Theorem 2.10].

3.1 The symplectic group

Let V be a vector space over F (possibly infinite dimensional) and V ∗ its algebraic

dual. We denote (v∗, v) = (v, v∗) = v∗(v), for v ∈ V and v∗ ∈ V ∗. This pairing defines

a symmetric, non-degenerate bilinear form. Furthermore, the space X = V ⊕V ∗ also

has the following bilinear form : 〈v1 + v∗1, v2 + v∗2〉 = (v1, v∗2) − (v2, v

∗1). It is easy to

see that this form is antisymmetric and non-degenerate.

We denote by Sp(X) the isometry group of X with respect to the form 〈., .〉 and

by Sp2n(F) the classical symplectic group over F, i.e. the set of 2n × 2n matrices A

satisfying ATΩA = Ω where Ω =

[0 In

−In 0

]. If V happens to be finite-dimensional,

then we have Sp(X) ∼= Sp2n(F) where n = dimV .

We will denote the action of an element of Sp(X) on X on the right. If g ∈ Sp(X),

v ∈ V and v∗ ∈ V ∗, we have (v + v∗)g = vg + v∗g = vα + vβ + v∗γ + v∗δ, where

18

CHAPTER 3. THE WEIL REPRESENTATION 19

vα ∈ V, vβ ∈ V ∗, v∗γ ∈ V, v∗δ ∈ V ∗. Then we can think of g as a 2× 2 matrix :[α β

γ δ

]

where α : V → V , β : V → V ∗, γ : V ∗ → V , δ : V ∗ → V ∗, and the action of g on

v + v∗ ∈ X can be seen as right multiplication of this matrix on the vector[v v∗

].

Definition 3.1.1. Suppose g =

[α β

γ δ

]as above. Then α∗ : V ∗ → V ∗ is a linear map

such that (v1α, v∗2) = (v1, v

∗2α∗), β∗ : V → V ∗ is a linear map such that (v1β, v2) =

(v1, v2β∗), γ∗ : V ∗ → V is a linear map such that (v∗1γ, v

∗2) = (v∗1, v

∗2γ∗), and δ∗ : V →

V is a linear map such that (v∗1δ, v∗2) = (v∗1, v

∗2δ∗). It also follows directly from the

definition that the same properties apply to the form 〈., .〉.

The above definition is valid because of Lemma 3.1.2.

Lemma 3.1.2. α∗, β∗, γ∗, δ∗ exist and are unique.

Proof. We start with uniqueness. The four proofs follow the exact same idea, so we

only do the proof for α∗. Suppose v1, v2 ∈ V and that α′∗ is another map satisfying

(v1α, v∗2) = (v1, v

∗2α′∗). Then

0 = (v1α, v∗2)− (v1α, v

∗2)

= (v1, v∗2α∗)− (v1, v

∗2α′∗)

= (v1, v∗2(α∗ − α′∗)).

But the form (., .) is non-degenerate, hence we must have v∗2(α∗ − α′) = 0 for all v∗2

and therefore α∗ = α′.

The existence of α and β are well-known facts. By definition of Sp(X), we know

that for all v1, v2 ∈ V

0 = 〈v1, v2〉 = 〈v1g, v2g〉 = 〈v1α + v1β, v2α + v2β〉

= 〈v1β, v2α〉+ 〈v1α, v2β〉 = −(v1β, v2α) + (v1α, v2β)

= −(v1βα∗, v2) + (v1αβ

∗, v2) = (v1(−βα∗ + αβ∗), v2).


By non-degeneracy of (., .), we obtain

βα∗ = αβ∗ (5)

We also have

(v, v∗) = 〈v, v∗〉 = 〈vg, v∗g〉 = 〈vα + vβ, v∗γ + v∗δ〉 = −(vβ, v∗γ) + (vα, v∗δ)

= (v,−v∗γβ∗) + (v, v∗δα∗) = (v, (v∗)(δα∗ − γβ∗)).

Therefore, we obtain

δα∗ − γβ∗ = 1V ∗ . (6)

Where 1V ∗ is the identity on V ∗. In particular, this means that we have,[α β

γ δ

][−β∗

α∗

]=

[0

1V ∗

]

Now write

g−1 =

[α′ β′

γ′ δ′

].

Then we also have, by replacing the role of g with g−1[α′ β′

γ′ δ′

][−β′∗

α′∗

]=

[0

1V ∗

].

Multiplying this last equation by g on the left yields[−β′∗

α′∗

]=

[β

δ

].

Then δ = α′∗ and we conclude that δ∗ = α′. This shows that δ∗ exists. Now note

that for all v∗1, v∗2 ∈ V ∗,

0 = 〈v∗1, v∗2〉 = 〈v∗1g, v∗2g〉 = 〈v∗1γ + v∗1δ, v∗2γ + v∗2δ〉

= −(v∗1δ, v∗2γ) + (v∗1γ, v

∗2δ)

= −(v∗1, v∗2γδ

∗) + (v∗1γδ∗, v∗2).

Therefore γδ∗ is self-adjoint. Also, we know that γβ∗ = δα∗ − 1V ∗ from Eq. (6), and

we can check that (γβ∗)∗ = αδ∗ − 1V . Indeed,


(v∗γβ∗, v) = (v∗(δα∗ − 1V ), v) = (v∗δα∗, v)− (v∗, v) = (v∗δ, vα)− (v∗, v)

= (v∗, vαδ∗)− (v∗, v) = (v∗, v(αδ∗ − 1V ))

as desired.

For each v∗ ∈ V ∗, note that by matrix multiplication of g−1 with g, we have

v∗(γ′β + δ′δ) = v∗. Now, define v∗γ∗ = v∗(γ′(γβ∗)∗) + v∗(δ′γδ∗). This is obviously

linear in V ∗ and by combining the last few facts, we obtain

(w∗γ, v∗) = (w∗γ, v∗γ′β + v∗δ′δ)

= (w∗γβ∗, v∗γ′) + (w∗γδ∗, v∗δ′)

= (w∗, v∗γ′(γβ∗)∗) + (w∗, v∗δ′γδ∗)

= (w∗, v∗(γ′(γβ∗)∗ + δ′γδ∗)).

Therefore, γ∗ exists, which completes the proof. We also obtain the equation

γδ∗ = δγ∗ (7)

because γδ∗ is self-adjoint.

Lemma 3.1.3. The following relations hold :

αβ∗ = βα∗, γδ∗ = δγ∗, δα∗ − γβ∗ = 1V ∗

Proof. These relations are Eq. (5), (7) and (6) in Lemma 3.1.2.

Corollary 3.1.4.

g−1 =

[δ∗ −β∗

−γ∗ α∗

]Proof. We multiply the two matrices together

gg−1 =

[α β

γ δ

][δ∗ −β∗

−γ∗ α∗

]=

[αδ∗ − βγ∗ −αβ∗ + βα∗

γδ∗ − δγ∗ −γβ∗ + δα∗

]=

[1V 0

0 1V ∗

].

where aδ∗ − βγ∗ = 1V from taking the transpose on each side of δα∗ − γβ∗ = 1∗V .

Therefore, g−1 is a right inverse for g. However, Sp(X) is a group, hence a right

inverse is also a left inverse, which completes the proof.


Proposition 3.1.5. H = X × F with operations

(x1, k1)(x2, k2) = (x1 + x2,1

2〈x1, x2〉+ k1 + k2)

and identity (0, 0) is a group.

Remark 3.1.6. The group H is known as Heisenberg group.

Proof. Associativity can be verified by a direct computation. Furthermore, suppose

(x1, k1) ∈ H. Then, we have (0, 0)(x1, k1) = (0 + x1,12〈0, x1〉+ 0 + k1) = (x1, k1) and

similarly, (x1, k1)(0, 0) = (x1, k1). We also have (x1, k1)(−x1,−k1) = (0, 〈x1,−x1〉) =

(0, 0) where the last equality follows from antisymmetry of 〈., .〉 and the fact that

charF 6= 2.

We obtain an action of Sp(X) on H by (x, k) · g = (xg, k). Indeed, suppose

g1, g2 ∈ Sp(X). Then

(x, k) · (g1g2) = (xg1g2, k) = (xg2, k) · g2 = (x, k) · g1 · g2

as desired.

Proposition 3.1.7. Any element (v+v∗, k) of H can be written uniquely as a product

(0, k′)(v, 0)(v∗, 0) where k′ = −12〈v, v∗〉+ k. The group H acts on S(V ), where S(V )

was defined in Section 2.2 for an arbitrary local field, by

1. (k · f)(x) = ψ(k)f(x)

2. (v · f)(x) = f(x+ v)

3. (v∗ · f)(x) = ψ(〈x, v∗〉)f(x)

Proof. First, we prove that every element of H can be written as a product of elements

of V , V ∗ and F in a unique way. Indeed, suppose (v + v∗, k) ∈ H. Then note that

(0,−1

2〈v, v∗〉+ k)(v, 0)(v∗, 0) = (v,−1

2〈v, v∗〉+ k)(v∗, 0) = (v + v∗, k).

Now, suppose we can write (v+v∗, k) as a product of (v′, 0), (v′∗, 0) and (0, k′). Then

(0, k′)(v′, 0)(v′∗, 0) = (v′, k′)(v′∗, 0) = (v′ + v′∗,1

2〈v′, v′∗〉+ k′).


Therefore, v′ = v, v′∗ = v∗ and k′ = k − 12〈v, v∗〉.

Define the map p : H → End(S(V )) by p(v + v∗, k)f = k′ · (v · (v∗ · f)). It can be

seen that p(h) is a composition of three linear maps each preserving S(V ) (the only

non-obvious part is to prove that v∗ · f ∈ S(V ). If F is non-archimedean, then the

restriction of v∗ · f to any finite-dimensional space still has compact support and is a

product of locally constant functions. If F = R or C however, it is an easy exercise),

hence p(h) is a linear map preserving S(V ). Therefore, p is well-defined. We still need

to show that p is a group homomorphism, but this can be done directly by computing

the actions of elements (v1 + v∗1, k1), (v2 + v∗2, k2) ∈ H on an arbitrary function f and

then compare it to the action of their product

(v1 + v2 + v∗1 + v∗2,1

2〈v1 + v∗1, v2 + v∗2〉k1 + k2)

on f again. In both cases, one will obtain as final result

ψ(〈x, v∗1 + v∗2〉+ 〈v1, v∗2〉+

1

2〈v1, v

∗1〉+

1

2〈v2, v

∗2〉+ k1 + k2)f(x+ v1 + v2),

hence p is a group homomorphism.

3.2 The Weil representation

One would naturally expect that one can define the Weil representation over Sp(X).

However, this appears to be impossible, hence we shall restrict ourselves to defining

it over a subgroup.

Let Sp(X, V ∗) be the subset of Sp(X) containing the elements g =

[α β

γ δ

]satis-

fying dim Im γ <∞. We will see that Sp(X, V ∗) is indeed a subgroup of Sp(X), but

we first prove a technical lemma.

Lemma 3.2.1. Suppose γ : V ∗ → V and n = dim Im γ <∞. Then dim Im γ∗ <∞.

Proof. Suppose by contradiction that dim Im γ∗ > dim Im γ. Then let w1, ..., wn+1 be

linearly independent elements of Im γ∗ and let v∗1, ..., v∗n+1 be elements of V ∗ such that

v∗i γ∗ = wi. Extend the wi’s to a basis wii∈N of V and choose f ∗1 , ..., f

∗n, f

∗n+1 to be

elements of V ∗ such that (f ∗i , wj) = δi,j. This can be done by simply defining f ∗i to


be 1 on wi and 0 on wj for j 6= i, and then extending f ∗i by linearity. Now, suppose

a1γ(f ∗1 ) + ...+ an+1γ(f ∗n+1) = 0. Then

0 = (a1f∗1γ + · · ·+ an+1f

∗n+1γ, v

∗i ) = (a1f

∗1 + · · ·+ anf

∗n+1, v

∗i γ∗)

= (a1f∗1 + · · ·+ anf

∗n+1, wi) = ai.

Therefore, the f ∗i γ’s form a linearly independent set, a contradiction. We conclude

that dim Im γ∗ ≤ dim Im γ <∞.

Proposition 3.2.2. Sp(X, V ∗) is a subgroup of Sp(X).

Proof. We need to show that Sp(X, V ∗) is closed under multiplication and inversion.

Write g1 =

[α1 β1

γ1 δ1

]and g2 =

[α2 β2

γ2 δ2

]∈ Sp(X, V ∗). Then we have the product

g1g2 =

[α1α2 + β1γ2 α1β2 + β1δ2

γ1α2 + δ1γ2 γ1β2 + δ1δ2

].

But now,

dim Im(γ1α2 + δ1γ2) ≤ dim Im(γ1α2) + dim Im(δ1γ2) ≤ dim Im γ1 + dim Im γ2 <∞

as desired.

For inversion, write g =

[α β

γ δ

]. Then g−1 =

[δ∗ −β∗

−γ∗ α∗

]by Corollary 3.1.4.

But now, by Lemma 3.2.1, we know that dim Im−γ∗ < ∞ and therefore g−1 ∈Sp(X, V ∗). Therefore Sp(X, V ∗) is indeed a subgroup of Sp(X).

Lemma 3.2.3. Suppose g =

[α β

γ δ

]∈ Sp(X) and x∗1, x

∗2 ∈ V ∗. If x∗1γ = x∗2γ, then

〈x∗1γ, x∗1δ〉 = 〈x∗2γ, x∗2δ〉.

Proof. Suppose x∗1γ = x∗2γ. Then using Lemma 3.1.3, we obtain

〈x∗1γ, x∗1δ〉 = 〈x∗2γ, x∗1δ〉 = 〈x∗2, x∗1δγ∗〉 = 〈x∗2, x∗1γδ∗〉

= 〈x∗2, x∗2γδ∗〉 = 〈x∗2, x∗2δγ∗〉 = 〈x∗2γ, x∗2δ〉


The following family of operators, parametrized by g ∈ Sp(X), will yield theWeil

representation upon restriction to a suitable subgroup.

Proposition 3.2.4. Suppose g =

[α β

γ δ

]∈ Sp(X, V ∗). Fix a Haar measure for

Im γ. Then define

(Tgf)(x) =

∫Vg

Sg(x+ x∗)f(xα + x∗γ)d(x∗γ)

where Vg = Im γ and

Sg(x+ x∗) = ψ(1

2〈xα, xβ〉+

1

2〈x∗γ, x∗δ〉+ 〈x∗γ, xβ〉).

Then h · (Tgf) = Tg((h · g) · f) for all f ∈ S(V ) and h ∈ H, where h · g is the action

of Sp(X) defined after Remark 3.1.6.

Remark : We can choose a Haar measure on Im γ because dim Im γ < ∞. Fur-

thermore, by Lemma 3.2.3, Sg(x+ x∗) is a function of x and x∗γ only, and therefore

the integration makes sense. We also have

|(Tgf)(x)| ≤∫Vg

|f(xα + x∗γ)| d(x∗γ) <∞

because ((xα, 0) · f)|Vg ∈ S(Vg). We will prove in Theorem 3.2.16 that Tgf ∈ S(V ),

but for now we will be content with the fact that Tgf is a well-defined function.

Proof. To check that h · (Tgf) = Tg((h · g) · f), it is enough to verify equality in the

three cases where h = (0, k) for k ∈ F, h = (v, 0) for v ∈ V and h = (v∗, 0) for

v∗ ∈ V ∗. This follows from the fact that (h1 · g)(h2 · g) = (h1h2) · g for all h1, h2 ∈ H.

