Kron

eckerP

roducts

and

Com

bin

atorics

ofsom

e

Rem

arkable

Diop

han

tine

System

Joint

work

ofA

.M.G

arsia,G

.M

usiker,

N.W

allach,G

.X

inan

dM

.Zab

rock

i

The

Kron

eckerco

efficien

ts

For

kirred

ucib

leS

nch

aractersχ

λ(1)χ

λ(2)···χ

λ(k)w

eset

cλ

(1),λ

(2),...,λ

(k)

=∑

b1+

2b2+···+

nbn=

n

χλ

(1)

bχ

λ(2)

b···χ

λ(k)

b

zb

where

forb

=1b12

b2···n

bn

zb

=1b12

b2···n

bn

b1 !b

2 !···bn !

Fundam

ental

Prob

lems:

1G

iveeffi

cientalgorithm

s

2Find

combinatorial

interpretation(such

asthe

LR

rule)

3Find

generatingfunctions

forfam

iliesof

Kronecker

coeffi

cients

The

Kron

eckerP

roduct

The

Kron

eckerP

roduct

ofth

eSch

ur

function

s

Sλ

(1) ,S

λ(2) ,...,S

λ(k) ,

(λ(i)�

n)

isdefi

ned

as

Sλ

(1) ∗

Sλ

(2) ∗

···∗S

λ(k)

=∑

b1+

2b2+···+

nbn=

n

χλ

(1)

bχ

λ(2)

b···χ

λ(k)

b

zb

pb1

1p

b2

2···p

bn

n

Itfollow

sfrom

the

orthogon

alityof

the

pow

ersu

ms

that

forall

1≤

i≤k

cλ

(1),λ

(2),...,λ

(k)

=⟨S

λ(1) ∗

···∗S

λ(i)

,S

λ(i+

1) ∗

···∗S

λ(k) ⟩

asw

ellas

cλ

(1),λ

(2),...,λ

(k)

=⟨S

λ(1) ∗

···∗S

λ(k),S

(n) ⟩

Ach

allengin

gquestion

(Wallach

)From

quan

tum

computin

g:

Construct

thegenerating

function

Wk (q)

=∑d≥

0

qd ⟨S

d,d

∗S

d,d

∗···∗

Sd

,d,S

2d ⟩

k≥

2

(th

eterm

inth

escalar

product

has

kfactors)

Easy

answ

ers

W2 (q)

=∑d≥

0

qd ⟨S

d,d

,S

d,d ⟩

=∑d≥

0

qd

=1

1−

q

W3 (q)

=∑d≥

0

qd ⟨S

d,d

∗S

d,d

,S

d,d ⟩

=∑deven

qd

=1

1−

q2

Afirst

non-triv

ialre

sult

Mike

Zab

rock

iusin

gsy

mm

etricfu

nction

iden

titiesan

dJeff

Rem

mel

usin

gcom

bin

atorialtrickery

were

able

tosh

owth

at

Theore

m

Sd

,d∗S

d,d (x)

=∑α�2d

αw

ith

4even

parts≥

0

Sα (x)

+∑β�

2d

βw

ith

4odd

parts

Sβ (x)

(∗)

Sin

ce⟨S

d,d

∗S

d,d

∗S

d,d

∗S

d,d

,S

2d ⟩

=⟨S

d,d

∗S

d,d

,S

d,d

∗S

d,d ⟩

from(*)

we

derive

that

W4 (q)

=∑α�2d

αw

ith

4even

parts≥

0

q|α

|/2

+∑β�

2d

βw

ith

4odd

parts

This

givesW4 (q)

=1

+q2

(1−

q)(1−

q2)(1

−q3)(1

−q4)

=1

(1−

q)(1−

q2)

2(1−

q3)

.

The

next

challe

nge:

Com

putin

gW

5 (q)

anyguesses

?

Wait

andsee

Itis

afascinating

story

Guessin

gth

eansw

er

Ifχ

d,d

den

otesth

eY

oung

irreducib

lech

aracterof

the

sym

metric

group

S2d ,

then

fork≥

2factors

we

have

⟨Sd

,d∗S

d,d

∗···∗

Sd

,d,S

2d ⟩

=cd (k)

=1

(2d)! ∑α

∈S2d

χd

,d(α)k

(∗)

We

wan

tW

5 (q)=

1+ ∑d

≥1

cd (5)q

d.

Sym

metric

function

mach

inery

quick

lyyield

sth

efirst

fewterm

sof

this

series.

But

note

that

ford

=20

the

defi

nition

requires

work

ing

with

S40

and

since

40!=

815915283247897734345611269596115894272000000000

Som

etrickery

isnecessary

here.

Pro

ceedin

gby

the

naiv

eappro

ach

Usin

gavailab

lesy

mm

etricfu

nction

(MA

PLE)

packages

we

get

W5 (q)

=1

+5q

2+

q3

+36q

4+

15q5

+228q

6+

231q7

+1313q

8+

1939q9

+6971q

10+

11899q11

+33118q

12

+59543q

13

+140620q

14

+254476q

15

+538042q

16+

959028q17

+1871808q

18

+3258512q

19

+5981444q

20

+···

Wallach

:“not

enough”

Usin

gnew

formulas

forSn

characters

(fromjoin

tw

orkw

ithA

lainG

oupil)

we

obtain

ed7

more

coeffi

cients

···+10140360q

21

+17726166q

22

+29257848q

23+

49127549q24

+79032258q

25+

128267727q26

+201437596q

27

+···

Wallach

:“V

eryim

pressivebut

notenough,

we

need52

coeffi

cientsto

predictW

5 (q)!”

Me:

“forgetit!

”104!=

1029901674514562762384858386476504428305377245499907218232549177688787173

247528717454270987168388800323596570414163837769517974197917558872

4736000000000000000000000000

The

Mira

cle

One

morn

ing

Wallach

wakes

up

toth

erealization

that

usin

gfan

cyLie

Grou

ptrickery

he

could

compute

coeffi

cients

ofW

5 (q)w

aybeyon

dd

=27.

After

severalhou

rsof

computer

time

the

need

ed52

coeffi

cients

were

obtain

edan

dW

allacchw

asab

leto

constru

cta

rational

expression

forW

5 (q).

The

Mira

cle

One

morn

ing

Wallach

wakes

up

toth

erealization

that

usin

gfan

cyLie

Grou

ptrickery

he

could

compute

coeffi

cients

ofW

5 (q)w

aybeyon

dd

=27.

After

severalhou

rsof

computer

time

the

need

ed52

coeffi

cients

were

obtain

edan

dW

allacchw

asab

leto

constru

cta

rational

expression

forW

5 (q).A

reyou

ready

forit?

The

Mira

cle

One

morn

ing

Wallach

wakes

up

toth

erealization

that

usin

gfan

cyLie

Grou

ptrickery

he

could

compute

coeffi

cients

ofW

5 (q)w

aybeyon

dd

=27.

After

severalhou

rsof

computer

time

the

need

ed52

coeffi

cients

were

obtain

edan

dW

allacchw

asab

leto

constru

cta

rational

expression

forW

5 (q).A

reyou

ready

forit?

Here

itis!

The

Mira

cle

One

morn

ing

Wallach

wakes

up

toth

erealization

that

usin

gfan

cyLie

Grou

ptrickery

he

could

compute

coeffi

cients

ofW

5 (q)w

aybeyon

dd

=27.

After

severalhou

rsof

computer

time

the

need

ed52

coeffi

cients

were

obtain

edan

dW

allacchw

asab

leto

constru

cta

rational

expression

forW

5 (q).A

reyou

ready

forit?

Here

itis!

W5 (q)

= (q54

+q52

+16q

50

+9q

49

+98q

48

+154q

47

+465q

46

+915q

45

+2042q

44

+3794q

43

+7263q

42

+12688q

41

+21198q

40

+34323q

39

+52205q

38

+77068q

37

+108458q

36

+147423q

35

+191794q

34

+241863q

33

+292689q

32

+342207q

31

+386980q

30

+421057q

29

+443990q

28

+451398q

27

+443990q

26

+421057q

25

+386980q

24

+342207q

23

+292689q

22

+241863q

21

+191794q

20

+147423q

19

+108458q

18

+77068q

17

+52205q

16

+34323q

15

+21198q

14

+12688q

13

+7263q

12

+3794q

11

+2042q

10

+915q

9+

465q8

+154q

7+

+98q

6+

9q5

+16q

4+

q2

+1 )

(1−

q2)

4(1−

q3)(1

−q4)

6(1−

q5)(1

−q6)

5

The

Mira

cle

One

morn

ing

Wallach

wakes

up

toth

erealization

that

usin

gfan

cyLie

Grou

ptrickery

he

could

compute

coeffi

cients

ofW

5 (q)w

aybeyon

dd

=27.

After

severalhou

rsof

computer

time

the

need

ed52

coeffi

cients

were

obtain

edan

dW

allacchw

asab

leto

constru

cta

rational

expression

forW

5 (q).A

reyou

ready

forit?

