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 anyone knowsthere is a naturalcorresponding condition one could put on Lusztig's construction to achieve this?
Anyway, here is how the construction sogoes--Start with a graph $\Gamma$ with vertices $\Gamma_0$ and edges $\Gamma_1$. Assume you have an automorphism such that the quotient graph $\gcd(d_1,\ldots,d_n)=1$$\Gamma / a$ has no loops (that is, if $(i,j)\in\Gamma_1$, then $i$ and $j$ are in different $a$-orbits). For $i\in\Gamma_0$, set $\alpha_i$ equal to the $a$-orbit of $i$.
Regard these orbits as a basis of a vector space (or free $\mathbb{Z}$-module) and define a bilinear form as follows. Set $d_i=|\alpha_i|,$, $(\alpha_i,\alpha_i)=2d_i,$ and let $(\alpha_i,\alpha_j)$ be the number of edges `connecting the orbits.' Specifically, $(\alpha_i,\alpha_j)$ is the number of edges $(i',j')\in\Gamma_1$ satisfying $i'\in\alpha_i$ and $j'\in\alpha_j$.
Lusztig proves that setting $$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!