Alphabeta Math
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First-row padding below the threshold is undefined

Statement refuted

Refuted claim: for every partition α and every integer n≥∣α∣ the displayed sequence (n−∣α∣,α1,α2,… ) is a partition of n, so that the padded family and the padded Kronecker coefficient gn(α,β,γ) would be defined as soon as the size condition n≥∣α∣ holds (First-row padding of a partition).

The witness is α=(2,1), of size ∣α∣=3 and first part α1=2. For n=3 and n=4 the size condition n≥∣α∣ holds, but the displayed sequences are (3−∣α∣,2,1)=(0,2,1),(4−∣α∣,2,1)=(1,2,1), and neither is a partition: the first has a nonpositive leading entry, and the second is not weakly decreasing. Only from the threshold ∣α∣+α1=3+2=5 on does the family become a family of partitions, starting with α[5]=(2,2,1). Hence the padded sequence is undefined below the threshold, and for this shape the stable family begins at n=5.

Facts & Assumptions

Given: The partition α=(2,1) and the integers n=3,4,5 (Partitions, English diagrams, and conjugation).

[F1]

Read ∣∅∣=∅1=0. For a partition α and an integer n≥∣α∣+α1 one sets α[n]=(n−∣α∣,α1,α2,… ), except that ∅[0]=∅; this padding is a partition of n. Below the threshold α[n] is undefined (First-row padding of a partition).

[F2]

A partition of n is a finite weakly decreasing sequence of positive integers with sum n; trailing zeros are not parts, so a displayed sequence that is not weakly decreasing, or that contains a nonpositive entry, is not a partition (Partitions, English diagrams, and conjugation).

[F3]

α=(2,1) has ∣α∣=2+1=3 and α1=2, so its padding threshold is ∣α∣+α1=5. [F1, given]

[F4]

The size-indexed padded coefficient gn(α,β,γ)=g(α[n],β[n],γ[n]) is defined exactly for n≥max⁡{∣α∣+α1,∣β∣+β1,∣γ∣+γ1}; below this common threshold it is undefined (Padded Kronecker coefficient sequences).

Counterexample

technique · direct
1.1F2F3given

For n=3 the displayed sequence is (3−3,2,1)=(0,2,1). Its leading entry 0 is not a positive integer and 0<2 shows that the sequence is not weakly decreasing, so by [F2] it is not a partition of 3, although 3=∣α∣ and hence n≥∣α∣ holds.

1.2F2F3given

For n=4 the displayed sequence is (4−3,2,1)=(1,2,1). Its entries are positive and sum to 4, but 1<2, so it is not weakly decreasing and by [F2] it is not a partition of 4, although 4≥∣α∣ holds.

2.1F1F2F3step 1.1step 1.2

For n=5, which is the threshold ∣α∣+α1 computed in [F3], the displayed sequence is (5−3,2,1)=(2,2,1): the entries are positive, weakly decreasing and sum to 5, so it is a partition of 5. By [F1] the padding α[5]=(2,2,1) is defined, and by step 1.1 and step 1.2 it is the first member of the family; the threshold is therefore exact.

3.1step 1.1step 1.2step 2.1F1F4∎

Consequently the refuted claim fails at n=3 and n=4: the size condition n≥∣α∣ is not enough, and the padded family of α=(2,1) is a family of partitions only from n=5 on. Any padded Kronecker coefficient gn(α,β,γ) with α=(2,1) and n<5 is likewise undefined, because one of its three entries is not a partition.

Remarks

  • The failed threshold ∣α∣ is exactly weaker than the true one ∣α∣+α1; the extra term α1 is what keeps the padded partition weakly decreasing, since the new first part n−∣α∣ must be at least the old first part α1.

  • The example also shows why the size-indexed padded Kronecker coefficients of Padded Kronecker coefficient sequences are defined only from the maximum of the three thresholds: a single unpadded entry makes the whole expression undefined.

Depends on

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