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In chapter 1 of Kac's book "Infinite dimensional Lie algebras" it is mentioned that two Kac-Moody algebras are isomorphic if and only if their Cartan matrices are isomorphic (i.e. they are the same up to permutation). A reference to the article "Infinite flag varieties and conjugacy theorems" by Peterson and Kac is given, where it is proven that Cartan subalgebras are conjugate (which implies this result). But in this article it is assumed that the Cartan matrix is symmetrizable! Now, is this result true for any Cartan matrix or just for symmetrizable ones?

At the end of the article it is mentioned that all results which do not use the bilinear form also hold for a general Cartan matrix but I don't want to go through this whole article.

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    $\begingroup$ Do non-symmetrisable Cartan matrices also give rise to Kac-Moody algebras? i.e., does the construction in terms of Serre relations still go through? And what sort of Lie algebra do you get? Sorry, this seems like a question, more than a comment... I'm just curious. $\endgroup$
    – José Figueroa-O'Farrill
    Commented Feb 25, 2010 at 12:23
  • $\begingroup$ I think so. Kac gives a general construction of a Lie algebra g(A) for a complex matrix A. If A is a (generalized) Cartan matrix, then he calls g(A) a Kac-Moody algebra. Cartan matrices of finite and affine type are symmetrizable, so here everything is okay. But I don't think this holds for Cartan matrices of indefinite type... $\endgroup$
    – user717
    Commented Feb 25, 2010 at 12:29
  • $\begingroup$ Ah, and if A is a symmetrizable Cartan matrix, then it follows from a theorem by Gabber-Kac that g(A) has a presentation like finite-dimensional semisimple lie algebras, i.e. Serre relations and stuff. $\endgroup$
    – user717
    Commented Feb 25, 2010 at 12:31

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