12
$\begingroup$

Usually computers can calculate the global dimension of a finite dimensional quiver algebra much faster than humans. But in this case a high end computer (calculating for 3 weeks) was not able to determine whether this algebra has global dimension 3 or not.

Let $A=K\langle a,b\rangle/I$ with $I$ the ideal generated by $\langle a^2, ab+b^2-aba, ab^2, bab, ab+b^2+b^2a, b^3\rangle$ over a field $K$ of characteristic not two. Let $D=\operatorname{Hom}_K(-,K)$ the natural duality.

This algebra is a local non-Gorenstein algebra that was found by Jan Geuenich as a rare algebra with $\operatorname{Ext}_A^1(D(A),A)=0$. Let $\tau_2 := \tau \Omega^1$. Now let $M:=A \oplus D(A) \oplus \tau_2(D(A)) \oplus \tau_2^2(D(A)) \oplus \tau_2^3(D(A))$ and $B:=\operatorname{End}_A(M)$.

The module $M$ has vector space dimension 33 and the algebra $B$ has vector space dimension 165.

It can be shown that $M$ is a precluster tilting object in the sense of Iyama and Solberg - Auslander-Gorenstein algebras and precluster tilting and that the algebra has dominant dimension equal to the Gorenstein dimension equal to three. But the computer was not able to determine whether $B$ has finite global dimension (the global dimension is either 3 or infinite).

Thus the question:

Does $B$ have finite global dimension?

In case the answer is positive it would be the first 2-cluster tilting object for a local algebra in history! (at least to my knowledge)

I can think of two ways to determine the answer. The first is to check whether $M$ is a 2-cluster tilting object directly but $A$ is representation-infinite and one needs good knowledge of the module category of $A$ for that. The other way would be to calculate the quiver and relations of $B$ but this looks like a cruel torture when even a high end computer can not do it. So I hope there might be a good trick. $B$ has Cartan determinant 1, which makes it look like the global dimension could really be finite.

A positive answer would also answer this old question: Cluster-tilting object for a local non-selfinjective algebra .

Improve this question
$\endgroup$
2
  • 3
    $\begingroup$ I don’t understand “The other way would be to calculate the endomorphism ring of B”. Typo? $\endgroup$
    – Jeremy Rickard
    Jan 14, 2019 at 13:58
  • 3
    $\begingroup$ @JeremyRickard I meant the quiver and relations of $B$. That is what also the computer can not do. $\endgroup$
    – Mare
    Jan 14, 2019 at 14:02

1 Answer 1

Reset to default
12
$\begingroup$

Let $M = P \oplus I \oplus \tau_2 \oplus \tau_2^2 \oplus \tau_2^3$, where the notation is the obvious one. One way of computing the global dimension of $B=\operatorname{End}(M)$ is to find the projective resolution of all the simple $B$-modules. The simple $B$-modules are given by for each indecomposable direct summand $M_i$ of $M$ finding a "radical map" $M_i\xrightarrow{f} M(i)$ such that the cokernel of the induced map $\operatorname{Hom}(M(i),M)\xrightarrow{\operatorname{Hom}(f,M)} \operatorname{Hom}(M_i,M)$ is a simple $B$-module, where $M(i)$ is in $\operatorname{add}M$. All the maps from all the indecomposable direct summands of $M$ different from $M_i$ are "radical maps". Hence a left approximation $f'\colon M_i\to \widehat{M_i}^{M_i}$ by all the indecomposable direct summands of $M$ different from $M_i$ is part of $f$. In some cases it might be everything, for instance when cokernel of the induced map $\operatorname{Hom}(f',M)$ is one dimensional. Then we would have $$\operatorname{Hom}(\widehat{M_i}^{M_i},M) \to \operatorname{Hom}(M_i,M) \to S_{M_i}\to 0,$$ where $S_{M_i}$ is the simple $B$-module associated to the indecomposable projective module $\operatorname{Hom}(M_i,M)$. We can continue this projective resolution by finding a left $\operatorname{add}M$-approximation $f_2\colon \operatorname{Coker}(f')\to M^{\operatorname{Coker}(f')}$ of $\operatorname{Coker}(f')$. Before doing this we can remove any direct summands of $\operatorname{Coker}(f')$ isomorphic to an indecomposable direct summand of $M$. Then if $K_2 = \operatorname{Coker}(f_2)$ is in $\operatorname{add}M$, then $\operatorname{pd}S_{M_i}\leq 3$. Carrying out these computations, as far as I can see, one gets that $\operatorname{pd}S_{M_i}$ is equal to $3$ for $M_i = P, \tau_2, \tau_2^2,\tau_2^3$ and equal to $2$ for $M_i = I$.

