Drinfeld twists of Koszul algebras

Abstract Given a Hopf algebra H and a counital 2-cocycle μ on H, Drinfeld introduced a notion of twist which deforms an H-module algebra A into a new algebra Aμ. We show that when A is a quadratic algebra, and H acts on A by degree-preserving endomorphisms, then the twist Aμ is also quadratic. Furthermore, if A is a Koszul algebra, then Aμ is a Koszul algebra. As an application, we prove that the twist of the q-quantum plane by the quasitriangular structure of the quantum enveloping algebra Uq(sl2) is a quadratic algebra equal to the q−1-quantum plane.


Introduction
Let k be an arbitrary field, and consider all objects to be linear over k, with ⊗ = ⊗ k .If H is a Hopf algebra over k, then let △ : H − → H ⊗ H denote the coproduct and ǫ : H − → k the counit.For an H-module A we denote the action of h ∈ H on a ∈ A as h a.An H-module algebra A is an algebra for which the product map m : A ⊗ A − → A is an H-module homomorphism, so h m(a ⊗ b) = m(△(h) a ⊗ b), and the unit 1 A satisfies h 1 A = ǫ(h)1 A .
In [3], Drinfeld introduced a notion of twist for Hopf algebras and their module algebras.We will briefly recall here how to perform this twist using the terminology of Majid [7,Section 2.3].See also Etingof and Gelaki [4,Section 5.14] for more.
A 2-cocycle is said to be counital if I wish to thank Yuri Bazlov for providing numerous helpful ideas and comments.Research for this paper was supported by the Engineering and Physical Sciences Research Council, UK.
The Drinfeld twist of H by µ is defined to be the Hopf algebra H µ which shares the same algebra structure and counit as H, and has coproduct △ µ : H µ − → H µ ⊗ H µ , h → µ • △(h) • µ −1 .We shall refer to H µ as a "Hopf algebra twist" in order to differentiate it from the following, which is also called a Drinfeld twist.
Definition 1.2.The Drinfeld twist of an H-module algebra (A, m) by µ is the H µ -module algebra (A µ , m µ ) where A µ = A as k-vector spaces, with the product map m µ : In Section 2 we consider the case when A is a quadratic H-module algebra, and show that when the action of H on A is degree-preserving, then any Drinfeld twist of A by a counital 2-cocycle of H is also a quadratic algebra.Using this, we prove in Section 3 that if A is also a Koszul algebra, then the Drinfeld twist A µ must be Koszul too.Finally in Section 4 we give several examples which apply these results.

