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9
ON THE CLASSIFICATION OF TWISTING MAPS BETWEEN Kn AND Km
P. JARA, J. L ´OPEZ PE ˜NA, G. NAVARRO, AND D. S¸TEFAN
ABSTRACT. We define the notion of admissible pair for an algebra A, consisting on a
couple (Γ, R), where Γ is a quiver and R a unital, splitted and factorizable representation
of Γ, and prove that the set of admissible pairs for A is in one to one correspondence
with the points of the variety of twisting maps T nA := T (Kn, A). We describe all these
representations in the case A = Km.
Twisted tensor products of algebras, also known as factorization structures, constitute a
particular instance of the notion of distributive law (cf. [4]) appeared in [17] and [21] and
were since studied in different contexts for various purposes, including their applications
to braided geometry [17, 18, 19], their realization as noncommutative analogues of princi-
pal bundles (cf. [6, 7]), their relation with the (quantum) Yang-Baxter and other nonlinear
equations (cf. [22]), and as a suitable replacement for cartesian products in noncommuta-
tive geometry (cf. [10, 13, 15]).
From a purely algebraic point of view, the notion of twisted tensor product comes di-
rectly from the factorization problem:
Given some kind of (algebraic) object, is it possible to find two suitable
subobjects, having minimal intersection and such that they generate our
original object?
The factorization problem has been intensively studied in the case of groups, coalge-
bras, Hopf algebras and algebras (cf. for instance [1, 8, 9, 20]). In the particular case of
algebras, a well known result (independently proven many times) establishes a one-to-one
correspondence between the set of factorization structures admitting two given algebras A
and B as factors and the set of so-called twisting maps, see Section 1, which allow us to
construct the twisted tensor product A⊗τ B associated to τ .
Henceforth, the problem of constructing factorization structures with given factors boils
down to the problem of finding all the existing twisting maps for those factors. Under
suitable, very mild, conditions (for instance, whenever A and B are affine algebras), the
set T (A,B) of all the twisting maps τ : B ⊗ A → A⊗B is an algebraic variety, and two
interesting problems arise:
PROBLEM 0.1. Is it possible to describe explicitly the variety T (A,B)?
PROBLEM 0.2. Once the variety T (A,B) is known, is it possible to determine which
points of the variety give rise to isomorphic algebras?
These two problems, even in the simplest cases, turn out to be very hard. Though there
are many different methods that produce twisted tensor products of two given algebras, not
a single one produces all the existing ones is known, let alone describing the properties of
the algebraic variety. Even harder is the problem of the determination of the isomorphism
classes of algebras obtained from the same factors through different tensor products, or
finding any description of these isomorphism classes in terms of the variety T (A,B).
Date: September 24, 2009.
1991 Mathematics Subject Classification. 16S35, 16G20.
Key words and phrases. Twisting maps, twisted tensor product, representation, quiver.
Research partially supported by MTM2007-66666 and FQM-266 (Junta de Andalucı`a Research Group). J.
Lo´pez Pen˜a was supported by Max-Planck Institut fu¨r Mathematik in Bonn and EU Marie-Curie fellowship
PIEF-GA-2008-221519. D. S¸tefan was financially supported by Contract 560/2009 (CNCSIS code ID 69).
1

2 P. JARA, J. L ´OPEZ PE ˜NA, G. NAVARRO, AND D. S¸TEFAN
Recently, Cibils showed in [11] that the set T (K2, A) of twisted tensor products be-
tween any algebra A and the commutative, semisimple algebra K2 (also called the set of
2–interlaces) is in one-to-one correspondence with couples of linear endomorphisms of the
algebra A satisfying certain conditions. If we take A = Kn, these couples of linear maps
can be described by combinatorial means using certain families of colored quivers, and
this description gives a simple way to describe all the twisted tensor products Kn ⊗τ K2,
up to isomorphism (cf. [11, 16]). Some other partial steps in the classification problem for
factorization structures have been undertaken in [5] and the final sections of [12].
In the present paper, we extend the combinatorial techniques developed by Cibils, de-
veloping the notion of an admissible pair for an algebra A, consisting on a couple (Γ, R)
where Γ is a quiver, and R a representation of Γ (cf. for instance [2, 3]) satisfying certain
restrictions (namely, to be unital, splitted and factorizable), and prove (cf. Theorem 1.8)
that the variety of twisting maps T nA := T (Kn, A) is in one to one correspondence with
the set of admissible pairs for A. Twisted tensor product of the form A ⊗τ Kn can be
reinterpreted as deformations of the usual tensor product A ⊗ Kn ∼= A × A ×
(n)· · · × A,
which is nothing but the direct product of n copies of the algebra A. If A is the algebra of
functions defined over the configuration space Q of some mechanical system, then A⊗Kn
is the algebra of functions defined over the configuration space Qn of the system consist-
ing on n disjoint copies of Q. The noncommutative deformations of this algebra obtained
as twisted tensor products A ⊗τ Kn are proposed to serve as toy-model for quantizations
of this situation when we assume that the disjoint copies of the physical states are close
enough so that they interact with each other.
The paper is structured as follows. In Section 1, we study the basic properties of admis-
sible pairs, and introduce the numerical invariants of rank and reduced rank as a measure
of the complexity of an admissible pair. We use this notion to characterize the connected
components of a quiver in an admissible pair of reduced rank one, proving that splitted,
unital and factorizable representations of a quiver Γ of reduced rank one are uniquely
determined by a set (Mi)i∈Γ0 of two-sided ideals of A, and a set (Bi)i∈Γ0 of (unital)
subalgebras of A such that
(1) For each vertex i ∈ Γ0, A = Bi ⊕Mi,
(2) For each arrow α which is not a loop, Ms(α)Mt(α) = 0.
A particular example of this setting can be obtained by means of Hochschild extensions
of an algebra B with given kernel M . Finally, we classify all the splitted, unital, and
factorizable representations associated to quivers of reduced rank one, without cycles of
length 2, when we take the algebra A to be equal to Km.
In Section 3 we introduce the notion of absolutely reducible (finite dimensional) rep-
resentations of an algebra A, and obtain a canonical form for defining the action of A on
an absolutely reducible representation of dimension n by means of a normalized invert-
ible matrix. We pay especial attention to the particular case of two-dimensional absolutely
reducible representations, characterizing them all in Theorem 3.16.
Using the aforementioned results, in Section 4, we describe all the splitted, unital and
factorizable representations of a quiver consisting in a cycle of length two, given a more
detailed description for the case of two-dimensional absolutely reducible representations.
These results are all merged together in Theorem 4.2, in which we classify all the repre-
sentations of a connected quiver of reduced rank one containing a cycle of length 2, thus
concluding the classification of all admissible representations of a quiver of reduced rank
one.

ON THE CLASSIFICATION OF TWISTING MAPS BETWEEN Kn AND Km 3
1. TWISTING MAPS BETWEEN Kn AND A
Let A be an algebra over a field K . The canonical basis of Kn will be denoted by
{e1, . . . , en}. Observe that
∑n
i=1 ei is 1Km , the unit of Kn. Thus a K-linear map
τ : Kn ⊗A −→ A⊗Kn
is uniquely determined by Eτ = (Eij)i,j=1,...,n, a set of K-linear endomorphisms of A,
satisfying
(1) τ(ei ⊗ a) =
n
∑
j=1
Eij(a)⊗ ej , ∀a ∈ A, ∀i = 1, . . . , n.
Our first aim is to identify the properties that Eτ has to verify in order to get a twisting
map between the algebras Kn and A. Recall that τ : Kn ⊗ A −→ A⊗Kn is a twisting
map if, and only if, the following four conditions hold (cf. [4, 10]):
τ(mKn ⊗A) = (A⊗mKn)(τ ⊗Kn)(Kn ⊗ τ).(2)
τ(Kn ⊗mA) = (mA ⊗Kn)(A ⊗ τ)(τ ⊗A).(3)
τ(1Kn ⊗ a) = a⊗ 1Kn , ∀a ∈ A.(4)
τ(x ⊗ 1A) = 1A ⊗ x, ∀x ∈ Kn.(5)
In the identities above, mKn and mA denote the multiplication on Kn and A respectively,
whilst 1Kn and 1A denote the units of these algebras.
If we evaluate both sides of relation (2) at ei ⊗ ej ⊗ a, in view of relation (1), we get
τ(eiej ⊗ a) =
n
∑
p=1
(A⊗mKn)(τ ⊗Kn)(ei ⊗ Ejp(a)⊗ ep)
=
n
∑
p=1
n
∑
q=1
Eiq(Ejp(a))⊗ epeq.
Therefore
n
∑
p=1
δi,jEip(a)⊗ ep =
n
∑
p=1
(Eip ◦ Ejp)(a)⊗ ep.
In conclusion, relation (2) implies
(6) Eip ◦ Ejp = δi,jEip, ∀i, j, p = 1, . . . , n.
Obviously, the converse holds too. Therefore equations (2) and (6) are equivalent.
The other identities in the definition of twisting maps can be written in a similar way.
By evaluating (3) at ei ⊗ a⊗ b we get that this relation is equivalent to
(7) Eij(ab) =
n
∑
p=1
Eip(a)Epj(b), ∀i, j = 1, . . . , n, ∀a, b ∈ A.
Since the unit of Kn is∑ni=1 ei, we obtain
(8)
n
∑
i=1
Eij = IdA, ∀j = 1, . . . , n.
Finally, by taking x := ei in (5), we deduce that this relation is equivalent to
(9) Eij(1A) = δi,j1A, ∀i, j = 1, . . . , n.
In conclusion, if T nA denotes the set of all twisting maps between Kn and A, and EnA
denotes the set of systems of endomorphisms (Eij)i,j=1,...,n satisfying relations (6)–(9),
we proved the following theorem.