Here is a copy of the GAP-session computing this (note that this is using a new function LeftApproximationByAddM added to QPA today) for $P$. The computations for the other direct summands are similar.

gap> Q := Quiver( 1, [[1,1,"a"],[1,1,"b"]] );
<quiver with 1 vertices and 2 arrows>
gap> kQ := PathAlgebra( GF( 3 ), Q );
<GF(3)[<quiver with 1 vertices and 2 arrows>]>
gap> AssignGeneratorVariables( kQ );
#I  Assigned the global variables [ v1, a, b ]
gap> $, b^3 ];                                                                 
[ (Z(3)^0)*a^2, (Z(3)^0)*a*b+(Z(3)^0)*b^2+(Z(3))*a*b*a, (Z(3)^0)*a*b^2, 
  (Z(3)^0)*b*a*b, (Z(3)^0)*a*b+(Z(3)^0)*b^2+(Z(3)^0)*b^2*a, (Z(3)^0)*b^3 ]
gap> A := kQ/relations;
<GF(3)[<quiver with 1 vertices and 2 arrows>]/
<two-sided ideal in <GF(3)[<quiver with 1 vertices and 2 arrows>]>, 
  (6 generators)>>
gap> P := IndecProjectiveModules(A)[1];
<[ 6 ]>
gap> I := IndecInjectiveModules(A)[1]; 
<[ 6 ]>
gap> OI := NthSyzygy(I,1);
<[ 6 ]>
gap> tau2 := DTr(OI,1);
Computing step 1...
<[ 8 ]>
gap> tau22 := DTr(NthSyzygy(tau2, 1),1);
Computing step 1...
<[ 5 ]>
gap> tau23 := DTr(NthSyzygy(tau22, 1),1);
Computing step 1...
<[ 8 ]>
gap> M := DirectSumOfQPAModules([P,I,tau2,tau22,tau23]);
<[ 33 ]>
gap> N := DirectSumOfQPAModules([I,tau2,tau22,tau23]);
<[ 27 ]>
gap> U := P;
<[ 6 ]>
gap> f := LeftApproximationByAddM(U,N);
<<[ 6 ]> ---> <[ 54 ]>>
gap> test := List(HomOverAlgebra(Range(f),M), s -> f*s);;
gap> n := Length(vectors[1]);
198
gap> V := Subspace(GF(3)^n, vectors);   
<vector space over GF(3), with 67 generators>
gap> Dimension(V); 
32
gap> Length( HomOverAlgebra(U,M) );   
33
gap> K1 := CoKernel(f);
<[ 48 ]>
gap> CommonDirectSummand(P,K1);
false
gap> CommonDirectSummand(I,K1);
[ <[ 6 ]>, <[ 0 ]>, <[ 6 ]>, <[ 42 ]> ]
gap> K1 := last[4];
<[ 42 ]>
gap> CommonDirectSummand(I,K1);
[ <[ 6 ]>, <[ 0 ]>, <[ 6 ]>, <[ 36 ]> ]
gap> K1 := last[4];            
<[ 36 ]>
gap> CommonDirectSummand(I,K1);
false
gap> CommonDirectSummand(tau2,K1);
[ <[ 8 ]>, <[ 0 ]>, <[ 8 ]>, <[ 28 ]> ]
gap> K1 := last[4];               
<[ 28 ]>
gap> CommonDirectSummand(tau2,K1);
false
gap> CommonDirectSummand(tau22,K1);
[ <[ 5 ]>, <[ 0 ]>, <[ 5 ]>, <[ 23 ]> ]
gap> K1 := last[4];                
<[ 23 ]>
gap> CommonDirectSummand(tau22,K1);
[ <[ 5 ]>, <[ 0 ]>, <[ 5 ]>, <[ 18 ]> ]
gap> K1 := last[4];                
<[ 18 ]>
gap> CommonDirectSummand(tau22,K1);
false
gap> CommonDirectSummand(tau23,K1);
[ <[ 8 ]>, <[ 0 ]>, <[ 8 ]>, <[ 10 ]> ]
gap> K1 := last[4];                
<[ 10 ]>
gap> CommonDirectSummand(tau23,K1);
false
gap> f2 := LeftApproximationByAddM( K1, M );
K2 := CoKernel(f2);
<<[ 10 ]> ---> <[ 99 ]>>
gap> K2 := CoKernel(f2);
<[ 89 ]>
gap> CommonDirectSummand(P,K2);
[ <[ 6 ]>, <[ 0 ]>, <[ 6 ]>, <[ 83 ]> ]
gap> K2 := last[4];            
<[ 83 ]>
gap> CommonDirectSummand(P,K2);
[ <[ 6 ]>, <[ 0 ]>, <[ 6 ]>, <[ 77 ]> ]
gap> K2 := last[4];            
<[ 77 ]>
gap> CommonDirectSummand(P,K2);
false
gap> CommonDirectSummand(I,K2);
[ <[ 6 ]>, <[ 0 ]>, <[ 6 ]>, <[ 71 ]> ]
gap> K2 := last[4];            
<[ 71 ]>
gap> CommonDirectSummand(I,K2);
false
gap> CommonDirectSummand(tau2,K2);
[ <[ 8 ]>, <[ 0 ]>, <[ 8 ]>, <[ 63 ]> ]
gap> K2 := last[4];               
<[ 63 ]>
gap> CommonDirectSummand(tau2,K2);
[ <[ 8 ]>, <[ 0 ]>, <[ 8 ]>, <[ 55 ]> ]
gap> K2 := last[4];               
<[ 55 ]>
gap> CommonDirectSummand(tau2,K2);
[ <[ 8 ]>, <[ 0 ]>, <[ 8 ]>, <[ 47 ]> ]
gap> K2 := last[4];               
<[ 47 ]>
gap> CommonDirectSummand(tau2,K2);
false
gap> CommonDirectSummand(tau22,K2);
[ <[ 5 ]>, <[ 0 ]>, <[ 5 ]>, <[ 42 ]> ]
gap> K2 := last[4];                
<[ 42 ]>
gap> CommonDirectSummand(tau22,K2);
[ <[ 5 ]>, <[ 0 ]>, <[ 5 ]>, <[ 37 ]> ]
gap> K2 := last[4];                
<[ 37 ]>
gap> CommonDirectSummand(tau22,K2);
[ <[ 5 ]>, <[ 0 ]>, <[ 5 ]>, <[ 32 ]> ]
gap> K2 := last[4];                
<[ 32 ]>
gap> CommonDirectSummand(tau22,K2);
false
gap> CommonDirectSummand(tau23,K2);
[ <[ 8 ]>, <[ 0 ]>, <[ 8 ]>, <[ 24 ]> ]
gap> K2 := last[4];                
<[ 24 ]>
gap> CommonDirectSummand(tau23,K2);
[ <[ 8 ]>, <[ 0 ]>, <[ 8 ]>, <[ 16 ]> ]
gap> K2 := last[4];                
<[ 16 ]>
gap> CommonDirectSummand(tau23,K2);
[ <[ 8 ]>, <[ 0 ]>, <[ 8 ]>, <[ 8 ]> ]
gap> K2 := last[4];                
<[ 8 ]>
gap> CommonDirectSummand(tau23,K2);
[ <[ 8 ]>, <[ 0 ]>, <[ 8 ]>, <[ 0 ]> ]

