Symmetric Functions, Alternating Sign Matrices, and Tokuyama’s Identity
Angèle Hamel Wilfrid Laurier University Discrete Math Days/OCW May 23, 2015
Tuesday, 2 June, 15
Symmetric Functions, Alternating Sign Matrices, and Tokuyamas - - PowerPoint PPT Presentation
Symmetric Functions, Alternating Sign Matrices, and Tokuyamas Identity Angle Hamel Wilfrid Laurier University Discrete Math Days/OCW May 23, 2015 Tuesday, 2 June, 15 Symmetric Functions Tuesday, 2 June, 15 Definition A function is
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Tuesday, 2 June, 15
Tuesday, 2 June, 15
1 + x2 2 + x2 3 + x2 4 + x1x2 + x1x3
1 + x3 2 + x3 3 + x2 1x2 + x2 1x3 + x2 2x1
2x3 + x2 3x1 + x2 3x2 + x1x2x3
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Tuesday, 2 June, 15
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Tuesday, 2 June, 15
1x2 2x2 3x3 4x5
1 xw2 2 · · · xwn n
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Tuesday, 2 June, 15
1 1 2
1 1 3 1 2 2 1 2 3
1 3 2
1 3 3
2 2 3
2 3 3
1x2
1x3
2
3
2x3
3
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Tuesday, 2 June, 15
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13
Tuesday, 2 June, 15
14
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1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 −1 1 1
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m−1
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Tuesday, 2 June, 15
Y
1≤i<j≤n
B B B B @
1
1
2
2
n
n
1 C C C C A
X
σ∈Sn
σ(1)xn−2 σ(2) . . . xσ(n−1)
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Tuesday, 2 June, 15
x2
1x2
Tuesday, 2 June, 15
x2
1x2
x2
1x3
x2
2x1
x2
2x3
x2
3x1
x2
3x2
x1x2x3
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x2
1x2
x2
1x3
x2
2x1
x2
2x3
x2
3x1
x2
3x2
x1x2x3
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x2
1x2
x2
1x3
x2
2x1
x2
2x3
x2
3x1
x2
3x2
σ = 1 2 3 · · · n σ(1) σ(2) σ(3) · · · σ(n)
!
σ(1)xn−2 σ(2) . . . xσ(n−1)
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Y
1≤i<j≤n
B B B B @
1
1
2
2
n
n
1 C C C C A
X
σ∈Sn
σ(1)xn−2 σ(2) . . . xσ(n−1)
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Tuesday, 2 June, 15
Y
1≤i<j≤n
X
σ∈Sn
σ(1)xn−2 σ(2) . . . xσ(n−1)
Y
1≤i<j≤n
X
A∈An
n
Y
i=1
i
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Tuesday, 2 June, 15
Tuesday, 2 June, 15
n
xi
(xi + tx j) sλ(x) =
thgt(ST )(1 + t)str(ST )−n xwgt(ST )
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1 1 1 2 2 2 3 3 5 2 2 3 3 4 5 5 6 3 3 4 4 5 6 4 5 5 5 5 6 6 6
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ST =
1 1 1 2 2 2 3 3 5 2 2 3 3 4 5 5 6 3 3 4 4 5 6 4 5 5 5 5 6 6 6
∈ ST 986431(6) with wgt(ST)=(3, 5, 6, 4, 8, 5) str(ST)=12, hgt(ST )=6.
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Tuesday, 2 June, 15
A = 1 1 1 −1 1 1 −1 1 1 −1 1 1 −1 1
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N E N E W E NW NW NW NW NW NW N E N E SE W E NW NW NW NW NW W E NW N S SE N E W E NW NW NW SE N E N E SE W E N S N E N E W E SE N E W E N S SE N E N E W E SW SE N E SE W E N S W E NW SW SW
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Y
1≤i<j≤n
(xi+txj) sλ(x) = X
A∈Aµ(n) n
Y
k=1
tSEk(A)(1+t)NSk(A)xNEk(A)+SEk(A)+NSk(A)
k
=
(xi + y j) sλ(x) =
n
x N Ek(A)
k
ySEk(A)
k
(xk + yk)N Sk(A).
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1 1 1 2 2 2 3 3 5 2 2 3 3 4 5 5 6 3 3 4 4 5 6 4 5 5 5 5 6 6 6
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PST =
1 1 1 2 3 3 4 4 4 2 2 2 3 4 5 5 5 3 4 4 4 5 6 4 5 5 6 5 6 6 6
= ⇒ M(PST )= 1 1 1 0 0 0 0 2 2 2 2 0 0 0 0 3 0 0 3 3 3 0 0 0 4 4 4 4 4 0 4 4 4 5 5 5 0 5 5 5 5 0 6 6 6 6 0 6 0 0 0
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PST =
1 1 1 2 3 3 4 4 4 2 2 2 3 4 5 5 5 3 4 4 4 5 6 4 5 5 6 5 6 6 6
= ⇒ M(PST )= 1 1 1 0 0 0 0 2 2 2 2 0 0 0 0 3 0 0 3 3 3 0 0 0 4 4 4 4 4 0 4 4 4 5 5 5 0 5 5 5 5 0 6 6 6 6 0 6 0 0 0 A = 1 1 1 −1 1 1 −1 1 1 −1 1 1 −1 1
= ⇒
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1 2 1 4 5 6 1 2 3 2 3 2 5 2 3 5 5 3 4 3 3 4 6 4 5 6 5 5 5 6 6
1 2 1 4 5 6 2 3 2 5 2 3 4 3 3 4 5 6 5 5 6
1 2 3 3 5 5 4 6 5 6
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Tuesday, 2 June, 15
↑ ↓ ↑ ↓ ↑ ↓ → · ← ← · → → · → ← · ← ← · ← → · → ↓ ↑ ↑ ↓ ↑ ↓
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Tuesday, 2 June, 15
A = 1 1 1 −1 1 1 −1 1 1 −1 1 1 −1 1
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Tuesday, 2 June, 15
1 1 1 1 1 1 1 1 1 1
SE SE SE SE WE SE SE SE NS SW WE SE WE SE NE SE NE SE NE SE NS SW NW WE WE SE NE NE NS WE NE NE WE NE NE NE
2 3 4 4 3 2 1
A∈Aδ
Xn
Xn(x; t) =
A∈Aλ
Xn
A∈Aδ
Xn
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Tuesday, 2 June, 15
Tuesday, 2 June, 15
(x1 − a1) (x1 − a2) (x2 − a4) (x4 − a7) (x2 − a1) (x3 − a3) (x3 − a4) (x4 − a2) (x4 − a3) (x5 − a5)
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16i6n
16i<j6n
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Tuesday, 2 June, 15
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Tuesday, 2 June, 15