# Characterisations of trivial extensions

Elsa Fernandez  [EF] Facultad de Ingeniería, Universidad Nacional de la
Patagonia San Juan Bosco. (9120) Puerto Madryn, Argentina<EMAIL_ADDRESS>,  Sibylle Schroll  [SS] Mathematical Institut University of Cologne Weyertal
86-90 (50931) Köln Germany and Institutt for matematiske fag, NTNU, N-7491
Trondheim, Norway<EMAIL_ADDRESS>,  Hipolito Treffinger  [HT]
Institut Mathématique Jussieu - Paris Rive Gauche Univeristé Paris Cité,
Bâtiment Sophie Germain 5, rue Thomas Mann. (75013) Paris, France.
<EMAIL_ADDRESS>,  Sonia Trepode  [ST] Centro Marplatense de
Investigaciones Matemáticas. FCEyN, Universidad Nacional de Mar del Plata,
CONICET. Funes 3350. (7600) Mar del Plata, Argentina<EMAIL_ADDRESS>and
Yadira Valdivieso  [YV] Departamento de Actuaría, Física y Matemáticas,
Universidad de las Américas Puebla, Ex Hacienda Sta. Catarina Mártir S/N, San
Andrés Cholula, 72810 Puebla, México<EMAIL_ADDRESS>
###### Abstract.

In this paper we give a characterisation of trivial extension algebras in
terms of quivers with relations. This result is based on a explicit
description of the ideal of relations of the trivial extension of an algebra,
given by the first author in the appendix. We also give a new proof of
Wakamatsu’s theorem in terms of their quiver and relations, which determines
when two given algebras have isomorphic trivial extensions.

##  1\. Introduction

Split extensions of a ring by a bimodule are classical constructions that
appear in many different contexts. For example, Hochschild [14] showed that
the trivial extension of a ring $R$ by an $R$-$R$-bimodule $M$ corresponds to
the zero element in the second cohomology group $H^{2}(R,M)$. In commutative
algebra, Nagata [18] used split-extensions to show that any module over a
commutative ring can be thought of as an ideal. More recently, split
extensions have played a central role in the study of cluster-tilted algebras
[1, 5], in the connections between gentle algebras and symmetric special
biserial algebras [12, 19], in higher homological algebra [10] and the
Hochschild (co)homology of split-extensions has been studied, for example, in
[3, 4, 6].

Although the theory of trivial extensions has been greatly developed, in
general, given an explicit algebra in terms of quiver and relations, there has
not been an explicit construction of the structure of the trivial extension as
an algebra, except for certain cases [7, 8, 13]. Note that by the trivial
extension of an algebra $A$, we mean the algebra $T(A)=A\ltimes D(A)$ where
$D(A)$ is $A$-$A$-bimodule given by the dual of $A$. In the appendix of this
paper, an explicit construction of the trivial extension is developed by the
first author (see Theorem 1.1). Namely given an algebra $A=KQ_{A}/I_{A}$ the
ideal of relations of the trivial extension $T(A)$ is explicitly constructed,
thus completing the description of $T(A)$ since in [7] the quiver of $T(A)$
was already constructed.

Furthermore, it has been an open question to know which algebras are
isomorphic to the trivial extension of a finite dimensional algebra. Since
trivial extensions are symmetric, the question reduces to asking when is a
symmetric algebra isomorphic to a trivial extension. We address this question
for all finite dimensional symmetric $K$-algebras, see Theorem 1.2. Finally,
we turn to the question of when two algebras have isomorphic trivial
extensions, a question already considered in the abstrac setting by Wakamatsu
in [20]. Lastly, in Theorem 1.3, we give a precise characterisation of the
algebras that have isomorphic trivial extensions in the language of admissible
cuts which were introduced in [9].

