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Valentin Féray

joint work (in progress) with Per Alexandersson (Zürich)

Institut für Mathematik, Universität Zürich

Séminaire Lotharingien de Combinatoire Strobl, Austria, September 9th, 2014

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1 Symmetric functions and Jack polynomials

2 Knop Sahi combinatorial formula

3 Lassalle’s dual approach

4 Unifying both ? Two new conjectures. . .

5 Partial results

V. Féray (with P.A.) Multirectangular shifted Jack(I-Math, UZH) SLC, 2014–09 2 / 16

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Symmetric functions

=“polynomials” in infinitely many variables x1,x2,x3, . . . that are invariant by permuting indices

Augmented monomial basis:

λ = X

i1,...,i`≥1 distinct

xiλ1

1 · · ·xiλ`

`

Example: m˜(2,1,1)=2x12x2x3+2x1x22x3+2x1x2x32+2x12x2x4+. . . Power-sumbasis:

pr =x1r +x2r +. . . , pλ =pλ1· · ·pλ`

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Symmetric functions and Jack polynomials

Schur functions

(sλ) is another basis of the symmetric function ring.

Several equivalent definitions:

sλ =P

TxT, sum over semi standard Young tableaux; orthogonalbasis (for Hall scalar product) +triangularover (augmented) monomial basis ;

with determinants. . .

-> Encode irreduciblecharacters of symmetric and general linear groups.

V. Féray (with P.A.) Multirectangular shifted Jack(I-Math, UZH) SLC, 2014–09 4 / 16

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Jack polynomials

Deformation of Schur functions with a positive realparameterα.

(Jλ(α)) basis, Jλ(1) =cstλ·sλ Several equivalent definitions:

Jλ=P

TψT(α)xT, sum oversemi standard Young tableaux ; orthogonalbasis (for adeformationof Hall scalar product) + triangularover (augmented) monomial basis.

For α=1/2,2, they also have arepresentation-theoreticalinterpretation (in terms of Gelfand pairs) but not in general !

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Knop Sahi combinatorial formula

Polynomiality in α with non-negative coefficients

Both definitions involve rational functionsin α. Nevertheless, . . . Macdonald-Stanley conjecture (∼90)

The coefficients of Jack polynomials in augmented monomial basis are polynomials in α with non-negative integer coefficients.

Notation: [ ˜mτ]Jλ.

KS give a combinatorial interpretation of [ ˜mτ]Jλ as a weighted enumeration of admissibletableaux.

V. Féray (with P.A.) Multirectangular shifted Jack(I-Math, UZH) SLC, 2014–09 6 / 16

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Polynomiality in α with non-negative coefficients

Both definitions involve rational functionsin α. Nevertheless, . . . Knop-Sahi theorem (97)

The coefficients of Jack polynomials in augmented monomial basis are polynomials in α with non-negative integer coefficients.

Notation: [ ˜mτ]Jλ.

KS give a combinatorial interpretation of [ ˜mτ]Jλ as a weighted enumeration of admissibletableaux.

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Lassalle’s dual approach

A function on the set of all Young diagrams

Definition

Let µbe a partition of k (without part equal to 1). Define Ch(α)µ (λ) =

( n−k+m

1(µ) m1(µ)

·zµ·[pµ1n−k]Jλ(α) ifn=|λ| ≥k;

0 otherwise.

Ch(α)µ is a function of all Young diagrams.

zµ: standard explicit numerical factor.

Specialization: if |µ|<|λ|,

Ch(1)µ (λ) = |λ|!

(|λ| − |µ|)!· χλµ1n−k

dim(Vλ).

Introduced by S. Kerov, G. Olshanski in the caseα=1 (to study random diagrams with Plancherel measure), by M. Lassalle in the general case.

V. Féray (with P.A.) Multirectangular shifted Jack(I-Math, UZH) SLC, 2014–09 7 / 16

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A function on the set of all Young diagrams

Definition

Let µbe a partition of k (without part equal to 1). Define Ch(α)µ (λ) =

( n−k+m

1(µ) m1(µ)

·zµ·[pµ1n−k]Jλ(α) ifn=|λ| ≥k;

0 otherwise.

Ch(α)µ is a function of all Young diagrams.

Specialization: if |µ|<|λ|,

Ch(1)µ (λ) = |λ|!

(|λ| − |µ|)!· χλµ1n−k

dim(Vλ).

Introduced by S. Kerov, G. Olshanski in the caseα=1 (to study random diagrams with Plancherel measure), by M. Lassalle in the general case.

