Young tableaux and snakes Yuliy Baryshnikov joint work with Dan - - PowerPoint PPT Presentation

Young tableaux and snakes

Yuliy Baryshnikov joint work with Dan Romik (Hebrew University)

Motivation

• A Young tableaux is a filling of Young diagram consisting of n boxes

with numbers 1, . . . , n increasing top-to-down and left-to-right.

1 2 4 5 6 7 8 n=9 3 9

Motivation

• A Young tableaux is a filling of Young diagram consisting of n boxes

with numbers 1, . . . , n increasing top-to-down and left-to-right.

1 2 4 5 6 7 8 n=9 3 9

• The number of YTs with given shape λ has various interpretations

(dimension of the representation λ of Sn, for example).

Motivation

• A Young tableaux is a filling of Young diagram consisting of n boxes

with numbers 1, . . . , n increasing top-to-down and left-to-right.

1 2 4 5 6 7 8 n=9 3 9

• The number of YTs with given shape λ has various interpretations

(dimension of the representation λ of Sn, for example).

• Asymptotic regime is of interest:

Motivation (cont’d)

• Consider a large shape tλ, t → ∞:

Motivation (cont’d)

• Consider a large shape tλ, t → ∞: and a typical Young tableaux

filling it:

• The natural question arises:

Motivation (cont’d)

• Consider a large shape tλ, t → ∞: and a typical Young tableaux

filling it:

• The natural question arises:

Conjecture A typical YT, considered as a function on the Young diageam tλ is close to some deterministic limiting function.

Motivation (cont’d)

• Consider a large shape tλ, t → ∞: and a typical Young tableaux

filling it:

• The natural question arises:

Conjecture A typical YT, considered as a function on the Young diageam tλ is close to some deterministic limiting function.

• How one would prove it?

Motivation (cont’d)

• By finding the rate function and then solving variational problem.

Motivation (cont’d)

• By finding the rate function and then solving variational problem.
• Rate function will count the (normalized, per unit area) number of

YT filling the shapes approximating a strip

Motivation (cont’d)

• Hence we have to compute the number of Young tableaux filling the

strips like

Motivation (cont’d)

• Hence we have to compute the number of Young tableaux filling the

strips like

Motivation (cont’d)

• Hence we have to compute the number of Young tableaux filling the

strips like

• We start with the simplest task: finding the number of YT filling the

strip of width 2 and slope 1.

Up-down permutations

• A permutation σ ∈ Sn is called an up-down permutation (also

zig-zag permutation, alternating permutation) if it satisfies σ(1) < σ(2) > σ(3) < σ(4) > . . .

Up-down permutations

• A permutation σ ∈ Sn is called an up-down permutation (also

zig-zag permutation, alternating permutation) if it satisfies σ(1) < σ(2) > σ(3) < σ(4) > . . . Equivalent to “2-strip” tableaux:

Up-down permutations

• A permutation σ ∈ Sn is called an up-down permutation (also

zig-zag permutation, alternating permutation) if it satisfies σ(1) < σ(2) > σ(3) < σ(4) > . . . Equivalent to “2-strip” tableaux:

6 10 7 2 12 9 14 1 5 4 11 3 13

Up-down permutations

• A permutation σ ∈ Sn is called an up-down permutation (also

zig-zag permutation, alternating permutation) if it satisfies σ(1) < σ(2) > σ(3) < σ(4) > . . . Equivalent to “2-strip” tableaux:

6 10 7 2 12 9 14 1 5 4 11 3 13

• Theorem (D. Andr´

e 1881): Let An = # of n-element up-down

• permutations. Then

∞

• n=0

Anxn n! = tan x + sec x.

