Timeline for Classification of commutative Frobenius algebras
Current License: CC BY-SA 4.0
14 events
| when toggle format | what | by | license | comment | |
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| Jul 10, 2021 at 11:59 | comment | added | Mare | It might be (by computer experiments) that without the length 3 condition the following is true: The algebra is finite dimensional if and only if the rank of the matrix is larger than 1. And then the algebra is Frobenius if and only if the rank is full as in your proof. | |
| Jul 10, 2021 at 11:54 | comment | added | Benjamin Steinberg | Maybe when A is invertible you can use the inverse of A to get to the case of the identity. That's why I said you might be able to change the quiver presentation to get one that works over all fields | |
| Jul 10, 2021 at 11:52 | comment | added | Benjamin Steinberg | The vector space dimensional depends only on n because all the monoids have the same cardinality. You have the identity the n loops and the element z. The monoids are not isomorphic. If course it's possible the algebras are. | |
| Jul 10, 2021 at 11:51 | comment | added | Mare | I forgot to say that I just looked at the case where the determinant is non-zero. In the other cases, one needs the length 3 zero condition. | |
| Jul 10, 2021 at 11:49 | comment | added | Benjamin Steinberg | For the specific case of $J_n-I_n$ the paths of length 3 are a consequence of the other relations. I guess this should be true as long as you have at least one zero in each row.if you have the all ones matrix it is not a consequence. | |
| Jul 10, 2021 at 11:47 | comment | added | Mare | Also interesting: If instead of $x_i x_j=0$ we impose the relations $x_i x_j x_i=0$, then we will get algebras that are not always finite dimensional (but most of the times) and not always Frobenius. | |
| Jul 10, 2021 at 11:46 | comment | added | Mare | I entered your construction in the computer but I am confused about the output. They are indeed all Frobenius, but the computer suggests that it is not necessary to impose the condition that length 3 paths vanish (this is automatic). Furthermore all algebras have the same vector sapce dimension so most of them seem to be isomorphic. (of course I might have a programming mistake, I will check it later again). | |
| Jul 10, 2021 at 11:04 | comment | added | Benjamin Steinberg | In general of you have a quiver presentation with integral relations, if the algebra is Frobenius over the rationals it will be frobenius over all but finitely many positive characteristics because frobenius's paratrophic determinant is a polynomial over the integers and so either it vanishes in characteristic 0 or only over finitely many primes characteristics. If it is not frobenius in characteristic 0 it is never frobenius | |
| Jul 10, 2021 at 11:00 | comment | added | Mark Wildon | Nice construction! | |
| Jul 10, 2021 at 10:47 | comment | added | Benjamin Steinberg | The case i=j is allowed. | |
| Jul 10, 2021 at 10:46 | comment | added | Mare | Thanks, I missunderstood that condition. | |
| Jul 10, 2021 at 10:41 | comment | added | Benjamin Steinberg | No you also have the relations $x_1^2=x_2^2$. All nonzero paths of length 2 are equal. That element gives your simple socle | |
| Jul 10, 2021 at 8:41 | comment | added | Mare | Thanks, so for the 2x2 identity matrix, we get the commutative algebra with relations $x_1 x_2=0, x_2 x_1=0$ and all of length 3 are 0. But this is not a Frobenius algebra even over the rationals. | |
| Jul 10, 2021 at 1:32 | history | answered | Benjamin Steinberg | CC BY-SA 4.0 |