The graded Cartan matrix of the dual numbers #
Let k be a field and grade the dual numbers A = k[ε] with ε in degree two
(TauCeti.dualNumberIntegerGrade). The degree-zero piece of A is k, spanned by the idempotent
1, so up to internal shift A has one indecomposable finite graded projective, A itself, and
one graded simple, the residue field S = A/(ε) concentrated in degree zero
(TauCeti.dualNumberGradedResidue). The regular module has the graded composition factors S and
S{2}, so the graded Cartan matrix of A is the one-by-one matrix 1 + q². These dual numbers
are the factor of the zigzag algebra at an isolated vertex, where
TauCeti.gradedCartanMatrix_zigzagAlgebra has the same diagonal entry 1 + q².
Its determinant 1 + q² is not a unit of ℤ[q,q⁻¹]: at q = 1 it becomes 2, the ungraded
Cartan matrix of k[ε] (TauCeti.cartanMatrix_dualNumber). Graded Cartan unimodularity
(TauCeti.isUnit_det_gradedCartanMatrix_gradedIdempotentProjectiveBasis) therefore fails for the
only pair of graded simples: (S, S) is not graded Euler-admissible, so the q-Euler characteristic
χ_q(S, S) is not defined and no totalised fallback value is exposed for it. This is the graded
counterpart of the dual-numbers rejection TauCeti.not_isEulerAdmissible_dualNumberResidue. It
also shows that S has no finite resolution by finite graded projectives, the graded counterpart
of TauCeti.not_admitsFiniteResolution_dualNumberResidue.
Main definitions #
TauCeti.dualNumberGradedProjectiveBasisandTauCeti.dualNumberGradedResidueBasis: the bases[A]ofK₀^gr(proj A)and[S]ofG₀^gr(mod A)overℤ[q,q⁻¹].
Main results #
TauCeti.gradedCartanMatrix_dualNumber: the graded Cartan matrix ofk[ε]is1 + q².TauCeti.not_isUnit_det_gradedCartanMatrix_dualNumber: its determinant is not a unit.TauCeti.not_isGradedEulerAdmissible_dualNumberGradedResidue: the pair(S, S)is not graded Euler-admissible.TauCeti.not_admitsFiniteResolution_dualNumberGradedResidue:Shas no finite resolution by finite graded projectives.
Implementation notes #
The integer grading is installed with attribute [local instance] dualNumberIntegerGradedAlgebra,
since TauCeti.dualNumberIntegerGradedAlgebra is a definition rather than a global instance.
Importing modules activate it in the same way to restate these results.
References #
- Zsuzsanna Dancso and Anthony Licata, "Koszul algebras and flow lattices", Journal of
Combinatorial Theory, Series A 185 (2022), Section 2.2, for the graded Cartan matrix and the
q-Euler form over
ℤ[q,q⁻¹]. - Ibrahim Assem, Daniel Simson and Andrzej Skowroński, Elements of the Representation Theory of Associative Algebras I, Chapter III, Section 3, for the ungraded Cartan matrix and the role of finite global dimension in unimodularity.
The idempotent data of the graded dual numbers #
The canonical bases #
The basis [A] of K₀^gr(proj A) over ℤ[q,q⁻¹], for the dual numbers A = k[ε] with
ε in degree two.
Instances For
The basis [S] of G₀^gr(mod A) over ℤ[q,q⁻¹] given by the graded residue field, for the
dual numbers A = k[ε] with ε in degree two.
Instances For
The vector of TauCeti.dualNumberGradedProjectiveBasis is the class [A] of the regular
module, as the graded ideal generated by 1.
The vector of TauCeti.dualNumberGradedResidueBasis is the class [S] of the graded residue
field.
The graded Cartan matrix #
The graded Cartan matrix of the dual numbers is 1 + q², with ε in degree two, in the
bases [A] and [S]: the regular module has the graded composition factors S and S{2}. At
q = 1 this is the ungraded Cartan matrix 2 of TauCeti.cartanMatrix_dualNumber.
The graded Cartan matrix of the dual numbers has determinant 1 + q².
The graded Cartan matrix of the dual numbers is not unimodular: its determinant 1 + q²
is not a unit of ℤ[q,q⁻¹], because it evaluates to 2 at q = 1.
The graded dual-numbers rejection #
The graded dual-numbers rejection. The graded residue field S of k[ε], with ε in
degree two, is not graded Euler-admissible against itself. Because TauCeti.gradedExtEuler
consumes a proof of TauCeti.IsGradedEulerAdmissible, it cannot be instantiated for this pair.
The graded residue field of k[ε] has no finite graded projective resolution: such a
resolution would make (S, S) graded Euler-admissible.