Let us consider the matrix $A$ defined as follows: $A_{i i} =2$, $A_{i j} = - \frac{2 d_i}{d - 1 - d_{j}}, \qquad i \neq j$; $i, j = 1, \dots, n$. Here $d_1,...,d_n$ are natural numbers, $n > 1$ and $d = \sum_{j=1}^n d_{j}$. We exclude $(d_1,d_2) = (1,k),(k,1)$. The problem is to find all $d_i$ such that $A$ is the Cartan matrix of some Kac-Moody (KM) algebra. The problem can be drastically simplified by using the fact that it should be a symmetrizable generalized Cartan matrix for hyperbolic KM algebra. This is by the origin of this matrix (which I do not explain here). At the Dynkin diagram any two nodes should be connected by some line since $A_{i j} \neq 0, \quad i \neq j$ . According to the most complete (and probably, correct and final) classification of hyperbolic KM algebras (see arxiv: 1003.0564) the only ranks n=2,3,4 should be considered (for $n > 4$ there are no diagrams where all nodes are connected by lines) . I have found three solutions for n=2: $(d_1,d_2) = (2,2),(2,3),(3,3)$, two solutions for n=3: $d_1 = d_2 = d_3 =1$, and $d_1 = d_2 = 1$, $d_3 = 2$ and one solution for n= 4: $d_1 = d_2 = d_3 = d_4 = 1$. But I am not sure that this is the final list. (Of course, this is up to permutations.)

  • $\begingroup$ To my memory I had a simple proof (in summer or spring) that this is the final list. (The proof is of the school olympiade level). I will present it when I will have a time and/or I will prepare the paper which will use this example. $\endgroup$ – Vladimir Jan 12 '17 at 21:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.