Page 132 - Textos de Matemática Vol. 39

122 JOA˜O FILIPE QUEIRO´
and (6.2) coming from
γ1 ≤α1+β1
both in Horn’s list.
Now the sufficiency. We have to exhibit γ1 and γ4 such that γ1≥γ2≥γ3≥γ4
and these numbers satisfy Horn’s inequalities with the 4-tuples (α1, α2, α3, α4) and (β1, β2, β3, β4).
For the record, the inequalities in Horn’s list involving only γ2, γ3 and γ2 + γ3 are the following:
α2+β4,α3+β3,α4+β2 ≤γ2 ≤α1+β2,α2+β1 α3+β4,α4+β3 ≤γ3 ≤α1+β3,α2+β2,α3+β1
α2 +α3 +β3 +β4⎫⎬ ⎧⎨α1 +α2 +β2 +β3 α2+α4+β2+β4⎭≤γ2+γ3 ≤⎩α1+α3+β1+β3 α3 +α4 +β2 +β3 α2 +α3 +β1 +β2
We first define γ1 as the minimum of the following five numbers: α1 + β1
α1 +α2 +β1 +β2 −γ2
α1 +α2 +β1 +β3 −γ3
α1 +α3 +β1 +β2 −γ3
α1 +α2 +α3 +β1 +β2 +β3 −γ2 −γ3
This ensures two things: that γ1 ≥ γ2, as γ2 is easily seen, using the hypothesis, to be less than or equal to all five numbers, and that γ1 satisfies all inequalities in Horn’s list that give upper bounds for it or for sums of it with the given γ2 and γ3.
Next, of course, we put
γ4 =α1 +α2 +α3 +α4 +β1 +β2 +β3 +β4 −γ1 −γ2 −γ3.
The rest of the proof consists of a tedious checking that γ4 ≤ γ3 and that γ4 satisfies all remaining relevant conditions in Horn’s list: upper bounds for γ4, γ2 +γ4, γ3 +γ4, and γ2 +γ3 +γ4.
The others (upper bounds for γ1 +γ4, γ1 +γ2 +γ4, and γ1 +γ3 +γ4) follow directly from the hypothesis (lower bounds for γ2 +γ3, γ3, and γ2, respectively). It is worth remarking, although not unexpected, that (6.1) is only used in proving that γ1 ≥ γ2, and (6.2) is only used in proving that γ4 ≤ γ3.
The following illustrates the polygon of realizable γ when the given spec- tra are α = (6,4,3,2), β = (7,5,4,1). We have marked the side coming from inequality (6.1). Condition (6.2) does not restrict the set in this example.