Post on 19-Jul-2020
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!