25March2015 u rMathematik,Universit a tZ u rich ¨ ¨ ¨ (jointworkwithGuo-NiuHAN)HuanXiongInstitutf Differenceoperatorsforpartitionsandsomeapplications

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Introduction Difference operators Some applications on partition formulas

Difference operators for partitions and some applications

(joint work with Guo-Niu HAN)

Huan Xiong

Institut für Mathematik, Universität Zürich

S´eminaire Lotharingien de Combinatoire(74) Ellwangen, Germany

25 March 2015

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Plan of Talk

1 Introduction

2 Difference operators

3 Some applications on partition formulas

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Definition

Apartitionis a finite weakly decreasing sequence of positive integers λ= (λ1, λ2, . . . , λr). The integer|λ|= P

1≤i≤r

λiis called thesizeof theλ.

A partitionλcould be identical with itsYoung diagram, which is a collection of boxes arranged in left-justified rows withλiboxes in thei-th row.

hook lengthh: the number of boxes exactly to the right, or exactly below, oritself.

hook product ofλ:H(λ) = Q

∈λ

h.

7 5 2 1 4 2 3 1 1

Figure:The Young diagram of the partition(4,2,2,1), together with the hook lengths of the corresponding boxes.

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Definition

1≤i≤r

∈λ

h.

7 5 2 1 4 2 3 1 1

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Definition

1≤i≤r

∈λ

h.

7 5 2 1 4 2 3 1 1

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Definition

1≤i≤r

∈λ

h.

7 5 2 1 4 2 3 1 1

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Definition

1≤i≤r

∈λ

h.

7 5 2 1 4 2 3 1 1

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standard Young tableau (SYT): Obtained by filling in the boxes of the Young diagram with distinct entries 1 tonsuch that the entries in each row and each column are increasing.

fλ: the number of SYTs of shapeλ.

f_λ/µ: the number of SYTs of skew shapeλ/µ. 1 4 5 9 2 6 3 7 8

Figure:An SYT of shape(4,2,2,1).

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f_λ/µ: the number of SYTs of skew shapeλ/µ.

1 4 5 9 2 6 3 7 8

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f_λ/µ: the number of SYTs of skew shapeλ/µ.

1 4 5 9 2 6 3 7 8

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Theorem (Frame, Robinson and Thrall) fλ= n!

Hλ

wheren=|λ|.

RSK algorithm or representation of finite groups⇒ X

|λ|=n

f_λ²=n! and therefore

1 n!

X

|λ|=n

fλ²=1.

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Theorem (Frame, Robinson and Thrall) fλ= n!

Hλ

wheren=|λ|.

RSK algorithm or representation of finite groups⇒ X

|λ|=n

f_λ²=n!

and therefore

1 n!

X

|λ|=n

fλ²=1.

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Theorem (Nekrasov and Okounkov 2003, Han 2008)

X

n≥0



 X

|λ|=n

fλ²

Y

∈λ

(t+h²)



 xⁿ n!² =Y

i≥1

(1−xⁱ)^−1−t.

First proved by Nekrasov and Okounkov. Rediscovered and generalized by Han with a more elementary proof.

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Theorem (Nekrasov and Okounkov 2003, Han 2008)

X

n≥0



 X

|λ|=n

fλ²

Y

∈λ

(t+h²)



 xⁿ n!² =Y

i≥1

(1−xⁱ)^−1−t.

First proved by Nekrasov and Okounkov. Rediscovered and generalized by Han with a more elementary proof.

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1 n!

X

|λ|=n

f_λ²g(λ)=??.

Han

1 n!

P

|λ|=n

f_λ² P

∈λ

h²=³ⁿ²₂⁻ⁿ.

1 n!

P

|λ|=n

f_λ² P

∈λ

h⁴=⁴⁰ⁿ³⁻⁷⁵ⁿ₆ ²⁺⁴¹ⁿ.

1 n!

P

|λ|=n

f_λ² P

∈λ

h⁶=¹⁰⁵⁰ⁿ⁴⁻⁴⁰⁶⁰ⁿ³₂₄⁺⁵⁵⁸⁶ⁿ²⁻²⁵⁵²ⁿ.

Conjecture (Han 2008)

P(n) = 1 n!

X

|λ|=n

f_λ²X

∈λ

h^2k is always a polynomial ofnfor everyk∈N. Proved and generalized by Stanley.

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1 n!

X

|λ|=n

f_λ²g(λ)=??.