Indeed, assuming the statement holds in these three cases, we have

(0, k) · ((v∗, 0) · ((v, 0) · Tgf)) = (0, k) · ((v∗, 0) · (Tg(((v, 0) · g) · f)))

= (0, k) · Tg(((v∗, 0) · g) · ((v, 0) · g)) · f))

= (0, k) · Tg(((v∗, 0)(v, 0) · g) · f)

= Tg(((0, k) · g) · (((v∗, 0)(v, 0) · g) · f))

= Tg(((0, k)(v∗, 0)(v, 0) · g) · f)


The case h = (0, k) is easy to check. Suppose that h = (v, 0) for v ∈ V . Then

h · (Tgf)(x) = (Tgf)(x+ v)

=

∫Vg

Sg(x+ v + x∗)f((x+ v)α + x∗γ)d(x∗γ).

However, we also have

Tg(((v, 0) · g) · f)(x) = Tg((vα + vβ, 0) · f)(x)

=

∫Vg

Sg(x+ x∗)(vα + vβ · f)(xα + x∗γ)d(x∗γ)

=

∫Vg

ψ(〈(xα + x∗γ +1

2vα, vβ〉)Sg(x+ x∗)f((x+ v)α + x∗γ)d(x∗γ)

where the last line follows from (vα+vβ, 0) = (0,−12〈vα, vβ〉)(vα, 0)(vβ, 0). However,

ψ(〈(x+1

2v)α + x∗γ, vβ〉)Sg(x+ x∗)

= ψ(〈(x+1

2v)α + x∗γ, vβ〉)ψ(

1

2〈xα, xβ〉+

1

2〈x∗γ, x∗δ〉+ 〈x∗γ, xβ〉)

= ψ(〈(x+1

2v)α, vβ〉+

1

2〈xα, xβ〉+

1

2〈x∗γ, x∗δ〉+ 〈x∗γ, (x+ v)β〉)

= Sg(x+ v + x∗)ψ(1

2(〈xα, vβ〉 − 〈vα, xβ〉))

= Sg(x+ v + x∗)ψ(1

2(〈xα, vβ〉+ 〈xβ, vα〉))

= Sg(x+ v + x∗)ψ(1

2(〈xα + xβ, vβ〉+ 〈xα + xβ, vα〉))

= Sg(x+ v + x∗)ψ(1

2(〈xα + xβ, vα + vβ〉))

= Sg(x+ v + x∗)ψ(1

2(〈xg, vg〉)) = Sg(x+ v + x∗)ψ(

1

2〈x, v〉) = Sg(x+ v + x∗).

We conclude that h · (Tgf)(x) = Tg((h · g) · f)(x). Next, suppose h = (v∗, 0) for

v∗ ∈ V ∗. Then

h · (Tgf)(x) = ψ(〈x, v∗〉)(Tgf)(x)

and h · g = (v∗γ + v∗δ, 0) = (0,−12〈v∗γ, v∗δ〉)(v∗γ, 0)(v∗δ, 0), hence

Tg((v∗, 0)g · f)(x)

=

∫Vg

Sg(x+ x∗)((v∗g, 0) · f)(xα + x∗γ)d(x∗γ)

=

∫Vg

Sg(x+ x∗)ψ(〈xα + x∗γ, v∗δ〉+1

2〈v∗γ, v∗δ〉)f(xα + x∗γ + v∗γ)d(x∗γ)


The Haar measure is translation invariant, hence we can replace x∗ by x∗ − v∗ to

obtain

Tg(((v∗, 0) · g) · f)(x) =

∫Vg

Sg(x+ x∗ − v∗)ψ(〈xα + x∗γ − 1

2v∗γ, v∗δ〉)f(xα + x∗γ)d(x∗γ)

However,

Sg(x+ x∗ − v∗)ψ(〈xα + x∗γ − 1

2v∗γ, v∗δ〉)

= ψ(1

2〈xα, xβ〉+

1

2〈x∗γ − v∗γ, x∗δ − v∗δ〉+ 〈x∗γ − v∗γ, xβ〉+ 〈xα + x∗γ − 1

2v∗γ, v∗δ〉)

= Sg(x+ x∗)ψ(−1

2〈v∗γ, x∗δ〉 − 1

2〈x∗γ, v∗δ〉 − 〈v∗γ, xβ〉+ 〈xα + x∗γ, v∗δ〉)

= Sg(x+ x∗)ψ(〈xα + xβ, v∗γ + v∗δ〉)ψ(1

2(〈x∗γ, v∗δ〉+ 〈x∗δ, v∗γ〉))

= Sg(x+ x∗)ψ(〈xα + xβ, v∗γ + v∗δ〉)ψ(1

2(〈x∗γ + x∗δ, v∗δ〉+ 〈x∗γ + x∗δ, v∗γ〉))

= Sg(x+ x∗)ψ(〈xα + xβ, v∗γ + v∗δ〉)ψ(1

2(〈x∗γ + x∗δ, v∗γ + v∗δ〉))

= Sg(x+ x∗)ψ(〈xα + xβ, v∗γ + v∗δ〉)ψ(1

2(〈x∗g, v∗g〉))

= Sg(x+ x∗)ψ(〈xα + xβ, v∗γ + v∗δ〉) = Sg(x+ x∗)ψ(〈x, v∗〉)

which proves that (v∗, 0)·(Tgf)(x) = Tg(((v∗, 0)·g)·f)(x) and completes the proof.

Our goal for the next few pages is to show that Tg is in fact a projective repre-

sentation of Sp(X, V ∗) on S(V ). We will need many lemmas before we can get there.

The general strategy is to reduce to the Weil representation in the finite-dimensional

case, which is known to be a projective representation [7, Theorem 4.1 (5)].

Lemma 3.2.5. For v = x + x∗ ∈ X, define Q(v) = Q(x + x∗) = ψ( 〈x,x∗〉

2). Then we

have

Sg(x+ x∗) = Q((x+ x∗)g)Q(x+ x∗)−1,

and

Sg1g2(x+ x∗) = Sg1(x+ x∗)Sg2((x+ x∗)g1).


Proof. We prove the result with a direct computation.

Q((x+ x∗)g)Q(x+ x∗)−1 = ψ

(〈xα + x∗γ, xβ + x∗δ〉

2

)ψ

(−〈x, x

∗〉2

)= ψ

(〈xα + x∗γ, xβ + x∗δ〉

2− 〈xg, x

∗g〉2

)= ψ

(〈xα + x∗γ, xβ + x∗δ〉

2− 〈xα + xβ, x∗γ + x∗δ〉

2

)= ψ

(〈xα + x∗γ, xβ + x∗δ〉

2− 〈xα, x

∗δ〉2

− 〈xβ, x∗γ〉

2

)= ψ

(〈xα, xβ〉

2+〈x∗γ, x∗δ〉

2+ 〈x∗γ, xβ〉

)= Sg(x+ x∗).

For the second claim, we have

Sg1g2(x+ x∗) = Q((x+ x∗)g1g2)Q(x+ x∗)−1

= Q(((x+ x∗)g1)g2)Q((x+ x∗)g1)−1Q((x+ x∗)g1)Q(x+ x∗)−1

= Sg2((x+ x∗)g1)Sg1(x+ x∗) = Sg1(x+ x∗)Sg2((x+ x∗)g1).

This completes the proof.

Set P = g ∈ Sp(X) : γ = 0.

Proposition 3.2.6. P is a subgroup of Sp(X, V ∗).

Proof. It is clear that P is a subset of Sp(X, V ∗). We must show that it is closed

under multiplication and inversion. Suppose p1 =

[α1 β1

0 δ1

]and p2 =

[α2 β2

0 δ2

]are

in P . Then we have

p1p2 =

[α1α2 α1β2 + β1δ2

0 δ1δ2

]

which is clearly in P . Next, write p =

[α β

0 δ

]. From Corollary 3.1.4, we have

p−1 =

[δ∗ −β∗

−γ∗ α∗

].


Since γ = 0, we see that γ∗ = 0, hence

p−1 =

[δ∗ −β∗

0 α∗

]which is clearly in P .

Lemma 3.2.7. Suppose p =

[α β

0 δ

]∈ P . Then α and δ are invertible.

Proof. Write

p−1 =

[δ∗ −β∗

0 α∗

]Then

pp−1 =

[αδ∗ −αβ∗ + βα∗

0 δα∗

]=

[1V 0

0 1V ∗

]Then α∗ is a right inverse of δ. By computing p−1p, we can easily find that α∗ is also

a left inverse of δ. Therefore, δ is invertible. Similarly, α is the inverse of δ∗.

Lemma 3.2.8. Suppose

g =

[α1 β1

γ1 δ1

]∈ Sp(X, V ∗) and p =

[α2 β2

0 δ2

]∈ P

Then

1. Vg = Vpg. Furthermore, if we choose the same Haar measure in Vg and Vpg and

the counting measure for Vp = 0, then TpTg = Tpg.

2. Vgα2 = Vgp. Furthermore, the Haar measures can be chosen so that TgTp = Tgp.

Proof. First, note that

pg =

[α3 β3

γ3 δ3

]=

[α2α1 + β2γ1 α2β1 + β2δ1

δ2γ1 δ2δ1

].

By Lemma 3.2.7, we know that V ∗δ2 = V ∗ , hence V ∗δ2γ1 = V ∗γ1. Therefore,

Vg = Vpg. Now, we have

(TpTgf)(x) = Sp(x)(Tgf)(xα2)

=

∫Vg

Sp(x)Sg(xα2 + x∗)f(xα2α1 + x∗γ1)d(x∗γ1).


Because the Haar measure is translation invariant, we can replace x∗ by x∗ + xβ2

to obtain

(TpTgf)(x) =

∫Vg

Sp(x)Sg(xα2 + xβ2 + x∗)f(xα2α1 + x∗γ1 + xβ2γ1)d(x∗γ1). (8)

By Lemma 3.2.5, and Lemma 3.2.7 we have

Spg(x+ x∗δ−12 ) = Sg((x+ x∗δ−1

2 )p)Sp(x+ x∗δ−12 )

= Sg(xα2 + xβ2 + x∗)Sp(x+ x∗δ−12 )

= Sg(xα2 + xβ2 + x∗)ψ(1

2〈xα, xβ〉+

1

2〈0x∗, x∗〉+ 〈0x∗, xβ〉)

= Sg(xα2 + xβ2 + x∗)Sp(x) = Sp(x)Sg(xα2 + xβ2 + x∗).

Furthermore, γ3 = δ2γ1 implies that γ1 = δ−12 γ3. Then equation 8 becomes

(TpTgf)(x) =

∫Vpg

Spg(x+ x∗δ−12 )f(xα3 + x∗δ−1

2 γ3)d(x∗δ−12 γ3)

which is exactly (Tpgf)(x) with integration variable x∗δ−12 instead of x∗.

The second statement is verified very similarly. However, we would like to make it

clear how we choose the Haar measures. If v1, ..., vn is a basis for Vg, then v1α2, ..., vnα2

is also a basis for Vgα2 because α2 is invertible. Then if we take the product measure

dv1...dvn, we obtain a Haar measure for Vg. But then, d(v1α2)...d(vnα2) is also a Haar

measure for Vpg.

The following is an immediate corollary to Lemma 3.2.9 :

Corollary 3.2.9. Suppose g and p are given as in the previous lemma. Then we have

for any choice of Haar measures

1. TpTg ∼ Tpg

2. TgTp ∼ Tgp

where a ∼ b means that a = cb for some c ∈ F.

Now, suppose g ∈ Sp(X). We let Xg = v ∈ X : vg = v. It is clear that Xg is a

subspace of X. Then let Spfin(X) = g ∈ Sp(X) : Xg has finite co-dimension in X.


Lemma 3.2.10. Suppose g ∈ Sp(X) and that U ⊆ X is a subspace of X of dimension

at least k such that for all v ∈ U \ 0, vg 6= v. Then, Xg has co-dimension in X at

least k.

Proof. We wish to show that dimX/Xg ≥ k. Let v1, ..., vk be a linearly independent

subset of U . Suppose that

a1(v1 +Xg) + a2(v2 +Xg) + ...+ ak(vk +Xg) = 0.

Then a1v1 + a2v2 + ... + akvk ∈ Xg, hence a1v1 + a2v2 + ... + akvk = 0. Therefore,

a1 = a2 = ... = ak = 0 by linear independence and v1 + Xg, ..., vk + Xg are linearly

independent.

Lemma 3.2.11. Spfin(X) is a subgroup of Sp(X, V ∗).

Proof. First, we must prove Spfin(X) is a subset of Sp(X, V ∗). Suppose g ∈ Spfin(X)

and as usual write

g =

[α β

γ δ

].

We need to show that dim Im γ <∞. Suppose by contradiction that dim Im γ =∞.

Then we can choose infinitely many linearly independent elements f ∗1 , f∗2 , ... of V ∗

such that f ∗i γ are linearly independent. In particular, U = spanf ∗1 , f ∗2 , ... satisfies

the property that if v ∈ U \ 0, then vg 6= v, hence Xg has infinite co-dimension in

X.

Next, we wish to prove that Spfin(X, V ∗) is a subgroup by the subgroup test.

Suppose g1, g2 ∈ Spfin(X). Now, it is clear that Xg1 ∩Xg2 ⊆ Xg1g2 . Then

codim(Xg1g2) ≤ codim(Xg1 ∩Xg2) ≤ codim(Xg1) + codim(Xg2) <∞.

Therefore, Xg1g2 has finite co-dimension as well. Now, suppose g ∈ Spfin(X). We

claim that Xg = Xg−1. Indeed, if vg = v, then v = vg−1 by applying g−1 on each

side. But then Xg ⊆ Xg−1and the other inclusion follows by substituting g by g−1

into Xg ⊆ Xg−1.

The following lemma is a standard exercise in linear algebra and will be useful in

the next two lemmas.


Lemma 3.2.12. Suppose f1, ..., fn are linearly independent elements of V ∗. Then

there exists x1, ..., xn ∈ V such that 〈fi, xj〉δi,j.

Proof. This can be found in [2, p. 53, Exercise 9]. The proof is also available at the

end of the book.

For an arbitrary subspace U of V ∗, define U⊥ = v ∈ V : 〈v, U〉 = 0. Similarly,

for a subspace U of V , define U⊥ = v ∈ V ∗ : 〈v, U〉 = 0.

Definition 3.2.13. We say that a subspace W of X is a Lagrangian subspace if for

all v ∈ X, 〈v,W 〉 = 0 if and only if v ∈ W .

Lemma 3.2.14. Suppose g ∈ Sp(X, V ∗). Then g = pg′, for p ∈ P and g′ ∈ Spfin(X).

Proof. Denote by Gr(X, V ∗) the set of all Lagrangian subspaces W of X such that

W ∩ V ∗ has finite co-dimension in both W and V ∗. We will first prove that Spfin(X)

acts on Gr(X, V ∗) transitively. Write πV : X → V , πV ∗ : X → V ∗ as the canonical

projection operators with respect to the direct sum decomposition X = V ⊕ V ∗ and

suppose W ∈ Gr(X, V ∗).

Step 1 : We claim that there exists a subspace K ⊆ X such that W = W ∩V ∗⊕Kand

(πV ∗K) ∩ (W ∩ V ∗) = 0. (9)

To prove this, first choose finite-dimensional subspaces of L ⊆ W and J ⊆ V ∗ such

that W = W ∩ V ∗ ⊕L and V ∗ = W ∩ V ∗ ⊕ J . Denote by πW∩V ∗ : V ∗ → W ∩ V ∗ the

canonical projection with respect to the decomposition V ∗ = W ∩ V ∗ ⊕ J . Now, set

K = l − πW∩V ∗πV ∗(l) : l ∈ L.

