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The addable-minus-removable node count is the level-one Cartan pairing

Statement

For every partition λ, every residue i∈I, and Ni(λ):=#{addable i-nodes of λ}−#{removable i-nodes of λ} one has

Ni(λ)=δi0−∑j∈Iaij mj(λ),

where mj(λ) is the number of residue-j cells of λ and A=(aij) is the cyclic affine type A Cartan matrix. Consequently ∑i∈INi(λ)=1; the toral operator acts by Ki∣λ⟩=qNi(λ)∣λ⟩=q⟨hi,wt(∣λ⟩)⟩, the central product K0K1⋯Ke−1 acts by q, and wt(∣λ⟩)=Λ0−∑jmjαj has level 1. In particular Ni(∅)=δi0, and adding a node of residue k changes Ni by −aik, hence the weight by −αk.

Facts & Assumptions

Given: The partition Fock space with residues c−r mod e, weight Λ0−∑jmjαj, the node formulas, and the Cartan datum with row sums ∑jaij=0.

[F1]

An addable node of λ is (r,λr+1) with r=1 or λr−1>λr, and a removable node is (r,λr) with λr>λr+1 (Removable and addable nodes).

[F2]

The residues of the nodes around a corner are determined by the row and column indices as c−r; in particular the new bottom row (k,1) has residue 1−k and the node (r,λr+1) has residue λr+1−r (The level-one partition Fock space and node residues).

[F3]

⟨αj,hi⟩=aij, ⟨c,Λi⟩=1, and ∑jaij=0 (The cyclic affine type A Cartan datum and its residues).

Proof

technique · verify the identity on the empty partition and then induct on the number of cells by the local corner change
1.1F1F2givenalgebra

The empty partition. The empty partition has exactly one addable node, the cell (1,1) of residue 0, and no removable nodes; hence Ni(∅)=δi0. The right-hand side is δi0−∑jaij⋅0=δi0.

2.1F1F2step 1.1algebra

The local step. Encode λ by its occupied integer indices Sλ={λr−r+1:r≥1}, with trailing zero rows understood, and let χm be the indicator of m∈Sλ. A pair (m,m+1) of occupancy (1,0) is exactly an addable node of content m; occupancy (0,1) is exactly a removable node of content m. Indeed increasing or decreasing the corresponding row length by one is admissible exactly when the neighbouring occupied index does not collide, by [F1]. Hence Ni(λ)=∑m≡i(χm−χm+1), a finite sum because all sufficiently negative indices are occupied and all sufficiently positive indices are empty. Adding a node of residue k changes χm from 1 to 0 and χm+1 from 0 to 1 at an index m≡k. Thus the count changes by −2δik+δi,k−1+δi,k+1=−aik; at e=2 the last two contributions coincide and give the double edge. Also mj(μ)=mj(λ)+δjk, so the claimed right side changes by the same −aik.

3.1step 2.1F3givenalgebra

The identity and its consequences. Induction on the number of cells from step 1.1 with the local step 2.1 gives Ni(λ)=δi0−∑jaijmj(λ) for every partition. Summing over i and using ∑iaij=0 and ∑iδi0=1 gives ∑iNi(λ)=1. Therefore the toral action Ki∣λ⟩=qNi(λ)∣λ⟩ equals qδi0−∑jaijmj=q⟨hi,Λ0−∑jmjαj⟩, and the product over i acts by q∑iNi=q, so wt(∣λ⟩)=Λ0−∑jmjαj has ⟨c,wt⟩=∑iNi=1 and level one; adding a node of residue k replaces mk by mk+1, changing the weight by −αk and Ni by −aik.

4.1step 1.1step 2.1step 3.1∎

Conclusion. The addable-minus-removable count is the level-one Cartan pairing, the sum of the counts is 1, the central product acts by q, and the declared toral action of the partition Fock space is the one required for a level-one module.

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