Han

1 n!

P

|λ|=n

f_λ² P

∈λ

h²=³ⁿ²₂⁻ⁿ.

1 n!

P

|λ|=n

f_λ² P

∈λ

h⁴=⁴⁰ⁿ³⁻⁷⁵ⁿ₆ ²⁺⁴¹ⁿ.

1 n!

P

|λ|=n

f_λ² P

∈λ

h⁶=¹⁰⁵⁰ⁿ⁴⁻⁴⁰⁶⁰ⁿ³₂₄⁺⁵⁵⁸⁶ⁿ²⁻²⁵⁵²ⁿ.

P(n) = 1 n!

X

|λ|=n

f_λ²X

∈λ

h^2k is always a polynomial ofnfor everyk∈N.

Proved and generalized by Stanley.

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1 n!

X

|λ|=n

f_λ²g(λ)=??.

Han

1 n!

P

|λ|=n

f_λ² P

∈λ

h²=³ⁿ²₂⁻ⁿ.

1 n!

P

|λ|=n

f_λ² P

∈λ

h⁴=⁴⁰ⁿ³⁻⁷⁵ⁿ₆ ²⁺⁴¹ⁿ.

1 n!

P

|λ|=n

f_λ² P

∈λ

h⁶=¹⁰⁵⁰ⁿ⁴⁻⁴⁰⁶⁰ⁿ³₂₄⁺⁵⁵⁸⁶ⁿ²⁻²⁵⁵²ⁿ.

P(n) = 1 n!

X

|λ|=n

f_λ²X

∈λ

h^2k is always a polynomial ofnfor everyk∈N.

Proved and generalized by Stanley.

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Theorem (Stanley 2010)

LetF be a symmetric function. Then P(n) = 1

n!

X

|λ|=n

f_λ²F(h²:∈λ) is a polynomial ofn.

Remark.Han-Stanley Theorem is a corollary of our main result.

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Definition

Letλbe a partition andgbe a function defined on partitions. Difference operatorsDandD⁻are defined by

Dg(λ) =X

λ⁺

g(λ⁺)−g(λ) and

D⁻g(λ) =|λ|g(λ)−X

λ⁻

g(λ⁻),

whereλ⁺ranges over all partitions obtained by adding a box toλandλ⁻ ranges over all partitions obtained by removing a box fromλ.

LetD⁰g=gandD^k+1g=D(D^kg)fork≥0.

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Definition

Letλbe a partition andgbe a function defined on partitions. Difference operatorsDandD⁻are defined by

Dg(λ) =X

λ⁺

g(λ⁺)−g(λ) and

D⁻g(λ) =|λ|g(λ)−X

λ⁻

g(λ⁻),

whereλ⁺ranges over all partitions obtained by adding a box toλandλ⁻ ranges over all partitions obtained by removing a box fromλ.

LetD⁰g=gandD^k+1g=D(D^kg)fork≥0.

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Main result (Han and Xiong 2015)

Suppose thatFis a symmetric function. Then there exists somer∈Nsuch thatD^r(^F(h

2 :∈λ)

H_λ ) =0 for every partitionλ.

Theorem (Han and Xiong 2015)

Suppose thatgis a function defined on partitions andµis a given partition. Then we have

X

|λ/µ|=n

fλ/µg(λ) =

n

X

k=0

n k

! D^kg(µ) and

Dⁿg(µ) =

n

X

k=0

(−1)^n+k n k

! X

|λ/µ|=k

fλ/µg(λ).

Our results⇒Han-Stanley theorem, (skew) marked hook formula, Okada-Panova hook length formula, and Fujii-Kanno-Moriyama-Okada content formula...

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2 :∈λ)

Suppose thatgis a function defined on partitions andµis a given partition.

Then we have

X

|λ/µ|=n

fλ/µg(λ) =

n

X

k=0

n k

! D^kg(µ) and

Dⁿg(µ) =

n

X

k=0

(−1)^n+k n k

! X

|λ/µ|=k

fλ/µg(λ).

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2 :∈λ)

Suppose thatgis a function defined on partitions andµis a given partition.

Then we have

X

|λ/µ|=n

fλ/µg(λ) =

n

X

k=0

n k

! D^kg(µ) and

Dⁿg(µ) =

n

X

k=0

(−1)^n+k n k

! X

|λ/µ|=k

fλ/µg(λ).

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When Dg = 0 or D

⁻

g = 0?

