Is this a counterexample to a conjecture on weakly periodic Coxeter matrices?

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.

• 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. – Timothy Chow Nov 21 at 19:34
• In particular are you sure there are no further conditions on the poset that you've overlooked somewhere along the line? – Timothy Chow Nov 21 at 19:36
• 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. – LSpice Nov 21 at 19:56
• @LSpice Yes, I meant the dimension of the Hom-space. Thank you. – Mare Nov 21 at 20:46
• @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). – Mare Nov 21 at 20:48