# Does a finite dimensional algebra having a Cartan matrix with determinant 1 imply finite global dimension (possibly with more hypotheses)?

Let $A$ be your favorite finite dimensional algebra, and $P_i$ be a sets of representatives for the indecomposible projectives (or PIMs, if you like). Then we have the Cartan matrix $C$ of the algebra, whose entries are $\dim Hom(P_i, P_j)$. You can think of this as the matrix of the Euler form on the Grothendieck group $K^0(A-pmod)$ of projective $A$-modules.

Now, if the algebra $A$ has finite global dimension, then we can define the classes of the simple heads of these $L_i$ in $K^0(A-pmod)$ as integer linear combinations of the $P_i$'s, and $[P_i]$ and $[L_i]$ are dual bases in the Euler form. That is, the matrix $C$ is integer valued and has integer-valued inverse, i.e. it has determinant 1.

To what degree is the converse of this true? Is there a weaker hypothesis than finite global dimension itself such that $det(C)=1$ and that hypothesis will imply finite global dimension?

The application I have in mind for this is a little more complex. I'd like to consider a graded version of this question. So, let $A$ be a graded algebra such that each degree is finite dimensional (and let say the appearing gradings are bounded below). The the graded version of $C$ is well-defined in $\mathbb{Z}((q))$, and similarly, if every simple has a resolution by projectives where only finitely many projectives generated in a given degree appear, this implies that this matrix has an inverse in $\mathbb{Z}((q))$, that is determinant with leading coefficient 1.

The same question as above: can I use a hypothesis like the graded Cartan matrix having determinant with integral leading coefficient to conclude the existence of such a projective resolution?

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I know this is an old post, but since I have been working on similar questions it is too tempting to leave a comment. Graded Cartan determinant=1 does not imply finite global dimension, counterexample in arxiv.org/abs/0803.3550. (Interestingly, the algebra in the counterexample can be regraded in a way that makes the determinant different from 1.) For finite dimensional Koszul algebras, however, graded Cartan determinant=1 is equivalent to finite global dimension, proof in the same paper. – Dag Oskar Madsen Oct 30 '11 at 21:56