Up-down permutations

• A permutation σ ∈ Sn is called an up-down permutation (also

zig-zag permutation, alternating permutation) if it satisfies σ(1) < σ(2) > σ(3) < σ(4) > . . . Equivalent to “2-strip” tableaux:

6 10 7 2 12 9 14 1 5 4 11 3 13

• Theorem (D. Andr´

e 1881): Let An = # of n-element up-down

• permutations. Then

∞

• n=0

Anxn n! = tan x + sec x.

• Up-down permutations were named snakes and studied by V. Arnold

to enumerate morsifications of real singularities.

Up-down permutations (continued)

• Reminder:

sechx =

∞

• n=0

Enxn n! , (En)n≥0 – Euler numbers, tan x =

∞

• n=1

Tnx2n−1 (2n − 1)!, (Tn)n≥0 – Tangent numbers, x ex − 1 =

∞

• n=0

Bnxn n! , (Bn)n≥0 – Bernoulli numbers.

Up-down permutations (continued)

• Reminder:

sechx =

∞

• n=0

Enxn n! , (En)n≥0 – Euler numbers, tan x =

∞

• n=1

Tnx2n−1 (2n − 1)!, (Tn)n≥0 – Tangent numbers, x ex − 1 =

∞

• n=0

Bnxn n! , (Bn)n≥0 – Bernoulli numbers.

• In this notation:

A2n = |E2n|, A2n−1 = Tn = (−1)n−14n(4n−1)

2n

B2n.

Transfer operators

Many proofs of Andr´ e’s theorem, mostly algebraic. Proof using transfer

• perators (due to N. Elkies, 2003):

Transfer operators

Many proofs of Andr´ e’s theorem, mostly algebraic. Proof using transfer

• perators (due to N. Elkies, 2003):
• Let Pn =
• (x1, x2, . . . , xn) ∈ [0, 1]n : x1 < x2 > x3 < x4 > . . .
• .

Transfer operators

Many proofs of Andr´ e’s theorem, mostly algebraic. Proof using transfer

• perators (due to N. Elkies, 2003):
• Let Pn =
• (x1, x2, . . . , xn) ∈ [0, 1]n : x1 < x2 > x3 < x4 > . . .
• .
• Compute vol(Pn) in two ways:

Transfer operators

Many proofs of Andr´ e’s theorem, mostly algebraic. Proof using transfer

• perators (due to N. Elkies, 2003):
• Let Pn =
• (x1, x2, . . . , xn) ∈ [0, 1]n : x1 < x2 > x3 < x4 > . . .
• .
• Compute vol(Pn) in two ways: First, vol(Pn) = An

n! ;

Transfer operators

Many proofs of Andr´ e’s theorem, mostly algebraic. Proof using transfer

• perators (due to N. Elkies, 2003):
• Let Pn =
• (x1, x2, . . . , xn) ∈ [0, 1]n : x1 < x2 > x3 < x4 > . . .
• .
• Compute vol(Pn) in two ways: First, vol(Pn) = An

n! ;

• Second,

vol(Pn) =

Transfer operators

Many proofs of Andr´ e’s theorem, mostly algebraic. Proof using transfer

• perators (due to N. Elkies, 2003):
• Let Pn =
• (x1, x2, . . . , xn) ∈ [0, 1]n : x1 < x2 > x3 < x4 > . . .
• .
• Compute vol(Pn) in two ways: First, vol(Pn) = An

n! ;

• Second,

x 1

vol(Pn) = 1 dx1

Transfer operators

Many proofs of Andr´ e’s theorem, mostly algebraic. Proof using transfer

• perators (due to N. Elkies, 2003):
• Let Pn =
• (x1, x2, . . . , xn) ∈ [0, 1]n : x1 < x2 > x3 < x4 > . . .
• .
• Compute vol(Pn) in two ways: First, vol(Pn) = An

n! ;

• Second,

x 1 x 2

vol(Pn) = 1 dx1 1

x1

dx2

Transfer operators

Many proofs of Andr´ e’s theorem, mostly algebraic. Proof using transfer

• perators (due to N. Elkies, 2003):
• Let Pn =
• (x1, x2, . . . , xn) ∈ [0, 1]n : x1 < x2 > x3 < x4 > . . .
• .
• Compute vol(Pn) in two ways: First, vol(Pn) = An

n! ;

