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ON ITERATED TWISTED TENSOR PRODUCTS OF ALGEBRAS
PASCUAL JARA MART´INEZ, JAVIER L ´OPEZ PE ˜NA, FLORIN PANAITE, AND FRED VAN OYSTAEYEN
ABSTRACT. We introduce and study the definition, main properties and applications of iterated
twisted tensor products of algebras, motivated by the problem of defining a suitable representa-
tive for the product of spaces in noncommutative geometry. We find conditions for constructing
an iterated product of three factors, and prove that they are enough for building an iterated prod-
uct of any number of factors. As an example of the geometrical aspects of our construction,
we show how to construct differential forms and involutions on iterated products starting from
the corresponding structures on the factors, and give some examples of algebras that can be
described within our theory. We prove a certain result (called “invariance under twisting”) for a
twisted tensor product of two algebras, stating that the twisted tensor product does not change
when we apply certain kind of deformation. Under certain conditions, this invariance can be it-
erated, containing as particular cases a number of independent and previously unrelated results
from Hopf algebra theory.
INTRODUCTION
The difficulty of constructing concrete, nontrivial examples of noncommutative spaces start-
ing from simpler ones is a common problem in all different descriptions of noncommutative
geometry. If we think of the commutative situation, we have an easy procedure, the cartesian
product, which allows us to generate spaces of dimension as big as we want from lower di-
mensional spaces. Thinking in terms of the existing dualities between the categories of spaces
and the categories of (commutative) algebras, the natural replacement for the cartesian product
of commutative spaces turns out to be the tensor product of commutative algebras. The tensor
product has often been considered a replacement for the product of spaces represented by non-
commutative algebras. As it was pointed out in [CSV95], this is a very restricted approach.
If the “axiom” of noncommutative geometry consists in considering noncommutative algebras
as the representatives for the algebras of functions over certain “quantum” spaces, hence as-
suming that two different measurements (or functions) on this kind of spaces do not commute
to each other, then why should we assume that the measurements on the product commute to
each other? There is no reason for imposing this artificial commutation, hence what we need
is a “noncommutative” replacement of the tensor product of two algebras, which is supposed
Pascual Jara and Javier Lo´pez have been partially supported by projects MTM2004-08125 and FQM-266
(Junta de Andalucı´a Research Group). Javier Lo´pez has also been supported by the Spanish MEC FPU-
grant AP2003-4340 and the European Science Foundation Programme on NONCOMMUTATIVE GEOMETRY
(NOG). Florin Panaite and Fred Van Oystaeyen have been partially supported by the EC programme LIEGRITS,
RTN 2003, 505078, and by the bilateral project “New techniques in Hopf algebras and graded ring theory”, of
the Flemish and Romanian Ministries of Research. Florin Panaite has also been partially supported by the CEEX
programme of the Romanian Ministry of Education and Research, contract nr. CEx05-D11-11/2005 .
1

2 PASCUAL JARA MART´INEZ, JAVIER L ´OPEZ PE ˜NA, FLORIN PANAITE, AND FRED VAN OYSTAEYEN
to fit better as an analogue of the product of two noncommutative spaces and in the same time
to be a useful tool for overcoming the lack of examples formerly mentioned.
When we impose the natural restrictions a product should have, namely that it contains the
factors in a natural way and having linear size equal to the product of the linear sizes of the
factors, we arrive precisely at the definition of a twisted tensor product formerly studied by
many people, either for the particular case of algebras (cf. [Tam90], [CSV95], [VDVK94])
or aiming to define similar structures for discrete groups, Lie groups, Lie algebras and Hopf
algebras (as in [Tak81], [Maj90] and [Mic90]). Often, this structure appears in the so-called
factorization problem of studying under what conditions we may write an object as a product of
two subobjects having minimal intersection (see for instance the early paper [Maj90b]) . From
a purely algebraic point of view, twisted tensor products arise as a tool for building algebras
starting with simpler ones, and also, as shown in [VDVK94], in close relation with certain
nonlinear equations. Historically, the starting point of this theory was the “braided geometry”
developed by Majid in the early 1990’s, including the “braided tensor product” of algebras in a
braided monoidal category, of which the twisted tensor product of algebras is a sort of “local”
version.
Whatever the chosen approach to twisted tensor products is, a number of examples of both
classical and recently defined objects fits into this construction. Ordinary and graded tensor
products, crossed products, Ore extensions and skew group algebras are just some examples
of well-known constructions in classical ring theory that can be described as twisted tensor
products. In the Hopf algebras and quantum groups area we find smash products, Drinfeld and
Heisenberg doubles, and diagonal crossed products. With a more geometrical flavour, quantum
planes and tori may be realized as noncommutative products of commutative spaces. And last,
but not least, we may also find some physical models for which this structure is particularly
well suited, such as the Fock space representations of a particle system with generalized sta-
tistics, which is studied in [BM00] using techniques which arise directly from the realisation
of certain crossed enveloping algebras as twisted tensor products.
In the present work, our aim is to look at the twisted tensor product structure from a more
geometrical point of view, regarding it as the natural representative for the cartesian product
of noncommutative spaces. When we think of this construction geometrically, it becomes
unnatural to restrict ourselves to take the product of only two spaces, so it appears the problem
of finding suitable conditions that allow us to iterate the construction, and, whenever this is
possible, to check that the obtained iterated product is “associative” in the same sense in which
the usual tensor product is. Also, we will be interested in analyzing whether we may lift
geometrical invariants that we are able to calculate on the single factors to the iterated twisted
product and how to do this, if possible.
Being such an ubiquitous construction, there are several equivalent definitions of the twisted
tensor product appearing in the literature, often using different names and notation. In the
Preliminaries we recall some of the results we will use later on, fixing a unified notation.
Concretely, we introduce the definition of a twisted tensor product A ⊗R B of two algebras
A and B by means of a twisting map R : B ⊗ A → A ⊗ B, whose existence is sufficient

ON ITERATED TWISTED TENSOR PRODUCTS OF ALGEBRAS 3
for the existence of a deformed product in the tensor product vector space A ⊗ B, and is also
necessary when we impose unitality conditions.
In Section 2, we deal with the problem of iterating the twisted tensor products, and the
lifting of several structures to the iteration, finding that for three given algebras A, B and C,
and twisting maps R1 : B ⊗ A → A ⊗ B, R2 : C ⊗ B → B ⊗ C, R3 : C ⊗ A → A ⊗ C, a
sufficient condition for being able to define twisting maps T1 : C⊗(A⊗R1B) → (A⊗R1B)⊗C
and T2 : (B ⊗R2 C) ⊗ A → A ⊗ (B ⊗R2 C) associated to R1, R2 and R3 and ensuring that
the algebras A ⊗T2 (B ⊗R2 C) and (A ⊗R1 B) ⊗T1 C are equal, can be given in terms of the
twisting maps R1, R2 and R3 only. Namely, they have to satisfy the compatibility condition
(A⊗ R2) ◦ (R3 ⊗ B) ◦ (C ⊗ R1) = (R1 ⊗ C) ◦ (B ⊗ R3) ◦ (R2 ⊗ A).
This relation may be regarded as a “local” version of the hexagonal relation satisfied by the
braiding of a (strict) braided monoidal category. We also prove that whenever the algebras
and the twisting maps are unital, the compatibility condition is also necessary. As it happens
for the classical tensor product, and for the twisted tensor product, the iterated twisted tensor
product also satisfies a Universal Property, which we will state formally in Theorem 2.7. Once
the conditions needed to iterate the construction of the twisted tensor product are fulfilled, we
will prove the Coherence Theorem, stating that whenever one can build the iterated twisted
product of any three factors, it is possible to construct the iterated twisted product of any
number of factors, and that all the ways one might do this are essentially the same. This result
will allow us to lift to any iterated product every property that can be lifted to three-factors
iterated products. As applications of the former results we will characterize the modules over
an iterated twisted tensor product, also giving a method to build some of them from modules
given over each factor. As a first step towards our aim of building geometrical invariants over
these structures, we will show how to build the algebras of differential forms and how to lift
the involutions of ∗–algebras to the iterated twisted tensor products.
In Section 3, we illustrate our theory by presenting some examples of different structures
that arose in different areas of mathematics and can be constructed using our method. Two of
them (the generalized smash products and diagonal crossed products) come from Hopf algebra
theory, while the other two (the noncommutative 2n–planes defined by Connes and Dubois–
Violette, and the observable algebra A of Nill–Szlacha´nyi) appear in a more geometrical or
physical context. In particular, we show that the algebras defined by Connes and Dubois–
Violette can be seen as (iterated) noncommutative products of commutative algebras (as it
happens for the quantum planes and tori), and give a new proof of the fact that the algebra A
is an AF–algebra, proof which does not imply calculating any representation of it. We would
like to point out that the earliest nontrivial example of an iterated twisted tensor product of
algebras was given by Majid in [Maj90c], in the form of an iterated sequence of double cross
products of certain bialgebras.
Section 4 (together with several results from Section 2), illustrates the fact that Hopf algebra
theory represents not only a rich source of examples for the theory of twisted tensor products
of algebras, but also a valuable source of inspiration for it. In this section we prove a result,
called “invariance under twisting”, for a twisted tensor product of two algebras, which arose

4 PASCUAL JARA MART´INEZ, JAVIER L ´OPEZ PE ˜NA, FLORIN PANAITE, AND FRED VAN OYSTAEYEN
as a generalization of the invariance under twisting for the Hopf smash product (hence the
name). It states that if we start with a twisted tensor product A ⊗R B together with a certain
kind of datum corresponding to it, we can deform the multiplication of A to a new algebra
structure Ad, we can deform R to a new twisting map Rd : B ⊗ Ad → Ad ⊗ B, so that
the twisted tensor products Ad ⊗Rd B and A ⊗R B are isomorphic. It turns out that our
result is general enough to include as particular cases some more independent results from
Hopf algebra theory: the well-known theorem of Majid stating that the Drinfeld double of a
quasitriangular Hopf algebra is isomorphic to an ordinary smash product, a recent result of
Fiore–Steinacker–Wess from [FSW03] concerning a situation where a braided tensor product
can be “unbraided”, and also a recent result of Fiore from [Fi02] concerning a situation where
a smash product can be “decoupled” (this result in turn contains as a particular case the well–
known fact that a smash product corresponding to a strongly inner action is isomorphic to the
ordinary tensor product). We also prove that, under certain circumstances, our theorem can be
iterated, containing thus, as a particular case, the invariance under twisting of the two-sided
smash product from [BPVO].
Though we are mainly interested in results of geometrical nature, and hence most algebras
we would like to work with are defined over the field C of complex numbers, most of the
results can be stated with no change for algebras over a field or commutative ring k, that we
assume fixed throughout all the paper. All algebras will be supposed to be associative, and
usually unital, k–algebras. The term linear will always mean k–linear, and the unadorned
tensor product ⊗ will stand for the usual tensor product over k. We will also identify every
object with the identity map defined on it, so that A ⊗ f will mean IdA⊗f . For an algebra A
we will write µA to denote the product in A and uA : k → A its unit, and for an A–module M
we will use λM to denote the action of A on M . For bialgebras and Hopf algebras we use the
Sweedler-type notation ∆(h) = h1 ⊗ h2.
It is worth noting that the proofs of most of our main results are still valid if instead of
considering algebras over k we take algebras in an arbitrary monoidally closed category.
1. PRELIMINARIES
1.1. Twisted tensor products of algebras. The notion of twisted tensor product of algebras
has been independently discovered a number of times, and can be found in the literature un-
der different names and notation. In this section we collect some results that will be used
later, fixing a unified notation. Main references for definitions and proofs are [CSV95] and
[VDVK94].
When dealing with spaces that involve a number of tensor products, notation often becomes
obscure and complex. In order to overcome this difficulty, especially when dealing with iter-
ated products, we will use a graphical braiding notation in which tangle diagrams represent
morphisms in monoidal categories. For this braiding notation we refer to [RT90], [Maj94] and
[Kas95].
In this notation, a linear map f : A → B is simply represented by
A
f
B
. The composition of
morphisms can be written simply by placing the boxes corresponding to each morphism along

ON ITERATED TWISTED TENSOR PRODUCTS OF ALGEBRAS 5
the same string, being the topmost box the corresponding to the map that is applied in the first
place. Several strings placed aside will represent a tensor product of vector spaces (usually
algebras), and a tensor product of two linear maps, f ⊗ g : A⊗B → C ⊗D will be written as
A
f
C
B
g
D
.
With this notation, some well-known properties of morphisms on tensor products become
very intuitive. For instance, the identity f ⊗ g = (f ⊗D) ◦ (A⊗ g) = (C ⊗ g) ◦ (f ⊗ B) is
written in braiding notation as
A
f
C
B
g
D
≡
A
f
C
B
g
D
≡
A
f
C
B
g
D
There are several special classes of morphisms that will receive a particular treatment.
Namely, the identity will be simply written as a straight line (without any box on it), the
algebra product will be denoted by
A A
A
. With this notation, the associativity of the algebra
product can be written as:
A A A
A
≡
A A A
A
and the fact that f : A→ B is an algebra morphism may be drawn as
A A
f
B
≡
A A
f f
B
We will also adopt the convention of not writing the base field (or ring) whenever it appears as
a factor (representing the fact that scalars can be pushed in or out every factor). According to
this convention, the unit map of an algebra A is represented by A '!&"%#$
A
, and the compatibility of the
unit with the product and with algebra morphisms are respectively written as
A
A '!&"%#$
A
≡
A
A
≡
A
A '!&"%#$
A
and
A '!&"%#$
f
B
≡
B '!&"%#$
B
This conventions may also be applied to module morphisms. If M is a left A–module, we will
denote by
A M
M
the module action. Note that, in spite of the fact that the drawing is the same,
there is no risk of confusing the module action with the algebra product, since the strings are
labeled. Note that, for a morphism f : M → N of left A–modules, the module morphism

6 PASCUAL JARA MART´INEZ, JAVIER L ´OPEZ PE ˜NA, FLORIN PANAITE, AND FRED VAN OYSTAEYEN
property is not written the same way as the algebra morphism property, but as
A M
f
M
≡
A M
f
M
Recall that given two algebras A, B over k and R : B ⊗ A → A ⊗ B a k–linear map such
that
R ◦ (B ⊗ µA) = (µA ⊗B) ◦ (A⊗R) ◦ (R⊗A),(1.1)
R ◦ (µB ⊗ A) = (A⊗ µB) ◦ (R⊗ B) ◦ (B ⊗ R),(1.2)
then the application µR := (µA ⊗ µB) ◦ (A⊗ R⊗ B) is an associative product on A⊗ B. In
this case, the map R is said to be a twisting map, and we will denote by A ⊗R B the algebra
(A ⊗ B, µR) that has A ⊗ B as underlying vector space, endowed with the product µR. If,
using a Sweedler-type notation, we denote by R(b ⊗ a) = aR ⊗ bR = ar ⊗ br, for a ∈ A,
b ∈ B, then (1.1) and (1.2) may be rewritten as:
(aa′)R ⊗ bR = aRa′r ⊗ (bR)r,(1.3)
aR ⊗ (bb′)R = (aR)r ⊗ brb′R.(1.4)
In braiding notation, we will represent a twisting map R : B ⊗ A → A ⊗ B by a crossing
B A
R
A B
, where we will omit the label R when there is no risk of confusion, and equations (1.1)
and (1.2) are represented respectively by
B A A
A B
≡
B A A
A B
and
B B A
A B
≡
B B A
A B
For further use, we record the following consequence of (1.3) and (1.4):
(aa′)R ⊗ (bb′)R = (aR)R(a′r)r ⊗ (bR)r(b′R)r,(1.5)
for all a, a′ ∈ A and b, b′ ∈ B, where R and r are two more copies of R; in braiding notation
this last identity is written as:
B B A A
A B
≡
B B A A
A B

