Classification of algebras of finite global dimension via determinants of certain 0-1-matrices

Question

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?

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

3. (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.

Answer 1

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.

Answer 2

(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.