Here

itis!

W5 (q)

= (q54

+q52

+16q

50

+9q

49

+98q

48

+154q

47

+465q

46

+915q

45

+2042q

44

+3794q

43

+7263q

42

+12688q

41

+21198q

40

+34323q

39

+52205q

38

+77068q

37

+108458q

36

+147423q

35

+191794q

34

+241863q

33

+292689q

32

+342207q

31

+386980q

30

+421057q

29

+443990q

28

+451398q

27

+443990q

26

+421057q

25

+386980q

24

+342207q

23

+292689q

22

+241863q

21

+191794q

20

+147423q

19

+108458q

18

+77068q

17

+52205q

16

+34323q

15

+21198q

14

+12688q

13

+7263q

12

+3794q

11

+2042q

10

+915q

9+

465q8

+154q

7+

+98q

6+

9q5

+16q

4+

q2

+1 )

(1−

q2)

4(1−

q3)(1

−q4)

6(1−

q5)(1

−q6)

5

Com

binatorialinterpretation?

Work

ing

Tow

ard

sa

Pro

of

Atruly

surprisingidentity

Theore

mIfF(a

1,a

2,...,a

k ;q)=

∏ki=

1 (1−

ai )

∏S⊆

[1,k

] (1−

q ∏i∈

Sai )

=∑r1 ≥

0 ∑r2 ≥

0

··· ∑rk ≥

0 ∑m≥

0

cm

;r1,r

2,...,r

kqm

ar1

1ar2

2···a

rk

k

thencm

;r1,r

2,...,r

k= ⟨S

m−

r1,r

1 ∗S

m−

r2,r

2 ∗···∗

Sm

−rk,r

k,S

m ⟩∀

m≥

2max(r

1,r

2,...,r

k )

Inparticu

larc2d;d

,d,...,d

= ⟨Sd

,d∗S

d,d

∗···∗

Sd

,d,S

2d ⟩

Thus

the

desired

seriesW

k (q)is

obtain

edby

selecting

the

terms

where

m=

2d,

r1

=d

,r2

=d

,...,r

k=

d,

Sele

cting

asu

bse

ries

Pro

positio

nIf

F(a

1,a

2,...a

k ;q)= ∑r

1 ≥0 ∑r

2 ≥0 ··· ∑r

k ≥0 ∑m

≥0

cm

;r1,r

2,...,r

kqm

ar1

1ar2

2···a

rk

k

then

Wk (q

2)= ∑d

≥0

c2d;d

,d,...d

q2d

=F

(a−

21

,a−

22

,···,a−

2k

;qa1 a

2 ···ak ) ∣∣∣a

01,a

02,...,a

02,

Pro

of

F(a

−2

1,a

−2

2,···,a

−2

k;q

a1 a

2 ···ak )

=

=∑r1 ≥

0 ∑r2 ≥

0 ··· ∑rk ≥

0

cm

;r1,r

2,...,r

kqm

am

−2r1

1am

−2r2

2···a

m−

2rk

k

QED

Kro

neck

er

Pro

ducts

and

Consta

nt

term

sid

entitie

s

Recallin

gth

at

F(a

1,a

2,...,a

k ;q)=

k∏i=

1 (1−

ai )

∏S⊆

[1,k

] (1−

q ∏i∈S

ai )

we

derive

that

F(a

−2

1,a

−2

2,···,a

−2

k;q

a1 a

2 ···ak )

=

k∏i=

1 (1−

1a2i )

∏S⊆

[1,k

] (1−

qa1 a

2 ···ak

∏i∈

Sa2i

)=

k∏i=

1 (1−

1a2i )

∏S⊆

[1,k

] (1−

q ∏i/∈

Sai

∏i∈

Sai )

Kro

neck

er

Pro

ducts

and

Consta

nt

term

sid

entitie

s

Recallin

gth

at

F(a

1,a

2,...,a

k ;q)=

k∏i=

1 (1−

ai )

∏S⊆

[1,k

] (1−

q ∏i∈S

ai )

we

derive

that

F(a

−2

1,a

−2

2,···,a

−2

k;q

a1 a

2 ···ak )

=

k∏i=

1 (1−

1a2i )

∏S⊆

[1,k

] (1−

qa1 a

2 ···ak

∏i∈

Sa2i

)=

k∏i=

1 (1−

1a2i )

∏S⊆

[1,k

] (1−

q ∏i/∈

Sai

∏i∈

Sai )

We

canth

us

state

Theore

m

Wk (q

2)=

∑d≥

0 ⟨Sd

,d∗S

d,d

∗···∗

Sd

,d,S

2d ⟩

q2d

=

k∏i=

1 (1−

1a2i )

∏S⊆

[1,k

] (1−

q ∏i/∈

Sai

∏i∈

Sai ) ∣∣∣∣∣a

01a02 ···a

0k

00 011110

Som

eR

em

ark

able

Dio

phantin

eSyste

ms

Agenera

lP

roble

mD

etermine

allpossible

ways

toassign

≥0

integralweights

tothe

verticesof

thek-dim

ensionalhyp

ercube

sothat

eachhyp

erfacehas

thesam

eweight .

Exam

ple

k=

2(see

attached

figu

re)w

em

ust

have

p00

+p

01

=p

10

+p

11

p00

+p

10

=p

01

+p

11

Insim

pler

notation

we

wan

tall

composition

sp

=(p

1,p

2,p

3,p

4 )w

hich

aresolu

-tion

sof

S2

= ∥∥∥p

1+

p2−

p3−

p4

=0

p1−

p2

+p

3−

p4

=0

Here

“all

solution

s”m

eans

the

generatin

gfu

nction

GF

S2 (x

1,x

2,x

3,x

4 )=

∑p∈S2 x

p1

1xp2

2xp3

3xp4

4

MacM

ahon

“Partitio

nA

naly

sis”

To

solveth

esy

stem

S2

=∥∥∥∥

p1

+p

2−

p3−

p4

=0

p1−

p2

+p

3−

p4

=0

(with

pi ≥

0)

MacM

ahon

“Partitio

nA

naly

sis”

To

solveth

esy

stem

S2

=∥∥∥∥

p1

+p

2−

p3−

p4

=0

p1−

p2

+p

3−

p4

=0

(with

pi ≥

0)

we

constru

ctth

ecorresp

ondin

g“O

mega

Function

”

ΩS2

=∑p1 ≥

0 ∑p2 ≥

0 ∑p3 ≥

0 ∑p4 ≥

0

xp1

1xp2

2xp3

2xp4

4ap1+

p2−

p3 −

p4

1ap1 −

p2

+p3 −

p4

2

=1

1−

x1 a

1 a2

11−

x2 a

1/a

2

11−

x3 a

2/a

1

11−

x4/a

1 a2

MacM

ahon

“Partitio

nA

naly

sis”

To

solveth

esy

stem

S2

=∥∥∥∥

p1

+p

2−

p3−

p4

=0

p1−

p2

+p

3−

p4

=0

(with

pi ≥

0)

we

constru

ctth

ecorresp

ondin

g“O

mega

Function

”

ΩS2

=∑p1 ≥

0 ∑p2 ≥

0 ∑p3 ≥

0 ∑p4 ≥

0

xp1

1xp2

2xp3

2xp4

4ap1+

p2−

p3 −

p4

1ap1 −

p2

+p3 −

p4

2

=1

1−

x1 a

1 a2

11−

x2 a

1/a

2

11−

x3 a

2/a

1

11−

x4/a

1 a2

and

then

GF

S2 (x

1,x

2,x

3,x

4 )=

ΩS ∣∣∣a

01a02

MacM

ahon

“Partitio

nA

naly

sis”

To

solveth

esy

stem

S2

=∥∥∥∥

p1

+p

2−

p3−

p4

=0

p1−

p2

+p

3−

p4

=0

(with

pi ≥

0)

we

constru

ctth

ecorresp

ondin

g“O

mega

Function

”