ON ITERATED TWISTED TENSOR PRODUCTS OF ALGEBRAS 7
Whenever A and B are unital, if R is a twisting map that satisfies the extra conditions
(1.6) R(1⊗ a) = a⊗ 1R(b⊗ 1) = 1⊗ b
}
then the canonical maps iA : A→ A⊗R B and iB : B → A⊗R B defined by iA(a) := a⊗ 1,
iB(b) := 1 ⊗ b, are algebra morphisms, and A ⊗R B is a unital algebra, with unit 1 ⊗ 1. In
this case, we say that R is a unital twisting map. Most of the twisting maps we will study are
unital; however, it is worth noting that associativity constraints do not depend on the unitality
of the twisting map. In braiding notation, the unitality conditions read
A
B '!&"%#$
A B
≡
A
B '!&"%#$
A B
and
B
A '!&"%#$
A B
≡
B
A '!&"%#$
A B
A special family of examples of twisting maps involves bijective maps. Concerning this
situation, we can state the following result from [CMZ02], which will be used later:
Proposition 1.1. Let A ⊗R B be a twisted tensor product of algebras such that the map R is
bijective, and denote by V : A ⊗ B → B ⊗ A its inverse. Then V is also a twisting map and
R is an algebra isomorphism between B ⊗V A and A⊗R B.
In classical homological algebra, the usual tensor product is commonly introduced by means
of its universal property, where the commutation between elements belonging to the first factor
and elements belonging to the second one is implicitly required. In this property, we have to
consider the canonical algebra monomorphisms iA : A →֒ A⊗B and iB : B →֒ A⊗B given
by iA(a) := a⊗1 and iB(b) := 1⊗b respectively. Because of the twisting map conditions, these
maps are still algebra morphisms when we consider a twisted tensor productA⊗RB instead of
A ⊗ B; moreover, twisted tensor products may be characterized as algebra structures defined
on A ⊗ B such that the above maps are algebra inclusions and satisfying a ⊗ b = iA(a)iB(b)
for all a ∈ A, b ∈ B. As a consequence, with a slight modification, that essentially involves
replacing the usual flip by the twisting map, one may also state a universal property for twisted
tensor products, as shown in [CIMZ00]:
Theorem 1.2. Let A, B be two k–algebras, and let R : B ⊗ A → A ⊗ B be a unital twisting
map. Given a k–algebra X , and algebra morphisms u : A→ X , v : B → X such that
(1.7) µX ◦ (v ⊗ u) = µX ◦ (u⊗ v) ◦R,
then we can find a unique algebra map ϕ : A⊗R B → X such that
ϕ ◦ iA = u,(1.8)
ϕ ◦ iB = v.(1.9)
IfA and B are ∗–algebras with involutions jA and jB , and R : B⊗A→ A⊗B is a twisting
map such that
(1.10) (R ◦ (jB ⊗ jA) ◦ τ) ◦ (R ◦ (jB ⊗ jA) ◦ τ) = A⊗ B,

8 PASCUAL JARA MART´INEZ, JAVIER L ´OPEZ PE ˜NA, FLORIN PANAITE, AND FRED VAN OYSTAEYEN
then A ⊗R B is a ∗–algebra with involution R ◦ (jB ⊗ jA) ◦ τ , where τ : A ⊗ B → B ⊗ A
denotes the usual flip. Moreover, if R is unital, then iA and iB become ∗–morphisms. This
involutive condition is written down in braiding notation in the following way:
A B
τ
jB jA
R
τ
jB jA
R
A B
≡
A B
A B
When we have a left A–moduleM , a leftB–moduleN , a twisting mapR : B⊗A→ A⊗B
and a linear map τM,B : B ⊗M → M ⊗B such that
τM,B ◦ (µB ⊗M) = (M ⊗ µB) ◦ (τM,B ⊗ B) ◦ (B ⊗ τM,B),(1.11)
τM,B ◦ (B ⊗ λM) = (λM ⊗B) ◦ (A⊗ τM,B) ◦ (R⊗M),(1.12)
then the map λτM,B : (A ⊗R B) ⊗ (M ⊗ N) → M ⊗ N defined by λτM,B := (λM ⊗ λN) ◦
(A ⊗ τM,B ⊗ N) yields a left (A ⊗R B)–module structure on M ⊗ N , which furthermore is
compatible with the inclusion of A. In this case, we say that τM,B is a (left) module twisting
map. If we denote by
B M
M B
the module twisting map, the module twisting conditions look the
same as the twisting conditions for algebra twisting maps (replacing A by M). Unlike what
happens for algebra twisting maps, usually is not enough to have a left (A ⊗R B)–module
structure on M ⊗N in order to recover a module twisting map. Some sufficient conditions for
this to happen can be found in [CSV95]
Besides module lifting conditions, in [CSV95] is shown how to lift twisting maps to algebras
of differential forms on them. More precisely:
Theorem 1.3. Let A, B be two algebras. Then any twisting map R : B⊗A→ A⊗B extends
to a unique twisting map R˜ : ΩB ⊗ ΩA→ ΩA⊗ ΩB which satisfies the conditions
R˜ ◦ (dB ⊗ ΩA) = (εA ⊗ dB) ◦ R˜,(1.13)
R˜ ◦ (ΩB ⊗ dA) = (dA ⊗ εB) ◦ R˜,(1.14)
where dA and dB denote the differentials on ΩA and ΩB, and εA, εB stand for the gradings
on ΩA and ΩB, respectively. Moreover, ΩA ⊗R˜ ΩB is a graded differential algebra with
differential d(ϕ⊗ ω) := dAϕ⊗ ω + (−1)|ϕ|ϕ⊗ dBω.

ON ITERATED TWISTED TENSOR PRODUCTS OF ALGEBRAS 9
Conditions (1.13) and (1.14) can be translated, in braiding notation, to the equalities
ΩB ΩA
dB
R˜
ΩAΩB
≡
ΩB ΩA
R˜
εA dB
ΩAΩB
and
ΩB ΩA
dA
R˜
ΩAΩB
≡
ΩB ΩA
R˜
dA εB
ΩAΩB
respectively.
1.2. The noncommutative planes of Connes and Dubois–Violette. The original definition
of noncommutative 4–planes (and 3–spheres) arises from some K–theoretic equations, in-
spired by the properties of the Bott projector on the cohomology of classical spheres. We
do not need this interpretation here, so we adopt directly the equivalent definition given by
means of generators and relations. Any reader interested in full details on the construction and
properties of noncommutative planes and spheres should look at [CDV02]. Our study will be
centered on the noncommutative planes associated to critical points of the scaling foliation,
as the definition of the noncommutative plane in these points is easily generalized to higher
dimensional frameworks.
Let us then consider θ ∈ Mn(R) an antisymmetric matrix, θ = (θµν), θνµ = −θµν , and let
Calg(R2nθ ) be the associative algebra generated by 2n elements {zµ, z¯µ}µ=1,...,n with relations
(1.15)
zµzν = λµνzνzµ
z¯µz¯ν = λµν z¯ν z¯µ
z¯µzν = λνµzν z¯µ


∀µ, ν = 1, . . . , n, being λ
µν := eiθµν .
Note that λνµ = (λµν)−1 = λµν for µ 6= ν, and λµµ = 1 by antisymmetry.
We can now endow the algebra Calg(R2nθ ) with the unique involution of C–algebras x 7→ x∗
such that (zµ)∗ = z¯µ. This involution gives a structure of ∗–algebra on Calg(R2nθ ). As a
∗–algebra, Calg(R2nθ ) is a deformation of the commutative algebra Calg(R2n) of complex poly-
nomial functions on R2n, and it reduces to it when we take θ = 0. The algebra Calg(R2nθ ) will
be then referred to as the (algebra of complex polynomial functions on the) noncommutative
2n–plane R2nθ . In fact, former relations define a deformation Cnθ of Cn, so we can identify the
noncommutative complex n–plane Cnθ with R2nθ by writing Calg(Cnθ ) := Calg(R2nθ ).
We define Ωalg(R2nθ ), the algebra of algebraic differential forms on the noncommutative
plane R2nθ , to be the complex unital associative graded algebra
Ωalg(R2nθ ) :=
⊕
p∈N
Ωpalg(R2nθ )
generated by 2n elements zµ, z¯µ of degree 0, with relations:
zµzν = λµνzνzµ
z¯µz¯ν = λµν z¯ν z¯µ
z¯µzν = λνµzν z¯µ


∀µ, ν = 1, . . . , n, being λ
µν := eiθµν ,

10 PASCUAL JARA MART´INEZ, JAVIER L ´OPEZ PE ˜NA, FLORIN PANAITE, AND FRED VAN OYSTAEYEN
and by 2n elements dzµ, dz¯µ of degree 1, with relations:
(1.16)
dzµdzν + λµνdzνdzµ = 0,
dz¯µdz¯ν + λµνdz¯νdz¯µ = 0,
dz¯µdzν + λνµdzνdz¯µ = 0,
zµdzν = λµνdzνzµ,
z¯µdz¯ν = λµνdz¯ν z¯µ,
z¯µdzν = λνµdzν z¯µ,
zµdz¯ν = λνµdz¯νzµ,



∀ µ, ν = 1, . . . , n.
In this setting, there exists a unique differential d of Ωalg(R2nθ ) (that is, an antiderivation
of degree 1 such that d2 = 0) which extends the mapping zµ 7→ dzµ, z¯µ 7→ dz¯µ. Indeed,
such a differential is obtained by extending the definition on the generators according to the
Leibniz rule. With this differential, Ωalg(R2nθ ) becomes a graded differential algebra. It is also
possible to extend the mapping zµ 7→ z¯µ, dzµ 7→ dz¯µ =: (dzµ) to the whole algebra Ωalg(R2nθ )
as an antilinear involution ω 7→ ω such that ωω′ = (−1)pqω′ω for any ω ∈ Ωpalg(R2nθ ), ω′ ∈
Ωqalg(R2nθ ). For this extension we have that dω = dω.
Our interest in these algebras arises from the fact that the noncommutative 4–plane can eas-
ily be realized as a twisted tensor product of two commutative algebras (namely as a twisted
product of two copies of C[x, x¯], which is nothing but the algebra of polynomial functions
on the complex plane), hence looking like the algebra representing a sort of noncommuta-
tive cartesian product of two commutative spaces. Our original interest in iterated twisted
tensor products came when we asked ourselves about the possibility of looking at the 2n–
noncommutative plane as a certain product of commutative algebras.
2. ITERATED TWISTED TENSOR PRODUCTS
In this section, our aim is to study the construction of iterated twisted tensor products. If we
think of twisted tensor products as natural noncommutative analogues for the usual cartesian
product of spaces, it is natural to require that the product of three or more spaces still respects
every single factor.
Morally, the construction of a twisting map boils down to giving a rule for exchanging
factors between the algebras involved in the product. A natural way for doing this would be
to perform a series of two factors twists, that should be related to the already given notion of
twisting map, and afterwards to apply algebra multiplication in each factor.
Suppose that A, B and C are algebras, let
R1 : B ⊗ A −→ A⊗ B,
R2 : C ⊗B −→ B ⊗ C,
R3 : C ⊗A −→ A⊗ C
(unital) twisting maps, and consider now the application
T1 : C ⊗ (A⊗R1 B) −→ (A⊗R1 B)⊗ C
given by T1 := (A⊗R2) ◦ (R3 ⊗ B). We can also build the map
T2 : (B ⊗R2 C)⊗ A −→ A⊗ (B ⊗R2 C)

ON ITERATED TWISTED TENSOR PRODUCTS OF ALGEBRAS 11
given by T2 = (R1 ⊗ C) ◦ (B ⊗ R3). It is a natural question to ask if these maps are twisting
maps. In general, this is not the case, as we will show in (Counter)example 2.2. In the following
Theorem, we state necessary and sufficient conditions for this to happen.
Theorem 2.1. With the above notation, the following conditions are equivalent:
(1) T1 is a twisting map.
(2) T2 is a twisting map.
(3) The maps R1, R2 and R3 satisfy the following compatibility condition (called the
hexagon equation):
(2.1) (A⊗ R2) ◦ (R3 ⊗ B) ◦ (C ⊗ R1) = (R1 ⊗ C) ◦ (B ⊗ R3) ◦ (R2 ⊗ A),
that is, the following diagram is commutative.
C ⊗ A⊗ B R3⊗B // A⊗ C ⊗B
A⊗R2
((P
PP
PP
PP
PP
PP
P
C ⊗ B ⊗A
C⊗R1
66nnnnnnnnnnnn
R2⊗A ((PP
PP
PP
PP
PP
PP
A⊗ B ⊗ C
B ⊗ C ⊗ A B⊗R3 // B ⊗ A⊗ C
R1⊗C
66nnnnnnnnnnnn
Moreover, if all the three conditions are satisfied, then the algebras A ⊗T2 (B ⊗R2 C) and
(A⊗R1 B)⊗T1 C are equal. In this case, we will denote this algebra by A⊗R1 B ⊗R2 C.
PROOF We prove only the equivalence between (1) and (3), being the equivalence between
(2) and (3) completely analogous.
3⇒1 Suppose that the hexagon equation is satisfied. In order to prove that T1 is a twisting
map, we have to check the conditions (1.1) and (1.2) for T1, namely, we have to check the
relations
T1 ◦ (C ⊗ µR1) = (µR1 ⊗ C) ◦ (A⊗ B ⊗ T1) ◦ (T1 ⊗ A⊗B),(2.2)
T1 ◦ (µC ⊗ A⊗ B) = (A⊗B ⊗ µC) ◦ (T1 ⊗ C) ◦ (C ⊗ T1).(2.3)
To prove this we use braiding notation. Taking into account that the hexagon equation is
written as:
C B A
A B C
≡
C B A
A B C

12 PASCUAL JARA MART´INEZ, JAVIER L ´OPEZ PE ˜NA, FLORIN PANAITE, AND FRED VAN OYSTAEYEN
the proof of condition (2.2) is given by:
C A B A B
A B C
[1]≡
C A B A B
A B C
[2]≡
C A B A B
A B C
[3]≡
C A B A B
A B C
where in [1] we use the twisting condition for R3, in [2] we use the twisting condition for R2,
and in [3] we use the hexagon equation. On the other hand, condition (2.3) is proven as follows:
C C A B
A B C
[1]≡
C C A B
A B C
[2]≡
C C A B
A B C
where now [1] is due to the twisting conditions for R3, and [2] to twisting conditions for R2.
This proves that T1 satisfies the pentagonal equations. Furthermore, if R2 and R3 are unital,
then we have that
T1(c⊗ 1⊗ 1) = (A⊗ R2)(R3 ⊗ B)(c⊗ 1⊗ 1) = (A⊗ R2)(1⊗ c⊗ 1) = 1⊗ 1⊗ c,
T1(1⊗ a⊗ b) = (A⊗ R2)(R3 ⊗ B)(1⊗ a⊗ b) = (A⊗ R2)(a⊗ 1⊗ b) = a⊗ b⊗ 1,
so T1 is also a unital twisting map.
1⇒3 Now we assume (2.2) and (2.3). It is enough to apply (2.2) to an element of the form
c⊗ 1⊗ b⊗ a⊗ 1 in order to recover the hexagon equation for a generic element c⊗ b⊗ a of
the tensor product C ⊗ B ⊗ A.
To finish the proof, assume that the three equivalent conditions are satisfied. To see that
the algebras A ⊗T2 (B ⊗R2 C) and (A ⊗R1 B) ⊗T1 C are equal, it is enough to expand the
expressions of the products
µT2 = (µA ⊗ µR2) ◦ (A⊗ T2 ⊗B ⊗ C),
µT1 = (µR1 ⊗ µC) ◦ (A⊗B ⊗ T1 ⊗ C),
and realize that they are exactly the same application, for which we only have to observe that
(A⊗ B ⊗ R2) ◦ (R1 ⊗ C ⊗B) = R1 ⊗R2 = (R1 ⊗B ⊗ C) ◦ (B ⊗A⊗ R2). 
When three twisting maps satisfy the hypotheses of Theorem 2.1, we will say either that
they are compatible twisting maps, or that the twisting maps satisfy the hexagon (or braid)
equation. If the twisting maps Ri are not unital, the hexagon equation is still sufficient for
getting associative products associated to T1 and T2, but in general we need unitality to recover
the compatibility condition from the associativity of the iterated products.

