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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.
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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].
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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
CHAPTER 1. INTRODUCTION 3
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
CHAPTER 1. INTRODUCTION 4
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.
CHAPTER 2. PRELIMINARIES 7
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.
CHAPTER 2. PRELIMINARIES 8
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 ).
CHAPTER 2. PRELIMINARIES 9
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
CHAPTER 2. PRELIMINARIES 10
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).
CHAPTER 2. PRELIMINARIES 11
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 :
CHAPTER 2. PRELIMINARIES 12
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.
CHAPTER 2. PRELIMINARIES 13
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.
CHAPTER 2. PRELIMINARIES 14
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
CHAPTER 2. PRELIMINARIES 15
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
CHAPTER 2. PRELIMINARIES 16
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) =
CHAPTER 2. PRELIMINARIES 17
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).
CHAPTER 3. THE WEIL REPRESENTATION 20
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,
CHAPTER 3. THE WEIL REPRESENTATION 21
(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.
CHAPTER 3. THE WEIL REPRESENTATION 22
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′).
CHAPTER 3. THE WEIL REPRESENTATION 23
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
CHAPTER 3. THE WEIL REPRESENTATION 24
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〉
CHAPTER 3. THE WEIL REPRESENTATION 25
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)
CHAPTER 3. THE WEIL REPRESENTATION 26
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)
CHAPTER 3. THE WEIL REPRESENTATION 27
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).
CHAPTER 3. THE WEIL REPRESENTATION 28
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∗
].
CHAPTER 3. THE WEIL REPRESENTATION 29
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).
CHAPTER 3. THE WEIL REPRESENTATION 30
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.
CHAPTER 3. THE WEIL REPRESENTATION 31
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.
CHAPTER 3. THE WEIL REPRESENTATION 32
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.
CHAPTER 3. THE WEIL REPRESENTATION 33
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.
CHAPTER 3. THE WEIL REPRESENTATION 34
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.
CHAPTER 3. THE WEIL REPRESENTATION 35
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,
CHAPTER 3. THE WEIL REPRESENTATION 36
(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
CHAPTER 3. THE WEIL REPRESENTATION 37
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.
CHAPTER 3. THE WEIL REPRESENTATION 38
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 .
CHAPTER 3. THE WEIL REPRESENTATION 39
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 ∗.
CHAPTER 3. THE WEIL REPRESENTATION 40
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
CHAPTER 3. THE WEIL REPRESENTATION 41
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).
CHAPTER 3. THE WEIL REPRESENTATION 42
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.
CHAPTER 4. CONTINUITY OF THE REPRESENTATIONS 45
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.
CHAPTER 4. CONTINUITY OF THE REPRESENTATIONS 46
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.
CHAPTER 4. CONTINUITY OF THE REPRESENTATIONS 47
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.
CHAPTER 4. CONTINUITY OF THE REPRESENTATIONS 48
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
CHAPTER 4. CONTINUITY OF THE REPRESENTATIONS 49
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.
CHAPTER 4. CONTINUITY OF THE REPRESENTATIONS 50
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.
CHAPTER 4. CONTINUITY OF THE REPRESENTATIONS 51
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〉).
CHAPTER 4. CONTINUITY OF THE REPRESENTATIONS 52
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).
CHAPTER 4. CONTINUITY OF THE REPRESENTATIONS 53
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 ′).
CHAPTER 4. CONTINUITY OF THE REPRESENTATIONS 54
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
CHAPTER 4. CONTINUITY OF THE REPRESENTATIONS 55
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
CHAPTER 5. A SPECIFIC CENTRAL EXTENSION 58
(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)
).
CHAPTER 5. A SPECIFIC CENTRAL EXTENSION 59
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)
CHAPTER 5. A SPECIFIC CENTRAL EXTENSION 60
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.
Lemma 5.1.4. Suppose g =
[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).
CHAPTER 5. A SPECIFIC CENTRAL EXTENSION 61
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.
Lemma 5.1.5. Suppose g =
[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
],
where µ ∈ U , we have
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).
CHAPTER 5. A SPECIFIC CENTRAL EXTENSION 62
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
]
where µ ∈ U , we have
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
,
CHAPTER 5. A SPECIFIC CENTRAL EXTENSION 63
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).
CHAPTER 5. A SPECIFIC CENTRAL EXTENSION 64
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.
CHAPTER 5. A SPECIFIC CENTRAL EXTENSION 65
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
]
where µ ∈ U , we have
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].
CHAPTER 5. A SPECIFIC CENTRAL EXTENSION 66
Lemma 5.1.8. Suppose g =
[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.
Lemma 5.1.10. Suppose g =
[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.
CHAPTER 5. A SPECIFIC CENTRAL EXTENSION 67
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)
CHAPTER 5. A SPECIFIC CENTRAL EXTENSION 68
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.
CHAPTER 5. A SPECIFIC CENTRAL EXTENSION 69
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).
CHAPTER 5. A SPECIFIC CENTRAL EXTENSION 70
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.
Lemma 5.2.2. Suppose 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]]∗
CHAPTER 5. A SPECIFIC CENTRAL EXTENSION 71
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.
CHAPTER 5. A SPECIFIC CENTRAL EXTENSION 72
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)
CHAPTER 5. A SPECIFIC CENTRAL EXTENSION 73
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).
CHAPTER 5. A SPECIFIC CENTRAL EXTENSION 74
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.
CHAPTER 5. A SPECIFIC CENTRAL EXTENSION 75
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.
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