By W.V.D. Hodge

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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). Proof. [Serre], Chapitre VI, Appendice Serre’s theorem states that the map “Lie algebra (modulo ∼ =) → Cartan ∼ matrix (modulo =)” considered at the beginning is surjective.

Clearly one has ai,j = αi (hj ) (1 ≤ i, j ≤ n). Note that the αi and the hj are linearly independent. We set ∆ = {α1 , . . , αn }. 31 32 CHAPTER 4. KAC-MOODY ALGEBRAS The αi are supposed to play the analogue of a root basis of a semisimple Lie algebra, the hj the role of the corresponding hαj from that theory, and A corresponds to the Cartan matrix. ) Accordingly, we shall set hαj := hj . Note also that, for having the hj and αi linearly independent, we had to define them in a 2n − l-dimensional space H.

The second identity of (2) is true since [xα yβ ] ∈ Lα−β , and since α − β is not a root (recall that any root is linear combination of the α with all coefficients either negative or either positive). (3) is clear from the definition of the xα , yα , hα . (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.