PART I: SYMPLECTIC ICE. PART II: GLOBAL AND …sd054cv7359/...Hamel and King [7] showed how characters of irreducible representations times the deformed Weyl denominators are equal

PART I: SYMPLECTIC ICE.

PART II: GLOBAL AND LOCAL KUBOTA SYMBOLS.

A DISSERTATION

SUBMITTED TO THE DEPARTMENT OF MATHEMATICS

AND THE COMMITTEE ON GRADUATE STUDIES

OF STANFORD UNIVERSITY

IN PARTIAL FULFILLMENT OF THE REQUIREMENTS

FOR THE DEGREE OF

DOCTOR OF PHILOSOPHY

Dmitriy Ivanov

August 2010

http://creativecommons.org/licenses/by-nc/3.0/us/

This dissertation is online at: http://purl.stanford.edu/sd054cv7359

© 2010 by Dmitriy Ivanov. All Rights Reserved.

Re-distributed by Stanford University under license with the author.

This work is licensed under a Creative Commons Attribution-Noncommercial 3.0 United States License.

ii



http://purl.stanford.edu/sd054cv7359

I certify that I have read this dissertation and that, in my opinion, it is fully adequatein scope and quality as a dissertation for the degree of Doctor of Philosophy.

Daniel Bump, Primary Adviser


Akshay Venkatesh


Anthony Licata

Approved for the Stanford University Committee on Graduate Studies.

Patricia J. Gumport, Vice Provost Graduate Education

This signature page was generated electronically upon submission of this dissertation in electronic format. An original signed hard copy of the signature page is on file inUniversity Archives.

iii

Preface

Hamel and King [7] showed how characters of irreducible representations times the

deformed Weyl denominators are equal to the partition functions of certain ice models,

and Brubaker, Bump and Fne]riedberg [6] showed how to use the Yang - Baxter

equation to investigate these ice models.

In the first chapter of my thesis I use the Yang - Baxter equation to investigate

a certain ice model (a 6-vertex model). We consider a graph in the shape of a grid

with the caps on the right end. A state of this system consists of assignment of signs

+ or − to each edge. To each vertex we assign a number (called Boltzmann weight);

the Boltzmann weight of a vertex depends on the signs adjacent to this vertex. A

Boltzmann weight of a state is equal to the product of the Boltzmann weights of all

vertices, and the partition function is the sum of the Boltzmann weights of all possible

states. I show that the partition function of this ice model is equal to the product

of an irreducible character of the symplectic group Sp(2n,C) and a deformation of

the Weyl denominator. A similar result was originally proved by Hamel and King

[7], but the Boltzmann weights (for the vertices at the caps) that I use are different

then the ones that are used by Hamel and King [7]. Also, my proof of this result

uses Yang-Baxter equation (while the proof of Hamel and King does not). This

gives us a 6-vertex models for characters of Sp(2n,C), but this result can also be

interpreted as an example of an exactly solved model, that is, an ice model whose

partition function can be computed explicitly (this is of interest to people who work

in statistical mechanics).

Ice models can also be used to describe Whittaker functions. Let F be a locally

compact field and G be a split reductive group over F . Let T denote the maximal

iv

torus of G, B the positive Borel subgroup and U the unipotent radical of B = TU . Let

ψ denote a non-trivial character of U . Given an irreducible representation (π, V ) of G,

a Whittaker model of this representation is a space of functions Wπ on G satisfying

W (ug) = ψ(u)W (g) which is closed under right translations and such that Wπ is

isomorphic to V (as G-modules). It is known that a Whittaker model is unique (if it

exists). Casselman-Shalika formula relates the characters of the L-group with values

of the Whittaker functions on a p-adic group. It is known that a Whittaker function

can be described as the partition function of a statistical system in the 6-vertex model.

For example, Brubaker, Bump and Friedberg [6] used a certain statistical system in

the 6-vertex model and Yang-Baxter equation to study p-adic Whittaker functions of

type A. The ice model studied here is similar to the U-turn models used by Kuperberg

[9] to enumerate classes of alternating sign matrices. After the work in this paper was

done, Brubaker, Bump, Chinta, Gunnells [5] followed my arguments (using the same

ice model but with different Boltzmann weights) to represent a Whittaker function

on the metaplectic double cover of Sp(2n, F ) where F is a nonarchimedean local field.

In chapters 2 and 3 of my thesis I study the global and local Kubota symbols and

their properties. Kubota has proven the following theorems:

Theorem 1. Let F be a totally complex number field and O be its ring of integers; let

n be a positive integer. We assume that F contains a primitive nth root of unity, and

let µn denote the (multiplicative) group of roots of unity. For M ∈ O we shall define

Γ(M) to be the subgroup of SL2(O) which consists of matrices that are congruent to

the identity matrix modulo M . Define a map κ : Γ(M) → µn by

κ(γ) =( cd

),

where γ =

(a b

c d

)and

(cd

)is the power residue symbol. (κ is called the global

Kubota symbol.) Then there exists M such that κ is a homomorphism from Γ(M) to

µn.

A value of M for which this theorem is true is a called a level for the global Kubota

symbol. Kubota has shown that M = n2 is a level for the global Kubota symbol, and

v

later Bass, Milnor and Serre [3] have improved this result by finding a smaller value

of M which is a level for the Kubota symbol (see proposition 3.1 on page 14 of this

paper).

Then Kubota proved the following theorem about local fields:

Theorem 2. Let Fp be a nonarchimedean local field, and p be a finite prime of F .

We assume that F contains a primitive nth root of unity for some integer n, and

we assume that p does not divide n. Define χ(σ) = c, if c 6= 0 and χ(σ) = 0

otherwise; here σ ∈ SL2(Fp), σ =

(a b

c d

). Then define a(σ, τ) = (χ(σ), χ(τ)) ·

(−χ(σ)−1χ(τ), χ(στ)), where (∗, ∗) is the Hilbert - Hasse norm residue symbol of

degree n of Fp. Then a(σ, τ) is a 2-cocycle which determines a topological covering

group Gp of Gp such that Gp is central as a group extension

Using the previous theorem. Kubota proved the following theorem

Theorem 3. Assume p does not divide n. Let Op be the ring of integers of Fp

and Kp = GL2(Op). Then a(γ, γ′) splits on Kp. More precisely, the claim is that

a(γ, γ′) = s(γ)s(γ′)s(γγ′)−1 for γ, γ′ ∈ Kp. Here s(γ) = (c, d)−1 if cd 6= 0 and ordp c

is not divisible by n. Otherwise s(γ) = 1.

The global Kubota symbol has the following important properties:

1) If c ≡ c′ mod d, then(

cd

)=(

cd′

).

2) If c and d are relatively prime, then(

cd

)=(

dc

)·∏

v∈S(c, d)v where S is a certain

set of places of F . This property is usually called the reciprocity law.

In chapter 2 of my thesis I introduce the local Kubota symbol. A local Kubota

symbol is defined as follows: Let F be a local field. The local Kubota symbol is then

defined by(

dc

)F

= (c, d) if n does not divide ordp c and(

dc

)F

= 1 if n divides ordp c.

Equivalently, (dc)F = ($ord(c), d) where $ denotes an irreducible element of Op; $ is

unique (up to multiplication by a unit).

I show that this local symbol has properties that are similar to these two proper-

ties, namely

3)If x ≡ y mod z, then(

xz

)F

=(

yz

)F

and

4)If c, d are coprime elements of Op then we have(

cd

)F

=(

dc

)F· (c, d).

vi

After that I use parallel between properties 1) and 2) of the global Kubota symbol

and properties 3) and 4) of the local Kubota symbol to give alternative proofs of

theorems 1 and 3 based on these properties. My proofs are shorter and simpler than

Kubota’s proofs. I also generalize theorem 3 above to the case when p divides n (and

give a proof of this generalization).

In the last chapter of my thesis I work with the global Kubota symbol. I investigate

the levels for the global Kubota symbols for particular fields, and I obtain new levels

for these fields which improve the results of Kubota, and of Bass, Milnor and Serre.

vii

Acknowledgments

It is a pleasure to thank all the people who made this thesis possible.

First, I would like to thank my advisor Daniel Bump for his continuous support

during all the time I was at Stanford. His encouragement, guidance and ability to

explain things clearly and simply have been invaluable, and have helped me to develop

an understanding of the subject. I could not have imagined having a better advisor

and mentor for my Ph.D study.

I also would like to thank everyone in the Department of Mathematics at Stanford

University for their constant help and support for the last six years.

Finally, I would like to thank my parents Gennadiy Ivanov and Lidiya Pozd-

nyakova who also have helped me a lot and supported me in any respect while I was

in the Ph.D program. This thesis is dedicated to them.

This work was supported in part by NSF grant DMS-0652817.

viii

Contents

Preface iv

Acknowledgments viii

1 Symplectic ice 1

2 Global and local Kubota symbols 22

3 Levels for the global Kubota symbol 29

References 40

ix

Chapter 1

Symplectic ice

In this chapter we shall give a new proof of a result due to Hamel and King [7]

based on the Yang - Baxter equation. It represents the product of the character of

the symplectic group times a deformation of the Weyl denominator as a partition

function of a certain type of ice. Our proof is similar to the proof of a similar result

for the general linear group in Brubaker, Bump, Friedberg [6]. Our proof will also use

Proctor patterns, and a good reference for this topic is Beineke, Brubaker, Frechette

[4].

