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Given a finite dimensional algebra $A$ with finite global dimension such that there are only finitely many basic tilting modules. Then every selforthogonal indecomposable module $M$ ( that is a module with $Ext_A^i(M,M)=0$ for all $i \geq 1$) is a direct summand of a tilting module and thus there are only finitely many such indeocmposable selforthogonal modules. See 3.9. in https://arxiv.org/abs/1803.10707  .

Question 1: Is there an effective method to obtain every indecomposable selforthogonal module in a nice way? For example using the GAP-package QPA.

(A nice method would for example be an "Auslander-Reiten" theory for indecomposable selforthogonal modules in such algebras, where one can obtain every such module using certain exact sequences starting from just one such module)

Question 2: Is there an example of an Auslander algebra with infinitely many selforthogonal indecomposable modules?

Given a finite dimensional algebra $A$ with finite global dimension such that there are only finitely many basic tilting modules. Then every selforthogonal indecomposable module $M$ (that is a module with $\mathrm{Ext}_A^i(M,M)=0$ for all $i \geq 1$) is a direct summand of a tilting module and thus there are only finitely many such indecomposable selforthogonal modules. See 3.9 in https://arxiv.org/abs/1803.10707.

Question 1: Is there an effective method to obtain every indecomposable selforthogonal module in a nice way? For example using the GAP-package QPA.

(A nice method would for example be an "Auslander-Reiten" theory for indecomposable selforthogonal modules in such algebras, where one can obtain every such module using certain exact sequences starting from just one such module.)

Question 2: Is there an example of an Auslander algebra with infinitely many selforthogonal indecomposable modules?

Given a finite dimensional algebra $A$ with finite global dimension such that there are only finitely many basic tilting modules. Then every selforthogonal indecomposable module $M$ ( that is a module with $Ext_A^i(M,M)=0$ for all $i \geq 1$) is a direct summand of a tilting module and thus there are only finitely many such indeocmposable selforthogonal modules. See 3.9. in https://arxiv.org/abs/1803.10707 .

Question 1: Is there an effective method to obtain every indecomposable selforthogonal module in a nice way? For example using the GAP-package QPA.

(A nice method would for example be an "Auslander-Reiten" theory for indecomposable selforthogonal modules in such algebras, where one can obtain every such module using certain exact sequences starting from just one such module)

Question 2: Is there an example of an Auslander algebra with infinitely many selforthogonal indecomposable modules?

Im especially interested how to obtain the indecomposable selforthogonal modules quickly in the case of the Auslander algebra of $K[x]/(x^n)$ using QPA.

Given a finite dimensional algebra $A$ with finite global dimension such that there are only finitely many basic tilting modules. Then every selforthogonal indecomposable module $M$ ( that is a module with $Ext_A^i(M,M)=0$ for all $i \geq 1$) is a direct summand of a tilting module and thus there are only finitely many such indeocmposable selforthogonal modules. See 3.9. in https://arxiv.org/abs/1803.10707 .

Question 1: Is there an effective method to obtain every indecomposable selforthogonal module in a nice way? For example using the GAP-package QPA.

(A nice method would for example be an "Auslander-Reiten" theory for indecomposable selforthogonal modules in such algebras, where one can obtain every such module using certain exact sequences starting from just one such module)

Question 2: Is there an example of an Auslander algebra with infinitely many selforthogonal indecomposable modules?

Given a finite dimensional algebra $A$ with finite global dimension such that there are only finitely many basic tilting modules. Then every selforthogonal indecomposable module $M$ ( that is a module with $Ext_A^i(M,M)=0$ for all $i \geq 1$) is a direct summand of a tilting module and thus there are only finitely many such indeocmposable selforthogonal modules. See 3.9. in https://arxiv.org/abs/1803.10707 .

Question 1: Is there an effective method to obtain every indecomposable selforthogonal module in a nice way? For example using the GAP-package QPA.

(A nice method would for example be an "Auslander-Reiten" theory for indecomposable selforthogonal modules in such algebras, where one can obtain every such module using certain exact sequences starting from just one such module)

Question 2: Is there an example of an Auslander algebra with infinitely many selforthogonal indecomposable modules?

Im especially interested how to obtain the indecomposable selforthogonal modules in the case of the Auslander algebra of $K[x]/(x^n)$ using QPA.

Given a finite dimensional algebra $A$ with finite global dimension such that there are only finitely many basic tilting modules. Then every selforthogonal indecomposable module $M$ ( that is a module with $Ext_A^i(M,M)=0$ for all $i \geq 1$) is a direct summand of a tilting module and thus there are only finitely many such indeocmposable selforthogonal modules. See 3.9. in https://arxiv.org/abs/1803.10707 .

Question 1: Is there an effective method to obtain every indecomposable selforthogonal module in a nice way? For example using the GAP-package QPA.

(A nice method would for example be an "Auslander-Reiten" theory for indecomposable selforthogonal modules in such algebras, where one can obtain every such module using certain exact sequences starting from just one such module)

Question 2: Is there an example of an Auslander algebra with infinitely many selforthogonal indecomposable modules?

Im especially interested how to obtain the indecomposable selforthogonal modules quickly in the case of the Auslander algebra of $K[x]/(x^n)$ using QPA.