Quadraticity and Koszulity for Graded Twisted Tensor Products

Quadraticity and Koszulity for Graded Twisted Tensor Products Peter Goetz Humboldt State University Arcata, CA 95521 September 9, 2017

Outline Outline I. Preliminaries II. Quadraticity III. Koszulity IV. Separable twisting maps V. Examples VI. References Joint work with Andrew Conner (Saint Mary s College of California)

Preliminaries Let K be a field. All objects and maps are in the category K-Vect. All tensor products are over K. We work with N-graded, connected, associative K-algebras. R = R n, R 0 = K n 0 R + = n>0 R n R/R + = K, trivial module We will say R is graded from now on. All algebras are associative.

Internal twisted products Let A and B be algebras with multiplication maps µ A and µ B. Definition An internal twisted tensor product of A and B is a triple (C, i A, i B ) where C is an algebra and i A : A C and i B : B C are injective algebra homomorphisms such that the linear map A B C given by a b i A (a)i B (b) is an isomorphism of vector spaces. Definition We say two internal twisted tensor products (C, i A, i B ) and (C, i A, i B ) of (graded) A and B are isomorphic if there exist (graded) algebra homomorphisms α : A A, β : B B, and γ : C C such that γi A = i A α, γi B = i Bβ, and γ is an isomorphism.

Twisting maps Definition We call a map τ : B A A B an algebra twisting map if τ(1 B a) = a 1 B, τ(b 1 A ) = 1 A b, and τ(µ B µ A ) = (µ A µ B )(1 A τ 1 B )(τ τ)(1 B τ 1 A ). The last equation asserts an equality of maps B B A A A B. If A and B are graded, A B admits an N-grading by the Künneth formula (A B) m = A k B l. k+l=m The map τ : B A A B is graded if τ((b A) n ) (A B) n for all n 0.

Internal twisted products and twisting maps Proposition (ČSV, C-G) Let A and B be (graded) algebras. Let τ : B A A B be a (graded) map. Define a map µ τ : A B A B A B by µ τ = (µ A µ B )(1 A τ 1 B ). The map τ : B A A B is an (graded) algebra twisting map if and only if µ τ defines an associative multiplication giving A B the structure of a (graded) algebra. The algebra determined by τ is denoted A τ B, and is called the external twisted tensor product of A and B.

Internal twisted products and twisting maps Proposition (ČSV, C-G) Let (C, i A, i B ) be a (graded) internal twisted tensor product of (graded) algebras A and B. Then there is a unique (graded) twisting map τ such that C is isomorphic to A τ B as a (graded) twisted tensor product.

Types of graded twisting maps If A and B are graded, a graded twisting map τ satisfies τ(b + A + ) (A + B 0 ) (A + B + ) (A 0 B + ). We say τ is strongly-graded if We say τ is one-sided if or τ(b i A j ) A j B i for all i, j 0. τ(b + A + ) (A + B 0 ) (A + B + ) τ(b + A + ) (A + B + ) (A 0 B + ). Many references to graded twisting maps in the literature in fact refer to the more restrictive cases of strongly-graded, or one-sided twisting maps. We sometimes refer to the most general type of twisting map as two-sided.

Example Example The commutative tensor product: A τ B, τ(b a) = a b. This twisting map is invertible and strongly-graded.

Example Example An Ore extension: A τ K[x], τ(x a) = σ(a) x + δ(a) 1, where σ End(A), δ is a σ-derivation. This twisting map is one-sided, and is invertible if σ is invertible.

Example Example A graded Hopf smash product: let H be a graded Hopf algebra; let A be a graded left Hopf H-module algebra. A#H = A τ H, τ(h a) = (h (1).a) h (2), where (h) = h (1) h (2). This twisting map is one-sided.

Quadratic algebras Let A be a graded algebra which is locally finite dimensional, dim K A n < for all n 0. A is one-generated if the canonical multiplication map π : T (A 1 ) A is surjective. Let I = (ker π) A 1 A 1 be the ideal of T (A 1 ) generated by the degree 2 elements of the kernel of π. If A is one-generated, the quadratic part of A is the algebra q(a) = T (A 1 )/I. If A = q(a) then A is called quadratic.