We now state the main results of this paper. For this we recall that in [7]
the quiver $Q_{T(A)}$ of the trivial extension $T(A)$ of a finite dimensional
algebra $A=KQ_{A}/I_{A}$ has been described. Namely, the set of vertices
$(Q_{A})_{0}$ and $(Q_{T(A)})_{0}$ coincide. Then given a $K$-basis
$\mathbb{M}=\\{p_{1},\dots,p_{n}\\}$ of $\mbox{soc}_{A^{e}}A$, the set of
arrows $(Q_{T(A)})_{1}$ is the disjoint union of $(Q_{A})_{1}$ and
$\\{\beta_{p_{1}},\dots,\beta_{p_{n}}\\}$ where $s(\beta_{p_{i}})=t(p_{i})$
and $t(\beta_{p_{i}})=s(p_{i})$ for every $p_{i}\in\mathbb{M}$. Given a path
$p$, we call a path $q$ such that $pq$ is a cycle a supplement of $p$ (see
Definition 2.2. For the notion of elementary cycle, we refer to Definition
2.1. In the appendix of this paper the following is shown.

######  Theorem 1.1 (Theorem A.1).

Let $A=KQ_{A}/I_{A}$ be a finite-dimensional algebra and let
$T(A)=KQ_{T(A)}/I_{T(A)}$ be its trivial extension. Then the quiver $Q_{T(A)}$
is as above and the ideal $I_{T(A)}$ is generated by the union of the
following sets.

  1. (1)

A generating set of the ideal of relations $I_{A}$ of $A$.

  2. (2)

The paths that are not contained in an elementary cycle.

  3. (3)

For any vertices $x$ and $y$ in $Q_{T(A)}$, the linear combinations of paths
$\rho\in e_{x}KQ_{T(A)}e_{y}$ such that $q\rho\in I_{x}^{\prime}$ or $\rho
q\in I_{y}^{\prime}$ for any supplement path $q$ in an elementary cycle $C$.

Having this complete description of the ideal of relations of a trivial
extension, we are able to give a complete characterisation of when a symmetric
algebra is isomorphic to a trivial extension of some finite-dimensional
$K$-algebra.

######  Theorem 1.2 (Theorem 3.6).

Let $A=KQ/I$ be an algebra. Then $A$ is isomorphic to the trivial extension of
some finite-dimensional $K$-algebra if and only if:

  1. (a)

There is a presentation of $A$ for which there exists a set $\mathcal{E}$ of
distinguished cycles in $Q_{A}$ with weight function $\omega:\mathcal{E}\to K$
and

  2. (b)

There is an allowable cut $\mathcal{D}=\\{\gamma_{1},\cdots,\gamma_{t}\\}$ of
$A$ such that the quotient $B=A/\langle\mathcal{D}\rangle$ verifies the
following:

    1. (i)

$A$ is a split-by-nilpotent extension of $B$.

    2. (ii)

The supplements of the cut arrows in the cycles in $\mathcal{E}$ are in one-
to-one correspondence with the elements of a basis of $\mbox{soc}_{B^{e}}B$.

In this case the two-sided ideal $\langle\mathcal{D}\rangle$ is isomorphic to
$DB$ as a $B$-$B$-bimodule and $A$ is isomorphic to $T(B)$.

We note that the previous result reduces the problem of identifying if an
algebra is a trivial extension to the problem of determining whether an
algebra is a split-by-nilpotent extension, a problem that has been solved in
[2]. Thus our result gives a complete answer which in combination with [2]
should be implementable in terms of a computer algorithm.

In the 70’s, Müller [17], Green and Reiten [11], Iwanaga and Wakamatsu [16],
and Hughes and Waschbüsch [15] studied the relationship between the
representation type of the algebra $T(A)$ and the algebra $A$. In particular,
in [15], they gave a complete description of the representation-finite trivial
extension of algebras which relies on whether two trivial extensions are
isomorphic or not. This result motivated Wakamatsu to study when two trivial
extensions are isomorphic, see [20]. He gave, for two Artin algebras,
necessary and sufficient conditions for having isomorphic trivial extensions.
In more precise terms, he shows that two algebras $A$ and $A^{\prime}$ have
isomorphic trivial extensions if and only if they are split-by-nilpotent
extensions of a common subalgebra $S$ by a $S$-$S$-bimodule $M$ and its dual
$D(M)$, respectively.

