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A GCM is square matrix $A = (a_{ij})$ satisfying: (1) $a_{ij} \in \mathbf{Z}$ (2) $a_{ii} = 2$ for all $i$. (3) $a_{ij} \leq 0$ for $i \neq j$. (4) $a_{ij} = 0$ iff $a_{ji} = 0$. There is a standard notion of irreducibility among GCMs and the common term for it is "indecomposable". Now if you look at most books there is a standard basic classification of GCMs: An indecomposable GCM is of 3 kinds: finite, affine and another category which can be described as all-other-GCMs-that-I-know-little-about-so-I-will-bundle-them-together. There is also another classification based on whether the GCM is "symmetrizable" or not (the finite and affine GCMs are symmetric and hence symmetrizable).

What I am looking for is a smart classification of the GCMs that are not finite or affine. For example here is a number of questions:

    [1] Is the number of indefinite GCMs (not finite, not affine but symmetrizable) finite? As far as I know indefinite GCMs is a larger set than hyperbolic GCMs which are only finitely many.
    [2] Are all indefinite GCMs invertible?
    [3] If $A$ is indefinite and of size $N$ what is the biggest matrix of finite type that can appear as a principal minor in $A$?

EDIT: corrected a mathematical mistake pointed out by Jim Humphreys, also improved the format as to make things more visible. Again suggested by Jim Humphreys.

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  • $\begingroup$ Two quick comments: 1) I don't understand the parenthetic remark at the end of the first paragraph, which on face of it looks incorrect. 2) Several distinct questions come up in your second paragraph and might be numbered/displayed for clarity. (The answer to the first question is almost certainly no.) $\endgroup$ Commented Feb 15, 2011 at 13:42
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    $\begingroup$ Finite Cartan matrices are not all symmetric. Take e.g. $B_2$. They are all symmetrizable. Also, it is quite easy to see that the number of non-affine symmetric Cartan matrices is infinite. $\endgroup$ Commented Feb 15, 2011 at 13:48
  • $\begingroup$ [RE-1]: Sorry I forgot about finite/affine Cartan matrices that are not symmetric! But both classes are symmetrizable. Will fix that mistake. [RE-2]: I will separate them. And can you point me to references for the no answer to the first question? $\endgroup$
    – Najdorf
    Commented Feb 15, 2011 at 13:53
  • $\begingroup$ P.S. It's important to be aware of different conventions in the literature about the definition of a GCM. For instance, in the 2005 Cambridge text Lie Algebras of Finite and Affine Type by R.W. Carter, the off-diagonal entries of a GCM are allowed at first to be arbitrary complex numbers. See Carter's Chapters 14 and 15 for his viewpoint on classification of GCMs and related Lie algebras. Other sources, such as the 1995 Wiley book Lie Algebras with Triangular Decompositions by Moody-Pianzola, have different emphases. But GCMs mainly come up in Lie algebra theory. $\endgroup$ Commented Feb 15, 2011 at 14:27
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    $\begingroup$ I see so many "classification" results into the Good, the Bad, and the Ugly. The most interesting case tends to be the Bad. $\endgroup$ Commented Feb 15, 2011 at 14:56

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