ON ITERATED TWISTED TENSOR PRODUCTS OF ALGEBRAS 13
One could wonder whether the braid relation is automatically satisfied for any three unital
twisting maps. This is not the case, as shown in the following example:
Example 2.2. Take H a noncocommutative (finite dimensional) bialgebra, A = B = H∗,
C = H . Consider the left regular action of H on H∗ given by (h ⇀ p)(h′) := p(h′h); with
this action, H∗ becomes a left H–module algebra, so we can define the twisting map induced
by the action as:
σ : H ⊗H∗ −→ H∗ ⊗H
h⊗ p 7−→ (h1 ⇀ p)⊗ h2.
If we consider now the twisting maps R1 : B ⊗ A −→ A ⊗ B, R2 : C ⊗ B −→ B ⊗ C,
R3 : C ⊗ A −→ A ⊗ C, defined as R1 := τ , R2 = R3 := σ, being τ the usual flip, then the
braid relation among R1, R2 and R3 boils down to the equality
(h1 ⇀ q)⊗ (h2 ⇀ p)⊗ h3 = (h2 ⇀ q)⊗ (h1 ⇀ p)⊗ h3,
for all h ∈ H , p, q ∈ H∗, but this relation is false, as we chose H to be noncocommutative.
Remark 2.3. The multiplication in the algebra A ⊗R1 B ⊗R2 C can be given, using the
Sweedler-type notation recalled before, by the formula:
(a⊗ b⊗ c)(a′ ⊗ b′ ⊗ c′) = a(a′R3)R1 ⊗ bR1b
′
R2 ⊗ (cR3)R2c
′.(2.4)
The next natural question that arises is whether whenever we have a twisting map T : C ⊗
(A ⊗R B) → (A ⊗R B) ⊗ C, it splits as a composition of two suitable twisting maps. Once
again, this is not possible in general.
Theorem 2.4 (Right splitting). Let A, B, C be algebras, R1 : B ⊗ A → A ⊗ B and T :
C ⊗ (A⊗R1 B) → (A⊗R1 B)⊗ C unital twisting maps. The following are equivalent:
(1) There exist R2 : C ⊗ B → B ⊗ C and R3 : C ⊗A→ A⊗ C twisting maps such that
T = (A⊗ R2) ◦ (R3 ⊗ B).
(2) The map T satisfies the (right) splitting conditions:
T (C ⊗ (A⊗ 1)) ⊆ (A⊗ 1)⊗ C,(2.5)
T (C ⊗ (1⊗ B)) ⊆ (1⊗ B)⊗ C.(2.6)
PROOF
1⇒2 It is trivial.
2⇒1 Because of the conditions imposed to T , the map R2 : C ⊗ B → B ⊗ C given as
the only k–linear map such that (uA ⊗ R2) ◦ (τ ⊗ B) = T ◦ (C ⊗ (uA ⊗ B)) is well defined.
From the fact that T is a twisting map it is immediately deduced that also R2 is a twisting
map. Analogously, we can define R3 : C ⊗ A → A ⊗ C as the only k–linear map such that
(A ⊗ τ) ◦ (R3 ⊗ uB) = T ◦ (C ⊗ (A ⊗ uB)), which is also a well defined twisting map. We

14 PASCUAL JARA MART´INEZ, JAVIER L ´OPEZ PE ˜NA, FLORIN PANAITE, AND FRED VAN OYSTAEYEN
only have to check that T = (A⊗R2) ◦ (R3 ⊗B). Using braiding notation we have
C A B
A B C
≡
C A B
B '!&"%#$ A '!&"%#$
A B C
[1]≡
C A B
B '!&"%#$ A '!&"%#$
A B C
[2]≡
C A B
A '!&"%#$
B '!&"%#$
A B C
[3]≡
[3]≡
C A B
B '!&"%#$ A '!&"%#$
A B C
≡
C A B
A '!&"%#$ B '!&"%#$
A B C
≡
C A B
A B C
as we wanted to show, and where in [1] we are using that T is a twisting map, and in [2] and
[3] the definitions of R3 and R2 respectively. 
Again, we can ask ourselves whether the condition we required for the twisting map T to
split might be trivial. The following example shows a situation in which an iterated twisted
tensor product cannot be split:
Example 2.5. We give an example of twisting maps R : B ⊗A→ A⊗B and T : C ⊗ (A⊗R
B) → (A⊗R B)⊗C for which it is not true that T (c⊗ (a⊗ 1)) ∈ (A⊗ 1)⊗C for all a ∈ A,
c ∈ C.
Let H be a finite dimensional Hopf algebra with antipode S. Recall that the Drinfeld double
D(H) is a Hopf algebra having H∗cop ⊗H as coalgebra structure and multiplication
(p⊗ h)(p′ ⊗ h′) = p(h1 ⇀ p′ ↼ S−1(h3))⊗ h2h′,
for all p, p′ ∈ H∗ and h, h′ ∈ H , where ⇀ and ↼ are the left and right regular actions of H on
H∗ given by (h ⇀ p)(h′) = p(h′h) and (p ↼ h)(h′) = p(hh′). The Heisenberg double H(H)
is the smash productH#H∗, where H∗ acts on H via the left regular action p ⇀ h = p(h2)h1.
Recall from [Lu94] that H(H) becomes a left D(H)-module algebra, with action
(p⊗ h) ⇀ (h′ ⊗ q) = p2(h′2)q2(h)(h′1 ⊗ p3q1S∗−1(p1)),
for all p, q ∈ H∗ and h, h′ ∈ H , which is just the left regular action of D(H) on H(H)
identified as vector space with D(H)∗.
Now, we takeA = H ,B = H∗, C = D(H),R : H∗⊗H → H⊗H∗, R(p⊗h) = p1 ⇀ h⊗p2
(hence H ⊗R H∗ = H#H∗ = H(H)) and
T : D(H)⊗H(H) → H(H)⊗D(H),
T ((p⊗ h)⊗ (h′ ⊗ q)) = (p⊗ h)1 ⇀ (h′ ⊗ q)⊗ (p⊗ h)2

ON ITERATED TWISTED TENSOR PRODUCTS OF ALGEBRAS 15
(hence H(H)⊗T D(H) = H(H)#D(H), so T is a twisting map). Now we can see that
T ((p⊗ h)⊗ (h′ ⊗ 1)) = p3(h′2)(h′1 ⊗ p4S∗−1(p2))⊗ (p1 ⊗ h),
which in general does not belong to (H ⊗ 1)⊗D(H).
Of course, there exists an analogous left splitting theorem, that we state for completeness,
and whose proof is analogous to the former one.
Theorem 2.6 (Left splitting). Let A, B, C be algebras, R2 : C ⊗ B → B ⊗ C and T :
(B ⊗R2 C)⊗A→ A⊗ (B ⊗R2 C) twisting maps. The following are equivalent:
(1) There exist R1 : B ⊗ A→ A⊗ B and R3 : C ⊗ A→ A⊗ C twisting maps such that
T = (R1 ⊗ C) ◦ (B ⊗R3).
(2) The map T satisfies the (left) splitting conditions:
T ((1⊗ C)⊗ A) ⊆ (A⊗ 1)⊗ C,(2.7)
T ((B ⊗ 1)⊗ A) ⊆ A⊗ (B ⊗ 1).(2.8)
The universal property (Theorem 1.2) formerly stated can be easily extended to the iterated
setting, as we show in the following result:
Theorem 2.7. Let (A,B,C,R1, R2, R3) be as in Theorem 2.1. Assume that we have a k–
algebra X and algebra morphisms u : A→ X , v : B → X , w : C → X , such that
(2.9) µX ◦ (w ⊗ v ⊗ u) = µX ◦ (u⊗ v ⊗ w) ◦ (A⊗R2) ◦ (R3 ⊗B) ◦ (C ⊗R1).
Then there exists a unique algebra map ϕ : A ⊗R1 B ⊗R2 C → X such that ϕ ◦ iA = u,
ϕ ◦ iB = v, ϕ ◦ iC = w.
PROOF Assume the we have a map ϕ satisfying the conditions in the theorem, then we may
write
ϕ(a⊗ b⊗ c) = ϕ((a⊗ 1⊗ 1)(1⊗ b⊗ 1)(1⊗ 1⊗ c))
= ϕ(a⊗ 1⊗ 1)ϕ(1⊗ b⊗ 1)ϕ(1⊗ 1⊗ c))
= ϕ(iA(a))ϕ(iB(b))ϕ(iC(c))
= u(a)v(b)w(c),
and so ϕ is uniquely defined.
For the existence part, define ϕ(a⊗ b ⊗ c) := u(a)v(b)w(c), and let us check that this map
is indeed an algebra morphism. Using formula (2.4), we have
ϕ((a⊗ b⊗ c)(a′ ⊗ b′ ⊗ c′)) = ϕ(a(a′R3)R1 ⊗ bR1b
′
R2 ⊗ (cR3)R2c
′)
= u(a)u((a′R3)R1)v(bR1)v(b
′
R2)w((cR3)R2)w(c
′).
On the other hand, we have
ϕ(a⊗ b⊗ c)ϕ(a′ ⊗ b′ ⊗ c′) = u(a)v(b)w(c)u(a′)v(b′)w(c′)
= u(a)v(b)u(a′R3)w(cR3)v(b
′)w(c′)
= u(a)u((a′R3)R1)v(bR1)v(b
′
R2)w((cR3)R2)w(c
′),

16 PASCUAL JARA MART´INEZ, JAVIER L ´OPEZ PE ˜NA, FLORIN PANAITE, AND FRED VAN OYSTAEYEN
and thus we conclude that ϕ is an algebra morphism. The fact that ϕ satisfies the required
relations with u, v and w is immediately deduced from its definition. 
To reach completely the aim of defining an analogue for the product of spaces, one should
be able to construct a product of any number of factors. In order to construct the three–factors
product, we had to add one extra condition, namely the hexagon equation, to the conditions that
were imposed for building the two–factors product (the twisting map conditions). Fortunately,
in order to build a general n–factors twisted product of algebras one needs no more conditions
besides the ones we have already met. Morally, this just means than having pentagonal (twist-
ing) and hexagonal (braiding) conditions, we can build any product without worrying about
where to put the parentheses. The way to prove this is using induction. As our induction hy-
pothesis, we assume that whenever we have n− 1 algebras B1, . . . , Bn−1, with a twisting map
Sij : Bj⊗Bi → Bi⊗Bj for every i < j, and such that for any i < j < k the maps Sij , Sjk and
Sik are compatible, then we can build the iterated product B1⊗S12 B2⊗S23 · · ·⊗Sn−1 nBn with-
out worrying about parentheses. Let then A1, . . . , An be algebras, Rij : Aj ⊗ Ai → Ai ⊗ Aj
twisting maps for every i < j, such that for any i < j < k the maps Rij , Rjk and Rik are
compatible. Define now for every i < n− 1 the map
T in−1,n : (An−1 ⊗Rn−1 n An)⊗ Ai → Ai ⊗ (An−1 ⊗Rn−1 n An)
by T in−1,n := (Ri n−1 ⊗ An) ◦ (An−1 ⊗ Ri n), which are twisting maps for every i, as we can
directly apply Theorem 2.1 to the maps Ri n−1, Ri n and Rn−1n. Furthermore, we have the
following result:
Lemma 2.8. In the above situation, for every i < j < n − 1, the maps Rij , T in−1,n and T jn−1,n
are compatible.
PROOF Using braiding notation the proof can be written as:
An−1 An Aj Ai
Ai Aj An−1 An
≡
An−1 An Aj Ai
Ai Aj An−1 An
[1]≡
An−1 An Aj Ai
Ai Aj An−1 An
[2]≡
An−1 An Aj Ai
Ai Aj An−1 An
≡
≡
An−1 An Aj Ai
Ai Aj An−1 An

ON ITERATED TWISTED TENSOR PRODUCTS OF ALGEBRAS 17
where in [1] we use the compatibility condition for Rij , Ri n−1 and Rj n−1, and in [2] we use
the compatibility condition for Rij , Rin and Rjn. 
So we can apply the induction hypothesis to the n−1 algebrasA1, . . . , An−2, and (An−1⊗Rn−1 n
An), and we obtain that we can build the twisted product of these n− 1 factors without worry-
ing about parentheses, so we can build the algebra
A1 ⊗R12 · · · ⊗ An−2 ⊗Tn−2n−1,n (An−1 ⊗Rn−1 n An).
Simply observing that
An−2 ⊗Tn−2n−1,n (An−1 ⊗Rn−1 n An) = (An−2 ⊗Rn−2 n−1 An−1)⊗Tnn−2 n−1 An,
we see that we could have grouped together any two consecutive factors. Summarizing, we
have sketched the proof of the following theorem (which we will not write formally to avoid
the cumbersome notation it would involve):
Theorem 2.9 (Coherence Theorem). Let A1, . . . , An be algebras, Rij : Aj ⊗ Ai → Ai ⊗ Aj
(unital) twisting maps for every i < j, such that for any i < j < k the maps Rij , Rjk and Rik
are compatible. Then the maps
T ij−1,j : (Aj−1 ⊗Rj−1 j Aj)⊗ Ai → Ai ⊗ (Aj−1 ⊗Rj−1 j Aj)
defined for every i < j − 1 by T ij−1,j := (Ri j−1 ⊗Aj) ◦ (Aj−1 ⊗ Ri j), and the maps
T ij−1,j : Ai ⊗ (Aj−1 ⊗Rj−1 j Aj) → (Aj−1 ⊗Rj−1 j Aj)⊗ Ai
defined for every i > j by T ij−1,j := (Aj−1 ⊗ Rj i) ◦ (Rj−1 i ⊗ Aj), are twisting maps with
the property that for every i, k /∈ {j − 1, j} the maps Rik, T in−1,n and T kn−1,n are compatible.
Moreover, for any i the (inductively defined) algebras
A1 ⊗R12 · · · ⊗Ri−3 i−2 Ai−2 ⊗T i−2i−1,i (Ai−1 ⊗Ri−1 i Ai)⊗T i+1i−1,i Ai+1 ⊗Ri+1 i+2 · · · ⊗Rn−1 n An
are all equal.
As a consequence of this theorem, any property that can be lifted to iterated twisted tensor
products of three factors can be lifted to products of any number of factors. One of the most
interesting consequences of the Coherence Theorem, or more accurately, of the former lemma,
is that we can state a universal property, analogous to Theorems 1.2 and 2.7. In order to state
the result it is convenient to introduce some notation. Let us first define the maps
T1 : An ⊗ · · · ⊗A1 −→ A1 ⊗ An ⊗ · · · ⊗ A2,
T1 := (R1n ⊗ IdAn−1⊗···⊗A2) ◦ · · · ◦ (IdAn⊗···⊗A3 ⊗R12),
T2 : A1 ⊗An ⊗ · · · ⊗A2 −→ A1 ⊗ A2 ⊗An ⊗ · · · ⊗ A3,
T2 := (A1 ⊗ R2n ⊗ IdAn−1⊗···⊗A3) ◦ · · · ◦ (IdA1⊗An⊗···⊗A4 ⊗R23),
.
.
.
Tn−1 : A1 ⊗ · · · ⊗ An−2 ⊗ An ⊗An−1 −→ A1 ⊗ · · · ⊗ An−2 ⊗ An−1 ⊗ An,
Tn−1 := A1 ⊗ · · · ⊗ An−2 ⊗ Rn−1n,