ΩS2

=∑p1 ≥

0 ∑p2 ≥

0 ∑p3 ≥

0 ∑p4 ≥

0

xp1

1xp2

2xp3

2xp4

4ap1+

p2−

p3 −

p4

1ap1 −

p2

+p3 −

p4

2

=1

1−

x1 a

1 a2

11−

x2 a

1/a

2

11−

x3 a

2/a

1

11−

x4/a

1 a2

and

then

GF

S2 (x

1,x

2,x

3,x

4 )=

ΩS ∣∣∣a

01a02

Inparticu

larGF

S2 (q

,q,q

,q)= ∑p

∈S2 q

p1+

p2+

p3+

p4

=1

∏S⊆

[1,2

] (1−

q ∏i∈

Sai

∏i/∈

Sai ) ∣∣∣∣∣a

01a02

011001

101111

010

110100

000

The

Dio

phantin

eSyste

mfo

rk

=3

Equatin

gth

ew

eights

ofop

posite

facesgives

p000

+p

001

+p

010

+p

011

=p

100

+p

101

+p

110

+p

111

p000

+p

001

+p

100

+p

101

=p

010

+p

011

+p

110

+p

111

p000

+p

010

+p

100

+p

110

=p

001

+p

011

+p

101

+p

111

011001

101111

010

110100

000

The

Dio

phantin

eSyste

mfo

rk

=3

Equatin

gth

ew

eights

ofop

posite

facesgives

p000

+p

001

+p

010

+p

011

=p

100

+p

101

+p

110

+p

111

p000

+p

001

+p

100

+p

101

=p

010

+p

011

+p

110

+p

111

p000

+p

010

+p

100

+p

110

=p

001

+p

011

+p

101

+p

111

and

we

getth

esy

stem

S3

=

∥∥∥∥∥p

000

+p

001

+p

010

+p

011−

p100−

p101−

p110−

p111

=0

p000

+p

001−

p010−

p011

+p

100

+p

101−

p110−

p111

=0

p000−

p001

+p

010−

p011

+p

100−

p101

+p

110−

p111

=0

011001

101111

010

110100

000

The

Dio

phantin

eSyste

mfo

rk

=3

Equatin

gth

ew

eights

ofop

posite

facesgives

p000

+p

001

+p

010

+p

011

=p

100

+p

101

+p

110

+p

111

p000

+p

001

+p

100

+p

101

=p

010

+p

011

+p

110

+p

111

p000

+p

010

+p

100

+p

110

=p

001

+p

011

+p

101

+p

111

and

we

getth

esy

stem

S3

=

∥∥∥∥∥p

000

+p

001

+p

010

+p

011−

p100−

p101−

p110−

p111

=0

p000

+p

001−

p010−

p011

+p

100

+p

101−

p110−

p111

=0

p000−

p001

+p

010−

p011

+p

100−

p101

+p

110−

p111

=0

orin

simpler

notation

S3

=

∥∥∥∥∥p

1+

p2

+p

3+

p4−

p5−

p6−

p7−

p8

=0

p1

+p

2−

p3−

p4

+p

5+

p6−

p7−

p8

=0

p1−

p2

+p

3−

p4

+p

5−

p6

+p

7−

p8

=0

q-C

ountin

gby

“to

talw

eig

ht”

for

k=

3

Apply

ing

MacM

ahon

Partition

analy

sisto

the

system

S3

=

∥∥∥∥∥p

1+

p2

+p

3+

p4−

p5−

p6−

p7−

p8

=0

p1

+p

2−

p3−

p4

+p

5+

p6−

p7−

p8

=0

p1−

p2

+p

3−

p4

+p

5−

p6

+p

7−

p8

=0

q-C

ountin

gby

“to

talw

eig

ht”

for

k=

3

Apply

ing

MacM

ahon

Partition

analy

sisto

the

system

S3

=

∥∥∥∥∥p

1+

p2

+p

3+

p4−

p5−

p6−

p7−

p8

=0

p1

+p

2−

p3−

p4

+p

5+

p6−

p7−

p8

=0

p1−

p2

+p

3−

p4

+p

5−

p6

+p

7−

p8

=0

we

obtain

that

GF

S3 (q)

= ∑p∈S3 q

p1+

p2+

p3+

p4+

p5+

p6+

p7+

p8

= ∑p1 ≥

0 ∑p2 ≥

0 ··· ∑p8 ≥

0

qp1+

p2+

p3+

p4+

p5+

p6+

p7+

p8

ap1+

p2+

p3+

p4 −

p5 −

p6 −

p7 −

p8

1×

×ap1+

p2 −

p3 −

p4+

p5+

p6 −

p7 −

p8

2ap1 −

p2+

p3 −

p4+

p5 −

p6+

p7 −

p8

3

∣∣∣∣∣a01a02a03

q-C

ountin

gby

“to

talw

eig

ht”

for

k=

3

Apply

ing

MacM

ahon

Partition

analy

sisto

the

system

S3

=

∥∥∥∥∥p

1+

p2

+p

3+

p4−

p5−

p6−

p7−

p8

=0

p1

+p

2−

p3−

p4

+p

5+

p6−

p7−

p8

=0

p1−

p2

+p

3−

p4

+p

5−

p6

+p

7−

p8

=0

we

obtain

that

GF

S3 (q)

= ∑p∈S3 q

p1+

p2+

p3+

p4+

p5+

p6+

p7+

p8

= ∑p1 ≥

0 ∑p2 ≥

0 ··· ∑p8 ≥

0

qp1+

p2+

p3+

p4+

p5+

p6+

p7+

p8

ap1+

p2+

p3+

p4 −

p5 −

p6 −

p7 −

p8

1×

×ap1+

p2 −

p3 −

p4+

p5+

p6 −

p7 −

p8

2ap1 −

p2+

p3 −

p4+

p5 −

p6+

p7 −

p8

3

∣∣∣∣∣a01a02a03

and

sum

min

gth

eseries

we

get

GF

S3 (q)

=1

∏S⊆

[1,2

,3] (

1−

q ∏i∈

Sai

∏i/∈

Sai ) ∣∣∣∣∣a

01a02a03

000

001

010

011

100

101

111

110

A1

A2

A3

Com

bin

ato

ricsofV

enn

Dia

gra

ms

of3

sets

Three

subsets

ofcard

inality

dof

aset

Bof

cardin

ality2d

Let

p000

=|A

1∩

A2∩

A3 |

,p

001

=|A

1∩

A2∩

Ac3 |

,

p010

=|A

1∩

Ac2∩

A3 |

,p

011

=|A

1∩

Ac2∩

Ac3 |

p100

=|A

c1∩

A2∩

A3 |

,p

101

=|A

c1∩

A2∩

Ac3 |

,

p110

=|A

c1∩

Ac2∩

A3 |

,p

111

=|A

c1∩

Ac2∩

Ac3 |

000

001

010

011

100

101

111

110

A1

A2

A3

Com

bin

ato

ricsofV

enn

Dia

gra

ms

of3

sets

Three

subsets

ofcard

inality

dof

aset

Bof

cardin

ality2d

Let

p000

=|A

1∩

A2∩

A3 |

,p

001

=|A

1∩

A2∩

Ac3 |

,

p010

=|A

1∩

Ac2∩

A3 |

,p

011

=|A

1∩

Ac2∩

Ac3 |

p100

=|A

c1∩

A2∩

A3 |

,p

101

=|A

c1∩

A2∩

Ac3 |

,

p110

=|A

c1∩

Ac2∩

A3 |

,p

111

=|A

c1∩

Ac2∩

Ac3 |

Sam

eD

iophantin

eSyste

m!

|A1 |=

p000

+p

001

+p

010

+p

011

=|A

c1 |=p

100

+p

101

+p

110

+p

111

|A2 |=

p000

+p

001

+p

100

+p

101

=|A

c2 |=p

010

+p

011

+p

110

+p

111

|A3 |=

p000

+p

010

+p

100

+p

110

=|A

c3 |=p

001

+p

011

+p

101

+p

111

Som

eR

epre

senta

tion

Theory

(1)T

heaction

ofS

2d

onthe

dsubsets

Aof

a2d

setB

isequivalent

tothe

actionof

S2d

onthe

leftcosets

ofthe

Young

subgroupS

d ×S

d .

Som

eR

epre

senta

tion

Theory

(1)T

heaction

ofS

2d

onthe

dsubsets

Aof

a2d

setB

isequivalent

tothe

actionof

S2d

onthe

leftcosets

ofthe

Young

subgroupS

d ×S

d .

(2)T

husthe

Frob

eniuscharateristic

ofthis

actionis

thesym

metric

functionh

d hd

Som

eR

epre

senta

tion

Theory

(1)T

heaction

ofS

2d

onthe

dsubsets

Aof

a2d

setB

isequivalent

tothe

actionof

S2d

onthe

leftcosets

ofthe

Young

subgroupS

d ×S

d .

(2)T

husthe

Frob

eniuscharateristic

ofthis

actionis

thesym

metric

functionh

d hd

(3)It

follows

thenthat

theFrob

eniuscharacteristic

ofthe

actionof

S2d

ontriplets

(A1,A

2,A

3 )of

d-subsets

ofB

isthe

givenby

theK

ronecherpro

duct

hd h

d∗h

d hd∗h

d hd

Som

eR

epre

senta

tion

Theory

(1)T

heaction

ofS

2d

onthe

dsubsets

Aof

a2d

setB

isequivalent

tothe

actionof

S2d

onthe

leftcosets

ofthe

Young

subgroupS

d ×S

d .