• Second,

x 1 x 2 x 3

vol(Pn) = 1 dx1 1

x1

dx2 x2 dx3

Transfer operators

Many proofs of Andr´ e’s theorem, mostly algebraic. Proof using transfer

• perators (due to N. Elkies, 2003):
• Let Pn =
• (x1, x2, . . . , xn) ∈ [0, 1]n : x1 < x2 > x3 < x4 > . . .
• .
• Compute vol(Pn) in two ways: First, vol(Pn) = An

n! ;

• Second,

x 1 x 2 x 3 x 4

vol(Pn) = 1 dx1 1

x1

dx2 x2 dx3 1

x3

dx4

Transfer operators

Many proofs of Andr´ e’s theorem, mostly algebraic. Proof using transfer

• perators (due to N. Elkies, 2003):
• Let Pn =
• (x1, x2, . . . , xn) ∈ [0, 1]n : x1 < x2 > x3 < x4 > . . .
• .
• Compute vol(Pn) in two ways: First, vol(Pn) = An

n! ;

• Second,

x 1 x 2 x 3 x 4

vol(Pn) = 1 dx1 1

x1

dx2 x2 dx3 1

x3

dx4 . . . =

Transfer operators

Many proofs of Andr´ e’s theorem, mostly algebraic. Proof using transfer

• perators (due to N. Elkies, 2003):
• Let Pn =
• (x1, x2, . . . , xn) ∈ [0, 1]n : x1 < x2 > x3 < x4 > . . .
• .
• Compute vol(Pn) in two ways: First, vol(Pn) = An

n! ;

• Second,

x 1 x 2 x 3 x 4

vol(Pn) = 1 dx1 1

x1

dx2 x2 dx3 1

x3

dx4 . . . = . . . ◦ T ◦ S ◦ T ◦ S1, 1L2[0,1],

Transfer operators

Many proofs of Andr´ e’s theorem, mostly algebraic. Proof using transfer

• perators (due to N. Elkies, 2003):
• Let Pn =
• (x1, x2, . . . , xn) ∈ [0, 1]n : x1 < x2 > x3 < x4 > . . .
• .
• Compute vol(Pn) in two ways: First, vol(Pn) = An

n! ;

• Second,

x 1 x 2 x 3 x 4

vol(Pn) = 1 dx1 1

x1

dx2 x2 dx3 1

x3

dx4 . . . = . . . ◦ T ◦ S ◦ T ◦ S1, 1L2[0,1], where (Tf )(x) = x f (y)dy, (Sg)(x) = 1

x

g(y)dy

Transfer operators

Many proofs of Andr´ e’s theorem, mostly algebraic. Proof using transfer

• perators (due to N. Elkies, 2003):
• Let Pn =
• (x1, x2, . . . , xn) ∈ [0, 1]n : x1 < x2 > x3 < x4 > . . .
• .
• Compute vol(Pn) in two ways: First, vol(Pn) = An

n! ;

• Second,

x 1 x 2 x 3 x 4

vol(Pn) = 1 dx1 1

x1

dx2 x2 dx3 1

x3

dx4 . . . = . . . ◦ T ◦ S ◦ T ◦ S1, 1L2[0,1], where (Tf )(x) = x f (y)dy, (Sg)(x) = 1

x

g(y)dy (continuous analogue of using adjacency matrix to count paths in graphs).

Transfer operators (continued)

• Note that S = C ◦ T ◦ C where (Cg)(x) = g(1 − x), so we have

shown that An = n!Rn−11, 1L2[0,1], where R = C ◦ T, (Rf )(x) = 1−x f (y)dy.