18 PASCUAL JARA MART´INEZ, JAVIER L ´OPEZ PE ˜NA, FLORIN PANAITE, AND FRED VAN OYSTAEYEN
and now define the map
S : An ⊗An−1 ⊗ · · · ⊗A1 −→ A1 ⊗ A2 ⊗ · · · ⊗An,
S := Tn−1 ◦ · · · ◦ T2 ◦ T1.
With this notation, we can state the Universal Property for iterated twisted tensor products as
follows:
Theorem 2.10 (Universal Property). Let A1, . . . , An be algebras, Rij : Aj ⊗ Ai → Ai ⊗ Aj
(unital) twisting maps for every i < j, such that for any i < j < k the maps Rij , Rjk and
Rik are compatible. Suppose that we have an algebra X together with n algebra morphisms
ui : Ai → X such that
(2.10) µX ◦ (un ⊗ · · · ⊗ u1) = µX ◦ (u1 ⊗ · · · ⊗ un) ◦ S.
Then there exists a unique algebra morphism
ϕ : A1 ⊗R12 A2 ⊗R23 · · · ⊗Rn−1 n An −→ X
such that
ϕ ◦ iAj = uj, for all j = 1, . . . , n.
PROOF Following the same procedure as in the proof of Theorem 2.7, it is easy to see that any
map ϕ verifying the conditions of the theorem must satisfy
ϕ(a1 ⊗ · · · ⊗ an) = u1(a1) · · · · · un(an),
and hence it must be unique. Whence it suffices to define ϕ as above. The checking of the
multiplicative property is also similar to the one done in the proof of Theorem 2.7, and thus is
left to the reader. 
A further step in the study of the iterated twisted tensor products is the lifting of module
structures on the factors. Again, if we have M a left A–module, N a left B–module, and P a
left C–module, the natural way in order to define a left (A⊗R1 B⊗R2 C)–module structure on
M ⊗N ⊗P is looking for module twisting maps τM,C : C ⊗M →M ⊗C, τM,B : B⊗M →
M ⊗B and τN,C : C ⊗N → N ⊗ C, and defining
λM⊗N⊗P := (λM ⊗ λN ⊗ λP ) ◦ (A⊗ τM,B ⊗ τN,C ⊗ P ) ◦ (A⊗B ⊗ τM,C ⊗N ⊗ P ).
However, as it happened with the iterated product of algebras, in order to have a left module
action it is not enough that τM,C , τN,C and τM,B are module twisting maps. Realize that, using
theA⊗R1 B–module structure induced on M ⊗N by τM,B , we can also write the above action
as
λM⊗N⊗P = (λM⊗N ⊗ λP ) ◦ (A⊗ B ⊗M ⊗ τN,C ⊗ P ) ◦ (A⊗ B ⊗ τM,C ⊗N ⊗ P )
= (λM⊗N ⊗ λP ) ◦ (A⊗ B ⊗ σC ⊗ P ),
where σC : C⊗ (M ⊗N) → (M ⊗N)⊗C is defined by σC := (M ⊗ τN,C) ◦ (τM,C ⊗N), so
proving that the three module twisting maps induce a left module structure on M ⊗N ⊗ P is
equivalent to prove that the map σC is a module twisting map, thus giving a left (A⊗R1 B)⊗T1

ON ITERATED TWISTED TENSOR PRODUCTS OF ALGEBRAS 19
C–module structure on (M ⊗N)⊗ P . We give sufficient conditions for this to happen in the
following result.
Theorem 2.11. With the above notation, suppose that the module twisting maps τM,C , τM,B
and the twisting map R2 satisfy the compatibility relation (also called the module hexagon
condition)
(2.11) (M ⊗ R2) ◦ (τM,C ⊗B) ◦ (C ⊗ τM,B) = (τM,B ⊗ C) ◦ (B ⊗ τM,C) ◦ (R2 ⊗M),
that is, the following diagram
C ⊗M ⊗ B
τM,C⊗B
// M ⊗ C ⊗ B
M⊗R2
((Q
QQ
QQ
QQ
QQ
QQ
Q
C ⊗B ⊗M
C⊗τM,B
66mmmmmmmmmmmm
R2⊗M ((QQ
QQ
QQ
QQ
QQ
QQ
M ⊗B ⊗ C
B ⊗ C ⊗M
B⊗τM,C
// B ⊗M ⊗ C
τM,B⊗C
66mmmmmmmmmmmm
is commutative; then:
(1) The map σC : C⊗(M⊗N) → (M⊗N)⊗C given by σC := (M⊗τN,C)◦(τM,C⊗N)
is a module twisting map.
(2) The map σB⊗C : (B ⊗ C) ⊗M → M ⊗ (B ⊗ C) given by σB⊗C := (τM,B ⊗ C) ◦
(B⊗ τM,C) is a module twisting map (giving a left A⊗T2 (B⊗R2 C)–module structure
on M ⊗ (N ⊗ P )).
Moreover, the module structures induced on M ⊗N ⊗ P by σC and σB⊗C are equal.
PROOF
1 We have to check that σC satisfies the conditions (1.11) and (1.12). For the first one, we
have that
C C M N
M N C
≡
C C M N
M N C
[1]≡
C C M N
M N C
[2]≡
C C M N
M N C
≡
C C M N
M N C

20 PASCUAL JARA MART´INEZ, JAVIER L ´OPEZ PE ˜NA, FLORIN PANAITE, AND FRED VAN OYSTAEYEN
where in [1] we are using the first module twisting condition for τM,C , and in [2] the first module
twisting condition for τN,C . For the second one, we have
C A B M N
M N C
≡
C A B M N
M N C
[1]≡
C A B M N
M N C
[2]≡
C A B M N
M N C
[3]≡
[3]≡
C A B M N
M N C
≡
C A B M N
M N C
where in [1] and [2] we use again the module twisting conditions and in [3] the module hexagon
condition.
The proof of (2) is very similar and left to the reader. 
Remark 2.12. Note that in this case we cannot prove the module hexagon condition from the
twisting conditions on the maps. The situation is similar to what happens for the case of the
existence of module twisting maps. It is reasonable to think that some sufficient conditions on
the modules and the algebras can be given in order to recover the converse. For instance, if the
modules are free, the situation is analogous to the iterated twisting construction for algebras,
and the converse result can easily be stated.
We recall that it is possible to give an explicit description of modules over various twisted
tensor products of algebras arising in Hopf algebra theory. The same holds in general, as the
next result shows.
Proposition 2.13. Let A, B be associative unital algebras and R : B ⊗ A → A ⊗ B a unital
twisting map. If M is a vector space, then M is a left A⊗R B-module if and only if it is a left
A-module and a left B-module satisfying the compatibility condition
(2.12) b · (a ·m) = aR · (bR ·m), for all a ∈ A, b ∈ B, m ∈M ,
where we denote by · both the actions of A and B.
PROOF IfM is a leftA⊗RB-module, it becomes a leftA-module with action a·m = (a⊗1)·m
and a leftB-module with action b ·m = (1⊗b) ·m. Conversely, it becomes an A⊗RB-module
with action (a⊗ b) ·m = a · (b ·m), details are left to the reader. 
This result can be iterated, generalizing thus the description of modules over a two-sided
smash product from [Pan02].

ON ITERATED TWISTED TENSOR PRODUCTS OF ALGEBRAS 21
Proposition 2.14. Assume that the hypotheses of Theorem 2.1 are satisfied, such that all alge-
bras and twisting maps are unital. If M is a vector space, then M is a left A ⊗R1 B ⊗R2 C-
module if and only if it is a left A-module, a left B-module, a left C-module (all actions are
denoted by ·) satisfying the compatibility conditions
b · (a ·m) = aR1 · (bR1 ·m),(2.13)
c · (b ·m) = bR2 · (cR2 ·m),(2.14)
c · (a ·m) = aR3 · (cR3 ·m),(2.15)
for all a ∈ A, b ∈ B, c ∈ C, m ∈ M (by Proposition 2.13, these conditions tell that M is a
left module over A⊗R1 B, B ⊗R2 C and A⊗R3 C).
PROOF We only prove that M becomes a left A ⊗R1 B ⊗R2 C-module with action (a ⊗ b ⊗
c) ·m = a · (b · (c ·m)). We compute (using formula (2.4)):
((a⊗ b⊗ c)(a′ ⊗ b′ ⊗ c′)) ·m = a(a′R3)R1 · (bR1b′R2 · ((cR3)R2c′ ·m))
(2.14) = a(a′R3)R1 · (bR1 · (cR3 · (b′ · (c′ ·m))))
(2.13) = a · (b · (a′R3 · (cR3 · (b′ · (c′ ·m)))))
(2.15) = a · (b · (c · (a′ · (b′ · (c′ ·m)))))
= (a⊗ b⊗ c) · ((a′ ⊗ b′ ⊗ c′) ·m),
finishing the proof. 
Our next result arises as a generalization of the fact from [HN99], [BPVO] that a two-sided
smash product over a Hopf algebra is isomorphic to a diagonal crossed product.
Proposition 2.15. Let (A,B,C,R1, R2, R3) be as in Theorem 2.1, and assume thatR2 is bijec-
tive with inverse V : B ⊗ C → C ⊗ B. Then (A,C,B,R3, V, R1) satisfy also the hypotheses
of Theorem 2.1, and the map A ⊗ R2 : A ⊗R3 C ⊗V B → A ⊗R1 B ⊗R2 C is an algebra
isomorphism.
PROOF By Proposition 1.1, V is a twisting map, and it is obvious that the hexagon condition
for (R3, V, R1) is equivalent to the one for (R1, R2, R3). ObviouslyA⊗R2 is bijective, we only
have to prove that it is an algebra map. This can be done either by direct computation or, more
conceptually, as follows. Denote T2 = (R1 ⊗ C) ◦ (B ⊗R3) and T˜2 = (R3 ⊗B) ◦ (C ⊗R1),
hence A ⊗R3 C ⊗V B = A ⊗T˜2 (C ⊗V B) and A ⊗R1 B ⊗R2 C = A ⊗T2 (B ⊗R2 C). By
Proposition 1.1 we know that R2 : C ⊗V B → B ⊗R2 C is an algebra map, and we obviously
have (A⊗R2)◦ T˜2 = T2 ◦ (R2⊗A), because this is just the hexagon condition. Now it follows
that A⊗ R2 is an algebra map, using the following general fact from [BM00]: if A⊗R B and
C ⊗T D are twisted tensor products of algebras and f : A → C and g : B → D are algebra
maps satisfying the condition (f ⊗ g) ◦R = T ◦ (g ⊗ f), then f ⊗ g : A⊗R B → C ⊗T D is
an algebra map. 
As our main motivations aimed at applications of our construction to the field of noncom-
mutative geometry, we are especially interested in finding processes that allow us to lift con-
structions associated to geometrical invariants of the algebras to their (iterated) twisted tensor

22 PASCUAL JARA MART´INEZ, JAVIER L ´OPEZ PE ˜NA, FLORIN PANAITE, AND FRED VAN OYSTAEYEN
products. Among these geometrical invariants, the first one to be taken into account is of
course the algebra of differential forms. For the case of the twisted product of two algebras,
a twisted product of the algebras of universal differential forms is build in a unique way, as it
is shown in Theorem 1.3; there, the construction of these extended twisting maps is deduced
from the universal property of the first order universal differential calculus. This extension
is compatible with our extra condition for constructing iterated products, as we show in the
following result:
Theorem 2.16. LetA,B, C be algebras, and letR1 : B⊗A −→ A⊗B,R2 : C⊗B −→ B⊗C,
R3 : C ⊗ A −→ A ⊗ C be twisting maps satisfying the hexagon equation, then the extended
twisting maps R˜1, R˜2 and R˜3 also satisfy the hexagon equation. Moreover, ΩA⊗R˜1ΩB⊗R˜2ΩC
is a differential graded algebra, with differential
d = dA ⊗ ΩB ⊗ ΩC + εA ⊗ dB ⊗ ΩC + εA ⊗ εB ⊗ dC .
PROOF For proving that the extended twisting maps satisfy the hexagon equation, we use a
standard technique when dealing with algebras of differential forms.
Firstly, observe that when restricted to the zero degree part of the algebras of differential
forms, the extended twisting maps coincide with the original ones, and hence they trivially
satisfy the hexagon equation.
Now, suppose that we have elements ω ∈ ΩA, η ∈ ΩB, θ ∈ ΩC such that the hexagon
equation is satisfied when evaluated on ω ⊗ η ⊗ θ, and let us show that then the hexagon
equation is also satisfied when evaluated in dAω⊗ η⊗ θ, ω⊗dBη⊗ θ and ω⊗ η⊗dCθ, that is,
we will show that the hexagon condition is stable under application of any of the differentials
dA, dB and dC .
Let us start proving that the condition holds for ω⊗ η⊗dCθ. Using again braiding notation,
we have
ΩC ΩB ΩA
dC
ΩA ΩB ΩC
[1]≡
ΩC ΩB ΩA
εB dC
ΩA ΩB ΩC
[2]≡
ΩC ΩB ΩA
εB
εA dC
ΩA ΩB ΩC
[3]≡
ΩC ΩB ΩA
dC
εA εB
ΩA ΩB ΩC
[4]≡
ΩC ΩB ΩA
εA εB dC
ΩA ΩB ΩC
[5]≡
[5]≡
ΩC ΩB ΩA
dC
εA
ΩA ΩB ΩC
[6]≡
ΩC ΩB ΩA
dC
ΩA ΩB ΩC

ON ITERATED TWISTED TENSOR PRODUCTS OF ALGEBRAS 23
where in [1], [2], [5] and [6] we are using the property (1.13) for dC with respect to R2 and
R3 respectively, in [3] the (obvious) fact that the gradings commute with the extended twisting
maps (since they are homogeneous), and in [4] we are using the hexagon equation for ω⊗η⊗θ.
The corresponding proofs for ω ⊗ dη ⊗ θ and dω ⊗ η ⊗ θ are almost identical. Summarizing,
the hexagon condition is stable under differentials in ΩA, ΩB and ΩC.
Finally, suppose that we have elements ω ∈ ΩA, η ∈ ΩB, θ1, θ2 ∈ ΩC such that the
hexagon equation is satisfied when evaluated on ω ⊗ η ⊗ θ1 and ω2 ⊗ η ⊗ θ2, and let us show
that in this case the hexagon condition also holds on ω ⊗ η ⊗ θ1θ2:
ΩC ΩC ΩB ΩA
ΩA ΩB ΩC
[1]≡
ΩC ΩC ΩB ΩA
ΩA ΩB ΩC
[2]≡
ΩC ΩC ΩB ΩA
ΩA ΩB ΩC
[3]≡
ΩC ΩC ΩB ΩA
ΩA ΩB ΩC
[4]≡
ΩC ΩC ΩB ΩA
ΩA ΩB ΩC
[5]≡
[5]≡
ΩC ΩC ΩB ΩA
ΩA ΩB ΩC
[6]≡
ΩC ΩC ΩB ΩA
ΩA ΩB ΩC
where in [1], [2], [5] and [6] we use the pentagon equations (1.2) for the twisting maps R˜2 and
R˜3, and in [3] and [4] we use the hexagon condition for ω⊗ η⊗ θ1 and ω⊗ η⊗ θ2 respectively.
In a completely analogous way we can prove that the hexagon condition holds for ω⊗η1η2⊗θ
and ω1ω2⊗η⊗θ, that is: the hexagon condition remains stable under products in ΩA, ΩB and
ΩC.
Now, taking into account that ΩA, ΩB and ΩC are generated as graded differential algebras
by the elements of degree 0, we may conclude that the hexagon condition holds completely.
In order to prove that ΩA ⊗R˜1 ΩB ⊗R˜2 ΩC is a graded differential algebra, it is enough to
observe that ΩA⊗R˜1 ΩB⊗R˜2 ΩC = (ΩA⊗R˜1 ΩB)⊗T˜2 ΩC, the last being (because of Theorem
1.3) a graded differential algebra with differential
d = dA⊗R1B ⊗ ΩC + εA⊗R1B ⊗ dC,

