Quiver and relations for a quotient algebra via QPA

Question

Given a finite dimensional quiver algebra $B$ and a basic idempotent $e$. Is there a quick way to obtain the algebra B/BeB by quiver and relations using the GAP-package QPA?

One way to do this is to obtain the module $B/BeB$ using CoKernel(TraceOfModule(eB,B)) and the calculate the endomorphism ring of $B/BeB$. This is however very slow in most cases so I wonder whether there is a quicker way.

Answer 1

Assume that $B = kQ/I$ and $e$ is a sum of vertices for a quiver $Q$ and admissible relations $I$. Let $Q'$ be the full subquiver of $Q$ consisting of all the vertices in $Q$ except those occurring in $e$. Then $kQ = kQ' \oplus kQ e kQ$ as a vectorspace over $k$. Consider the natural algebra homomorphism $kQ' \to B/BeB \simeq kQ/(I + kQekQ)$. Then we can write $I = I' \oplus V$ for some subspace $I'$ of $kQ'$ and $V$ of $kQekQ$. Using this I think one can show that $B/BeB \simeq kQ'/I'$. There is now a function in QPA that does this as follows:

gap> Q:= Quiver(3,[[1,2,"a"],[1,2,"b"],[2,2,"c"],[2,3,"d"],[3,1,"e"]]);
<quiver with 3 vertices and 5 arrows>
gap> KQ:= PathAlgebra(Rationals, Q);
<Rationals[<quiver with 3 vertices and 5 arrows>]>
gap> AssignGeneratorVariables(KQ);
#I  Assigned the global variables [ v1, v2, v3, a, b, c, d, e ]
gap> relations := [d*e,c^2,a*c*d-b*d,e*a];
[ (1)*d*e, (1)*c^2, (-1)*b*d+(1)*a*c*d, (1)*e*a ]
gap> A:= KQ/relations;
<Rationals[<quiver with 3 vertices and 5 arrows>]/
<two-sided ideal in <Rationals[<quiver with 3 vertices and 5 arrows>]>, (5 generators)>>
gap> B := QuiverAlgebraOfAmodAeA(A, [ 2 ] );
<Rationals[<quiver with 2 vertices and 1 arrows>]>
gap> Dimension(B);
3
gap> C := QuiverAlgebraOfAmodAeA(A, [ 1, 3 ] );
<Rationals[<quiver with 1 vertices and 1 arrows>]/
<two-sided ideal in <Rationals[<quiver with 1 vertices and 1 arrows>]>, (1 generators)>>
gap> Dimension(C);
2