Quantum Cartan matrices and Coxeter elements Let $\Gamma$ be a bipartite graph, with the vertices partitioned into disjoint sets $\Pi_1$ and $\Pi_2$.  Let $W$ be the associated Weyl group, with Coxeter generators $\{s_i\}_{i\in \Gamma}$.  Let $\{\alpha_i\}$ be the simple roots of the associated root system.  Here are two ways to associate a polynomial to $\Gamma$:
1) Let $c = (\prod_{i\in\Pi_1} s_i)(\prod_{j\in \Pi_2} s_j)$ be a bicolored Coxeter element.
Let $charCox_\Gamma(t)$ denote the characteristic polynomial of c (acting in the reflection representation).
2) Let $C_\Gamma$ denote the Cartan matrix, whose entries are $(\langle \alpha_i,\alpha_j \rangle)_{ij}$ (so in this case the entries are all 2,-1,or 0).  Let $QC_\Gamma$ denote the quantum Cartan matrix, where all $2$'s have been changed to $1+q^2$, and all $-1$'s to $-q$'s.  Then we have a polynomial $detQC_\Gamma(q)$, the determinant of the quantum Cartan matrix.  (In fact this is a polynomial in $q^2$).
It appears to be the case that
$$ charCox_\Gamma(q^2) = detQC_\Gamma(q). $$
I suspect this, or something very close, is already known, in which case it would be great to have a reference.  One straightforward way to prove it, at least when $\Gamma$ is a tree, is by showing that both polynomials satisfy the same recursion for joining.  (Probably something similar will prove the statement for more general bipartite graphs.) But that proof is not particularly enlightening.  Is there a conceptual explanation for this fact?  
 A: Notice that
$$\det(c - q^2) = \pm \det\left( \prod_{s \in \Pi_1} s - q^2 \prod_{s \in \Pi_2} s \right).$$
We write the reflection representation in the basis of simple roots, and we order our roots so that $\Pi_1$ comes before  $\Pi_2$. The product $\prod_{s \in \Pi_1} s$ has block form
$$\begin{pmatrix} -\mathrm{Id} & A \\ 0 & \mathrm{Id} \end{pmatrix}$$
where $A$ is the adjacency matrix between $\Pi_1$ and $\Pi_2$. Computing the other product similarly gives
$$\prod_{s \in \Pi_1} s - q^2 \prod_{s \in \Pi_2} s = \begin{pmatrix} -\mathrm{Id} & A \\ 0 & \mathrm{Id} \end{pmatrix} - q^2 \begin{pmatrix} \mathrm{Id} & 0 \\ A^T & -\mathrm{Id} \end{pmatrix} = \begin{pmatrix} -1-q^2 & A \\ -q^2 A^T & 1+q^2 \end{pmatrix}.$$
Left multiplying by the block matrix $\left( \begin{smallmatrix} -q & 0\\0  & 1 \end{smallmatrix} \right)$ and right multiplying by $\left( \begin{smallmatrix} q^{-1} & 0\\0  & 1 \end{smallmatrix} \right)$ turns this into the quantum Cartan matrix you propose.
No insight into any higher context, I'm afraid.