We begin this chapter by explaining the notation we will be using later. By ice we

shall mean a lattice (or a more general graph), to each edge of which we may assign

a sign (either + or −). We shall consider six different types of ice, called delta ice,

gamma ice, delta-delta ice, delta-gamma ice, gamma-delta ice and gamma-gamma ice.

To each vertex we assign a weight (called Boltzmann weight); for each type of ice the

weight depends on the signs of the four adjacent edges. Let z1, ..., zn, t be complex

numbers with zi 6= 0 for all i. We shall refer to zi as spectral parameters, and t a

deformation parameter. In the table below we include Boltzmann weights for each

type of ice.

1

CHAPTER 1. SYMPLECTIC ICE 2

Gamma

Icei i i i i i

Boltzmann

weight1 zi(t+ 1) 1 t zi zi

Delta

Icei i i i i i

Boltzmann

weightzi zi(t+ 1) 1 tzi 1 1

Delta-Delta

Ice

j i

i j

j i

i j

j i

i j

j i

i j

j i

i j

j i

i j

Boltzmann

r weighttzi + zj zj(t+ 1) t(zj − zi) zi − zj (t+ 1)zi zi + tzj

Gamma-Delta

Ice

j i

i j

j i

i j

j i

i j

j i

i j

j i

i j

j i

i j

Boltzmann

weightt2zj − zi zj(t+ 1) tzj + zi tzj + zi (t+ 1)zi zi − zj

Gamma-Gamma

Ice

j i

i j

j i

i j

j i

i j

j i

i j

j i

i j

j i

i j

Boltzmann

weighttzi + zj tzj + zi t(zj − zi) zi − zj (t+ 1)zi (t+ 1)zj

Delta-Gamma

Ice

j i

i j

j i

i j

j i

i j

j i

i j

j i

i j

j i

i j

Boltzmann

weightzi − zj zi(t+ 1) tzi + zj tzi + zj (t+ 1)zj −t2zi + zj

By an admissible configuration (or admissible state) we shall mean a labeling of

the edges of the graph by signs + or − such that each vertex on the graph is listed in


the table above. The Boltzmann weight of an admissible state is simply the product

of the Boltzmann weights of all vertices in this state. The partition function of an

ice is the sum of Boltzmann weights of all admissible states. (This definition of the

partition function comes from statistical mechanics).

Yang-Baxter equation will be an important tool in our proof. Here is the version

of the Yang-Baxter equation that we will use:

Lemma 1. Let X, Y ∈ {Γ,∆}, and let S be a vertex of type X, T of type Y and R of

type XY . We use the spectral parameter zi at the vertex S, we use zj at the vertex T

and the spectral parameters zi, zj at the vertex R. The Boltzmann weights are given

in the table on the previous page. Then the partition function of

τ

σ µ

ν β

γ

α

θ

ρR

S

T

is equal to the partition function of

σ

τ β

δ

α

ψ

φ

θ

ρR

T

S

for every fixed combination of signs σ, τ, α, β, ρ, θ.

We refer the reader to Brubaker, Bump, Friedberg [6] for the proof of this proposi-

tion. Theorem 4 on page 13 of that paper gives a proof of this proposition in the case

X = Y = Γ and a similar argument shows that the proposition is true in the other

three cases. Informally, this means that if the signs on the outer edges are fixed, then

we can push the braid (i.e. a vertex of type XY ) to the right without changing the

partition function. We emphasize that the vertices R,S, T have the same Boltzmann


weights in both pictures. A consequence of this proposition is that it allows us to

switch the spectral parameters without changing the partition function. We see that

zi corresponds to the top row and zj corresponds to the bottom row while in the

second picture zi corresponds to the bottom row and zj corresponds to the top one.

We may apply the Yang-Baxter equation for delta ice three times, we see that the

partition functions of the following ice are equal:

∆

∆

A

i i i

j j j

and

∆

∆

A

j j j

i i i

Note that the braid in this example connects two rows of delta ice (in both cases),

so the vertex A is a vertex of delta - delta ice. The spectral parameters corresponding

to these rows have been switched; in other words, if the spectral parameters of the

rows of the first ice are zi(top row) and zj(bottom row), then after applying the

Yang-Baxter equation zj will correspond to the top row and zi will correspond to the

bottom row.

Yang-Baxter equation can be even applied to ice which contains rows of both

delta ice and gamma ice; in this case the braid will have more than one vertex. For

example, if we apply the Yang-Baxter equation twenty times, we see that the partition

functions of the following ice are equal:


∆

Γ

∆

Γ

A

B

CD

and

∆

Γ

∆

Γ

A

B

D C

Here the vertex A is a vertex of delta - delta ice, the vertex B is a vertex of

gamma - gamma ice, the vertex C is a vertex of gamma - delta ice and the vertex D

is a vertex of delta - gamma ice. Also, the spectral parameters of the delta rows and

of the gamma rows have been switched. This means that if the spectral parameters

corresponding to delta rows are zi (upper delta row) and zj (lower delta row), then

after applying the Yang-Baxter equation, the spectral parameter corresponding to

the upper delta row is zj and the spectral parameter corresponding to the lower one

is zi. The same is true for the rows of gamma ice. This example will be used later in

our proof.

Yang-Baxter equation will be a useful tool later when we will be proving that the

partition function of certain ice is invariant under interchanging spectral parameters.

Let λ = (λ1, λ2, ..., λn) ∈ Zn; we assume that λ1 ≥ λ2 ≥ λ3... ≥ λn ≥ 0. We


may view λ as a dominant weight of the symplectic group Sp(2n,C). We recall that

Sp(2n,C) consists of all 2n by 2n invertible matrices g satisfying gJgT = J where J =(0 −II 0

); here 0 denotes the zero matrix and I denotes the identity matrix. Let

χλ be the character of an irreducible representation of Sp(2n,C) with highest weight

λ. By the Weyl character formula there exists a Laurent polynomial in n variables

sspλ (x1, x2, .., xn) ∈ C[x1, x

−11 , x2, x

−12 , .., xn, x

−1n ] such that χλ(g) = ssp

λ (z1, z2, .., zn)

where z1, z2, .., zn, z−11 , z−1

2 , .., z−1n are the eigenvalues of g ∈ Sp(2n,C). Define

D(z1, z2, .., zn, t) =∏

i

(1 + tz2i ) ·∏i<j

(1 + tzizj)(1 + tziz−1j ) · z−ρ

where z−ρ = z−n1 z−n+1

2 ...z−1n . We will show that ssp

λ ·D equals the partition function

of a six-vertex model system. We recall that the Weyl character fromula states that

χλ(g) =

∑w∈W (−1)l(w)zw(λ+ρ)∏α∈Φ+(zα/2 − z−α/2)

.

In this formula W is the Weyl group, l(w) is the length function and Φ+ is the

set of positive roots. Here z(λ+ρ) means zλ1+n1 zλ2+n−1

2 ...zλn+1n and zw(λ+ρ) means

(w(z1))λ1+n(w(z2))

λ2+n−1...(w(zn))λn+1 We now observe that the denominator in this

formula can be written as

∏α∈Φ+

(zα/2 − z−α/2) =∏

α∈Φ+

(z−α/2 − zα/2) =

z−ρ∏

α∈Φ+

(1− zα) = z−ρ∏

i

(1− z2i ) ·∏i<j

(1− zizj)(1− ziz−1j ).

Hence we see that D(z1, z2, .., zn,−1) is the denominator in the Weyl character for-

mula, so this result is a deformation of the Weyl character formula. This deformation

is due to Hamel and King [7] (even though their Boltzmann weights for the vertices

on the caps are different than ours), but here we give a new proof of this using Yang

- Baxter equation.

We consider the following problem. We consider a domain which consists of a


rectangular lattice with 2n rows; each odd numbered row is a row of delta ice and

each even numbered row is a row of gamma ice. Hence there are n rows of delta ice

and n rows of gamma ice. Let zi be the spectral parameter corresponding to the ith

row of delta ice and z−1i be the spectral parameter corresponding to the ith row of

gamma ice. We assume that λ = (λ1, .., λn) ∈ Zn be a partition (this means that

λ1 ≥ λ2 ≥ . . . ≥ λn ≥ 0). We assign signs to each edge as follows: on the left column,

we assign − to each row of delta ice (i.e. to each odd numbered row), and a + to

each even numbered row. On the bottom, we assign + to each edge. On the top,

we assign − to each column labeled λi + n − i (for 1 ≤ i ≤ n), and we assign +

to each remaining column. The columns of this lattice are numbered in descending

order from λ1 + n to 1. On the right side we connect each row of delta ice with the

following row of gamma ice with a cap (see picture).

z1

z−11

z2

z−12

12345

The signs at the ends of a row of delta ice and the row of gamma ice right below

it must be the same (thus, we can have either two + signs or two − signs). The

Boltzmann weights for the cap are as shown in the picture below


+

+

−

−

tzi

z−1i

We consider the partition function of the ice described above. Our result is the

following theorem:

Theorem 4. Let λ = (λ1, ..., λn) ∈ Zn be a partition. Then the partition function of

the ice described above is given by D · sspλ (z1, z2, .., zn).