24 PASCUAL JARA MART´INEZ, JAVIER L ´OPEZ PE ˜NA, FLORIN PANAITE, AND FRED VAN OYSTAEYEN
and taking into account that
dA⊗R1B = dA ⊗ ΩB + εA ⊗ dB,
εA⊗R1B = εA ⊗ εB,
we obtain
d = dA ⊗ ΩB ⊗ ΩC + εA ⊗ dB ⊗ ΩC + εA ⊗ εB ⊗ dC ,
as we wanted to show. 
As most of our motivation comes from some algebras used in Connes’ theory, in order to
deal properly with ∗–algebras we would like to find a suitable extension of condition (1.10) to
our framework. As the definition of the involution in a twisted tensor product also involves the
usual flip τ , before extending the conditions to an iterated product, we need a technical (and
easy to prove) result:
Lemma 2.17. Let A, B, C be algebras, and let R : B ⊗ A → A ⊗ B be a twisting map.
Consider also the usual flips τBC : B ⊗ C → C ⊗ B and τAC : A ⊗ C → C ⊗ A, then the
maps τAC , R and τBC satisfy the hexagon condition (in B ⊗A⊗ C).
PROOF Just write down both sides of the equation and realize they are equal. 
Remark 2.18. In general, we can say that any twisting map is compatible with a pair of usual
flips, regardless the ordering of the factors. As the inverse of a usual flip is also a usual flip,
we may also use this result when one of the flips is inverted.
Similarly to what happened with differential forms, in order to be able to extend the invo-
lutions to the iterated product, it is enough that condition (1.10) is satisfied for every pair of
algebras. More concretely, we have the following result:
Theorem 2.19. Let A, B, C be ∗–algebras with involutions jA, jB and jC respectively, R1 :
B⊗A→ A⊗B, R2 : C ⊗B → B⊗C and R3 : C ⊗A→ A⊗C compatible twisting maps
such that
(R1 ◦ (jB ⊗ jA) ◦ τAB) ◦ (R1 ◦ (jB ⊗ jA) ◦ τAB) = A⊗B,(2.16)
(R2 ◦ (jC ⊗ jB) ◦ τBC) ◦ (R2 ◦ (jC ⊗ jB) ◦ τBC) = B ⊗ C,(2.17)
(R3 ◦ (jC ⊗ jA) ◦ τAC) ◦ (R3 ◦ (jC ⊗ jA) ◦ τAC) = A⊗ C.(2.18)
Then A⊗R1 B ⊗R2 C is a ∗–algebra with involution
j = (R1 ⊗C) ◦ (B ⊗R3) ◦ (R2 ⊗A) ◦ (jC ⊗ jB ⊗ jA) ◦ (C ⊗ τAB) ◦ (τAC ⊗B) ◦ (A⊗ τBC),
where τAB : A ⊗ B → B ⊗ A, τBC : B ⊗ C → C ⊗ B, and τAC : A ⊗ C → C ⊗ A denote
the usual flips.
PROOF Consider j defined as above, and let us check that it is an involution, i. e., that
j2 = A ⊗ B ⊗ C. Firstly, observe that, if we denote by τ all the usual flips and by τ¯ their

ON ITERATED TWISTED TENSOR PRODUCTS OF ALGEBRAS 25
inverses, we have that
A B C
τ
τ
τ
jC jB jA
R2
R3
R1
τ
τ
τ
jC jB jA
R2
R3
R1
A B C
[1]≡
A B C
τ
τ
τ
jC jB jA
R2
R3
R1
τ
τ
τ
jC jB jA
R1
R3
R2
A B C
[2]≡
A B C
τ
τ
τ
jC jB jA
R2
R3
R1
τ
jB jA
R1
τ
τ
jC
R3
R2
A B C
[3]≡
A B C
τ
τ
τ
jC jB jA
R2
R3
τ¯
jA jB
τ
jB jA
R1
τ
jB jA
R1
τ
τ
jC
R3
R2
A B C
[4]≡
A B C
τ
τ
τ
jC jB jA
R2
R3
τ¯
jA jB
τ
τ
jC
R3
R2
A B C
[5]≡
A B C
τ
τ
τ
jC jB jA
R2
R3
jB jA
τ
τ
τ¯
jC
R3
R2
A B C
where in [1] we use the hexagon conditions for the flips (which is obvious) and the hexagon
conditions for R1, R2, R3, in [2] we use the fact that the involutions jA and jB commute with
the flips, and the hexagon condition for R1 and two flips (as stated in the former lemma).
Equivalence [3] is due to the fact that both the square of the involutions, and the composition
of a flip with its inverse are the identity. In [4] we apply (2.16), and in [5] we use again that the
involutions commute with the flips, plus the hexagon condition for τ−1AB and two usual flips. To
conclude the proof, observe that

26 PASCUAL JARA MART´INEZ, JAVIER L ´OPEZ PE ˜NA, FLORIN PANAITE, AND FRED VAN OYSTAEYEN
A B C
τ
τ
τ
jC jB jA
R2
R3
jB jA
τ
τ
τ¯
jC
R3
R2
A B C
[6]≡
A B C
τ
τ
τ
jC jB jA
R2
R3
jB
τ
jC jA
R3
τ¯
τ
R2
A B C
[7]≡
A B C
τ
τ
τ
jC jB jA
R2
jB
τ¯
jA jC
τ
jC jA
R3
τ
jC jA
R3
τ¯
τ
R2
A B C
[8]≡
A B C
τ
τ
τ
jC jB jA
R2
jB
τ¯
jA jC
τ¯
τ
R2
A B C
[9]≡
A B C
τ
jC jB
jA
R3
jA
τ
jC jB
R3
A B C
[10]≡
A B C
A B C
where in [6] we apply (twice) the commutation of jC with the flips, plus the hexagon for R3
and two flips (again because of the former lemma). Equality [7] holds again because we are
just adding a term (two squared involutions, a flip, and its inverse) that equals the identity,
while [8] holds by applying (2.17). [9] is due to the fact that in the last diagram the element of
A is not modified at all, since all the crossings are usual flips, and we get [10] using (2.17) and
the fact that jA is an involution. 
3. EXAMPLES
3.1. Generalized smash products. We begin by recalling the construction of the so-called
generalized smash products. Let H be a bialgebra. For a right H-comodule algebra (A, ρ) we
denote ρ(a) = a<0> ⊗ a<1>, for any a ∈ A. Similarly, for a left H-comodule algebra (B, λ),
if b ∈ B then we denote λ(b) = b[−1] ⊗ b[0].

ON ITERATED TWISTED TENSOR PRODUCTS OF ALGEBRAS 27
Let A be a left H-module algebra and B a left H-comodule algebra. Denote by A◮<B the
k-vector space A⊗ B with newly defined multiplication
(3.19) (a◮<b)(a′◮<b′) = a(b[−1] · a′)◮<b[0]b′,
for all a, a′ ∈ A and b, b′ ∈ B. Then A◮<B is an associative algebra with unit 1A◮<1B. If
we take B = H then A◮<H is just the ordinary smash product A#H , whose multiplication
is
(a#h)(a′#h′) = a(h1 · a′)#h2h′.
The algebra A◮<B is called the (left) generalized smash product of A and B.
Similarly, if B is a right H-module algebra and A is a right H-comodule algebra, then we
denote by A >◭ B the k-vector space A⊗ B with the newly defined multiplication
(3.20) (a >◭ b)(a′ >◭ b′) = aa′<0> >◭ (b · a′<1>)b′,
for all a, a′ ∈ A and b, b′ ∈ B. Then A >◭ B is an associative algebra with unit 1A >◭ 1B ,
called also the (right) generalized smash product of A and B.
We recall some facts from [BPVO]. Let H be a bialgebra, A a left H-module algebra, B
a right H-module algebra and A an H-bicomodule algebra. Then A◮<A becomes a right
H-comodule algebra with structure
A◮<A → (A◮<A)⊗H, a◮<u 7→ (a◮<u<0>)⊗ u<1>,
and A >◭ B becomes a left H-comodule algebra with structure
A >◭ B → H ⊗ (A >◭ B), u >◭ b 7→ u[−1] ⊗ (u[0] >◭ b).
Moreover, we have:
Proposition 3.1. ([BPVO]) (A◮<A) >◭ B ≡ A◮<(A >◭ B) as algebras. If A = H , this
algebra is denoted by A#H#B and is called a two-sided smash product.
This result is a particular case of Theorem 2.1. Indeed, define the maps
R1 : A ⊗A→ A⊗ A, R1(u⊗ a) = u[−1] · a⊗ u[0],
R2 : B ⊗ A → A ⊗B, R2(b⊗ u) = u<0> ⊗ b · u<1>,
R3 : B ⊗ A→ A⊗ B, R3(b⊗ a) = a⊗ b,
which obviously are twisting maps because A⊗R1 A = A◮<A and A ⊗R2 B = A >◭ B are
associative algebras. Moreover, if we define the maps
T1 : B ⊗ (A⊗ A) → (A⊗ A)⊗B, T1 := (A⊗ R2) ◦ (R3 ⊗ A),
T2 : (A ⊗ B)⊗ A→ A⊗ (A ⊗B), T2 := (R1 ⊗B) ◦ (A ⊗R3),
then one can see that
(A◮<A)⊗T1 B = (A◮<A) >◭ B, A⊗T2 (A >◭ B) = A◮<(A >◭ B).

28 PASCUAL JARA MART´INEZ, JAVIER L ´OPEZ PE ˜NA, FLORIN PANAITE, AND FRED VAN OYSTAEYEN
3.2. Generalized diagonal crossed products. We recall the construction of the so-called
generalized diagonal crossed product, cf. [BPVO], [HN99]. Let H be a Hopf algebra with
bijective antipode S, A an H-bimodule algebra and A an H-bicomodule algebra. Then the
generalized diagonal crossed product A ⊲⊳ A is the following associative algebra structure on
A⊗ A:
(3.21) (ϕ ⊲⊳ u)(ϕ′ ⊲⊳ u′) = ϕ(u{−1} · ϕ′ · S−1(u{1})) ⊲⊳ u{0}u′,
for all ϕ, ϕ′ ∈ A and u, u′ ∈ A, where
u{−1} ⊗ u{0} ⊗ u{1} := u<0>[−1] ⊗ u<0>[0] ⊗ u<1> = u[−1] ⊗ u[0]<0> ⊗ u[0]<1>.
We recall some facts from [PVO]. Let H be a Hopf algebra with bijective antipode S, A an
H-bimodule algebra and A an H-bicomodule algebra. Let also A be an algebra in the Yetter-
Drinfeld category HHYD, that is,A is a leftH-module algebra, a leftH-comodule algebra (with
left H-comodule structure denoted by a 7→ a(−1) ⊗ a(0) ∈ H ⊗ A) and the Yetter-Drinfeld
compatibility condition holds:
h1a(−1) ⊗ h2 · a(0) = (h1 · a)(−1)h2 ⊗ (h1 · a)(0), ∀ h ∈ H, a ∈ A.(3.22)
Consider first the generalized smash product A◮<A, as associative algebra. From the condi-
tion (3.22), it follows that A◮<A becomes an H-bimodule algebra, with H-actions
h · (ϕ◮<a) = h1 · ϕ◮<h2 · a,
(ϕ◮<a) · h = ϕ · h◮<a,
for all h ∈ H , ϕ ∈ A and a ∈ A, hence we may consider the algebra (A◮<A) ⊲⊳ A.
Then, consider the generalized smash product A◮<A, as associative algebra. Using the con-
dition (3.22), one can see that A◮<A becomes an H-bicomodule algebra, with H-coactions
ρ : A◮<A → (A◮<A)⊗H, ρ(a◮<u) = (a◮<u<0>)⊗ u<1>,
λ : A◮<A → H ⊗ (A◮<A), λ(a◮<u) = a(−1)u[−1] ⊗ (a(0)◮<u[0]),
for all a ∈ A and u ∈ A, hence we may consider the algebra A ⊲⊳ (A◮<A).
A similar computation to the one in the proof of Proposition 3.4 in [PVO] shows:
Proposition 3.2. We have an algebra isomorphism (A◮<A) ⊲⊳ A ≡ A ⊲⊳ (A◮<A), given by
the trivial identification.
This result is also a particular case of Theorem 2.1. Indeed, define the maps:
R1 : A⊗A → A⊗ A, R1(a⊗ ϕ) = a(−1) · ϕ⊗ a(0),
R2 : A ⊗ A→ A⊗ A, R2(u⊗ a) = u[−1] · a⊗ u[0],
R3 : A ⊗A → A⊗ A, R3(u⊗ ϕ) = u{−1} · ϕ · S−1(u{1})⊗ u{0},
which are all twisting maps because A⊗R1 A = A◮<A, A⊗R2 A = A◮<A and A⊗R3 A =
A ⊲⊳ A are associative algebras. Now, if we define the maps
T1 : A ⊗ (A⊗ A) → (A⊗A)⊗ A, T1 := (A⊗ R2) ◦ (R3 ⊗ A),
T2 : (A⊗ A)⊗A → A⊗ (A⊗ A), T2 := (R1 ⊗ A) ◦ (A⊗ R3),

ON ITERATED TWISTED TENSOR PRODUCTS OF ALGEBRAS 29
then one can check that we have
(A◮<A)⊗T1 A = (A◮<A) ⊲⊳ A, A⊗T2 (A◮<A) = A ⊲⊳ (A◮<A),
hence indeed we recover Proposition 3.2.
3.3. The noncommutative 2n–planes. Recall from section 1 that the noncommutative plane
associated to an antisymmetric matrix, θ = (θµν) ∈ Mn(R), is the associative algebra
Calg(R2nθ ) generated by 2n elements {zµ, z¯µ}µ=1,...,n with relations
zµzν = λµνzνzµ
z¯µz¯ν = λµν z¯ν z¯µ
z¯µzν = λνµzν z¯µ