Koszul algebras Let A be a graded, locally finite dimensional, one-generated algebra. A is Koszul if the trivial module A/A + admits a resolution P 3 P 2 P 1 P 0 A/A + 0 such that each P i is a graded free left A-module generated in degree i. Two well-known facts follow immediately from this definition: 1 every Koszul algebra is quadratic, 2 every one-generated, free algebra is Koszul. Many important quadratic algebras arising naturally in mathematics are Koszul, and Koszul algebras satisfy a powerful duality property that sometimes underlies deep connections between apparently unrelated problems.

Natural questions 1 When are graded twisted tensor products of quadratic algebras quadratic? 2 When are graded twisted tensor products of Koszul algebras Koszul?

Unique extension property Definition If A and B are graded algebras, we say a graded twisting map τ : B A A B has the unique extension property if for all graded twisting maps τ : B A A B and all n N such that τ i = τ i for i < n, then τ n = τ n. Theorem (C-G) Every one-sided graded twisting map has the unique extension property.

Quadraticity for twisted tensor products Theorem (C-G) Assume that A and B are quadratic algebras, and let τ : B A A B be a graded twisting map. Then τ has the unique extension property if and only if A τ B is quadratic. If we drop the hypothesis that A and B are quadratic, we have an example of a graded twisting map τ : B A A B which has the unique extension property, the algebra A τ B is Koszul and yet A is not quadratic.

Koszulity for one-sided twisted tensor products Theorem (C-G) Let τ : B A A B be a one-sided graded twisting map. 1 If A and B are quadratic algebras, then A τ B is quadratic. 2 If A and B are Koszul algebras, then A τ B is Koszul. Statement (1) is immediate. For (2) we use the following result. Theorem (PP, Example 2, p. 90) Let f : A C be a homomorphism of quadratic algebras such that C 1 f(a 1 ) f(a 1 )C 1. Assume that A and C/f(A 1 )C are Koszul algebras and the left action of A on C is free in degrees 3. Then C is a Koszul algebra.

Twisting free algebras Theorem (C-G) If A and B are one-generated free algebras and τ : B A A B is a graded twisting map, then A τ B is a Koszul algebra whenever it is quadratic. Proof. Suppose that A = K x 1,..., x m and B = K y 1,..., y n are one-generated free algebras. Assume that τ : B A A B is a graded twisting map. Let C = A τ B and suppose that C is quadratic. The Hilbert series of A and B are respectively, h A (t) = (1 mt) 1 and h B (t) = (1 nt) 1, so h C (t) = (1 (m + n)t + mnt 2 ) 1. That C is Koszul follows immediately from [PP] Chapter 2, Proposition 2.3.

Twisted tensor products need not be Koszul Example Let A = K[x], B = K[y]. Define τ : B A A B by { τ(y i x j x j y i if i or j is even ) = x j+1 y i 1 x j y i + x j 1 y i+1 otherwise. Then: τ is a graded twisting map τ does not have the unique extension property consequently, A τ B is not quadratic, and hence A τ B is not Koszul. Additionally, it is easy to prove that A τ B is isomorphic to the algebra K x, y / x 2, y 2 x xy 2. This algebra has infinite global dimension, so A τ B is not Artin-Schelter regular.

Problem Find an example of a graded twisting map between two Koszul algebras which has the unique extension property, and the associated twisted tensor product is not Koszul, or prove that no such example exists; that is, prove that, for twisted tensor products, quadraticity is the only obstruction to Koszulity. Of course, if such an example exists, it is necessarily two-sided.

Separable twisting maps Definition A graded twisting map τ : B A A B is called separable if there exists a decomposition B 1 = B 1 B 1 such that τ(b 1 A 1 ) A 2 B 0 A 1 B 1 τ(b 1 A 1 ) A 1 B 1 A 0 µ B (B 1 B 1 ). Proposition (C-G) Every separable graded twisting map has the unique extension property. Hence, if A and B are quadratic algebras and τ : B A A B is separable, then A τ B is quadratic.

Constructing separable twisting maps Let A 1 be the vector space with basis {x 1,..., x l }. Let B 1 be the vector space with basis {u 1,..., u m, d 1,..., d n }. Let A = T (A 1 ) and B = T (B 1 ). Suppose that t : B 1 A 1 A B is any linear map such that t(b 1 A 1 ) A 1 B 1 A 2 B 0 and t(b 1 A 1 ) A 1 B 1 A 0 µ B (B 1 B 1 ). Theorem (C-G) There exists a unique graded twisting map τ : B A A B such that τ B1 A 1 = t. To construct examples of separable twisting maps on algebras with relations, one can check that the defining ideals are preserved by the twisting map, in an appropriate sense. See Theorem 3.7 of our paper.