The original motivation for this paper was to give an explicit description of
the relationship of two algebras $A$ and $A^{\prime}$ that have isomorphic
trivial extensions. This is achieved in our next result, which gives an
explicit description of the algebra $S$ and the extending $S$-$S$-bimodules in
the case where $A$ and $A^{\prime}$ are given by quiver and relations. We also
give an explicit characterisation of when the trivial extensions of two
algebras are isomorphic. Our result is shown by a proof which is independent
of Wakamatsu’s proof.

######  Theorem 1.3 (Theorem 4.1).

Let $A=KQ_{A}/I_{A}$ be a finite-dimensional algebra with trivial extension
$T(A)$ and set $(Q_{A})_{1}=\\{\alpha_{1},\ldots,\alpha_{n}\\}$ and
$\\{\beta_{1},\dots,\beta_{t}\\}=(Q_{T(A)})_{1}\setminus(Q_{A})_{1}$. The
following are equivalent

  1. (a)

$T(A)\cong T(A^{\prime})$.

  2. (b)

There exists an admisible cut $\mathcal{D}$ of the form

$\mathcal{D}\;=\;\\{\alpha_{1},\ldots,\alpha_{r},\beta_{1},\ldots,\beta_{s}:\alpha_{i}\in(Q_{A})_{1}\text{
and }\beta_{j}\not\in(Q_{A})_{1}\\}$

with $A^{\prime}\cong T(A)/\langle\mathcal{D}\rangle$, $T(A)$ is a split-by-
nilpotent extension of $A^{\prime}$ and the supplements in the elementary
cycles of the cut arrows are in one-to-one correspondence with the elements of
$\mbox{soc}_{A^{\prime e}}A^{\prime}$.

  3. (c)

$A\cong S\ltimes M$ and $A^{\prime}=S\ltimes D(M)$, where

    1. (i)

$S$ is the subalgebra of $T(A)$ generated by $\sum_{x\in(Q_{T_{A}})_{0}}e_{x}$
the identity of $T(A)$ and $\alpha_{r+1},\ldots,\alpha_{n}$,

    2. (ii)

$M=S\left\langle\alpha_{1},\ldots,\alpha_{r}\right\rangle S$,

    3. (iii)

and $D(M)=S\left\langle\beta_{s+1},\ldots,\beta_{t}\right\rangle S$.

##  2\. Background

In this paper, by an algebra we mean a basic connected finite-dimensional
algebra over an algebraically closed field $K$. It is well-known that every
such algebra $A$ is Morita equivalent to the path algebra of a (finite) quiver
$Q$ modulo an admissible ideal $I$ of $KQ$. For the purposes of this paper, we
always identify $A$ with $KQ_{A}/I_{A}$. If $A=KQ_{A}/I_{A}$, we say that
$Q_{A}$ is the quiver of $A$. In addition, every element of a set of
generators of $I_{A}$ is called a relation of $A$ and we refer to $I_{A}$ as
the ideal of relations of $A$.

Given a quiver $Q$, we denote by $Q_{0}$ the set of vertices of $Q$ and
$Q_{1}$ the set of arrows of $Q$. For every arrow $\alpha\in Q_{1}$,
$s(\alpha)$ and $t(\alpha)$ denote the source and the target of $\alpha$,
respectively. A path $p$ of length $t$ from the vertex $x$ to the vertex $y$
in $Q$ is an ordered set of arrows $\\{\alpha_{1},\dots,\alpha_{t}\\}$ such
that $s(\alpha_{1})=x$, $t(\alpha_{t})=y$ and $t(\alpha_{i})=s(\alpha_{i+1})$
for all $1\leq i\leq t-1$. By abuse of notation we write
$p=\alpha_{1}\alpha_{2}\dots\alpha_{t}$ and we define $s(p):=s(\alpha_{1})$
and $t(p):=t(\alpha_{t})$. Also, for every vertex $x\in(Q_{A})_{0}$ we denote
by $e_{x}$ the stationary path at vertex $x$, that is the unique path of
length $0$ from $x$ to $x$. We say that a path $p$ is a cycle if $p$ is of
strictly positive length and $s(p)=t(p)$. A cycle $p$ is said to be non-zero
if $p\not\in I_{A}$.