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Lassalle’s dual approach

A function on the set of all Young diagrams

Definition

Let µbe a partition of k (without part equal to 1). Define Ch(α)µ (λ) =

( n−k

+m1(µ) m1(µ)

·zµ·[pµ1n−k]Jλ(α) ifn=|λ| ≥k;

0 otherwise.

Proposition (Kerov/Olshanski forα =1, Lassalle in general) For any r, the application

1, . . . , λr)7→Ch(α)µ1, . . . , λr)

is a polynomial in λ1, . . . ,λr. Besides, it is symmetric in λ1−1/α, . . . , λr −r/α.

In other words, Ch(α)µ is a shifted symmetric function.

V. Féray (with P.A.) Multirectangular shifted Jack(I-Math, UZH) SLC, 2014–09 7 / 16

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Multirectangular coordinates (R. Stanley)

Consider two lists pandq of positive integers of the same size, withq non-decreasing.

We associate to them the partition λ(p,q) = q1, . . . ,q1

| {z }

p1 times

,q2, . . . ,q2

| {z }

p2times

, . . . .

Young diagram ofλ(p,q)

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Lassalle’s dual approach

Multirectangular coordinates (R. Stanley)

Consider two lists pandq of positive integers of the same size, withq non-decreasing.

We associate to them the partition λ(p,q) = q1, . . . ,q1

| {z }

p1 times

,q2, . . . ,q2

| {z }

p2times

, . . . .

Proposition

Let µbe a partition of k. Ch(α)µ (λ(p,q))is a polynomial in p1,p2, . . . ,q1,q2, . . . , α

V. Féray (with P.A.) Multirectangular shifted Jack(I-Math, UZH) SLC, 2014–09 8 / 16

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Multirectangular coordinates (R. Stanley)

Consider two lists pandq of positive integers of the same size, withq non-decreasing.

We associate to them the partition λ(p,q) = q1, . . . ,q1

| {z }

p1 times

,q2, . . . ,q2

| {z }

p2times

, . . . .

Conjecture (M. Lassalle)

Let µbe a partition of k. (−1)kCh(α)µ (λ(p,q)) is a polynomial in p1,p2, . . . ,−q1,−q2, . . . , α−1

with non-negative integercoefficients.

Still open. . .

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Unifying both ? Two new conjectures. . .

Link between the two questions ?

Knop-Sahi theorem and Lassalle conjecture do not seem related.

Two (main) differences:

monomial coefficients vs power-sum coefficients ; look at some Jλ(α) vs seen as a function ofλ.

Idea: look at monomial coefficients as functions on Young diagrams.

V. Féray (with P.A.) Multirectangular shifted Jack(I-Math, UZH) SLC, 2014–09 9 / 16

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Link between the two questions ?

Knop-Sahi theorem and Lassalle conjecture do not seem related.

Two (main) differences:

monomial coefficients vs power-sum coefficients ; look at some Jλ(α) vs seen as a function ofλ.

Idea: look at monomial coefficients as functions on Young diagrams.

(16)

Unifying both ? Two new conjectures. . .

Monomial coefficients as shifted symmetric functions

Definition

Let µbe a partition of k (without part equal to 1). Define Ko(α)µ (λ) =

( n−k+m

1(µ) m1(µ)

·zµ·[m˜µ1n−k]Jλ(α) if n=|λ| ≥k;

0 otherwise.

Proposition

Ko(α)µ is a shifted symmetric function.

Proof: Uses Ko(α)µ = P

ν`k

Lµ,νCh(α)ν and Lassalle proposition.

(Lµ,ν is defined by pν = P

µ`k

Lµ,νµ).

V. Féray (with P.A.) Multirectangular shifted Jack(I-Math, UZH) SLC, 2014–09 10 / 16

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Unifying both ? Two new conjectures. . .

A new conjecture

Proposition

Ko(α)µ (p×q) is a polynomial inp,qandα.

In thefalling factorial basisin p andq, Ko(α)µ (p×q) has non-negative integer coefficients.

falling factorial: (n)k :=n(n−1). . .(n−k+1). falling factorial basis:

(p1)i1(p2)i2. . .(q1)j1(q2)j2. . . αk

.

It is stronger than positivity in Knop-Sahi theorem (and does not follow from their combinatorial interpretation) !

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Unifying both ? Two new conjectures. . .

A new conjecture

Proposition

Ko(α)µ (p×q) is a polynomial inp,qandα.

