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The Hecke Jucys-Murphy elements through rank four

Example

Let R be a commutative ring and q∈R×. In H=HR,q(Sn) the Hecke Jucys-Murphy elements (Hecke Jucys-Murphy elements and q-contents) for n≥4 are L1=0,L2=q−1T(1,2),L3=q−1T(2,3)+q−2T(1,3),L4=q−1T(3,4)+q−2T(2,4)+q−3T(1,4), and in general Lk=∑j<kqj−kT(j,k). For the standard tableau t=1342 of shape (3,1) the contents of the entries k=1,2,3,4 are 0,−1,1,2 (The content of a node and the content vector of a standard tableau), so the diagonal eigenvalues of L1,…,L4 on the Murphy vector mt are [0]q=0, [−1]q=−q−1, [1]q=1 and [2]q=1+q; the triangular correction terms ∑v⊳tavmv are present in general (The Hecke Jucys-Murphy action is dominance triangular) in the Murphy basis, even in the semisimple range. In the separated range, replacing mt by the seminormal vector ft removes these correction terms, as proved in Separated Hecke residues give seminormal idempotents and bases. At q=1 the four elements specialize to the ordinary Jucys-Murphy elements X1=0, X2=(1 2), X3=(1 3)+(2 3), X4=(1 4)+(2 4)+(3 4) and the eigenvalues to 0,−1,1,2, the standard contents of t.

Facts & Assumptions

Given: A commutative ring R, q∈R×, the Hecke algebra HR,q(Sn), the Jucys-Murphy elements and q-contents (Hecke Jucys-Murphy elements and q-contents), the content vector of a standard tableau (The content of a node and the content vector of a standard tableau), and the q=1 specialization HR,1(Sn)≅R[Sn] (Hecke Jucys-Murphy elements and q-contents).

[F1]

(Hecke Jucys-Murphy elements and q-contents) Lk=∑j<kqj−kT(j,k); [m]q=1+q+⋯+qm−1 for m≥0, [−m]q=−q−m[m]q for m>0, and [m]1=m for all m∈Z.

[F2]

(The Hecke Jucys-Murphy action is dominance triangular) mtLk=[content⁡t(k)]q mt+∑v⊳tavmv on the Murphy cell module, so the displayed values are the diagonal entries and the correction terms are as stated.

Verification

1.1givenF1

Substitution: in [F1], for k=1 the sum is empty, giving L1=0; for k=2 it has the single term q1−2T(1,2)=q−1T(1,2); for k=3 it has q−1T(2,3)+q−2T(1,3); and for k=4 it has q−1T(3,4)+q−2T(2,4)+q−3T(1,4), matching the displayed list.

1.2givenF1

Contents: reading the tableau t with rows (1,3,4) and (2), the node of 1 is (1,1) with content 1−1=0, the node of 2 is (2,1) with content 1−2=−1, the node of 3 is (1,2) with content 2−1=1 and the node of 4 is (1,3) with content 3−1=2; by definition [content⁡t(k)]q is the q-integer of these contents.

2.1step 1.1step 1.2F1F2

Eigenvalues: by [F1], [0]q=0, [−1]q=−q−1[1]q=−q−1, [1]q=1 and [2]q=1+q; substituting into the triangularity formula [F2] gives the stated diagonal eigenvalues 0,−1,1,2 up to the q-powers −q−1,1,1+q. The correction terms are exactly the sum ∑v⊳tavmv of [F2] and are not asserted to vanish outside the semisimple range.

3.1step 1.2step 2.1F1∎

Specialization: setting q=1 in the displayed formulas turns Lk into ∑j<k(j k)=Xk by the specialization isomorphism of Hecke Jucys-Murphy elements and q-contents and [F1], giving X1=0, X2=(1 2), X3=(1 3)+(2 3) and X4=(1 4)+(2 4)+(3 4); the eigenvalues [m]1=m become 0,−1,1,2, the contents of t.

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