Timeline for Rep infinite, but $\tau$-tilting finite
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 11, 2023 at 20:44 | vote | accept | It'sMe | ||
| May 11, 2023 at 20:17 | comment | added | Dave Benson | To expand, for a local algebra the only brick is the one dimensional module. | |
| May 11, 2023 at 20:06 | comment | added | Dave Benson | Those aren't bricks. | |
| May 11, 2023 at 20:05 | comment | added | It'sMe | But even then I don't think this is going to be $\tau$-tilting finite, because by the result of Demonet, Iyama and Jasso, $\tau$-tilting finite is same as brick-finite (i.e., $\mathrm{mod}-A$ has finitely many bricks $B$ ($\mathrm{End}_{A}(B)\cong\mathbb{K}$), up to isomorphism. But in the above quiver if you take the dimension vector to be $(2)$, then you have an infinite family of non-isomorphic bricks, and hence it is brick infinite, and therefore $\tau$-tilting infinite. | |
| May 11, 2023 at 19:56 | comment | added | Dave Benson | No, it isn't gentle. Read the introduction to Plamondon. Second paragraph. | |
| May 11, 2023 at 19:46 | comment | added | It'sMe | So your example is basically the algebra given by one vertex and two loops quiver, with radical square zero relation. But this algebra is representation infinite and hence by the result of Plamondon it is $\tau$-tilting infinite as well. But I'm looking for an example that is rep infinite, but $\tau$-tilting finite. | |
| May 11, 2023 at 19:03 | vote | accept | It'sMe | ||
| May 11, 2023 at 19:24 | |||||
| May 11, 2023 at 18:26 | history | undeleted | Dave Benson | ||
| May 11, 2023 at 18:26 | history | deleted | Dave Benson | via Vote | |
| May 11, 2023 at 18:26 | history | answered | Dave Benson | CC BY-SA 4.0 |