4 P. JARA, J. L ´OPEZ PE ˜NA, G. NAVARRO, AND D. S¸TEFAN
PROPOSITION 1.1. There is a one-to-one correspondence φA : T nA −→ EnA that maps a
twisting map τ : Kn ⊗A −→ A⊗Kn to the set of endomorphisms Eτ = (Eij)i,j=1,...,n
which is uniquely defined such that relation (1) holds true.
From now on, instead of working with twisting maps we shall work with systems of
endomorphisms in EnA. To such a system E = (Eij)i,j=1,...,n we are going to associate
two invariants: a quiver ΓE and a representation RE of ΓE .
DEFINITIONS 1.2. Let Γ be a quiver (cf. [2, 3]). Let Γ0 and Γ1 be, respectively, the set
of vertices and the set of arrows of Γ. The source and the target maps will be respectively
denoted by s : Γ1 → Γ0 and t : Γ1 → Γ0.
(1) A path in Γ is a sequence p = (α1, α2, . . . , αn) of arrows such that t(αi) = s(αi+1),
for any i = 1, . . . , n− 1.
p : ◦ α1 // ◦ α2 // · · · αn−1 // ◦ αn // ◦
The set of paths of length n in Γ is denoted by Γn.
(2) For a path p = (α1, α2, . . . , αn) we define the source of p by s(p) := s(α1). Similarly,
the target of p is given by t(p) := t(αn). The set of paths p of length n with s(p) = i
and t(p) = j is denoted by Γn(i, j). Note that Γ0(i, i) can be identified with the vertex
i, while Γ1(i, j) is equal to the set of arrows a such that s(a) = i and t(a) = j. In
particular, Γ1(i, i) consists of all loops having the vertex i as a source. The source of
a loop will be called a loop vertex.
(3) An oriented cycle is a path p with s(p) = t(p).
DEFINITION 1.3. Let E := (Eij)i,j=1,...,n be in EnA. The quiver ΓE is defined as follows:
the set of vertices of ΓE is {v1, . . . , vn}. The vertices vi and vj are joined by an arrow
with the source in vi if, and only if, Eij 6= 0. The vertex vi shall be represented simply by
i as well.
For future references we state the following proposition.
PROPOSITION 1.4. The quiver ΓE associated to E ∈ EnA has no multiple arrows and, for
every vertex vi ∈ Γ0E , there is a loop having the source (and therefore the target) in vi.
Proof. By construction, ΓE has no multiple arrows. Let vi ∈ Γ0E . Since Eii(1A) = 1A,
it follows that there is an arrow α such that s(α) = t(α) = vi. This arrow obviously is
unique, as ΓE has no multiple arrows. 
Now we are going to construct the second invariant of E, namely a certain linear
representation of ΓE . Recall from [2, 3] that a representation of a quiver Γ is given
by a family of vector spaces (Vi)i∈Γ0 and a family of K-linear maps (ϕα)α∈Γ1 , where
ϕα : Vs(α) −→ Vt(α).
DEFINITION 1.5. Let ΓE be the quiver associated to E = (Eij)i,j=1,...,n, a system of
endomorphisms in EnA. The representation RE of ΓE is defined by the family of vector
spaces (Vi)i∈Γ0E and the family of K-linear maps (ϕα)α∈Γ1E , where
Vi := A and ϕα := Es(α),t(α),
for every i ∈ Γ0E and α ∈ Γ1E .
Let E be an element in EnA. In the next proposition relations (6)–(9) are rewritten using
the maps (ϕα)α∈Γ1E that define RE .
PROPOSITION 1.6. The maps (ϕα)α∈Γ1E satisfy the following properties.
(1) For any i ∈ Γ0E , the set {ϕα | t(α) = i} is a complete set of non-zero orthogonal
idempotents on EndK(A). Hence, for any α′, α′′ ∈ Γ1E such that t(α′) = t(α′′) = i,
(Si) ϕα′ ◦ ϕα′′ = δs(α′),s(α′′)ϕα′ and
∑
{α|t(α)=i}
ϕα = IdA.

ON THE CLASSIFICATION OF TWISTING MAPS BETWEEN Kn AND Km 5
(2) For any arrow α ∈ Γ1E ,
(U ) ϕα(1A) = δs(α),t(α)1A.
(3) For any i, j ∈ Γ0E and any a, b ∈ A, we have
(Fi,j)
∑
(α1,α2)∈Γ2E(i,j)
ϕα1(a)ϕα2(b) =
{
ϕα(ab), if Γ1E(i, j) = {α};
0, if Γ1E(i, j) = ∅.
Proof. The relations in (Si) follow by (6), (8) and the fact that Ekh = 0, whenever there
is no α ∈ Γ1E such that s(α) = k and t(α) = h. In a similar way (U) follows from (9).
Finally, it is not difficult to see that (7) is equivalent to the fact that (Fi,j) holds for every
pair (i, j) of vertices. 
DEFINITION 1.7. Let A be an algebra. We say that (Γ, R) is an admissible pair of order
n for A (shortly an admissible pair) if
(1) Γ is a quiver with n vertices such that every vertex is a loop vertex and there are no
multiple arrows.
(2) R is a representation of Γ such that the vector spaces associated to its vertices are all
equal to A. The family (ϕα)α∈Γ1 that define R satisfies the following conditions:
(a) The representation R is splitted, i.e. relation (Si) holds true for every i ∈ Γ0.
(b) The representation R is unital, that is, relation (U ) holds true.
(c) The representation R is factorizable, i.e. relation (Fi,j) holds true for every pair
(i, j) of vertices in Γ.
The set of all admissible pairs (Γ, R) of order n for A will be denoted by RnA.
Summarizing, for every E ∈ EnA, we got an admissible pair (ΓE , RE) for A and E 7→
(ΓE , RE) defines a map from EnA to RnA. By the proof of Proposition 1.6, one can see
easily that this map is bijective. In fact, it is sufficient to notice that the inverse maps an
admissible pair (Γ, R) to the set E = (Eij)i,j=1,...,n, where Es(α),t(α) = ϕα, for any
arrow α ∈ Γ1, and all other morphisms Eij are zero. In conclusion we have proved the
following.
THEOREM 1.8. For an arbitrary K-algebra A the sets T nA , EnA and RnA are in one-to-one
correspondence.
2. SOME BASIC PROPERTIES OF ADMISSIBLE PAIRS
Let A be a K-algebra. Throughout this section, (Γ, R) will denote an element in
RnA. For the family of maps that defines the representation R we shall use the notation
(ϕα)α∈Γ1 . The maps associated to the loops of Γ will play an important roˆle, so we shall
use a special notation for them. Namely, ϕi will denote the morphism corresponding to the
unique loop α such that s(α) = i.
Our purpose now is to investigate some basic properties of Γ and R. First, let us notice
that the existence of the representation R imposes some restrictions on Γ.
PROPOSITION 2.1. Let α be an arrow of Γ. Then ϕα = IdA if, and only if, α is a loop and
there is no other arrow with target t(α).
Proof. Since ϕα is the unique non-zero morphism corresponding to an arrow whose target
is t(α), we deduce the required equality by using the second relation in (Si). Conversely,
let us assume that ϕα = IdA. First we prove that α is the unique arrow having the target
in t(α). Let us assume that α′ is another arrow such that t(α′) = t(α). Since Γ has no
multiple arrows, s(α) 6= s(α′). We get
ϕα′ = ϕα′ ◦ ϕα = δs(α′),s(α)ϕα = 0.

