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Let $(Q, I)$ be a bound quiver such that $A=KQ/I$ has infinite global dimension.

I want to ask the following questionss:

(1) Is the Cartan matrix $C_A$ of $A$ invertible in the matrix ring $M_n(Z)$?

(2) what are the relations between the Cartan matrix ( or Coxeter matrix) of a finite dimensional algebra $B$ and the global dimension of $B$? ( Here $B$ may have infinite global dimension)

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(1) No. Example: $$ \begin{array}{ccc} & \alpha & \\ 1 & \rightleftarrows & 2 \\ & \beta & \end{array}$$ with all paths of length 3 = 0. This is a symmetric algebra wigh Loewy structure $$ \begin{array}{ccc} 1 & & 2\\ 2 & \oplus & 1 \\ 1 & & 2 \end{array}$$ and has infinite global dimension. Its Cartan matrix is $$ \left(\begin{array}{cc} 2 & 1 \\ 1 & 2 \end{array}\right) $$ whose determinant is not 1, hence not inveritble in $\mathrm{Mat}_2(\mathbb{Z})$.

(2) In general, I don't know. But this is (partly) known for cellular algebras. (And I think BGG algebras, which always is of finite gloabl dimension, always have Cartan matrix with determinant 1, but I am not 100% sure, nor do I have any reference) See C.C.Xi's lecture notes. The result is: the global dimension of a cellular algebra is finite if and only if the Cartan matrix has determinant 1.

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Thank you very much, Aaron Is there a finite dimensional algebra $A$ satisfying the following conditions: (1) The Cartan matrix of $A$ is not invertible (2) $A$ has finite globle dimension. – Aimin Xu Nov 28 '12 at 7:31
I can't come up with any example yet, so I would say I don't know. For most interesting algebras (to me), i.e. the class of directed algebras, hereditary algebras, quasi-hereditary algebras. They all have finite global dimension with invertible Cartan matrix. (Also explain why it is hard to come up with one such example...) – Aaron Nov 28 '12 at 14:33

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