Theorem

For any partitionλ, we have

D( 1 Hλ

) =0 and

D⁻( 1 Hλ

) =0.

(|λ|+1)fλ=X

λ⁺

f_λ+⇒X

λ⁺

1 H_λ+ − 1

Hλ

=0⇒D( 1 Hλ

) =0.

fλ=X

λ⁻

f_λ−⇒ |λ| Hλ

−X

λ⁻

1 H_λ−

=0⇒D⁻( 1 Hλ

) =0.

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When Dg = 0 or D

⁻

g = 0?

Theorem

D( 1 Hλ

) =0 and

D⁻( 1 Hλ

) =0.

(|λ|+1)fλ=X

λ⁺

f_λ+⇒X

λ⁺

1 H_λ+ − 1

Hλ

=0⇒D( 1 Hλ

) =0.

fλ=X

λ⁻

f_λ−⇒ |λ| Hλ

−X

λ⁻

1 H_λ−

=0⇒D⁻( 1 Hλ

) =0.

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When Dg = 0 or D

⁻

g = 0?

Theorem

D( 1 Hλ

) =0 and

D⁻( 1 Hλ

) =0.

(|λ|+1)fλ=X

λ⁺

f_λ+⇒X

λ⁺

1 H_λ+ − 1

Hλ

=0⇒D( 1 Hλ

) =0.

fλ=X

λ⁻

f_λ−⇒ |λ| Hλ

−X

λ⁻

1 H_λ−

=0⇒D⁻( 1 Hλ

) =0.

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When Dg = 0 or D

⁻

g = 0?

Theorem

D⁻g(λ) =0for everyλ⇒g(λ) = _H^a

λ for some constant a.

Remark.WhenDg(λ) =0 for everyλ, it is not easy to determineg(λ).For example, actually we can show

D( P

∈λ

(h² −1)−3 ^|λ|₂ Hλ

) =0.

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When Dg = 0 or D

⁻

g = 0?

Theorem

D⁻g(λ) =0for everyλ⇒g(λ) = _H^a

λ for some constant a.

Remark.WhenDg(λ) =0 for everyλ, it is not easy to determineg(λ).For example, actually we can show

D(

P

∈λ

(h² −1)−3 ^|λ|₂ Hλ

) =0.

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Some properties of D and D

⁻

Theorem

Letλbe a partition. Suppose that g1,g2are functions defined on partitions and a1,a2∈R. Then we have

D(a1g1+a2g2)(λ) =a1Dg1(λ) +a2Dg2(λ) and

D⁻(a1g1+a2g2)(λ) =a1D⁻g1(λ) +a2D⁻g2(λ).

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Some properties of D and D

⁻

Theorem

For any function g defined on partitions, we have D(g(λ)

Hλ

) =X

λ⁺

g(λ⁺)−g(λ) H_λ+

and

D⁻(g(λ) Hλ

) =X

λ⁻

g(λ)−g(λ⁻) H_λ−

.

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Some properties of D and D

⁻

For product of two functions:

Theorem

D(g1(λ)g2(λ) Hλ

) = g1(λ)D(g2(λ) Hλ

) +g2(λ)D(g1(λ) Hλ

)

+X

λ⁺

(g1(λ⁺)−g1(λ))(g2(λ⁺)−g2(λ)) H_λ+

and

D⁻(g1(λ)g2(λ) Hλ

) = g1(λ)D⁻(g2(λ) Hλ

) +g2(λ)D⁻(g1(λ) Hλ

)

−X

λ⁻

(g1(λ)−g1(λ⁻))(g2(λ)−g2(λ⁻)) H_λ−

.

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Some properties of D and D

⁻

For product of several functions:

Theorem

Suppose that g1,g2,· · ·,grare functions defined on partitions. Let

[r] ={1,2,· · ·,r}and∆j(λ, µ) =gj(µ)−gj(λ)for1≤j≤r . Then we have D(

Q

1≤j≤rgj(λ) Hλ

) =X

λ⁺

X

A∪B=[r]

A∩B=∅

A6=∅

Q

k∈A∆k(λ, λ⁺)Q

l∈Bgl(λ) Hλ⁺

and D⁻(

Q

1≤j≤rgj(λ) Hλ

) =−X

λ⁻

X

A∪B=[r]

A∩B=∅

A6=∅

Q

k∈A∆k(λ, λ⁻)Q

l∈Bgl(λ) H_λ−

.