6 P. JARA, J. L ´OPEZ PE ˜NA, G. NAVARRO, AND D. S¸TEFAN
Since ϕα′ 6= 0, by the definition of admissible pair, it follows that α is the unique arrow
having the target in t(α). Again, by the definition of admissible pair, every vertex is a loop
vertex. In the view of the foregoing, α has to be a loop. 
By the above Proposition, for every vertex of Γ, there are two possibilities, as it is
indicated in the two pictures below.
◦
IdA
◦
◦
◦
◦
::
55kkkkkkkkkkkk
11ccccccccccccc
,,YYYYY
YYYYYY
YY
((PP
PPP
PPP
PPP
P ◦
◦
◦
◦
◦
◦
◦
◦
33hhhhhhhhhhh //
++VVVV
VVVVV
VV))
SSSS
SSSS
SSSS
--[[[[[[[[[
[[[[
66nnnnnnnnnnnn
22eeeeeeeeeeeee
ϕ 6=IdA

In the first one it is represented the case when the morphism corresponding to the loop is
the identity of A. In the second picture, the morphism ϕ, associated to the loop, is not the
identity of A. In what follows in the pictures we only draw the loops α with ϕ = IdA.
In order to measure the complexity of Γ we introduce a numerical invariant, the rank of
a vertex. Let i ∈ Γ0. We set
(10) rank(i) :=
∑
j
#Γ1(j, i) = #{α ∈ Γ1 | t(α) = i}.
We also define the reduced rank of i as
(11) rrank(i) :=
∑
j 6=i
#Γ1(j, i) = #{α ∈ Γ1 | t(α) = i and s(α) 6= i}.
Note that, in the quiver of an admisible pair, rrank(i) = 0 if, and only if, there is only
one arrow α with t(α) = i. Of course, in this case we also have s(α) = i and ϕα = IdA.
If rrank(i) = r, then there are exactly r arrows having their target in i and which are not
loops.
DEFINITION 2.2. The rank of Γ is defined by
(12) rank(Γ) := max{rank(i) | i ∈ Γ0}.
In the same way, we may define the reduced rank of Γ.
PROPOSITION 2.3. Let A be a K-algebra of dimension m. If E ∈ EnA, then rrank(ΓE) ≤
min(n− 1,m− 1).
Proof. Obviously, rrank(ΓE) ≤ n − 1, as there are at most n − 1 arrows α such that
t(α) = i and s(i) 6= i, for any vertex i in ΓE . On the other hand, for i ∈ Γ0E , let X =
{α ∈ Γ1E | t(α) = i}, then {ϕα | α ∈ X} is a complete set of orthogonal idempotents.
Therefore A decomposes as a direct sum of non zero vector subspaces A = ⊕α∈X Wα,
where Wα = Imϕα. Hence
#X ≤
∑
α∈X
dimWα = dimA.
It results rrank(i) ≤ dimA− 1. Thus the proposition is proved. 
The simplest non-trivial quivers (i.e. containing arrows that are not loops) are those of
reduced rank 1. Their connected components are described in the following proposition.
Note that when we speak about the connected components of Γ, we mean the connected
components of the undirected graph obtained by removing the orientation of all arrows in
Γ. On the other hand, each connected component can be seen as a quiver with respect to
the orientation of its edges that is inherited from Γ.
PROPOSITION 2.4. Let Γ be an arbitrary finite quiver of reduced rank one, then any con-
nected component contains at most a unique cycle which is not a loop.

ON THE CLASSIFICATION OF TWISTING MAPS BETWEEN Kn AND Km 7
Proof. Clearly we may assume that the connected component Λ is not reduced to a unique
vertex. Let us show that in the same component cannot exist two different cycles. We pick
up a vertex i in the first cycle and a vertex j 6= i in the second one. Thus, as Λ is connected,
there is a sequence of vertices i = i1, . . . , ih+1 = j and a sequence of arrows α1, . . . , αh
in Λ such that either s(αk) = ik and t(αk) = ik+1 or s(αk) = ik+1 and t(αk) = ik, for
any k = 1, . . . , h.
◦
◦
◦
◦
◦
◦
◦
◦
◦
i1 ◦
i2 ◦
ih ◦
ih+1??
??
??
 


77oooo
ggOOOO
77oooo
ggOOOO
α1
//
αh
oo
?
??
??
?
 


Since rrank(i) = 1 it follows that s(α1) = i1 = i and t(α1) = i2. By induction, as
rrank(ik) = 1, we get s(αk) = ik and t(αk) = ik+1, for any k = 1, . . . , n. It follows that
t(αh) = ih+1 = j, so rrank(ih+1) ≥ 2, which is impossible.

PROPOSITION 2.5. With the same assumption as in Proposition 2.4, every vertex in the
possible oriented cycle is the root of a (naturally) oriented tree which can be degenerated
(i.e., with only one vertex) and such that two different of these trees are disjoint.
Proof. Let us denote by Λ′ the quiver obtained removing all the arrows α1, . . . , αr in the
cycle with vertices i1, . . . , ir such that s(αk) = ik for any k = 1, . . . , r. See the figure
below.
◦
◦ ◦
i1
i3
i2
◦ ◦
◦ ◦ ◦
◦
◦ ◦
◦
α1

α2
ii
α3
NN
TT))))
JJ
TT))))
OO JJ
OO
RR%%%%
LL
OO
⇒ ◦
◦ ◦
◦
i1
i3
i2
◦
◦ ◦ ◦
◦
◦ ◦
◦
TT))))
JJ
TT))))
OO JJ
OO
RR%%%%
LL
OO
We fix k ∈ {1, . . . , r}. The connected component Λ′k ⊆ Λ′ that contains ik has no cycles,
otherwise there would be two different cycles in Λ. Hence, Λ′k is a tree (recall that a graph
is a tree if, and only if, it is connected and does not contains cycles, or equivalently, any
two vertices are connected by a unique path). Since the rank of Λ is 1, the arrows of
Λ′k are oriented in the canonical way, that is, the one such that the arrows at the root are
outgoing. 
LEMMA 2.6. Let (Γ, R) be an admissible pair.
(1) If i ∈ Γ0 is not a vertex in a cycle of length 2 then condition (Fi,i) is equivalent to the
fact that ϕi is an algebra map (recall that ϕi denotes the morphism corresponding to
the loop having the source in i).
(2) Let (α1, α2) ∈ Γ2 be a path such that α2 and α1 are as in the following picture,
◦i ◦k◦
jα1 // α2 //
If rrank(j) = rrank(k) = 1, then conditions (Fi,k) and (Fj,k) are equivalent. More-
over, these conditions are also equivalent to
(13) Ker (ϕs(α2))Ker (ϕt(α2)) = 0.

8 P. JARA, J. L ´OPEZ PE ˜NA, G. NAVARRO, AND D. S¸TEFAN
Proof. By assumption i is not a vertex in a cycle of length 2. Then, either Γ1(i, k) = ∅ or
Γ1(k, i) = ∅, for any k 6= i. Therefore, in this case, equation (Fi,i) becomes
ϕi(a)ϕi(b) = ϕi(ab), ∀a, b ∈ A.
Let us prove the second claim. We first show that k is not a vertex in a cycle of length
2. Indeed, if l were the second vertex of such a cycle and l 6= j then we would have
rrank(k) ≥ 2. Thus l = j. On the other hand, this equality implies rrank(j) > 1, which
is also impossible. Since k is not the vertex of a cycle of length 2, it follows that ϕk is an
algebra map. Since Γ2(i, k) = {(α1, α2)} and there is no arrow having the source in i and
the target in k, relation (Fi,k) is equivalent to
ϕα1(a)ϕα2(b) = 0, ∀a, b ∈ A.
By hypothesis, the reduced rank of j and of k is one. Therefore, ϕα1 = IdA − ϕj and
ϕα2 = IdA − ϕk. Then the above equality is equivalent to
(14) [a− ϕj(a)][b − ϕk(b)] = 0, ∀a, b ∈ A.
Obviously, Γ2(j, k) = {(λ1, α2), (α2, λ2)}, where λ1 and λ2 are the unique loops such
that s(λ1) = j and s(λ2) = k. As ϕλ1 = ϕj and ϕλ2 = ϕk, relation (Fj,k) can be written
as follows
(15) ϕj(a)ϕα2(b) + ϕα2(a)ϕk(b) = ϕα2(ab), ∀x, y ∈ A.
Since ϕα2 = IdA − ϕk one can prove easily that (14) and (15) are equivalent. In conclu-
sion, (Fi,k) and(Fj,k) are equivalent too. To conclude it is enough to prove that both are
equivalent to (13). Relation (14) can be written as:
Im (IdA − ϕs(α2))Im (IdA − ϕt(α2)) = 0.
On the other hand, Im (IdA−ϕs(α2)) = Ker (ϕs(α2)), as ϕs(α2) is an idempotent K-linear
map. A similar equality holds for ϕt(α2), so the lemma is proved. 
THEOREM 2.7. Let A be a K-algebra and let Γ be a quiver such that its vertices are loop
vertices and it has no multiple arrows. Assume that rrank(Γ) = 1 and that Γ does not
contain any cycle of length 2. Then to give a splitted, unital and factorizable representation
of Γ over A is equivalent to give a set of idempotent algebra endomorphisms (ϕi)i∈Γ0 such
that, for any arrow α : i → j which is not a loop,
(16) KerϕiKerϕj = 0.
Proof. Let R be a splitted, unital and factorizable representation of Γ. Let (ϕα)α∈Γ1 de-
note the family of K-linear endomorphisms that defines R. For i ∈ Γ0 we take ϕi to be the
morphism corresponding to the loop α that has the source in i. Since Γ does not contain
cycles of length 2, by the previous lemma, it follows that ϕi is an algebra map, for any
i ∈ Γ0. Let α ∈ Γ1 which is not a loop. If ϕs(α) = IdA we have nothing to prove.
Let us consider the case when ϕs(α) 6= IdA. Thus there is an arrow β, which is not a
loop, such that t(β) = s(α). Hence relation (16) follows by the second part of Lemma 2.6.
Conversely, let (ϕi)i∈Γ0 be a family of idempotent algebra morphisms satisfying rela-
tion (16). We want to construct a splitted, unital and factorizable representation R of Γ.
For every arrow α which is not a loop, we set ϕα = IdA − ϕt(α). Obviously, {ϕα, ϕt(α)}
is a complete set of orthogonal idempotents, so the representation R defined by (ϕα)α∈Γ1
is splitted. Trivially R is unital, as ϕi is a morphism of algebras, for any i ∈ Γ0. It remains
to prove that R is factorizable. Since rrank(Γ) = 1 and Γ has not cycles of length 2, the
non-trivial relations that can occur are (Fi,k) and (Fj,k), where the vertices i, j and k are
as in the picture below.
◦i ◦k◦
jβ // α //
By Lemma 2.6 (2) these relations are equivalent to (16), so the Theorem is proved. 

