Using the notion of a graph with compatible automorphism, Lusztig constructs all symmetrizable Cartan data (i.e. Cartan matrices $A$ for which there is a diagonal matrix $D=\mathrm{diag}(d\_1,\ldots,d\_n)$ with integer entries such that $DA$ is symmetric). 


It is sometimes useful to insist that $\gcd(d\_1,\ldots,d\_n)=1$, and I am wondering if there is a corresponding condition one could put on Lusztig's construction to achieve this?


Anyway, here is how the construction goes--Start with a graph $\Gamma$ with vertices $\Gamma\_0$ and edges $\Gamma\_1$. Assume you have an automorphism such that the quotient graph $\Gamma / a$ has no loops (that is, if $(i,j)\in\Gamma\_1$, then $i$ and $j$ are in different $a$-orbits). For $i\in\Gamma\_0$, set $\alpha\_i$ equal to the $a$-orbit of $i$. 


Regard these orbits as a basis of a vector space (or free $\mathbb{Z}$-module) and define a bilinear form as follows. Set
$d\_i=|\alpha\_i|,$, $(\alpha\_i,\alpha\_i)=2d\_i,$
and let $(\alpha\_i,\alpha\_j)$ be the number of edges `connecting the orbits.' Specifically, $(\alpha\_i,\alpha\_j)$ is the number of edges $(i',j')\in\Gamma\_1$ satisfying $i'\in\alpha\_i$ and $j'\in\alpha\_j$.


Lusztig proves that setting $$a\_{ij}=2(\alpha\_i,\alpha\_j)/d\_i$$ defines a symmetrizable Cartan matrix $A$ and all such Cartan matrices arise in this way.


Now, my question is if there is a condition on $a$ that would force the set of $d\_i$ to be relatively prime (not pairwise, but overall). One obvious thing to do is to make $a$ have fixed points (so at least one of the $d\_i$ is 1). This doesn't give you everything, but I beleive that you can construct all affine Kac-Moody data this way. (I don't have the table of graphs in front of me but I'd be sort of surprised if there were a counterexample. Do all hyperbolic Kac-Moody have this property?)


Thanks!



A question on Lusztig's `graph with automorphism' construction?
I think what you're looking for is the observation that if you do have a non-trivial gcd, then you can replace $\Gamma$ and $a$ with another pair that give the same answer and have gcd 1. 


After all, let $\ell$ be the quotient $\operatorname{lcm}(d\_1,\dots, d\_n)/\operatorname{gcd}(d\_1,\dots, d\_n)$, and consider the automorphism $a^\ell$; this generates an automorphism of order $\operatorname{gcd}$ which acts freely. If you consider the quotient $\Gamma/a^\ell$, with the automorphism induced by $a$, that will give you the same Cartan matrix (since both $d\_i$ and $(\alpha\_i,\alpha\_j)$ get divided by $ \operatorname{gcd}$), and has gcd 1.


