When is a generalized Cartan matrix invertible?

Let $A=(a_{ij})$ be a generalized Cartan matrix, i.e. $a_{ij} \in Z, a_{ii}=2$, $a_{ij}\leq 0$ for $i \neq j$ and $a_{ij}=0$ iff $a_{ji}=0.$ If $A$ is classical Cartan matrix or hyperbolic, it is known that $A$ is invertible, while if $A$ is affine it has a 1-dimensional kernel.

What is known about general (indecomposable) $A$ of indefinite type?

EDIT: If $A \in Z^n$ is invertible, then one clearly can find $v,w \in Z^n$ such that $$A':= \begin{pmatrix} A & v \\\ w^t & 2 \end{pmatrix}$$

is again an indecomposable gCM. So the question should be: Given $A$ invertible, how do you produce $A'$ such that $A$ is the upper-left corner of $A'$ and $rank(A)=rank(A')$?

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Have you looked at V.Kac's Infinite dimensional Lie algebras? May be he has something to say. On another note, apparently very little is known about Kac-Moody algebras of indefinite type (ref en.wikipedia.org/wiki/Kac%E2%80%93Moody_algebra). –  Somnath Basu Feb 19 '10 at 15:34

2 Answers

I think the answer is that we can say very little about a generalized Cartan matrix of indefinite type in general. Most of them will be invertible in the reals (because invertible matrices form a dense open algebraic subset of all matrices), but many of them will not. If you want to produce a singular matrix A' from a nonsingular A in the way you describe, you have to find a nonpositive integer solution to a system of quadratic equations with coefficients given by minors of A. If we let $A(i,j)$ denote the determinant of the submatrix of A given by deleting the ith row and jth column, the equation is $\sum_{i,j}(-1)^{i+j} v_i w_j A(i,j) = -2\operatorname{det}(A)$. I am having difficulty imagining a situation in which a solution $(v,w)$ does not exist, but I don't have a solution.

If you allow the more general Cartan matrices attached to generalized Kac-Moody algebras, this becomes much simpler, since diagonal entries are allowed to be zero.

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