Quadratic algebras
Suppose A is a connected N-graded k-algebra, meaning A = i∈N A i with A i •A j ⊆ A i+j and A 0 = k.We will assume throughout this paper that A is locally finite-dimensional, meaning that each grading component A i is finite-dimensional.Now A is also a quadratic algebra if it is generated by its degree 1 elements V := A 1 , and has quadratic relations R ⊆ V ⊗ V , so that A = T (V )/(R), where T (V ) is the tensor algebra over V and (R) is the 2-sided ideal generated by R inside T (V ).
Suppose A = T (V )/(R) is both a quadratic algebra, and an H-module algebra, for some Hopf algebra H acting by degree-preserving endomorphisms (i.e.h a ∈ A i ∀h ∈ H, a ∈ A i ).Then, it easy to see that the degree 1 subspace V := A 1 is an H-submodule of A. Additionally, this action of H on V extends naturally by means of the coproduct of H to make T (V ) an H-module algebra.Now a fact that will be very useful throughout the paper is that the space of relations R is an H-submodule of T (V ).This can be seen as follows.Note that a quotient of an H-module algebra by an ideal can only be an H-module algebra too if the ideal is also an H-submodule.Since the H-module algebra structure of A arises from quotienting the H-module algebra T (V ) by the ideal (R), we see that (R) must also be an H-submodule of T (V ).Additionally, since R is precisely the subspace of degree 2 elements in (R), it must be fixed under the degree-preserving endomorphisms of H.And so R is an H-submodule of T (V ).
We now give our first main result.Proof.Firstly let us show A µ is a connected N-graded algebra under the same grading as on A. Since A µ = A as vector spaces, A µ forms an N-graded vector space under the grading of A. Let m and m µ denote the products on A and A µ respectively.Now A µ is a graded algebra if m µ (v ⊗ w) ∈ A i+j ∀v ∈ A i , w ∈ A j .But this follows since m µ := m(µ −1 ), and the action of H is degree-preserving so µ Note, since A = A µ as N-graded vector spaces, V := A 1 is additionally the subspace of degree 1 elements of A µ .Also, as the action of H is degree-preserving, we see that R µ is a subspace of V ⊗ V .Therefore, by showing A µ = T (V )/(R µ ), we will have proven that A µ is a quadratic algebra, as required.
First we prove A µ is generated by V by induction on the degree of elements of A µ .Consider a degree 2 element Since the action of H is degree preserving, µ v i ⊗ w i ∈ V ⊗ V , and so x can be expressed as a linear combination of m µ -products of elements of V , as required.Now let k ≥ 2, and suppose every element of degree k in A µ can be expressed as a linear combination of m µ -products of elements of V .We show that this implies the elements of degree k + 1 have a similar such decomposition.Consider x ∈ A k+1 .Since A is generated by V , we may decompose x as a linear combination of terms which are the m-products Now µ 2 m(v 2 ⊗ m(. . .)) ∈ A k , and so by the induction hypothesis this can be decomposed into a linear combination of m µ -products of k-elements of V .Therefore m(v 1 ⊗ m(v 2 ⊗ m(. . .))) can be decomposed using m µ -products of k + 1-elements, and this implies the same holds for x too.This concludes the induction argument proving A µ is generated by V .
There now exists a natural surjective algebra homomorphism φ : T (V ) − → A µ .We show ker(φ) = (R µ ) to complete the proof that A µ is a quadratic algebra.Note that for the natural embedding i : , and therefore A µ = T (V )/ ker(φ) is seen to be a quotient of T (V )/(R µ ).
To show that A µ is equal to T (V )/(R µ ), we will apply a dimensional argument.Note that all dimensions in the following are finite due to our assumption that A is locally finite-dimensional.Now, since A µ is a quotient of T (V )/(R µ ), the dimension of each grading component of A µ must be less than or equal to the dimension of the same degree component of Next we will consider the Drinfeld twist of T (V )/(R µ ) by the cocycle µ −1 , which leads us to another dimensional inequality (see (5)).First we must check that we can indeed take such a twist.Note that this twist is over the Hopf algebra H µ , rather than the Hopf algebra H that we have used so far.It is a simple exercise to check µ −1 is a counital 2-cocycle of the Hopf algebra twist H µ , and we show next that is an H µ -module algebra.
Since H µ = H as algebras, the H-action on V defines an H µ -action on V .This extends, by means of the twisted coproduct △ µ , to make T (V ) an H µ -module algebra.Now R µ is an H µ -submodule of T (V ) since, if h ∈ H µ and r ∈ R µ , where r = µ r ′ for some r ′ ∈ R, we have where we use the fact that △(h) r ′ ∈ R, since we proved above that R is an H-submodule of the H-module algebra T (V ).So we have established T (V ) is an H µ -module algebra, and R µ is an H µ -submodule, and therefore T (V )/(R µ ) is an H µ -module algebra as we required.
We can now consider the Drinfeld twist of T (V )/(R µ ) by µ −1 .In direct analogy to the argument used for A µ , one may show that V generates (T (V )/(R µ )) µ −1 , i.e. use the fact T (V )/(R µ ) is generated by V to express elements of (T (V )/(R µ )) µ −1 as linear combinations of T (V )/(R µ )-products of elements of V .Then one rewrites each T (V )/(R µ )-product as a linear combination of (T (V )/(R µ )) µ −1 -products.
We therefore have a surjective map φ ′ : It is easy to show that (R) ⊆ ker(φ ′ ), and so we find (T (V )/(R µ )) µ −1 is a quotient of A. This implies the following inequality in the dimensions of the grading components of degree i, But as discussed at the start of the proof, twisting preserves the grading on algebras, so Combining (in)equalities ( 4), ( 5) and ( 6) we find that dim(A µ , we find that the algebras must be equal. 3 Koszul algebras 3.1.Statement of the main theorem.Let A be a connected, N-graded, and locally finite-dimensional, k-algebra.Define A + := i>0 A i , so A/A + ∼ = k, and we call the corresponding quotient map ǫ ′ : A − → k the augmentation map.Now k is an A-module via ǫ ′ , and A is called a Koszul algebra if there is a linear graded free resolution of k as an A-module (see Witherspoon [9,Definition 3.4.3]).If A is Koszul, then it is a standard fact that it is also quadratic, so The next result establishes that Koszulity is preserved under Drinfeld twists.This generalises to arbitrary Hopf algebras a result of Davies [2, Proposition 4.25] who proved this for the case when the Hopf algebra H is the group algebra of a finite abelian group.The rest of the section is dedicated to proving this result.
Theorem 3.1.Let A = T (V )/(R) be a Koszul H-module algebra, where H is a Hopf algebra acting by degree-preserving endomorphisms.If µ is a counital 2-cocycle of H, then the Drinfeld twist A µ is a Koszul algebra given by T (V )/(R µ ) where R µ := {µ r | r ∈ R}.