∀µ, ν = 1, . . . , n, being λ
µν := eiθµν ,
and endowed with the ∗–operation induced by (zµ)∗ := z¯µ (cf. [CDV02]).
Observe that as θ is antisymmetric, we have that zµz¯µ = z¯µzµ, so for every µ = 1, . . . , n
the algebra Aµ generated by the elements zµ and z¯µ is commutative, so Aµ ∼= C[zµ, z¯µ].
We have then n commutative algebras (indeed, n copies of the polynomial algebra in two
variables) contained in the noncommutative plane. Consider, for µ < ν, the mappings defined
on generators by
Rµν : C[zν , z¯ν ]⊗ C[zµ, z¯µ] −→ C[zµ, z¯µ]⊗ C[zν , z¯ν ],
zν ⊗ zµ 7−→ λνµzµ ⊗ zν ,
z¯ν ⊗ z¯µ 7−→ λνµz¯µ ⊗ z¯ν ,
z¯ν ⊗ zµ 7−→ λµνzµ ⊗ z¯ν ,
zν ⊗ z¯µ 7−→ λµν z¯µ ⊗ zν .
Obviously these formulae extend in a unique way to (unital) twisting maps Rµν . Condition
(1.10) is trivially satisfied, so every possible twisted tensor product is still a ∗–algebra. As
on the algebra generators our twisting map is just the usual flip multiplied by a constant, the
hexagon condition is also satisfied in a straightforward way. The iterated twisted tensor product
C[z1, z¯1]⊗R12 C[z2, z¯2]⊗R23 · · · ⊗Rn−1 n C[zn, z¯n]
is isomorphic to the noncommutative 2n–planeCalg(R2nθ ). Furthermore, for every µ = 1, . . . , n,
let Ωµ := Ωalg(R2) be the graded differential algebra of algebraic differential forms build over
the algebra C[zµ, z¯µ], and observe that for µ < ν the map Rµν : Ων ⊗Ωµ −→ Ωµ⊗Ων defined
on generators by
zν ⊗ zµ 7−→ λνµzµ ⊗ zν , z¯ν ⊗ z¯µ 7−→ λνµz¯µ ⊗ z¯ν ,
z¯ν ⊗ zµ 7−→ λµνzµ ⊗ z¯ν , zν ⊗ z¯µ 7−→ λµν z¯µ ⊗ zν ,
dzν ⊗ dzµ 7−→ −λνµdzµ ⊗ dzν , dz¯ν ⊗ dz¯µ 7−→ −λνµdz¯µ ⊗ dz¯ν ,
dz¯ν ⊗ dzµ 7−→ −λµνdzµ ⊗ dz¯ν , dzν ⊗ dz¯µ 7−→ −λµνdz¯µ ⊗ dzν ,
zν ⊗ dzµ 7−→ λνµdzµ ⊗ zν , z¯ν ⊗ dz¯µ 7−→ λνµdz¯µ ⊗ z¯ν ,
z¯ν ⊗ dzµ 7−→ λµνdzµ ⊗ z¯ν , zν ⊗ dz¯µ 7−→ λµνdz¯µ ⊗ zν ,
extends in a unique way to a twisting map defined on Ων⊗Ωµ. This twisting map satisfies con-
ditions (1.13) and (1.14), hence, by the uniqueness of the twisting map extension to the algebras

30 PASCUAL JARA MART´INEZ, JAVIER L ´OPEZ PE ˜NA, FLORIN PANAITE, AND FRED VAN OYSTAEYEN
of differential forms given by Theorem 1.3, the mapsRµν coincide with the maps R˜µν obtained
in the theorem. So, by applying Theorem 2.16 it follows that they are compatible twisting
maps. It is then easy to check that the iterated twisted tensor product Ω1 ⊗R12 · · ·⊗Rn−1n Ωn is
isomorphic, as a graded (involutive) differential algebra, to the algebra Ωalg(R2nθ ) of algebraic
differential forms on the noncommutative 2n–plane.
3.4. The Observable Algebra of Nill–Szlacha´nyi. In [NS97], Nill and Szlacha´nyi construct,
given a finite dimensional C∗–Hopf algebra H and its dual Hˆ , the algebra of observables,
denoted by A, by means of the smash products defined by the natural actions existing between
H and Hˆ. Their interest in studying such an algebra arises as it turns out to be the observable
algebra of a generalized quantum spin chain with H–order and Hˆ–disorder symmetries, and
they also observe that when H = CG is a group algebra this algebra A becomes an ordinary
G–spin model. We do not need here the physical interpretation of this algebra, our aim is to
show that the construction of the algebra A carried out in [NS97] fits inside our framework of
iterated twisted tensor products.
We start by fixing H a finite dimensional C∗–Hopf algebra, that is, a C∗–algebra endowed
with a comultiplication ∆ : H → H ⊗ H , a counit ε : H → C and an antipode S : H → H
satisfying the usual compatibility relations required for defining Hopf algebras, and with the
extra assumptions that ∆ and ε are ∗–algebra morphisms, and such that S(S(x)∗)∗ = x for
all x ∈ H (see [Kas95, Section IV.8] for details). If H is a ∗–Hopf algebra, it follows that
S−1 = S¯ = ∗◦S ◦∗ is the antipode of the opposite Hopf algebraHop (see [Swe69] for details).
The dual Hopf algebra of a ∗–Hopf algebra is also a ∗–Hopf algebra, with involution given
by ϕ∗ := S(ϕ∗), where ϕ 7→ ϕ∗ is the antilinear involutive algebra automorphism given by
ϕ∗(x) := ϕ(x∗). We have canonical pairings between H and Hˆ given by
〈, 〉 : H ⊗ Hˆ → C, a⊗ ϕ 7→ 〈a, ϕ〉 := ϕ(a),
〈, 〉 : Hˆ ⊗H → C, ϕ⊗ a 7→ 〈ϕ, a〉 := ϕ(a),
that give a structure of dual pairing of Hopf algebras between H and Hˆ . Associated to this
pairing we have the natural actions
⊲: H ⊗ Hˆ → Hˆ, a⊗ ϕ 7→ ϕ1 〈a, ϕ2〉 ,
⊳: Hˆ ⊗H → Hˆ, ϕ⊗ a 7→ 〈ϕ1, a〉ϕ2.
Now, for every i ∈ Z, let us take Ai := Hˆ if i is odd and Ai := H if i is even, and define the
maps:
R2k 2k+1 : A2k+1 ⊗ A2k −→ A2k ⊗A2k+1,
ϕ⊗ a 7−→ (ϕ1 ⊲ a)⊗ ϕ2 = a1 〈a2, ϕ1〉 ⊗ ϕ2,
R2k−1 2k : A2k ⊗ A2k−1 −→ A2k−1 ⊗A2k,
a⊗ ϕ 7−→ (a1 ⊲ ϕ)⊗ a2 = ϕ1 〈ϕ2, a1〉 ⊗ a2,
Rij : Aj ⊗ Ai −→ Ai ⊗Aj ,
a⊗ b 7−→ b⊗ a, whenever j − i > 2.

ON ITERATED TWISTED TENSOR PRODUCTS OF ALGEBRAS 31
As all the maps Rij are either usual flips or the maps induced by a module algebra action, it
is clear that all of them are twisting maps. Furthermore, it is easy to check that they satisfy
condition (1.10), so they define an involution on every twisted tensor product. Let us now
check that these maps are compatible. More precisely, let i < j < k, and consider the three
maps Rij , Rjk, and Rik, and let us check that they satisfy the hexagon equation. We have to
distinguish among several cases:
• If both |j − i| , |k − j| ≥ 2, all three maps are just usual flips, and thus the hexagon
condition is trivially satisfied.
• If |j − i| = 1, |k − j| ≥ 2, then we have that both Rik and Rjk are usual flips, and so
the compatibility of Rij with them follows from Lemma 2.17. The same thing happens
if |k − j| = 1, |j − i| ≥ 2.
• If j = i + 1, k = i + 2, then only the map Ri i+2 is a flip. Then we face two possible
situations.
If i = 2n− 1 is odd, then, describing explicitly the maps, we have that
R2n−1 2n(a⊗ ϕ) = 〈ϕ2, a1〉ϕ1 ⊗ a2,
R2n 2n+1(ϕ⊗ b) = 〈b2, ϕ1〉 b1 ⊗ ϕ2.
Hence, applying the left-hand side of the hexagon equation to a generator a⊗ϕ⊗ b of
A2n+1 ⊗A2n ⊗ A2n−1 = H ⊗ Hˆ ⊗H , we have
(A2n−1 ⊗ R2n 2n−1)(τ ⊗A2n)(A2n−1 ⊗ R2n−1 2n)(a⊗ b⊗ c)
= (A2n−1 ⊗ R2n 2n−1)(τ ⊗A2n)(〈b2, ϕ1〉 a⊗ b1 ⊗ ϕ2)
= (A2n−1 ⊗ R2n 2n−1)(〈b2, ϕ1〉 a⊗ ϕ2 ⊗ b1)
= 〈b2, ϕ1〉 〈ϕ3, a1〉 b1 ⊗ ϕ2 ⊗ a2.
On the other hand, for the right hand side we get
(R2n−1 2n ⊗ A2n+1)(A2n ⊗ τ)(R2n 2n+1 ⊗ A2n−1)(a⊗ ϕ⊗ b)
= (R2n−1 2n ⊗ A2n+1)(A2n ⊗ τ)(〈ϕ2, a1〉ϕ1 ⊗ a1 ⊗ b)
= (R2n−1 2n ⊗A2n+1)(〈ϕ2, a1〉ϕ1 ⊗ b⊗ a1)
= 〈b2, ϕ1〉 〈ϕ3, a1〉 b1 ⊗ ϕ2 ⊗ a2,
where for both expressions we are using the coassociativity of Hˆ . This proves the
hexagon condition for i odd. For i even, the proof is very similar.
Now, once proven that any three twisting maps chosen from the above ones are compatible,
we can apply the Coherence Theorem and build any iterated twisted tensor product of these
algebras. In particular, for any n ≤ m ∈ Z we may define the algebras
An,m := An ⊗Rnn+1 An+1 ⊗ · · · ⊗Rm−1m Am.
It is easy to see that if n′ ≤ n and m ≤ m′, then An,m ⊆ An′,m′ and hence the inclusions
give us a direct system of algebras {An,m}n,m∈Z, being its direct limit lim−→An,m precisely the
observable algebra A defined in [NS97]. Furthermore, as the action that defines the twisting
map is a ∗–Hopf algebra action, we have an involution defined on any of these products, and

32 PASCUAL JARA MART´INEZ, JAVIER L ´OPEZ PE ˜NA, FLORIN PANAITE, AND FRED VAN OYSTAEYEN
all the involved algebras being of finite dimension, we have no problem involving nuclearity
nor completeness, and henceforth all the algebras An,m are well defined, finite dimensional
C∗–algebras (a fact that was proven in [NS97] using representations of these algebras on some
Hilbert spaces). In particular, we get a new proof of the fact that the algebra A is an AF–
algebra.
4. INVARIANCE UNDER TWISTING
We begin this section with a result which does not involve a twisted tensor product of alge-
bras and which is of independent interest.
Proposition 4.1. LetA,B be two algebras andR : B⊗A→ A⊗B a linear map, with notation
R(b ⊗ a) = aR ⊗ bR, for all a ∈ A and b ∈ B. Assume that we are given two linear maps,
µ : B ⊗ A → A, µ(b ⊗ a) = b · a, and ρ : A → A ⊗ B, ρ(a) = a(0) ⊗ a(1), and denote
a ∗ a′ := a(0)(a(1) · a′), for all a, a′ ∈ A. Assume that the following conditions are satisfied:
ρ(1) = 1⊗ 1, 1 · a = a, a(0)(a(1) · 1) = a, for all a ∈ A, and
b · (a ∗ a′) = a(0)R(bRa(1) · a′),(4.1)
ρ(a ∗ a′) = a(0)a′(0)R ⊗ a(1)Ra
′
(1),(4.2)
for all a, a′ ∈ A and b ∈ B. Then (A, ∗, 1) is an associative unital algebra, denoted in what
follows by Ad.
PROOF Obviously 1 is the unit, so we only prove the associativity of ∗; we compute:
(a ∗ a′) ∗ a′′ = (a ∗ a′)(0)((a ∗ a′)(1) · a′′)
(4.2)
= a(0)a′(0)R(a(1)Ra
′
(1) · a′′),
a ∗ (a′ ∗ a′′) = a(0)(a(1) · (a′ ∗ a′′))
(4.1)
= a(0)a′(0)R(a(1)Ra
′
(1) · a′′),
and we see that the two terms are equal. 
Remark 4.2. The datum in Proposition 4.1 is a generalization of the left-right version of a
so-called left twisting datum in [FST99], which is obtained if B is a bialgebra and the map R
is given by R(b⊗ a) = b1 · a⊗ b2.
As a consequence of Proposition 4.1 we can obtain the following result from [BCZ96]:
Corollary 4.3. ([BCZ96]) Let H be a bialgebra and A a right H-comodule algebra with co-
module structure A → A ⊗ H , a 7→ a(0) ⊗ a(1), together with a linear map H ⊗ A → A,
h⊗ a 7→ h · a, satisfying 1 · a = a, h · 1 = ε(h)1, for all h ∈ H , a ∈ A, and
(h2 · a)(0) ⊗ h1(h2 · a)(1) = h1 · a(0) ⊗ h2a(1),(4.3)
h · (a ∗ a′) = (h1 · a(0))(h2a(1) · a′),(4.4)
where we denoted a ∗ a′ = a(0)(a(1) · a′). Then (A, ∗, 1) is an associative algebra.

ON ITERATED TWISTED TENSOR PRODUCTS OF ALGEBRAS 33
PROOF We takeB = H andR : H⊗A→ A⊗H ,R(h⊗a) = h1 ·a⊗h2. Then (4.1) is exactly
(4.4) and (4.2) is an easy consequence of (4.3) and of the fact thatA is a comodule algebra. 
Theorem 4.4. Assume that the hypotheses of Proposition 4.1 are satisfied, such that moreover
R is a twisting map. Assume also that we are given a linear map λ : A→ A⊗B, with notation
λ(a) = a[0] ⊗ a[1], such that λ(1) = 1⊗ 1 and the following relations hold:
λ(aa′) = a[0] ∗ (a′R)[0] ⊗ (a′R)[1](a[1])R,(4.5)
a(0)[0] ⊗ a(0)[1]a(1) = a⊗ 1,(4.6)
a[0](0) ⊗ a[0](1)a[1] = a⊗ 1,(4.7)
for all a, a′ ∈ A. Define the map
Rd : B ⊗ Ad → Ad ⊗ B, Rd(b⊗ a) = (a(0)R)[0] ⊗ (a(0)R)[1]bRa(1).(4.8)
Then Rd is a twisting map and we have an algebra isomorphism
Ad ⊗Rd B ≃ A⊗R B, a⊗ b 7→ a(0) ⊗ a(1)b.
PROOF It is easy to see that Rd satisfies (1.6). We check (1.3) for Rd; we compute (denoting
also R = r = R = r copies of R):
(a ∗ a′)Rd ⊗ bRd = ((a ∗ a′)(0)R)[0] ⊗ ((a ∗ a′)(0)R)[1]bR(a ∗ a′)(1)
(4.2)
= ((a(0)a′(0)r)R)[0] ⊗ ((a(0)a′(0)r )R)[1]bRa(1)ra′(1)
(1.3)
= (a(0)R(a′(0)r)R)[0] ⊗ (a(0)R(a′(0)r)R)[1](bR)Ra(1)ra′(1)
(4.5)
= (a(0)R)[0] ∗ (((a′(0)r)R)r)[0] ⊗ (((a′(0)r)R)r)[1]((a(0)R)[1])r
(bR)Ra(1)ra′(1),
aRd ∗ a′rd ⊗ (bRd)rd = (a(0)R)[0] ∗ a′rd ⊗ ((a(0)R)[1]bRa(1))rd
= (a(0)R)[0] ∗ (a′(0)r)[0] ⊗ (a′(0)r)[1]((a(0)R)[1]bRa(1))ra′(1)
(1.4)
= (a(0)R)[0] ∗ (((a′(0)r)R)r)[0] ⊗ (((a′(0)r)R)r)[1]((a(0)R)[1])r
(bR)Ra(1)ra′(1),
and we see that the two terms coincide. Now we check (1.4) for Rd; we compute:
aRd ⊗ (bb′)Rd = (a(0)R)[0] ⊗ (a(0)R)[1](bb′)Ra(1)
(1.4)
= ((a(0)R)r)[0] ⊗ ((a(0)R)r)[1]brb′Ra(1),
(aRd)rd ⊗ brdb′Rd = ((a(0)R)[0])rd ⊗ brd(a(0)R)[1]b′Ra(1)
= ((((a(0)R)[0])(0))r)[0] ⊗ ((((a(0)R)[0])(0))r)[1]br
((a(0)R)[0])(1)(a(0)R)[1]b′Ra(1)
(4.7)
= ((a(0)R)r)[0] ⊗ ((a(0)R)r)[1]brb′Ra(1),
and we see that the two terms coincide, hence indeed Rd is a twisting map. We prove now
that the map ϕ : Ad ⊗Rd B → A ⊗R B, ϕ(a ⊗ b) = a(0) ⊗ a(1)b, is an algebra isomorphism.