(2)T

husthe

Frob

eniuscharateristic

ofthis

actionis

thesym

metric

functionh

d hd

(3)It

follows

thenthat

theFrob

eniuscharacteristic

ofthe

actionof

S2d

ontriplets

(A1,A

2,A

3 )of

d-subsets

ofB

isthe

givenby

theK

ronecherpro

duct

hd h

d∗h

d hd∗h

d hd

(4)In

particularwe

seethat

thescalar

product

⟨hd h

d∗h

d hd∗h

d hd

,S

2d ⟩

(∗)

givesthe

number

ofoccurrences

ofthe

trivialrepresentation

underthis

actionand

thusit

alsogives

thenum

ber

ofdistinct

orbits.

Som

eR

epre

senta

tion

Theory

(1)T

heaction

ofS

2d

onthe

dsubsets

Aof

a2d

setB

isequivalent

tothe

actionof

S2d

onthe

leftcosets

ofthe

Young

subgroupS

d ×S

d .

(2)T

husthe

Frob

eniuscharateristic

ofthis

actionis

thesym

metric

functionh

d hd

(3)It

follows

thenthat

theFrob

eniuscharacteristic

ofthe

actionof

S2d

ontriplets

(A1,A

2,A

3 )of

d-subsets

ofB

isthe

givenby

theK

ronecherpro

duct

hd h

d∗h

d hd∗h

d hd

(4)In

particularwe

seethat

thescalar

product

⟨hd h

d∗h

d hd∗h

d hd

,S

2d ⟩

(∗)

givesthe

number

ofoccurrences

ofthe

trivialrepresentation

underthis

actionand

thusit

alsogives

thenum

ber

ofdistinct

orbits.

(5)T

heorbit

ofa

triplet(A

1,A

2,A

3 )is

completely

characterisedby

thecardinalities

|A1∩

A2∩

A3 |

,|A

1∩

A2∩

Ac3 |

,|A

1∩

Ac2∩

A3 |

,|A

1∩

Ac2∩

Ac3 |

|Ac1∩

A2∩

A3 |

,|A

c1∩

A2∩

Ac3 |

,|A

c1∩

Ac2∩

A3 |

,|A

c1∩

Ac2∩

Ac3 |

Som

eR

epre

senta

tion

Theory

(1)T

heaction

ofS

2d

onthe

dsubsets

Aof

a2d

setB

isequivalent

tothe

actionof

S2d

onthe

leftcosets

ofthe

Young

subgroupS

d ×S

d .

(2)T

husthe

Frob

eniuscharateristic

ofthis

actionis

thesym

metric

functionh

d hd

(3)It

follows

thenthat

theFrob

eniuscharacteristic

ofthe

actionof

S2d

ontriplets

(A1,A

2,A

3 )of

d-subsets

ofB

isthe

givenby

theK

ronecherpro

duct

hd h

d∗h

d hd∗h

d hd

(4)In

particularwe

seethat

thescalar

product

⟨hd h

d∗h

d hd∗h

d hd

,S

2d ⟩

(∗)

givesthe

number

ofoccurrences

ofthe

trivialrepresentation

underthis

actionand

thusit

alsogives

thenum

ber

ofdistinct

orbits.

(5)T

heorbit

ofa

triplet(A

1,A

2,A

3 )is

completely

characterisedby

thecardinalities

|A1∩

A2∩

A3 |

,|A

1∩

A2∩

Ac3 |

,|A

1∩

Ac2∩

A3 |

,|A

1∩

Ac2∩

Ac3 |

|Ac1∩

A2∩

A3 |

,|A

c1∩

A2∩

Ac3 |

,|A

c1∩

Ac2∩

A3 |

,|A

c1∩

Ac2∩

Ac3 |

(6)In

conclusionthe

Kronecker

coeffi

cientin

(*)is

equalto

thenum

ber

ofsolutions

ofthe

Diophantine

system

|A1 |=

p000

+p

001

+p

010

+p

011

=d

,|A

c1 |=p

100

+p

101

+p

110

+p

111

=d

|A2 |=

p000

+p

001

+p

100

+p

101

=d

,|A

3 |=p

000

+p

010

+p

100

+p

110

=d

Kro

neck

er

coeffi

cients

and

Dio

phantin

esy

stem

s.

The

results

we

obtain

edfor

k=

3estab

lisha

connection

betw

eensolu

tions

ofou

rD

iophan

tine

prob

leman

dth

eR

epresen

tationT

heory

ofS

n .T

he

general

result

is:

Kro

neck

er

coeffi

cients

and

Dio

phantin

esy

stem

s.

The

results

we

obtain

edfor

k=

3estab

lisha

connection

betw

eensolu

tions

ofou

rD

iophan

tine

prob

leman

dth

eR

epresen

tationT

heory

ofS

n .T

he

general

result

is:

Theore

mDenoting

byad (k)

thenum

ber

ofways

ofplacing

integralweights

onthe

verticesof

thehyp

ercube

sothat

eachhyp

erfacehas

weight

dwe

have

ad (k)

=⟨h

d hd∗h

d hd∗···∗

hd h

d,S

2d ⟩

(kfactors)

Kro

neck

er

coeffi

cients

and

Dio

phantin

esy

stem

s.

The

results

we

obtain

edfor

k=

3estab

lisha

connection

betw

eensolu

tions

ofou

rD

iophan

tine

prob

leman

dth

eR

epresen

tationT

heory

ofS

n .T

he

general

result

is:

Theore

mDenoting

byad (k)

thenum

ber

ofways

ofplacing

integralweights

onthe

verticesof

thehyp

ercube

sothat

eachhyp

erfacehas

weight

dwe

have

ad (k)

=⟨h

d hd∗h

d hd∗···∗

hd h

d,S

2d ⟩

(kfactors)

Moreover

we

have∑d≥

0

ad (k)q

2d

=1

∏S⊆

[1,k

] (1−

q ∏i∈

Sai

∏j/∈Saj ) ∣∣∣∣∣a

01a22 ···a

0k

Kro

neck

er

coeffi

cients

and

Dio

phantin

esy

stem

s.

The

results

we

obtain

edfor

k=

3estab

lisha

connection

betw

eensolu

tions

ofou

rD

iophan

tine

prob

leman

dth

eR

epresen

tationT

heory

ofS

n .T

he

general

result

is:

Theore

mDenoting

byad (k)

thenum

ber

ofways

ofplacing

integralweights

onthe

verticesof

thehyp

ercube

sothat

eachhyp

erfacehas

weight

dwe

have

ad (k)

=⟨h

d hd∗h

d hd∗···∗

hd h

d,S

2d ⟩

(kfactors)

Moreover

we

have∑d≥

0

ad (k)q

2d

=1

∏S⊆

[1,k

] (1−

q ∏i∈

Sai

∏j/∈Saj ) ∣∣∣∣∣a

01a22 ···a

0k

Recall

that

we

proved

that

if

cd (k)

=⟨S

d,d

∗S

d,d

∗···∗

Sd

,d,S

2d ⟩

(kfactors)

Kro

neck

er

coeffi

cients

and

Dio

phantin

esy

stem

s.

The

results

we

obtain

edfor

k=

3estab

lisha

connection

betw

eensolu

tions

ofou

rD

iophan

tine

prob

leman

dth

eR

epresen

tationT

heory

ofS

n .T

he

general

result

is:

Theore

mDenoting

byad (k)

thenum

ber

ofways

ofplacing

integralweights

onthe

verticesof

thehyp

ercube

sothat

eachhyp

erfacehas

weight

dwe

have

ad (k)

=⟨h

d hd∗h

d hd∗···∗

hd h

d,S

2d ⟩

(kfactors)

Moreover

we

have∑d≥

0

ad (k)q

2d

=1

∏S⊆

[1,k

] (1−

q ∏i∈

Sai

∏j/∈Saj ) ∣∣∣∣∣a

01a22 ···a

0k

Recall

that

we

proved

that

if

cd (k)

=⟨S

d,d

∗S

d,d

∗···∗

Sd

,d,S

2d ⟩

(kfactors)

then

∑d≥

0

cd (k)q

2d

=

k∏i=

1 (1−

1a2i )

∏S⊆

[1,k

] (1−

q ∏i∈

Sai

∏j/∈Saj ) ∣∣∣∣∣a

01a22 ···a

0k

Kro

neck

er

coeffi

cients

and

Dio

phantin

esy

stem

s.

The

results

we

obtain

edfor

k=

3estab

lisha

connection

betw

eensolu

tions

ofou

rD

iophan

tine

prob

leman

dth

eR

epresen

tationT

heory

ofS

n .T

he

general

result

is:

Theore

mDenoting

byad (k)

thenum

ber

ofways

ofplacing

integralweights

onthe

verticesof

thehyp

ercube

sothat

eachhyp

erfacehas

weight

dwe

have

ad (k)

=⟨h

d hd∗h

d hd∗···∗

hd h

d,S

2d ⟩

(kfactors)

Moreover

we

have∑d≥

0

ad (k)q

2d

=1

∏S⊆

[1,k

] (1−

q ∏i∈

Sai

∏j/∈Saj ) ∣∣∣∣∣a

01a22 ···a

0k

Recall

that

we

proved

that

if

cd (k)

=⟨S

d,d

∗S

d,d

∗···∗

Sd

,d,S

2d ⟩

(kfactors)

then

∑d≥

0

cd (k)q

2d

=

k∏i=

1 (1−

1a2i )

∏S⊆

[1,k

] (1−

q ∏i∈

Sai

∏j/∈Saj ) ∣∣∣∣∣a

01a22 ···a

0k

Notice

the

similarity

?