We claim that K satisfies the required properties. First, suppose v ∈ W ∩ V ∗ and

v ∈ K. Then v = l−πW∩V ∗πV ∗(l) for some l. But then, v ∈ W ∩V ∗, hence v = πV ∗(v)

and v = πW∩V ∗πV ∗(v) and therefore

v = πW∩V ∗πV ∗(v) = πW∩V ∗πV ∗(l)− πW∩V ∗πV ∗(l) = 0.

Now, suppose v ∈ W . Then v = w + l, w ∈ W ∩ V ∗, l ∈ L. But then, we also have

v = (w + πW∩V ∗πV ∗(l)) + (l − πW∩V ∗πV ∗(l)) ∈ W ∩ V ∗ +K.


Therefore, W = W ∩ V ∗ ⊕K. It remains to verify Eq.(9). To do this, suppose that

v ∈ (πV ∗K) ∩ (W ∩ V ∗). Then v = πV ∗(l − πW∩V ∗πV ∗(l)) for some l. But then,

v ∈ W ∩ V ∗, hence πW∩V ∗v = v. Therefore, we have

v = πW∩V ∗(v) = πW∩V ∗πV ∗(l)− πW∩V ∗πV ∗(l) = 0,

as desired.

Step 2 : We wish to show that (W ∩ V ∗)⊥ = πV (W ). Firstly, suppose v ∈ πV (W ).

Then v = πV (u) for some u ∈ W . Write u = v + v′, where v ∈ V, v′ ∈ V ∗. Now

suppose that w ∈ W ∩ V ∗. Then because W is Lagrangian, we must have that

〈w, u〉 = 0. Therefore,

0 = 〈w, u〉 = 〈w, v + v′〉 = 〈w, v〉

hence v = πV (u) ∈ (W ∩ V ∗)⊥.

Secondly, suppose that v ∈ (W ∩ V ∗)⊥. Choose a basis k1, ..., ks of K (recall that

K has finite dimension). For each i, write ki = vi + v∗i , for vi ∈ V, v∗i ∈ V ∗. Now,

choose w∗1, ..., w∗s ∈ V ∗ such that (w∗i , vj) = δi,j as in the proof of Lemma 3.2.1. We

claim that v +s∑i=1

〈v∗i , v〉w∗i is in W . Indeed, if w ∈ W ∩ V ∗, we have

〈v +s∑i=1

(v∗i , v)w∗i , w〉 = 0

because v ∈ (W ∩ V ∗)⊥ and w∗i ∈ V ∗. If w ∈ K, write w =s∑j=1

ajkj. Then

〈v +s∑i=1

(v∗i , v)w∗i ,s∑j=1

ajkj〉 = 〈v,s∑j=1

ajkj〉 − 〈s∑j=1

ajkj,

s∑i=1

(v∗i , v)w∗i 〉

= (v,s∑j=1

ajv∗j )− (

s∑j=1

ajvj,s∑i=1

(v∗i , v)w∗i )

=s∑j=1

aj(v∗j , v)−

s∑i,j=1

(v∗i , v)(w∗i , ajvj)

=s∑j=1

aj(v∗j , v)−

s∑j=1

aj(v∗j , v) = 0.


Therefore, 〈v +∑i

(v∗i , v)w∗i , w〉 = 0 for all w ∈ W , hence, v +∑

(v∗i , v)w∗i ∈ W

(because W is Lagrangian) and consequently v = πV (v +∑

(v∗i , v)w∗i ) ∈ πV (W ), as

claimed.

Step 3 : We extend πV ∗(K) to a finite-dimensional subspace M of V ∗ so that

V ∗ = W ∩ V ∗ ⊕M . We can do this because W ∩ V ∗ has finite co-dimension in V ∗

and because of Eq. (9). We claim that V = (W ∩ V ∗)⊥ ⊕M⊥.

Firstly, suppose that v ∈ (W∩V ∗)⊥∩M⊥. Then 〈v, u〉 = 0 for all u ∈ W∩V ∗+M .

But then, (u, v) = 0 for all u ∈ V ∗, hence v = 0.

Secondly, we show that V = (W ∩ V ∗)⊥ +M⊥. We do so by dimension counting.

Supppose M = spanv∗1, ..., v∗k. Consider the map φ : V → Fk defined by

φ(v) = (v∗1(v), ..., v∗k(v)).

The kernel of this map is precisely M⊥ by definition. Furthermore, φ is surjective

because of Lemma 3.2.12. Therefore, dimV/M⊥ = k = dimM and we conclude that

M⊥ has co-dimension k. We wish to show that dim(W ∩ V ∗)⊥ = k. The dimension

of (W ∩ V ∗)⊥ is at most k because otherwise the sum (W ∩ V ∗)⊥ ∩M⊥ 6= 0. Now,

suppose dim(W ∩ V ∗)⊥ < k and let v1, ..., vs be a basis for (W ∩ V ∗)⊥. Consider

the map, ψ : V ∗ → Fs, defined by ψ(f) = (f(v1), ..., f(vs)). ψ is surjective because

v1, ..., vs are linearly independent. Furthermore, it is easy to see that W ∩V ∗ ⊆ kerψ.

We claim that we actually have equality.

Suppose h ∈ ker g. Then 〈h, v〉 = 0 for all v ∈ (W ∩ V ∗)⊥ = πV (W ) by Step

2. Now suppose v ∈ W . Then write v = u + u∗, u ∈ πV (W ), u∗ ∈ πV ∗(W ). Then

〈h, v + v∗〉 = 〈h, v〉 = 0 by assumption. Therefore, h ∈ W because W is Lagrangian,

and then h ∈ W ∩ V ∗, as desired. Therefore, kerψ = W ∩ V ∗ and this shows that

s = dimV ∗/ kerψ = dimV ∗/(W ∩ V ∗) = k, hence k = s.

Combining the two steps, we conclude that V = (W ∩ V ∗)⊥ ⊕M⊥.

Step 4 : The last three steps prove that

X = (W ∩ V ∗ ⊕M⊥)⊕ (M ⊕ πV (W ))

We claim that Spfin(X) acts on Gr(X, V ∗) transitively.


It is easy to see that M ⊕πV (W ) is a finite dimensional symplectic space with the

restriction of the bilinear form 〈., .〉. (The restriction is clearly antisymmetric and non-

degeneracy follows from knowing that M⊥ ∩ πV (W ) = 0 and W ∩ V ∗ ∩M = 0).In particular, both M and πV (W ) are isotropic subspaces of dimension k, hence are

Lagrangian subspaces of M ⊕ πV (W ). Also, K ⊆M ⊕ πV (W ) because πV ∗(K) ⊆M ,

and πV (K) = πV (W ). Since K ⊆ W , we have that K is isotropic. In fact, we claim

that K is Lagrangian in M ⊕ πV (W ).

Indeed, suppose that a ∈ M + πV (W ) such that 〈a,K〉 = 0. Since both M ⊆ V ∗

and W ∩ V ∗, we see that 〈M,W ∩ V ∗〉 = 0. Furthermore,

〈πV (W ),W ∩ V ∗〉 = 〈(W ∩ V ∗)⊥,W ∩ V ∗〉 = 0

Then we have

〈a,K +W ∩ V ∗〉 = 〈a,W ∩ V ∗〉 ∈ 〈M + πV (W ),W ∩ V ∗〉 = 0

and then 〈a,W 〉 = 0. Therefore, a ∈ W because W is Lagrangian. Then write

a = u+ v where u ∈ W ∩ V ∗ and v ∈ K. Then u = a− v ∈ W and a ∈M ⊕ πV (W )

by assumption. But we know that W ∩ (M ⊕ πV (W )) = 0, hence u = 0 and

a = v ∈ K. Therefore, K is a Lagrangian subspace.

Now, from the theory of finite dimensional symplectic spaces [7, Lemma 2.1],

there exists g′′ ∈ Sp(M + πVW ) such that Mg′′ = K. Extend g′′ to a map g′ on X

linearly such that it restricts to the identity on W ∩V ∗⊕M⊥. It is easy to check that

g′ ∈ Sp(X). In particular, g′ ∈ Spfin(X). Then we have V ∗g′ = W and the action of

Spfin(X) on Gr(X, V ∗) is transitive.

Step 5 : We complete the proof of the the lemma. Suppose g ∈ Sp(X, V ∗). Then we

claim that V ∗g = V ∗(γ+ δ) ∈ Gr(X, V ∗). Indeed, it is clear that V ∗g is isotropic. To

see why it is Lagrangian, suppose v ∈ X such that 〈v, V ∗g〉 = 0. Then 〈vg−1, V ∗〉 = 0

implies that vg−1 ∈ V ∗ or equivalently that v ∈ V ∗g. It remains to prove that

V ∗g ∩ V ∗ has finite co-dimension in both V ∗ and V ∗g. It is enough to prove that

[V ∗g ∩ V ∗ : V ∗g] < ∞ because [V ∗g ∩ V ∗ : V ∗] = [V ∗ ∩ V ∗g−1 : V ∗g−1]. Suppose

that x = v + v∗ = yg ∈ V ∗g. Choose elements f ∗1 , ..., f∗n such that f ∗1γ, ..., f

∗nγ is

a basis for Im γ. Then, v = (a1f∗1 + ... + anf

∗n)γ for some a1, ..., an. In particular,


(y− a1f∗1 − ...− anf ∗n)g ∈ V ∗ ∩V ∗g. Therefore, V ∗ ∩V ∗g+ spanf ∗1 g, ..., f ∗ng = V ∗g,

hence [V ∗ ∩ V ∗g : V ∗g] <∞.

Now, by step 4, there exists g′ ∈ Spfin(X) such that V ∗g = V ∗g′ by transitivity.

Therefore, V ∗gg′−1 = V ∗ or equivalently gg′−1 ∈ P . We conclude that g = pg′ for

some p ∈ P, g′ ∈ Spfin(X). This completes the proof of this lemma.

The next lemma is the last one we will need before we can prove the main result

of this chapter.

Lemma 3.2.15. If g1, g2 are in Spfin(X), then there exists V1, V2 such that V =

V1 ⊕ V ⊥2 , V ∗ = V2 ⊕ V ⊥1 , V ⊥1 ⊕ V ⊥2 is fixed pointwise by g1, g2 and V1 ⊕ V2 is finite-

dimensional and g1, g2 invariant.

Furthermore, if J ⊆ V is an arbitrary finite-dimensional subspace, then we can

choose V1 and V2 so that J ⊆ V1.

Proof. Write g1 =

[α1 β1

γ1 δ1

]and g2 =

[α2 β2

γ2 δ2

]. Similarly to Proposition 3.2.11, it

can be proven that rk(β1), rk(β2), rk(α1−IV ), rk(α2−IV ), rk(γ1), rk(γ2), rk(δ1−IV ∗),rk(δ2 − IV ∗) are all finite. We claim that we can choose M,N ⊆ V such that V =

M ⊕N ,

M1 := Im(α1 − IV ) + Im(α2 − IV ) + Im(γ1) + Im(γ2) ⊆M,

and

M2 := Im(β1) + Im(β2) + Im(δ1 − IV ∗) + Im(δ2 − IV ∗) ⊆ N⊥.

In fact, we prove the following more general statement : we can choose M,N such

that V = M ⊕ N and for any finite-dimensional subspaces M1 ⊆ V and M2 ⊆ V ∗,

we have that M1 ⊆M and M2 ⊆ N⊥.

Choose a basis w∗1, ..., w∗n for M2. For n = 0, there is nothing to show (one

simply takes A = M1). Choose a complementary subspace N for M1. If M2 ⊆ N⊥,

there is nothing to show, hence assume this is not the case. From Lemma 3.2.12 there

exists linearly independent vectors v1, ..., vn ∈ V such that 〈w∗i , vj〉 = δi,j. Suppose

that nii∈I is a basis for N . Then for each i ∈ I, let

n′i = ni −n∑j=1

〈w∗j , ni〉vj


and set M ′ = C1 ⊕ Fv1 ⊕ ... ⊕ Fvn as well as N ′ = spanb′ii∈I . M ′ is clearly still

finite-dimensional, M1 ⊆ M ′, and we still have M ′ + N ′ = V . Furthermore, we now

have wi ∈ N ′⊥ by construction. To obtain a direct sum M ′ ⊕ N ′, it is enough to

extend a basis of the intersection of M ′ ∩ N ′ to a basis of M ′, and then remove the

elements from the intersection of M ′∩N ′. What we have left is a basis for a subspace

N ′′ which still satisfies M2 ⊆ N ′′⊥ (because N ′′ ⊆ N ′ implies that N ′⊥ ⊆ N ′′⊥), but

also satisfies M ′ ⊕N ′′ = V .

Now, one can write g1 in the following way

T =

A 0 C 0

A′ 1N C ′ 0

B 0 D 0

B′ 0 D′ 1M⊥

where A : M → N , A′ : N → M , C : M → N⊥, C ′ : N → N⊥, B : N⊥ → M ,

B′ : N⊥ → M , D : N⊥ → N⊥ and D′ : M⊥ → N⊥. We obtain zeroes in the other

entries because M and N⊥ contain the images of the maps β1, γ1, α1−IV and δ1−IV ∗ .If v ∈ X, then v = v1 + v2 + v∗1 + v∗2 for v1 ∈M, v2 ∈ N, v∗1 ∈M⊥, v∗2 ∈ N⊥ and

vg1 =[v1 v2 v∗2 v∗1

]T.

Clearly, A,B,C,D are maps between finite dimensional spaces, hence their trans-

poses exist and are unique. In fact, the transpose of all the other maps also exist :

for example, D′∗ = (δ∗ −D∗)πN : M → N . Indeed, if v ∈M⊥ and w ∈M , we have

〈vD′, w〉 = 〈v(D′+D)−vD,w〉 = 〈vδ−vD,w〉 = 〈v, w(δ∗−D∗)〉 = 〈v, w(δ∗−D∗)πN〉.

Therefore, D′∗ exist. A similar argument holds for all the other transposes.

Now, we can also write g−11 using Corollary 3.1.4 and uniqueness of the transpose

to obtain D∗ D′∗ −C∗ −C ′∗

0 1N 0 0

−B∗ −B′∗ A∗ A′∗

0 0 0 1M⊥

.Taking the product g−1

1 g1 = I, we obtain A′ = 0, C ′ = 0, B′ = 0, D′ = 0.


Therefore, g1 is actually given by the matrix

T =

A 0 C 0

0 1N 0 0

B 0 D 0

0 0 0 1M⊥

and V1 ⊕ V2 is then g1-invariant. A similar proof holds for g2.

The second statement of the lemma follows, but with

M1 := Im(α1 − IV ) + Im(α2 − IV ) + Im(γ1) + Im(γ2) + J ⊆M

instead.

Theorem 3.2.16. The map g 7→ Tg is a projective representation of Sp(X, V ∗) on

S(V ).

Proof. First, we claim that Tgf ∈ S(V ) for any g ∈ Sp(X, V ∗) and f ∈ S(V ). By

Lemma 3.2.14, we have that g−1 = p−1g′−1 for p ∈ P and g′ ∈ Spfin(X) and then

g = g′p. Therefore, by Corollary 3.2.9, Tgf ∼ Tg′Tpf . Write g′ =

[α1 β1

γ1 δ1

]and

p =

[α2 β2

0 δ2

]. Then

(Tpf)(x) = Sp(x)f(xα2) = ψ(1

2〈xα2, xβ2〉)f(xα2) = ((xβ2, 0) · f)(xα2).

However, α2 is invertible and linear by Lemma 3.2.7 and (xβ2, 0) · f ∈ S(V ), hence

Tpf ∈ S(V ). Next, suppose W ⊆ V is finite-dimensional and h ∈ S(V ). By Lemma

3.2.15, choose V1 and V2 such that V = V1 ⊕ V ⊥2 , V ∗ = V2 ⊕ V ⊥1 , V ⊥1 ⊕ V ⊥2 is fixed

pointwise by g′ and V1 ⊕ V2 is finite-dimensional, g′-invariant and W ⊆ V1 (take

g1 = g′ and g2 to be the identity). Define h′ = h|V1 . Clearly, h′ ∈ S(V1) by definition.