I hope that these comments are helpful.

The QPA-team.

Improve this answer
$\endgroup$
8
  • $\begingroup$ Thanks. Do I understand it correctly that like this it would not be possible to check for a general generator-cogenerator $M$ over an algebra $A$ whether $End_A(M)$ has finite global dimension, but one has to be lucky and all the cokernel of the induced map $Hom_A(f',M)$ has to be 1-dimensional? (this condition should be equivalent to the condition that for every simple $B$-module $S$, one has a minimal projective presentation $P_1 \rightarrow P_0 \rightarrow S \rightarrow 0$ such that $P_0$ is not a direct summand of $P_1$ and should be rather rare.) $\endgroup$
    – Mare
    Jan 15, 2019 at 10:29
  • 5
    $\begingroup$ So who won? Human or computer? $\endgroup$
    – Jeremy Rickard
    Jan 15, 2019 at 11:23
  • 2
    $\begingroup$ I agree with your comments. Finding the quiver of $\operatorname{End}(M)$ is not so expensive. But determining the relations are more costly. Here is what you can do to find the quiver: $\endgroup$
    – Oeyvind Solberg
    Jan 15, 2019 at 11:29
  • 2
    $\begingroup$ endo := EndOverAlgebra(M); radendo := RadicalOfAlgebra(endo); radendo2 := ProductSpace(radendo,radendo); f := NaturalHomomorphismByIdeal(endo,radendo2); idemps := IdempotentsForDecomposition(endo); B := BasisVectors(Basis(radendo));; I := Ideal(Range(f), List(B, b->ImageElm(f, b))); mat := List( [1..5], i -> List([1..5], j -> Dimension( Subspace( Range(f), ImageElm(f,idemps[i])*BasisVectors(Basis(I))*ImageElm(f,idemps[j]))))); Display(mat); $\endgroup$
    – Oeyvind Solberg
    Jan 15, 2019 at 11:29
  • 4
    $\begingroup$ @JeremyRickard This time it is probably a draw. $\endgroup$
    – Mare
    Jan 15, 2019 at 12:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.