Proof. Let I1 denote the given ice (see picture below)

z1

z−11

z2

z−12

12345

Let Z(I) denote the partition function of the ice I. Consider Z(I1)/D. We first

show that this quotient is invariant under the action of the Weyl group (which is

generated by permutations of z1, .., zn and also by the functions that take zi to z−1i

(for some i)).

Lemma 2. Z(I1)/D is invariant under any permutation of z1, ..., zn.

Proof. Since the group Sn is generated by transpositions of the form (k, k + 1), it

suffices to show that Z(I1)/D is invariant under the interchange i↔ i+1 (for any i).


We attach a braid to I1 (the picture below is the part of the ice which corresponds

to spectral parameters with indices i and i + 1, namely two rows of delta ice witrh

spectral parameters zi and zi+1 and two rows of gamma ice with spectral parameters

z−1i and z−1

i+1. The picture (ice) extends above and below this part.).

i i i i i

i i i i i

i+ 1 i+ 1 i+ 1 i+ 1 i+ 1

i+ 1 i+ 1 i+ 1 i+ 1 i+ 1

Let us denote this new ice by I2. The only admissible configuration of spins here is

shown in the picture below.

−

+

−

+

−

+

−

+

−

+

−

+

Therefore, every state of this new boundary problem (with ice I2) determines a

unique state of the original problem (with ice I1). Hence the partition function of I2

is the partition function of the ice I1 times the partition function of the braid. The

partition function of the braid is equal to (zi+tzi+1)(tz−1i +z−1

i+1)(tzi+z−1i+1)(tzi+1+z

−1i ),

and therefore

Z(I2) = (zi + tzi+1)(tz−1i + z−1

i+1)(tzi + z−1i+1)(tzi+1 + z−1

i )Z(I1).


An application of the Yang-Baxter equation tells us that the partition function of

I2 is equal to the partition of the ice I3 (see picture below) with zi, zi+1 switched. We

will denote this by Z∗(I3); ∗ means that zi, zi+1 are switched.

Note that Z∗(I3) =∑

eiZ∗(I4)(e1, e2, e3, e4) · Z(I5)(e1, e2, e3, e4) where the sum is

taken over all possible spins e1, e2, e3, e4, and the pictures of I4, I5 are shown below.

Here is a picture of I4 :

e4

e3

e2

e1

z−1i

zi

z−1i+1

zi+1

And here is a picture of I5:


e4

e3

e2

e1

At this point we used the computer program SAGE[14] to check that the following

equality holds:

Lemma 3. For every choice of e1, e2, e3, e4 ∈ {±} we have

Z(I5)(e1, e2, e3, e4) = z−2i z−2

i+1 ·Z∗(I6)(e1, e2, e3, e4) · (tzi +zi+1)(tzi+1 +zi)(tzizi+1 +1)2.

Here I6 is the ice that consists of two caps (see picture).

Note, in particular, that the expression z−2i z−2

i+1 ·(tzi +zi+1)(tzi+1 +zi)(tzizi+1 +1)2

does not depend on the choice of spins e1, e2, e3, e4, and therefore, we get

Z(I2) = Z∗(I3) =∑

eiZ∗(I4)(e1, e2, e3, e4) · z−2

i z−2i+1 · Z∗(I6) · (tzi + zi+1)(tzi+1 +

zi)(tzizi+1 + 1)2, or, equivalently,

(zi + tzi+1)(tz−1i + z−1

i+1)(tzi + z−1i+1)(tzi+1 + z−1

i )Z(I1) =

∑ei

Z∗(I4)(e1, e2, e3, e4) · z−2i z−2

i+1 · Z∗(I6) · (tzi + zi+1)(tzi+1 + zi)(tzizi+1 + 1)2.


We observe that∑

eiZ∗(I4) · Z∗(I6) = Z∗(I1), and hence we obtain the following

equality

(zi + tzi+1)(tz−1i + z−1

i+1)(tzi + z−1i+1)(tzi+1 + z−1

i )Z(I1) =

Z∗(I1) · z−2i z−2

i+1 · (tzi + zi+1)(tzi+1 + zi)(tzizi+1 + 1)2.

After simplifying it, we obtain

(tzi + zi+1)Z(I1) = (tzi+1 + zi)Z∗(I1),

which means that

Z(I1) =(tzi+1 + zi)

(tzi + zi+1)Z(I1)

∗

.

Recall that we defined D as

z−ρ ·∏

i

(1 + tz2i ) ·∏i<j

(1 + tzizj)(1 + tziz−1j ).

If we let D∗ denote the same expression, but with zi and zi+1 switched (i.e.

D∗(z1, z2, .., zi, zi+1, .., zn) = D(z1, z2, .., zi+1, zi, .., zn))

then it is easy to check that DD∗ = (tzi+1+zi)

(tzi+zi+1), and therefore we conclude that

Z(I1) =D

D∗Z(I1)∗,

which is to say that

Z(I1)/D = Z(I1)∗/D∗


which is exactly what we needed to show.

Lemma 4. Z(I1)/D is invariant under the change zi ↔ z−1i .

Proof. Since we have already shown that this expression is invariant under the change

zi ↔ zi+1 for any i, it suffices to show that Z(I1)/D is invariant under the change zn ↔z−1

n . We transform the last row of the given ice (the row of gamma ice corresponding

to spectral parameter z−1n ) into a row of delta ice as follows: we change the sign on

the left edge from + to − and we also change the signs of all the entries on the edges

in the last row. We observe that in the last row of horizontal edges only the following

types of Gamma ice can appear:

Gamma

Icei i i

Boltzmann

weight1 1 zi

These change to:

Delta

Icei i i

Boltzmann

weight1 1 zi

We observe that the Boltzmann weights are unchanged, and therefore, the partition

function is also unchanged by this transformation. Let us denote this new ice by I7;

hence Z(I7) = Z(I1). (We note that this will work only in the last row because it is

essential that there be no − signs on the bottom edges). We also define a new cap

which connects two rows of delta ice. The signs on the ends of this new cap must be

opposite (thus, one of them must be a + and the other one must be a −), and the

Boltzmann weights for this cap are shown in the picture below.


−

+

+

−

tzi

z−1i

So now, the last two rows are rows of delta ice. We attach a braid to this ice, as

shown in the picture below. We denote this new ice by I8.

∆

∆

zn

z−1n

The only admissible configuration of spins here is shown in the picture below.

−

−

−

−

Therefore, every state of this new boundary problem (with ice I8) determines a

unique state of the original problem (with ice I7). Hence the partition function of

each state of I8 is the partition function of the corresponding state of ice I7 times

the partition function of the braid. The partition function of the braid is equal to

(zn + tz−1n ).

An application of the Yang-Baxter equation tells us that the partition function of

I8 is equal to the partition of the ice I9 (see picture below) with zn and z−1n switched.


∆

∆

z−1n

zn

We will denote this by Z∗(I9); ∗ means that zn and z−1n are switched.

We have Z∗(I9)) =∑

e1,e2Z∗(I10)(e1, e2) · Z∗(I11)(e1, e2) where the sum is taken

over all possible spins e1, e2, and the pictures of I10 and I11 are shown below:

Here is a picture of I10:r

e1

e2

And here is a picture of I11:e1

e2

We checked that the following equality holds:

Z∗(I11) = (tzn + z−1n )Z∗(I12).

where I12 is the ice that only consists of the new cap (see picture)

This result is very similar to the fish equation (lemma 9 on page 14 of Kuperberg

[9]), although the Boltzmann weights that are used in the fish equation and our

Boltzmann weights are different.

Note in particular that the factor (tzn + z−1n ) does not depend on the choice of

spins e1, e2, and therefore, we get

Z∗(I9) =∑

e1,e2Z∗(I10)(e1, e2) · Z∗(I11)(e1, e2) =

(tzn + z−1n ) ·

∑e1,e2

Z∗(I9)(e1, e2)Z∗(I12)(e1, e2) = (tzn + z−1

n )Z∗(I7).


Now we may change the last row back to the gamma ice. As mentioned before,

the partition function is not affected by this change; this means that Z∗(I7) = Z∗(I1).

Hence we obtain the following equality:

(zn + tz−1n )Z(I1) = (tzn + z−1

n )Z∗(I1).

From this we conclude that Z(I1)Z(I1)∗

= (tzn+z−1n )

(zn+tz−1n )

. But we also have DD∗ = (tzn+z−1

n )

(zn+tz−1n )

,

so that Z(I1)Z(I1)∗

= DD∗ , and so this proves that Z(I1)/D is invariant under the change

zi ↔ z−1i .

Lemma 5. Z(I1)/D is a polynomial in t (with coefficients in

C[z1, z2, , .., zn, z−11 , z−1

2 , .., z−1n ]),

and it is independent of t.

Proof. We observe that we may write D = z−ρ ·∏

α∈Φ+(1 + tzα), where Φ+ denotes

the set of positive roots. Let D′ = zρ ·∏

α∈Φ−(1 + tzα).

We see that DD′ is invariant under the action of the Weyl group (since the Weyl

group simply permutes all the roots), and therefore Z(I)D′ = (Z(I1)/D) · (DD′) is

also invariant under the action of the Weyl group.