Then (Tg′h)|W = T ′g′(h′)|W where T ′ is the finite-dimensional Weil representation on

the finite dimensional symplectic space V1 ⊕ V2. However, by [7, Lemma 3.2 (3)],

T ′g′h′ ∈ S(V1), hence Tg′h|W ∈ S(W ). Therefore, Tg′h ∈ S(V ). We conclude that

Tgf ∈ S(V ).

Next, we claim that Tg is a projective representation. Suppose g1, g2 ∈ Sp(X, V ∗).From Lemma 3.2.14, we have g′1 ∈ Spfin(X) such that g1 = p1g

′1 for some p1 ∈ P .


From the same lemma, we have (g′2)−1 ∈ Spfin(X) such that g−12 = p−1

2 (g′2)−1 for

some p2 ∈ P . Equivalently, g2 = g′2p2. Now, from Corollary 3.2.9, we have Tg1Tg2 ∼Tp1Tg′1Tg′2Tp2 . But now, from Lemma 3.2.15, there exists V1, V2 such that V = V1⊕V ⊥2 ,

V ∗ = V2⊕V ⊥1 , V ⊥1 ⊕V ⊥2 is fixed pointwise by g1 and g2 and V1⊕V2 is finite-dimensional

and g1,g2 invariant.

Suppose that f ∈ S(V ). For each v2 ∈ V ⊥2 , define fv2 : V1 → C by fv2(v1) =

f(v1 + v2). It is not hard to see that fv2 ∈ S(V1). Then Tg′1f(v1 + v2) = T ′g′1fv2(v1)

where T ′g′1is the Weil representation on the finite dimensional symplectic space V1⊕V2.

Now, we know from the finite-dimensional case ([7], Theorem 4.1 (5)) that T ′g′1T ′g′2

=

c(g′1, g′2)T ′g′1g′2

for some number c. Therefore, for every f ∈ S(V ), v1 ∈ V1 and v2 ∈ V ⊥2 ,

we have

Tg′1Tg′2f(v1 + v2) = T ′g′1T′g′2fv2(v1) = c(g′1, g

′2)T ′g′1g′2fv2(v1) = c(g′1, g

′2)Tg′1g′2f(v1 + v2).

In particular, Tg′1Tg′2 ∼ Tg′1g′2 . We conclude that

Tg1Tg2 ∼ Tp1Tg′1Tg′2Tp2 ∼ Tp1Tg′1g′2Tp2 ∼ Tp1g′1g′2p2 = Tg1g2

where the last ∼ follows from Corollary 3.2.9 applied twice. Therefore, g 7→ Tg is a

projective representation.

The group Sp(X, V ∗) is difficult to work with. In most cases, we prefer to restrict

to a specific subgroup. Consider V = (t−1F[t−1])2n, the direct sum of 2n copies of the

vector space of polynomials in t−1 with 0 constant term. For an element p(t) ∈ R((t)),

denote by Res(p) the coefficient of t−1. Then we have the following :

Lemma 3.2.17. Suppose V = (t−1F[t−1])2n. Consider W = F[[t]]2n, 2n copies of the

vector space of formal power series. For v =2n∑i=1

t−1pi(t−1)ei where pi are polynomials

and also for w =2n∑i=1

qi(t)ei where qi are formal power series, define a pairing

(v, w)W =n∑i=1

Res(piqi+n − pi+nqi)

Then, the pairing results in an isomorphism W ∼= V ∗.


Proof. Clearly, we have a map ψ : W → V ∗, defined by (ψ(w), v) = (w, v)W . Linearity

of ψ is easy. We show that ψ is injective and surjective.

Suppose ψ(w) = 0. Then (w, v) = 0 for all v ∈ V . Write w =2n∑i=1

∑j≥0

ai,jtjei. Then

0 = (w, t−jei) = ±ai,j−1.

Since we can do this for all i ∈ Z, j ≥ 1, we obtain that w = 0.

Now suppose f ∈ V ∗. Then set ai+n,j = −(f, t−j−1ei) for i ≤ n, bi−n,j =

(f, t−j−1ei) for i ≥ n + 1 and w =n∑i=1

∑j≥0

bi,jtjei +

2n∑i=n+1

∑j≥0

ai,jtjei. It is enough to

check that (ψ(w), t−j−1ei) = (f, t−j−1ei), because the set t−j−1eij≥0 forms a basis

for V .

For i ≤ n, we have

(ψ(w), t−j−1ei) = Res(−∑k≥0

ai+n,ktkt−j−1) = −ai+n,j = (f, t−j−1ei).

For i ≥ n+ 1, we have

(ψ(w), t−jei) = Res(∑k≥0

bi−n,ktkt−j−1) = bi−n,j = (f, t−j−1ei)

as desired. We conclude that ψ is an isomorphism.

Now, recall that X = V ⊕ V ∗, hence X = R((t))2n, a direct sum of 2n copies of

the field discussed in Section 2.1. It will be useful to have an explicit formula for the

symplectic form on X. Recall that for a pairing (., .) between a vector space V and

its dual V ∗,

〈v + v∗, w + w∗〉 = (v, w∗)− (v∗, w).

Corollary 3.2.18. Suppose v =2n∑i=1

pi(t)ei and w =2n∑i=1

qi(t)ei where pi, qi ∈ R((t)).

Then

〈v, w〉 =n∑i=1

Res(piqi+n − pi+nqi).

Proof. Write pi(t) = ai(t) + bi(t) and qi(t) = ci(t) + di(t) where ai, ci ∈ t−1R[t−1] and


bi, di ∈ R[[t]]. Then

〈v, w〉 =

(2n∑i=1

aiei,

2n∑j=1

diei

)W

−

(2n∑i=1

biei,

2n∑j=1

ciei

)W

= Res

(n∑i=1

aidi+n − ai+ndi

)−

(n∑i=1

cibi+n − ci+nbi

)

=n∑i=1

Res (aidi+n − ai+ndi − cibi+n + ci+nbi)

=n∑i=1

Res (aici+n + aidi+n − ai+nci − ai+ndi − dibi+n − cibi+n + di+nbi + ci+nbi)

=n∑i=1

Res (aiqi+n − ai+nqi − qibi+n + biqi+n)

=n∑i=1

Res (piqi+n − pi+nqi) .

Consider the R((t))-valued form on X⟨2n∑i=1

piei,2n∑j=1

qjej

⟩R((t))

=n∑i=1

piqi+n − pi+nqi

Then, it is clear by Corollary 3.2.18 that any isometry of 〈., .〉R((t)) is also an isometry of

〈., .〉, hence Sp2n(R((t))) ⊆ Sp(X). However, we actually have the following stronger

statement :

Lemma 3.2.19. Sp2n(F((t))) is a subgroup of Sp(X, V ∗)

Proof. It is well known that Sp2n(F((t))) is a group. We wish to show that Sp2n(F((t)))

is a subset of Sp(X, V ∗). Suppose that A ∈ Sp2n(F((t))). Then we can write

A =2n∑i=1

2n∑j=1

pi,jei,j

where pi,j ∈ F((t)).

The product of A with a vector v∗ =2n∑k=1

qkek in V ∗ is2n∑i=1

(2n∑j=1

pi,jqj)ei. However,

v∗ is in V ∗, hence v(qj) ≥ 0 for all j. Then v(pi,jqj) ≥ v(pi,j). Let m = minv(pi,j).


Therefore, we have v∗A ∈ (Ft−m ⊕ ... ⊕ Ft−1 ⊕ F[[t]])2n for any v∗ ∈ V ∗. But this

means that

Im γ ⊆ (Ft−m ⊕ ...⊕ Ft−1)2n

hence dim Im γ ≤ 2nm <∞, as desired.

We can define the Weil representation on Sp2n(F((t))) to be the restriction of

g 7→ Tg on Sp2n(F((t))). By Theorem 3.2.16, we know that Tg is a projective repre-

sentation.

We will use the following fact about symplectic matrices, a proof of which can be

found in [13, p .167] using a polynomial known as the Pfaffian.

Lemma 3.2.20. Suppose g ∈ Sp2n(F((t))). Then det(g) = 1.

Chapter 4

Continuity of the representations

In this chapter, we will establish some continuity results of the Heisenberg repre-

sentation and the projective representation defined in the previous chapter. We will

restrict ourselves to the case F = R. Let V be an arbitrary vector space. We start by

dealing with dimV <∞ and then we will switch to V = t−1R[t−1]. Henceforth, ψ is

a fixed continuous unitary character as in Chapter 3.

For k ∈ Z(H) ∼= R, v ∈ V , v∗ ∈ V ∗ and f ∈ S(V ), recall the H-action

1. (k · f)(x) = ψ(k)f(x),

2. (v · f)(x) = f(x+ v),

3. (v∗ · f)(x) = ψ(〈x, v∗〉)f(x).

If h = (v + v∗, k) ∈ H, recall that h can be written as

(0, k − 1

2〈v, v∗〉)(v, 0)(v∗, 0)

in a unique way, as proven in Proposition 3.1.7. Therefore, we can think of the action

of an element of H as three actions from three subgroups done in succession. This

will be the key idea used in Section 3.1 to prove that the action of H on S(V ) is

continuous.

4.1 Finite dimensional case

Assume that dimV <∞, so that V ∼= Rn. We shall equip H = V ⊕ V ∗⊕R with the

standard Euclidean topology and S(V ) with the Frechet topology on S(V ) defined in

43

CHAPTER 4. CONTINUITY OF THE REPRESENTATIONS 44

Section 2.2. We consider the l∞ norm on H defined by∥∥∥∥∥(

n∑i=1

aiei +n∑i=1

bif∗i , k

)∥∥∥∥∥ = maxai, bi, k : 1 ≤ i ≤ n

for a basis e1, ..., en of V , f ∗1 , ..., f∗n of V ∗ and (0, 1) for Z(H).

Lemma 4.1.1. Suppose ‖(v1 + v∗1, k1)− (v2 + v∗2, k2)‖ < δ and A > 0 ∈ R. Suppose

further that max‖(v1 + v∗1, k1)‖ , ‖(v2 + v∗2, k2)‖ < A. Then we have ‖v1 − v2‖ < δ,

‖v∗1 − v∗2‖ < δ and∣∣(k1 − 1

2〈v1, v

∗1〉)− (k2 − 1

2〈v2, v

∗2〉∣∣ < δ(An+ 1).

Proof. Write vj =n∑i=1

aijei, v∗j =

n∑i=1

bijf∗i , then we also have |ai1 − ai2| < δ, and

|bi1 − bi2| < δ and |k1 − k2| < δ. The first two claims follow. For the third claim,∣∣∣∣(k1 −1

2v∗1(v1))− (k2 −

1

2v∗2(v2))

∣∣∣∣ =

∣∣∣∣∣(k1 − k2) +1

2

n∑i=1

(bi1ai1 − bi2ai2)

∣∣∣∣∣≤ |k1 − k2|+

1

2

n∑i=1

∣∣bi1ai1 − bi2ai2∣∣ ≤ |k1 − k2|+1

2

n∑i=1

∣∣bi1ai1 − bi1ai2 + bi1ai2 − bi2ai2

∣∣≤ |k1 − k2|+

1

2

n∑i=1

∣∣bi1ai1 − bi1ai2∣∣+∣∣bi1ai2 − bi2ai2∣∣

≤ |k1 − k2|+A

2

n∑i=1

∣∣ai1 − ai2∣∣+∣∣bi1 − bi2∣∣ < δ + Anδ = δ(An+ 1).

Lemma 4.1.2. Suppose f ∈ S(V ). Then for all multi-indices α = (α1, ..., αn) and

β = (β1, ...βn), we have

xα11 x

α22 ...x

αnn

∂β1

∂xβ11

...∂βn

∂xβnnf ∈ S(V ).

Proof. This is an easy exercise.

Lemma 4.1.3. Suppose v ∈ V , A > 0 ∈ R and consider the map φv : S(V )→ S(V )

defined by φv(f)(x) = f(x + v). Then there exists a Frechet open set U ⊆ S(V )

around the origin, such that, for all v with ‖v‖ < A and for all ε > 0, if f − g ∈ U ,

then pα,β(φv(f)− φv(g)) < ε.


Proof. Fix α, β and suppose f, g ∈ S(V ). Then from Lemma 4.1.2, the functions

f = ∂β1

∂xβ11

... ∂βn

∂xβnnf and g = ∂β1

∂xβ11

... ∂βn

∂xβnng are in S(V ). Define an ordering

α′ ≤ α⇔ α′1 ≤ α1, ..., α′n ≤ αn

of multi-indices. Write v = (v1, ..., vn). Then we have

xα11 x

α22 ...x

αnn = (x1 + v1 − v1)α1(x2 + v2 − v2)α2 ...(xn + vn − vn)αn

=∑α′≤α

Aα′(x1 + v1)α′1 ...(xn + vn)α

′n

for some Aα′ . Note that ‖v‖ < A implies that |Aα′ | < A|α| (where |α| =n∑i=1

αi). Let

M = |α′ : α′ ≤ α| (note that M < ∞). Let B = A|α|. Finally, let N = BM and

suppose that ∥∥∥f − g∥∥∥α′,0

=∥∥∥xα′11 x

α′22 ...x

α′nn (f − g)

∥∥∥∞<

ε

N(10)

for all α′ ≤ α. Therefore, we obtain

pα,β(φv(f)− φv(g)) = supx∈V

∣∣∣xα11 x

α22 ...x

αnn (f(x+ v)− g(x+ v))

∣∣∣= sup

x∈V

∣∣∣∣∣∑α′≤α

Aα′(x1 + v1)α′1 ...(xn + vn)α

′n(f(x+ v)− g(x+ v))

∣∣∣∣∣≤∑α′≤α

|Aα′ | supx∈V

∣∣∣(x1 + v1)α′1 ...(xn + vn)α

′n(f(x+ v)− g(x+ v))

∣∣∣=∑α′≤α

|Aα′ | supx∈V

∣∣∣(x1)α′1 ...(xn)α

′n(f(x)− g(x))

∣∣∣<∑α′≤α

|Aα′ |ε

BM≤∑α′≤α

ε

M< ε.

Thus, if U is chosen to be x ∈ V : pα′,β(x) < εN

: α′ ≤ α, then f and g satisfy

Eq. (10). This proves the lemma.

The proof of the following lemma can be found in [3, p. 89]

Lemma 4.1.4. Let ψ : R → C× be a continuous additive unitary character. Then

we have ψ(k) = eiαk for some α.

Lemma 4.1.5. Suppose k ∈ R. Then the map φk : S(V )→ S(V ) defined by φk(f) =

ψ(k)f is an isometry for any norm pα,β.


Proof. This is clear from the fact that ψ is a unitary character.

Lemma 4.1.6. Suppose f ∈ S(V ). Then the map ψf : R→ S(V ) defined by ψf (k) =

ψ(k)f is continuous with respect to all norms pα,β, hence continuous.

Proof. Suppose k ∈ R and fix pα,β. Then from Lemma 4.1.2, we have that g =

xα11 x

α22 ...x

αnn

∂β1

∂xβ11

... ∂βn

∂xβnnf ∈ S(V ). Let A > 0 such that ‖g‖∞ ≤ A. Then we have

pα,β(ψ(k)f − ψ(k′)f) = ‖(ψ(k)− ψ(k′))g‖∞= |ψ(k)− ψ(k′)| ‖g‖∞ ≤

∣∣∣eiαk − eiαk′∣∣∣A.However, eiαk is a periodic function of k with period 2π

α, hence it is uniformly contin-

uous. Therefore, we can choose δ such that if |k − k′| < δ, then∣∣eiαk − eiαk′∣∣ < ε

A,

which completes the proof.