3.2.
Plan for the proof.It follows immediately from Theorem 2.1 that A µ is a quadratic algebra of the form T (V )/(R µ ).Using this we can construct a complex K • (A µ ), which, by Witherspoon [9, Theorem 3.4.6], is a resolution of k as an A µ -module precisely when A µ is a Koszul algebra.We therefore work to show K • (A µ ) is indeed a resolution to complete the proof.We start by considering the Koszul resolution K • (A) of A, and twist this resolution using a functor of Giaquinto and Zhang.This produces a new resolution of k as an A µ -module, which we denote K • (A) µ .We then construct an isomorphism of complexes between ) is a resolution k as an A µ -module, as required.

The Koszul resolution of A.
Let us start by defining the Koszul resolution of k as an A-module.

It is given by K
The differentials d n are induced by the canonical embedding of K for a 0 , . . ., a n ∈ A and ǫ ′ is the augmentation map of A. K n (A) is a left A-module under multiplication on the leftmost tensor leg.It is also a left H-module by restricting the natural action of H on B n (A) = A ⊗(n+1) (using the coproduct △ of H) onto K n (A).K n (A) is closed under this H-action since we showed above that V and R are H-submodules of T (V ), so can also be seen as H-submodules of A and A ⊗ A respectively.Therefore K n (A) is an intersection of H-submodules of B n (A), so is an H-submodule itself.
3.4.The Giaquinto and Zhang twisting functor.Let A-Mod be the category of all left A-modules.Take the category (H, A)-Mod to be the subcategory of A-Mod whose objects M are also left H-modules where . The morphisms are those of A-Mod that are also H-module homomorphisms.
So far, Drinfeld twists have provided a mechanism for deforming the Hopf algebra H and the H-module algebra A. Giaquinto and Zhang [6, Theorem 1.7] extends this by defining a way to twist a module in (H, A)-Mod into a module in (H µ , A µ )-Mod.This new twist defines an equivalence of categories (H, A)-Mod − → (H µ , A µ )-Mod.We show next that the Koszul resolution K • (A) can be defined within (H, A)-Mod, and describe its image under this twisting functor of Giaquinto and Zhang.
Let us first check that (9) holds on K n (A), ∀n ≥ 0. For K 0 So ( 9) is satisfied and K n (A) is an object in (H, A)-Mod.It is standard that the differentials d n of the Koszul resolution are A-module homomorphisms, but we note that they are also H-module homomorphisms.Indeed, on inspecting (8), we see that as a map, where m denotes the product map of A and ǫ ′ is the augmentation map of A. Since m and ǫ ′ are H-module homomorphisms, it follows that d n also is.Therefore the whole Koszul resolution K • (A) can be defined within the category (H, A)-Mod.
We now apply the functor of Giaquinto and Zhang [6,Theorem 1.7].Under this functor the A-modules K n (A) are twisted into A µ -modules K n (A) µ in the following way: let K n (A) µ = K n (A) as k-vector spaces, and equip K n (A) µ with the action Here we use the fact that A and K n (A) are H-modules, so A ⊗ K n (A) is naturally an H ⊗ H-module.
Therefore µ −1 defines an endomorphism of A ⊗ K n (A), and also of A µ ⊗ K n (A) µ , since these are equal as k-vector spaces.
We must also apply the functor to the differentials in the Koszul resolution, but Giaquinto and Zhang do not explicitly describe how their functor behaves on morphisms.For completeness we prove here that we can take it as mapping each morphism in (H, A)-Mod to itself.Suppose V, W ∈ (H, A)-Mod with the actions of A on V and W denoted by V and W respectively.Let φ : V − → W be an A-module homomorphism, so where we use the fact id A ⊗ φ is an H-module homomorphism in the first equality.φ is also an H µ -module homomorphism V µ − → W µ , since H µ acts on V µ and W µ in exactly the same way that H acts on V and W respectively.Therefore φ can also be viewed as a morphism in (H µ , A µ )-Mod from V µ to W µ , and it makes sense to let the functor of Giaquinto and Zhang act identically on morphisms.Therefore each differential d n of the Koszul resolution K • (A) will be sent by the functor of Giaquinto and Zhang to itself.