But Z(I1)D′ is clearly divisible by D′ (here we view both terms as polynomials

in t), and therefore, since Z(I1)D′ is invariant under the action of the Weyl group, it

must also be divisible by D. Now, since D and D′ are relatively prime in a unique

factorization domain C[z1, z−11 , z2, z

−12 , .., zn, z

−1n ][t] (note that this ring is a localization

of a UFD C[z1, z2, , .., zn, ][t], so it is indeed a UFD), it follows that D must divide

Z(I1), as desired.

To show that Z(I1)/D is independent of t, we observe that D has degree n2 (as a

polynomial in t). Hence it suffices to show that Z(I1) has degree n2 as a polynomial

in t. In fact, since we already know that Z(I1)/D is a polynomial in t, it suffices to

show that the degree of Z(I1)) is ≤ n2. Z(I1) is the sum of Boltzmann weights of

each state, so it suffices to show that the partition function of each state has degree

≤ n2.


So consider an admissible state. We observe that the only (possible) powers of t

come from the following types of vertices:

Gamma

Icei i

Boltzmann

weightzi(t+ 1) t

Delta

Icei i

Boltzmann

weightzi(t+ 1) zit

We see that we need to count the number of − signs in the rows of edges between

the rows of delta ice and gamma ice.

We recall lemma 2 on page 17 of Brubaker, Bump, Friedberg [6], which counts

the number of − signs in the rows of edges above and below a row of gamma ice. We

state this lemma here for the convenience of the reader:

Lemma 6. Suppose we have a row of gamma ice and that the sign of the left edge

is +. Let m be the number of − signs in the vertical row of edges above this row of

gamma ice and m′ be the number of − signs in the row of vertical edges below this

row. Then m = m′ if and only if the sign of the right edge is − and m = m′ − 1 if

and only if the sign of the right edge is +. Moreover, if α1, α2, .., are the locations

of the − signs in the row of edges above this row of gamma ice and β1, β2, .. are the

locations of − signs in the row of edges below this row, the sequences α1, α2, .. and

β1, β2... interleave.

By symmetry of the delta ice and gamma ice, we obtain the following lemma:

Lemma 7. Suppose we have a row of delta ice and that the sign of the left edge is −.

Let m be the number of − signs in the row of edges above this row of delta ice and

m′ be the number of − signs in the row of edges below this row. Then m = m′ if and

only if the sign of the right edge is − and m = m′ − 1 if and only if the sign of the


right edge is +. Moreover, if α1, α2, .., are the locations of the − signs in the row of

edges above this row of delta ice and β1, β2, .. are the locations of − signs in the row

of edges below this row, the sequences α1, α2, .. and β1, β2... interleave.

Since the number of − signs in the top row of edges is n, we see that the number

of − signs in the rows of edges between the rows of delta ice and gamma ice is

≤ 2C(n, 2) = n(n− 1).

Moreover, since there are n caps, the degree of t coming from the Boltzmann

weights of these caps is ≤ n. We see that if the signs on the cap are ++, then the

Boltzmann weight is tzi, so it will contribute only one power of t to the Boltzmann

weight of a state; if the signs are −−, then the Boltzmann weight is z−1i , which does

not contribute any powers of t.

Hence the degree of the partition function of each state is ≤ 2C(n, 2) + n = n2,

and as explained above, this means that Z(I1)/D is independent of t. This completes

the proof of lemma 4.

By a Proctor pattern we shall mean a sequence of rows of nonnegative integers

ai,j and bi,j

a0,1 a0,2 . . . a0,r

b1,1 b1,2 . . . b1,r

a1,2 . . . a1,r...

. . . . . .

br,r

such that the rows interleave. This last condition means that

min{a(i−1,j), a(i,j)} ≥ bi,j ≥ max{a(i−1,j+1), a(i,j+1)}

and also

min{b(i+1,j−1), b(i,j−1)} ≥ ai,j ≥ max{b(i+1,j), b(i,j)}.


We now would like to establish a bijection between the set of all admissible states and

the Proctor patterns. Consider a row of delta ice and a row of gamma ice right below

it. Let ai denote the locations of − signs in the top row of vertical edges, bi denote

the locations of − signs of middle row of vertical edges and ci denote the locations of

− signs of bottom row of vertical edges. We consider two cases:

Case I: the signs on the cap that connects these two rows are −−. Then (using

lemmas 6 and 7) we have the following inequalities:

a1 ≥ b1 ≥ a2 ≥ . . . ≥ bm

b1 ≥ c1 ≥ b2 . . . ≥ bm.

Case II: the signs on the cap that connects these two rows are ++. Then we have

the following inequalities:

a1 ≥ b1 ≥ a2 ≥ . . . ≥ bm−1 ≥ am

b1 ≥ c1 ≥ b2 . . . ≥ bm−1 ≥ cm−1.

In this case we define bm = 0.

We see that the locations of − signs in the rows of vertical edges gives us a Proctor

pattern. Here are three rows of a Proctor pattern.

a1 a2 . . . am

b1 b2 . . . bm

c1 . . . cm−1

Moreover, we see that this is a one-to-one correspondence, meaning that given any

Proctor pattern, there exists an admissible state such that the locations of − signs


are exactly as described by this pattern. This also follows from the lemmas 6 and 7

above.

Since Z(I1)/D is independent of t, we may choose t = −1. This way, the Boltz-

mann weight of the vertices

Gamma

Icei

Boltzmann

weightzi(t+ 1)

Delta

Icei

Boltzmann

weightzi(t+ 1)

is equal to 0, so we consider all possible states which omit both of these vertices. .

We observe that there exists a bijection between the Weyl group and the set of

all Proctor patterns of the particular type, in which each entry in the pattern (except

for entries in the top row) is equal to one of the entries above it (or to 0 in the case

of the last entry in an even numbered row). This bijection is given as follows: given

a Proctor pattern, let di denote the sum of the entries of the (2i− 1)th row; in other

words, it denotes the sum of the entries in each odd numbered row; here 1 ≤ i ≤ n.

Also let dn+1 = 0. Note that the numbers di− di+1 for 1 ≤ i ≤ n all distinct. In fact,

since each entry in the pattern (except for entries in the top row) is equal to one of

the entries above it (or to 0 in the case of the last entry in an even numbered row),

it follows that di− di+1 is equal to λj + ρj for some j (1 ≤ j ≤ n). In other words, we

get a permutation of n distinct elements λ1 +ρ1, λ2 +ρ2, ..., λn +ρn. And for any such

permutation and any choice of the signs on the caps, we can construct a Proctor pat-

tern corresponding to it. We see that w(z1), w(z2)...w(zn) is equal to zε1σ(1), z

ε2σ(2), z

εn

σ(n)

for some σ ∈ Sn and each εi ∈ {±1}; the choice of εi depends on whether an admis-

sible state corresponds to case I or case II described above. Hence we see that we

indeed get a bijection between the Weyl group and the set of all Proctor patterns


of the particular type decribed above. Since we’ve already observed that there is a

bijection between the set of all Proctor patterns of the particular type and the set of

all admissible states, by composing these two bijections we obtain a bijection between

the set of all admissible states and the Weyl group. An explicit formula for this bijec-

tion is given by w → (−1)l(w)zw(λ+ρ), where (−1)l(w)zw(λ+ρ) represents the Boltzmann

weight of the corresponding admissible state. Here z(λ+ρ) means zλ1+n1 zλ2+n−1

2 ...zλn+1n

and zw(λ+ρ) means (w(z1))λ1+n(w(z2))

λ2+n−1...(w(zn))λn+1.

Note that the partition function is equal to∑

w∈W (−1)l(w)zw(λ+ρ) which is equal

to the numerator in the Weyl character formula. Also note that with t = −1, we

have D =∏

i(1− z2i ) ·∏

i<j(1− zizj)(1− tziz−1j ) · z−ρ which is the denominator in the

Weyl character formula. Hence we’ve obtained a deformation of the Weyl character

formula (which, as we mentioned before, is due to Hamel and King [7]), and the proof

of the theorem is complete.

Chapter 2

Global and local Kubota symbols

We begin by recalling the definition of a power residue symbol. Let n be a positive

integer and F be a number field containing a primitive nth root of unity, and O be its

ring of integers. Then the power residue symbol is defined as follows: If p is a prime

ideal and c ∈ p, then(

cp

)= 0. If c /∈ p, then

(cp

)is defined to be the unique nth root

of unity such that c(Np−1)/n ≡(

cp

)mod p. Here N is the norm of an ideal.

Now, if I is any ideal in O, we may factor I as a product of prime ideals I =∏

Pi,

and then we define(

cI

)=∏(

cPi

).

Finally, if d is any element in O, then we define(

cd

)to be equal to

(cI

)where I is

the principal ideal generated by d.

We also define(

01

)= 1.

We recall the following properties of the power residue symbol:

1) If ε is a unit in O then(

cεd

)=(

cd

).

2) If c ≡ c′ mod d, then(

cd

)=(

cd′

).

3) We have(

cc′

d

)=(

cd

) (c′

d

).

4) We have(

cdd′

)=(

cd

) (cd′

).

5)(−1

d

)= 1.

6) If p is prime (and prime to n), then(

cp

)= 1 if and only if c is an n-th power

residue mod p.

After reviewing the properties of a power residue symbol, we will consider the

following problem:

22

CHAPTER 2. GLOBAL AND LOCAL KUBOTA SYMBOLS 23

Let Fp be a nonarchimedean local field containing a primitive nth root of unity;

here p is a finite prime of F . Let Gp denote the group GL2(Fp). Let Op denote and

let Kp = GL2(Op); Kp is a compact subgroup of Gp.

