Timeline for Is the Cartan permanent odd for finite global dimension?
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
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| Mar 14, 2018 at 14:19 | vote | accept | Mare | ||
| Mar 14, 2018 at 14:18 | answer | added | Philipp Lampe | timeline score: 6 | |
| Mar 14, 2018 at 13:33 | comment | added | Mare | @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. | |
| Mar 14, 2018 at 13:31 | history | edited | Mare | CC BY-SA 3.0 |
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| Mar 14, 2018 at 9:36 | comment | added | Philipp Lampe | 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. | |
| Mar 14, 2018 at 9:29 | comment | added | Fedor Petrov | Modulo 2 the determinant and permanent are the same, or what do I miss? | |
| Mar 14, 2018 at 9:16 | history | asked | Mare | CC BY-SA 3.0 |