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Corners of partitions

For a partitionλ, theouter cornersare the boxes which can be removed to get a new partitionλ⁻.Let(α1, β1), . . . ,(αm, βm)be the coordinates of outer corners such thatα1> α2>· · ·αm.Letyj =βj−αjbe the contents of outer corners for 1≤j≤m.We setαm+1=β0=0 and call

(α1, β0),(α2, β1). . . ,(αm+1, βm)theinner cornersofλ. Letxi=βi−αi+1be the contents of inner corners for 0≤i≤m.

Theorem P

0≤i≤m

xi = P

1≤j≤m

yj. P

0≤i≤m

x_i²− P

1≤j≤m

y_j²=2|λ|.

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Theorem P

0≤i≤m

xi = P

1≤j≤m

yj.

P

0≤i≤m

x_i²− P

1≤j≤m

y_j²=2|λ|.

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Theorem P

0≤i≤m

xi = P

1≤j≤m

yj. P

0≤i≤m

x_i²− P

1≤j≤m

y_j²=2|λ|.

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An example

Figure:The Young diagrams of the partition(4,2,2,1).

Outer corners:(4,1), (3,2), (1,4). {yj}={−3,−1,3}.

inner corners:(4,0),(3,1), (1,2), (0,4). {xi}={−4,−2,1,4}.

P

0≤i≤m

xi =−1= P

1≤j≤m

yj. P

0≤i≤m

x_i²− P

1≤j≤m

y_j²=18=2·9.

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An example

Outer corners:(4,1), (3,2), (1,4).

{yj}={−3,−1,3}.

inner corners:(4,0),(3,1), (1,2), (0,4). {xi}={−4,−2,1,4}.

P

0≤i≤m

xi =−1= P

1≤j≤m

yj. P

0≤i≤m

x_i²− P

1≤j≤m

y_j²=18=2·9.

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An example

Outer corners:(4,1), (3,2), (1,4).

{yj}={−3,−1,3}.

inner corners:(4,0),(3,1), (1,2), (0,4).

{xi}={−4,−2,1,4}.

P

0≤i≤m

xi =−1= P

1≤j≤m

yj. P

0≤i≤m

x_i²− P

1≤j≤m

y_j²=18=2·9.

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An example

Outer corners:(4,1), (3,2), (1,4).

{yj}={−3,−1,3}.

inner corners:(4,0),(3,1), (1,2), (0,4).

{xi}={−4,−2,1,4}.

P

0≤i≤m

xi =−1= P

1≤j≤m

yj.

P

0≤i≤m

x_i²− P

1≤j≤m

y_j²=18=2·9.

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An example

Outer corners:(4,1), (3,2), (1,4).

{yj}={−3,−1,3}.

inner corners:(4,0),(3,1), (1,2), (0,4).

{xi}={−4,−2,1,4}.

P

0≤i≤m

xi =−1= P

1≤j≤m

yj. P

0≤i≤m

x_i²− P

1≤j≤m

y_j²=18=2·9.

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Onqγ(λ) Define

qk(λ) = X

0≤i≤m

xik

− X

1≤j≤m

yjk

and

qγ(λ) = Y

1≤l≤t

qγ_l(λ) for the partitionγ= (γ1, γ2, . . . , γt).

q0(λ) =1,q1(λ) =0,q2(λ) =2|λ|.

The idea to studyxi,yjandqγ(λ)comes from Kerov, Okounkov and Olshanski.

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Onqγ(λ) Define

qk(λ) = X

0≤i≤m

xik

− X

1≤j≤m

yjk

and

qγ(λ) = Y

1≤l≤t

q0(λ) =1,q1(λ) =0,q2(λ) =2|λ|.

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Onqγ(λ) Define

qk(λ) = X

0≤i≤m

xik

− X

1≤j≤m

yjk

and

qγ(λ) = Y

1≤l≤t

q0(λ) =1,q1(λ) =0,q2(λ) =2|λ|.

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Outline of proof of D

^r

(

^F(h

2 :∈λ)

Hλ

) = 0.

LetS(∅,r) =0 andS(λ,r) = P

∈λ r

Q

i=1

(h²−i²)for|λ|≥1.

Step1:We want to show that there exist somebγ∈Qsuch that S(λ,r) = X

|γ|≤2r+2

bγqγ(λ) for every partitionλ.

Suppose thatfis a function defined on integers. Let F1(n) =

n

X

k=1

f(k) and

F2(n) =

n

X

k=1

F1(k).