ON THE CLASSIFICATION OF TWISTING MAPS BETWEEN Kn AND Km 9
COROLLARY 2.8. Let A be a finite dimensionalK-algebra and let Γ be a quiver of reduced
rank one without cycles of length 2. A splitted, unital and factorizable representation of
Γ is uniquely defined by a set (Mi)i∈Γ0 of two-sided ideals and a set (Bi)i∈Γ0 of unital
subalgebras of A that satisfy the following two conditions:
(1) For each vertex i ∈ Γ0, A = Bi
⊕Mi;
(2) For each arrow α which is not a loop, Ms(α)Mt(α) = 0.
Proof. For a vertex i, take in the previous theorem Bi := Imϕi and Mi := Kerϕi.
Conversely, for a family of ideals (Mi)i∈Γ0 , satisfying the conditions in the corollary,
we define ϕi := σi ◦ pii, where σi is the inclusion Bi ⊆ A and pii is the projection of
A = Bi
⊕Mi onto Bi. 
REMARK 2.9. Let A be a K-algebra. We assume that there is a two-sided ideal M ≤ A
such that M2 = 0. Let Γ be a quiver of reduced rank one and let (ϕi)i∈Γ0 be a family of
idempotent algebra endomorphisms of A. If
Kerϕi ⊆ M
for any i ∈ Γ0, then condition (16) is automatically satisfied. Hence the family (ϕi)i∈Γ0
induces a splitted, unital and factorizable representation of Γ.
In order to give examples of families (ϕi)i∈Γ0 satisfying the assumptions in the above
remark, we recall the construction of Hochschild extensions (also called in the literature
abelian extension, cf. [14, Sec. 1.5.3]) of an algebra B with kernel a given B-bimodule
M . By definition, such an extension is given by a normalized Hochschild 2-cocycle, that
is, a K-linear map ω : B ⊗B → M that verifies the following equalities:
• ω(a⊗ 1B) = ω(1B ⊗ a) = 0, for all a ∈ B.
• a · ω(b⊗ c)− ω(ab⊗ c) + ω(a⊗ bc)− ω(a⊗ b) · c = 0, for all a, b, c ∈ B.
To these data one can associate a unital and associative algebra A as follows. As a
vector space A := B ⊕M . The multiplication on A is defined by
(b,m) · (b′,m′) = (bb′, bm′ +mb′ + ω(b⊗ b′))
and the unit is (1, 0). Note that M is a two-sided ideal in A and M2 = 0 (of course M can
be identified with a subset of A via the map m 7→ (0,m)).
LEMMA 2.10. Let f : M → M be a K-linear map.
(1) The K-linear map ϕ : A → A, defined by ϕ(b,m) = (b, f(m)) for any b ∈ B
and m ∈ M , is an algebra map if, and only if, f is a morphism of B-bimodules and
f ◦ ω = ω.
(2) ϕ is idempotent if, and only if, f is so.
(3) The kernel of ϕ is contained into M .
Proof. ϕ is a morphism of algebras if, and only, if
(17) f(bm′ +mb′) + (f ◦ ω)(b⊗ b′) = bf(m′) + f(m)b′ + ω(b⊗ b′)
for any b, b′ ∈ B and m,m′ ∈ M .
Let us assume that ϕ is a morphism of algebras, hence the above equation holds. By
taking b′ = 0 we get that f is a morphism of left B-modules. Similarly, we deduce that f
is a map of right B-modules. In particular,
f(mb′ + bm′) = f(m)b′ + bf(m′),
so f ◦ ω = ω holds too. Conversely, if f is a morphism of B-bimodules and f ◦ ω = ω,
then f obviously satisfies the equation (17), so ϕ is a morphism of algebras.
The second and the third part of the Lemma are obvious. 

10 P. JARA, J. L ´OPEZ PE ˜NA, G. NAVARRO, AND D. S¸TEFAN
THEOREM 2.11. Let A be the Hochschild extension associated to (B,M,ω), where B is
a K-algebra, M is a B-bimodule and ω is a 2-cocycle of B with coefficients in M . Let
Γ be a quiver of reduced rank one without cycles of length two. If (fi)i∈Γ0 is a family
of idempotent endomorphisms of M and Im ω ⊆ ⋂i∈Γ0 Im fi then (ϕ)i∈Γ0 induces a
splitted, unital and factorizable representation of Γ, where ϕi is constructed from fi as in
the previous Lemma.
Proof. It is a direct consequence of Theorem 2.7 and the previous Lemma. See also Re-
mark 2.9. 
Throughout the remaining of this section we take A = Km. Our purpose is to classify
all splitted, unital and factorizable representations (over A) of a quiver Γ of reduced rank
one and without cycles of length two.
Let θ : Km → Km be an algebra endomorphism of Km. If {f1, . . . , fm} is the
canonical basis on Km, then θ(fp) is an idempotent of Km. Therefore there is a set
Θp ⊆ {1, . . . ,m} such that
θ(fp) =
∑
q∈Θp
fq.
Note that if Θp = ∅ then θ(fp) = 0. Thus the kernel of θ is the vector subspace of Km
generated by all elements fp such that Θp = ∅. Moreover, (Θp)p=1,...,m is a partition of
{1, . . . ,m} in m (possibly empty) subsets. Indeed, if p 6= q then fpfq = 0, so
0 = θ(fpfq) =
∑
r∈Θp
∑
s∈Θq
frfs =
∑
r∈Θp
T
Θq
fr.
Hence Θp and Θq are disjoint. On the other hand,
m
∑
p=1
fp =
m
∑
p=1
θ(fp) =
m
∑
p=1
∑
q∈Θp
fq =
∑
q∈
Sm
p=1 Θp
fq.
Thus
⋃m
p=1 Θp = {1, . . . ,m}.
The partition (Θp)p=1,...,m defines a unique function u : {1, . . . ,m} → {1, . . . ,m}
given by u(q) = p for all q ∈ Θp. Hence, for an arbitrary p ∈ {1, . . . ,m}, we have
(18) θ(fp) =
m
∑
q=1
δp,u(q)fq.
LEMMA 2.12. The algebra endomorphism θ is idempotent if, and only if, the function u is
so. Moreover, the kernel of θ is given by
Ker θ = 〈fp | p /∈ Imu〉.
Proof. For p ∈ {1, . . . ,m}, we have θ2(fp) = θ(fp). By relation (18) we get
(19) δp,u2(r) = δp,u(r), for all p, r ∈ {1, . . . ,m}.
Conversely, if (19) holds, then θ is idempotent. We deduce that θ is an idempotent if, and
only if, u is so.
The partition associated to θ is given by Θp = u−1(p). Thus Ker θ is generated, as a
vector space, by all fp such that u−1(p) = ∅. 
THEOREM 2.13. Let A = Km and let Γ be a quiver of reduced rank one without cycles
of length two. A splitted, unital and factorizable representation of Γ is uniquely defined by
a set of idempotent functions ui : {1, . . . ,m} → {1, . . . ,m} with i ∈ Γ0, satisfying the
following condition: if α ∈ Γ1 is not a loop and p ∈ {1, . . . ,m}, then us(α)(p) = p or
ut(α)(p) = p.

