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A level-two cyclotomic polynomial from the highest-weight filtration

Example

Assume the inherited Axiom of Choice. Take source-ordered row lengths p=(1,2), so the column heights are q=(2,1), and let N=3, n=2, c=(0,0). Number the boxes down the columns: 1=(1,1), 2=(2,1), 3=(2,2). The pyramid nilpotent is e=E23, and on the one-step deformed tensor space Vc⊗1=V=C3 the operator x1 is x1=E23+diag⁡(0,0,−1). Its eigenvalue 0 has multiplicity 2 and eigenvalue −1 has multiplicity 1, with −1-eigenspace C(v3−v2); its minimal polynomial is x(x+1), with roots a=(0,−1), and Λ=Λ0+Λ−1. For every d≥1 the raw tensor operator x1 on Vc⊗d still has exactly this minimal polynomial, by the column-filtration and leading-string argument. Consequently Hd(Λ) has dimension 2dd!, and at d=1 its image splits the two stated eigenspaces.

Facts & Assumptions

Given: The pyramid with columns q=(2,1), n=2, N=3, origin c=(0,0), the numbers a=(0,−1), and the operator x1 on Vc⊗d.

[F1]

On the raw tensor space the operator x1 is (e+∑j(ccol(j)+qcol(j)−n)ej,j)⊗1⊗(d−1) minus its off-diagonal cross-column corrections; its minimal polynomial on Vc⊗d is exactly ∏j(x−aj) with aj=cj+qj−n, and the tensor action of Hd factors through Hd(a) (The parabolic highest-weight truncation factors through the cyclotomic quotient, Arakawa-Suzuki Casimir operators on tensor functors).

[F2]

The cyclotomic quotient Hd(a) of the degenerate affine Hecke algebra at the roots a=(0,−1) has the bounded monomial basis x1r1⋯xdrdw, 0≤ri<2, w∈Sd, so that dim⁡Hd(a)=2dd!; the pyramid roots are aj=cj+qj−n, and the integral-root dictionary gives Λ=Λ0+Λ−1 (The degenerate cyclotomic Hecke algebra).

Proof

technique · direct

Given: The boxes 1=(1,1), 2=(2,1), 3=(2,2), the nilpotent e=E23, and the roots a1=0+2−2=0, a2=0+1−2=−1.

1.1F1algebra

The operators on V. For d=1 the cross-column correction in [F1] is absent and ∑j(ccol(j)+qcol(j)−n)ej,j=diag⁡(0,0,−1), so x1=E23+diag⁡(0,0,−1): explicitly x1v1=0, x1v2=0 and x1v3=v2−v3. Hence x1(x1+1)=0, the eigenspaces of x1 are Cv1⊕Cv2 for the eigenvalue 0 and C(v3−v2) for the eigenvalue −1, and the minimal polynomial is exactly x(x+1): x1≠0 since x1v3=v2−v3≠0, and x1+1≠0 since (x1+1)v1=v1≠0.

2.1F1F2step 1.1algebra∎

All degrees and the cyclotomic quotient. By the minimal-polynomial clause of [F1], the operator x1 on Vc⊗d has minimal polynomial ∏j(x−aj)=x(x+1) for every d≥1, so the tensor action factors through Hd(a) with a=(0,−1), i.e. through Hd(Λ) for Λ=Λ0+Λ−1; and by [F2] that quotient has dimension 2dd!. At d=1 the quotient is C[x]/(x(x+1))≅C×C, acting on V through x1 as the two-dimensional algebra of operators that are scalar on each of the two eigenspaces of step 1.1; hence the image is semisimple and V is the direct sum of those two eigenspaces.

Remarks

  • The two roots 0 and −1 are exactly the column scalars cj+qj−n of the two columns; the example shows how the highest-weight (column) filtration of the ground module produces the level-two cyclotomic polynomial x(x+1).
  • The ordered-row dictionary matters here: the row lengths are read in the source order p=(1,2) with column heights q=(2,1), so box 3 is the rightmost box of the second row and e=E23; a different box numbering would produce a different matrix for x1.

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