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Outline of proof of D

^r

(

^F(h

2 :∈λ)

Hλ

) = 0.

∈λ r

Q

i=1

(h²−i²)for|λ|≥1.

|γ|≤2r+2

n

X

k=1

f(k) and

F2(n) =

n

X

k=1

F1(k).

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Outline of proof of D

^r

(

^F(h

2 :∈λ)

Hλ

) = 0.

∈λ r

Q

i=1

(h²−i²)for|λ|≥1.

|γ|≤2r+2

n

X

k=1

f(k) and

F2(n) =

n

X

k=1

F1(k).

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Outline of proof of D

^r

(

^F(h

2 :∈λ)

Hλ

) = 0.

For 0≤j<i≤m, let

Bij ={(a,b)∈λ:αi+1+1≤a≤αi, βj+1≤b≤βj+1}.

Then

λ= [

0≤j<i≤m

Bij

and thus

X

∈λ

f(h) = X

0≤j<i≤m

X

∈B_ij

f(h). The multiset of hook lengths ofBij are

x_i−x_j−1

[

a=x_i−y_j+1

{a,a−1,a−2, . . . ,a−(xi−yi−1)}.

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Outline of proof of D

^r

(

^F(h

2 :∈λ)

Hλ

) = 0.

Then

λ= [

0≤j<i≤m

Bij

and thus

X

∈λ

f(h) = X

0≤j<i≤m

X

∈B_ij

f(h).

The multiset of hook lengths ofBij are

x_i−x_j−1

[

a=x_i−y_j+1

{a,a−1,a−2, . . . ,a−(xi−yi−1)}.

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Outline of proof of D

^r

(

^F(h

2 :∈λ)

Hλ

) = 0.

Then

λ= [

0≤j<i≤m

Bij

and thus

X

∈λ

f(h) = X

0≤j<i≤m

X

∈B_ij

f(h).

The multiset of hook lengths ofBijare

x_i−x_j−1

[

a=x_i−y_j+1

{a,a−1,a−2, . . . ,a−(xi−yi−1)}.

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Outline of proof of D

^r

(

^F(h

2 :∈λ)

Hλ

) = 0.

This means that

X

∈B_ij

f(h) =

x_i−x_j−1

X

a=x_i−y_j+1 x_i−y_i−1

X

b=0

f(a−b)

=

x_i−x_j−1

X

a=x_i−y_j+1

(F1(a)−F1(a−xi+yi))

= F2(xi−xj−1) +F2(yi−yj+1−1)

−F2(xi−yj+1−1)−F2(yi−xj−1).

ReplaceP

∈λf(h)byS(λ,r).There exist somebk ∈Qsuch that for every partitionλ,we have

S(λ,r) = X

1≤k≤r+1

bk( X

0≤i≤j≤m

(xi−xj)^2k+ X

1≤i≤j≤m

(yi−yj)^2k

− X

0≤i≤m

X

1≤j≤m

(xi−yj)^2k).

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Outline of proof of D

^r

(

^F(h

2 :∈λ)

Hλ

) = 0.

This means that

X

∈B_ij

f(h) =

x_i−x_j−1

X

a=x_i−y_j+1 x_i−y_i−1

X

b=0

f(a−b)

=

x_i−x_j−1

X

a=x_i−y_j+1

(F1(a)−F1(a−xi+yi))

= F2(xi−xj−1) +F2(yi−yj+1−1)

−F2(xi−yj+1−1)−F2(yi−xj−1).

ReplaceP

∈λf(h)byS(λ,r).There exist somebk ∈Qsuch that for every partitionλ,we have

S(λ,r) = X

1≤k≤r+1

bk( X

0≤i≤j≤m

(xi−xj)^2k+ X

1≤i≤j≤m

(yi−yj)^2k

− X

0≤i≤m

X

1≤j≤m

(xi−yj)^2k).

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Outline of proof of D

^r

(

^F(h

2 :∈λ)

Hλ

) = 0.

Compare the coefficients ofz^2k(1≤k≤r+1)on both sides of

( X

0≤i≤m

exp(xiz)− X

1≤j≤m

exp(yjz))( X

0≤i≤m

exp(−xiz)− X

1≤j≤m

exp(−yjz))

= X

0≤i≤m

X

0≤j≤m

exp((xi−xj)z) + X

1≤i≤m

X

1≤j≤m

exp((yi−yj)z)

− X

0≤i≤m

X

1≤j≤m

exp((xi−yj)z)− X

0≤i≤m

X

1≤j≤m

exp((yj−xi)z).

