I restrict to the elementary problem that is equivalent to give a classification when Morita-Nakayama algebras have finite global dimension (see the end of this post for some background).
A Morita-Nakayama algebra is a tuple $(w,v)$, where $v$ is a non-zero $n$-vector with entries either 0 or 1 and $w$ is a natural number with $2 \leq w \leq n$. We say that two such Morita-Nakayama algebras $(w_1,v_1)$ and $(w_2,v_2)$ are isomorphic in case $w_1=w_2$ and $v_1$ is a cyclic shift of $v_2$. The size $r$ of $v$ is the number of non-zero entries of $v$ and let $x_i$ for $i=1,2,..,r$ be the position of the $i$-th non-zero entry in $v$. For example $[0,1,0,1]$ has size 2 and $x_1=2 , x_2=4$.
Let $T=(w,v)$ be a Morita-Nakayama algebra. We associate 3 matrices to $T$. The first $n \times n$ matrix $A_T=(a_{i,j})$ is defined as follows: We have $a_{i,j}=1$ for $j=i,i+1,...,i+w-1$ modulo $n$ and $a_{i,j}=0$ else.
The second $n \times r$ matrix $B_T$ is defined as having as $l$-th column the 0-1-vector with a 1 in position $x_l$ and zeros else.
The third $r \times n$ matrix $C_T=(c_{i,j})$ is defined by $c_{i,j}=1$, if $j=x_i+w-1$ modulo $n$ and $c_{i,j}=0$ else.
The Cartan matrix $M_T$ of $T$ is then defined as the $(n+r) \times (n+r)$ matrix $M_T:= \left[\begin{matrix} A_T & B_T \\C_T & E_r\end{matrix}\right]$. Here $E_r$ is the identity $r \times r$-matrix.
In general one has $det(M_t)=det(A_T - B_T C_T)$ ,see for example http://djalil.chafai.net/blog/2012/10/14/determinant-of-block-matrices/ , so the problem to calculate the determinant of $M_T$ is reduced to the calculation of a determinant of an $n \times n$ 0-1-matrix of the form $A_T - B_T C_T$.
Here an example: Let $n=4, w=3$ and $v=[0,1,0,1]$. Then $A_T=\left[\begin{matrix}1 & 1 & 1 & 0\\0 & 1 &1 &1 \\ 1 & 0 & 1 & 1 \\ 1 & 1 & 0 & 1\end{matrix}\right]$, $B_T=\left[\begin{matrix} 0 & 0 \\ 1 & 0 \\ 0 & 0 \\ 0 & 1\end{matrix}\right]$ and $C_T=\left[\begin{matrix} 0 & 0 & 0 & 1\\ 0 & 1 & 0 &0 \end{matrix}\right]$. One can check that in this case $M_T$ has determinant equal to one.
We say that $T=(w,v)$ has finite global dimension in case $M_T$ has determinant equal to one. Note that the determinant of $M_T$ is always positive and thus this is equivalent to the condition that $M_T$ is invertible over $\mathbb{Z}$.
Questions: 1. Is there a nice condition/classification when $T=(w,v)$ has finite global dimension for a given $w$? Is there a closed formula for the determinant of $M_T$ in general?
- How many such tuples of finite global dimension exist for a given $n$ up to isomorphism?
- (edited, special case of 1.) Given n, for which $w<n$ (we can assume $w<n$ since for $w=n$ it always exists) does exist a tuple $(w,v)$ such that $M_T$ has determinant 1? Computer experiments suggest for $n \geq 4$ the answer is that such $T$ exists if and only if $w \in \{2,3,..., \frac{n+1}{2} \}$ in case $n$ is odd and $w \in \{2,3,..., \frac{n+2}{2} \}$ in case $n$ is even.
Here are some partial results:
-For general $v$ and $w=2$ all tuples have finite global dimension and thus we can assume $w>2$.
-For $w>2$ and $v$ having size 1, $(w,v)$ has finite global dimension if and only if $w$ divides $n+1$.
-For $w=n \geq 3$, the only $v$ with finite global dimension are those with exactly one 0 as an entry.
-For $n \geq 5$ and $w=n-1$, there seem to be no $v$ with finite global dimension.
-For $w=3$ the problem seems already more complicated. Here is the list of $v$ for $n=6$ up to isomorphism where the global dimension is finite: [ [ 0, 0, 0, 0, 1, 1 ], [ 0, 0, 0, 1, 0, 1 ], [ 0, 0, 1, 0, 1, 1 ], [ 0, 0, 1, 1, 0, 1 ], [ 0, 1, 0, 1, 0, 1 ], [ 0, 1, 0, 1, 1, 1 ], [ 0, 1, 1, 0, 1, 1 ], [ 0, 1, 1, 1, 1, 1 ] ]
I did not really calculate determinants since I had linear algebra many years ago, so I am not very experienced. Maybe this problem has an easy solution that I miss.
Partial solutions would also be interesting, for example the case $w=3$ in general or a general solution for vectors $v$ having size 2.
Background: I noted that the classification of Morita-Nakayama algebras (=Nakayama algebras that are also Morita algebras in the sense of https://www.sciencedirect.com/science/article/pii/S0021869313001002 ) with finite global dimension reduces to a nice problem on determinants of 0-1-matrices using a derived equivalence together with the main result in https://www.ams.org/journals/proc/1985-095-02/S0002-9939-1985-0801315-7/S0002-9939-1985-0801315-7.pdf that the global dimension of such algebras is finite iff their Cartan determinant is equal to one.
The tuple $T=(w,v)$ corresponds to a selfinjective Nakayama algebra algebra $A$ with Loewy length $w$ and the generator $N=A \oplus e_{x_1} J^{w-1} \oplus ... \oplus e_{x_r} J^{w-1}$ when $J$ is the Jacobson radical of $A$. The matrix $M_T$ is then the Cartan matrix of $B=End_A(M)$.
edit: I made a bounty for this question. In case there is no complete answer at the end of the bounty period, I can also award it for some interesting special cases like $w=3$ or $v$ having exactly two non-zero entries.