To summarise, the result of applying the twisting functor to K • (A) is a complex K • (A) µ of A µ -modules which share the same underlying vector spaces, and differentials, as those on K • (A).Since K • (A) is a resolution of k as an A-module, we see that the twisted complex K • (A) µ must be a resolution of k as an A µ -module.
3.5.The Koszul complex of A µ .Using Theorem 2.1 we know that A µ is a quadratic algebra of the form T (V )/(R µ ).We can therefore construct a complex K • (A µ ) completely analogously to the Koszul resolution for A given in (7), and we describe this explicitly next.Let K • (A µ ) = n≥0 K n (A µ ) where This is a complex of A µ -modules, where the differentials are inherited from the Bar resolution B • (A µ ), where where we use the same augmentation map ǫ ′ as on A, since A µ and A have the same grading.
Note that by Witherspoon [9, Theorem 3.4.6],K • (A µ ) is a resolution of k as an A µ -module precisely when A µ is a Koszul algebra.We use this to prove the theorem, showing that K • (A µ ) is indeed a resolution.To do so we construct an isomorphism of complexes from K • (A) µ to K • (A µ ), and use the fact established in the last section that K • (A) µ is a resolution of k as an A µ -module.To construct this isomorphism of complexes we first define a sequence of elements which turn out to generalise counital 2-cocycles.
3.6.Higher counital 2-cocycles.First we introduce some notation: for i ∈ {1, . . ., n}, let △ (n) i : , where △ appears in the i-th position.By convention also let Recall that our counital 2-cocycle µ satisfies the 2-cocycle equation in (1).This equation can be rewritten in the above notation as: It turns out that µ can be viewed as just one step in a sequence of "higher counital 2-cocycles", where the next two lemmas justify this terminology.For n ≥ 0, let f n ∈ H ⊗(n+1) be defined as f 0 = 1, f 1 = µ and for n ≥ 2, where • denotes composition, and all products are taken in the algebra H ⊗(n+1) .Notice that , which is just the left hand side of (12).The following lemma can be viewed as saying that each of the elements f n satisfies a kind of "higher" version of the 2-cocycle equation ( 12): Lemma 3.2.For all n ∈ N and i ∈ {1, . . ., n}, for n ≥ 2: To finish proving the lemma we show that, for each n ≥ 2, f i n = f i+1 n ∀i ∈ {1, . . ., n − 1}.We do so by performing induction on n.The base case n = 2 holds since f 1 2 = f 2 2 is precisely equation (12).Now suppose the hypothesis holds for n = k − 1, and let us show it holds for n = k.Take some i ∈ {1, . . ., n − 1}.We know that f k−1 = f 1 k−1 , and now by the induction hypothesis we have Therefore, where in the first equality we use the definition of f i+1 k , and insert the expression (14) for f k−1 .Next we use the fact that △ (k) i+1 is an algebra homomorphism.In the 3rd equality we take the product of the first two terms in the previous line, and in the 4-th equality we apply the 2-cocycle equation (12).Finally notice that △ by coassociativity of H, and therefore the final expression above reduces to Recall that µ also satisfies the counital equation ( 2), i.e. (ǫ ⊗ id)(µ) = 1 = (id ⊗ ǫ)(µ), where ǫ is the counit of H.The following lemma proves that the elements f n satisfy a generalised notion of counitality, Lemma 3.3.For all n ≥ 1 and i ∈ {0, . . ., n}, Proof.The n = 1 case of this is precisely the counital axiom (2).For arbitrary n ≥ 1 and i ∈ {0, . . ., n}, we can apply Lemma 3.2 to express f n as ( so by the counitality of µ, this reduces to 1 ⊗n .Also, and this is equal to f n−1 since by the counit axiom for H, (id ⊗ ǫ) • △ = id.

