The Yoneda algebra of a K_2 algebra need not be another K_2 algebra
aa r X i v : . [ m a t h . R A ] N ov THE YONEDA ALGEBRA OF A K ALGEBRA NEED NOTBE ANOTHER K ALGEBRA
Thomas Cassidy † , Christopher Phan ‡ and Brad Shelton ‡† Department of MathematicsBucknell UniversityLewisburg, Pennsylvania 17837 ‡ Department of MathematicsUniversity of OregonEugene, Oregon 97403-1222
Abstract.
The Yoneda algebra of a Koszul algebra or a D -Koszul al-gebra is Koszul. K algebras are a natural generalization of Koszulalgebras, and one would hope that the Yoneda algebra of a K algebrawould be another K algebra. We show that this is not necessarily thecase by constructing a monomial K algebra for which the correspondingYoneda algebra is not K . Introduction
Let A be a connected graded algebra over a field K . Correspondences be-tween A and its bigraded Yoneda algebra E ( A ) = L n,m Ext n,mA ( K, K ) havebeen studied in many contexts (e.g. [4], [5], [6] and [10]). In particular thereare very interesting classes of algebras where E ( A ) inherits good propertiesfrom A . Perhaps the most famous and intently studied of such classes ofalgebra is the class of Koszul algebras.An algebra is Koszul [10] if its Yoneda algebra is generated as an algebraby cohomology degree one elements. Koszul algebras will always have qua-dratic defining relations and given such an algebra, A , the Yoneda algebra isisomoprhic to the quadratic dual algebra A ! . In particular, one has Koszulduality: If A is Koszul then E ( A ) is Koszul and E ( E ( A )) = A .The following natural generalization of Koszul was introduced in [2] andalso investigated in [7] and [8]. We write E n ( A ) for L p Ext n,mA ( K, K ). Key words and phrases. graded algebra, Koszul algebra, Yoneda algebra.
CASSIDY, PHAN AND SHELTON
Definition 1.1.
The graded algebra A is said to be K if E ( A ) is generatedas an algebra by E ( A ) and E ( A ).Koszul algebras are simply quadratic K algebras. K algebras sharemany of the nice properties of Koszul algebras, including stability undertensor products, regular normal extensions and graded Ore extensions, (cf.[2]). Every graded complete intersection is a K algebra.Another important class of algebras is the class of D -Koszul algebrasintroduced by Berger in [1]. This is the class defined by: Ext n,mA ( K, K ) = 0unless m = δ ( n ), where δ (2 n ) = nD and δ (2 n + 1) = nD + 1. These algebrasarise naturally in certain contexts and all such D -Koszul algebras are easilyseen to be K . A remarkable theorem in [3] states that if A is D -Koszulalgebra, then E ( A ) is a K algebra, and furthermore, it is possible to regrade E ( A ) in such a way that E ( A ) becomes a Koszul algebra. In particular onegets a “delayed” duality: E ( E ( A )) = E ( A ) ! and E ( E ( E ( A ))) = E ( A ).Based on the above theorem of [3], Koszul duality, and calculations ofmany other K -examples, it seems reasonable to hope that the Yoneda alge-bra of any K algebra would also be K , perhaps even Koszul. Unfortunately,this is not always the case, and the purpose of this article is to exhibit anexample of a K algebra for which the corresponding Yoneda algebra is notKoszul nor even K . Our example has 13 generators and 9 monomial definingrelations. We believe that such a monomial algebra cannot be constructedwith fewer generators and relations.We wish to thank Jan-Erik Roos for pointing out an error in an earlierversion of this paper.2. The algebras
A, E ( A ) and E ( E ( A ))Let K be a field. Let { m, n, p, q, r, s, t, u, v, w, x, y, z } be a basis for avector space V . We define A to be the K -algebra T ( V ) /I where I is theideal generated by this list of monomial tensors: R = { mn p, n pqr, npqrs, pqrst, stu, tuvwx, uvwxy, vwxy , xy z } . Theorem 2.1.
The algebra A is K , but the algebra E ( A ) is not K . Proof .
We use the algorithm given in section 5 of [2] to prove that A is K .From the set R one can calculate that S = { m, n, p, q, r, s, t, u, v, w, x, y, z } , S = { mn , n pq, npqr, pqrs, st, tuvw, uvwx, vwxy, xy } , S = { pqr, vw } , S = { n } and S = ∅ . One easily verifies that for every b ∈ S with minimalleft annihilator a we have either deg( a ) = 1 or ab ∈ R , and hence A is K .Let B = E ( A ). In what follows we consider only the cohomology gradingon B . Following section 5 of [2] we can construct a minimal projectiveresolution P • of A K and see that the Hilbert Series of the algebra B is1 + 13 t + 9 t + 8 t + 4 t + 3 t + t .It is possible (although laborious) to describe B in terms of generators andrelations and then construct a minimal resolution of B K and apply Theorem4.4 of [2] to show that B is not K . However B ’s failure to be K is apparentalready in Ext B ( K, K ) and consequently there is a more efficient way for usto illustrate this.
HE YONEDA ALGEBRA OF A K ALGEBRA NEED NOT BE ANOTHER K ALGEBRA 3
Let ¯ m and ¯ z denote the basis elements in B dual to m and z in A . Thevector space B has a basis dual to the elements of the list of relations R .We will use α, β and γ to denote the dual basis elements corresponding tothe monomials n pqr, stu and vwxy . From the maps in the resolution P • one can see that ¯ mα , γ ¯ z and βγ are nonzero in B , while ¯ mαβ and βγ ¯ z areeach zero.Recall that T or B ( K, K ) can be calculated using the bar-complex [9] where B ar i ( K, B, K ) = K ⊗ B ⊗ B ⊗ B + ⊗· · ·⊗ B + ⊗ K = B ⊗ i + . Let ζ = ¯ mα ⊗ βγ ⊗ ¯ z ∈ B ⊗ . The differential on the bar-complex gives us ∂ ( ζ ) = ¯ mαβγ ⊗ ¯ z − ¯ mα ⊗ βγ ¯ z = 0. ζ is not in the image of B ⊗ because ∂ ( ¯ m ⊗ α ⊗ βγ ⊗ ¯ z ) =¯ mα ⊗ βγ ⊗ ¯ z − ¯ m ⊗ αβγ ⊗ ¯ z while ∂ ( ¯ mα ⊗ β ⊗ γ ⊗ ¯ z ) = − ¯ mα ⊗ βγ ⊗ ¯ z + ¯ mα ⊗ β ⊗ γ ¯ z .Thus ζ represents a non-zero homology class in T or B .In contrast the element ¯ mα ⊗ βγ = ∂ ( − ¯ mα ⊗ β ⊗ γ ) represents zero in T or B and ¯ mα = ∂ ( ¯ m ⊗ α ) represents zero in T or B . Therefore under theco-multiplication map∆ : T or B ( K, K ) → T or B ( K, K ) ⊗ T or B ( K, K ) ⊕ T or B ( K, K ) ⊗ T or B ( K, K )we have ∆( ζ ) = 0. This failure of ∆ to be injective is equivalent to themultiplication map E ( B ) ⊗ E ( B ) ⊕ E ( B ) ⊗ E ( B ) → E ( B )not being surjective. Hence E ( B ) is not generated by E ( B ) and E ( B ),and so B is not a K algebra. (cid:3) References
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