Hook Lengths and Contents
Richard P. Stanley
M.I.T.
Hook Lengths and Contents – p.
Standard Young tableau
standard Young tableau (SYT) of shape
λ = (4, 4, 3, 1):
<
<
1
3
4
2
6
9 11
5
7 12
8
10
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fλ : number of SYT of shape λ
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fλ : number of SYT of shape λ
f3,2 = 5:
123
45
124
35
134
25
135
24
125
34
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Hook length formula
For u ∈ λ, let hu be the hook length at u, i.e.,
the number of squares directly below or to
the right of u (counting u once)
Hook Lengths and Contents – p.
Hook length formula
For u ∈ λ, let hu be the hook length at u, i.e.,
the number of squares directly below or to
the right of u (counting u once)
7
5
4
2
6
4
3
1
4
2
1
1
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Theorem (Frame-Robinson-Thrall). Let λ ` n.
Then
n!
fλ = Q
.
u∈λ hu
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Theorem (Frame-Robinson-Thrall). Let λ ` n.
Then
n!
fλ = Q
.
u∈λ hu
f (4,4,3,1) =
12!
7·6·5·43 ·3·22 ·13
= 2970
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Nekrasov-Okounkov identity
RSK algorithm (or representation theory) ⇒
X
fλ2 = n!
λ`n
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Nekrasov-Okounkov identity
RSK algorithm (or representation theory) ⇒
X
fλ2 = n!
λ`n
Nekrasov-Okounkov (2006), G. Han (2008):
!
n
Y
X X
x
fλ2
(t + h2u )
2
n!
n≥0 λ`n
u∈λ
=
Y
(1 − xi )−t−1
i≥1
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A corollary
ek : kth elementary SF
X
xi1 xi2 · · · x ik
ek (x1 , x2 , . . . ) =
1≤i1 <i2 <···<ik
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A corollary
ek : kth elementary SF
X
xi1 xi2 · · · x ik
ek (x1 , x2 , . . . ) =
1≤i1 <i2 <···<ik
Corollary. Let
gk (n) =
1 X
n! λ`n
fλ2 ek (h2u : u ∈ λ).
Then gk (n) ∈ Q[n].
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Conjecture of
Conjecture (Han). Let j ∈ P. Then
X
X
1
2
2j
f
h
λ
u ∈ Q[n].
n!
λ`n
u∈λ
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Conjecture of
Conjecture (Han). Let j ∈ P. Then
X
X
1
2
2j
f
h
λ
u ∈ Q[n].
n!
λ`n
u∈λ
True for j = 1 by above.
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Conjecture of
Conjecture (Han). Let j ∈ P. Then
X
X
1
2
2j
f
h
λ
u ∈ Q[n].
n!
λ`n
u∈λ
True for j = 1 by above.
Stronger conjecture. For any symmetric
function F ,
X
1
2
2
fλ F (hu : u ∈ λ) ∈ Q[n].
n!
λ`n
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Examples
Let dk (n)=
1 X
n! λ`n
fλ2
X
h2k
u .
u∈λ
d1 (n) =
1
n(3n
2
− 1)
d2 (n) =
1
n(n
24
− 1)(27n2 − 67n + 74)
d3 (n) =
1
n(n −
48
3
1)(n − 2)
(27n − 174n2 + 511n − 552).
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Open variants
1 X
n! λ`n
1 X
n! λ`n
1 X
n! λ`n
2
fλ
X
hu = ?
X
hu = ?
X
h2u = ?
u∈λ
fλ
u∈λ
fλ
u∈λ
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Contents
For u ∈ λ let c(u)= i − j, the content of
square u = (i, j).
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Contents
For u ∈ λ let c(u)= i − j, the content of
square u = (i, j).
0 1 2 3
−1 0 1 2
−2 −1 0
−3
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Semistandard tableaux
semistandard Young tableau (SSYT) of shape
λ = (4, 4, 3, 1):
<
<
1
2
2
4
2
3
6
6
4
4
7
7
xT = x1 x32 x34 x5 x26 x27
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Schur functions
sλ =
X
xT ,
T
summed over all SSYT of shape λ
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Schur functions
sλ =
X
xT ,
T
summed over all SSYT of shape λ
s3 =
X
xi xj xk
X
xi xj xk = e 3 .
i≤j≤k
s1,1,1 =
i<j<k
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Hook-content formula
f (1t )= f (1, 1, . . . , 1) (t 10 s)
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Hook-content formula
f (1t )= f (1, 1, . . . , 1) (t 10 s)
sλ (1t ) = #{SSYT of shape λ, max ≤ t}.
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Hook-content formula
f (1t )= f (1, 1, . . . , 1) (t 10 s)
sλ (1t ) = #{SSYT of shape λ, max ≤ t}.
Theorem. sλ (1t ) =
Y t + cu
u∈λ
hu
.