Let $A$ be an algebra and $M$ be a $A$-$A$-bimodule equipped with a
multiplication map $\mu:M\otimes_{A}M\to M$. Then the $K$-vector space
$B=A\oplus M$ has the structure of an algebra with the following
multiplication

$(a_{1},m_{1})(a_{2},m_{2})=(a_{1}a_{2},a_{1}m_{2}+m_{1}a_{2}+\mu(m_{1}\otimes
m_{2})),$

where $a_{1},a_{2}\in A$ and $m_{1},m_{2}\in M$. If the map $\sigma:A\to B$
defined by $\sigma(a)=(a,0)$ is a morphism of algebras we say that $B$ is a
split extension of $A$ by $M$. Moreover, if $M$ is nilpotent for $\mu$ we say
that $B$ is a split-by-nilpotent extension of $A$ by $M$ and we denote it as
$B=A*M$. If $\mu(m_{1},m_{2})=0$ for all $m_{1},m_{2}\in M$ we say that $B$ is
the trivial extension of $A$ by $M$ and we denote it by $A\ltimes M$. Finally,
if $M=D(A)$ with its natural bimodule structure, where
$D=\mbox{Hom}_{A}(-,K)$, we say that $B=A\ltimes D(A)$ is simply the trivial
extension of $A$ and we denote it by $T(A)$.

We recall from [7] an explicit description of the ordinary quiver of $T(A)$
based on the quiver of $A$. This description depends on a $K$-basis
$\mathbb{M}=\\{p_{1},\dots,p_{n}\\}$ of $\mbox{soc}_{A^{e}}A$. The quiver
$Q_{T(A)}$ is then defined as follows. The set of vertices $(Q_{A})_{0}$ and
$(Q_{T(A)})_{0}$ coincide. The set of arrows $(Q_{T(A)})_{1}$ is the disjoint
union of $(Q_{A})_{1}$ and $\\{\beta_{p_{1}},\dots,\beta_{p_{n}}\\}$ where
$s(\beta_{p_{i}})=t(p_{i})$ and $t(\beta_{p_{i}})=s(p_{i})$ for every
$p_{i}\in\mathbb{M}$. A set of generators of the ideal of relations $I_{T(A)}$
of $T(A)$ is given in Theorem A.1.

Finally, we recall the notion of admissible cut and elementary cycles,
introduced in [7] to study when two algebras of a certain family have
isomorphic trivial extension.

######  Definition 2.1.

Let $\mathbb{M}=\\{p_{1},\dots,p_{n}\\}$ be a basis of $\mbox{soc}_{A^{e}}A$.
An oriented cycle $C=\alpha_{1}\dots\alpha_{t}\beta_{p_{j}}$ in $KQ_{T(A)}$ is
said to be elementary if $\alpha_{1}\dots\alpha_{n}$ is a path in $KQ_{A}$,
$p_{j}\in\mathbb{M}$ and $p_{j}^{*}(\alpha_{1}\dots\alpha_{t})\neq 0$, where
$p_{j}^{*}$ is the dual of $p_{j}$. In this case, we say that
$\omega(C)=p_{j}^{*}(\alpha_{1}\dots\alpha_{t})\in K$ is the weight of $C$.

######  Definition 2.2.

Let $Q$ be a quiver, $C$ a cycle of $Q$, and $q$ a path contained in $C$. The
_supplement_ of $q$ in $C$ is the path $p$ such that $C=qp$ up to rotation.
Note that if $C=q$ then $p$ is the stationary path $e_{(s(q))}$.

######  Definition 2.3.

Let $A=KQ/I$ be a finite-dimensional algebra. Then an _admissible cut
$\mathcal{D}$ of $T(A)$_ is a subset of arrows of $Q_{T(A)}$ containing
exactly one arrow in each elementary cycle of $T(A)$ in such a way that if an
arrow in $\mathcal{D}$ appears in an elementary cycle $C$ then it appears only
once in $C$.