We shall assume that p does not divide n for now. We shall deal with the case

when p divides n later.

Define χ(σ) = c, if c 6= 0 and χ(σ) = 0 otherwise; here σ ∈ SL2(Fp), σ =(a b

c d

). Then define a(σ, τ) = (χ(σ), χ(τ)) · (−χ(σ)−1χ(τ), χ(στ)), where (∗, ∗) is

the Hilbert - Hasse norm residue symbol of degree n of Fp. We recall the definition

of the Hilbert-Hasse norm here: let K be a local field containing a primitive nth root

of unity and L be the maximal abelian extension with Galois group G of exponent

n, and let K∗ denote the group of units of K and K∗n denote the subgroup of K∗

generated by the nth powers of elements of K∗. The local class field theory gives

us an isomorphism θ : K∗/K∗n → G. Then the Hilbert-Hasse norm of the pair

(x, y) ∈ K∗×K∗ is given by (x, y) = θ(y)(x1n )

x1n

; (x, y) is an nth root of unity. In Kubota

[8] Kubota has proved that a(σ, τ) is a 2 - cocycle which determines a topological

covering group Gp of Gp such that Gp is central as a group extension.

Define the local Kubota symbol as follows:(

dc

)F

= (c, d) if n does not divide ordp c

and(

dc

)F

= 1 if n divides ordp c. Equivalently, (dc)F = ($ord(c), d) where $ denotes an

irreducible element of Op; $ is unique (up to multiplication by a unit). To see that

these two definitions are equivalent, note that if n divides ordp c then ($ord(c), d) = 1.

If n does not divide ordp c, then (c,d)

($ord(c),d)= (c($−ord(c)), d)) = 1 because c($−ord(c))

is a unit.

We emphasize that this local symbol is not the same as the global power residue

symbol. Later in this chapter we shall establish a relationship between the global

power residue symbol and this local symbol.

We now show that this local symbol has properties analogous to the properties of

the global power residue symbol.

Lemma 8. If x ≡ y mod z, then(

xz

)F

=(

yz

)F.

Proof. If p does not divide z then z is a unit and so(

xz

)F

=(

yz

)F

= 1. If p does divide

z, then proposition 8 on page 210 in Serre [13] shows that the value(

xz

)F

depends


only on the residue x mod p; in particular, the condition x ≡ y mod z implies that

x ≡ y mod p and therefore, we may conclude that(

xz

)F

=(

yz

)F.

Lemma 9. If c, d are coprime elements of Op then we have(cd

)F

=(

dc

)F· (c, d).

Proof. Using the formula(

dc

)F

= ($ord(c), d) we have(cd

)F

(dc

)−1

F= ($ord(c), d) ·($−ord(d), c) = ($, dord(c)c−ord(d)). Proposition 8 on page

210 in Serre [13] tells us that this is equal to (c, d), as was to be shown.

Kubota also proved the following theorem:

Theorem 5. Assume p does not divide n. Then a(γ, γ′) splits on Kp. More precisely,

the claim is that a(γ, γ′) = s(γ)s(γ′)s(γγ′)−1 for γ, γ′ ∈ Kp = GL2(Op). Here s(γ) =

(c, d)−1 if cd 6= 0 and ordp c is not divisible by n. Otherwise s(γ) = 1.

Kubota proves it by considering several cases. Below we give an alternative proof

of this theorem without having to consider different cases.

Proof. We let γ, γ′, γ′′ be elements of Kp with γ′′ = γγ′. Let’s write

γ =

(a b

c d

), γ′ =

(a′ b′

c′ d′

), γ′′ =

(a′′ b′′

c′′ d′′

).

Let g denote the gcd of d′ and d′′; we write d′ = gd′0 and d′′ = gd′′0 where d′0 and

d′′0 are coprime. Since c = c′′d′ − c′d′′ we see that g divides c and so we may write

c = c0g for some c0.

We have c0 = c′′d′0 − c′d′′0 (*). Then(d′′

c′′

)F

=( gc′′

)F

(d′′0c′′

)F

=( gc′′

)F

(d′′0d′0c

′′

)F

(d′′0d′0

)−1

F

=

(by reciprocity)

( gc′′

)F

(d′0c′′, d′′0)

(d′′0d′0c

′′

)F

(d′′0d′0

)−1

F


= (d′0, c′′, d′′0)

( gc′′

)F

(d′′0c0

)F

(d′′0d′0

)−1

F

(d′′0, c0) =

(by reciprocity)

(d′0c′′, d′′0)

(g

d′0c′′

)F

(d′0c0

)F

(d0

c0

)F

(d′′0d′0

)−1

F

(d′′0c0) =

(by (*))

(d′0c′′, d′′0)

( gc′′

)F

(d′0c0

)F

(d0

c0

)F

(d′′0d′0

)−1

F

(d′′0, c0) =

(d′0c′′, d′′0)

( gc′′

)F

(d′0c0

)F

(d0

c0

)F

(d′′0d′0

)−1

F

(d′′0, c0)

(by reciprocity).

So the conclusion will follow if we can show that

(d′0c′′0, d

′′0)(d

′′0, c0)(c0, d

′0)(c

′′, g)(c′, g)−1 = σ(g, g′).

Recall that for all x we have (x, 1− x) = 1. By taking x = c0c′d′′0

we get

1 = ( c0c′d′′0

, 1− c0c′d′′0

) = ( c0c′d′′0

,d′0c′′

c′d′′0).

Therefore

σ(g, g′) = ( c′′

c, c′′

c) = ( c′′

gc0, c′′

c′)( c0

c′d′′0,

d′0c′′

c′d′′0) =

(d′0c′′, d′′0)(d

′′0, c0)(c0, d

′0)(d

′0, c

′)(c′′.g)(c′, g)−1 · (c′, c′)(c′′, c′′) =

(d′0c′′0, d

′′0)(d

′′0, c0)(c0, d

′0)(c

′′, g)(c′, g)−1

So the theorem is proved.

We now assume that p divides n.

We begin by developing properties of a local symbol analogous to the properties

in lemmas 8 and 9 above.

Lemma 10. If x ≡ y mod z and x ≡ y mod n2, then(

xz

)F

=(

yz

)F.

Proof. If p does not divide z, then z is a unit and so(

xz

)F

=(

yz

)F

= 1. Suppose p

divides z. We see that it suffices to show that(

xz

)F

= 1 if and only if(

yz

)F

= 1, and


therefore, we may (and do) assume that both x and y are units. We must show that(xy−1

z

)F

= 1. To do this, it suffices to show that xy−1 is an nth power in O (note

that since x, y are units, so is xy−1).

We shall use the fact that every unit of O sufficiently close to 1 has an nth root.

This is a consequence of Hensel’s lemma, and this is explained in Lang [10] on page

43. We include this proof below.

We consider an equation tn − xy−1 = 0. Hensel’s lemma shows that this equation

has a root in O provided |xy−1 − 1| < |n|2 where | | denotes the p - adic absolute

value. But our assumption that x ≡ y mod n2 clearly implies |xy−1−1| < |n|2 (recall

that we are assuming that p divides n), and therefore we conclude that xy−1 is an

nth power in O.

Therefore xy−1 is an nth power in O and so(

xy−1

z

)F

= 1, as was to be shown.

Next we prove the reciprocity law.

Lemma 11. If x and y are coprime elements of O then(

xy

)F

=(

yx

)F· (x, y).

Proof. We observe that since x and y are coprime, they can’t both be non-units, so

at least one of them is a unit. We consider two cases:

i) y is a unit. Then, by definition,(

yx

)F

= 1, and so in this case the claim is that(xy

)F

= (x, y) which is true by the definition of a local symbol(

xy

)F.

ii) x is a unit. In this case(

xx

)F

= 1, and so the claim we need to prove is that(yx

)F· (x, y) = 1. Using the fact that (x, y) = (y, x)−1 (this is one of the properties

of the Hilbert symbol), we see that the claim we need to prove is equivalent to the

claim(

yx

)F

= (y, x) which again is true by the definition of a local symbol(

yx

)F.

We can now prove a theorem that is analogous to theorem 5 above.

Theorem 6. Assume p divides n. Then a(γ, γ′) splits on Sp; Sp consists of elements

of GL2(Op) which are congruent to I mod n2). More precisely, the claim is that

a(γ, γ′) = s(γ)s(γ′)s(γγ′)−1 for γ, γ′ ∈ Sp. Here s(γ) = (c, d)−1 if cd 6= 0 and ordp c

is not divisible by n. Otherwise s(γ) = 1.


Proof. The proof of this theorem is the same as the proof of theorem 2 above. The

proof of theorem 2 relied on lemmas 8 and 9, and lemmas 10 and 11 above are

analogous to lemmas 8 and 9 in the case when p divides n.

Now we will establish the relationship between the global power residue symbol

and the local symbol defined above. Our goal is to prove the following theorem:

Theorem 7. Let F be a totally complex number field; we assume that −1 is an nth

power in F . Let S be a finite set of places of F containing the archimedean ones, and

containng all places that divide n. Then we have(d

c

)=∏v/∈S

(d

c

)Fv

,

provided that c and d are relatively prime.