ON THE CLASSIFICATION OF TWISTING MAPS BETWEEN Kn AND Km 11
Proof. By Theorem 2.7, a splitted, unital and factorizable representation of Γ is given by
a family (ϕi)i∈Γ0 of idempotent algebra endomorphisms of A = Km such that, for any
arrow α ∈ Γ1 that is not a loop, we have
Kerϕs(α)Kerϕt(α) = 0.
By Lemma 2.12, for every i ∈ Γ0 the algebra map ϕi corresponds to an idempotent func-
tion ui : {1, . . . ,m} → {1, . . . ,m}. Let α be an arrow that is not a loop. Since
Kerϕi = 〈fp | p /∈ Im ui〉
we deduce that
Kerϕt(α)Kerϕs(α) = 〈fpfq | p /∈ Imus(α), q /∈ Imut(α)〉
= 〈fp | p /∈ Imus(α) ∪ Imut(α)〉.
Therefore Kerϕt(α)Kerϕs(α) = 0 if, and only if, Imus(α) ∪ Imut(α) = {1, . . . ,m}.
Thus, for any p ∈ {1, . . . ,m}, p ∈ Imus(α) or p ∈ Imut(α). Since ui is idempotent for
every i ∈ Γ0, it follows that Imui = {p | ui(p) = p}, so the theorem is now proved. 
3. ABSOLUTELY REDUCIBLE REPRESENTATIONS OF AN ALGEBRA
We fix a finite-dimensional algebra A over a field K and a vector space W of dimension
n. Our aim in this section is to find all A-module structures on W which are absolutely
reducible in the following sense:
DEFINITION 3.1. An A-module W is called absolutely reducible (or diagonalizable) if
there are n submodules W1, . . . , Wn of dimension one such that W =
⊕n
i=1 Wi.
Let us fix a basis {w1, . . . , wn} on W . Thus a module structure on W is given by some
K-linear maps ωij : A −→ K , uniquely defined such that
(20) awi =
n
∑
j=1
ωji(a)wj , ∀i = 1, . . . , n and ∀a ∈ A.
The maps ωij are subject to the following relations:
n
∑
k=1
ωik(a)ωkj(b) = ωij(ab).(21)
ωij(1A) = δi,j .(22)
It is not difficult to see that, conversely, maps ωij satisfying (21) and (22) define a module
structure on W .
EXAMPLE 3.2. Let τ : Kn ⊗ A −→ A⊗Kn be a twisting map, where A = Km. Let
Eτ = {Eij}i,j=1,...,n be the corresponding set of K-linear endomorphisms of A. We fix
a basis {f1, . . . , fm} on Km and write
Eij(a) =
m
∑
p=1
Epij(a)fp, ∀i, j = 1, . . . , n and ∀a ∈ A.
Hence, the set {Epij}i,j,=1,...,n satisfies (21) and (22), for any p = 1, . . . ,m.
We now assume that W is absolutely reducible. Let W = W1 ⊕ · · · ⊕ Wn be the
corresponding decomposition. For each i = 1, . . . , n, we may choose a non-zero element
w′i ∈ Wi. Since Wi is an one-dimensional submodule of W , there is an algebra map
αi : A −→ K such that
aw′i = αi(a)w′i, ∀a ∈ A.

12 P. JARA, J. L ´OPEZ PE ˜NA, G. NAVARRO, AND D. S¸TEFAN
As {w′1, . . . , w′n} is another basis on W , there is an invertible matrix X ∈ GLn(K) such
that X = (aij)i,j=1,...,n and
w′i =
n
∑
j=1
ajiwj .
If {ωij}i,j=1,...,n are the maps that define the module structure on W , then we get
(ωij(a))i,j=1,...,n = X ·



α1(a) · · · 0
.
.
.
.
.
.
.
.
.
0 · · · αn(a)



·X−1, ∀a ∈ A.
For simplicity, we shall denote the matrix (ωij)i,j=1,...,n by ω. Note that the elements of
ω are in HomK(A,K). Hence the above equality can be written as
(23) ω = X ·



α1 · · · 0
.
.
.
.
.
.
.
.
.
0 · · · αn



·X−1.
Thus we have just proved the following Lemma.
LEMMA 3.3. Let A be a K-algebra. For every module structure on W which is absolutely
reducible, there are X ∈ GLn(K) and α1, . . . , αn ∈ AlgK(A,K) such that the matrix ω
defining the action of A on W satisfies (23).
In view of this lemma, the set of characters of A plays an important roˆle in the descrip-
tion of absolutely reducible representations of an algebra A. By Dedekind’s theorem on
linear independence of characters, any set of distinct algebra morphisms from A to K is
linearly independent. In particular AlgK(A,K) is a finite set. We shall denote it by
(24) AlgK(A,K) = {θ1, . . . , θr}.
REMARK 3.4. Lemma 3.3 can be rephrased as follows: if W is an absolutely reducible
module then there are X ∈ GLn(K) and a set map u : {1, . . . , n} −→ {1, . . . , r} such
that
(25) ω = X ·



θu(1) · · · 0
.
.
.
.
.
.
.
.
.
0 · · · θu(n)



·X−1,
where ω is the set of K-linear maps associated to W . We shall say that W is defined by the
matrix X and the map u. Of course, X and u are not uniquely determined by W . Given
such a map u, our aim is to find a matrix X0 ∈ GLn(K) such that, together with u, it
defines W and has as many zero elements as possible.
Throughout the remaining of this section we fix an n-dimensional absolutely reducible
representation W . Let ω denote the corresponding set of K-linear maps and let u :
{1, . . . , n} −→ {1, . . . , r} be a function that defines W . Let
Im (u) = {i1, . . . , is},
where 1 ≤ i1 < i2 < . . . < is ≤ r. We denote u−1(ik), the fiber of u over ik, by Fk.
Hence F = {F1, . . . , Fs} is a partition of {1, . . . , n}.
DEFINITION 3.5. Let F = {F1, . . . , Fs} be a partition of {1, . . . , n}. For every matrix
X ∈ Mn(K), we define the F -blocks of X by
X ij = (xp,q)p∈Fi,q∈Fj ,
where xp,q are the elements of X and i, j are in {1, . . . , s}.

ON THE CLASSIFICATION OF TWISTING MAPS BETWEEN Kn AND Km 13
EXAMPLE 3.6. Let u : {1, 2, 3, 4} −→ {1, 2, 3, 4} given by u(1) = 4, u(2) = u(4) = 2,
u(3) = 1. Hence F1 = {3}, F2 = {2, 4} and F3 = {1}. Thus, for any X ∈ M4(K) there
are nine F -blocks (Xkl)k,l=1,2,3. For example
X21 =
(
x23
x43
)
, X22 =
(
x22 x24
x42 x44
)
, X32 =
(
x12 x14
)
.
LEMMA 3.7. Let X ∈ GLn(K), u : {1, . . . , n} → {1, . . . , r} be a set map and F be the
set of fibers of u.
(1) The set
Hu = {Y ∈ GLn(K) | Y kl = 0, for k 6= l}
is a subgroup of GLn(K).
(2) If X ∈ GLn(K) and Y ∈ Hu, then X and XY define, together with u, the same
representation. Moreover,
(XY )kl = XklY ll, ∀k, l ∈ {1, . . . , s}.
(3) Let X , Y ∈ GLn(K). If X and Y define the same representation, then there is
Z ∈ Hu such that Y = XZ .
Proof. (1) Let us show that Hu contains all matrices in GLn(K) that commute with
θ =



θu(1) · · · 0
.
.
.
.
.
.
.
.
.
0 · · · θu(n)



.
Indeed, for Y ∈ GLn(K), we have Y θ = θY if, and only if,
yij(θu(i) − θu(j)) = 0, ∀i, j = 1, . . . , n.
These equalities are equivalent to yij = 0 for any pair (i, j) such that i and j are not in
the same fiber of u. In conclusion, Y commutes with θ if, and only if, Y ∈ Hu. Since the
set of commuting matrices with θ is a subgroup of GLn(K), the first part of the lemma is
proved.
(2) Let Y ∈ Hu. Since Y θY −1 = θ, then obviously XY and Y define the same
representation. It remains to prove the formula that describes the F -blocks of XY . Let k,
and l be arbitrary in {1, . . . , s}. We choose i ∈ Fk and j ∈ Fl. If zij is the element of XY
in the spot (i, j), then
zij =
n
∑
p=1
xipxpj =
s
∑
l′=1
∑
p∈Fl′
xipypj .
By definition of Hu, for l′ 6= l we have ypj = 0, as ypj is an element in Y l
′l
. So
zij =
∑
p∈Fl
xipypj .
Obviously the element in the (i, j)-spot of XklY ll is ∑p∈Fl xipxpj , so this part of the
lemma is also proved.
(3) If X and Y , together with u, define the same representation, then Z = X−1Y
commutes with θ. Then Z ∈ Hu and Y = XZ . 
THEOREM 3.8. We keep the notation from Lemma 3.7. Then
Hu ∼= GLn1(K)× · · · ×GLns(K),
where nk = #Fk for k = 1, . . . , s. The right action of Hu on GLn(K) given by right
matrix multiplication is defined on F -blocks by the formula
(26) (XY )kl = XklY ll, ∀k, l = 1, . . . , s.
for any X ∈ GLn(K) and Y ∈ Hu.