Conjecture (F., Alexandersson)

In thefalling factorial basisin p andq, Ko(α)µ (p×q) has non-negative integer coefficients.

falling factorial: (n)k :=n(n−1). . .(n−k+1).

falling factorial basis:

(p1)i1(p2)i2. . .(q1)j1(q2)j2. . . αk

.

It is stronger than positivity in Knop-Sahi theorem (and does not follow from their combinatorial interpretation) !

V. Féray (with P.A.) Multirectangular shifted Jack(I-Math, UZH) SLC, 2014–09 11 / 16

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A new conjecture

Proposition

Ko(α)µ (p×q) is a polynomial inp,qandα.

Conjecture (F., Alexandersson)

In thefalling factorial basisin p andq, Ko(α)µ (p×q) has non-negative integer coefficients.

falling factorial: (n)k :=n(n−1). . .(n−k+1).

falling factorial basis:

(p1)i1(p2)i2. . .(q1)j1(q2)j2. . . αk

.

It is stronger than positivity in Knop-Sahi theorem (and does not follow from their combinatorial interpretation) !

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Unifying both ? Two new conjectures. . .

Another conjecture

Another interesting family of shifted symmetric function Shifted Jack polynomials (Okounkov, Olshanski, 97)

J](α)µ is the unique shifted symmetric function whose highest degree component is the Jack polynomialJµ.

Conjecture (F., Alexandersson)

In thefalling factorial basisin p andq,α`(µ)J](α)µ (p×q) has non-negative integer coefficients.

For a fixed α, FF-positivity ofα`(µ)J](α)µ (p×q)implies FF-positivity of Ko(α)µ (p×q).

V. Féray (with P.A.) Multirectangular shifted Jack(I-Math, UZH) SLC, 2014–09 12 / 16

(21)

Another conjecture

Another interesting family of shifted symmetric function Shifted Jack polynomials (Okounkov, Olshanski, 97)

J](α)µ is the unique shifted symmetric function whose highest degree component is the Jack polynomialJµ.

Conjecture (F., Alexandersson)

In thefalling factorial basisin p andq,α`(µ)J](α)µ (p×q) has non-negative integer coefficients.

For a fixed α, FF-positivity ofα`(µ)J](α)µ (p×q)implies FF-positivity of Ko(α)µ (p×q).

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Partial results

Case α = 1 (1/2)

For α=1, there is a combinatorial formula for Ch(1)µ : Theorem (F. 2007; F., Śniady 2008 ; conj. by Stanley 2006) Let µa partition ofk. Fix a permutationπ inSk of type µ. Then

(−1)kChµ(p×q) = X

σ,τ∈Sk στ=π

Nσ,τ(p,−q).

Nσ,τ : combinatorial polynomial with non-negative integer coefficients.

⇒ Lassalle conjecture holds forα=1.

Similar formula for α=2: replace permutations by pairings of [2n] (F., Śniady, 2011).

V. Féray (with P.A.) Multirectangular shifted Jack(I-Math, UZH) SLC, 2014–09 13 / 16

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Case α = 1 (1/2)

For α=1, there is a combinatorial formula for Ch(1)µ : Theorem (F. 2007; F., Śniady 2008 ; conj. by Stanley 2006) Let µa partition ofk. Fix a permutationπ inSk of type µ. Then

(−1)kChµ(p×q) = X

σ,τ∈Sk στ=π

Nσ,τ(p,−q).

Proposition

Fixa set-partition Πwhose block size are given byµ.

(−1)kKo(1)µ (p×q) = X

σ,τ∈Sk στ∈SΠ

Nσ,τ(p,−q).

(−1)ksλ]µ(p×q) = X

σ,τ∈Sk

χµ(σ τ)Nσ,τ(p,−q)

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Partial results

Case α = 1 (2/2)

. . . use explicit expression ofNσ,τ(p,q) + sum manipulations . . . It is enough to prove

Question 1

For any three set partitions T,U andΠof the same set, X

σ∈ST,τ∈SU στ∈SΠ

ε(τ)≥0.

Question 2

For any two set partitions T,U of[n]and integer partitionµof n, X

σ∈ST,τ∈SU

ε(τ)χµ(σ τ)≥0.

Proof: representation theory + group algebra manipulation.

V. Féray (with P.A.) Multirectangular shifted Jack(I-Math, UZH) SLC, 2014–09 14 / 16

(25)

Case α = 1 (2/2)

. . . use explicit expression ofNσ,τ(p,q) + sum manipulations . . . It is enough to prove

Conjecture

For any three set partitions T,U andΠof the same set, X

σ∈ST,τ∈SU στ∈SΠ

ε(τ)≥0.