000

001

010

011

100

101

111

110

A1

A2

A3

Clo

sely

rela

ted

Dio

phantin

eSyste

ms

Three

subsets

A1,A

2,A

3of

aset

Bof

cardin

ality2d

with

|A1 |=

d+

1,

|A2 |=

d+

1,

|A3 |=

d.

This

gives|A1 |−

|Ac1 |=

2,

|A2 |−

|Ac2 |=

2,

|A3 |−

|Ac3 |=

0

With

the

prev

ious

notation

the

new

Diop

han

tine

system

is

S∗3

=

∥∥∥∥∥p

1+

p2

+p

3+

p4−

p5−

p6−

p7−

p8

=2

p1

+p

2−

p3−

p4

+p

5+

p6−

p7−

p8

=2

p1−

p2

+p

3−

p4

+p

5−

p6

+p

7−

p8

=0

and

we

obtain

that

GF

S∗3 (q)

= ∑p∈S3 q

p1+

p2+

p3+

p4+

p5+

p6+

p7+

p8

= ∑p1 ≥

0 ∑p2 ≥

0 ··· ∑p8 ≥

0

qp1+

p2+

p3+

p4+

p5+

p6+

p7+

p8

ap1+

p2+

p3+

p4 −

p5 −

p6 −

p7 −

p8−

21

×

×ap1+

p2 −

p3 −

p4+

p5+

p6 −

p7 −

p8−

22

ap1 −

p2+

p3 −

p4+

p5 −

p6+

p7 −

p8

3

∣∣∣∣∣a01a02a03

Sum

min

gG

FS

∗3 (q)=

1a21

1a22

∏S⊆

[1,2

,3] (

1−

q ∏i∈

Sai

∏i/∈

Sai ) ∣∣∣∣∣a

01a02a03

More

Repre

senta

tion

Theory

(1)T

heaction

ofS2don

thed

+1

subsetsA

ofa

setB

ofcardinality

2dset

isequivalent

tothe

actionof

S2d

onthe

leftcosets

ofthe

Young

subgroupS

d+

1 ×S

d−1 .

More

Repre

senta

tion

Theory

(1)T

heaction

ofS2don

thed

+1

subsetsA

ofa

setB

ofcardinality

2dset

isequivalent

tothe

actionof

S2d

onthe

leftcosets

ofthe

Young

subgroupS

d+

1 ×S

d−1 .

(2)T

husthe

Frob

eniuscharateristic

ofthis

actionis

thesym

metric

functionh

d+

1 hd−

1

More

Repre

senta

tion

Theory

(1)T

heaction

ofS2don

thed

+1

subsetsA

ofa

setB

ofcardinality

2dset

isequivalent

tothe

actionof

S2d

onthe

leftcosets

ofthe

Young

subgroupS

d+

1 ×S

d−1 .

(2)T

husthe

Frob

eniuscharateristic

ofthis

actionis

thesym

metric

functionh

d+

1 hd−

1

(3)It

follows

thenthat

theFrob

eniuscharacteristic

ofthe

actionof

S2d

ontriplets

(A1,A

2,A

3 )w

ith|A

1 |=

d+

1,|A

2 |=

d+

1and

|A3 |

=d

isthe

givenby

theK

ronecherpro

duct

hd+

1 hd−

1∗h

d+

1 hd−

1∗h

d hd

More

Repre

senta

tion

Theory

(1)T

heaction

ofS2don

thed

+1

subsetsA

ofa

setB

ofcardinality

2dset

isequivalent

tothe

actionof

S2d

onthe

leftcosets

ofthe

Young

subgroupS

d+

1 ×S

d−1 .

(2)T

husthe

Frob

eniuscharateristic

ofthis

actionis

thesym

metric

functionh

d+

1 hd−

1

(3)It

follows

thenthat

theFrob

eniuscharacteristic

ofthe

actionof

S2d

ontriplets

(A1,A

2,A

3 )w

ith|A

1 |=

d+

1,|A

2 |=

d+

1and

|A3 |

=d

isthe

givenby

theK

ronecherpro

duct

hd+

1 hd−

1∗h

d+

1 hd−

1∗h

d hd

(4)In

particularwe

seethat

thescalar

product

ad (2

,2,0)

=⟨h

d+

1 hd−

1∗h

d+

1 hd−

1∗h

d hd

,S

2d ⟩

(∗)

givesthe

number

ofoccurrences

ofthe

trivialrepresentation

underthis

actionand

thusit

alsogives

thenum

ber

ofdistinct

orbits.

More

Repre

senta

tion

Theory

(1)T

heaction

ofS2don

thed

+1

subsetsA

ofa

setB

ofcardinality

2dset

isequivalent

tothe

actionof

S2d

onthe

leftcosets

ofthe

Young

subgroupS

d+

1 ×S

d−1 .

(2)T

husthe

Frob

eniuscharateristic

ofthis

actionis

thesym

metric

functionh

d+

1 hd−

1

(3)It

follows

thenthat

theFrob

eniuscharacteristic

ofthe

actionof

S2d

ontriplets

(A1,A

2,A

3 )w

ith|A

1 |=

d+

1,|A

2 |=

d+

1and

|A3 |

=d

isthe

givenby

theK

ronecherpro

duct

hd+

1 hd−

1∗h

d+

1 hd−

1∗h

d hd

(4)In

particularwe

seethat

thescalar

product

ad (2

,2,0)

=⟨h

d+

1 hd−

1∗h

d+

1 hd−

1∗h

d hd

,S

2d ⟩

(∗)

givesthe

number

ofoccurrences

ofthe

trivialrepresentation

underthis

actionand

thusit

alsogives

thenum

ber

ofdistinct

orbits.

(5)From

thisit

follows

that

∑d≥

0

ad (2

,2,0)q

2d

=

1a21

1a22

∏S⊆

[1,k

] (1−

q ∏i∈

Sai

∏j/∈Saj ) ∣∣∣∣∣a

01a22 ···a

0k

Dio

phantin

esy

stem

sand

Sch

ur

functio

ns

Note

that

the

Jacobi-T

rudiid

entity

gives

Sd

,d=

hd h

d−

hd−

1 hd+

1

Dio

phantin

esy

stem

sand

Sch

ur

functio

ns

Note

that

the

Jacobi-T

rudiid

entity

gives

Sd

,d=

hd h

d−

hd−

1 hd+

1

Thus

cd (3)

=⟨S

d,d

∗S

d,d

∗S

d,d

,S

2d ⟩

=

=⟨(h

d hd−

hd−

1 hd+

1 )∗(h

d hd−

hd−

1 hd+

1 )∗(h

d hd−

hd−

1 hd+

1 ),S

2d ⟩

=⟨h

d hd∗

hd h

d∗

hd h

d,S

2d ⟩

−3 ⟨h

d hd∗

hd h

d∗h

d−

1 hd+

1,S

2d ⟩

+3 ⟨h

d hd∗

hd−

1 hd+

1∗

hd−

1 hd+

1,S

2d ⟩

⟨h

d−

1 hd+

1∗

hd−

1 hd+

1∗

hd−

1 hd+

1 ∗,S

2d ⟩

Dio

phantin

esy

stem

sand

Sch

ur

functio

ns

Note

that

the

Jacobi-T

rudiid

entity

gives

Sd

,d=

hd h

d−

hd−

1 hd+

1

Thus

cd (3)

=⟨S

d,d

∗S

d,d

∗S

d,d

,S

2d ⟩

=

=⟨(h

d hd−

hd−

1 hd+

1 )∗(h

d hd−

hd−

1 hd+

1 )∗(h

d hd−

hd−

1 hd+

1 ),S

2d ⟩

=⟨h

d hd∗

hd h

d∗

hd h

d,S

2d ⟩

−3 ⟨h

d hd∗

hd h

d∗h

d−

1 hd+

1,S

2d ⟩

+3 ⟨h

d hd∗

hd−

1 hd+

1∗

hd−

1 hd+

1,S

2d ⟩

⟨h

d−

1 hd+

1∗

hd−

1 hd+

1∗

hd−

1 hd+

1 ∗,S

2d ⟩

fromw

hich

we

derive

acom

bin

atorial-represen

tationth

eoreticalpro

ofof

the

iden

tity

Dio

phantin

esy

stem

sand

Sch

ur

functio

ns

Note

that

the

Jacobi-T

rudiid

entity

gives

Sd

,d=

hd h

d−

hd−

1 hd+

1

Thus

cd (3)

