prove that v= (- m/M)v
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Answer:
8.4. Finite-Dimensional Irreducible Representations
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Proof. Let M be an irreducible subquotient of V. We will show that M V(u) for some u eh with us and u e W. A. To prove this, note that every weight u of M is also a weight of V, because Vlus is completely reducible. Hence Usa (Proposition 8.3) and so d-u Eaea Zaa with unique za e Z. Choosing a weight u of M such that EeeA Z, is minimal, we see that no u+ a with a € is a weight of M. Therefore, 1.My = 0 by (8.10) and any 0 M E M is a maximal vector. Since M = Ug.m by irreducibility, M is an irreducible highest weight representation with highest weight u, and so M V ) (Corollary 8.8).
Recall that the center (Ug) acts on V, and hence also on the subquotient M, via the central character X and on V() via x The isomorphism M V(4) forces X = Xw, which in turn is equivalent to u e W . A by the fact mentioned above (Corollary 8.24). This proves our assertions about irreducible subquotients of V.
In particular, since M = {u h lusAnwl is a finite set and dim <00 for each u e b (Proposition 8.3), we may define l(V) = verdim VEZ and similarly for any subrepresentation of V. Note that V is noetherian, being a cyclic module over the noetherian ring Ug. Thus, we may choose a series of with irreducible quotients V,/V+1 subrepresentation V = V ɔ V > V 5 Since V/V+1 = V(H) with H E M by the first paragraph of this proof, we see that (V) > E(V) > e(V2) > .. Therefore, the series must terminate at 0 after at most ((V) steps, which shows that length V S ((V).
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Finally, the statement that V(2) occurs exactly once as a composition factor of V is clear, because dim, V 1 (Proposition 8.3), and the statement about the other composition factors has been verified in the first paragraph of this proof.