Transfer operators (continued)

• Note that S = C ◦ T ◦ C where (Cg)(x) = g(1 − x), so we have

shown that An = n!Rn−11, 1L2[0,1], where R = C ◦ T, (Rf )(x) = 1−x f (y)dy.

• Therefore

An = n!

∞

• k=1

λn−1

k

1, φk2

L2[0,1],

where (φk)k≥1 is the orthonormal system of eigenfunctions of the self-adjoint operator R, with corresponding eigenvalues (λk)k≥1.

Transfer operators (continued)

• Note that S = C ◦ T ◦ C where (Cg)(x) = g(1 − x), so we have

shown that An = n!Rn−11, 1L2[0,1], where R = C ◦ T, (Rf )(x) = 1−x f (y)dy.

• Therefore

An = n!

∞

• k=1

λn−1

k

1, φk2

L2[0,1],

where (φk)k≥1 is the orthonormal system of eigenfunctions of the self-adjoint operator R, with corresponding eigenvalues (λk)k≥1.

• It remains to diagonalize the operator R.

Snake calculus

• We can refine the enumeration An by splitting the number of

up-down permutations according to the last number

Snake calculus

• We can refine the enumeration An by splitting the number of

up-down permutations according to the last number

• Then the numbers An,k form the snake triangle:

Snake calculus

• We can refine the enumeration An by splitting the number of

up-down permutations according to the last number

• Then the numbers An,k form the snake triangle:

1 1 1 1 1 2 2 5 5 4 2 5 10 14 16 16 . . .

Snake calculus

• We can refine the enumeration An by splitting the number of

up-down permutations according to the last number

• Then the numbers An,k form the snake triangle:

1 1 1 1 1 2 2 5 5 4 2 5 10 14 16 16 . . .

• Plotting the last line one already can guess the base eigenfunction!

Diagonalizing the transfer operator

• Looking for an eigenfunction:

λf (x) = (Rf )(x) = 1−x f (y)dy,

Diagonalizing the transfer operator

• Looking for an eigenfunction:

λf (x) = (Rf )(x) = 1−x f (y)dy, λf ′(x) = −f (1 − x),

Diagonalizing the transfer operator

• Looking for an eigenfunction:

λf (x) = (Rf )(x) = 1−x f (y)dy, λf ′(x) = −f (1 − x), λf ′′(x) = f ′(1 − x) = − 1 λf (x).

Diagonalizing the transfer operator

• Looking for an eigenfunction:

λf (x) = (Rf )(x) = 1−x f (y)dy, λf ′(x) = −f (1 − x), λf ′′(x) = f ′(1 − x) = − 1 λf (x).

• So f solves the Sturm-Liouville problem:

f ′′(x) = − 1 λ2 f (x),

Diagonalizing the transfer operator

• Looking for an eigenfunction:

λf (x) = (Rf )(x) = 1−x f (y)dy, λf ′(x) = −f (1 − x), λf ′′(x) = f ′(1 − x) = − 1 λf (x).

• So f solves the Sturm-Liouville problem:

f ′′(x) = − 1 λ2 f (x), f (1) = 0,

Diagonalizing the transfer operator

• Looking for an eigenfunction:

λf (x) = (Rf )(x) = 1−x f (y)dy, λf ′(x) = −f (1 − x), λf ′′(x) = f ′(1 − x) = − 1 λf (x).

• So f solves the Sturm-Liouville problem:

f ′′(x) = − 1 λ2 f (x), f (1) = 0, f ′(0) = 0.

Diagonalizing the transfer operator

• Looking for an eigenfunction:

λf (x) = (Rf )(x) = 1−x f (y)dy, λf ′(x) = −f (1 − x), λf ′′(x) = f ′(1 − x) = − 1 λf (x).

• So f solves the Sturm-Liouville problem:

f ′′(x) = − 1 λ2 f (x), f (1) = 0, f ′(0) = 0.

