# Derived equivalences and the Coxeter polynomial

Let $$A$$ be a quiver algebra with an acyclic quiver and primitive idempotents $$e_i$$. The Cartan matrix $$C_A$$ of $$A$$ is defined as the matrix with entries $$dim(e_i A e_j)$$ and the Coxeter matrix $$\phi_A$$ of $$A$$ is defined as $$\phi_A=-C_A^{-1} C_A^T$$. The Coxeter polynomial of $$A$$ is defined as characteristic polynomial of $$\phi_A$$. The Coxeter polynomial is a derived invariant and thus derived equivalent algebras share the same Coxeter polynomial.

Is the following true:

A is derived equivalent to a path algebra $$KQ$$ of Dynkin type if and only if it has the same Coxeter polynomial as $$KQ$$?

In case the answer is no, is this true in case $$A$$ is additionally a representation-finite or a Nakayama algebra with a linear quiver?

I think this is at least true when $$A$$ is a Nakayama algebra with a linear quiver (corresponding to a Dyck path) and there is computational way using trivial extensions and representation-finiteness of those trivial extensions to test it, but it gets very ugly when $$Q$$ is of type $$D_n$$ ($$E_n$$ can be done with the computer and indeed, for $$E_6$$ it is true. It is also true for Dynkin type $$A_n$$ and $$n \leq 8$$ and Dynkin type $$D_n$$ for $$n \leq 6$$).

• You definitely need to restrict to Nakayama algebras (or at least have some extra conditions on $A$). For example, if $A$ is the radical square zero algebra with the same Cartan matrix as $KQ$, then it’s not usually derived equivalent to $KQ$: for $Q=A_n$ with all arrows in the same direction, $A$ has infinite representation type if $n>3$. – Jeremy Rickard Sep 26 '18 at 7:51
• @JeremyRickard Thanks, maybe being representation-finite is a good condition? At least it might exclude the radical square zero example. – Mare Sep 26 '18 at 7:59
• The radical square zero example for $Q=A_3$ has finite representation type but is not derived equivalent to $KQ$. – Jeremy Rickard Sep 26 '18 at 8:02
• @JeremyRickard Ah right because the quiver of this radical square zero algebra is not a tree. – Mare Sep 26 '18 at 8:41