Is there a classification of selfinjective algebras having $Ext^{1}(X,X)=0$ for every indecomposable module X? Examples include trivial extensions of representation-finite hereditary algebras. One might assume algebraically closed field and Im mostly interested in a classification up to stable equivalence then.
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$\begingroup$ 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. $\endgroup$– Jim HumphreysCommented Feb 9, 2017 at 17:18
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$\begingroup$ 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. $\endgroup$– MareCommented Feb 9, 2017 at 18:26
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$\begingroup$ [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? $\endgroup$– Jim HumphreysCommented Feb 11, 2017 at 20:26
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$\begingroup$ 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. $\endgroup$– MareCommented Feb 11, 2017 at 21:06
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