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Example text

Proof. Let χ1 = χq (Vωi¯ (aq −r h )) and χ2 is obtained from χ (Vωi (a)) by replacing ±1 ∓1 Yi,aq n by Yi,aq −n . 12. 17. 19. One can define a similar procedure for obtaining the q-character of the dual to any irreducible Uq g-module V . 3, χq (V ) is a subsum in the product of q-characters of fundamental representations. 18 tells us what to do with each m(i) . This procedure is consistent because χq ((V ⊗ W )∗ ) = χq (V ∗ ) · χq (W ∗ ). Note that under this procedure the dominant monomials go to the antidominant monomials and vice versa.

5. A Characterization of q-Characters in Terms of the Screening Operators In this section we prove Conjecture 2 from [FR2]. 1. Definition of the screening operators. First we recall the definition of the screening ±1 ]i∈I ;a∈C× from [FR2] and state the main result. operators on Y = Z[Yi,a Consider the free Y-module with generators Si,x , x ∈ C× , Yi = ⊕ Y · Si,x . x∈C× Let Yi be the quotient of Yi by the relations Si,xq 2 = Ai,xqi Si,x . 1) Y · Si,x , and so Yi is also a free Y-module. Define a linear operator Si : Y → Yi by the formula Si (Yj,a ) = δij Yi,a Si,a and the Leibniz rule: Si (ab) = bSi (a) + a Si (b).

Hence if the highest weight monomial of χq (V ) has positive lattice support with base a, then so do all monomials in χq (V ). Now consider the case of general Uq g. Suppose there exists a monomial in χ = χq (Vωi (a)), which does not have positive lattice support with base a. Let m be a highest among such monomials (with respect to the partial ordering by weights). 5, the monomial m is not dominant. In other words, if we rewrite m ±1 as a product of Yi,b , we will have at least one generator in negative power, say Yi−1 .

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Communications in Mathematical Physics - Volume 216 by M. Aizenman (Chief Editor)

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