A filtration for separable twisting maps Let Γ = N 3. Then Γ is a commutative monoid under component-wise addition. We equip Γ with a grading by total degree; that is, we let g : Γ N be the grading homomorphism defined by g(a, b, c) = a + b + c. We order the fibers g 1 (n) lexicographically, with (0, 0, 1) < (0, 1, 0) < (1, 0, 0) in g 1 (1). This determines the structure of a graded, ordered monoid on Γ. In particular, if α, β g 1 (n) and γ g 1 (m), then α < β implies α + γ < β + γ.

A filtration for separable twisting maps Let τ : B A A B be a separable twisting map, with B 1 = B 1 B 1. We define a one-generated, Γ-valued filtration F on A τ B as follows: F (0,0,0) = A 0 B 0 F (0,0,1) = A 0 B 1 + F (0,0,0) F (0,1,0) = A 1 B 0 + F (0,0,1) F (1,0,0) = A 0 B 1 + F (0,1,0) and for all α Γ such that g(α) > 1, F α = F i1 F i2 F ik. i 1+ +i k =α Note that by restricting our attention to the subalgebra B we obtain a filtration by the commutative monoid N 2, viewed as the submonoid of Γ generated by (1, 0, 0) and (0, 0, 1). We denote this filtration by F B.

Koszulity for separable twisting maps Theorem (C-G) Let A and B be quadratic algebras with spaces of quadratic relations I 2, J 2, respectively. Let τ : B A A B be a separable graded twisting map. Assume τ satisfies 1 π A0 A 2 B 1 (1 τ)(τ B 1)(B 1 I 2 ) = 0, and 2 π A1 B 2 B 0 (τ 1)(1 τ A )(J 2 A 1 ) = 0. Let B = gr F B (B). If A is Koszul, the quadratic part of B is Koszul and B has no defining relations in degree 3, then A τ B is Koszul.

Koszulity for separable twisting maps Corollary (C-G) Let A be a Koszul algebra with quadratic relation space I 2, let B be a free algebra, and suppose τ : B A A B is a separable graded twisting map. Assume that π A0 A 2 B 1 (1 τ)(τ B 1)(B 1 I 2 ) = 0. Then A τ B is Koszul.

Example Example Let A = K[x, y], B = K d, u. Define a separable graded twisting map τ by τ(d x) = x d + 1 d 2, τ(d y) = y d + 1 d 2. τ(u x) = x u + x 2 1, τ(u y) = y u + y 2 1. Then A τ B is Koszul, and gldim(a τ B) = 3.

Separable maps need not be one-sided Example Let A = K[x], B = K d, u, and let τ : B A A B be the separable, graded twisting map uniquely determined by τ(d x) = x d + 1 d 2, τ(u x) = x 2 1 + x u. There do not exist non-trivial, one-generated, graded K-algebras C and D and a one-sided graded twisting map τ : D C C D such that A τ B = C τ D as graded algebras.

Example Here is an example that shows that A τ B Koszul does not imply that A and B are quadratic. Example Let U(g 3 ) be the quotient of the free algebra Q x ij 1 i j 3 by the homogeneous ideal generated by [x ij, x ik + x jk ] i, j, k distinct [x ik, x jk ] i, j, k distinct. In [CG1] the following were proved: 1 U(g 3 ) is Koszul, 2 U(g 3 ) is isomorphic to A τ Q x 12, x 21 where A is isomorphic to the subalgebra of U(g 3 ) generated by {x 13, x 23, x 31, x 32 }; the twisting map τ in this case is one-sided so the subalgebra A is normal, 3 the algebra A is not finitely presented.

References [CG1] Andrew Conner and Peter Goetz. Some non-koszul algebras from rational homotopy theory. Bull. Lond. Math. Soc., 47(3):473-482, 2015. [CG2] Andrew Conner and Peter Goetz. The Koszul property for graded twisted tensor products. Submitted. arxiv:1708.02514. [CSV] Andreas Čap, Hermann Schichl, and Jiri Vanžura. On twisted tensor products of algebras. Comm. Algebra, 23(12):4701-4735, 1995. [PP] Alexander Polishchuk and Leonid Positselski. Quadratic algebars, volume 37 of University Lecture Series. American Mathematical Society, Providence, RI, 2005.

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