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Lecture notes in arithmetic No.947

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Indeed, applying the diagonalization lemma to the root space decomposition of L with respect to H, one finds I ∩ Lα = 0 for some root α of L. Either Lα = H, or else Lα is one dimensional, hence Lα ⊂ I. But in the latter case the one dimensional subspace [Lα , L−α ] ⊂ H is contained in I. It is thus reasonable to remove the ideals I with I ∩ H = 0 from L(A) by setting Definition. L(A) := L(A)/I(A). Note that the kernel of the canonical map π : L(A) → L(A) intersects H trivially. Thus we shall henceforth identify H with its image under π.

It follows that [Yα1 , [Yα2 [. . Yαn ] . . ] ∈ N− −α1 −α1 −···−αn . Moreover, any element of N− is a sum of such elements (use the antisymmetry of the Lie bracket to write an arbitrary Lie product of n elements in this form). Hence N− is H-diagonalizable (with respect to the adjoint action of H) with finite dimensional weight spaces (N− )λ , and (N− )λ = 0 unless −λ ∈ Q+ . The analogue statement for N+ (and with Xα instead of Yα ) then shows that L(A) is H-diagonalizable with dim(L(A) where we use deg( α∈∆ cα α) = λ ≤ deg(λ), cα .

4) follows from the fact that the element on the left is in Lγ , where γ = β + (−β(hα ) + 1)α = σα (β − α). with respect to H. But σα (β − α) is not a root since β − α is not. (5) follows similarly. Let now R be an arbitrary root system with Cartan matrix A. Let ∆ be a basis. –5. (with xα , . . replaced by Xα , . . ). We explain in the next section what this precisely means. Then one has Theorem. ) The Lie algebra L(A) is semisimple, and its Cartan matrix is A (up to permutation of indices).