Page 99 - Textos de Matemática Vol. 39

SPECTRALLY ARBITRARY FACTORIZATION: THE NONDEROGATORY CASE 89
number of equal values among β′s, or among the γ′s, are counted.) By the total multiplicity, TM(βi,γj), of a λ−pair βi,γj we mean m(βi)+m(γj). For at most one λ−pair could TM(βi,γj) exceed n. We observe the following
Lemma 1.9. Let β1,...,βn;γ1,...γn and λ ∈ C be given. It is possible to re- order the β’s and γ’s: βj1 , ..., βjn ; γk1 , ...γkn so that
(i) equal β’s (resp. γ’s) occur consecutively; and (ii) βjiγki ̸= λ, i = 1,...,n
if and only if there is no pair βp,γq such that TM(βp,γq) > n, with respect to λ.
Proof. The necessity of the condition is obvious. So, we turn to the proof of sufficiency.
We assume, without loss of generality, that there are k distinct β’s β1, ...,βk with respective multiplicities m1, ..., mk and k distinct γ’s γ1, ..., γk with re- spective multiplicities n1, ..., nk such that βiγi = λ, i = 1, ..., k. In this event k
TM(βi,γi) = 2n, the maximum possible. (Otherwise, unmatched β’s or γ’s
i=1
may be identified or grouped as matched λ−pairs.) The case k = 1 cannot
occur, by hypothesis, and the case k = 2 is easily checked. An ordering is easily constructed, given the assumption.
We next show that the required implication follows inductively, given the case k = 3. Suppose k ≥ 4, and assume without loss of generality that the β’s and γ’s are ordered by descending total multiplicity. In this event TM(βk−1,γk−1)+TM(βk,γk) ≤ n, and if we identify βk (γk) with βk−1 (γk−1), our assumption on total multiplicity remains satisfied. When the renamed β’s and γ’s are properly ordered according to the inductive case k − 1, we are at liberty to order within the βk−1,βk (γk−1,γk) block so that requirement (i) is met. Requirement (ii) is obviously met.
Thus, the proof of sufficiency rests upon the case k = 3. In this event,
Now, suppose there is a distinct pair i, j such that both mi + mj + ni ≤ n and ni + nj + mj ≤ n. Then, we may place first in the β list all the βj’s, then βi’s and in the γ list last all the γi’s and next to last all the γj’s. This determines the placement of the remaining β’s and γ’s in their respective lists and, given the stated inequalities, it is easily checked that for no l, 1 ≤ l ≤ 3, is a βl in the same position in the β list as γl in the γ list. Requirement (i) is also met by design.
3 i=1
note that since TM(βi,γi) ≤ n and
TM(βj,γj) ≥ n for each pair i ̸= j. Also, for each distinct pair i,j, 1 ≤ i,j ≤ 3,
TM(βi,γi) ≤ 2n we have TM(βi,γi)+ mi+mj+ni≤n or mk+nj+nk≤n (∗)
and k ̸= i, j, 1 ≤ k ≤ 3, either