14 P. JARA, J. L ´OPEZ PE ˜NA, G. NAVARRO, AND D. S¸TEFAN
Proof. Let Y ∈ Hu. By Lemma 3.7(1) the F -blocks Y kl are trivial, for any k 6= l. Hence
the map Φ : Hu −→ GLn1(K)× · · · ×GLns(K) given by
Φ(Y ) := (Y 11, . . . , Y ss)
is well-defined. Note that Y kk is invertible, for any k = 1, . . . , s. Indeed, up to a permuta-
tion of the rows and columns of Y , we have
Y =



Y11 · · · 0
.
.
.
.
.
.
.
.
.
0 · · · Yss



.
By the formula in Lemma 3.7(2), it follows that Φ is an isomorphism of groups. By the
same formula, we can describe the action of Hu on GLn(K) as in (26). 
Recall that our aim is to find a “normalized” matrix X ∈ GLn(K) that defines, together
with u, a given absolutely reducible representation W . We show that the action of Hu
on Mn(K) can be splitted in s actions Mn,nk(K) × GLnk(K) → Mn,nk(K), for k =
1, . . . , s.
Let X = (xij)i,j=1,...,n and let u : {1, . . . , n} −→ {1, . . . , r}. Recall that F denotes
the partition {F1, . . . , Fs} associated to u. For 1 ≤ k ≤ s, we define Xk ∈ Mn,nk(K) to
be the matrix whose elements are xij , where i ∈ {1, . . . , n} and j ∈ Fk .
Each Xk is made of the F -blocks X1k, X2k, . . . , Xsk. Hence we have
(27) (XY )k = XkY kk
for any Y ∈ Hu.
The group GLnk(K) acts on Mn,nk(K) by right multiplication. Thus, the action of Hu
on GLn(K) can be recovered from these s actions by the relation (27).
DEFINITION 3.9. The symmetric groupSn acts on Mn,m(K) by row permutation, (σ,X) 7→
Xσ, where
Xσ :=





xσ(1)1 · · · xσ(1)m
xσ(2)1 · · · xσ(2)m
.
.
.
.
.
.
xσ(n)1 · · · xσ(n)m





THEOREM 3.10. Let X ∈ GLn(K) and let u : {1, . . . , n} −→ {1, . . . , r} be an arbitrary
set map. The orbit of X with respect to the action of Hu on GLn(K) contains a matrix X
such that
Xk =
(
Ink
Zk
)
σk
,
where Zk is a certain matrix in Mn−nk,nk(K) and σk ∈ Sn for any k ∈ {1, . . . , s}.
Proof. Since X ∈ GLn(K), the blocks X1, . . . , Xs are matrices of rank n1, . . . , ns
respectively (the columns of X are linearly independent, so the columns of each Xk are
so). Thus there are nk rows of Xk that are linearly independent. There is τk ∈ Sn such
that Xkτk has the first nk rows linearly independent. It follows that
Xkτk =
(
Yk
X ′k
)
with Yk ∈ GLnk(K) and X ′k ∈ Mn−nk,nk(K). Hence
Xk =
(
Ink
X ′kY −1k
)
τ−1k
· Yk,
so we may take Zk = X ′kY −1k and σk = τ−1k . 

ON THE CLASSIFICATION OF TWISTING MAPS BETWEEN Kn AND Km 15
REMARK 3.11. The matrices Z1, . . . , Zs and the permutations σ1, . . . , σs are not uniquely
determined.
DEFINITION 3.12. The matrix X as in Theorem 3.10 will be called a normalized invertible
matrix.
COROLLARY 3.13. Let W be an absolutely reducible representation of A of dimension
n. Then there are a normalized invertible matrix X ∈ GLn(K) and a set map u :
{1, . . . , n} −→ {1, . . . , r} that define the action of A on W .
Proof. Any absolutely reducible representation is defined by a matrix X0 and a map u :
{1, . . . , n} −→ {1, . . . , r}. By the Theorem 3.10, there is a normalized invertible matrix
X in the orbit of X0, with respect to the action of Hu on GLn(K). We have already noticed
that two matrices in the same orbit define the same representation. So W is defined by X
and u. 
EXAMPLE 3.14. We keep the notation from Example 3.6. An arbitrary element in Hu has
the following form
Y =




y11 0 0 0
0 y22 0 y24
0 0 y33 0
0 y42 0 y44




,
where y11 and y33 are non-zero and y22 y44 6= y24 y42. For an normalized invertible matrix
X ∈ GL4(K) we have
X1 =




1
x21
x31
x41




σ1
, X2 =




1 0
0 1
x32 x34
x42 x44




σ2
, X3 =




1
x23
x33
x43




σ3
.
The permutations σ1, σ2, σ3 ∈ S4 have to be chosen such that det(X) 6= 0. For example
X =




x21 0 x43 1
x41 x42 1 x44
1 x32 x23 x34
x21 1 x33 0




is a normalized invertible matrix if, and only if, det(X) 6= 0.
We end this section by analyzing in detail the case of two-dimensional absolutely re-
ducible representations.
PROPOSITION 3.15. Let X ∈ GL2(K) and let u : {1, 2} −→ {1, . . . , r} be a set map.
Then there are x, y ∈ K , with xy 6= 1 such that the orbit of X with respect to the action
of Hu on GL2(K) contains one of the matrices
X1 =
(
1 x
y 1
)
or X2 =
(
x 1
1 y
)
.
Proof. We start by considering the case u(1) = u(2) = i1 ∈ {1, . . . , r}. Hence the
partition F associated to u has one set F1 = {1, 2} and Hu = GL2(K). In conclusion, X
and I2 = XX−1 are in the same orbit, so in this case we can take x = y = 0 in X1.
Let now take u : {1, 2} −→ {1, . . . , r} such that u(1) 6= u(2). If Im (u) = {i1, i2},
with 1 ≤ i1 < i2 ≤ r, then the partition F is given either by F1 = {1} and F2 = {2} or
F1 = {2} and F2 = {1}. In both cases we get that a normalized invertible matrix is of the
following type:
Z1 =
(
1 1
x1 y1
)
, Z2 =
(
x2 y2
1 1
)
, Z3 =
(
1 x3
y3 1
)
, Z4 =
(
x4 1
1 y4
)
.

16 P. JARA, J. L ´OPEZ PE ˜NA, G. NAVARRO, AND D. S¸TEFAN
Since Zi is invertible we have xi 6= yi for i ∈ {1, 2}, and xiyi 6= 1 for i ∈ {3, 4}. We
know that in the orbit of X there is a matrix Zi, for a certain i ∈ {1, 2, 3, 4}. If either i = 3
or i = 4, one can take X1 = Z3 or X2 = Z4.
Let us assume that i = 1. Note that
Hu =
{(
y1 0
0 y2
)
| y1y2 6= 0
}
and either x1 6= 0 or y1 6= 0. If x1 6= 0 we have
Z1 ·
(
x−11 0
0 1
)
=
(
x−11 1
1 y1
)
,
so in the orbit of Z1 (and hence of X) there is a matrix of type
(
x 1
1 y
)
. In the case
when y1 6= 0 one can show that the orbit of X contains a matrix of the same type. In the
case i = 2 we can proceed similarly to show that the orbit of X contains a matrix X2 as in
the statement of the theorem. 
Let W be a vector space of dimension 2. We fix a basis {w1, w2} on W. We want to
classify all A-module structures on W which are absolutely reducible.
THEOREM 3.16. Let A be a K-algebra and let W be a two-dimensional absolutely re-
ducible A-module. Then there are x, y ∈ K and α1, α2 ∈ AlgK(A,K) such that xy 6= 1
and for any a ∈ A
aw1 =
1
1− xy [α1(a)− xyα2(a)]w1 +
y
1− xy [α1(a)− α2(a)]w2,(28)
aw2 =
−x
1− xy [α1(a)− α2(a)]w1 +
1
1− xy [−xyα1(a) + α2(a)]w2,(29)
Proof. The representation W is defined by a function u : {1, 2} −→ {1, . . . , r} and
a normalized invertible matrix X . Since two matrices in the same orbit with respect to
the action of Hu define the same representation we may assume, in view of the previous
proposition, that
X =
(
1 x
y 1
)
or X =
(
x 1
1 y
)
.
Let ω := (ωij)i,j=1,2 be the matrix associated to the A-module W as in (20). There are
θu(1) and θu(2) in AlgK(A,K) such that
ω = X
(
θu(1) 0
0 θu(2)
)
X−1.
We first consider the case X =
(
1 x
y 1
)
. If α1 := θu(1) and α2 := θu(2), then a
straightforward computation shows us that
(30) ω =
(
(1− xy)−1 [α1 − xyα2] −x(1− xy)−1 [α1 − α2]
y(1− xy)−1 [α1 − α2] (1− xy)−1 [−xyα1 + α2]
)
It is easy to see that the module structure defined by ω in this case is as in (28) and (29).
If X =
(
x 1
1 y
)
, then, for α1 = θu(2) and α2 = θu(1), one can show that the
corresponding action also satisfies relations (28) and (29). 