There exist somebγ∈Qsuch that S(λ,r) = X

|γ|≤2r+2

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Outline of proof of D

^r

(

^F(h

2 :∈λ)

Hλ

) = 0.

Step2:Suppose thatδis a partition. Then there exist somebγ ∈Qsuch that D(qδ(λ)

Hλ

) = X

|γ|≤|δ|−2

bγ

qγ(λ) Hλ

for every partitionλ.

For convenience, we just show the caseδ= (k)here. Letλ^k+=λ∪(αk+1+1, βk+1). First we have

Q

∈λ^k+g(h) Q

∈λg(h) =g(1) Y

0≤i≤k−1

g(xk −xi) g(xk−yi+1)

Y

k+1≤i≤m

g(xi−xk) g(yi−xk). In particular, we have

H_λk+

Hλ

= Q

0≤i≤m i6=k

(xk−xi) Q

1≤j≤m

(xk−yj).

(54)

Outline of proof of D

^r

(

^F(h

2 :∈λ)

Hλ

) = 0.

Hλ

) = X

|γ|≤|δ|−2

bγ

qγ(λ) Hλ

For convenience, we just show the caseδ= (k)here.

Letλ^k+=λ∪(αk+1+1, βk+1). First we have Q

∈λ^k+g(h) Q

∈λg(h) =g(1) Y

0≤i≤k−1

Y

k+1≤i≤m

H_λk+

Hλ

= Q

0≤i≤m i6=k

(xk−xi) Q

1≤j≤m

(xk−yj).

(55)

Outline of proof of D

^r

(

^F(h

2 :∈λ)

Hλ

) = 0.

Hλ

) = X

|γ|≤|δ|−2

bγ

qγ(λ) Hλ

For convenience, we just show the caseδ= (k)here.

Letλ^k+=λ∪(αk+1+1, βk +1). First we have Q

∈λ^k+g(h) Q

∈λg(h) =g(1) Y

0≤i≤k−1

Y

k+1≤i≤m

H_λk+

Hλ

= Q

0≤i≤m i6=k

(xk−xi) Q

1≤j≤m

(xk−yj).

(56)

Outline of proof of D

^r

(

^F(h

2 :∈λ)

Hλ

) = 0.

Let

g1(λ) = X

0≤i≤m

g(xi)− X

1≤j≤m

g(yj).

Then

D(g1(λ) Hλ

) = X

0≤i≤m

g(xi+1) +g(xi−1)−2g(xi) H_λi+

.

In particular, letg(z) =z^k andg1(λ) =qk(λ),then we obtain D(qk(λ)

Hλ

) = X

0≤i≤m

2P

1≤l≤^k 2

k 2l

xik−2l

H_λi+

.

(57)

Outline of proof of D

^r

(

^F(h

2 :∈λ)

Hλ

) = 0.

Let

g1(λ) = X

0≤i≤m

g(xi)− X

1≤j≤m

g(yj).

Then

D(g1(λ) Hλ

) = X

0≤i≤m

g(xi+1) +g(xi−1)−2g(xi) H_λi+

.

In particular, letg(z) =z^k andg1(λ) =qk(λ),then we obtain D(qk(λ)

Hλ

) = X

0≤i≤m

2P

1≤l≤^k 2

k 2l

xik−2l

H_λi+

.

(58)

Outline of proof of D

^r

(

^F(h

2 :∈λ)

Hλ

) = 0.

Let

g(z) = Y

1≤j≤m

(1−yjz)− X

0≤i≤m

Hλ

H_λi+

Y

0≤l≤m l6=i

(1−xlz) be a polynomial ofzwith degreem.

Then we obtain g(1

xt

) = Y

1≤j≤m

(1−yj

xt

)− Hλ

H_λt+

Y

0≤l≤m l6=t

(1−xl

xt

)

= Y

1≤j≤m

(1−yj

xt

)− Q

1≤j≤m

(xt−yj) Q

0≤l≤m l6=t

(xt−xl)· Y

0≤l≤m l6=t

(1−xl

xt

)

= 0.

This means thatg(z)has at leastm+1 roots and thereforeg(z) =0. Now we have

X

0≤i≤m

Hλ

H_λi+

· 1 1−xiz =

Q

1≤j≤m(1−yjz) Q

0≤i≤m(1−xiz).