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A curiosity
Let κ(w) denote the number of cycles of
w ∈ Sn . Then
1
X
n! u,v∈S
q κ(uvu
n
v
−1 −1
)
=
XY
(q + cu ).
λ`n u∈λ
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A curiosity
Let κ(w) denote the number of cycles of
w ∈ Sn . Then
1
X
n! u,v∈S
q κ(uvu
n
v
−1 −1
)
=
XY
(q + cu ).
λ`n u∈λ
Restatement:
X
n!
ek (cu : u ∈ λ)
λ`n
= #{(u, v) ∈ Sn × Sn : κ(uvu−1 v −1 ) = n − k}
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Another curiosity
X
w∈Sn
q κ(w
2
)
=
X
λ`n
fλ
Y
(q + cu )
u∈λ
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Han’s conjecture for contents
Theorem. For any symmetric function F ,
1 X
n! λ`n
fλ2 F (cu : u ∈ λ) ∈ Q[n].
Idea of proof. By linearity, suffices to take
F = eµ .
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Power sum symmetric functions
x = (x1 , x2 , . . . )
P m
pm = pm (x) =
i xi
p λ = p λ1 p λ2 · · ·
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Power sum symmetric functions
x = (x1 , x2 , . . . )
P m
pm = pm (x) =
i xi
p λ = p λ1 p λ2 · · ·
Let `(λ) be the length (number of parts) of λ
pm (1t ) = t
pλ (1t ) = t`(λ)
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Some notation
x(1) , . . . , x(k) : disjoint sets of variables
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Some notation
x(1) , . . . , x(k) : disjoint sets of variables
ρ(w): cycle type of w ∈ Sn, e.g.,
w = (1, 3, 4)(2)(5) ⇒ pρ(w) = p(3,1,1) = p3 p21
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Some notation
x(1) , . . . , x(k) : disjoint sets of variables
ρ(w): cycle type of w ∈ Sn, e.g.,
w = (1, 3, 4)(2)(5) ⇒ pρ(w) = p(3,1,1) = p3 p21
Hλ =
Y
hu
u∈λ
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The fundamental tool
Theorem.
X
Hλk−2 sλ (x(1) ) · · · sλ (x(k) )
λ`n
=
1
n! w
X
pρ(w1 ) (x(1) ) · · · pρ(wk ) (x(k) )
1 ···wk =1
in Sn
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The fundamental tool
Theorem.
X
Hλk−2 sλ (x(1) ) · · · sλ (x(k) )
λ`n
=
1
n! w
X
pρ(w1 ) (x(1) ) · · · pρ(wk ) (x(k) )
1 ···wk =1
in Sn
Proof. Follows from representation theory:
the relation between multiplying conjugacy
class sums in Z(CG) (the center of the group
algebra of the finite group G) and the
irreducible characters of G.
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Set x
(i)
(i)
ti
= 1 , so sλ (x ) →
(i)
pρ(wi ) (x ) →
Q ti +cu
hu
κ(wi )
ti
.
Get
X
−2
Hλ
λ`n
Y
(t1 + cu ) · · · (tk + cu ) =
u∈λ
1
n! w
X
κ(w1 )
t1
κ(wk )
· · · tk
.
1 ···wk =1
in Sn
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Completion of proof
n−µk
1
Take coefficient of tn−µ
·
·
·
t
:
1
k
X
1
2
f
λ eµ (cu : u ∈ λ)
n!
λ`n
= #{(w1 , . . . , wk ) ∈ Skn : w1 · · · wk = 1,
c(wi ) = n − µi }.
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Completion of proof
n−µk
1
Take coefficient of tn−µ
·
·
·
t
:
1
k
X
1
2
f
λ eµ (cu : u ∈ λ)
n!
λ`n
= #{(w1 , . . . , wk ) ∈ Skn : w1 · · · wk = 1,
c(wi ) = n − µi }.
Elementary combinatorial argument shows
this is a polynomial in n. Hook Lengths and Contents – p. 2
Shifted parts
Let λ = (λ1 , . . . , λn ) ` n.
partitions of 3: 300, 210, 111
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Shifted parts
Let λ = (λ1 , . . . , λn ) ` n.
partitions of 3: 300, 210, 111
shifted parts of λ: λi + n − i, 1 ≤ i ≤ n
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Shifted parts
Let λ = (λ1 , . . . , λn ) ` n.
partitions of 3: 300, 210, 111
shifted parts of λ: λi + n − i, 1 ≤ i ≤ n
shifted parts of (3, 1, 1, 0, 0): 7, 4, 3, 1, 0
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Contents and shifted parts
Let F (x; y) be symmetric in x and y variables
separately, e.g.,
p1 (x)p2 (y) = (x1 + x2 + · · · )(y12 + y22 + · · · ).
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Contents and shifted parts
Let F (x; y) be symmetric in x and y variables
separately, e.g.,
p1 (x)p2 (y) = (x1 + x2 + · · · )(y12 + y22 + · · · ).