Proof. We shall use Proposition V.3.4 on page 335 of Neukirch [11] (however, his

Hilbert symbol is the inverse of ours). We have

∏v/∈S,πv |c

(πv, d)−ordv(c) =

∏v/∈S,πv |c

(d

πv

)ordv(c)

=

(d

c

).

Here πv denotes a prime element at the place v. We have used the fact that when

πv|c, d is a unit in Ov (because c and d are coprime); Qv is the ring of integers of Fv.

Let FS =∏

v∈S Fv where Fv is the completion of F at v. Let OS be the ring of

S - integers. We have an embedding OS → FS. We see that OS is discrete and that

the quotient FS/OS is compact. We have an embedding SL2(OS) → SL2(FS).

Theorem 8. The embedding SL2(OS) → SL2(FS) described above can be lifted to an

embedding SL2(OS) → ˜SL2(FS) (where ˜SL2(FS) denotes the metaplectic group).

Proof. For each place v we have a cocycle σv : SL2(Fv) × SL2(Fv) → C∗. We’d like

to define a cocycle σ on SL2(FS).


For g, g′ ∈ SL2(FS) we define

σ(g, g′) =∏v∈S

σv(gv, g′v).

Here gv denotes the component of g corresponding to v (and similarly for g′v).

By Hilbert reciprocity law (see theorem 8.1 on page 414 of Neukirch [11]) this is

equal to ∏v/∈S

σv(gv, g′v)−1

(for g, g′ ∈ SL2(F )).

(We recall that the Hilbert reciprocity law states that 1 =∏

v σv(g, g′). In this

case we have to consider several cases, depending on whether c, c′, c′′ are all non-zero,

or whether some of them (possibly all of them) are equal to 0. We observe that if an

element is equal to 0 at one place, then it is equal to 0 at every place).

Finally, we see that this is equal to

κ(gg′)

κ(g)κ(g′)

(for g, g′ ∈ SL2(OS)). Here κ(g) =∏

v/∈S s−1(g) (s(g) was defined in theorem 5

above).

We have a short exact sequence 1 → µn → ˜SL2(FS) → SL2(FS) → 1 where µn

denotes the group of n - th roots of unity. An element of ˜SL2(FS) has the form

(g, ε) where g ∈ SL2(FS) and ε ∈ µn. The group operation is (g, ε) · (g′, ε′) =

(gg′, εε′σ(g, g′)). (It’s easy to check that ˜SL2(FS) is indeed a group under this op-

eration. Perhaps the only non-obvious thing to check is that this operation is as-

sociative, but this follows from the fact that σ is a 2-cocycle, so that it satisfies

σ(g, g′)σ(gg′, g′′) = σ(g′, g′′)σ(g, g′g′′)).

The embedding SL2(OS) → SL2(FS) lifts to an embedding SL2(OS) → ˜SL2(FS)

and this embedding is given by γ → (γ, κ(γ)).

Chapter 3

Levels for the global Kubota

symbol

Let F be a global field, and O denote its ring of integers. For M ∈ O we shall define

Γ(M) to be the subgroup of SL2(O) which consists of matrices that are congruent to

the identity matrix modulo M . Kubota has defined a map κ : Γ(M) → µn by

κ(γ) =( cd

),

where µn denotes the group of roots of unity in F and γ =

(a b

c d

)(Note that since ad− bc = 1, c and d are relatively prime, so that

(cd

)6= 0).

In general, this map need not be a group homomorphism, but we’d like to deter-

mine the values of M for which this map is a homomorphism. Kubota has proved

that this map is a (non-degenerate) homomorphism for M = n2, but this is not the

best possible value of M . Bass, Milnor and Serre [3] improved Kubota’s result by

obtaining a smaller value of M for which κ is a homomorphism (see proposition 3.1

on page 14 of this paper). However, their result can be improved, and in this chapter

we shall determine the best(smallest) possible value of M for the cases n = 2, 3, 5, 7.

We will need the following theorem:

Theorem 9. Let n be a positive integer, let ζn denote a primitive nth root of unity

29

CHAPTER 3. LEVELS FOR THE GLOBAL KUBOTA SYMBOL 30

and let F be a totally complex number field containing Q(ζn) with ring of integers O.

We assume that O is a principal ideal domain. Let f be an ideal in O satisfying the

following conditions:

i) The map O∗ → (O/f)∗ is surjective.

ii) If a, b ∈ O satisfy a ≡ b ≡ 1 mod f then(

ab

)=(

ba

). We shall refer to this

condition as the reciprocity law.

iii) If a ∈ O satisfies a ≡ 1 mod f then(−1

a

)= 1. (Remark: In particular, this

condition is satisfied if −1 is an nth power in F )

Then the Kubota symbol κ : Γ(f) → µn defined by

κ

(a b

c d

)=( cd

)is a group homomorphism.

Proof. Let γ =

(a b

c d

), γ′ =

(a′ b′

c′ d′

), γ′′ =

(a′′ b′′

c′′ d′′

)be elements of Γ(f)

such that γ′′ = γ · γ′.Let t denote the greatest common divisor of d and d′. Since both d and d′ are

congruent to 1 mod f we see that t is coprime to f and therefore (after multiplying t

by a suitable unit, if necessary) we may assume that t ≡ 1 mod f. We write d′′ = d′′0t

and d′ = d′0t. Since c = c′′d′− c′d′′ we see that t divides c and therefore we may write

c = c0t for some c0. Also, the greatest common divisor of d′0 and d′′0 is 1. We have(c′′

d′′

)=

(c′′

d′′0

)(c′′

t

)=

(c′′

t

)(c′′d′0d′′0

)(c′′′0d′′0

)−1

=

(c′′

t

)(c0d′′0

)(c′′′0d′′0

)−1

.

The last equality follows from the fact that c0 = c′′d′0−c′d′′0 and therefore c0 ≡ c′′d′0

mod d′′0. Next, since d′′ = dd′ + cb′ we see that d′′ ≡ dd′ mod cb′ and thus d′′ ≡ dd′

mod f2. Since t is coprime to f, we see that d′′0 ≡ dd′0 mod f2. Also, we see that

d′′0 ≡ dd′0 mod c0. Thus,(c′′

d′′

)=

(c′′

t

)(c0d′′0

)(d′0d′′0

)−1

=

(c′′

t

)(d′0d′′0

)−1 (c0d

)( c0d′0

).


Since c0 ≡ −d′′0c′ mod d′0 we see that(

c0d′0

)=(−d′′0 c′

d′0

)=(

d′′0 c′

d′0

)and therefore

(c′′

d′′

)=

(c′′

t

)(d′0d′′0

)−1 (c0d

)(d′′0c′d′0

)=

(c′′

t

)(c0d

)( c′d′0

).

Since c′′ = ca′ + dc′ we have c′′ ≡ dc′ mod c and therefore c′′ ≡ dc′ mod t. Hence(c′′

d′′

)=(c0d

)( c′d′0

)(c′

t

)(d

t

).

Since both t and d are congruent to 1 mod f we see that(

td

)=(

dt

)and therefore

we have (c′′

d′′

)=(c0d

)( c′d′0

)(c′

t

)(t

d

)=

(c0t

d

)(c′

td′0

)=( cd

)( c′d′

),

and this completes the proof of the theorem.

We prove that in the case n = 2, F = Q(i),O = Z[i], λ = 1 + i we can choose the

level for the Kubota symbol f = λ2 = (1 + i)2. (It is known that f = λ3 works, so this

result will produce a better level for the Kubota symbol).

Lemma 12. For n = 2, F = Q(i),O = Z[i], λ = 1 + i we have the following:

i) the map (O)∗ → (O/f)∗ is surjective.

ii) The reciprocity law holds: if a, b ∈ O with a ≡ b ≡ 1 mod f then(

ab

)=(

ba

).

Proof. Note that (O/f)∗ has order 2, and we see that the image of −i has order 2 in

(O/f)∗. Therefore, the image of −i under the map (O)∗ → (O/f)∗ generates (O/f)∗

which means that the map (O)∗ → (O/f)∗ is surjective.

Next, we prove the reciprocity law. If a ≡ 1 mod (1 + i)2, then a ≡ 1 or 3 mod

(1 + i)3, so there exists u = 0 or 1 such that (−1)ua ≡ 1 mod (1 + i)3. Similarly,

there exists v = 0 or 1 such that (−1)vb ≡ 1 mod (1 + i)3. Then (since i2 = −1)

we have(

(−1)ua(−1)vb

)=(

ab

)and similarly

((−1)vb(−1)ua

)=(

ba

). Hence we must prove that(

(−1)ua(−1)vb

)=(

(−1)vb(−1)ua

). But using the law of biquadratic reciprocity and the fact that

the Legendre symbol is a square of the quartic residue symbol we get


((−1)ua

(−1)vb

)=

((−1)ua

(−1)vb

)2

4

=

((−1)vb

(−1)ua

)2

4

·(−1)N((−1)ua−1)N((−1)vb−1)) =

((−1)vb

(−1)ua

)2

4

·1 =

=

((−1)vb

(−1)ua

),

and so the claim is proved.

(Here(∗∗

)4

denotes the quartic residue symbol and N denotes the norm of an

element).

