Hasse Walk Length Formula (Theorem 8.6) #
This file formalizes Theorem 8.6 from Stanley's Algebraic Combinatorics:
the formula for the number of Hasse walks of length ℓ from ∅ to a
partition λ ⊢ n in Young's lattice.
Main results #
HasseWalkFormula.bijCoeff_n_0_eq: Specialization of Lemma 8.5 atj = 0:b_{n,0}(ℓ) = C(ℓ,n) · (ℓ-n-1)!!whenn ≤ ℓandℓ - nis even.HasseWalkFormula.hasseWalkCount_formula(Theorem 8.6):β(ℓ, λ) = C(ℓ,n) · (ℓ-n-1)!! · f^λ.
References #
- Stanley, Algebraic Combinatorics, Section 8 (Theorem 8.6)
Definitions #
The number of standard Young tableaux of shape λ, denoted f^λ in the book.
This equals the number of saturated chains from ∅ to λ in Young's lattice.
Instances For
The number of Hasse walks of length ℓ from ∅ to λ in Young's lattice,
denoted β(ℓ, λ) in the book. Each consecutive pair in the walk is related
by a covering relation (either up or down).
Instances For
Helper: in a Hasse walk from ⊥, the cardinality at position k has parity k.
Each step changes cardinality by ±1 (via card_covBy_succ), so starting from
card 0, the parity of walk(k).card equals the parity of k.
(F1) β(ℓ, λ) = 0 unless ℓ ≡ |λ| (mod 2). Each Hasse walk step changes the diagram's cardinality by ±1, so after ℓ steps from ∅ (card 0) the endpoint must have the same parity as ℓ.
The number of Hasse walks of length n from ∅ to λ that use only
up-steps (i.e., each step is a cover relation going up in Young's lattice).
Unlike numSYT, this is parameterized by an explicit walk length n and target
λ, so it counts up-walks of any specified length. When n = |λ| the count
coincides with numSYT λ.
Instances For
An up-walk of length i from ⊥ reaches a diagram of card i. If i ≠ λ.card,
there are no such walks, so upWalkCount i λ = 0.
upWalkCount lam.card lam and numSYT lam count the same set: saturated chains
from ⊥ to lam in Young's lattice. The definitions are identical when the walk
length parameter n equals lam.card.
The walk type for hasseWalkCount n lam.
Instances For
Cells of a bounded-card diagram lie in a bounded grid.
The walk type HasseWalkType m lam is finite for every m and lam.
(F2): β(ℓ, λ) is the coefficient of λ in (D+U)^ℓ(∅).
If ν ∈ μ.coversUp then μ ⋖ ν in Young's lattice. Each element of coversUp
is obtained by adding one cell, which produces a covering.
If μ ⋖ ν in Young's lattice then ν ∈ μ.coversUp. A covering in the poset
must correspond to adding a single cell.
μ ⋖ ν ↔ ν ∈ μ.coversUp in Young's lattice.
If μ ∈ ν.coversDown then μ ⋖ ν in Young's lattice. Each element of coversDown
is obtained by removing one cell, which produces a covering from below.
If μ ⋖ ν in Young's lattice then μ ∈ ν.coversDown. A covering from below
corresponds to removing a single cell.
μ ⋖ ν ↔ μ ∈ ν.coversDown in Young's lattice.
iterU i emptyBasis is supported on diagrams of cardinality i.
Walk decomposition by penultimate step #
Walks of length n+2 from ⊥ to lam biject with pairs (μ, walk of length n+1 from ⊥ to μ) where μ ⋖ lam. The book treats this as immediate from the definition of up-walks.
The walk type for upWalkCount n lam.
Instances For
The walk type UpWalkType m mu is finite for every m and mu.
Normal ordering decomposition: β(ℓ,λ) = b_{n,0}(ℓ) · (up-walks of length n to λ). The book's proof applies equation (44) (normal ordering of (D+U)^ℓ) to ∅, uses fact (F2) (β(ℓ,λ) is the coefficient of λ in (D+U)^ℓ(∅)), and the identity D^j(∅) = 0 for j > 0.
In a walk from ⊥ where each step is a covering relation (up or down), the
cardinality at any position is bounded above by the position index plus
the cardinality at the starting position offset. This upper bound is used to
derive a contradiction when a down-step occurs in a length-n walk to a
diagram of cardinality n.
When ℓ = |λ|, every Hasse walk from ⊥ to λ of length |λ| is an up-walk:
there is no room for any down-step. This is because each step changes cardinality
by ±1, and starting at 0, with n steps to reach n, every step must be +1.
Formally, the set of Hasse walks (allowing both up and down steps) of length n
from ⊥ to λ with |λ| = n is in bijection with the set of up-walks.
Specialization of Lemma 8.5 at j = 0 #
When ℓ = n, b_{n,0}(n) = 1.
Theorem 8.6: The Hasse walk formula #
Theorem 8.6. Let ℓ ≥ n and λ ⊢ n, with ℓ - n even. Then
β(ℓ, λ) = C(ℓ, n) · (1 · 3 · 5 · ⋯ · (ℓ - n - 1)) · f^λ.
Proof (from the book): Apply (D+U)^ℓ to ∅. Since U^i D^j(∅) = 0
unless j = 0, we get (D+U)^ℓ(∅) = ∑_i b_{i,0}(ℓ) U^i(∅).
Since U^n(∅) = ∑_{λ ⊢ n} f^λ · λ, by (F2) the coefficient of λ in
(D+U)^ℓ(∅) is b_{n,0}(ℓ) · f^λ. By Lemma 8.5,
b_{n,0}(ℓ) = C(ℓ, n) · (1 · 3 · 5 · ⋯ · (ℓ-n-1)), giving the result.
Theorem 8.6 (ℕ version). Let ℓ ≥ n and λ ⊢ n, with ℓ - n even. Then
β(ℓ, λ) = C(ℓ, n) · (1 · 3 · 5 · ⋯ · (ℓ - n - 1)) · f^λ as natural numbers.
When ℓ = n, β(n, λ) = f^λ.
β(ℓ, λ) = 0 when ℓ < |λ|.
β(ℓ, λ) = 0 when ℓ - |λ| is odd (algebraic version of F1).
Corollary 8.7: Closed walks counted by double factorial #
The number of standard Young tableaux of the empty shape is 1. There is exactly one saturated chain from ∅ to ∅: the trivial chain of length 0.
Corollary 8.7. The total number of Hasse walks in Young's lattice of length 2m
from ∅ to ∅ is given by β(2m, ∅) = 1 · 3 · 5 · ⋯ · (2m-1) = (2m-1)!!.
Proof (from the book): Simply substitute λ = ∅ (so n = 0) and ℓ = 2m
in Theorem 8.6. Then C(2m, 0) = 1, f^∅ = 1 (numSYT_bot), and the double
factorial product gives 1 · 3 · 5 · ⋯ · (2m-1).
Dependency chain: closedWalkCount_eq_doubleFactorial → hasseWalkCount_formula
(Theorem 8.6) → bijCoeff_n_0_eq (Lemma 8.5 at j=0) → bijCoeff_of_valid
(fully proved in NormalOrderCoeff).
Corollary 8.7 (ℕ version). The total number of Hasse walks in Young's lattice
of length 2m from ∅ to ∅ equals (2m-1)!! as natural numbers:
β(2m, ∅) = 1 · 3 · 5 · ⋯ · (2m-1).
β(0, λ) = if λ = ⊥ then 1 else 0. A walk of length 0 from ⊥ is just the
constant sequence at ⊥, so it exists iff λ = ⊥, and there is exactly one such walk.