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$\DeclareMathOperator{\Ext}{\operatorname{Ext}}$The first question is about group algebras:

Question 1: Let $A=kG$ be a group algebra (with $G$ finite) and let $M$ be an indecomposable $A$-module. Does $\Ext_A^1(M,M) \neq 0$ imply $\Ext_A^2(M,M) \neq 0$ or even $\Ext_A^i(M,M) \neq 0$ for all $i>0$? This is true in case $A$ is representation-finite.

The next question is about representation-finite symmetric algebras: Let $A$ be a finite dimensional symmetric quiver algebra. In https://folk.ntnu.no/oyvinso/Papers/symmetric.pdf corollary 2.6. it was noted that for a simple module $M$, we have $Ext^1(M,M) \neq 0$ implies $\Ext^2(M,M) \neq 0$. Now I noted that for a general indecomposable module $M$ over a representation-finite symmetric algebra $\Ext^1(M,M) \neq 0$ implies $\Ext^2(M,M) \neq 0$ and for Brauer tree algebras $\Ext^1(M,M) \neq 0$ even implies $\Ext^i(M,M) \neq 0$ for all $i >1$ (which is not true for general representation-finite symmetric algebras).

Question 2: I used the classification of representation-finite symmetric algebras up to stable equivalence to prove this (it was rather ugly). Is there a proof not using the classification, that might even work for general Artin algebras (not just quiver algebras)?

The result is not true for general symmetric algebras (see for example Liu-Schulz example algebras), but in case my question has a positive answer, maybe one expect this to hold for a slightly more general class than representation-finite symmetric algebras.

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    $\begingroup$ Are you assuming that $G$ is finite or something? Otherwise $G=\mathbb{Z}$ and $M=k$ with the trivial action seems like an easy counterexample to Question 1. $\endgroup$
    – Simon Wadsley
    Commented Oct 24, 2019 at 9:38
  • $\begingroup$ @SimonWadsley Yes, G is finite. $\endgroup$
    – Mare
    Commented Oct 24, 2019 at 11:04

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