3.7.
Defining the isomorphism of complexes F • .Using the higher counital 2-cocycles constructed in the last section we will now define an isomorphism of complexes between K • (A) µ and K • (A µ ).From this we will deduce that K • (A µ ) is a resolution, and therefore that A µ is a Koszul algebra.
Recall that the complex K • (A) embeds into the Bar resolution B • (A), so in particular K n (A) is a subspace of A ⊗n+1 , for all n ≥ 0. Additionally, the twisted module K n (A) µ is equal to K n (A) as a k-vector space, and therefore it is also a subspace of A ⊗n+1 .Since A is an H-module algebra, A ⊗n+1 is naturally an H ⊗n+1 -module.Therefore we may consider the action of f n ∈ H ⊗n+1 on the space K n (A) µ . Define ) to be the corresponding sequence of k-linear maps, i.e. for each n ≥ 0, F n is the map given by f n acting on K n (A) µ .We check in Section 3.8 that the image of F n is indeed inside K n (A µ ).The first few maps of F • are depicted in the diagram below, We will also check in Section 3.9 that the diagram above commutes, i.e. d µ n F n = F n−1 d n , ∀n ≥ 1, and in Section 3.10 that each F n has a k-linear inverse.These facts are sufficient to deduce that K • (A µ ) is a resolution of k as an A µ -module, as we will explain in Section 3.11.Remark 3.4.It is also possible to show that the k-linear maps F n are A µ -module homomorphisms, and therefore, in combination with the above properties, F • is an isomorphism of complexes.However it is not required that F n be an A µ -module homomorphism in order to prove K • (A µ ) is a resolution, so we won't include the proof of this fact here.For brevity though, we will still refer to F • as being a chain map, or isomorphism of complexes, throughout the paper.

Checking im(F
For the n = 0 case, F 0 = id, and as vector spaces, K 0 (A) µ = A = A µ = K 0 (A µ ), so the image of F 0 is equal to K 0 (A µ ), as required.
When n = 1, we have as vector spaces . Now V was defined to be the degree 1 component A 1 of A, and since the H-action on A is degree-preserving, V is an H-submodule of A.
Therefore A ⊗ V is an H ⊗ H-submodule of A ⊗ A, so the action of µ on K 1 (A) µ will be closed.Due to the equality of vector spaces K 1 (A) µ = K 1 (A µ ), we can view the image of µ as being in K 1 (A µ ).
In the first equality we use the definition of d µ n in (11).In the second we turn m µ into m, which results in the ( As A is an H-module algebra, we have h 1 = ǫ(h)1 where ǫ is the counit of H. Therefore the degree 0 component A 0 ∼ = k of A has the structure of a trivial H-module, i.e. h λ = ǫ(h)λ, ∀h ∈ H, λ ∈ A 0 .
Since the augmentation map ǫ ′ maps into k, we find h ǫ ′ (x) = ǫ(h)ǫ ′ (x).Using this, and the fact ǫ ′ is an H-module homomorphism, we find where we apply the "higher counitality" of f n (Lemma 3.3) in the final equality to say f n−1 = (id ⊗n ⊗ ǫ)(f n ).
3.10.The inverse chain map.Finally it is clear from the fact that µ is invertible that each map F n has a k-linear inverse.In particular, let Although this is not required, it can be checked that F −1 • also forms a chain map, and therefore that F • is an isomorphism of complexes, as claimed.
3.11.K n (A µ ) is a resolution of k as an A µ -module.Recall that the complex K • (A) µ , which arose from applying the functor of Giaquinto and Zhang to the Koszul resolution K n (A), is a resolution of k as an A µ -module.This means that the following complex is exact, We wish to show that K • (A µ ) is also a resolution of k as an A µ -module, as this is equivalent to A µ being a Koszul algebra.Hence we must show the following is exact, In particular we wish to show im(d µ n+1 ) = ker(d µ n ) for n ≥ 1 and im(d µ 1 ) = ker(ǫ ′ ).