34 PASCUAL JARA MART´INEZ, JAVIER L ´OPEZ PE ˜NA, FLORIN PANAITE, AND FRED VAN OYSTAEYEN
First, using (4.6) and (4.7), it is easy to see that ϕ is bijective, with inverse given by a ⊗ b 7→
a[0]⊗a[1]b. It is obvious that ϕ(1⊗1) = 1⊗1, so we only have to prove that ϕ is multiplicative.
We compute:
ϕ((a⊗ b)(a′ ⊗ b′)) = ϕ(a ∗ a′Rd ⊗ bRdb′)
= ϕ(a ∗ (a′(0)R)[0] ⊗ (a
′
(0)R)[1]bRa
′
(1)b′)
= (a ∗ (a′(0)R)[0])(0) ⊗ (a ∗ (a
′
(0)R)[0])(1)(a
′
(0)R)[1]bRa
′
(1)b′
(4.2)
= a(0)(((a′(0)R)[0])(0))r ⊗ a(1)r((a
′
(0)R)[0])(1)(a
′
(0)R)[1]bRa
′
(1)b′
(4.7)
= a(0)(a′(0)R)r ⊗ a(1)rbRa
′
(1)b′,
ϕ(a⊗ b)ϕ(a′ ⊗ b′) = (a(0) ⊗ a(1)b)(a′(0) ⊗ a′(1)b′)
= a(0)a′(0)R ⊗ (a(1)b)Ra
′
(1)b′
(1.4)
= a(0)(a′(0)R)r ⊗ a(1)rbRa
′
(1)b′,
finishing the proof. 
Let H be a bialgebra and F ∈ H ⊗H a 2-cocycle, that is F is invertible and satisfies
(ε⊗ id)(F ) = (id⊗ ε)(F ) = 1,
(1⊗ F )(id⊗∆)(F ) = (F ⊗ 1)(∆⊗ id)(F ).
We denote F = F 1 ⊗ F 2 and F−1 = G1 ⊗ G2. We denote by HF the Drinfeld twist of H ,
which is a bialgebra having the same algebra structure as H and comultiplication given by
∆F (h) = F∆(h)F−1, for all h ∈ H .
If A is a left H-module algebra (with H-action denoted by h ⊗ a 7→ h · a), the invariance
under twisting of the smash product A#H is the following result (see [Maj97], [BPVO00]).
Define a new multiplication on A, by a ∗ a′ = (G1 · a)(G2 · a′), for all a, a′ ∈ A, and denote
by AF−1 the new structure; then AF−1 is a left HF -module algebra (with the same action as for
A) and we have an algebra isomorphism
AF−1#HF ≃ A#H, a#h 7→ G1 · a#G2h.(4.9)
We prove that this result is a particular case of Theorem 4.4.
We takeB = H and R : H⊗A→ A⊗H , R(h⊗a) = h1 ·a⊗h2, hence A⊗RB = A#H .
Define the maps
µ : H ⊗ A→ A, µ(h⊗ a) = h · a,
ρ : A→ A⊗H, ρ(a) = G1 · a⊗G2,
λ : A→ A⊗H, λ(a) = F 1 · a⊗ F 2,
hence the corresponding product ∗ on A is given by
a ∗ a′ = a(0)(a(1) · a′) = (G1 · a)(G2 · a′),

ON ITERATED TWISTED TENSOR PRODUCTS OF ALGEBRAS 35
which is exactly the product of AF−1 . One can check, by direct computation, that all the
necessary conditions for applying Theorem 4.4 are satisfied, hence we have the twisting map
Rd : H ⊗AF−1 → AF−1 ⊗H , which looks as follows:
Rd(h⊗ a) = (a(0)R)[0] ⊗ (a(0)R)[1]hRa(1)
= (h1 · a(0))[0] ⊗ (h1 · a(0))[1]h2a(1)
= (h1G1 · a)[0] ⊗ (h1G1 · a)[1]h2G2
= F 1h1G1 · a⊗ F 2h2G2
= h(1) · a⊗ h(2),
where we denoted by ∆F (h) = h(1) ⊗ h(2) the comultiplication of HF . Hence, we obtain that
Ad ⊗Rd B = AF−1 ⊗Rd H = AF−1#HF , and it is obvious that the isomorphism Ad ⊗Rd B ≃
A⊗R B provided by Theorem 4.4 coincides with the one given by (4.9).
Let H be a finite dimensional Hopf algebra with antipode S. As before, we work with
the realization of the Drinfeld double on H∗cop ⊗ H . A well-known theorem of Majid (see
[Maj91]) asserts that if (H, r) is quasitriangular then the Drinfeld double of H is isomorphic
to an ordinary smash product. More explicitly, for the realization of D(H) we work with, the
isomorphism is given as follows. First, we have a left H-module algebra structure on H∗,
denoted by H∗, given by (we denote r = r1 ⊗ r2):
h · ϕ = h1 ⇀ ϕ ↼ S−1(h2),
ϕ ∗ ϕ′ = (ϕ ↼ S−1(r1))(r21 ⇀ ϕ′ ↼ S−1(r22)),
for all h ∈ H and ϕ, ϕ′ ∈ H∗, and then we have an algebra isomorphism
H∗#H ≃ D(H), ϕ#h 7→ ϕ ↼ S−1(r1)⊗ r2h.(4.10)
We prove now that this result is also a particular case of Theorem 4.4.
We take A = H∗, with its ordinary algebra structure, B = H , and R : H ⊗H∗ → H∗ ⊗H ,
R(h⊗ ϕ) = h1 ⇀ ϕ ↼ S−1(h3)⊗ h2, hence A⊗R B = D(H). Denoting r−1 = u1 ⊗ u2, we
define the maps:
µ : H ⊗H∗ → H∗, µ(h⊗ ϕ) = h · ϕ = h1 ⇀ ϕ ↼ S−1(h2),
ρ : H∗ → H∗ ⊗H, ρ(ϕ) = ϕ ↼ S−1(r1)⊗ r2,
λ : H∗ → H∗ ⊗H, λ(ϕ) = ϕ ↼ S−1(u1)⊗ u2,
hence the corresponding product ∗ on H∗ is given by
ϕ ∗ ϕ′ = ϕ(0)(ϕ(1) · ϕ′)
= (ϕ ↼ S−1(r1))(r2 · ϕ′)
= (ϕ ↼ S−1(r1))(r21 ⇀ ϕ′ ↼ S−1(r22)),
which is exactly the product of H∗. One can check, by direct computation, that all the nec-
essary conditions for applying Theorem 4.4 are satisfied, hence we have the twisting map

36 PASCUAL JARA MART´INEZ, JAVIER L ´OPEZ PE ˜NA, FLORIN PANAITE, AND FRED VAN OYSTAEYEN
Rd : H ⊗H∗ → H∗ ⊗H , which looks as follows:
Rd(h⊗ ϕ) = (ϕ(0)R)[0] ⊗ (ϕ(0)R)[1]hRϕ(1)
= ϕ(0)R ↼ S−1(u1)⊗ u2hRϕ(1)
= (ϕ ↼ S−1(r1))R ↼ S−1(u1)⊗ u2hRr2
= h1 ⇀ ϕ ↼ S−1(r1)S−1(h3)S−1(u1)⊗ u2h2r2
= h1 ⇀ ϕ ↼ S−1(u1h3r1)⊗ u2h2r2
= h1 ⇀ ϕ ↼ S−1(h2)⊗ h3
= h1 · ϕ⊗ h2
(for the sixth equality we used the fact that ∆cop(h)r = r∆(h)), hence we obtain that Ad ⊗Rd
B = H∗ ⊗Rd H = H∗#H , and it is obvious that the isomorphism Ad ⊗Rd B ≃ A ⊗R B
provided by Theorem 4.4 coincides with the one given by (4.10).
Proposition 4.1 and Theorem 4.4 admit right-left versions, whose proofs are similar to the
left-right versions above and therefore will be omitted:
Proposition 4.5. Let B,C be two algebras and R : C ⊗ B → B ⊗ C a linear map, with
notation R(c ⊗ b) = bR ⊗ cR, for all b ∈ B and c ∈ C. Assume that we are given two linear
maps, ν : C⊗B → C, ν(c⊗ b) = c · b, and θ : C → B⊗C, θ(c) = c<−1>⊗ c<0>, and denote
c ∗ c′ = (c · c′<−1>)c′<0>, for all c, c′ ∈ C. Assume that the following conditions are satisfied:
θ(1) = 1⊗ 1, c · 1 = c, (1 · c<−1>)c<0> = c, for all c ∈ C, and
(c ∗ c′) · b = (c · c′<−1>bR)c′<0>R,(4.11)
θ(c ∗ c′) = c<−1>c′<−1>R ⊗ c<0>Rc
′
<0>,(4.12)
for all c, c′ ∈ C and b ∈ B. Then (C, ∗, 1) is an associative unital algebra, denoted in what
follows by dC.
Theorem 4.6. Assume that the hypotheses of Proposition 4.5 are satisfied, such that moreover
R is a twisting map. Assume also that we are given a linear map γ : C → B⊗C, with notation
γ(c) = c{−1} ⊗ c{0}, such that γ(1) = 1⊗ 1 and the following relations hold:
γ(cc′) = c′{−1}R(cR){−1} ⊗ (cR){0} ∗ c
′
{0},(4.13)
c<−1>c<0>{−1} ⊗ c<0>{0} = 1⊗ c,(4.14)
c{−1}c{0}<−1> ⊗ c{0}<0> = 1⊗ c,(4.15)
for all c, c′ ∈ C. Define the map
dR : dC ⊗ B → B ⊗ dC, dR(c⊗ b) = c<−1>bR(c<0>R){−1} ⊗ (c<0>R){0}.(4.16)
Then dR is a twisting map and we have an algebra isomorphism
B ⊗dR dC ≃ B ⊗R C, b⊗ c 7→ bc<−1> ⊗ c<0>.
A particular case of Theorem 4.6 is the invariance under twisting of the right smash product
from [BPVO]. Namely, let H be a bialgebra, C a rightH-module algebra (with action denoted
by c⊗h 7→ c·h) and F ∈ H⊗H a 2-cocycle. The right smash productH#C has multiplication
(h#c)(h′#c′) = hh′1#(c · h′2)c′.

ON ITERATED TWISTED TENSOR PRODUCTS OF ALGEBRAS 37
If we define a new multiplication on C, by c∗c′ = (c ·F 1)(c′ ·F 2) and denote the new structure
by FC, then FC becomes a right HF -module algebra and we have an algebra isomorphism
HF# FC ≃ H#C, h#c 7→ hF 1#c · F 2, see [BPVO]. This result may be reobtained as a
consequence of Theorem 4.6, by taking B = H , R(c ⊗ h) = h1 ⊗ c · h2, ν(c ⊗ h) = c · h,
θ(c) = F 1 ⊗ c · F 2, γ(c) = G1 ⊗ c ·G2, where we denoted as before F−1 = G1 ⊗G2.
A careful look at the proof of Theorem 4.4 shows that actually it admits a more general
form, which we record here for further use (the same holds for Theorem 4.6).
Theorem 4.7. LetA⊗RB be a twisted tensor product of algebras, and denote the multiplication
of A by a⊗ a′ 7→ aa′. Assume that on the vector space A we have one more algebra structure,
denoted by A′, with the same unit as A and multiplication denoted by a ⊗ a′ 7→ a ∗ a′ (for
instance, A′ may be A itself or Ad as in Proposition 4.1). Assume that we are given two linear
maps ρ, λ : A→ A⊗B, with notation ρ(a) = a(0) ⊗ a(1) and λ(a) = a[0] ⊗ a[1], such that ρ is
an algebra map from A′ to A⊗R B, λ(1) = 1⊗ 1 and relations (4.5), (4.6), (4.7) are satisfied.
Then the map
R′ : B ⊗A′ → A′ ⊗ B, R′(b⊗ a) = (a(0)R)[0] ⊗ (a(0)R)[1]bRa(1),(4.17)
is a twisting map and we have an algebra isomorphism
A′ ⊗R′ B ≃ A⊗R B, a⊗ b 7→ a(0) ⊗ a(1)b.
Theorem 4.8. LetB⊗RC be a twisted tensor product of algebras and denote the multiplication
of C by c⊗ c′ 7→ cc′. Assume that on the vector space C we have one more algebra structure,
denoted by C ′, with the same unit as C and multiplication denoted by c ⊗ c′ 7→ c ∗ c′ (for
instance, C ′ may be C itself or dC as in Proposition 4.5). Assume that we are given two linear
maps θ, γ : C → B ⊗ C, with notation θ(c) = c<−1> ⊗ c<0> and γ(c) = c{−1} ⊗ c{0}, such
that θ is an algebra map from C ′ to B ⊗R C, γ(1) = 1⊗ 1 and relations (4.13), (4.14), (4.15)
are satisfied. Then the map
R′ : C ′ ⊗B → B ⊗ C ′, R′(c⊗ b) = c<−1>bR(c<0>R){−1} ⊗ (c<0>R){0},(4.18)
is a twisting map and we have an algebra isomorphism
B ⊗R′ C ′ ≃ B ⊗R C, b⊗ c 7→ bc<−1> ⊗ c<0>.
We recall the following result of G. Fiore from [Fi02], in a slightly modified (but equivalent)
form. Let H be a Hopf algebra with antipode S and A a left H-module algebra. Assume that
there exists an algebra map ϕ : A#H → A such that ϕ(a#1) = a for all a ∈ A. Define the
map
θ : H → A⊗H, θ(h) = ϕ(1#S(h1))⊗ h2.
Then θ is an algebra map from H to A#H and the smash product A#H is isomorphic to the
ordinary tensor product A⊗H .
We prove that this result is a particular case of Theorem 4.8, with B = A and C = C ′ = H
(in the notation of Theorem 4.8).
Define the map γ : H → A ⊗ H , γ(h) = ϕ(1#h1) ⊗ h2, and denote as above θ(h) =
h<−1>⊗h<0> and γ(h) = h{−1}⊗h{0}. The relations (4.14) and (4.15) are easy to check, so we