ON THE CLASSIFICATION OF TWISTING MAPS BETWEEN Kn AND Km 17
4. SPLITTED, UNITAL AND FACTORIZABLE REPRESENTATIONS OF CYCLES OF
LENGTH 2
In this section, for A = Km, we shall classify all splitted, unital and factorizable repre-
sentations of a cycle Γ of length 2. We use the standard quiver-notation Γ0 = {1, 2} and
Γ1 = {α1, α2} in order to represent the quiver below.
1 ◦ ◦ 2
α1
$$
α2
dd
A representation of Γ is defined by a set {ϕ1, ϕ2, ϕα1 , ϕα2} of K-linear maps. For i =
1, 2, there are ϕ1i , . . . , ϕmi , ϕ1αi , . . . , ϕmαi ∈ HomK(A,K) such that
ϕi(a) =
m
∑
p=1
ϕpi (a)fp, ∀a ∈ A,(31)
ϕαi(a) =
m
∑
p=1
ϕpαi(a)fp, ∀a ∈ A.(32)
Here {f1, . . . , fm} denotes the canonical basis on Km. If {f∗1 , . . . , f∗m} is the dual basis
of {f1, . . . , fm}, then
AlgK(Km,K) = {f∗1 , . . . , f∗m}.
Let us assume that {ϕ1, ϕ2, ϕα1 , ϕα2} defines a splitted, unital and factorizable represen-
tation of Γ. Since it is unital and factorizable, it is not difficult to see that the matrix
ωp = (ωpij)i,j=1,2 given by
(33) ωp =
(
ϕp1 ϕpα1
ϕpα2 ϕ
p
2
)
∀p ∈ {1, . . . ,m}
defines an A-module structure on a vector space W of dimension 2. If we fix a basis
{w1, w2} on W , the corresponding module structure is defined by
awi =
n
∑
j=1
ωpji(a)wj , ∀a ∈ A, ∀i ∈ {1, 2}.
To emphasize the fact that this module structure depends on p ∈ {1, . . . ,m}, we shall
denote it by Wp.
THEOREM 4.1. Let A = Km and let Γ be a cycle of length 2. If ωp is defined as in (33),
then there are a1, . . . , am ∈ K and u : {1, . . . ,m} −→ {1, . . . ,m} such that
(34) ωp =
(
apf∗p + (1− ap)f∗u(p) ap(f∗p − f∗u(p))
(1 − ap)(f∗p − f∗u(p)) (1− ap)f∗p + apf∗u(p)
)
.
Proof. Since A = Km is a semisimple algebra and every simple A-module is one dimen-
sional, it follows that every representation of A is absolutely reducible. In particular, Wp
is so, for any p ∈ {1, . . . ,m}. By the proof of Theorem 3.16, there are xp, yp ∈ K with
xpyp 6= 1, and αp1, αp2 ∈ AlgK(Km,K) such that
ωp =
(
(1 − xpyp)−1 [αp1 − xpypαp2] −xp(1− xpyp)−1 [αp1 − αp2]
yp(1 − xpyp)−1 [αp1 − αp2] (1− xpyp)−1 [−xpypαp1 + αp2]
)
.
On the other hand, since {ϕ1, ϕ2, ϕα1 , ϕα2} defines a splitted representation of Γ, we have
ϕ1 + ϕα2 = IdA and ϕ2 + ϕα1 = IdA. These relations are equivalent to
(35) ϕp1 + ϕpα2 = f∗p and ϕp2 + ϕpα1 = f∗p for any p ∈ {1, . . . , n},
therefore the sum of the elements of each column of ωp is equal to f∗p .

18 P. JARA, J. L ´OPEZ PE ˜NA, G. NAVARRO, AND D. S¸TEFAN
If αp1 = αp2, then the above condition imposed on the elements of the column in ωp
means that αp1 = αp2 = f∗p . In this case we take ap = 1 and we define u(p) = p.
Now let us assume that αp1 6= αp2. Then these algebra morphisms are linearly indepen-
dent. By imposing that ωp satisfies (35), we get
yp + 1
1− xpypα
p
1 −
yp(xp + 1)
1− xpyp α
p
2 = f∗p =
−xp(yp + 1)
1− xpyp α
p
1 +
xp + 1
1− xpypα
p
2.
These equalities are possible only in the following two cases:
First case: yp = −1 and αp2 = f∗p . As xpyp 6= 1, we have xp 6= −1. Since αp1
is an algebra map, there is u(p) ∈ {1, . . . ,m} such that αp1 = f∗u(p). We now define
ap := xp(1 + xp)−1. It is not difficult to see that ωp satisfies (34).
Second case: xp = −1 and αp1 = f∗p . As above, yp 6= −1 and there is 1 ≤ u(p) ≤ m
such that αp2 = f∗u(p). We conclude by defining ap := (1 + yp)−1. 
Now we can classify all splitted, unital and factorizable representations of a cycle of
length 2.
THEOREM 4.2. Let A = Km and let Γ be a cycle of length 2. The set of K-linear
endomorphisms ϕ := {ϕ1, ϕ2, ϕα1 , ϕα2} defines a splitted, unital and factorizable repre-
sentation of Γ if, and only if, there are a1, . . . , am ∈ K and a set map u : {1, . . . ,m} −→
{1, . . . ,m} such that
ϕ1 =
m
∑
p=1
[
apf∗p + (1− ap)f∗u(p)
]
fp, ϕα1 =
m
∑
p=1
[
ap(f∗p − f∗u(p))
]
fp,(36)
ϕ2 =
m
∑
p=1
[
(1− ap)f∗p + apf∗u(p)
]
fp, ϕα2 =
m
∑
p=1
[
(1− ap)(f∗p − f∗u(p))
]
fp,(37)
and, in addition, they satisfy the following conditions:
(1) If u(p) = p, then ap = 0.
(2) If u(p) 6= p, then ap + au(p) = 1. In addition, if u2(p) 6= p, then ap ∈ {0, 1}.
Proof. In view of Theorem 4.1, there are a1, . . . , am ∈ K and u such that the maps in ϕ
satisfy the four equalities in the statement. Note that if u(p) = p, then the p-th term of
each sum does not depend on ap. Therefore, in this case, we may normalize ap by setting
ap = 0. Thus we may choose a1, . . . , am such that the first condition is verified.
Note that the above description of the maps ϕ1, ϕ2, ϕα1 and ϕα2 was obtained as a
consequence of the fact that ϕ defines a unital and factorizable representation of Γ such
that ϕ1 + ϕα2 = ϕ2 + ϕα1 = IdA. In conclusion, the representation associated to ϕ is
also splitted if, and only if, ϕα1 and ϕα2 are idempotents. Remark that {ϕ1, ϕα2} and
{ϕ2, ϕα1} then are complete sets of orthogonal idempotents, as it is required.
Trivially, ϕα1 is an idempotent K-linear map if, and only if, ϕ2α1(fq) = ϕα1(fq), for
any q ∈ {1, . . . ,m}. By definition of ϕα1 , we get that these equalities hold if, and only if,
for arbitrary p, q ∈ {1, . . . ,m},
(38) ap
[
(δp,q − δu(p).q)ap − (δu(p),q − δu2(p),q)au(p)
]
= ap(δp,q − δu(p),q).
Let us assume that ap 6= 0. By the normalization condition, then u(p) 6= p.
First case: u2(p) = p. Since ap 6= 0, relation (38) is equivalent in this case to
(ap + au(p) − 1)(δp,q − δu(p),q) = 0, ∀q ∈ {1, . . . ,m}.
Since u(p) 6= p, we deduce that ap + au(p) = 1.
Second case: u2(p) 6= p. Hence, for q = p, relation (38) becomes ap = 1. We want
to prove that au(p) = 0. Note that if u2(p) = u(p), then this equality follows from the
fact that we have normalized the elements a1, . . . , am. So we may assume that u2(p) and

