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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?

  1. How many such tuples of finite global dimension exist for a given $n$ up to isomorphism?

  2. (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?

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.

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  • $\begingroup$ For w and n having a sufficiently large common divisor, A has determinant 0. If BC is sparse enough, A-BC will also have determinant 0. Can you say how many ones BC will have? Gerhard ""Hopefully Providing A Useful Fact" Paseman, 2019.03.05. $\endgroup$ – Gerhard Paseman Mar 5 at 16:15
  • $\begingroup$ In fact, one can extend this: Say kw=3n, so I can pick k rows, do row addition, and transform A to a matrix with one row being all 3's. Then A will have a determinant which is 0 mod 3. Because A is cyclic, if BC has fewer than n/k ones, the same will be true of det A-BC. This may help with some of your computations. Gerhard "Just Stay Away From K" Paseman, 2019.03.05. $\endgroup$ – Gerhard Paseman Mar 5 at 16:54
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I've decided to expand upon the observations on the comments.

I haven't wrapped my head yet around the idea that $A-BC$ is a 0-1 matrix. Thus I assume you or someone else has a proof for that part. Then this matrix differs from $A$ in at most $r$ rows. But if $r$ is less than $n/k$, where $k$ is smallest such that $kw$ is greater than and a multiple of $n,$ then (using that $A$ is cyclic) there is a set of $k$ rows of $A-BC$ which add up to a nontrivial multiple of the row of all ones, and thus $A$ and $A-BC$ have determinants which are not one. This handles some of the cases and reveals some of the number theory going on here.

Gerhard "Number Theory To The Rescue?" Paseman, 2019.03.05.

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  • $\begingroup$ If we have kw = n, then it will suffice for r+1 to be less than w, as then we can find 2 disjoint sets of k rows each adding up to the all ones vector, giving a zero determinant for A - BC. It might be useful to look at A for w coprime to n. In this case small r might lead to large determinants. Gerhard "Looking At Combinatorial Matrix Theory" Paseman, 2019.03.05. $\endgroup$ – Gerhard Paseman Mar 5 at 21:38
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(Not a solution, just a reformulation and a conjecture for $w=3$)

(1) Remark: the question above may equivalently be stated as follows:

Let $Z$ be the matrix of the cyclic shift (the companion matrix of $X^n-1$), and for $\mathbf{v}\in \{0,1\}^n$ let $\mathrm{diag}(\mathbf{v})$ be the diagonal matrix with $\mathbf{v}$ on the diagonal, and $M_\mathbf{v}:=I + Z+ \ldots + Z^{w-1}-\mathrm{diag}(\mathbf{v})$.
For which $\mathbf{v}$ is $\det(M_\mathbf{v})=(-1)^{(n-1)(w-1)}$?

Proof: let $^t$ denote transposition. By definition $A_T^t=I+Z+\dots+Z^{w-1}$. To each $v_j$ associate the column-vector $\mathbf{v}_j:=v_j\mathbf{e}_j$, where $\mathbf{e}_j$ is the $j$-th standard column vector. Then the columns of $B_t$ are $\mathbf{v}_{x_1},\ldots,\mathbf{v}_{x_r}$, and the rows of $C_T$ are $(Z^{w-1}\mathbf{v}_{x_1})^t,\ldots,(Z^{w-1}\mathbf{v}_{x_r})^t$. Therefore $C_T^tB_T^t=Z^{w-1}B_TB_T^t=Z^{w-1}\mathrm{diag}(\mathbf{v})$. Thus $$\det(A_T^t - C_T^t B_T^t)=\det\big(I+Z+\ldots+Z^{w-1}-Z^{w-1}\mathrm{diag}(\mathbf{v})\big)$$ and since $\det(Z)=(-1)^{n-1}$ and $Z^{-1}=Z^t$ this may equivalently be rewritten as $$\det(A_T - B_T C_T)=(-1)^{(n-1)(w-1)}\det\big(I+Z+\ldots+Z^{w-1}-\mathrm{diag}(\mathbf{v})\big)$$ End proof

(2) a conjecture for $w=3$

Notation: call a subset of $[n]:=\{1,\ldots,n\}$ separated if does not contain two (cyclically) adjacent elements, let $S_j(n):=\{ M\in [n]\;:\, |M|=j, M \mbox{ is separated}\}$ denote the set of separated subsets of $[n]$ with $j$ elements, and for $i=1,\ldots,n$ let $d_i:=1-v_i$.

Conjecture: for $w=3$ and $n\geq 3$ the determinant is $$\det(M_{\mathbf{v}})=\sigma_n(\mathbf{v})+\sum_{j=0}^{n-1} (-1)^{n-1-j} \sigma_j(\mathbf{v})$$ where $\sigma_0(\mathbf{v})=\det(Z+Z^2)=\bigg\{\begin{array}{cr} 2 & \mbox{ for odd } n\\ 0 & \mbox{ for even } n\end{array}$

$\sigma_1(\mathbf{v})=d_1+\ldots+d_n$, $\sigma_n(\mathbf{v})=\prod_{i=1} d_i$, and for $2\leq j \leq n$ $$\sigma_j(\mathbf{v})= \sum_{(k_1,\ldots,k_j)\in S_j(n)}\prod_{i=1}^j d_{k_i}$$

Comments:

(1) I have only checked it up to $n=12$. For determinant experts the proof is probably easy, but I don't see an elegant way to prove it.

(2) Thus if $\mathbf{v}\neq \mathbf{0}$ and $n$ is even the determinant $\chi(\mathbf{v}):=\det(A_T - B_T C_T)$ has conjecturally the form of an Euler characteristic

$$\chi(\mathbf{v})=A_0(\mathbf{v})- A_1(\mathbf{v})+A_2(\mathbf{v})-\ldots $$ where the vertices are the positions of the zeroes in $\mathbf{v}$, and the $j$-dimensional objects are the separated subsets of cardinality $j+1$ of these positions. The determinants may therefore have appeared elsewhere.

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  • $\begingroup$ Thank you, that looks very interesting. Maybe it would be a good idea to post the conjecture in a new thread, since here it likely wont get much attention? $\endgroup$ – Mare Mar 19 at 12:47
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    $\begingroup$ You're welcome. Yes, I think it will get more attention with the tags "linear algebra" and "determinants". Feel free to post it - in a sense it is your conjecture (as follow-up of this post). $\endgroup$ – esg Mar 19 at 18:04
  • $\begingroup$ It would feel a little weird to post your conjecture. So it is probably better when you post it ;) $\endgroup$ – Mare Mar 19 at 18:47
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    $\begingroup$ I have just corrected a sign, but nothing essential has changed. The structure of the formula is easy to explain, but I don't see why only the factors +1 and -1 appear. If I don't have a new idea tomorrow I'll post the conjecture. $\endgroup$ – esg Mar 20 at 19:38
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    $\begingroup$ Yes, you are right. Apologies for being sloppy. I'll correct it. $\endgroup$ – esg Mar 22 at 15:37

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