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Define the permanent of an algebra as the permanent of the Cartan matrix. For simplicity assume all algebras are connected quiver algebras.

Questions:

  1. Let $X_n$ be the set of all algebras with $n$ simple modules having finite global dimension. Is the permanent of algebras in $X_n$ bounded? If yes, what is the bound? Answer is no: See the answer by Pierre-Guy Plamondon.

I suggest the following modified question too:

1.* Let $X_n$ be the set of all representation-finite algebras with $n$ simple modules having finite global dimension. Is the permanent of algebras in $X_n$ bounded? If yes, what is the optimal bound?

  1. Let $Y(Q)$ be the set of algebras with quiver $Q$ having finite global dimension. Is the permanent of such algebras bounded for given $Q$? If yes, what are good bounds. For example, my conjecture in the thread Permanent of Nakayama algebras I says that for $Q$ a circle with $n \geq 2$ points, the answer is $\sum\limits_{k=0}^{\infty}{\frac{k^n}{2^{k+1}}}$.

Of course answers for special cases are welcome.

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    $\begingroup$ Question 2 seems related to an open problem of Happel and Zacharia: for a quiver $Q$, is there a bound on the dimension of all algebras in $Y(Q)$? $\endgroup$ – Pierre-Guy Plamondon Aug 28 '17 at 15:43
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    $\begingroup$ @Pierre-GuyPlamondon A positive answer to the problem of Happel and Zacharia should imply a positive answer for 2. But Im not sure for the other direction. $\endgroup$ – Mare Aug 28 '17 at 16:48
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The answer to question 1 is no, unless $n=1$, where the bound on the permanents is $1$.

For $n=1$, the algebra is either semisimple (so the permanent is $1$) or has infinite global dimension. Thus it suffices to give examples for $n=2$.

Let $A$ be the algebra given by the quiver with two vertices $1$ and $2$, with arrows $x_1, \ldots, x_r$ from $1$ to $2$ and arrow $y$ from $2$ to $1$, subject to the relations $x_iy=0$ for all $i\in \{1, \ldots, r\}$. Then $A$ is of global dimension $2$, and its Cartan matrix is $$ \begin{pmatrix} 1 & 1 \\ r & r+1 \end{pmatrix}, $$ so its permanent is $2r+1$.

I do not know the answer to question 2.

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  • $\begingroup$ Thanks, I added a similar modified question 1.* in my question. $\endgroup$ – Mare Aug 28 '17 at 14:17

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