Theorem 10. Let ζ denote the primitive 5th root of unity, Let O denote the ring

Z[ζ] (which is the ring of integers in Q(ζ)). We know that O∗ is generated by −ζand ζ + ζ−1. Let fr = (1 − ζ)r (here r = 1, 2, 3, ....) Then the map O∗ → (O/fr)

∗ is

surjective if r ≤ 3 and is not surjective if r > 3.

Proof. First, we show that this map is surjective if r ≤ 3.

If r = 1, then (O/f1) is isomorphic to Z/5Z. We know that the group of units

of Z/5Z is generated by 2. Now, ζ ≡ 1 mod (f1) and hence ζ−1 ≡ 1 mod(f1) and

therefore ζ + ζ−1 ≡ 2 mod (f1). In other words, the image of ζ + ζ−1 in (O/f1)

generates the group of units of (O/f1). So O∗ → (O/f1)∗ is surjective.

Now consider the case r = 2. Note that we have a surjective map (O/f2) → (O/f1)

which gives us a surjective map (O/f2)∗ → (O/f1)

∗. Now, (O/f2)∗ has order 20 and

(O/f1)∗ has order 4, so the kernel of the map (O/f2)

∗ → (O/f1)∗ has order 5 (so that

this kernel (which we’ll denote by N) is a cyclic group of order 5 and it is generated

by any non-identity element). We have already seen that the map O∗ → (O/f1)∗ is

surjective, so in order for the map O∗ → (O/f2)∗ to be surjective, it will suffice to

show that the map from the kernel of the map O∗ → (O/f1)∗ to N is surjective. So

we need to find an element in the kernel of the map O∗ → (O/f1)∗ whose image in

(O/f2)∗ generates N .

−(ζ−1 + 2ζ + ζ3) is such an element. We can see that this element is congruent


to 1 modulo (1 − ζ) and the norm of 1 + ζ−1 + 2ζ + ζ3 is 5. The first fact tells

us that the image of this element in (O/f2) is in the N and the second fact tells

us that this element is not divisible by f2 (so it is a nonzero element of O/f2). So

we have a surjective map O∗ → (O/f1)∗ and a surjective map from the kernel of

O∗ → (O/f1)∗ to N and so we get a surjection O∗ → (O/f2)

∗ because (O/f2)∗ is a

semidirect product of N and (O/f1)∗). The case r = 3 is similar to the case r = 2.

We have a surjection O∗ → (O/f2)∗ and also we have a surjection (O/f3)

∗ → (O/f2)∗.

Let N ′ be the kernel of the last surjection. If we can find an element of O∗ which

is congruent to 1 modulo (1 − ζ)2 and such that 1 minus that element does not

belong to the ideal generated by (1 − ζ)3 then we’ll be done because the image of

that element under the map O∗ → (O/f2)∗ will generate N ′. Then, since we also

have the surjection (O/f3)∗ → (O/f2)

∗ we would be able to conclude that the map

O∗ → (O/f3)∗ is surjective. −(ζ2 + 2 + ζ−2) is such an element. We see that it is

congruent to 1 modulo (1− ζ)2 and that the norm of (ζ2 +3+ ζ−2) is 25 which means

that (ζ2 +3+ζ−2) does not belong to the ideal generated by (1−ζ)3. So we’ve shown

that the map O∗ → (O/fr)∗ is surjective if r ≤ 3. Now let us find all a, b such that

(−ζ)a(ζ + ζ−1)b ≡ 1 mod (1 − ζ)3. We note that a could be 0, 1, 2, 3, ..., 9 mod 10

(since −ζ has order 10 in O∗) and b could be 0, 1, 2, 3 mod 4 . At this point we used

a computer to check that (−ζ)a(ζ + ζ−1)b ≡ 1 mod (1− ζ)3 implies that a ≡ 0 mod

10 (we computed the norm of (−ζ)a(ζ + ζ−1)b − 1 for different values of a and b and

checked whether this norm is divisible by 125). (ζ + ζ−1)b ≡ 1 mod (1− ζ)3 implies

that (ζ + ζ−1)b ≡ 1 mod (1− ζ)2. Since (ζ + ζ−1) ≡ 2 mod (1− ζ)2, we need to find

all b such that 2b ≡ 1 mod (1− ζ)2. We see that this equality holds only when b ≡ 0

mod 4. Since a = 0 mod 10 and b = 0 mod 4, we may write b = 4k and so we get

(ζ + ζ−1)4k = 1 mod (1− ζ)3. Then we made the following table:

k Order (ζ + ζ−1)4k − 1

1 2

5 6

25 10

125 14


(Order (ζ + ζ−1)4k − 1 means the highest power of 5 dividing the norm of (ζ +

ζ−1)4k − 1)

Now let us generalize the argument we’ve used before: we have a surjection

(O/fr)∗ → (O/fr−1)

∗. Let K denote the kernel of this map; we know that the order

of K is φ(5r)/φ(5r−1) = 5, where φ is the Euler φ function. So the map O∗ → (O/fr)∗

will be surjective provided that the map O∗ → (O/fr−1)∗ and the map from the kernel

of O∗ → (O/fr)∗ to K are surjective.

Let us now show that the map from O∗ to (O/f4)∗ is not surjective. We know

that the map from O∗ to (O/f3)∗ is surjective, and so the map from O∗ to (O/f4)

∗

will be surjective if and only if there exists x ∈ O∗ which is congruent to 1 modulo

(1− ζ)3 and such that 1−x does not belong to the ideal generated by (1− ζ)4. From

the argument above, we know that if x ∈ O∗ is congruent to 1 modulo (1− ζ)3 then

x = (ζ + ζ−1)4k for some integer k. So, in order to show that the map O∗ → (O/f4)∗

isn’t surjective, we need to show that if x ≡ 1 mod (1− ζ)3 then x ≡ 1 mod (1− ζ)4.

Let us now show that x ≡ 1 mod (1− ζ)3 implies that k must be a multiple of 5.

If this were false, then there would exist an integer s such that ks ≡ 1 mod 5. So if

x ≡ 1 mod (1 − ζ)3, then xs = (ζ + ζ−1)4ks ≡ (ζ + ζ−1)4 mod f3 (here we used the

fact that (ζ + ζ−1)4 has order 5 in (O/f3)∗)). However, we know that (ζ + ζ−1)4 is

not congruent to 1 mod f3, and so we conclude that k is a multiple of 5. This implies

that x is a power of (ζ + ζ−1)20, hence x is congruent to 1 mod f6 (from the table

above) and thus x is congruent to 1 mod f4.

Thus, the map O∗ → (O/f4)∗ isn’t surjective and thus the map O∗ → (O/fr)

∗ is

not surjective for r ≥ 4.

So we’ve proved the claim that the map O∗ → (O/fr)∗ is surjective if r ≤ 3 and

is not surjective if r > 3.

Next, we shall prove the reciprocity law. We need the following lemma:

Lemma 13. Let l be an odd prime, let ζ denote a primitive l - th root of unity and

let k = Ql(ζ) where Ql is the l - adic completion of the rationals. Let S denote the

trace from k to Ql. Let λ = 1− ζ. Let α, β be elements of k such that α ≡ 1 mod λ2


and β ≡ 1 mod λ.Then (α

β

)(β

α

)−1

= ζ(1/l)S(ζ log(α)D log(β))

Here by log we mean a power series expansion log(1 + x) = x− x2/2 + x3/3− ....

It is known that the series converges whenever x is divisible by l. Also, if x ≡ 1 mod

λ, we define log(x) = log(1 − (1 − x)) (if x is not congruent to 1 mod λ, log(x) is

undefined).

D log(β) is defined as follows: If β is a unit in k, we may write β as a power

series in λ: β =∑∞

j=0 ajλj where aj are rational integers. By D log(β) we mean β′

β

where β′ is the derivative of the power series β′ =∑∞

j=1 jajλj−1. Since k/Ql is tamely

ramified, the different of k/Ql is λl−2. Hence lemma 2 on page 152 of Artin, Tate

[1] shows that β′ is well defined modulo λl−2, so that D log(β) is also well defined

modulo λl−2. We emphasize that D log(β) is not necessarily equal to (log β)′ because

log(β) may not be a unit (in fact, log(β) may not even be defined because β may not

be congruent to 1 mod λ).

Proof. See example 2, p.172 of Artin, Tate [1].

We now prove that the reciprocity law holds for r = 3. More precisely, we prove

the following theorem:

Theorem 11. Let ζ denote the primitive 5th root of unity, let λ = 1− ζ. Let α and

β be two prime elements of the ring Z[ζ] (although the fact that they are prime will

not be used in the proof below) satisfying α ≡ β ≡ 1 mod λ3. Then(α

β

)(β

α

)−1

= 1.

Proof. Using lemma 13 (with l = 5), we see that our claim is equivalent to the claim

that S(log(α)D log(β)) is a multiple of 25. We may expand α and β as power series

in λ:

α = 1 + α3λ3 + ... and

β = 1 + β3λ3 + ....


Then using the formulas log(1 − x) = x + (x2/2) + (x3/3) + ... and log(x) =

log(1− (1− x)) we see that

log(α) = α3(λ3) + ..., in other words log(α) ≡ 0 mod λ3. Of course, we also have

that log(β) ≡ 0 mod λ3.

We also see that D (β) ≡ 0 mod λ2 and hence D(log(β)) ≡ 0 mod λ2. So finally

we see that (log(α)D log(β)) ≡ 0 mod λ5 and so we want to show that if t satisfies

t ≡ 0 mod λ5 then S(t) ≡ 0 mod 25.