Examples
Next we demonstrate Theorems 2.1 and 3.1 with several examples.In particular, in Section 4.1, we use Theorem 2.1 to determine the result of twisting the quantum plane by the quasitriangular structure of the quantum enveloping algebra U q (sl 2 ), whilst in Section 4.2 we apply Theorem 3.1 to provide a new proof of the Koszulity of the quantum symmetric and exterior algebras S −1 (V ) and −1 (V ).
4.1.The quantum plane and U q (sl 2 ).For a non-zero q ∈ k, the q-quantum plane is given by A = k x, y /(xy − qyx).In the following we explain why we can twist A by the quasitriangular structure R of the quantum enveloping algebra U q (sl 2 ), and we find via Theorem 2.1 that A R is the quadratic algebra equal to the q −1 -quantum plane k x, y /(xy − q −1 yx).
First recall that a quasitriangular structure on a Hopf algebra H is an invertible R ∈ H ⊗ H satisfying: where τ is the transposition map Let us introduce the quantum enveloping algebra U q (sl 2 ).We now suppose q also satisfies q 2 = 1.Then U q (sl 2 ) is defined to be the k-algebra generated by E, F, K, K −1 , and satisfying the relations U q (sl 2 ) may additionally be given the structure of a Hopf algebra via the following: Now the quantum plane A = k x, y /(xy − qyx) is, by construction, a quadratic algebra.Additionally, we may define a representation of U q (sl 2 ) on the degree 1 homogeneous subspace V = span k {x, y} of A as follows: It is well-known (see, for instance, [7, Exercise 9.1.13])that this action extends to make A a U q (sl 2 )-module algebra.Note that, by construction, this action is degree-preserving.
In the terminology of Vlaar [8, Theorem 6.7], U q (sl 2 ) has a quasitriangular structure R "up to completion" -meaning that R does not lie in U q (sl 2 ) ⊗ U q (sl 2 ), but rather in a completion of this space.Despite this technicality, R still satisfies axioms (1) and ( 2) of a counital 2-cocycle.However, since R is not in U q (sl 2 ) ⊗ U q (sl 2 ), we must check that there is still a well-defined action of R on A ⊗ A in order for us to define the twist of A by R.
Etingof [5,Remark 3.41] states that given two representations of U q (sl 2 ), say ρ : U q (sl 2 ) − → End(V ) and ρ ′ : U q (sl 2 ) − → End(W ), which are locally nilpotent (i.e.∀v ∈ V or W , ∃n ∈ N such that E n v = 0), we have that (ρ ⊗ ρ ′ )(R) is a well-defined operator on V ⊗ W .Therefore, if the action of U q (sl 2 ) on A is locally nilpotent then it will follow that R has a well-defined action on A ⊗ A. The action on A is indeed locally nilpotent, and this can be seen from how E acts on a general basis element of A: E x a y b = [b] q x a+1 y b−1 , where [b] q := q b − q −b q − q −1 (18) It is a simple exercise to check (18), first by showing E x a y b = x a (E y b ), and then using induction (in the degree b), to show E y b = [b] q xy b−1 .From (18), we see E b+1 x a y b = 0, and so U q (sl 2 ) indeed acts locally nilpotently on A. Hence R has a well-defined action on A ⊗ A and we may construct the Drinfeld twist A R .
Finally we can apply our first main result, Theorem 2.1.The conditions of the theorem are met since the action of U q (sl 2 ) on A is degree-preserving.We deduce that A R is a quadratic algebra, and is given by k x, y /(R (x ⊗ y − qy ⊗ x)).Vlaar [8, Equation 6.37] tells us how R acts on V ⊗ V explicity, and for the basis vectors x ⊗ y and y ⊗ x it is, R x ⊗ y = q −1/2 (x ⊗ y + (q − q −1 )y ⊗ x), R y ⊗ x = q −1/2 y ⊗ x Therefore R x ⊗ y − qy ⊗ x = q −1/2 (x ⊗ y − q −1 y ⊗ x) and so R (x ⊗ y − qy ⊗ x) is equal to the ideal (x ⊗ y − q −1 y ⊗ x).Hence A R is the q −1 -quantum plane.

Symmetric and Exterior algebras.
Here we will consider k = C, and in each example we will take the Hopf algebra H to be the group algebra CT , where T is the finite abelian group given by (C 2 ) n = t 1 , . . ., t n | t 2 i = 1, t i t j = t j t i for some n ≥ 2. The counit of H is given by ǫ(t) = 1, ∀t ∈ T , whilst the coproduct is △(t) = t ⊗ t ∀t ∈ T . Let, By [1, Lemma 4.5], µ is a counital 2-cocycle of H = CT .Note also that µ = µ −1 .
Consider an n-dimensional C-vector space V with a fixed basis x 1 , . . ., x n .For an n × n-matrix q = (q ij ) satisfying q ii = 1 and q ij q ji = 1, the corresponding quantum symmetric algebra is defined as Likewise the quantum exterior algebra is given as q (V ) := T (V )/(x i ⊗ x j + q ij x j ⊗ x i | 1 ≤ i, j ≤ n).
In the following examples we will be interested in the case when q = (−1), where The quantum symmetric and exterior algebras are known to be Koszul, but we show that this can also be deduced as an application of Theorem 3.1.

Theorem 2 . 1 .
Let A = T (V )/(R) be a quadratic H-module algebra, where H is a Hopf algebra acting by degree-preserving endomorphisms.If µ is a counital 2-cocycle of H, then the Drinfeld twist A µ is a quadratic algebra of the form T (V )/(R µ ) where R µ := {µ r | r ∈ R}.