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Distinct addable nodes of a partition have distinct contents

Statement

Let λ be a partition. If x≠y are addable nodes of [λ], then c(x)≠c(y); equivalently the content map is injective on the set Add⁡(λ) of addable nodes.

Facts & Assumptions

Given: A partition λ=(λ1,…,λk) with Young diagram [λ] and set of addable nodes Add⁡(λ) (Removable and addable nodes).

[F1]

A node (i,λi+1) with 1≤i≤k is addable if and only if i=1 or λi−1>λi; the node (k+1,1) is always addable; and these are all addable nodes. With the conventions λ0:=+∞ and λk+1:=0, the addable nodes of [λ] are exactly the nodes (i,λi+1) for the indices 1≤i≤k+1 satisfying λi−1>λi (Removable and addable nodes).

[F2]

The content of a node is c(r,c)=c−r; in particular c(i,λi+1)=λi+1−i (The content of a node and the content vector of a standard tableau).

Proof

technique · direct
1.1F1given

By [F1] every addable node has the form (i,λi+1) for a unique index i∈I:={1≤i≤k+1:λi−1>λi}, where we use λ0=+∞ and λk+1=0; indeed for i≤k the condition is exactly the addability criterion, and i=k+1 is the new-row node (k+1,1)=(k+1,λk+1+1) with λk>λk+1=0.

1.2F2given

For such an index i the content of the addable node is c(i,λi+1)=λi+1−i by [F2], and the partition is weakly decreasing, so λi≥λi+1≥⋯≥0.

2.1step 1.2algebra

Let i<i′ be two indices in I. By weak monotonicity λi≥λi′, and since i<i′ we get λi−i>λi′−i′, that is λi+1−i>λi′+1−i′.

3.1step 1.1step 2.1∎

Combined with step 1.1, distinct addable nodes (i,λi+1) and (i′,λi′+1) with i≠i′ have contents differing by the strict inequality of step 2.1; hence c is injective on Add⁡(λ).

Remarks

  • The content is the addable-node coordinate. For an addable node (i,λi+1) the content λi+1−i is the integer at which the interpolation factors of the projector recursion of Primitive tableau idempotents by Jucys-Murphy interpolation are evaluated; the lemma is what makes all their denominators nonzero.

  • Empty partition. For λ=∅ the only addable node is (1,1), of content 0, so injectivity is vacuous there; the argument above applies with k=0 and I={1}.

Depends on

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