We know that λ5 = −5ζ + 10ζ2 − 10ζ3 + 5ζ4. We also note that the trace of any

power of ζ is −1. This is because ζ4+ζ3+ζ2+ζ = −1. Hence S(λ5) = 5−10+10−5 =

0. Now let us consider S(ζ iλ5). If i = 0 then S(ζ iλ5) = Tr(λ5) ≡ 0 mod(25). If i is

nonzero, we shall consider 3 cases:

1) i = 1. Then S(ζ iλ5) = 25

2) i = 2. Then S(ζ iλ5) = −50

3) i = 3. Then S(ζ iλ5) = 50

Since any element of Z[ζ] can be written as a0 +a1ζ+a2ζ2 +a3ζ

3 for some integers

a0, a1, a2, a3 we conclude that

S(f · λ5) ≡ 0 mod 25 for any f ∈ Z[ζ]. This completes the proof of the theorem.

Let ζ denote a primitive 5th root of unity, let λ = 1 − ζ, let α be any element of

Z[ζ] such that α ≡ 1 mod λ3. We may expand α as a power series in λ, and write

α = 1 +α3λ3 +α4λ

4 +α5λ5 + ...( the formulas below will only involve the coefficients

α3, α4, α5). We will establish the formulas for the power residue symbols(

ζα

),(

λα

),(

ζ+ζ−1

α

)Lemma 14. We have the following formulas:(

ζα

)= ζα3+α4.(

λα

)= ζ−α5.(

ζ+ζ−1

α

)= ζ−α3

Proof. To prove the first claim, we use a formula from example 2 on p.173 of Artin,

Tate [1]:


(ζα

)= ζ(1/5)S(log α) where S denotes the trace. As we saw before, log α = α3λ

3 +

α4λ4 + .... Also, S(f · λ5) ≡ 0 mod 25 for any f ∈ Z[ζ]. Now we observe that

S(λ3) = S(λ4) = 5 and so we see that S(log α) ≡ (5α3 + 5α4) mod (25) and so using

the formula mentioned above proves the claim.

To prove the second claim, we use a formula from example 2 on p.173 of Artin,

Tate [1]:(λα

)= ζ(−1/5)S((ζ/λ) log α). As we saw before, log α = α3λ

3 + α4λ4 + .... Hence

(ζ/λ) logα) = α3ζλ2 + α4ζλ

3 + α5ζλ4 + .... Now, the trace of ζλ2 and of ζλ3 is equal

to 0 and the trace of ζλ4 is 5. So we see that S( ζλ· log(α)) ≡ 5α5 mod (25) and so

using the formula mentioned above proves the claim.

To prove the last claim, we begin by computing(

1+ζ2

α

). We use theorem 3 on

page 25 from Sen [12]. We need to express ζ and 1 + ζ2 as power series in λ. (In this

case both of these are actually polynomials in λ). ζ = g(λ) = 1− λ, so g′ = −1 and

1 + ζ2 = f(λ) = 2− 2λ+ λ2 so f ′ = −2 + 2λ. Then we need to compute the inverse

of 2− 2λ+ λ2. We write (2− 2λ+ λ2)(a0 + a1λ+ a2λ2 + a3λ

3 + a4λ4 + ...) = 1 and

solve for the coefficients ai. We get a0 = a1 = 1/2; a2 = 1/4; a3 = 0; a4 = −1/8...

Hence ζ(2−2λ)(2−2λ+λ2)

= ζ(1− λ2 − (1/2)λ3 − (1/4)λ4 + (1/4)λ5 + ...) and multiplying this

by log(α) = α3λ3 + α4λ

4 + ... we get ζ(2−2λ)(2−2λ+λ2)

· log (α) = ζ(α3λ3 + α4λ

4 + ...). Since

the trace of ζλ3 is 0, the trace of ζλ4 is 5 and the trace of any multiple of λ5 is a

multiple of 25 we conclude that the trace of ζ(2−2λ)(2−2λ+λ2)

· log(α) ≡ 5α4 mod 25. Then the

theorem mentioned above gives us(

1+ζ2

α

)= ζα4 . Now, using this fact and also the

fact that(

ζα

)= ζα3+α4 we may conclude that

(ζ+ζ−1

α

)= ζ−α3 . So the last formula is

proved.

Now we prove similar results for the case n = 7.

Theorem 12. Let ζ denote the primitive 7th root of unity, let λ denote the principal

ideal generated by 1− ζ. Let O = Z[ζ].

We shall prove that the map O∗ → (O/λ4)∗ is surjective.

Proof. First, let us show that the map O∗ → (O/λ)∗ is surjective. We know that

O/λ is (isomorphic to) a finite field with 7 elements. Since ζ ≡ 1 mod λ we see that


−(1+ζ2) ≡ 5 mod λ. But 5 generates the group of units of F7 so the image of −(1+ζ2)

under the map O∗ → (O/λ)∗ generates (O/λ)∗. Hence the map O∗ → (O/λ)∗ is

surjective. Next, let us establish the surjectivity of the map O∗ → (O/λ2)∗. We

know that it suffices to find an element e of O∗ such that e ≡ 1 mod λ and e is not

equivalent to 1 mod λ2. We claim that e = ζ is such an element. Clearly, e ≡ 1 mod

λ and since the norm of e− 1 is 7 we conclude that e is not equivalent to 1 mod λ2.

Next, let us establish the surjectivity of the map O∗ → (O/λ3)∗. We know that it

suffices to find an element e′ of O∗ such that e′ ≡ 1 mod λ2 and e′ is not equivalent

to 1 mod λ3. We claim that e′ = ζ4(1 + ζ2)3 is such an element. Since (1 + ζ2) ≡ 2ζ

mod λ2 we see that e′ ≡ ζ4(2ζ)3 ≡ ζ7 = 1 mod λ2 and since the norm of e′ − 1 is 49

we conclude that e′ is not equivalent to 1 mod λ3.

Finally, let us establish the surjectivity of the map O∗ → (O/λ4)∗. We know that

it suffices to find an element e′′ of O∗ such that e′′ ≡ 1 mod λ3 and e′ is not equivalent

to 1 mod λ4. We used a computer to produce such a unit. Let e′′ = (1+ζ)30(1+z2)153.

Then we checked that the highest power of 7 that divides the norm of e′′ − 1 is 3.

This means that e′′ ≡ 1 mod λ3 and e′′ is not equivalent to 1 mod λ4. This proves

that the map O∗ → (O/λ4)∗ is surjective.

(Remark: When we used a computer to find this unit, we defined a function

F [a, b, d] := Norm[(1 + z)a(1 + z2)b(−z)d − 1] and we checked that the highest power

of 7 that divides F (30, 153, 364) is 3. But (−ζ)14 = 1 and since 364 ≡ 0 mod 14 we

see that we can take d = 0 instead of 364).

Theorem 13. Let ζ denote the primitive 7th root of unity, let λ = 1 − ζ. Let α

and β be two prime elements of the ring Z[ζ] satisfying α ≡ β ≡ 1 mod λ4. Then(αβ

) (βα

)−1= 1.

Proof. We shall use lemma 13 with l = 7. We may expand α and β as power series

in λ:

α = 1 + α4λ4 + ... and

β = 1 + β4λ4 + ....

Then using the formulas log(1 − x) = x + (x2/2) + (x3/3) + ... and log(x) =

log(1− (1− x)) we see that


log(α) = α4(λ4) + ..., in other words log(α) ≡ 0 mod λ4. Of course, we also have

that log(β) ≡ 0 mod λ4.

We also see that D(β) ≡ 0 mod λ3 and hence D (log(β)) ≡ 0 mod λ3. So finally

we see that (log(α)D log(β)) ≡ 0 mod λ7 and so we want to show that if t satisfies

t ≡ 0 mod λ7 then S(t) ≡ 0 mod 49.

We know that λ7 = −7ζ + 21ζ2− 35ζ3 + 35ζ4− 21ζ5 + 7ζ6. We also note that the

trace of any power of ζ is −1. This is because ζ6 + ζ5 + ζ4 + ζ3 + ζ2 + ζ = −1. Hence

S(λ7) = 7 − 21 + 35 − 35 + 21 − 7 = 0. Now let us consider S(ζ iλ7). If i = 0 then

S(ζ iλ7) = S(λ7) ≡ 0 mod 49. If i is nonzero, we shall consider 5 cases:

1) i = 1. Then S(ζ iλ7) = 49.

2) i = 2. Then S(ζ iλ7) = −3 · 49.

3) i = 3. Then S(ζ iλ7) = 5 · 49.

4) i = 4. Then S(ζ iλ7) = −5 · 49.

5) i = 5. Then S(ζ iλ7) = 3 · 49.

Since any element of Z[ζ] can be written as a0 + a1ζ + a2ζ2 + a3ζ

3 + a4ζ4 + a5ζ

5

for some integers a0, a1, a2, a3, a4, a5, we conclude that

S(f · λ7) ≡ 0 mod 49 for any f ∈ Z[ζ]. This completes the proof.

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PART I: SYMPLECTIC ICE. PART II: GLOBAL AND …sd054cv7359/...Hamel and King [7] showed how characters of irreducible representations times the deformed Weyl denominators are equal