MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $rank(A)=rank(A')$?

share|cite|improve this question
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 – Somnath Basu Feb 19 '10 at 15:34

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.