Timeline for Algebras where all indecomposable modules are rigid
Current License: CC BY-SA 3.0
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Feb 11, 2017 at 21:52 | history | edited | YCor |
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Feb 11, 2017 at 21:06 | history | edited | Mare | CC BY-SA 3.0 |
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Feb 11, 2017 at 21:06 | comment | added | Mare | The condition on simples isnt so interesting I think, since its not invariant under derived or stable equivalence. So Im mostly interested in a classification up to stable equivalence and one might assume the field to be algebraically closed. I added this. | |
Feb 11, 2017 at 20:26 | comment | added | Jim Humphreys | [Small typo: My first word should have been "There".] Concerning your comment, keep in mind that the group algebra of a finite group often fails to be representation-finite even when your Ext condition is satisfied by simple modules. Maybe restate your question in a little more detail and with more examples? | |
Feb 9, 2017 at 18:26 | comment | added | Mare | Hmm, actually I expect any such algebra to be representation-finite. For group algebras the answer is then (in case this is true) easy: A block has this property iff it is stable equivalent to a Brauer tree algebra with multiplicity one. Another interesting question might be to restrict to simple modules with this property. | |
Feb 9, 2017 at 17:18 | comment | added | Jim Humphreys | The might or might not be such a classification, but it's worth asking. One instance which makes me a bit skeptical is the case of the group algebra in the defining prime characteristic of a finite group of Lie type. While the algebra itself is well-behaved, there are isolated special cases in which a simple module can have nontrivial self-extensions, e.g., finite symplectic groups. Aside from this, there are many selfinjective algebras whose indecomposables are mostly unknown. So it's not easy to visualize a classification of the type you ask about. | |
Feb 9, 2017 at 8:58 | history | asked | Mare | CC BY-SA 3.0 |