Lemma 4.1.7. Suppose f ∈ S(V ). Then the map ψf : V → S(V ) defined by

ψf (v)(x) = f(v + x) is continuous with respect to all norms pα,β, hence continuous.

Proof. Suppose v ∈ V and fix pα,β. From Lemma 4.1.2, we have that g = ∂β1

∂xβ11

... ∂βn

∂xβnnf ∈

S(V ). Choose R1 such that if ‖x‖ ≥ R1, then ‖xα11 x

α22 ...x

αnn g(x)‖ < ε

2and R2 > R1

such that if ‖x‖ ≥ R2, then ‖xα11 x

α22 ...x

αnn g(x)‖ < ε

4.

Next, set A = R|α|2 (where |α| =

∑αi). On the disk D = x : ‖x‖ ≤ R2, g

is uniformly continuous (because D is compact), so we can choose δ′ such that if

x, y ∈ D and ‖x− y‖ < δ′, then |g(x)− g(y)| < εA

. Now, let δ = minδ′, R2 −R1.Suppose v′ ∈ V such that ‖v − v′‖ < δ, and then ‖(x+ v)− (x+ v′)‖ < δ. We

wish to compute |xα11 x

α22 ...x

αnn (g(x+ v)− g(x+ v′))|. There are three cases.

1. Case 1: If x+ v, x+ v′ ∈ D, we have

|xα11 x

α22 ...x

αnn (g(x+ v)− g(x+ v′))| = |xα1

1 xα22 ...x

αnn | |(g(x+ v)− g(x+ v′)|

< Aε

A= ε.

2. Case 2 : If, x+ v, x+ v′ /∈ D, we have

|xα11 x

α22 ...x

αnn g(x+ v)− xα1

1 xα22 ...x

αnn g(x+ v′)| < ε

4+ε

4=ε

2< ε.


3. Case 3 : If x + v ∈ D but x + v′ /∈ D (or vice-versa), we must have that

‖x+ v‖ ≥ R1 by the choice of δ. Therefore,

|xα11 x

α22 ...x

αnn g(x+ v)− xα1

1 xα22 ...x

αnn g(x+ v′)| < ε

2+ε

4=

3ε

4< ε.

Note that δ does not depend on v, hence ψf is in fact uniformly continuous.

Lemma 4.1.8. The map ψf : V ∗ → S(V ) defined by ψf (v∗) = ψ(〈x, v∗〉)f is contin-

uous with respect to all norms pα,β, hence continuous.

Proof. The Fourier transform f → F(f) is a continuous map on S(V ) ([4, p. 250]).

Furthermore, if v∗ =n∑i=1

aiv∗i , we can compute using Fubini’s theorem and the appro-

priate normalization of Fourier tranform that

ψf (v∗) = F4(ψf (v

∗)) = F4(ψ(〈x, v∗〉)f) = F4(eiα(x·(

n∑i=1

aifi))f)) = F3(F(f)(x−

n∑i=1

aiei)).

Therefore, ψf is the composition of continuous maps F3 ψF(f)(−∑aiei)), where

ψF(f) is as in Lemma 4.1.7. Then ψf is continuous.

Proposition 4.1.9. Suppose f ∈ S(V ). Then the map φ : H → S(V ) defined by

φ(h) = h · f is continuous.

Proof. Fix a norm pα,β. Suppose h1 ∈ H. We will show that the map is continuous

around h1. Suppose h2 ∈ H also. Write h1 = (v1 + v∗1, k1) and h2 = (v2 + v∗2, k2).

Furthermore, write k′1 = k1 − 12v∗1(v1) and k′2 = k2 − 1

2v∗2(v2). First, we note that

‖(k′1v1v∗1 · f − k′2v2v

∗2 · f)‖α,β

= ‖(k′1v1v∗1 · f − k′2v1v

∗1 · f + k′2v1v

∗1 · f − k′2v2v

∗1 · f + k′2v2v

∗1 · f − k′2v2v

∗2 · f)‖α,β

≤ ‖(k′1v1v∗1 · f − k′2v1v

∗1 · f)‖α,β + ‖(k′2v1v

∗1 · f − k′2v2v

∗1 · f)‖α,β

+ ‖(k′2v2v∗1 · f − k′2v2v

∗2 · f)‖α,β .

We want to make each of these three terms less than ε3.

1. For the first term, we know from Lemma 4.1.6 that there exists δ1 > 0 such

that if |k′1 − k′2| < δ1, then pα,β((k′1 · v1 · v∗1 − k′2 · v1 · v∗1) < ε3.


2. For the second term, by Lemma 4.1.5, pα,β(g − h) = pα,β(k′2 · g − k′2 · h). Fur-

thermore, from Lemma 4.1.7, there exists δ2 such that if ‖v1 − v2‖ < δ2, then

pα,β(v1 · (v∗1 · f) − v2 · (v∗1 · f)) < ε3. The entire term will be less than ε

3with

g = v1 · v∗1 · f and h = v2 · v∗1 · f .

3. For the last term, by Lemma 4.1.5, pα,β(g − h) = pα,β(k′2 · g − k′2 · h). But now,

from Lemma 4.1.3, we also have that whenever ‖v2‖ < A for some A, there exists

a Frechet open set U such that g′−h′ ∈ U implies that pα,β(v2 · g′− v2 ·h′) < ε3.

Finally, from Lemma 4.1.8, we have that there exists δ3 s.t. if ‖v∗1 − v∗2‖ < δ3,

then v∗1 · f − v∗2 · f ∈ U . Now take g = v2 · v∗1 · f and h = v2 · v∗2 · f , g′ = v∗1 · fand h′ = v∗2 · f and then ‖(k′2v2v

∗1 · f − k′2v2v

∗2 · f)‖α,β will be less than ε

3.

Fix the Frechet open set U corresponding to A = 2 ‖v1‖+1 as described by Lemma

4.1.3. Choose B > ‖h1‖+‖v1‖+1 and δ = min‖v1‖+1, δ1Bn+1

, δ2, δ3. For all v2 such

that ‖v1 − v2‖ < δ ≤ ‖v1‖ + 1, we then have ‖v2‖ < 2 ‖v1‖ + 1 = A by the reverse

triangle inequality. Furthermore, with this choice of δ, ‖h2‖ ≤ ‖v1‖ + 1 < B. Then

by Lemma 4.1.1, ‖v1 − v2‖ < δ2, ‖v∗1 − v∗2‖ < δ3 and |k′1 − k′2| < δ1Bn+1

(Bn + 1) = δ1,

which is enough for ‖(k′1v1v∗1 · f − k′2v1v

∗1 · f)‖α,β, ‖(k′2v1v

∗1 · f − k′2v2v

∗1 · f)‖α,β and

‖(k′2v2v∗1 · f − k′2v2v

∗2 · f)‖α,β to all be less than ε

3, completing the proof.

4.2 Infinite-dimensional case

We now consider the case V = (t−1R[t−1])2n so that V ∗ = R[[t]]2n, X = R((t))2n and

H = X × R. Recall that by Proposition 2.1.2, a basis for the topology on R((t))

consists of the translations of the sets Un = x ∈ R((t)) : v(x) ≥ n, where v(x) is

the valuation map. The topology on H is then simply the product topology of 2n

copies of R((t)) with the Euclidean topology on R. For convenience, we will write the

2n components of X as R((ti)) for 1 ≤ i ≤ 2n to differentiate them from each other.

Lemma 4.2.1. H is a topological group.

Proof. We first prove that the group multiplication is continuous. Suppose U ⊆ H

is open and (h1, h2) ∈ H × H such that h1h2 ∈ U . Write h1 = (v1 + v∗1, k1), h2 =

(v2 + v∗2, k2). Similarly, suppose h′1 = (w1 + w∗1, k′1) and h′2 = (w2 + w∗2, k

′2) are other


elements of H. We know that v1 =∑

1≤j≤2n,i∈N+

ai,jt−ij and v2 =

∑1≤j≤2n,i∈N+

bi,jt−ij for

finitely many non-zero ai,j ∈ R and bi,j ∈ R. Because h1h2 ∈ U , there exists M, ε > 0

such that (h1h2 + x ∈ X : v(xi) ≥M : 1 ≤ i ≤ 2n)× (−ε, ε) ⊆ U .

Now, let N ′ = max(M ∪ i : ai,j 6= 0 or bi,j 6= 0 for some j), N = N ′ + 1 and

U ′ = x ∈ X : v(xi) ≥ N for all 1 ≤ i ≤ 2n. Suppose further that (v1 + v∗1)− (w1 +

w∗1) ∈ U ′ and (v2 + v∗2) − (w2 + w∗2) ∈ U ′. Then w∗1 = v∗1 + u1 where u1 ∈ U ′ and

v2 = w2, hence 〈w2, w∗1〉 = 〈v2, v

∗1〉. Similarly, 〈w1, w

∗2〉 = 〈v1, v

∗2〉. Then h′1h

′2 − h1h2

is equal to

(w − v, k +1

2(〈v2, v

∗1〉+ 〈v1, v

∗2〉+ 〈w2, w

∗1〉 − 〈w1, w

∗2〉)),

where w = w1 + w2 + w∗1 + w∗2, v = v1 − v2 − v∗1 − v∗2 and k = −k1 − k2 + k′1 − k′2.

After some simplification, we obtain that

h′1h′2 − h1h2 = (u1 + u2,−k1 − k2 + k′1 + k′2)

for some u2 satisfying w∗2 = v∗2 + u2 and u2 ∈ U ′. Next, write U ′ε = U ′ × (− ε2, ε

2) and

consider the set

(h1 + U ′ε)× (h2 + U ′ε) ⊆ H ×H.

This set contains (h1, h2), is open, and it follows from the above computation that

the group operation sends it into U , as desired. Therefore, the group operation is a

continuous map from H ×H into H.

To show that the map h 7→ h−1 is continuous, suppose h1 = (v1 + v∗1, k1) ∈ H and

remember that from Proposition 3.1.5

(h1)−1 = (−v1 − v∗1,−k1)

Set U = U1 × ... × U2n × V where U1, ..., U2n, V are open. In particular, it is easy

to prove that −U1, ...,−U2n,−V are all also open. But now, the inverse map sends

−U1 × −U2 × ... − U2n × −V into U (from the above formula). Therefore, it is

continuous.

Proposition 4.2.2. Suppose f ∈ S(V ). Then the map φ : H → S(V ) defined by

φ(h) = h · f is continuous.


Proof. As usual, it is enough to check continuity with respect to every seminorm. Fix

a seminorm pα,β,m and (v1+v∗1, k1) ∈ H. Then v1 =∑

1≤j≤2n,i∈N+

ai,jt−ij for finitely many

non-zero ai,j ∈ R. Let N ′ = maxi : ai,j 6= 0 for some j. Then let N = maxm,N ′.From Lemma 4.1.6 applied to V = Vm, there exists δ such that if |k′1 − k′2| < δ, then

pα,β,m(k′1 · g − k′2 · g) < ε. Write

U ′ = x ∈ X : v(xi) ≥ N for all 1 ≤ i ≤ 2n.

Consider the open set

U = (v1 + v∗1 + U ′)× (k1 − δ, k1 + δ).

Now suppose that (v2 + v∗2, k2) ∈ U . We make two observations.

|k′1 − k′2| < δ : Recall that k′1 = k1 − 12〈v1, v

∗1〉, k′2 = k2 − 1

2〈v2, v

∗2〉. Note that if

v2 + v∗2 ∈ (v1 + v∗1 + U ′), then v2 + v∗2 = v1 + v∗1 + v′ where v′ ∈ U ′. However, from

N ≥ N ′ we have that 〈v′, v1〉 = 0, hence 〈v2, v∗2〉 = 〈v1, v

∗1〉 and then k′1−k′2 = k1−k2.

Therefore, |k1 − k2| < δ implies that |k′1 − k′2| < δ. From this proof, we also see that

v1 = v2.

v∗1 · f(x) = v∗2 · f(x) for all x ∈ Vm : Note that by the choice of N , we must have

〈x, U ′〉 = 0 for all x ∈ Vm, hence 〈x, v∗2〉 = 〈x, v∗1 + v′〉 = 〈x, v∗1〉. It then follows that

v∗1 · f(x) = v∗2 · f(x) for all x ∈ Vm.

From these observations we obtain that

pα,β,m(k′1 · v1 · v∗1 · f − k′2 · v2 · v∗2 · f) = pα,β,m(k′1 · (v1 · v∗1 · f)− k′2 · (v2 · v∗1 · f))

= pα,β,m(k′1 · (v1 · v∗1 · f)− k′2 · (v1 · v∗1 · f)) < ε.

Since this is true for any seminorm, we conclude that φ is continuous.

We now wish to show that the map g → Tg as defined in 3.2.4 is continuous when

restricted to Sp2n(R((t))) with respect to the product topology of 4n2 copies of R((t)),

obtained from the subspace topology as a subset of M2n,2n(R((t))). However, for this

to make sense, first we should prove the following proposition.

Proposition 4.2.3. The group Sp2n(R((t))), with the product topology of 4n2 copies

of X, is a topological group.


Proof. Denote the standard basis for M2n,2n by ei,j for 1 ≤ i, j ≤ 2n. Let us look at

the formula for product of two matrices :( ∑1≤i,j≤2n

ai,jei,j

)( ∑1≤k,l≤2n

bk,lek,l

)=

∑1≤i,l≤2n

( ∑1≤k≤2n

ai,kbk,l

)ei,l.

As usual, it is enough to check that the map is continuous in every component. How-

ever, the map sending two matrices (ai,j) and (bi,j) to∑

1≤k≤2n

ai,kbk,l is a composition

of projections, additions and multiplications, all of which are continuous. The result

is then continuous as well.

We now wish to prove that the inverse map g 7→ g−1 is also continuous. Because

det(g) = 1 by Lemma 3.2.20, we know that this map is simply g 7→ C(g)T , where

C(g) is the cofactor matrix of g. However all the entries in the cofactor matrix of

g are polynomials in the entries of g, which can be thought of as compositions of

projections, additions and multiplications, all continuous maps.

This completes the proof.

Theorem 4.2.4. Suppose that Haar measures have been fixed for every possible finite-

dimensional subspace of V and f ∈ S(V ). Then the map φf : Sp2n(R((t))) 7→ S(V )

defined by φf (g) = Tgf is continuous.

Remark 4.2.5. If the topology on S(V ) was induced by a finite family of seminorms

pα,β,m, our proof would show that φf is locally constant, which corresponds to the

idea of smooth representation in p-adic representation theory.

Proof. For any seminorm pα,β,m, and any element g ∈ Sp2n(R((t))), our goal will be

to find an open set U ⊆ Sp2n(R((t))) containing g such that pα,β,m(Tg(f)−Tg′(f)) = 0

(which is clearly less than ε). We will prove that on this open set U , Tgf(x) = Tg′f(x)

for all x ∈ Vm.

Recall the definitions of Tg and Sg from Proposition 3.2.4 :

(Tgf)(x) =

∫Vg

Sg(x+ x∗)f(xα + x∗γ)d(x∗γ),

Sg(x+ x∗) = ψ(1

2〈xα, xβ〉+

1

2〈x∗γ, x∗δ〉+ 〈x∗γ, xβ〉).


Step 1 : Suppose g ∈ Sp2n(R((t))) and Vr ⊆ V for some r. Write

g =

[α β

γ δ

]as usual. Then we claim that there exists a neighbourhood W of g such that for all

g′ =

[α′ β′

γ′ δ′

]∈ W

we have γ′ = γ and α′|Vr = α|Vr .