ON THE CLASSIFICATION OF TWISTING MAPS BETWEEN Kn AND Km 19
u(p) are not equal. By taking q = u(p) in (38), it follows that au(p) = 0. Obviously,
ap + au(p) = 1 in this case too.
Summarizing, since ϕα1 is an idempotent K-linear map then, for any p such that ap 6=
0, one of the following conditions holds:
u2(p) = p, and then ap + au(p) = 1;(39)
u2(p) 6= p, and then ap = 1 and au(p) = 0.(40)
Since ϕα2 =
∑m
r=1
[
(1− ar)(f∗r − f∗u(r))
]
fp is an idempotent too, working with 1− ap
instead of ap, we can show that, for any p ∈ {1, . . . ,m} such that ap 6= 1, one of the
following two conditions holds:
u2(p) = p, and then ap + au(p) = 1;(41)
u2(p) 6= p, and then ap = 0 and au(p) = 1.(42)
Since either ap 6= 0 or ap 6= 1, the conditions (39)-(42) together imply that a1, . . . , am
and u verify the second property in the statement of the theorem.
Now let us prove that, for any a1, . . . , am and u satisfying (1) and (2), the set ϕ :=
{ϕ1, ϕ2, ϕα1 , ϕα2} defines a splitted, unital and factorizable representation of Γ. By the
proof of Theorem 4.1, ϕ defines a representation of Γ which is unital and factorizable.
Of course, by construction, ϕ1 + ϕα2 = IdA and ϕ2 + ϕα1 = IdA. By the proof of the
other implication, it results that conditions (1) and (2) imply that ϕ1 and ϕ2 are idempotent
maps. In conclusion, the representation defined by ϕ is splitted too. 
COROLLARY 4.3. Let τ : Kn ⊗ Km −→ Km ⊗ Kn be a twisting map such that the
connected components of Γ, the corresponding quiver, are all trivial excepting one which
is a cycle of length 2. We label the vertices of Γ as in the picture below.
◦1 ◦2 ◦3 ◦4 ◦5 · · · ◦
n
α1
##
α2
cc :: :: :: :::: dd
There are a1, . . . , am ∈ K and u : {1, . . . ,m} −→ {1, . . . ,m} that satisfy conditions (1)
and (2) in Theorem 4.2 such that
τ(e1 ⊗ x) = ϕ1(x) ⊗ e1 + ϕα1(x) ⊗ e2,
τ(e2 ⊗ x) = ϕα2(x)⊗ e1 + ϕ2(x) ⊗ e2,
τ(ei ⊗ x) = x⊗ ei, ∀i ≥ 3.
where ϕ1, ϕ2, ϕα1 , ϕα2 are given by the formulae (36) and (37).
REMARK 4.4. If n = 2, the corollary above gives the classification of noncommutative
duplicates of Km (cf. [11, 16]).
We are going to classify all splitted, unital, and factorizable representations of a con-
nected quiver of rank 1 that contains a cycle of length 2. We know that, removing the
arrows of the cycle from such a quiver, we get two rooted trees. These trees have their
roots in the vertices of the cycle and they are oriented. Note that they might be trivial, that
is they could have a unique vertex, namely their root.

20 P. JARA, J. L ´OPEZ PE ˜NA, G. NAVARRO, AND D. S¸TEFAN
Let us denote the vertices of this cycle by 1 and 2, being 3, 4, . . . , n the other vertices
of Γ, and denote by α1 and α2 the arrows of the cycle, as in the picture below.
◦ ◦1 2
◦3 ◦ 4 5
◦6 ◦7
◦
α1
##
α2
cc
TT))))
JJ
TT))))
JJ
OO
We take a splitted, unital and factorizable representation of Γ, which is defined by the K-
linear maps ϕ1, . . . , ϕn and {ϕα}, where α runs over the set of arrows which are not a
loop. Note that for i ≥ 3, ϕi is an idempotent algebra map, and thus, there are idempotent
maps ui : {1, . . . ,m} → {1, . . . ,m} such that
ϕi(fp) =
m
∑
q=1
δp,ui(q)fq, ∀ p = 1, . . . ,m.
Obviously, ϕ = {ϕ1, ϕ2, ϕα1 , ϕα2} defines a splitted, unital and factorizable representa-
tion of the unique cycle in Γ. Hence, there are a map u : {1, . . . ,m} → {1, . . . ,m} and
scalars a1, . . . , am ∈ K that satisfy conditions (1) and (2) in Theorem 4.2. Moreover, the
maps in ϕ are given by the formulae (36) and (37).
Note that, for an arrow α that is not a loop, we have ϕα = IdKm−ϕt(α), as rrank(Γ) =
1. Hence, the data that we need to define a representation of Γ is the following:
DEFINITION 4.5. The data D(u, a) is defined by
(1) The maps u1, u2, u3, . . . , un : {1, . . . ,m} → {1, . . . ,m}, where u1 = u2 = u and
for each i ≥ 3 the maps ui are idempotents.
(2) The elements a1, . . . , am ∈ K obeying conditions (1) and (2) in Theorem 4.2.
THEOREM 4.6. The data D(u, a) defines a splitted, unital and factorizable representation
of Γ if, and only if, for every α ∈ Γ1 \ {α1, α2} and every p ∈ {1, . . . ,m} such that
us(α)(p) 6= p and ut(α)(p) 6= p, one of the following conditions holds:
(1) s(α) = 1 and ap = 1.
(2) s(α) = 2 and ap = 0.
Proof. Let α ∈ Γ1 \ {α1, α2}, and let p ∈ {1, . . . ,m}. We assume that us(α)(p) 6= p and
that ut(α)(p) 6= p and we want to show that either (1) or (2) hold.
There is an arrow β such that t(β) = s(α), as in the following figure
◦i ◦k◦
jβ // α //
By Lemma 2.6(2), we get
(43) Ker (ϕs(α))Ker (ϕt(α)) = 0.
First let us prove that s(α) ∈ {1, 2}. If s(α) is not a vertex of the cycle, we may apply
Lemma 2.12 to show that Kerϕs(α) is generated by all fr with us(α)(r) 6= r. Similarly,
Kerϕt(α) is generated by all fr with ut(α)(r) 6= r. In particular, by our assumptions, we
deduce that fp is an element in Kerϕs(α)Kerϕt(α), which is not possible, in view of (43).
Let us now take an arrow such that s(a) = 1. We have to prove that ap = 1. Note that
(43) can be written as follows:
Imϕα2Kerϕk = 0.
Indeed, as rank(j) = 1, we have β = α2. Thus ϕj = IdA−ϕα2 , and the required relation
follows from the fact that ϕα2 is an idempotent K–linear map. We know that Ker (ϕk) is

ON THE CLASSIFICATION OF TWISTING MAPS BETWEEN Kn AND Km 21
spanned by all fq with uk(fq) 6= fq. On the other hand,
ϕa2 =
m
∑
q=1
[
(aq − 1)(f∗q − f∗u(q))
]
fq.
Thus, (43) holds if, and only if, ϕα2(x)fq = 0, for any q such that uk(fq) 6= fq. At its
turn, this property is equivalent to
(aq − 1)(f∗q − f∗u(q)) = 0
for any q such that uk(q) 6= q. As u(p) 6= p and uk(p) 6= p, we deduce, in particular, that
ap = 1. The case s(α) = 2 is handled in a similar way.
Conversely, let D(u, a) be some data as in Definition 4.5. We assume that D(u, a)
satisfy conditions (1) and (2) of the statement. We define ϕ = {ϕ1, ϕ2, ϕα1 , ϕα2} by (36)
and (37). By Theorem 4.2, ϕ defines a a splitted unital and factorizable representation of
the cycle in Γ. Furthermore, for i ≥ 3 we take ϕi to be the algebra endomorphism of Km
that corresponds to ui. Finally, for an arrow α ∈ Γ1 \ {α1, α2} that is not a loop, we set
ϕα := IdA − ϕt(α). Obviously, the maps constructed above define a unital and splitted
representation of Γ. We still have to check that this is factorizable. Since rrank(Γ) = 1, in
view of Lemma 2.6, we have to prove that
Ker (ϕs(α))Ker (ϕt(α)) = 0
for every arrowα ∈ Γ1\{α1, α2}.If s(α) ∈ {1, 2}, then this equality follows by conditions
(1) and (2) in the statement of the theorem (see the proof of the other implication). If
s(α) /∈ {α1, α2}, then the required relation follows from Lemma 2.12 and the proof of
Theorem 2.13, taking into account that either us(α)(p) = p or ut(α)(p) = p. 
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DEPARTMENT OF ALGEBRA, UNIVERSITY OF GRANADA, E-18071–GRANADA, SPAIN
E-mail address: pjara@ugr.es
MATHEMATICS RESEARCH CENTRE, QUEEN MARY UNIVERSITY OF LONDON, MILE END ROAD, LON-
DON E1 4NS, UNITED KINGDOM
E-mail address: jlopez@maths.qmul.ac.uk
DEPARTMENT OF COMPUTER SCIENCE AND IA, UNIVERSITY OF GRANADA, E-18071–GRANADA, SPAIN
E-mail address: gnavarro@ugr.es
UNIVERSITY OF BUCHAREST, FACULTY OF MATHEMATICS, STR. ACADEMIEI 14, RO-70109, BUCHAREST,
ROMANIA
E-mail address: dstefan@al.math.unibuc.ro