=⟨S

d,d

∗S

d,d

∗S

d,d

,S

2d ⟩

=

=⟨(h

d hd−

hd−

1 hd+

1 )∗(h

d hd−

hd−

1 hd+

1 )∗(h

d hd−

hd−

1 hd+

1 ),S

2d ⟩

=⟨h

d hd∗

hd h

d∗

hd h

d,S

2d ⟩

−3 ⟨h

d hd∗

hd h

d∗h

d−

1 hd+

1,S

2d ⟩

+3 ⟨h

d hd∗

hd−

1 hd+

1∗

hd−

1 hd+

1,S

2d ⟩

⟨h

d−

1 hd+

1∗

hd−

1 hd+

1∗

hd−

1 hd+

1 ∗,S

2d ⟩

fromw

hich

we

derive

acom

bin

atorial-represen

tationth

eoreticalpro

ofof

the

iden

tity

W3 (q)

=∑d≥

0

cd (3)q

2d

=

3∏i=

1 (1−

1a2i )

∏S⊆

[1,3

] (1−

q ∏i∈

Sai

∏j/∈Saj ) ∣∣∣∣∣a

01a22a03

Dio

phantin

esy

stem

sand

Sch

ur

functio

ns

Note

that

the

Jacobi-T

rudiid

entity

gives

Sd

,d=

hd h

d−

hd−

1 hd+

1

Thus

cd (3)

=⟨S

d,d

∗S

d,d

∗S

d,d

,S

2d ⟩

=

=⟨(h

d hd−

hd−

1 hd+

1 )∗(h

d hd−

hd−

1 hd+

1 )∗(h

d hd−

hd−

1 hd+

1 ),S

2d ⟩

=⟨h

d hd∗

hd h

d∗

hd h

d,S

2d ⟩

−3 ⟨h

d hd∗

hd h

d∗h

d−

1 hd+

1,S

2d ⟩

+3 ⟨h

d hd∗

hd−

1 hd+

1∗

hd−

1 hd+

1,S

2d ⟩

⟨h

d−

1 hd+

1∗

hd−

1 hd+

1∗

hd−

1 hd+

1 ∗,S

2d ⟩

fromw

hich

we

derive

acom

bin

atorial-represen

tationth

eoreticalpro

ofof

the

iden

tity

W3 (q)

=∑d≥

0

cd (3)q

2d

=

3∏i=

1 (1−

1a2i )

∏S⊆

[1,3

] (1−

q ∏i∈

Sai

∏j/∈Saj ) ∣∣∣∣∣a

01a22a03

How

do

we

com

pute

these

consta

nt

term

s

How

do

we

com

pute

these

consta

nt

term

sA

vailable

Tools

by

han

d:

How

do

we

com

pute

these

consta

nt

term

sA

vailable

Tools

by

han

d:

(a)Som

eof

MacM

ahon

’sau

xiliary

iden

tities

11−

xa

11−

y/a ∣∣∣a

0=

11−

xy

11−

xa

11−

ya1

1−

w/a ∣∣∣a

0=

1(1

−xw

)(1−

yw

)

11−

xa

11−

y/a

11−

w/a ∣∣∣a

0=

1(1

−xy)(1

−xw

)

11−

xa

11−

ya1

1−

w/a

11−

z/a ∣∣∣a

0=

1−

xyw

z(1

−xw

)(1−

xz)(1

−yw

)(1−

yz)

How

do

we

com

pute

these

consta

nt

term

sA

vailable

Tools

by

han

d:

(a)Som

eof

MacM

ahon

’sau

xiliary

iden

tities

11−

xa

11−

y/a ∣∣∣a

0=

11−

xy

11−

xa

11−

ya1

1−

w/a ∣∣∣a

0=

1(1

−xw

)(1−

yw

)

11−

xa

11−

y/a

11−

w/a ∣∣∣a

0=

1(1

−xy)(1

−xw

)

11−

xa

11−

ya1

1−

w/a

11−

z/a ∣∣∣a

0=

1−

xyw

z(1

−xw

)(1−

xz)(1

−yw

)(1−

yz)

(b)

The

more

general

partial

fractionalgorith

mof

Guoce

Xin

How

do

we

com

pute

these

consta

nt

term

sA

vailable

Tools

by

han

d:

(a)Som

eof

MacM

ahon

’sau

xiliary

iden

tities

11−

xa

11−

y/a ∣∣∣a

0=

11−

xy

11−

xa

11−

ya1

1−

w/a ∣∣∣a

0=

1(1

−xw

)(1−

yw

)

11−

xa

11−

y/a

11−

w/a ∣∣∣a

0=

1(1

−xy)(1

−xw

)

11−

xa

11−

ya1

1−

w/a

11−

z/a ∣∣∣a

0=

1−

xyw

z(1

−xw

)(1−

xz)(1

−yw

)(1−

yz)

(b)

The

more

general

partial

fractionalgorith

mof

Guoce

Xin

Availab

leTools

by

Com

puter:

How

do

we

com

pute

these

consta

nt

term

sA

vailable

Tools

by

han

d:

(a)Som

eof

MacM

ahon

’sau

xiliary

iden

tities

11−

xa

11−

y/a ∣∣∣a

0=

11−

xy

11−

xa

11−

ya1

1−

w/a ∣∣∣a

0=

1(1

−xw

)(1−

yw

)

11−

xa

11−

y/a

11−

w/a ∣∣∣a

0=

1(1

−xy)(1

−xw

)

11−

xa

11−

ya1

1−

w/a

11−

z/a ∣∣∣a

0=

1−

xyw

z(1

−xw

)(1−

xz)(1

−yw

)(1−

yz)

(b)

The

more

general

partial

fractionalgorith

mof

Guoce

Xin

Availab

leTools

by

Com

puter:

(a)T

he

“O

mega”

MA

TH

EM

AT

ICA

pagkage

ofA

ndrew

set

all

How

do

we

com

pute

these

consta

nt

term

sA

vailable

Tools

by

han

d:

(a)Som

eof

MacM

ahon

’sau

xiliary

iden

tities

11−

xa

11−

y/a ∣∣∣a

0=

11−

xy

11−

xa

11−

ya1

1−

w/a ∣∣∣a

0=

1(1

−xw

)(1−

yw

)

11−

xa

11−

y/a

11−

w/a ∣∣∣a

0=

1(1

−xy)(1

−xw

)

11−

xa

11−

ya1

1−

w/a

11−

z/a ∣∣∣a

0=

1−

xyw

z(1

−xw

)(1−

xz)(1

−yw

)(1−

yz)

(b)

The

more

general

partial

fractionalgorith

mof

Guoce

Xin

Availab

leTools

by

Com

puter:

(a)T

he

“O

mega”

MA

TH

EM

AT

ICA

pagkage

ofA

ndrew

set

all

(b)

The

“Partial

Fraction

”M

AP

LE

package

ofG

uoce

Xin

How

do

we

com

pute

these

consta

nt

term

sA

vailable

Tools

by

han

d:

(a)Som

eof

MacM

ahon

’sau

xiliary

iden

tities

11−

xa

11−

y/a ∣∣∣a

0=

11−

xy

11−

xa

11−

ya1

1−

w/a ∣∣∣a

0=

1(1

−xw

)(1−

yw

)

11−

xa

11−

y/a

11−

w/a ∣∣∣a

0=

1(1

−xy)(1

−xw

)

11−

xa

11−

ya1

1−

w/a

11−

z/a ∣∣∣a

0=

1−

xyw

z(1

−xw

)(1−

xz)(1

−yw

)(1−

yz)

(b)

The

more

general

partial

fractionalgorith

mof

Guoce

Xin

Availab

leTools

by

Com

puter:

(a)T

he

“O

mega”

MA

TH

EM

AT

ICA

pagkage

ofA

ndrew

set

all

(b)

The

“Partial

Fraction

”M

AP

LE

package

ofG

uoce

Xin

(Guo

ceXin

>>

>>

>>

>Andrew

set

all)

Pure

lyC

om

bin

ato

rially

by

hand

Pure

lyC

om

bin

ato

rially

by

hand

Say

S2

=∥∥∥∥

p1

+p

2−

p3−

p4

=0

p1−

p2

+p

3−

p4

=0

(with

pi ≥

0)

Set

e=

min

(p1,p

4 ),

f=

min

(p2,p

3 )

and

note

that

(e,f,f,e)∈

S2

Pure

lyC

om

bin

ato

rially

by

hand

Say

S2

=∥∥∥∥

p1

+p

2−

p3−

p4

=0

p1−

p2

+p

3−

p4

=0

(with

pi ≥

0)

Set

e=

min

(p1,p

4 ),

f=

min

(p2,p

3 )

and

note

that

(e,f,f,e)∈

S2

so(u

1,u

2,u

3,u

4 )=

(p1,p

2,p

3,p

4 )−(e

,f,f,e)∈S

Pure

lyC

om

bin

ato

rially

by

hand

Say

S2

=∥∥∥∥

p1

+p

2−

p3−

p4

=0

p1−

p2

+p

3−

p4

=0

(with

pi ≥

0)