(59)

Outline of proof of D

^r

(

^F(h

2 :∈λ)

Hλ

) = 0.

Let

g(z) = Y

1≤j≤m

(1−yjz)− X

0≤i≤m

Hλ

H_λi+

Y

0≤l≤m l6=i

(1−xlz) be a polynomial ofzwith degreem. Then we obtain

g(1 xt

) = Y

1≤j≤m

(1−yj

xt

)− Hλ

H_λt+

Y

0≤l≤m l6=t

(1−xl

xt

)

= Y

1≤j≤m

(1−yj

xt

)− Q

1≤j≤m

(xt−yj) Q

0≤l≤m l6=t

(xt−xl)· Y

0≤l≤m l6=t

(1−xl

xt

)

= 0.

This means thatg(z)has at leastm+1 roots and thereforeg(z) =0. Now we have

X

0≤i≤m

Hλ

H_λi+

· 1 1−xiz =

Q

0≤i≤m(1−xiz).

(60)

Outline of proof of D

^r

(

^F(h

2 :∈λ)

Hλ

) = 0.

Let

g(z) = Y

1≤j≤m

(1−yjz)− X

0≤i≤m

Hλ

H_λi+

Y

0≤l≤m l6=i

g(1 xt

) = Y

1≤j≤m

(1−yj

xt

)− Hλ

H_λt+

Y

0≤l≤m l6=t

(1−xl

xt

)

= Y

1≤j≤m

(1−yj

xt

)− Q

1≤j≤m

(xt−yj) Q

0≤l≤m l6=t

(xt−xl)· Y

0≤l≤m l6=t

(1−xl

xt

)

= 0.

This means thatg(z)has at leastm+1 roots and thereforeg(z) =0.

Now we have

X

0≤i≤m

Hλ

H_λi+

· 1 1−xiz =

Q

0≤i≤m(1−xiz).

(61)

Outline of proof of D

^r

(

^F(h

2 :∈λ)

Hλ

) = 0.

Let

g(z) = Y

1≤j≤m

(1−yjz)− X

0≤i≤m

Hλ

H_λi+

Y

0≤l≤m l6=i

g(1 xt

) = Y

1≤j≤m

(1−yj

xt

)− Hλ

H_λt+

Y

0≤l≤m l6=t

(1−xl

xt

)

= Y

1≤j≤m

(1−yj

xt

)− Q

1≤j≤m

(xt−yj) Q

0≤l≤m l6=t

(xt−xl)· Y

0≤l≤m l6=t

(1−xl

xt

)

= 0.

This means thatg(z)has at leastm+1 roots and thereforeg(z) =0.

Now we have

X

0≤i≤m

Hλ

H_λi+

· 1 1−xiz =

Q

0≤i≤m(1−xiz).

(62)

Outline of proof of D

^r

(

^F(h

2 :∈λ)

Hλ

) = 0.

This means that X

0≤i≤m

Hλ

H_λi+

(X

k≥0

(xiz)^k) = exp( X

1≤j≤m

ln(1−yjz)− X

0≤i≤m

ln(1−xiz)

= exp(X

k≥1

qk(λ) k z^k).

By comparing the coefficients ofz^k on both sides, we obtain there exist some bγ∈Qsuch that

X

0≤i≤m

Hλ

H_λi+

xik

= X

|γ|≤k

(63)

Outline of proof of D

^r

(

^F(h

2 :∈λ)

Hλ

) = 0.

This means that X

0≤i≤m

Hλ

H_λi+

(X

k≥0

(xiz)^k) = exp( X

1≤j≤m

ln(1−yjz)− X

0≤i≤m

ln(1−xiz)

= exp(X

k≥1

qk(λ) k z^k).

By comparing the coefficients ofz^k on both sides, we obtain there exist some bγ∈Qsuch that

X

0≤i≤m

Hλ

H_λi+

xik

= X

|γ|≤k

(64)

Outline of proof of D

^r

(

^F(h

2 :∈λ)

Hλ

) = 0.

But we have

D(qk(λ) Hλ

) = X

0≤i≤m

2P

1≤l≤^k₂ k 2l

xik−2l

H_λi+

.

Thus there exist somebγ∈Qsuch that D(qk(λ)

Hλ

) = X

|γ|≤k−2

bγ

qγ(λ) Hλ

Forqδ(λ), the proof is similar.