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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 $n \leq 6$).

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    $\begingroup$ 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 $n>3$. $\endgroup$
    – Jeremy Rickard
    Commented Sep 26, 2018 at 7:51
  • $\begingroup$ @JeremyRickard Thanks, maybe being representation-finite is a good condition? At least it might exclude the radical square zero example. $\endgroup$
    – Mare
    Commented Sep 26, 2018 at 7:59
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    $\begingroup$ The radical square zero example for $Q=A_3$ has finite representation type but is not derived equivalent to $KQ$. $\endgroup$
    – Jeremy Rickard
    Commented Sep 26, 2018 at 8:02
  • $\begingroup$ @JeremyRickard Ah right because the quiver of this radical square zero algebra is not a tree. $\endgroup$
    – Mare
    Commented Sep 26, 2018 at 8:41

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