Given a connected quiver algebra $A$ that is representation finite, the Auslander algebra $B_A$ of $A$ is defined as the endomorphism ring of the direct sum of each indecomposable $A$-module.

It is stated on page 232 in the book by Auslander-Reiten-Smalo that the quiver algebra $KQ/I$ of $B_A$ is given by $Q$ being the opposite Auslander-Reiten quiver of $A$ and $I$ are the mesh relations in case the field is algebraically closed of characteristic different from 2.

Question 1: Does this description still hold over finite fields of characteristic different from 2? (is there a reference in case the answer is yes?) Are there examples where this fails in case the field is not algebraically closed? Are then the relations still quadratic?

Question 2: Is there an existing method in the GAP-package QPA to obtain the Auslander algebra of an algebra $A$ quickly by quiver and relations in case one has a list of all indecomposable $A$-modules (which are easily obtainable for example for hereditary algebras of Dynkin type)?

Question 3: Is there a method to check in QPA whether a given quiver algebra is quadratic , that is isomorphic to $KQ/I$ with quadratic $I$ and obtain such $I$? For example calculating the Auslander algebra of the Nakayama algebra with Kupisch series $[3,3]$ as the endomorphism ring of all indecomposable modules gives a presentation $KQ/I$ with non-quadratic $I$ but it there should be one with quadratic $I$ and Im not sure how to obain it via QPA(although I know how to obtain it theoretically in this easy example).

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    $\begingroup$ "Crawley-Boevey: Matrix reductions for artinian rings and an application to rings of finite representation type" seems related to Question 1 (but is more general). $\endgroup$ – Julian Kuelshammer Apr 26 '19 at 10:52

Question 3: QPA can for a given admissible ideal $I$ in a path algebra $kQ$ find a minimal generating set and also check if these generators are quadratic. For the requested example it would look like this:

gap> A := NakayamaAlgebra(GF(3),[3,3]);;
gap> S1 := SimpleModules(A)[1];;
gap> pred := PredecessorsOfModule(S1,7)[1];;
gap> pred := Unique(Flat(pred));
[ <[ 1, 0 ]>, <[ 1, 1 ]>, <[ 0, 1 ]>, <[ 2, 1 ]>, <[ 1, 1 ]>, <[ 1, 2 ]> ]
gap> M := DirectSumOfQPAModules(pred);;
gap> B := EndOfModuleAsQuiverAlgebra(M)[3];;
gap> kQ := OriginalPathAlgebra(B);;
gap> I := Ideal(kQ, RelationsOfAlgebra(B));;
gap> IsAdmissibleIdeal(I);
gap> mingens := MinimalGeneratingSetOfIdeal(I);
[ (Z(3)^0)*a1*a7, (Z(3)^0)*a3*a1+(Z(3))*a4*a5, (Z(3)^0)*a6*a2+(Z(3))*a7*a8, (Z(3)^0)*a8*a3 ]
gap> IsQuadraticIdeal(mingens);

Note that this is using the latest bug fix of QPA.

For Question 2, see the above. I don't know of any other method.

Best regards, the QPA-team.

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Question 1 is not true in general. In characteristic 2 there are two symmetric representation-finite symmetric algebras with the same Auslander-Reiten quiver that are not isomorphic so one algebra has its Auslander algebra not given by mesh relations (and the relations are not even quadratic).

Question 2: There is now such an algorithm for standard representation-finite algebras, but not for non-standard. Not sure how to describe all non-standard ones and their Auslander algebra.

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