Define the Cartan permanent of a finite dimensional algebra as the permanent of the Cartan matrix. Is the Cartan permanent of a finite dimensional algebra with finite global dimension always an odd integer? This is true for acyclic quiver algebras and experiments suggest that it is also true for all Nakayama algebras.
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4$\begingroup$ Modulo 2 the determinant and permanent are the same, or what do I miss? $\endgroup$– Fedor PetrovCommented Mar 14, 2018 at 9:29
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4$\begingroup$ A theorem of Eilenberg asserts that the Cartan determinant of a finite-dimensional algebra of finite global dimension is equal to $\pm 1$. Hence determinant and permanent are congruent to $1$ modulo $2$. The reference is S. Eilenberg, Algebras of cohomologically finite dimension, Comment. Math. Helv. 28 (1958), 310–319. $\endgroup$– Philipp LampeCommented Mar 14, 2018 at 9:36
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$\begingroup$ @PhilippLampe Thanks (also to Fedor Petrov), can you turn this into an answer? I was aware that the cartan determinant is either 1 or -1 for finite global dimension (this can be for example also be found in the textbook of assem, simson and skowronski), but I was too blind to see the easy mod 2 argument. $\endgroup$– MareCommented Mar 14, 2018 at 13:33
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A theorem of Eilenberg asserts that the Cartan determinant of a finite-dimensional algebra of finite global dimension is equal to $\pm 1$. Hence the Cartan determinant and the Cartan permanent are both congruent to $1$ modulo $2$ in this situation.
The result is Proposition 21 in the following reference.
- Samuel Eilenberg, Algebras of cohomologically finite dimension, Comment. Math. Helv. 28 (1954), no.1, 310--319.
answered Mar 14, 2018 at 14:18