Proof of step 1 : We will first give a more explicit formula for α and γ. Recall that

X = V ⊕ V ∗ = R((t))2n. Denote by ei,j the standard basis for M2n,2n(R((t))), where

1 ≤ i, j ≤ 2n. Furthemore, this time we will differentiate the variables ti by writing

them as tei to be compatible with matrix multiplication. Assume g =∑

1≤i,j≤2n

ai,jei,j,

ai,j ∈ R((t)). Then ai,j =∑k

bki,jtk where 1 ≤ i, j ≤ 2n, k ∈ Z, bki,j ∈ R and with

the property that only finitely of the bki,j such that k ≤ −1 are non-zero. Assume∑l∈Z

altl ∈ R((t)). Now, we can compute by matrix multiplication :

(∑l∈Z

altlem

)g =

∑1≤i≤2n

∑k,l∈Z

bkm,ialtk+lei.

Therefore, we obtain(∑l≤−1

altlem

)α =

∑1≤i≤2n

∑k+l≤−1l≤−1

albkm,it

k+lei,

and (∑l≥0

altlem

)γ =

∑1≤i≤2n

∑k+l≤−1l≥0

albkm,it

k+lei.

Note that in both cases the sum over k+ l = p is a finite sum for each p ≤ −1 because

only finitely many of the bki,j such that k ≤ −1 are non-zero. From Lemma 3.2.19, we

have dimVg <∞, hence Vg ⊆ Vr′ for some r′. Then (Vr+Vg) ⊆ Vs for s = maxr, r′.Let U = x ∈ R((t)) : v(x) ≥ s and

W = (a1,1 + U)× (a1,2 + U)× ...× (a2n,2n + U).


Clearly, W is open. But then if g′ ∈ V , we must have b′ki,j = bki,j for all k ≤ s− 1.

Now we can prove the claims of Step 1. We start with γ′ = γ. When l ≥ 0,

(k + l) ≤ −1 implies that k ≤ −1 ≤ s− 1. Then(∑l≥0

altlem

)γ =

∑1≤i≤2n

∑k+l≤−1l≥0

albki,mt

k+lei =∑

1≤i≤2n

∑k+l≤−1l≥0

alb′ki,mt

k+lei = (tlem)γ′

as desired (it remains to expand γ by linearity over all 2n basis elements em). This

proves that γ = γ′.

For the second claim, note that whenever l ≥ −s, we have that l ≤ −1−k implies

that k ≤ s, and then( ∑−s≤l≤−1

altlem

)α =

∑1≤i≤2n

∑−s≤l≤−1−k

l≤−1

albki,mt

k+lei =∑

1≤i≤2n

∑−s≤l≤−1−k

l≤−1

alb′ki,mt

k+lei

= (tlem)γ′

as desired (once again expanding over all em). This proves the second claim because

Vr ⊆ Vs.

Step 2 : Suppose Vr ⊆ V . Then there exists a neighbourhood U ⊆ W of g such

that Sg|x+V ∗ = Sg′|x+V ∗ for all x ∈ Vr and g′ ∈ U .

Proof of step 2 : Fix Vr. From the previous step, we have a neighbourhood W of

g such that for all g′ ∈ W , γ′ = γ and α′|Vr = α|Vr . Clearly, there exists s ≥ 0 such

that x, α(x) : x ∈ Vr ⊆ Vs. In a similar way to the last step, we can compute(∑l≤−1

altlem

)β =

∑1≤i≤2n

∑k+l≥0l≤−1

albkm,it

k+lei

for l < 0. For l ≥ 0, we have(∑l≥0

altlem

)δ =

∑1≤i≤2n

∑k+l≥0l≥0

albkm,it

k+lei.

Once again, note that in both cases the sum∑

k+l=p

is a finite sum for each p ≥ 0. Let

W ′ = x ∈ R((t)) : v(x) ≥ 2s and W = (a1,1 +W ′)× (a1,2 +W ′)× ...× (a2n,2n+W ′).


If g′ ∈ U = W ∩ W , we must have b′ki,j = bki,j for all k ≤ 2s − 1. Now, note that for

any l ≥ −s we have that (k + l) ≤ s− 1 imply k ≤ s− l − 1 ≤ 2s− 1. Therefore, if

y ∈ Vs, we have

〈y,

(∑l≤−1

altlem

)β〉 = 〈y,

∑1≤i≤2n

∑0≤(k+l)l≤−1

albkm,it

k+lei〉 = 〈y,∑

1≤i≤2n

∑0≤(k+l)≤s−1

l≤−1

albkm,it

k+lei〉

= 〈y,∑

1≤i≤2n

∑0≤(k+l)≤s−1

l≤−1

alb′km,it

k+lei〉 = 〈y,∑

1≤i≤2n

∑0≤(k+l)l≤−1

alb′km,it

k+lei〉

= 〈y,

(∑l≤−1

altlem

)β′〉.

The second and fourth equality follows from 〈y, tk+lei〉 = 0 for all k + l ≥ s, y ∈ Vs.Replacing y by xα = xα′ (because U ⊆ W ) and extending by linearity to all basis

elements em, we obtain 〈xα, xβ〉 = 〈xα′, xβ′〉. Similarly, replacing y by x∗γ = x∗γ′

(because γ = γ′) and extending by linearity again, we obtain 〈x∗γ, xβ〉 = 〈x∗γ′, xβ′〉.The proof that 〈x∗γ, x∗δ〉 = 〈x∗γ′, x∗δ′〉 is almost exactly the same computation as

for β (the only difference being l ≥ 0 instead of l ≥ −1), using y = x∗γ = x∗γ′. By

adding those three terms (with a factor 12

in front of the first two), we see that with

the open set U chosen, Sg|x+V ∗ = Sg′|x+V ∗ for all x ∈ Vr and g′ ∈ U .

Step 3 : We prove the statement of the proposition.

Let g ∈ Sp2n(R((t)) and pα,β,m be a semi-norm and fix f ∈ S(V ). We want

to show that the map g 7→ Tgf is continuous. From the previous two steps with

Vr = Vm, there exists a neighbourhood U of g such that for all g′ ∈ U and x ∈ Vm,

Sg′ |x+V ∗ = Sg|x+V ∗ , γ = γ′, Vg = Im γ = Im γ′ = Vg′ and α|Vm = α′|Vm . In particular,

for any function f ∈ S(V ), we have f(xα + x∗γ) = f(xα′ + x∗γ′) for all x ∈ Vm.

Therefore, for this open set U , and for all x ∈ Vm,

(Tgf)(x) =

∫Vg

Sg(x+ x∗)f(xα + x∗γ)d(x∗γ) =

∫Vg′

Sg′(x+ x∗)f(xα′ + x∗γ′)d(x∗γ)

= (Tg′f)(x).

Therefore, pα,β,m(Tgf − Tg′f) = 0 for all g′ ∈ U . In particular, g 7→ Tg is continuous


with respect to the seminorm pα,β,m. Since this is true for any seminorm pα,β,m, we

conclude that the map g 7→ Tgf is continuous.

Chapter 5

A specific central extension

Let SL2(R((t))) be the group of 2× 2 matrices with coefficients in R((t)) of determi-

nant 1.

Lemma 4.0.1. Sp2(R((t))) = SL2(R((t)))

Proof. Write g =

[a b

c d

], v = (v1, v2) and w = (w1, w2). Then

〈v, w〉R((t)) = v1w2 − v2w1

and

〈vg, wg〉R((t)) = 〈(v1a+ v2c, v1b+ v2d), (w1a+ w2c, w1b+ w2d)〉R((t))

= v1w2(ad− bc)− v2w1(ad− bc)

Therefore, 〈v, w〉R((t)) = 〈vg, wg〉R((t)) if and only if ad− bc = 1.

In this chapter, we will then denote Sp2(R((t))) by SL2(R((t))) instead.

5.1 Kubota’s theorem

Define the tame symbol

t(tmσ1, tnσ2) = (−1)mncn1/c

m2

56

CHAPTER 5. A SPECIFIC CENTRAL EXTENSION 57

where σ1, σ2 ∈ R[[t]]∗ and c1, c2 are the constant terms of σ1, σ2. Note that t(tmσ1, tnσ2)

is never equal to 0. The following proposition gives the most important properties of

the tame symbol.

Proposition 5.1.1. Suppose a, b, b1, b2 ∈ R((t)) \ 0. Then

(i) t(a, b1)t(a, b2) = t(a, b1b2)

(ii) t(b1, a)t(b2, a) = t(b1b2, a)

(iii) t(a,−a) = 1

(iv) t(a, b) = 1 if a, b ∈ R[[t]]∗.

(v) t(a, 1− a) = 1 for a 6= 1

(vi) t(a, b)t(b, a) = 1

(vii) t(a, b)t(− ba, a+ b) = 1 for a 6= −b

(viii) t(1, a) = t(a, 1) = 1

Proof. Suppose σ1, σ2, σ3 are elements of R[[t]]∗ with constant terms c1, c2, c3. For

property (i), we have

t(tmσ1, tnσ2)t(tmσ1, t

kσ3) = (−1)mncn1/cm2 (−1)mkck1/c

m3

= (−1)m(n+k)cn+k1 /(c2c3)m = t(tmσ1, t

n+kσ2σ3).

Property (ii) is similar. For (iii), we have

t(tnσ1,−tnσ1) = (−1)n2

cn1/(−c1)n = (−1)n2−n = 1.

Property (iv) follows directly from the definition. Property (viii) follows from (i)

because

t(1, a) = t(1, a)t(1, a)

and then we obtain t(1, a) = 1 by cancellation. Property (v) depends on three cases.

If v(a) > 0, then the constant term of 1 − a is 1, hence t(a, 1 − a) = t(a, 1) = 1

by property (viii). If v(a) = 0, then v(1 − a) = 0 and then t(a, 1 − a) = 1 from

property (iv). Finally, if v(a) < 0, then t(a, 1 − a) = t(a,−a) = 1. Property


(vi) follows immediately from the definition. For property (vii), t(b, 1a)t(b, a) = 1

by property (viii), hence t(b, 1a) = t(a, b) from property (vi). But now, we obtain

t(− ba, 1a) = t(b, 1

a)t(− 1

a, 1a) = t(b, 1

a) = t(a, b) from property (iii). Then we have

t(a, b)t(− ba, a+ b) = t(− b

a, a+ b)t(− b

a,

1

a) = t(− b

a,a+ b

a) = t(− b

a, 1 +

b

a) = 1

where the last equality follows from property (iv).

We will frequently use these properties and will not always refer to them. Define

x

([a b

c d

])=

c : if c 6= 0

d : if c = 0

and for g1, g2 ∈ SL2(R((t))),

α(g1, g2) = t(x(g1g2)/x(g1),x(g1g2)/x(g2)) (11)

For

[a b

c d

]∈ SL2(R((t))), if c = 0, then we must have d 6= 0 (because ad−bc = 1),

hence α(g1, g2) is well-defined and never vanishes.

We now wish to verify the cocycle conditions for α as defined in Definition 2.3.5.

The first cocycle condition is easy. Indeed,

α(g, e) = t(x(g1)/x(g1),x(g1)) = t(1,x(g1)) = 1

by Proposition 5.1.1(viii), and α(e, g) = 1 similarly.

The challenge is to verify the second cocycle condition, which will be done in

Theorem 5.1.11. The main reference for this section is [5]. Since our definition of α is

in appearance different from [5], we should first verify that they are in fact the same.

Lemma 5.1.2.

α(g1, g2) = t (x(g1),x(g2)) t

(−x(g2)

x(g1),x(g1g2)

)(12)

Proof. Indeed, by applying Proposition 5.1.1 (i) and (ii) repeatedly to Eq.(11), we

obtain

α(g1, g2) = t (x(g1g2),x(g1g2)) t

(x(g1g2),

1

x(g2)

)t

(1

x(g1),x(g1g2)

)t

(1

x(g1),

1

x(g2)

).


However, t(x(g1g2),x(g1g2)) = t(−1,x(g1g2))t(−x(g1g2),x(g1g2)) = t(−1,x(g1g2))

by Proposition 5.1.1(ii) and Proposition 5.1.1(iii). Furthermore,

t

(x(g1g2),

1

x(g2)

)t (x(g1g2),x(g2)) = t(x(g1g2), 1) = 1

by Proposition 5.1.1(i) and (viii), hence t(x(g1g2), 1

x(g2)

)= t(x(g1g2),x(g2))−1 =

t(x(g2),x(g1g2)) by Proposition 5.1.1(vi). Therefore, we obtain

α(g1, g2) = t(−1,x(g1g2))t(x(g2),x(g1g2))t

(1

x(g1),x(g1g2)

)t

(1

x(g1),

1

x(g2)

).

But now, by combining the first three terms in this last expression with property (ii)

and noting that

t

(1

x(g1),

1

x(g2)

)= t

(x(g1),

1

x(g2)

)−1

= t(x(g1),x(g2))

from Proposition 5.1.1(i), (ii) and (viii), we obtain that Eq.11 is equivalent to Eq.12.

For the rest of this section, write gi =

[ai bi

ci di

]∈ SL2(R((t))).

Lemma 5.1.3. If g1, g2 ∈ SL(2,R((t))) such that c1 6= 0 but c2 = 0, then

α(g1, g2) = α(g2, g1) = t(x(g1),x(g2)).

If both c1 and c2 are zero, then

α(g1, g2) = t(x(g1),x(g2))−1.

Proof. In the first case, we have

g1g2 =

[a1a2 a1b2 + b1d2

c1a2 c1b2 + d1d2

]and

g2g1 =

[a2a1 + b2c1 a2b1 + b2d1

d2c1 d1d2

].

But note that det(g2) = a2d2 = 1, hence a2 = d−12 . Then we obtain

x(g1g2) = c1a2 =x(g1)

x(g2)(13)


and

x(g2g1) = d2c1 = x(g1)x(g2). (14)

Therefore,

α(g1, g2) = t(x(g1),x(g2))t

(−x(g2)

x(g1),x(g1g2)

)= t(x(g1),x(g2))t

(−x(g2)

x(g1),x(g1)

x(g2)

)= t(x(g1),x(g2))t

(−x(g2)

x(g1),x(g2)

x(g1)

)−1

= t(x(g1),x(g2)),

and

α(g2, g1) = t(x(g1g2)/x(g2),x(g1g2)/x(g1))

= t(x(g1)/x(g2)2,x(g2))

= t(x(g1),x(g2))t(−1/x(g2),x(g2))t(−1/x(g2),x(g2))

= t(x(g1),x(g2)).

This proves the first claim. For the second claim, we have x(g1g2) = x(g1)x(g2)

because c1 = c2 = 0 and then

α(g1, g2) = t(x(g1g2)/x(g1),x(g1g2)/x(g2))

= t(x(g2),x(g1))

= t(x(g1),x(g2))−1

as claimed.


[a b

c d

]∈ SL(2,R((t))) and c 6= 0. Then there exists a

neighbourhood U ⊆ R((t)) of 0 such that for all y ∈ R((t) \ 0 and

h = h(µ) =

[1 0

µ 1

],

where µ ∈ U , we have

t(x(hg), y) = t(x(g), y),

t(x(gh), y) = t(x(g), y).


In particular, we also have

t(y,x(hg)) = t(y,x(g)),

t(y,x(gh)) = t(y,x(g)).

Proof. Note that

hg =

[a b

µa+ c µb+ d

]and

gh =

[a+ bµ b

c+ dµ d

]Let M = maxv(c)− v(a) + 1, v(c)− v(d) + 1 and

U = UM = x ∈ R((t)) : v(x) ≥M

which is clearly non-empty. Then v(µa) = v(µ) + v(a) > v(c) by the choice of M .

