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?)


  • $\begingroup$ At first sight the question looks reasonable but could use an explicit reference. It might be useful to ask Lusztig directly. Maybe also a tag lie-algebras rather than rt? $\endgroup$ – Jim Humphreys Feb 24 '13 at 20:54
  • $\begingroup$ hyperbolic KM's don't always have the property that you can take one of the d_i equal to one - this doesn't even happen in rank two. $\endgroup$ – Peter McNamara Feb 24 '13 at 21:36

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.


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