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Given a matrix with natural numbers $\geq 0$ as entries and having determinant equal to one and positive diagonal entries. Is it the Cartan matrix of a finite dimensional algebra of finite global dimension? If not, can the matrices having this property be classified?

The Cartan matrix of a finite dimensional algebra is defined as the matrix having entries $c_{i,j}:=$dimension of $(Hom_{A}(P_i,P_j))$ over the divison ring $D_i$, where $D_i:=End_A(S_i)$. Here $P_i$ are the indecomposable projective modules and $S_i$ the simple modules which are the top of $P_i$.

We can look also at small cases for $n=1,2,3,...$ . I begin: For $n=1$ all matrices can be realised, since the only one is $[1]$, which is realised by any division ring.

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    $\begingroup$ You at least need the diagonal entries to be strictly positive. $\endgroup$
    – Jeremy Rickard
    Aug 30, 2017 at 14:07

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I don't know a classification, but as well as needing strictly positive entries on the main diagonal (as I pointed out in a comment), you also need to rule out matrices of the form $$\pmatrix{A&B\\0&C}$$ where $A$ and $C$ are not Cartan matrices of algebras with finite global dimension: for example, many (possibly all) matrices with determinant $-1$.

In particular, a complete classification would require settling the Cartan determinant conjecture.

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  • $\begingroup$ Thanks, I leave it open as the question might be still interesting when looking at small $n$ corresponding to $2 \times 2$ or $3 \times 3$-matrices. $\endgroup$
    – Mare
    Aug 30, 2017 at 15:21

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