Similarly, v(dµ) > v(c). Then v(µa + c) = v(c + dµ) = v(c) and the constant terms

of µa+ c, c+ dµ and c are clearly all the same, hence we obtain the statement of the

lemma.


[a b

c d

]∈ SL(2,R((t))). Then there exists a neighbour-

hood U ⊆ R((t)) of 0 such that for all

h = h(µ) =

[1 0

µ 1

],


α(h, g) = α(g, h) =

1 if c 6= 0,

t(x(h),x(g)) if c = 0.

Proof. If c = 0, but µ 6= 0 is arbitrary, we are done by Lemma 5.1.3. If both c = 0

and µ = 0, then again by Lemma 5.1.3,

α(h, g) = t(x(h),x(g))−1 = t(x(g), 1) = 1 = t(x(h),x(g)).

But we also have α(g, h) = t(x(g),x(h))−1 = t(x(h),x(g)). This completes the case

where c = 0 (any non-empty open set will do).


Next we assume c 6= 0. Choose U such that Lemma 5.1.4 holds. We obtain

α(h, g) = t(x(h),x(g))t(−x(g)

x(h),x(hg)) = t(x(h),x(g))t

(−x(g)

x(h),x(g)

)= t(x(h),x(g))t(−x(g),x(g))t

(1

x(h),x(g)

)= t(1,x(g)) = 1.

The proof that α(g, h) = 1 is done similarly.

Lemma 5.1.6. Suppose g1, g2 ∈ SL(2,R((t))). Then there exists a neighbourhood

U ⊆ R((t)) of 0 such that for all

h = h(µ) =

[1 0

µ 1

]


α(h, g1)α(hg1, g2) = α(h, g1g2)α(g1, g2), (15)

and

α(g1, g2)α(g1g2, h) = α(g1, g2h)α(g2, h). (16)

Proof. First, note that Eq. (15) is equivalent to

α(h, g1)α(hg1, g−11 g2) = α(h, g2)α(g1, g

−11 g2). (17)

This can be seen by replacing g2 with g−11 g2. Equivalently, we will prove

α(hg1, g−11 g2)α(g1, g

−11 g2)−1 = α(h, g2)α(h, g1)−1. (18)

We expand the left hand side to obtain

t(x(hg1),x(g−11 g2))t(x(g1),x(g−1

1 g2))−1t

(−x(g−1

1 g2)

x(hg1),x(hg2)

)t

(−x(g−1

1 g2)

x(g1),x(g2)

)−1

(19)

First we suppose that c1 6= 0 and c2 6= 0. From Lemma 5.1.5, choose U so that

Lemma 5.1.4 and Lemma 5.1.5 hold for g1 and g2. Then Expression (19) reduces to

t(x(g1),x(g−11 g2))t(x(g1),x(g−1

1 g2))−1t

(−x(g−1

1 g2)

x(g1),x(g2)

)t

(−x(g−1

1 g2)

x(g1),x(g2)

)−1

,


which simplifies to 1. However, we also have α(h, g2) = α(h, g1) = 1. This verifies

Eq. (18).

Secondly, suppose c1 6= 0 but c2 = 0. Then we can choose U so that Lemma

5.1.4 holds for g1 and Lemma 5.1.5 holds for g1 and g2. Then α(h, g1) = 1 and

α(h, g2) = t(x(h),x(g2)). Using Eq. (13) on x(hg2) and x(g−11 g2), we can simplify

Expression (19) to obtain

t

(− x(g−1

1 )

x(g1)x(g2),

x(h)

x(g2)2

).

Furthermore, from the formula for the inverse of a 2×2 matrix, we know that x(g−11 ) =

−x(g1), so we can simplify even further to obtain

t

(1

x(g2),

x(h)

x(g2)2

)= t

(1

x(g2),x(h)

)t

(1

x(g2),

1

x(g2)2

)= t(x(h),x(g2)).

This verifies Eq. (18). (Note that t(x, x2) = 1 for all x because, using Proposition

5.1.1 (i), t(x, x2) = t(x,−x)t(x,−x) = 1).

As third case, suppose c1 = 0 but c2 6= 0. Then we can choose U such that

Lemma 5.1.4 holds for g2 and Lemma 5.1.5 holds for g1 and g2. Then α(h, g2) = 1 and

α(h, g1) = t(x(h),x(g1)). Furthermore, from Eq. (13) and 14, we have that x(hg1) =x(h)x(g1)

and x(g−11 g2) = x(g−1

1 )x(g2). Then we can once again simplify Expression (19)

to obtain

t

(x(h)

x(g1),x(g−1

1 )x(g2)

)t

(1

x(g1),x(g−1

1 )x(g2)

)t

(−x(g1)

x(h)(x(g−1

1 )x(g2)),x(g2)

)t(−x(g1)x(g2)−1x(g−1

1 )−1,x(g2))

or equivalently, since in this case from the formula of the inverse of a 2× 2 matrix we

have x(g−11 ) = 1

x(g1)

t

(x(h)

x(g1)2,x(g2)

x(g1)

)t

(x(g1)2

x(h),x(g2)

)= t

(x(h)

x(g1)2,

1

x(g1)

)Now, similarly as in the second case, we obtain

t(x(h),x(g1))−1

which verifies Eq. (18).


In the fourth case, where c1 = c2 = 0, we apply Eq. (13) and (14) multiple times

on Expression (19) to obtain

t

(x(h)

x(g1),x(g−1

1 g2)

)t

(1

x(g1),x(g−1

1 g2)

)t

(−x(g1)

x(h)x(g−1

1 g2),x(h)

x(g2)

)t

(−x(g1)

x(g−11 g2)

,x(g2)

),

or equivalently

t

(x(h)

x(g1)2,x(g−1

1 g2)

)t

(−x(g1)

x(h)x(g−1

1 g2),x(h)

x(g2)

)t

(− x(g1)

x(g−11 g2)

,x(g2)

).

In this case, one can compute from matrix multiplication that x(g−11 g2) = x(g2)x(g1)−1,

simplifying this expression to

t

(x(h)

x(g1)2,x(g2)

x(g1)

)t

(−x(g2)

x(h),

x(h)

x(g2)

)t

(−x(g1)2

x(g2),x(g2)

)and the middle term equals 1, hence can be removed to yield

t

(x(h)

x(g1)2,x(g2)

x(g1)

)t

(−x(g1)2

x(g2),x(g2)

)= t

(x(h)

x(g1)2,x(g2)

x(g1)

)t

(x(g2),− x(g2)

x(g1)2

)= t

(x(h)

x(g1)2,x(g2)

x(g1)

)t

(x(g2),

x(g2)2

x(g1)2

)= t

(x(h)x(g2)2

x(g1)2,x(g2)

x(g1)

)= t

(x(g2)2

x(g1)2,x(g2)

x(g1)

)t

(x(h),

x(g2)

x(g1)

)= t

(x(h),

x(g2)

x(g1)

),

which verifies Eq. (18) using Lemma 5.1.5 on the right side.

Note that Eq. (16) is equivalent to

α(g1g−12 , g2)α(g1, h) = α(g1g

−12 , g2h)α(g2, h). (20)

This can be seen by replacing g1 with g1g−12 . Equivalently, we can prove

α(g1g−12 , g2h)α(g1g

−12 , g2)−1 = α(g2, h)−1α(g1, h) (21)

The computations are very similar to the ones for Eq. (18). One expands the

left side of this equation and then deals with four cases using Lemmas 5.1.3, 5.1.4,

5.1.5.


Lemma 5.1.7. Let g1, g2, g3 ∈ SL2(R((t))) and

β(g1, g2, g3) = α(g1, g2)α(g1g2, g3)α(g1, g2g3)−1α(g2, g3)−1

Then there exists a neighbourhood U ⊆ R((t)) of 0 such that for all

h = h(µ) =

[1 0

µ 1

]


β(hg1, g2, g3) = β(g1, g2, g3h) = β(g1, g2, g3)

Proof. Using Lemma 5.1.6, choose U1 such that the following three equations hold

simultaneously.

α(hg1, g2) = α(h, g1)−1α(h, g1g2)α(g1, g2),

α(hg1g2, g3) = α(h, g1g2)−1α(h, g1g2g3)α(g1g2, g3),

α(hg1, g2g3)−1 = α(h, g1)α(h, g1g2g3)−1α(g1, g2g3)−1.

Then we have

β(hg1, g2, g3) = α(hg1, g2)α(hg1g2, g3)α(hg1, g2g3)−1α(g2, g3)−1

By replacing α(hg1, g2), α(hg1g2, g3) and α(hg1, g2g3)−1 with the above three equations

and cancelling the terms, we obtain

α(g1, g2)α(g1g2, g3)α(g1, g2g3)−1α(g2, g3)−1 = β(g1, g2, g3),

hence β(hg1, g2, g3) = β(g1, g2, g3).

The claim that β(g1, g2, g3h) = β(g1, g2, g3) is similar, but uses Eq.(16) instead

and a different open set U2. Taking U = U1 ∩U2 as the open set for the statement of

the lemma, this completes the proof.

Let

N =

[1 b

0 1

]: b ∈ R((t))

Clearly, N is a subgroup of SL2(R((t))).

The following lemma is only briefly explained in [5].



[a b

c d

]∈ SL2(R((t))). Then there exist v, w ∈ N such

that either

vgw =

[a 0

0 a−1

]or vgw =

[0 −c−1

c 0

]Proof. We deal with two cases.

1. Case 1 : Suppose c 6= 0. Then let v =

[1 −a

c

0 1

]. Now, the product vg is

[0 b− ad

c

c d

]

Letting w =

[1 −d

c

0 1

], we obtain the second form after multiplication (note

that −c−1 = b− adc

).

2. Case 2 : Suppose c = 0. Then d = a−1 6= 0. Therefore, we can choose w = I

and v =

[1 − b

d

0 1

]and we obtain the first form, as desired.

Notation 5.1.9. If A = (ai,j)1≤i,j≤m is a matrix, denote πi,j(A) = ai,j.


[a b

c d

]∈ SL2(R((t))) and π2,1(g) 6= 0. Then there

exist v, v′ ∈ N such that

vg =

[0 b′

c′ d′

]and gv′ =

[a′′ b′′

c′′ 0

]

for some a′, b′, c′, a′′, b′′, c′′ ∈ R((t)).

Proof. The proof is very similar to Lemma 5.1.8 and is an easy exercise.

We now prove the following theorem, which is the main goal of this section.

Theorem 5.1.11. ([5], Section 2) Let β be defined as in Lemma 5.1.7. Then we

have β(g1, g2, g3) = 1. In particular, α is a cocycle.


Proof. Let U be as in Lemma 5.1.7 and write

h =

[1 0

µ 1

]h′ =

[1 0

µ′ 1

]for µ, µ′ ∈ U . It is easy to see that every non-empty open set of R((t)) is infinite,

hence U also is. Consider the products

hg1 =

[a b

µa1 + c1 µb+ d

]and g1h =

[a1 + b1µ b1

c1 + µd1 d1

].

We cannot have both a1 = 0 and c1 = 0 (otherwise g1 would have determinant 0).

Therefore, if c1 = 0, then π2,1(hg1) = µa1 + c1 = 0 implies that µ = 0. If c1 6= 0, then

µa1 + c1 = 0 implies that µa1 = −c1 and then µ = −c1a−11 (note that in this case,

a1 6= 0 as well). In both cases, µ is uniquely determined by a1 and c1. Similarly, the

condition π2,1(g1h) = c1 + µd1 = 0 also determines µ uniquely. However, U is infinite

so for any finite collection ki’s and lj’s of elements of SL2(R((t))), we can choose

µ ∈ U such that π2,1(hki) 6= 0 and π2,1(ljh) 6= 0.

Then choose µ ∈ U such that both π2,1(hg1) 6= 0 and π2,1(hg1g2) 6= 0 hold.

Similarly, one can also choose µ′ ∈ U such that π2,1(g3h′) 6= 0, π2,1(g2g3h

′) 6= 0 and

π2,1(hg1g2g3h′) 6= 0. Now, note that from Lemma 5.1.7, we have

β(hg1, g2, g3h′) = β(hg1, g2, g3) = β(g1, g2, g3).

Therefore, we can assume that π2,1(g1) 6= 0, π2,1(g3) 6= 0, π2,1(g1g2) 6= 0, π2,1(g2g3) 6= 0

and π2,1(g1g2g3) 6= 0 by replacing g1 with hg1 and g3 with g3h′.

It is easy to see from a simple computation that for all v, w ∈ N , x(vgw) = x(g).

Then we have that for any g, g′ ∈ SL2(R((t))),

α(vg, g′) = t(x(vg),x(g′))t(−x(vg)−1x(g′),x(vgg′))

= t(x(g),x(g′))t(−x(g)−1x(g′),x(gg′))

= t(g, g′)

(22)

andα(g, g′v) = t(x(g),x(g′v))t(−x(g)−1x(g′v),x(gg′v))

= t(x(g),x(g′))t(−x(g)−1x(g′),x(gg′))

= t(g, g′)

(23)


andα(gv, g′) = t(x(gv),x(g′))t(−x(gv)−1x(g′),x(gvg′))

= t(x(g),x(vg′))t(−x(v)−1x(vg′),x(gvg′))

= α(g, vg′)

(24)

Then we obtain for all v1, v2, v3 ∈ N , g1, g2, g3 ∈ SL2(R((t))) that

β(v1g1v2, v−12 g2v3, v

−13 g3v4)

= α(v1g1v2, v−12 g2v3)α(v1g1g2v3, v

−13 g3v4)α(v1g1v2, v

−12 g2g3v4)−1α(v−1

2 g2v3, v−13 g3v4)−1

= α(g1v2, v−12 g2)α(g1g2v3, v

−13 g3)α(g1v2, v

−12 g2g3)−1α(g2v3, v

−13 g3)−1

= α(g1, g2)α(g1g2, g3)α(g1, g2g3)−1α(g2, g3)−1

= β(g1, g2, g3)

Therefore, we can choose v−12 and v3 such that v−1

2 g2v3 is in one of the two forms of

Lemma 5.1.8. Then we also have π2,1(g1v2) 6= 0 and π2,1(v−13 g3) 6= 0, so we can choose

v1 and v4 so that v1g1v2 and v−13 g3v4 are in the forms of Lemma 5.1.10. Therefore, it

is enough to verify the theorem in the following two cases :

1. π2,1(g2) 6= 0 :

g1 =

[0 −a−1

a d

], g2 =

[0 −b−1

b 0

], g3 =

[e −c−1

c 0

]We obtain from our assumptions that x(g1g2) = bd, x(g2g3) = be, and x(g1g2g3) =

bde− ab−1c. Therefore, β(g1, g2, g3) equals

α(g1, g2)α(g1g2, g3)α(g1, g2g3)−1α(g2, g3)−1

= t(a, b)t(−a−1b, bd)t(bd, c)t(−b−1d−1c, bde− ab−1c)

t(a, be)−1t(−a−1be, bde− ab−1c)−1t(b, c)−1t(−b−1c, be)−1

= t(a, e−1)t(−a−1b, bd)t(−ab−1c(bde)−1, bde− ab−1c)t(d, c)t(−b−1c, be)−1

= t(a−1, e)t(−a−1b, bd)t(d, c)t(−b−1c, be)−1t(−ab−1c, bde)

= t(a−1, e)t(c, bd)t(d, c)t(−bc−1, be)t(−ab−1c, e)

= t(c, b)t(c, d)t(d, c)t(−bc−1, be)t(−b−1c, e)

= t(c, b)t(−bc−1, b) = t(−b, b) = 1

as desired.