Theorem. Let
1 X 2
r(n)=
fλ F (cu , u ∈ λ; λi + n − i, 1 ≤ i ≤ n
n! λ`n
Then r(n) ∈ Z (n fixed), and r(n) ∈ Q[n].
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Similar results are false
Note. Let
P (n)=
1 X
n! λ`n
2
fλ
X
i
2
λi
!
.
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Similar results are false
Note. Let
P (n)=
1 X
n! λ`n
2
fλ
X
i
2
λi
!
.
Then
P (3) =
16
3
6∈ Z
P (n) 6∈ Q[n]
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ϕ
ΛQ = {symmetric functions over Q}
Define a linear transformation
ϕ : ΛQ → Q[t]
by
ϕ(sλ ) =
n
Y
(t + λi + n − i)
i=1
Hλ
.
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Key lemma for shifted parts
Lemma. Let µ ` n, ` = `(µ). Then
ϕ(pµ ) =
m X
m
n−`
t(t + 1) · · · (t + i − 1),
(−1)
i
i=0
where m= m1 (µ), the number of parts of µ
equal to 1.
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Proof of main result
Theorem. For any symmetric function F ,
X
1
2
2
f
F
(h
λ
u : u ∈ λ) ∈ Q[n].
n!
λ`n
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Proof of main result
Theorem. For any symmetric function F ,
X
1
2
2
f
F
(h
λ
u : u ∈ λ) ∈ Q[n].
n!
λ`n
Proof based on the multiset identity
{hu : u ∈ λ} ∪ {λi − λj − i + j :
1 ≤ i < j ≤ n}
= {n + cu : u ∈ λ} ∪ {1n−1 , 2n−2 , . . . , n − 1}.
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Example. λ = (3, 1, 0, 0),
λ − (1, 2, 3, 4) = (2, −1, −3, −4):
{4, 2, 1, 1} ∪ {3, 5, 6, 2, 3, 1}
= {3, 4, 5, 6} ∪ {1, 1, 1, 2, 2, 3}
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{hu : u ∈ λ} ∪ {λi − λj − i + j :
1 ≤ i < j ≤ n}
= {n + cu : u ∈ λ} ∪ {1n−1 , 2n−2 , . . . , n − 1}.
Hook Lengths and Contents – p. 3
{hu : u ∈ λ} ∪ {λi − λj − i + j :
1 ≤ i < j ≤ n}
= {n + cu : u ∈ λ} ∪ {1n−1 , 2n−2 , . . . , n − 1}.
Note.
λi − λj − i + j = (λi + n − i) − (λj + n − j)
Not symmetric in λi + n − i and λj + n − j,
but (λi − λj − i + j)2 is symmetric.
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Further results
Recall:
Nekrasov-Okounkov (2006), G. Han (2008):
!
n
Y
X X
x
fλ2
(t + h2u )
2
n!
n≥0 λ`n
u∈λ
=
Y
(1 − xi )−t−1
i≥1
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Content analogue
“Content Nekrasov-Okounkov identity”
!
n
X X
Y
x
fλ2
(t + c2u )
2
n!
u∈λ
n≥0 λ`n
= (1 − x)−t .
Proof: simple consequence of “dual Cauchy
identity" and the hook-content formula.
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Fujii et al. variant
1 X
n! λ`n
fλ2
=
X k−1
Y
(c2u − i2 )
u∈λ i=0
(2k)!
(k + 1)!
(n)
.
k+1
2
where (n)k+1 = n(n − 1) · · · (n − k).
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Okada’s conjecture
1 X
n! λ`n
=
fλ2
k
XY
u∈λ i=1
1
2(k +
(h2u − i2 )
1)2
2k 2k + 2
k
k+1
(n)k+1
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Proof of Okada’s conjecture
Proof (Greta Panova).
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Proof of Okada’s conjecture
Proof (Greta Panova).
Both sides are polynomials in n.
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Proof of Okada’s conjecture
Proof (Greta Panova).
Both sides are polynomials in n.
The two sides agree for 1 ≤ n ≤ k + 2.
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Proof of Okada’s conjecture
Proof (Greta Panova).
Both sides are polynomials in n.
The two sides agree for 1 ≤ n ≤ k + 2.
Key step: both sides have degree k + 1.
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Central factorial numbers
xn =
X
T (n, k)x(x − 12 )(x − 22 ) · · · (x − k2 )
k
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Central factorial numbers
xn =
X
T (n, k)x(x − 12 )(x − 22 ) · · · (x − k2 )
k
n\k
1
2
3
4
5
1 2
3
4 5
1
1 1
1 5
1
1 21 14 1
1 85 147 30 1
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Okada-Panova redux
Corollary.
1 X
n! λ`n
=
k+1
X
j=1
2
fλ
X
2k
hu
u∈λ
T (k + 1, j)
1 2(j − 1) 2j
2j 2
j−1
j
(n)j .
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