Set

e=

min

(p1,p

4 ),

f=

min

(p2,p

3 )

and

note

that

(e,f,f,e)∈

S2

so(u

1,u

2,u

3,u

4 )=

(p1,p

2,p

3,p

4 )−(e

,f,f,e)∈S

Possib

lepattern

s(u1,u

2,0

,0),

(u1,0

,u3,0)

,(0

,u2,0

,u4 )

,(0

,0,u

3,u

4 ),

01

111000

0110

00

011110

00

01

1000

Pure

lyC

om

bin

ato

rially

by

hand

Say

S2

=∥∥∥∥

p1

+p

2−

p3−

p4

=0

p1−

p2

+p

3−

p4

=0

(with

pi ≥

0)

Set

e=

min

(p1,p

4 ),

f=

min

(p2,p

3 )

and

note

that

(e,f,f,e)∈

S2

so(u

1,u

2,u

3,u

4 )=

(p1,p

2,p

3,p

4 )−(e

,f,f,e)∈S

Possib

lepattern

s(u1,u

2,0

,0),

(u1,0

,u3,0)

,(0

,u2,0

,u4 )

,(0

,0,u

3,u

4 ),

allim

possib

le

01

111000

0110

00

011110

00

01

1000

011110

00

011110

00

Pure

lyC

om

bin

ato

rially

by

hand

Say

S2

=∥∥∥∥

p1

+p

2−

p3−

p4

=0

p1−

p2

+p

3−

p4

=0

(with

pi ≥

0)

Set

e=

min

(p1,p

4 ),

f=

min

(p2,p

3 )

and

note

that

(e,f,f,e)∈

S2

so(u

1,u

2,u

3,u

4 )=

(p1,p

2,p

3,p

4 )−(e

,f,f,e)∈S

Possib

lepattern

s(u1,u

2,0

,0),

(u1,0

,u3,0)

,(0

,u2,0

,u4 )

,(0

,0,u

3,u

4 ),

allim

possib

le

Solu

tion

GF

S2 (x

1,x

2,x

3,x

4 )=

∑e≥0 ∑f≥

0

xe1 x

f2 xf3 x

e4=

1(1

−x1 x

4 )(1−

x2 x

3 )=

1(1

−)(1

−)

The

genera

lso

lutio

nfo

rk=

3by

Com

pute

r

The

genera

lso

lutio

nfo

rk=

3by

Com

pute

r

Apply

ing

MacM

ahon

Partition

Analy

sis

GF

S3 (x

1,x

2,...,x

8 )=

1(1

−x1 a

1 a2 a

3 )(1−

x2

a2a3

a1 )(1

−x3

a1a3

a2 )(1

−x4

a1a2

a3 )

x1

(1−

x5

a3

a1a2 )(1

−x6

a2

a1a3 )(1

−x7

a1

a2a3 )(1

−x8

1a1a2a3 ) ∣∣∣∣a

01a02a03

The

genera

lso

lutio

nfo

rk=

3by

Com

pute

r

Apply

ing

MacM

ahon

Partition

Analy

sis

GF

S3 (x

1,x

2,...,x

8 )=

1(1

−x1 a

1 a2 a

3 )(1−

x2

a2a3

a1 )(1

−x3

a1a3

a2 )(1

−x4

a1a2

a3 )

x1

(1−

x5

a3

a1a2 )(1

−x6

a2

a1a3 )(1

−x7

a1

a2a3 )(1

−x8

1a1a2a3 ) ∣∣∣∣a

01a02a03

and

the

MA

PLE

package

ofG

uoce

Xin

gives

GF

S (x1,x

2,...,x

8 )=

1(1

−x1 x

8 )(1−

x2 x

7 )(1−

x3 x

6 )(1−

x4 x

5 ) ×

× (1

+x1 x

4 x6 x

7

1−

x1 x

4 x6 x

7+

x2 x

3 x5 x

8

1−

x2 x

3 x5 x

8 )

The

genera

lso

lutio

nfo

rk=

3by

Com

pute

r

Apply

ing

MacM

ahon

Partition

Analy

sis

GF

S3 (x

1,x

2,...,x

8 )=

1(1

−x1 a

1 a2 a

3 )(1−

x2

a2a3

a1 )(1

−x3

a1a3

a2 )(1

−x4

a1a2

a3 )

x1

(1−

x5

a3

a1a2 )(1

−x6

a2

a1a3 )(1

−x7

a1

a2a3 )(1

−x8

1a1a2a3 ) ∣∣∣∣a

01a02a03

and

the

MA

PLE

package

ofG

uoce

Xin

gives

GF

S (x1,x

2,...,x

8 )=

1(1

−x1 x

8 )(1−

x2 x

7 )(1−

x3 x

6 )(1−

x4 x

5 ) ×

× (1

+x1 x

4 x6 x

7

1−

x1 x

4 x6 x

7+

x2 x

3 x5 x

8

1−

x2 x

3 x5 x

8 )

Visu

ally

0

111

000

11

00

The

genera

lso

lutio

nfo

rk=

3by

Com

pute

r

Apply

ing

MacM

ahon

Partition

Analy

sis

GF

S3 (x

1,x

2,...,x

8 )=

1(1

−x1 a

1 a2 a

3 )(1−

x2

a2a3

a1 )(1

−x3

a1a3

a2 )(1

−x4

a1a2

a3 )

x1

(1−

x5

a3

a1a2 )(1

−x6

a2

a1a3 )(1

−x7

a1

a2a3 )(1

−x8

1a1a2a3 ) ∣∣∣∣a

01a02a03

and

the

MA

PLE

package

ofG

uoce

Xin

gives

GF

S (x1,x

2,...,x

8 )=

1(1

−x1 x

8 )(1−

x2 x

7 )(1−

x3 x

6 )(1−

x4 x

5 ) ×

× (1

+x1 x

4 x6 x

7

1−

x1 x

4 x6 x

7+

x2 x

3 x5 x

8

1−

x2 x

3 x5 x

8 )

Visu

ally

1(

1−

)(1−

)(1−

)(1−

) (1

+(

1−

)+

(1−

) )

0

111

000

11

00

The

genera

lso

lutio

nfo

rk=

3by

Com

pute

r

Apply

ing

MacM

ahon

Partition

Analy

sis

GF

S3 (x

1,x

2,...,x

8 )=

1(1

−x1 a

1 a2 a

3 )(1−

x2

a2a3

a1 )(1

−x3

a1a3

a2 )(1

−x4

a1a2

a3 )

x1

(1−

x5

a3

a1a2 )(1

−x6

a2

a1a3 )(1

−x7

a1

a2a3 )(1

−x8

1a1a2a3 ) ∣∣∣∣a

01a02a03

and

the

MA

PLE

package

ofG

uoce

Xin

gives

GF

S (x1,x

2,...,x

8 )=

1(1

−x1 x

8 )(1−

x2 x

7 )(1−

x3 x

6 )(1−

x4 x

5 ) ×

× (1

+x1 x

4 x6 x

7

1−

x1 x

4 x6 x

7+

x2 x

3 x5 x

8

1−

x2 x

3 x5 x

8 )

Visu

ally

1(

1−

)(1−

)(1−

)(1−

) (1

+(

1−

)+

(1−

) )

Work

ing

by

hand

Work

ing

by

hand

Note

that

the

system

S3

=

∥∥∥∥∥p

1+

p2

+p

3+

p4−

p5−

p6−

p7−

p8

=0

p1

+p

2−

p3−

p4

+p

5+

p6−

p7−

p8

=0

p1−

p2

+p

3−

p4

+p

5−

p6

+p

7−

p8

=0

Work

ing

by

hand

Note

that

the

system

S3

=

∥∥∥∥∥p

1+

p2

+p

3+

p4−

p5−

p6−

p7−

p8

=0

p1

+p

2−

p3−

p4

+p

5+

p6−

p7−

p8

=0

p1−

p2

+p

3−

p4

+p

5−

p6

+p

7−

p8

=0

has

the

followin

gob

viou

ssolu

tions

(1,0

,0

,0

,0

,0

,0

,1)

(0,1

,0

,0

,0

,0

,1

,0)

(0,0

,1

,0

,0

,1

,0

,0)

(0,0

,0

,1

,1

,0

,0

,0)

Work

ing

by

hand

Note

that

the

system

S3

=

∥∥∥∥∥p

1+

p2

+p

3+

p4−

p5−

p6−

p7−

p8

=0

p1

+p

2−

p3−

p4

+p

5+

p6−

p7−

p8

=0

p1−

p2

+p

3−

p4

+p

5−

p6

+p

7−

p8

=0

has

the

followin

gob

viou

ssolu

tions

(1,0

,0

,0

,0

,0

,0

,1)