Proposition

For any two set partitions T,U of[n]and integer partitionµof n, X

σ∈ST,τ∈SU

ε(τ)χµ(σ τ)≥0.

Proof: representation theory + group algebra manipulation.

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Partial results

Case α = 1 (2/2)

. . . use explicit expression ofNσ,τ(p,q) + sum manipulations . . . It is enough to prove

Conjecture

For any three set partitions T,U andΠof the same set, X

σ∈ST,τ∈SU στ∈SΠ

ε(τ)≥0.

Proposition

For any two set partitions T,U of[n]and integer partitionµof n, X

σ∈ST,τ∈SU

ε(τ)χµ(σ τ)≥0.

Conclusion: Our second (and hence both) conjecture(s) hold(s) forα=1.

V. Féray (with P.A.) Multirectangular shifted Jack(I-Math, UZH) SLC, 2014–09 14 / 16

(27)

Partial results

Ko

(k)

is FF non-negative.

Observation: (−1)kKo(1)(k)(p×q) = P

σ,τ∈Sk no restriction

Nσ,τ(p,−q).

For a generalα,

(−1)kKo(α)(k)(p×q) = X

σ,τ∈Sk

αk−#(LR-max(σ))Nσ,τ(p,−q)

Proof: KS combinatorial interpretation + a new bijection. Corollary (special case of our first conjecture)

The coefficients of Ko(α)(k) in the falling factorial basis are non-negative.

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Partial results

Ko

(k)

is FF non-negative.

Observation: (−1)kKo(1)(k)(p×q) = P

σ,τ∈Sk no restriction

Nσ,τ(p,−q).

Proposition For a generalα,

(−1)kKo(α)(k)(p×q) = X

σ,τ∈Sk

αk−#(LR-max(σ))Nσ,τ(p,−q)

Proof: KS combinatorial interpretation + a new bijection.

Corollary (special case of our first conjecture)

The coefficients of Ko(α)(k) in the falling factorial basis are non-negative.

V. Féray (with P.A.) Multirectangular shifted Jack(I-Math, UZH) SLC, 2014–09 15 / 16

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Ko

(k)

is FF non-negative.

Observation: (−1)kKo(1)(k)(p×q) = P

σ,τ∈Sk no restriction

Nσ,τ(p,−q).

Proposition For a generalα,

(−1)kKo(α)(k)(p×q) = X

σ,τ∈Sk

αk−#(LR-max(σ))Nσ,τ(p,−q)

Proof: KS combinatorial interpretation + a new bijection.

Corollary (special case of our first conjecture)

The coefficients of Ko(α)(k) in the falling factorial basis are non-negative.

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Conclusion

Conclusion

A bridge between KS theorem and Lassalle’s conjecture:

Our conjecture involves shifted symmetric functions and multirectangular coordinates and implies KS theorem ; Our partial results use (partial) results to both questions.

Other partial results?

α=2 works similarly as α=1 with a bit more work ;

Case of rectangular Young diagram is perhaps tractable (Lassalle proved his conjecture in this case);

An extension?

What about (shifted) Macdonald polynomials and multirectangular coordinates?

V. Féray (with P.A.) Multirectangular shifted Jack(I-Math, UZH) SLC, 2014–09 16 / 16

(31)

Conclusion

Conclusion

A bridge between KS theorem and Lassalle’s conjecture:

Our conjecture involves shifted symmetric functions and multirectangular coordinates and implies KS theorem ; Our partial results use (partial) results to both questions.

Other partial results?

α=2 works similarly as α=1 with a bit more work ;

Case of rectangular Young diagram is perhaps tractable (Lassalle proved his conjecture in this case);

What about (shifted) Macdonald polynomials and multirectangular coordinates?

(32)

Conclusion

Conclusion

A bridge between KS theorem and Lassalle’s conjecture:

Our conjecture involves shifted symmetric functions and multirectangular coordinates and implies KS theorem ; Our partial results use (partial) results to both questions.

Other partial results?

α=2 works similarly as α=1 with a bit more work ;

Case of rectangular Young diagram is perhaps tractable (Lassalle proved his conjecture in this case);

An extension?

What about (shifted) Macdonald polynomials and multirectangular coordinates?

V. Féray (with P.A.) Multirectangular shifted Jack(I-Math, UZH) SLC, 2014–09 16 / 16

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