2. π2,1(g2) = 0 :

g1 =

[0 −a−1

a d

], g2 =

[b−1 0

0 b

], g3 =

[e −c−1

c 0

]In this case, x(g1g2) = ab−1, x(g2g3) = bc, x(g1g2g3) = ab−1e+ bcd. Then

β(g1, g2, g3) = α(g1, g2)α(g1g2, g3)α(g1, g2g3)−1α(g2, g3)−1

From Lemma 5.1.3 on the first and last terms, we have

β(g1, g2, g3) = t(a, b)t(ab−1, c)t(−a−1bc, ab−1e, ab−1e+ bcd)t(a, bc)−1

t(−a−1bc, ab−1e+ bcd)−1t(c, b)−1

= t(a, b)t(b, c)t(ab−1, c)t(bc, a)

= t(a, b)t(a, c)t(bc, a)

= t(a, bc)t(bc, a) = 1

which completes the proof of the theorem.

5.2 Splitting of the cocycle over a subgroup

The goal of this section is to show that the cocycle α splits over SL2(R[[t]]). We do

so by proving that it is equal to a coboundary. For g =

[a b

c d

], define

s(g) =

t(c, d) if v(c) > 0 and c 6= 0

1 if v(c) = 0,+∞and

γ(g1, g2) = s(g1)−1s(g2)−1s(g1g2).

Clearly, γ(., .) is a coboundary. Write N ′ = N ∩ SL2(R[[t]]). It is easy to see that

N ′ is a subgroup of SL2(R[[t]]).

Lemma 5.2.1. For all u, v, w ∈ N ′ and g1, g2 ∈ SL2(R[[t]]), we have

α(ug1v, v−1g2w) = α(g1, g2),

and

γ(ug1v, v−1g2w) = γ(g1, g2).


Proof. The first statement is proved more generally in Theorem 5.1.11, as a corollary

of Eq. (22), (23) and (24). For the second statement, it is enough to prove that

s(gu) = s(ug) = s(g) for all u ∈ N ′, g ∈ SL2(R[[t]]). Write u =

[1 x

0 1

]and

g =

[a b

c d

]. Then

ug =

[a+ xc b+ xd

c d

]and gu =

[a ax+ b

c cx+ d

]Clearly, s(ug) = s(g). If v(c) = 0, then we also have s(gu) = s(g) = 1. Therefore,

suppose v(c) > 0. Then ad − cd = 1 and v(cd) > 0 implies that v(ad) = 0, hence

v(d) = 0. But v(x) ≥ 0, hence v(cx) > 0 and then v(cx+ d) = 0. Then if c =∑i≥0

aiti,

d =∑i≥0

biti and cx+ d =

∑i≥0

citi, we have a0 = 0 and b0 = c0 6= 0. Therefore,

s(gu) = t(c, cx+ d) = (−1)0a−v(cx+d)0 c

v(c)0 = b

v(c)0 = t(c, d) = s(g),

as claimed.

The following is a more specialized variation of Lemma 5.1.8. The main difference

is that we have more information on g.


[a b

c d

]∈ SL2(R[[t]]). Then there exists w ∈ N ′ such

that

gw =

a 0

c a−1

if d ∈ R[[t]]∗a −c−1

c 0

if c ∈ R[[t]]∗

Similarly, there exists u ∈ N ′ such that

ug =

d−1 0

c d

if d ∈ R[[t]]∗0 −c−1

c d

if c ∈ R[[t]]∗


Proof. Since ad− bc = 1, we must have either both a, d ∈ R[[t]]∗ or b, c ∈ R[[t]]∗. In

the first case, we have

g

[1 − b

a

0 1

]=

[a 0

c − cba

+ d

]=

[a 0

c a−1

],

and [1 − b

d

0 1

]g =

[a− bc

d0

c d

]=

[d−1 0

c d

].

In the second case, we have

g

[1 −d

c

0 1

]=

[a −da

c+ b

c 0

]=

[a −c−1

c 0

],

and [1 −a

c

0 1

]g =

[0 −da

c+ b

c d

]=

[0 −c−1

c d

].

Theorem 5.2.3. For all g1, g2 ∈ SL2(R[[t]]), α(g1, g2) = γ(g1, g2). Consequently, α

is a coboundary.

Remark 5.2.4. The proof is done with a case by case argument. The main reason

for this is that the definition of x depends on whether or not π2,1(g) = 0 or not,

and the definition for s(g) depends on whether or not v(c) = 0 or not. Because the

definition of α and γ depend on g1, g2 and their product g1g2, we get a surprisingly

large amount of cases on the values they can take. We will use Lemmas 5.2.1 and

5.2.2 to help reduce the number of cases.

Proof. Write

g1 =

[a1 b1

c1 d1

]g2 =

[a2 b2

c2 d2

]g1g2 =

[a1a2 + b1c2 a1b2 + b1d2

c1a2 + d1c2 c1b2 + d1d2

].

From Lemma 5.2.1, we can replace g1 and g2 by ug1v and v−1g2w for u, v, w ∈ N ′.We start with two main cases.


Case 1 : Suppose c1 6= 0 and c2 6= 0. Then we deal with three more subcases

1. Suppose v(c1) = 0. Then by the proof of Lemma 5.1.8, we can choose u, v ∈ N ′

(because here v(c1) = 0) such that g′1 := ug1v =

[0 −c−1

1

c1 0

]. By Lemma 5.2.2,

there is w ∈ N ′ such that

g′2 := v−1g2w =

[a2 0

c2 a−12

]or g′2 := v−1g2w =

[a2 −c−1

2

c2 0

]

Assume the first case. Then v(a2) = 0 and g′1g′2 =

[−c−1

1 a2 −c−11 a−1

2

c1a2 0

]. Then

we must have x(g′1) = c1, x(g′2) = c2 and x(g′1g′2) = c1a2. Therefore,

α(g′1, g′2) = t(c1a2/c1, c1a2/c2) = t(a2, c1a2/c2) = t(a2, c

−12 ).

But also, s(g′1) = 1, s(g′1g′2) = 1. If v(c2) > 0, then s(g′2) = t(c2, a

−12 ) =

t(a2, c2) = t(a2, c−12 )−1. If v(c2) = 0, then both t(a2, c

−12 ) and s(g′2) equal 1.

Therefore, α(g′1, g′2) = γ(g′1, g

′2).

Now suppose the second case. Then v(c2) = 0 and g′1g′2 =

[−c−1

1 c2 0

c1a2 −c1c−12

]This yields three subcases :

(a) v(a2) = 0 : Clearly, we have s(g1) = s(g2) = s(g1g2) = 1 and we obtain

α(g′1, g′2) = t(a2,−c1a2c

−22 ) = 1 because v(c1) = v(c2) = v(a2) = 0.

(b) v(a2) > 0 : In this case, we still have s(g1) = 0 and s(g2) = 0, but

s(g1g2) = t(c1a2,−c1c−12 ). Furthermore,

α(g1, g2) = t(a2, c1a2/c2) = t(a2,−c1c−12 )t(a2,−a2) = s(g1g2).

(c) a2 = 0 : Clearly we have s(g1) = s(g2) = s(g1g2) = 1 and we obtain

α(g′1, g′2) = t(−c−1

2 ,−c1c−22 ) = 1 because v(c1) = v(c2) = 0.

In each of these subcases, we conclude that α(g′1, g′2) = γ(g′1, g

′2)


2. v(c2) = 0 : In this case, choose v−1, w such that g′2 = v−1g2w =

[0 −c−1

2

c2 0

].

Then by Lemma 5.2.2, there exists u ∈ N ′ such that

g′1 = ug1v =

[0 −c−1

1

c1 d1

]or g′1 = ug2v =

[d−1

1 0

c1 d1

]

If v(c1) = 0, then we can use the first case, hence we are done by the previous

result. Therefore, we can assume v(c1) > 0 and use g′1 =

[d−1

1 0

c1 d1

], hence

v(d1) = 0 and g′1g′2 =

[0 −(d1c2)−1

d1c2 −c1c−12

]. Then x(g′1) = c1, x(g′2) = c2 and

x(g′1g′2) = d1c2. Therefore, α(g′1, g

′2) = t(d1c2c

−11 , d1) = t(c−1

1 , d1). Furthermore,

s(g′1) = t(c1, d1) = t(c−11 , d1)−1, s(g′2) = 1 and s(g′1g

′2) = 1. Therefore, α(g′1g

′2) =

γ(g′1, g′2).

3. v(c1) > 0 and v(c2) > 0. In this case, we must have v(d) = v(d′) = 0. Therefore,

we can choose by Lemma 5.2.2 u,w such that

g′1 = ug1 =

[a1 0

c1 a−11

]and g′2 = g2w =

[a2 0

c2 a−12

]

Then g′1g′2 =

[a1a2 0

c1a2 + a−11 c2 (a1a2)−1

]. Clearly, we have v(a1) = v(a2) = 0 and

s(g′1) = t(a1, c1), s(g′2) = t(a2, c2). We deal with two cases.

(a) c1a2 +a−11 c2 = 0 : In this case, c2 = −(a1a2c1), x(g′1) = c1, x(g′2) = c2, and

x(g′1g′2) = (a1a2)−1, hence

α(g′1, g′2) = t((a1a2)−1c−1

1 , (a1a2)−1c−12 ) = t(a1a2c1, a1a2c2)

= t(−c2, a1a2c2) = t(−c2, a1a2).


But now, s(g′1g′2) = 1. Therefore,

s(g′1)−1s(g′2)−1 = t(c1, a1)t(c2, a2)

= t(−a−11 a−1

2 c2, a1)t(c2, a2)

= t(−a−11 a−1

2 , a1)t(c2, a1)t(c2, a2)

= t(a−12 , a2)t(−1, a1a2)t(−1, a1a2)t(c2, a1a2)

= t(−a−12 , a2)t(−1, a1)t(−c2, a1a2)

= t(−c2, a1a2).

(b) c1a2 +a−11 c2 6= 0 : In this case, s(g′1g

′2) = t(c1a2 +a−1

1 c2, (a1a2)−1). Indeed,

note that the equality holds in both cases where v(c1a2 + a−11 c2) = 0

(because v(a1a2) = 0) and v(c1a2 + a−11 c2) > 0. Furthermore, we have

x(g′1g′2) = c1a2 + a−1

1 c2 and we can compute

s(g′1)−1s(g′2)−1s(g′1g′2) = t(c1, a1)t(c2, a2)t(c1a2 + a−1

1 c2, (a1a2)−1)

= t(c1, a1)t(c2, a2)t(c1a2 + a−11 c2,−(a1a2)−1c2c

−11 )t(c1a2 + a−1

1 c2,−c1c−12 )

= t(c1, a1)t(c2, a2)t(c1a2, c2a−11 )t(c1a2 + a−1

1 c2, c1c−12 )t(c1a2 + a−1

1 c2,−1)

= t(c2, a2)t(a2, c2a−11 )t(c1, c2)t(c1a2 + a−1

1 c2, c1c−12 )t(c1a2 + a−1

1 c2, c1a2 + a−11 c2)

= t(a1, a2)t(c1, c2)t(c1a2 + a−11 c2, c1c

−12 )t(c1a2 + a−1

1 c2, c1a2 + a−11 c2)

= t(c−11 , c−1

2 )t(c1a2 + a−11 c2, c

−12 )t(c−1

1 , c1a2 + a−11 c2)t(c1a2 + a−1

1 c2, c1a2 + a−11 c2)

= t((c1a2 + a−11 c2)/c−1

1 , (c1a2 + a−11 c2)/c−1

2 ) = α(g′1, g′2).

Case 2 : Either c1 = 0 or c2 = 0 :

1. Suppose c1 = 0. Then v(d1) = 0.

(a) If c2 = 0, then by Lemma 5.1.3, we have α(g1, g2) = t(x(g2),x(g1)) =

t(d2, d1). But we must have v(d1) = v(d2) = 0, hence we also have

t(d2, d1) = 1 by property (iv). Furthermore, s(g1) = s(g2) = s(g1g2) = 1.

(b) If c2 6= 0, then by Lemma 5.1.3, we have α(g1, g2) = t(x(g1),x(g2)) =

t(d1, c2). Now, if v(c2) = 0, t(d1, c2) is just 1 by property (iv), but we also

have s(g1) = 1, s(g2) = 1 and s(g1g2) = 1.


If v(c2) > 0, we have v(d1) = v(d2) = 0. Furthermore, s(g1) = 1, s(g2) =

t(c2, d2) and s(g1g2) = t(d1c2, d1d2). Then

γ(g1, g2) = t(c2, d2)−1t(d1c2, d1d2)

= t(d2, c2)t(d1, d1)t(d1, d2)t(c2, d1)t(c2, d2)

= t(d1, c2) = α(g1, g2).

as claimed.

2. Suppose c2 = 0. Then v(a2) = v(d2) = 0. If c1 = 0, then we have already dealt

with this case. Therefore, assume c1 6= 0. Then by Lemma 5.1.3,

α(g1, g2) = t(x(g1),x(g2)) = t(c1, d2).

If v(c1) = 0, then t(c1, d2) is 1 by Proposition 5.1.1(iv), but we also have

s(g1) = s(g2) = s(g1g2) = 1.

If v(c1) > 0, we have v(d1) = 0. Furthermore, s(g1) = t(c1, d1). Set u =[d2 0

0 a2

]. Then

ug2 =

[1 d2b2

0 1

]∈ N ′.

But then

s(g1g2) = s(g1u−1ug2) = s(g1u

−1) = s

([a1a2 b1d2

c1a2 d1d2

])= t(c1a2, d1d2).

Therefore,

γ(g1, g2) = t(c1, d1)−1t(c1a2, d1d2)

= t(d1, c1)t(c1, d1)t(c1, d2)t(a2, d1)t(a2, d2)

= t(c1, d2) = α(g1, g2),

as desired. This completes every possible cases.

Bibliography

[1] D. Dummit, R. Foote, Abstract Algebra John Wiley and Sons, 2004

[2] Y. Eidelman, V. Milman, A. Tsolomitis Functional Analysis, An Introduction

American Mathemical Society, 2004.

[3] G. Folland, A course in Abstract Harmonic Analysis. CRC Press, 1994

[4] G. Folland, Real analysis, modern techniques and their applications. CRC Press,

1999

[5] T. Kubota Topological covering of SL(2) over a local field J. Math. Soc. Japan

Volume 19, Number 1 (1967), 114-121.

[6] C. Moeglin, M.-F.Vigneras, J.L.Waldspurger Correspondence de Howe sur un

corps p-adique, Springer (1987).

[7] R. Rao On some explicit formulas in the theory of Weil representation Pacific

Journal of Mathematics Volume 157, No. 2 (1993), 335-371.

[8] A. Weil Sur certaines groupes d’operateurs unitaires, Acta Math., 11 1964, 143-

211.

[9] I. E. Segal Foundations of the theory of dynamical systems of infinitely many

degrees of freedom (I) Mat.-Fys. Medd. Danske Vid. Selsk., 31, no 12 (1959),

1-39.

[10] I. E. Segal Transforms for operators and symplectic automorphisms over a locally

compact abelian group Math. Scand., 12-13 (1963), 31-43.

76

BIBLIOGRAPHY 77

[11] D. Shale Linear symmetries of free boson fields Trans. Amer. Math. Soc. 103

(1962), 149-167.

[12] A. Weil Basic Number Theory Springer-Verlag New York, 1967.

[13] H. Weyl The Classical Groups. Their Invariants and Representations Princeton

University Press, 1939.

[14] Y. Zhu, Theta functions and Weil Representations of Loop Symplectic Groups

Duke Mathematical Journal. 143 (2008), No. 1, 17-39

Documents

WEIL REPRESENTATION AND CENTRAL EXTENSION OF LOOP ... · preparation and writing style has been invaluable. I want to thank Erhard Neher for his never-ending source of wisdom that