• The (normalized) solutions are

φk(x) = √ 2 cos (2k − 1)πx 2

• ,

λk = (−1)k−1 (2k − 1) 2 π , k = 1, 2, . . . .

Diagonalizing the transfer operator

• Looking for an eigenfunction:

λf (x) = (Rf )(x) = 1−x f (y)dy, λf ′(x) = −f (1 − x), λf ′′(x) = f ′(1 − x) = − 1 λf (x).

• So f solves the Sturm-Liouville problem:

f ′′(x) = − 1 λ2 f (x), f (1) = 0, f ′(0) = 0.

• The (normalized) solutions are

φk(x) = √ 2 cos (2k − 1)πx 2

• ,

λk = (−1)k−1 (2k − 1) 2 π , k = 1, 2, . . . .

• So An = 2n+2n!

πn+1

∞

k=1 (−1)(k−1)(n−1) (2k−1)n+1 , which is equivalent to Andr´

e’s

• theorem. :)

The transfer operator for the 2m-strip

• c j−1

c j+1 c j c j+2

Figure: The coordinate filtration for the 4-strip.

The transfer operator for the 2m-strip

• c j−1

c j+1 c j c j+2

Figure: The coordinate filtration for the 4-strip.

• (main observation: better to cut tableau along diagonals!)

The transfer operator for the 2m-strip, continued

• The transfer operator works on the function space over the

m-dimensional simplex Ωm =

• (x1, . . . , xm) : 0 ≤ x1 ≤ x2 ≤ . . . ≤ xm ≤ 1
• ,

and is given by

The transfer operator for the 2m-strip, continued

• The transfer operator works on the function space over the

m-dimensional simplex Ωm =

• (x1, . . . , xm) : 0 ≤ x1 ≤ x2 ≤ . . . ≤ xm ≤ 1
• ,

and is given by (Tf )(x1, . . . , xm) = 1−xm 1−xm−1

1−xm

1−xm−2

1−xm−1

. . . 1−x1

1−x2

f (y1, . . . , ym)dym . . . dy1.

The transfer operator for the 2m-strip, continued

• Diagonalizing leads to boundary value problem:

∂2mf ∂2x1 . . . ∂2xm = (−1)m λ2 f , f ≡

• n: x1 = x2, x2 = x3, . . . , xm−1 = xm, xm = 1,

∂f ∂x1 ≡

• n: x1 = 0.

The transfer operator for the 2m-strip, continued

• Diagonalizing leads to boundary value problem:

∂2mf ∂2x1 . . . ∂2xm = (−1)m λ2 f , f ≡

• n: x1 = x2, x2 = x3, . . . , xm−1 = xm, xm = 1,

∂f ∂x1 ≡

• n: x1 = 0.
• Solutions are

φk1,...,km(x1, . . . , xm) = 2m/2 det

• cos

πkixj 2

i,j=1,...,m

, λk1,...,km = 2m(−1)

1 2

P(kj−1)

πmk1k2 . . . km , where 0 < k1 < k2 < . . . < km are odd integers.

The transfer operator for the 2m-strip, continued

• Diagonalizing leads to boundary value problem:

∂2mf ∂2x1 . . . ∂2xm = (−1)m λ2 f , f ≡

• n: x1 = x2, x2 = x3, . . . , xm−1 = xm, xm = 1,

∂f ∂x1 ≡

• n: x1 = 0.
• Solutions are

φk1,...,km(x1, . . . , xm) = 2m/2 det

• cos

πkixj 2

i,j=1,...,m

, λk1,...,km = 2m(−1)

1 2

P(kj−1)

πmk1k2 . . . km , where 0 < k1 < k2 < . . . < km are odd integers.

