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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 .

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  • 3
    $\begingroup$ I don’t understand “The other way would be to calculate the endomorphism ring of B”. Typo? $\endgroup$
    – Jeremy Rickard
    Commented 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
    Commented Jan 14, 2019 at 14:02

1 Answer 1

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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.

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  • $\begingroup$ 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
    Commented Jan 15, 2019 at 10:29
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    $\begingroup$ So who won? Human or computer? $\endgroup$
    – Jeremy Rickard
    Commented 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
    Commented Jan 15, 2019 at 11:29
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    $\begingroup$ 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
    Commented Jan 15, 2019 at 11:29
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    $\begingroup$ @JeremyRickard This time it is probably a draw. $\endgroup$
    – Mare
    Commented Jan 15, 2019 at 12:22

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