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The 3-cycle filter for ν=(3,2) on S_8

Example

For ν=(3,2)⊢5 and r=3 the filter The r-cycle virtual character filter gives the virtual character of S8 Ψ(3,2)=χ(6,2)−χ(4,4)−χ(3,2,2,1)+χ(3,2,1,1,1), because the four partitions λ⊢8 with λ/(3,2) a removable 3-border strip are (6,2) (height 0), (4,4) (height 1), (3,2,2,1) (height 1) and (3,2,1,1,1) (height 2), with signs +,−,−,+. It vanishes on every class of S8 whose cycle type has no part 3; the computations below verify this on the classes without 3-cycles that are displayed in the statement: the identity class, where Ψ(3,2) has degree 20−14−70+64=0, the class of 8-cycles, where all four characters vanish because none of the four shapes is a hook (The value of an irreducible character on a full cycle is supported on hooks), and the class (4,4), where the rule gives χ(6,2)=0, χ(4,4)=2, χ(3,2,2,1)=−2, χ(3,2,1,1,1)=0, so that Ψ(3,2)=0−2−(−2)+0=0.

Facts & Assumptions

Given: The partitions (3,2)⊢5 and (6,2),(4,4),(3,2,2,1),(3,2,1,1,1)⊢8, and for every partition of an integer the character χλ of the complex Specht module Sλ (Column antisymmetrizers, polytabloids, and Specht modules).

[F1]

Filter proposition: Ψ(3,2):=∑λ/(3,2)(−1)ht⁡(λ/(3,2))χλ∈R(S8) is the outer product χ(3,2)∘ψ3 with ψ3=χ(3)−χ(2,1)+χ(1,1,1), and it vanishes on every class of S8 whose cycle type has no part 3; ψ3 vanishes off the class of 3-cycles, so ψ3(1)=0 (The r-cycle virtual character filter).

[F2]

Removable border strip: a removable 3-strip of λ is a connected, 2×2-free skew diagram λ/ν with 3 nodes and ν=(3,2); its height is its number of rows minus one, equivalently 3−c where c is its number of columns (Removable border strips and their height, Hook, arm, leg, and hook length of a box).

[F3]

Murnaghan–Nakayama rule: χλ(σc)=∑λ/μ(−1)ht⁡(λ/μ)χμ(σ) for a cycle c of length r on a specified set of points fixed by σ (a specified fixed point for length 1), summed over the removable r-strips of λ (The Murnaghan–Nakayama character rule).

[F4]

Full-cycle values: for an 8-cycle ρ∈S8, χλ(ρ)=(−1)8−x if λ=(x,18−x), and 0 otherwise; in particular χλ(ρ)=0 whenever [λ] contains a 2×2 block (The value of an irreducible character on a full cycle is supported on hooks).

[F5]

Dimensions by the hook length formula: f(6,2)=8!/(7⋅6⋅4⋅3⋅2⋅1⋅2⋅1)=20, f(4,4)=14, f(3,2,2,1)=70, f(3,2,1,1,1)=64 (The hook length formula).

[F6]

Class language: a class contains a 3-cycle exactly when its cycle type has a part 3, and the identity class is the class of cycle type (18) (Support, fixed points, disjoint cycles, cycle length, disjoint-cycle decompositions, and cycle type).

Verification

technique · direct
1.1F1F2givenalgebra

A partition λ⊢8 contains (3,2) exactly when λ1≥3 and λ2≥2, leaving the eleven candidates (6,2),(5,2,1),(4,2,2),(4,2,1,1),(3,2,2,1),(3,2,1,1,1),(5,3),(4,3,1),(3,3,2),(3,3,1,1),(4,4); among their skew differences λ/(3,2), which all have 3 nodes, exactly four are edge-connected and contain no 2×2 block, namely the horizontal strip {(1,4),(1,5),(1,6)} with c=3 columns in λ=(6,2), the corners {(1,4),(2,3),(2,4)} and {(3,1),(3,2),(4,1)} with c=2 in λ=(4,4) and λ=(3,2,2,1), and the vertical strip {(3,1),(4,1),(5,1)} with c=1 in λ=(3,2,1,1,1); the other seven differences are disconnected, for instance λ=(5,3), where the node (2,3) is not adjacent to (1,4) or (1,5). The heights are 3−c, namely 0,1,1,2, so the four signs +,−,−,+ and the displayed expression for Ψ(3,2) follow.

2.1step 1.1F1F5algebra

The four dimensions of [F5] give Ψ(3,2)(1)=20−14−70+64=0; equivalently Ψ(3,2)=χ(3,2)∘ψ3 by [F1] and ψ3(1)=0 because the identity of S3 is not a 3-cycle, so the degree vanishes.

2.2F4step 1.1algebra

On the class of 8-cycles each of the four shapes contains a 2×2 block (rows 1 and 2, columns 1 and 2), so none is a hook and χλ=0 on that class for all four λ by the full-cycle lemma [F4]; hence Ψ(3,2) vanishes there.

2.3step 1.1F3algebra

On the class (4,4) the rule [F3] gives: χ(6,2)=0, since the unique 4-strip takes (6,2) to (2,2) with height 0 and (2,2) has no 4-strip; χ(4,4)=+1−(−1)=2 from the strips to (4) (height 0) and to (3,1) (height 1), with χ(4)(4)=1 and χ(3,1)(4)=−1; χ(3,2,2,1)=(−1)+(−1)=−2 from the strips to (3,1) and to (14), both of height 2, with χ(3,1)(4)=χ(14)(4)=−1; and χ(3,2,1,1,1)=0 because (3,2,1,1,1) has no 4-strip. Hence Ψ(3,2)=+0−2−(−2)+0=0 on the class (4,4), whose cycle type has no part 3.

3.1step 2.1step 2.2step 2.3F1F6algebra∎

Steps 2.1, 2.2 and 2.3 verify the vanishing predicted by [F1] on the identity class, the class of 8-cycles and the class (4,4), all of which are classes of S8 without a 3-cycle, confirming the displayed value Ψ(3,2)=0 there.

Remarks

  • Consistency of the four shapes with the hook-length list. The removable 3-strips found in step 1.1 correspond to the hook lengths of the four shapes: for instance the strip to (6,2) is the hook of the node (1,4), of length 3; the full hook-length multiset of (6,2) is 7,6,4,3,2,1,2,1, and the other strips arise from further hooks of length 3 in the respective shapes.

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