• In physicspeak: m-fermion systems

Other slopes

• The generalizations to other (rational) slopes are straightforward:

Other slopes

• The generalizations to other (rational) slopes are straightforward:
• One considers the “ribbon” shape

(SSTSTTSTTTS)

n

Other slopes

• The generalizations to other (rational) slopes are straightforward:
• One considers the “ribbon” shape

(SSTSTTSTTTS)

n

• Funds the eigenfunctions in the appropriate functional space

Other slopes

• The generalizations to other (rational) slopes are straightforward:
• One considers the “ribbon” shape

(SSTSTTSTTTS)

n

• Funds the eigenfunctions in the appropriate functional space
• For m-stack of ribbons, the eigenfunctions are m-fermionic states.

Other slopes (cont’d)

• For example, for slope 1/2,

Other slopes (cont’d)

• For example, for slope 1/2,

1 3 1 1 1 1 1 9 9 1 2 3 6 1 3 19 19 18 15 19

• One solves the boundary problem:

f ′′′(x) = 1/λf (x),

Other slopes (cont’d)

• For example, for slope 1/2,

1 3 1 1 1 1 1 9 9 1 2 3 6 1 3 19 19 18 15 19

• One solves the boundary problem:

f ′′′(x) = 1/λf (x), f (1) = 0,

Other slopes (cont’d)

• For example, for slope 1/2,

1 3 1 1 1 1 1 9 9 1 2 3 6 1 3 19 19 18 15 19

• One solves the boundary problem:

f ′′′(x) = 1/λf (x), f (1) = 0, f ′(0) = 0,

Other slopes (cont’d)

• For example, for slope 1/2,

1 3 1 1 1 1 1 9 9 1 2 3 6 1 3 19 19 18 15 19

• One solves the boundary problem:

f ′′′(x) = 1/λf (x), f (1) = 0, f ′(0) = 0, f ′′(0) = 0.

Other slopes (cont’d)

• For example, for slope 1/2,

1 3 1 1 1 1 1 9 9 1 2 3 6 1 3 19 19 18 15 19

• One solves the boundary problem:

f ′′′(x) = 1/λf (x), f (1) = 0, f ′(0) = 0, f ′′(0) = 0.

• For m-stack of ribbons, the eigenfunctions are m-fermionic states.

Finally...

• This leads to the a large deviation principle for Young tableaux:

given a continual Young diagram of shape λ, the typical growth profile of a randomly chosen Young tableau of shape λ maximizes the functional

Finally...

• This leads to the a large deviation principle for Young tableaux:

given a continual Young diagram of shape λ, the typical growth profile of a randomly chosen Young tableau of shape λ maximizes the functional J(g) = 1 ∞

−∞

• log

2 π cos π 2 ∂g ∂u

• − log ∂g

∂t

• ∂g

∂t du dt subject to being a feasible growth profile for the shape λ.

Conclusions

• Strip tableaux (and to a lesser extent their generalizations with

arbitrary slope) are an exactly solvable model.

Conclusions

• Strip tableaux (and to a lesser extent their generalizations with

arbitrary slope) are an exactly solvable model.

• Interesting determinantal formulas - connection to determinantal

point processes, random matrices?

Conclusions

• Strip tableaux (and to a lesser extent their generalizations with

arbitrary slope) are an exactly solvable model.

• Interesting determinantal formulas - connection to determinantal

point processes, random matrices?

• Connection to Euler and Bernoulli numbers and values of poly-zeta

functions.

Conclusions

• Strip tableaux (and to a lesser extent their generalizations with

arbitrary slope) are an exactly solvable model.

• Interesting determinantal formulas - connection to determinantal

point processes, random matrices?

• Connection to Euler and Bernoulli numbers and values of poly-zeta

functions.

• Connection to square ice model

Conclusions

• Strip tableaux (and to a lesser extent their generalizations with

arbitrary slope) are an exactly solvable model.

• Interesting determinantal formulas - connection to determinantal

point processes, random matrices?

• Connection to Euler and Bernoulli numbers and values of poly-zeta

functions.

• Connection to square ice model

Thank you!