38 PASCUAL JARA MART´INEZ, JAVIER L ´OPEZ PE ˜NA, FLORIN PANAITE, AND FRED VAN OYSTAEYEN
only have to prove (4.13) (here, the mapR : H⊗A→ A⊗H is given byR(h⊗a) = h1·a⊗h2).
We will need the following relation from [Fi02]:
ϕ(1#h)a = (h1 · a)ϕ(1#h2),(4.19)
for all h ∈ H , a ∈ A. Now we compute:
(h′{−1})R(hR){−1} ⊗ (hR){0}h′{0} = ϕ(1#h′1)Rϕ(1#(hR)1)⊗ (hR)2h′2
= (h1 · ϕ(1#h′1))ϕ(1#h2)⊗ h3h′2
(4.19)
= ϕ(1#h1)ϕ(1#h′1)⊗ h2h′2
= ϕ(1#h1h′1)⊗ h2h′2
= γ(hh′),
hence (4.13) holds. Theorem 4.8 may thus be applied, and we get the twisting map R′, which
looks as follows:
R′(h⊗ a) = h<−1>aR(h<0>R){−1} ⊗ (h<0>R){0}
= ϕ(1#S(h1))aR(h2R){−1} ⊗ (h2R){0}
= ϕ(1#S(h1))(h2 · a)(h3){−1} ⊗ (h3){0}
= ϕ(1#S(h1))(h2 · a)ϕ(1#h3)⊗ h4
(4.19)
= ϕ(1#S(h1))ϕ(1#h2)a⊗ h3
= ϕ(1#S(h1)h2)a⊗ h3
= a⊗ h,
so R′ is the usual flip, hence we obtain A#H ≃ A⊗H as a consequence of Theorem 4.8.
Remark 4.9. Let H be a Hopf algebra, let A be an algebra and u : H → A an algebra map;
consider the strongly inner action of H on A afforded by u, that is h · a = u(h1)au(S(h2)),
for all h ∈ H , a ∈ A. Then it is well-known (see for instance [Mon93], Example 7.3.3) that
the smash product A#H is isomorphic to the ordinary tensor product A ⊗ H . This result
is actually a particular case of Fiore’s theorem presented above (hence of Theorem 4.8 too),
because one can easily see that the map ϕ : A#H → A, ϕ(a#h) = au(h) is an algebra map
satisfying ϕ(a#1) = a for all a ∈ A.
We recall now the following result from [FSW03], with a different notation and in a slightly
modified (but equivalent) form, adapted to our purpose. Let (H, r) be a quasitriangular Hopf
algebra, H+ and H− two Hopf subalgebras of H such that r ∈ H+ ⊗ H− (we will denote
r = r1⊗r2 = R1⊗R2 ∈ H+⊗H−). Let B be a right H+-module algebra and C a right H−-
module algebra (actions are denoted by ·), and consider their braided product B⊗C, which is
just the twisted tensor product B ⊗R C, with twisting map given by
R : C ⊗ B → B ⊗ C, R(c⊗ b) = b · r1 ⊗ c · r2.
Assume that there exists an algebra map π : H+#B → B (where H+#B is the right smash
product recalled before) such that π(1#b) = b for all b ∈ B. Define the map
θ : C → B ⊗ C, θ(c) = π(r1#1)⊗ c · r2.

ON ITERATED TWISTED TENSOR PRODUCTS OF ALGEBRAS 39
Then θ is an algebra map from C to B⊗C and the braided tensor product B⊗C is isomorphic
to the ordinary tensor product B⊗C (hence the existence of π allows to “unbraid” the braided
tensor product; many examples where this happens may be found in [FSW03], especially
coming from quantum groups).
We prove now that this result is a particular case of Theorem 4.8, with C ′ = C (in the
notation of Theorem 4.8).
We first need to recall the axioms of a quasitriangular structure:
(∆⊗ id)(r) = r13r23,(4.20)
(id⊗∆)(r) = r13r12,(4.21)
∆cop(h)r = r∆(h), ∀ h ∈ H.(4.22)
Define the map γ : C → B ⊗C, γ(c) = π(u1#1)⊗ c · u2, where we denote r−1 = u1 ⊗ u2 =
U1 ⊗ U2 ∈ H+ ⊗H−. Denote as above θ(c) = c<−1> ⊗ c<0> and γ(c) = c{−1} ⊗ c{0}. The
relations (4.14) and (4.15) are easy to check, hence we only have to prove (4.13) (here, we
recall, ∗ coincides with the multiplication of C). We first record the relation:
c{−1}b⊗ c{0} = bR(cR){−1} ⊗ (cR){0}, ∀ b ∈ B, c ∈ C,(4.23)
which can be proved as follows:
bR(cR){−1} ⊗ (cR){0} = (b · r1)(c · r2){−1} ⊗ (c · r2){0}
= (b · r1)π(u1#1)⊗ c · r2u2
= π(1#b · r1)π(u1#1)⊗ c · r2u2
= π((1#b · r1)(u1#1))⊗ c · r2u2
= π(u11#b · r1u12)⊗ c · r2u2
(4.20)
= π(U1#b · r1u1)⊗ c · r2u2U2
= π(U1#b)⊗ c · U2
= π(U1#1)π(1#b)⊗ c · U2
= π(U1#1)b⊗ c · U2
= c{−1}b⊗ c{0}.
Now we compute:
γ(cc′) = π(u1#1)⊗ (c · u21)(c′ · u22)
(4.21)
= π(u1U1#1)⊗ (c · u2)(c′ · U2)
= π(u1#1)π(U1#1)⊗ (c · u2)(c′ · U2)
= c{−1}c′{−1} ⊗ c{0}c′{0}
(4.23)
= c′{−1}R(cR){−1} ⊗ (cR){0}c
′
{0},

40 PASCUAL JARA MART´INEZ, JAVIER L ´OPEZ PE ˜NA, FLORIN PANAITE, AND FRED VAN OYSTAEYEN
hence (4.13) holds. Theorem 4.8 may thus be applied, and we get the twisting map R′, which
looks as follows:
R′(c⊗ b) = c<−1>bR(c<0>R){−1} ⊗ (c<0>R){0}
= π(r1#1)bR((c · r2)R){−1} ⊗ ((c · r2)R){0}
= π(r1#1)(b · R1)(c · r2R2){−1} ⊗ (c · r2R2){0}
= π(r1#1)π(1#b · R1)π(u1#1)⊗ c · r2R2u2
= π((r1#1)(1#b · R1)(u1#1))⊗ c · r2R2u2
= π(r1u11#b · R1u12)⊗ c · r2R2u2
(4.20)
= π(r1U1#b · R1u1)⊗ c · r2R2u2U2
= π(1#b)⊗ c
= b⊗ c,
so R′ is the usual flip, hence we obtain B⊗C ≃ B ⊗ C as a consequence of Theorem 4.8.
A natural question that arises is to see whether Theorems 4.4 and 4.6 can be combined,
namely, if (A,B,C,R1, R2, R3) are as in Theorem 2.1 and we have a datum as in Theorem 4.4
between A and B and a datum as in Theorem 4.6 between B and C, under what conditions it
follows that (Ad, B, dC,Rd1, dR2, R3) satisfy again the hypotheses of Theorem 2.1.
Our first remark is that this does not happen in general, a counterexample may be obtained
as follows. Take B = H a bialgebra, A a left H-module algebra, C a right H-module algebra
and F ∈ H ⊗ H a 2-cocycle. Here R1(h ⊗ a) = h1 · a ⊗ h2, R2(c ⊗ h) = h1 ⊗ c · h2
and R3 = τCA, the usual flip, hence A ⊗R1 H ⊗R2 C = A#H#C, the two-sided smash
product. We consider the datum between A and H that allows us to define AF−1#HF , hence
Rd1(h⊗ a) = F 1h1G1 · a⊗F 2h2G2, and the trivial datum between H and C. One can see that
in general (Rd1, R2, R3) do not satisfy the hexagon condition.
Hence, the best we can do is to find sufficient conditions on the initial data ensuring that
(Rd1, dR2, R3) satisfy the hexagon condition. This is achieved in the next result. Note that the
conditions we found are not the most general one can imagine (in particular, we need to assume
that R3 is the flip), but they are general enough to include as a particular case the invariance
under twisting of the two-sided smash product from [BPVO], which was our guiding example
for this result.
Theorem 4.10. Let (A,B,C,R1, R2, R3) be as in Theorem 2.1, with R3 = τCA, the usual flip.
Assume that we have a datum between A and B as in Theorem 4.4 and a datum between B
and C as in Theorem 4.6, with notation as in these results. Assume also that the following
compatibility conditions hold:
a(0) ⊗ a(1)R2 (cR2){−1} ⊗ (cR2){0} = a(0)R1 ⊗ c{−1}R1a(1) ⊗ c{0},(4.24)
a[0] ⊗ c<−1>a[1]R2 ⊗ c<0>R2 = (aR1)[0] ⊗ (aR1)[1]c<−1>R1 ⊗ c<0>,(4.25)
a(0)R1 ⊗ c<−1>R1a(1)R2 ⊗ c<0>R2 = a(0) ⊗ a(1)c<−1> ⊗ c<0>,(4.26)

ON ITERATED TWISTED TENSOR PRODUCTS OF ALGEBRAS 41
for all a ∈ A, c ∈ C. Then (Ad, B, dC,Rd1, dR2, R3) satisfy also the hypotheses of Theorem
2.1, and we have an algebra isomorphism
Ad ⊗Rd1 B ⊗ dR2
dC ≃ A⊗R1 B ⊗R2 C, a⊗ b⊗ c 7→ a(0) ⊗ a(1)bc<−1> ⊗ c<0>.
PROOF We prove the hexagon condition for (Rd1, dR2, R3); we compute:
(A⊗ dR2) ◦ (R3 ⊗B) ◦ (C ⊗ Rd1)(c⊗ b⊗ a)
(4.8),(4.16)
= (a(0)R1 )[0] ⊗ c<−1>((a(0)R1 )[1]bR1a(1))R2(c<0>R2 ){−1} ⊗ (c<0>R2 ){0}
(1.3)
= (a(0)R1 )[0] ⊗ c<−1>((a(0)R1 )[1])R2(bR1)r2(a(1))R2(((c<0>R2 )r2)R2){−1}
⊗(((c<0>R2 )r2)R2){0}
(4.24)
= (((a(0))R1)R1)[0] ⊗ c<−1>((((a(0))R1)R1)[1])R2(bR1)r2
(((c<0>R2 )r2){−1})R1a(1) ⊗ ((c<0>R2 )r2){0},
(Rd1 ⊗ C) ◦ (B ⊗R3) ◦ (dR2 ⊗ A)(c⊗ b⊗ a)
(4.8),(4.16)
= (a(0)R1 )[0] ⊗ (a(0)R1 )[1](c<−1>bR2(c<0>R2 ){−1})R1a(1) ⊗ (c<0>R2 ){0}
(1.4)
= (((a(0)R1 )r1)R1)[0] ⊗ (((a(0)R1 )r1)R1)[1](c<−1>)R1(bR2)r1
((c<0>R2 ){−1})R1a(1) ⊗ (c<0>R2 ){0}
(4.25)
= ((a(0)R1 )r1)[0] ⊗ c<−1>(((a(0)R1 )r1)[1])r2(bR2)r1(((c<0>r2 )R2){−1})R1
a(1) ⊗ ((c<0>r2 )R2){0},
and the two terms are equal because of the hexagon condition for (R1, R2, R3):
aR1 ⊗ (bR1)R2 ⊗ cR2 = aR1 ⊗ (bR2)R1 ⊗ cR2 .
We prove now that the map
ψ : Ad ⊗Rd1 B ⊗ dR2
dC → A⊗R1 B ⊗R2 C,
ψ(a⊗ b⊗ c) = a(0) ⊗ a(1)bc<−1> ⊗ c<0>,
is an algebra isomorphism. First, using (4.6), (4.7), (4.14), (4.15), it is easy to see that ψ is
bijective, with inverse given by a ⊗ b ⊗ c 7→ a[0] ⊗ a[1]bc{−1} ⊗ c{0}. We prove now that ψ is
multiplicative. We compute (using (2.4)):

42 PASCUAL JARA MART´INEZ, JAVIER L ´OPEZ PE ˜NA, FLORIN PANAITE, AND FRED VAN OYSTAEYEN
ψ((a⊗ b⊗ c)(a′ ⊗ b′ ⊗ c′))
= ψ(a ∗ a′Rd1 ⊗ bRd1b
′
dR2 ⊗ c dR2 ∗ c
′)
(4.8),(4.16)
= ψ(a ∗ (a′(0)R1 )[0] ⊗ (a
′
(0)R1
)[1]bR1a′(1)c<−1>b′R2(c<0>R2 ){−1}
⊗(c<0>R2 ){0} ∗ c
′)
= (a ∗ (a′(0)R1 )[0])(0) ⊗ (a ∗ (a
′
(0)R1
)[0])(1)(a′(0)R1 )[1]bR1a
′
(1)c<−1>b′R2
(c<0>R2 ){−1}((c<0>R2 ){0} ∗ c
′)<−1> ⊗ ((c<0>R2 ){0} ∗ c
′)<0>
(4.2),(4.12)
= a(0)(((a′(0)R1 )[0])(0))r1 ⊗ a(1)r1 ((a
′
(0)R1
)[0])(1)(a′(0)R1 )[1]bR1a
′
(1)c<−1>b′R2
(c<0>R2 ){−1}((c<0>R2 ){0})<−1>(c
′
<−1>)r2 ⊗ (((c<0>R2 ){0})<0>)r2c
′
<0>
(4.7),(4.15)
= a(0)(a′(0)R1 )r1 ⊗ a(1)r1 bR1a
′
(1)c<−1>b′R2c
′
<−1>r2 ⊗ (c<0>R2 )r2c
′
<0>,
ψ(a⊗ b⊗ c)ψ(a′ ⊗ b′ ⊗ c′)
= (a(0) ⊗ a(1)bc<−1> ⊗ c<0>)(a′(0) ⊗ a′(1)b′c′<−1> ⊗ c′<0>)
= a(0)a′(0)R1 ⊗ (a(1)bc<−1>)R1(a
′
(1)b′c′<−1>)R2 ⊗ c<0>R2c
′
<0>
(1.3),(1.4)
= a(0)(((a′(0))R1)R1)r1 ⊗ a(1)r1 bR1(c<−1>)R1(a
′
(1))R2b′R2c
′
<−1>r2
⊗(((c<0>)R2)R2)r2c′<0>
(4.26)
= a(0)(a′(0)R1 )r1 ⊗ a(1)r1 bR1a
′
(1)c<−1>b′R2c
′
<−1>r2 ⊗ (c<0>R2 )r2c
′
<0>,
and we see that the two terms are equal. 
Let now H be a bialgebra, A a left H-module algebra, C a right H-module algebra and
F ∈ H ⊗ H a 2-cocycle. Then, by [BPVO], we have an algebra isomorphism (notation as
before):
AF−1#HF# FC ≃ A#H#C, a#h#c 7→ G1 · a#G2hF 1#c · F 2.
One can easily see that this result is a particular case of Theorem 4.10; indeed, the relations
(4.24), (4.25), (4.26) are easy consequences of the 2-cocycle condition for F .
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44 PASCUAL JARA MART´INEZ, JAVIER L ´OPEZ PE ˜NA, FLORIN PANAITE, AND FRED VAN OYSTAEYEN
DEPARTMENT OF ALGEBRA, UNIVERSITY OF GRANADA, AVDA. FUENTENUEVA S/N, E-18071, GRANADA,
SPAIN
E-mail address: pjara@ugr.es
DEPARTMENT OF ALGEBRA, UNIVERSITY OF GRANADA, AVDA. FUENTENUEVA S/N, E-18071, GRANADA,
SPAIN
E-mail address: jlopez@ugr.es
INSTITUTE OF MATHEMATICS OF THE ROMANIAN ACADEMY., PO-BOX 1-764, RO-014700 BUCHAREST
(ROMANIA)
E-mail address: Florin.Panaite@imar.ro
DEPARTMENT OF MATHEMATICS AND COMPUTER SCIENCES, UNIVERSITY OF ANTWERP., MIDDEL-
HEIMLAAN 1, B-2020 ANTWERP (BELGIUM)
E-mail address: Francine.Schoeters@ua.ac.be (secretary)