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So x13 generates a submodule isomorphic to v 1 recall

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Unformatted text preview: (1) = ⊕ L−α− β Claim: Lα+β ⊕ Vβ ∼ V (1). = Pf: Since xα ∈ Lα , ad xα sends Lα+β to L2α+β = 0. Therefore, x13 = xα+β is a maximal vector. But hα (x13 ) = (h1 − h3 )x13 = x13 . So, the weight is 1. So, x13 generates a submodule isomorphic to V (1). (Recall that V (λ) is generated by a maximal vector with maximal weight λ and that λ is a nonnegative integer.) Similarly, L−β ⊕ L−α−β ∼ V (1). This leaves only a one dimensional submodule V (0) = contained in H . 8.4.2. α-root string. We now return to the general case. Recall that we have a root space decomposition: ￿ L∼H⊕ Lα = α ∈Φ Definition 8.4.1. If α ∈ Φ and β ∈ Φ ∪ {0}, the α-root string through β is the Sα -module: ￿ M= Lβ + cα c∈ F (We will prove shortly that only integer values of c can occur in any root string.) Proposition 8.4.2. For any α ∈ Φ, L is a direct sum of α-root strings. From the example we know that the α-root string though 0 is not irreducible in general. Proposition 8.4.3. If α ∈ Φ then Lα is 1-dimensional. Furthermore, the only multiples of α which are roots are ±α. Proof. Consider the α-root string though 0: M =H⊕ ￿ Lα cα ∈ Φ Let K = ker(α : H → F ). Then H = K ⊕ F hα . Claim: K is an Sα -module. Pf: For any k ∈ K we have [xα k ] = −α(k )xα = 0. Similarly, [yα k ] = 0 and [hk ] = 0 since h, k ∈ H . So, every nonzero element of K generated an Sα -module isomorphic to V (0). This implies that ￿ M/K ∼ W = F hα ⊕ Lcα = cα ∈ Φ MATH 223A NOTES 2011 LIE ALGEBRAS 33 is an Sα -module. But the wei...
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