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

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 $$w (we can assume $$w 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.

• 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. – Gerhard Paseman Mar 5 '19 at 16:15
• 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. – Gerhard Paseman Mar 5 '19 at 16:54

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.

• 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. – Gerhard Paseman Mar 5 '19 at 21:38

(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 $$2\leq j \leq n$$ $$\sigma_j(\mathbf{v})= \sum_{(k_1,\ldots,k_j)\in S_j(n)}\prod_{i=1}^j d_{k_i}$$

(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.