(0,1

,0

,0

,0

,0

,1

,0)

(0,0

,1

,0

,0

,1

,0

,0)

(0,0

,0

,1

,1

,0

,0

,0)

So

we

sete

=m

in(p

1,p

8 ),

f=

min

(p2,p

7 ),

g=

min

(p3,p

6 ),

h=

min

(p4,p

5 ),

Work

ing

by

hand

Note

that

the

system

S3

=

∥∥∥∥∥p

1+

p2

+p

3+

p4−

p5−

p6−

p7−

p8

=0

p1

+p

2−

p3−

p4

+p

5+

p6−

p7−

p8

=0

p1−

p2

+p

3−

p4

+p

5−

p6

+p

7−

p8

=0

has

the

followin

gob

viou

ssolu

tions

(1,0

,0

,0

,0

,0

,0

,1)

(0,1

,0

,0

,0

,0

,1

,0)

(0,0

,1

,0

,0

,1

,0

,0)

(0,0

,0

,1

,1

,0

,0

,0)

So

we

sete

=m

in(p

1,p

8 ),

f=

min

(p2,p

7 ),

g=

min

(p3,p

6 ),

h=

min

(p4,p

5 ),

and

we

arered

uced

todeterm

ine

the

solution

s

(u1,u

2,u

3,u

4,u

5,u

6,u

7,u

8 )=

(p1,p

2,p

3,p

4,p

5,p

6,p

7,p

8 )−

(e,f

,g

,h

,h

,g

,f

,e)

Work

ing

by

hand

Note

that

the

system

S3

=

∥∥∥∥∥p

1+

p2

+p

3+

p4−

p5−

p6−

p7−

p8

=0

p1

+p

2−

p3−

p4

+p

5+

p6−

p7−

p8

=0

p1−

p2

+p

3−

p4

+p

5−

p6

+p

7−

p8

=0

has

the

followin

gob

viou

ssolu

tions

(1,0

,0

,0

,0

,0

,0

,1)

(0,1

,0

,0

,0

,0

,1

,0)

(0,0

,1

,0

,0

,1

,0

,0)

(0,0

,0

,1

,1

,0

,0

,0)

So

we

sete

=m

in(p

1,p

8 ),

f=

min

(p2,p

7 ),

g=

min

(p3,p

6 ),

h=

min

(p4,p

5 ),

and

we

arered

uced

todeterm

ine

the

solution

s

(u1,u

2,u

3,u

4,u

5,u

6,u

7,u

8 )=

(p1,p

2,p

3,p

4,p

5,p

6,p

7,p

8 )−

(e,f

,g

,h

,h

,g

,f

,e)

This

proves

GF

S3 (x

1,x

2,...,x

8 )=

1(

1−

)(1−

)(1−

)(1−

)x

(someth

ing)

Dete

rmin

ing

the

asy

mm

etric

solu

tions

The

diff

erences

(u1,u

2,u

3,u

4,u

5,u

6,u

7,u

8 )=

(p1,p

2,p

3,p

4,p

5,p

6,p

7,p

8 )−

(e,f

,g

,h

,h

,g

,f

,e)

have

the

prop

ertyth

atu1 u

8=

0;u

2 u7

=0;

u3 u

6=

0;u

4 u5

=0.

The

support

ofon

eof

these

solution

scou

ld,in

prin

ciple,

be

givenby

any

one

of24

=16

pattern

s:

(x,x

,x,x

,0,0

,0,0)

,(0

,x,x

,x,0

,0,0

,x),

(x,0

,x,x

,0,0

,x,0)

,(0

,0,x

,x,0

,0,x

,x),

(x,x

,0,x

,0,x

,0,0)

,(0

,x,0

,x,0

,x,0

,x),

(x,0

,0,x

,0,x

,x,0)

,(0

,0,0

,x,0

,x,x

,x),

(x,x

,x,0

,x,0

,0,0)

,(0

,x,x

,0,x

,0,0

,x),

(x,0

,x,0

,x,0

,x,0)

,(0

,0,x

,0,x

,0,x

,x),

(x,x

,0,0

,x,x

,0,0)

,(0

,x,0

,0,x

,x,0

,x),

(x,0

,0,0

,x,x

,x,0)

,(0

,0,0

,0,x

,x,x

,x),

How

ever,all

but

two

ofth

esepattern

scan

be

the

support

ofa

solution

,nam

ely

(0,x

,x,0

,x,0

,0,x)

and

(x,0

,0,x

,0,x

,x,0)

This

proves

the

iden

tity

(someth

ing)

= (1

+x1 x

4 x6 x

7

1−

x1 x

4 x6 x

7+

x2 x

3 x5 x

8

1−

x2 x

3 x5 x

8 )= (

1+ (

1−

)+ (

1−

) )

1 2

3

4

5

87

6

1 2

3

4

5

87

6

1 2

3

4

5

87

6

1 2

3

4

5

87

6

1 2

3

4

5

87

6

1 2

3

4

5

87

6

1 2

3

4

5

87

6

1 2

3

4

5

87

6

1 2

3

4

5

87

6

1 2

3

4

5

87

6

1 2

3

4

5

87

6

1 2

3

4

5

87

6

1 2

3

4

5

87

6

1 2

3

4

5

87

6

1 2

3

4

5

87

6

1 2

3

4

5

87

6

AV

enn

dia

gra

mvie

wofth

e14

exclu

ded

patte

rns

Here

ablack

dot

ina

subset

represen

tsan

assignm

ent

ofan

integer

givin

gth

ecard

inality

ofth

atsu

bset.

Rem

arkably,

allbut

two

ofth

eseassign

emts

canbe

elimin

atedregard

lessof

the

actual

assigned

values.

Connectio

ns

with

Invaria

nt

Theory

If

A1

= [a1

00

1/a

1 ]an

dA

2= [

a2

00

1/a

2 ]

then

A1⊗

A2

=

⎡⎢⎣a1 a

20

00

0a2/a

10

00

0a1/a

20

00

01/a

1 a2 ⎤⎥⎦

More

generally

givenk

matrices

A1

= [a1

00

1/a

1 ],

A2

= [a2

00

1/a

2 ],

...,A

2= [

ak

00

1/a

k ]

then

A1⊗

A2⊗

⊗···⊗

Ak

=diag ∥∥∥∥ ∏

i∈Sai

∏i/∈

Sai ∥∥∥∥

S⊆

[1,k

]

Don’t

youfind

theseeigenvalues

somew

hatfam

iliar?

MacM

ahon

strikes

again

Itfollow

sfrom

Molien

sT

heorem

that

the

Hilb

ertseries

ofth

erin

gof

invarian

tsof

the

group

Gk

={A

1⊗

A2⊗

⊗···⊗

Ak }

with

a1

=eiθ

1,a2

=eiθ

2,...,ak

=eiθ

kis

the

rational

function

FG

(q)=

1(2

π)k ∫

2π

0

∫2π

0

··· ∫2π

0

1∏S⊆

[1,k

] (1−

q ∏i∈

Sai

∏i/∈

Sai )

dθ1 d

θ2 ···d

θk

But

we

caneasily

seeth

atth

isis

same

as

FG

(q)=

1∏S⊆

[1,k

] (1−

q ∏i∈

Sai

∏i/∈

Sai ) ∣∣∣∣∣a

01a02 ···a

0k

This

shou

ldnot

be

surp

rising

since,

forin

stance

the

mon

omials

xp1

1xp2

2···x

p8

8th

atare

invarian

tunder

the

group

G3

areprecisely

those

forw

hich

(p1,p

2,...,p

8 )∈S3

The

final(S

chur

functio

n)

surp

rise

Inth

esam

evein

,w

em

ayask

forth

eH

ilbert

seriesof

the

group

Hk

={A

1⊗

A2⊗

···⊗A

k:A

i ∈SU

[2]}

Inth

iscase

Molien

sth

eoremgives

that

the

Hilb

ertSeries

ofth

ecorresp

ondin

grin

gof

invarian

tsis

(here

againa1

=eiθ

1,a2

=eiθ

2,...,ak

=eiθ

k,)

FH

k (q)=

1(π)

k ∫2π

0

∫2π

0

··· ∫2π

0

1∏S⊆

[1,k

] (1−

q ∏i∈

Sai

∏i/∈

Sai )

k∏i=

1 (sin2(θ

i ) )dθ1 d

θ2 ···d

θk

and

iseasily

seenth

atupon

mak

ing

the

substitu

tion

sin2(θ

i )=

1−

cos(2θi )

2=

1−

a2i+

1/a2i

2

2

that

this

isth

esam

easF

Hk (q)

=

k∏i=

1 (1−

1a2i )

∏S⊆

[1,k

] (1−

q ∏i∈

Sai

∏j/∈Saj ) ∣∣∣∣∣a

01a22 ···a

0k

TH

ESA

GA

IST

ILL

CO

NT

INU

ING

with

Diophantine

systems

andconstant

termidentities

galore!