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It seems I found an easy counterexample to a conjecture, but most often I did a stupid mistake or misunderstood the conjecture when this happens. Maybe someone can see whether this indeed is a counterexample. (You just need to know what a poset is to understand it and you can ignore finite dimensional algebra stuff.) Maybe a counterexample is also already known since the article is from 2005 in which case a reference would be interesting.

Let $A=kQ/I$ be a quiver algebra with acyclic quiver $Q$. The Cartan matrix $C=C_A$ of $A$ is defined as the matrix with entries $(c_{i,j}=\dim \operatorname{Hom}_A(P_j,P_i))$ when $P_i$ denote the indecomposable projective $A$-modules. $c_{i,j}$ is just the number of non-zero paths that start at $i$ and end at $j$ (modulo the relations $I$ of course). You can for simplicity think of $A$ as being the incidence algebra of a finite poset $P$, then $C$ is just the matrix with $c_{i,j}=1$ when $i \leq j$ and $c_{i,j}=0$ else with $i,j \in P$.

The Coxeter matrix $\Phi_A$ is defined as $-C^{-1}C^T$. According to Sato - Periodic Coxeter matrices and their associated quadratic forms, $\Phi_A$ is called weakly periodic if $(\Phi_A^m-\operatorname{id})$ is nilpotent for some $m>0$ and by Theorem 2.6. this is equivalent to the statement that all eigenvalues of $\Phi_A$ have absolute value 1. Define the quadratic form $q(x)=x^T C x$ associated to $A$ and say that $q$ is non-negative if $q(x) \geq 0$ for all $x \in \mathbb{Q}^n$ ($n$ being the number of simple $A$-modules). The Coxeter polynomial is defined as the characteristic polynomial of $\Phi_A$.

According to Theorem 3.4. in the article when $q(x)$ is non-negative definite, $\Phi_A$ is weakly periodic.

Now one can find the following conjecture in the article:

Conjecture: The quadratic form with respect to $C$ of $A$ is non-negative definite.

But the computer gives me many easy counterexample to the conjecture so I probably have a thinking error. Namely let $L$ be the poset (it is a lattice) with cover relations $1<5<6$, $1<2<6$, $1<3<6$, $1<4<6$, which has 6 elements, and let $A$ be the incidence algebra of $L$. According to GAP the Coxeter matrix is given by the matrix with rows:

[ [ 0, -2, -2, -2, -2, -3 ], [ 0, 0, 1, 1, 1, 1 ], [ 0, 1, 0, 1, 1, 1 ], [ 0, 1, 1, 0, 1, 1 ], [ 0, 1, 1, 1, 0, 1 ], 
  [ -1, -1, -1, -1, -1, -1 ] ]

The Coxeter polynomial is given by $x^6+x^5-5x^4-10x^3-5x^2+x+1$ and factors into the polynomials $[ x+1, x+1, x+1, x+1, x^2-3x+1 ]$. But the polynomial $x^2-3x+1$ has zeros which do not have absolute value 1. Thus it seems this gives a counterexample to the conjecture.

There are also representation-finite examples: Namely let $A$ be the Nakayama algebra with Kupisch series $[ 2, 3, 4, 3, 3, 2, 2, 1 ]$. It has irreducible Coxeter polynomial equal to $x^8+x^7-x^6-4x^5-5x^4-4x^3-x^2+x+1$, which has zeros with absolute value not equal to one.

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  • $\begingroup$ I don't think it's possible to understand what you've said if all I know is what a poset is. I personally happen to know what a quiver algebra and an indecomposable projective module is, but maybe you could rewrite your post so that indeed, someone who knows only what a poset is can assess your claim. Then you can add the representation-theoretic background at the end. $\endgroup$ – Timothy Chow Nov 21 at 19:34
  • $\begingroup$ In particular are you sure there are no further conditions on the poset that you've overlooked somewhere along the line? $\endgroup$ – Timothy Chow Nov 21 at 19:36
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    $\begingroup$ I proofread to fix what seemed like some obvious typos, but there was one place where I guessed. You put $c_{i, j} = Hom_A(P_i, P_j)$, but, since the matrix is surely meant to be integer-valued, I assumed you meant $c_{i, j} = \dim \operatorname{Hom}_A(P_i, P_j)$. I edited accordingly. $\endgroup$ – LSpice Nov 21 at 19:56
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    $\begingroup$ @LSpice Yes, I meant the dimension of the Hom-space. Thank you. $\endgroup$ – Mare Nov 21 at 20:46
  • $\begingroup$ @TimothyChow No there are no conditions on the poset other than being finite (since every incidence algbra is a triangulated artinian ring as in the conjecture in the article). $\endgroup$ – Mare Nov 21 at 20:48

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