######  Remark 2.4.

The set of new arrows $\mathcal{D}=\\{\beta_{p_{1}},\dots,\beta_{p_{j}}\\}$
added in the construction of the quiver $Q_{T(A)}$ form an admissible cut of
$T(A)$. Moreover, it is easy to see that $A$ is isomorphic to the quotient
$T(A)/\langle\mathcal{D}\rangle$ where $\langle\mathcal{D}\rangle$ is the two-
sided of $T(A)$ ideal generated by $\mathcal{D}$.

We now give three examples. In the first example, we show that the weights of
the elementary cycles are not always equal to one, in the second we show that
not every arrow in an elementary cycle is included in an admissible cut. In
the third example, we show that not every admissible cut $\mathcal{D}$ of
$T(A)$ gives rise to an algebra $B=T(A)/\langle\mathcal{D}\rangle$ such that
$T(A)$ is a split-by-nilpotent extension of $B$.

######  Example 2.5.

Consider the algebra $A=KQ_{A}/I_{A}$ with quiver

${2}$${{Q_{A}:=}}$${1}$${3}$${5}$${4}$$\scriptstyle{\alpha_{1}}$$\scriptstyle{\gamma_{1}}$$\scriptstyle{\epsilon_{1}}$$\scriptstyle{\gamma_{2}}$$\scriptstyle{\alpha_{2}}$$\scriptstyle{\epsilon_{2}}$

quotiented by the ideal
$I_{A}=\langle\lambda_{1}\alpha_{1}\alpha_{2}+\lambda_{2}\gamma_{1}\gamma_{2}+\lambda_{3}\epsilon_{1}\epsilon_{2}\rangle$.
Then soc${}_{A^{e}}A$ is two dimensional with basis
$\\{p_{1}=\gamma_{1}\gamma_{2},p_{2}=\epsilon_{1}\epsilon_{2}\\}$. Then
$T(A)=KQ_{T(A)}/I_{T(A)}$ where

${2}$${{Q_{T(A)}:=}}$${1}$${3}$${5}$${4}$$\scriptstyle{\alpha_{1}}$$\scriptstyle{\gamma_{1}}$$\scriptstyle{\epsilon_{1}}$$\scriptstyle{\gamma_{2}}$$\scriptstyle{\alpha_{2}}$$\scriptstyle{\epsilon_{2}}$$\scriptstyle{\beta_{p_{1}}}$$\scriptstyle{\beta_{p_{2}}}$

In this case we have four different elementary cycles,
$C_{1}=\beta_{p_{1}}\gamma_{1}\gamma_{2}$,
$C_{2}=\beta_{p_{2}}\epsilon_{1}\epsilon_{2}$,
$C_{3}=\beta_{p_{1}}\alpha_{1}\alpha_{2}$ and
$C_{4}=\beta_{p_{2}}\alpha_{1}\alpha_{2}$. We calculate

$w(C_{1})=p_{1}^{*}(\gamma_{1}\gamma_{2})=1$
$w(C_{2})=p_{2}^{*}(\epsilon_{1}\epsilon_{2})=1$
$w(C_{3})=p_{1}^{*}(\alpha_{1}\alpha_{2})=p_{1}^{*}(-\lambda_{2}\gamma_{1}\gamma_{2}-\lambda_{3}\epsilon_{1}\epsilon_{2})=-\frac{\lambda_{2}}{\lambda_{1}}$
$w(C_{4})=p_{2}^{*}(\alpha_{1}\alpha_{2})=p_{2}^{*}(-\lambda_{2}\gamma_{1}\gamma_{2}-\lambda_{3}\epsilon_{1}\epsilon_{2})=-\frac{\lambda_{3}}{\lambda_{1}}.$

######  Example 2.6.

Let $Q$ be the quiver

${1}$${2}$$\scriptstyle{\alpha}$$\scriptstyle{\beta}$

and $I$ the ideal generated by $\alpha^{3}$. Denote by $A$ the algebra $KQ/I$.
Then the set $\mathbb{M}=\\{\alpha^{2}\beta\\}$ is a $K$-basis for
$\mbox{soc}_{A^{e}}A$ and $Q_{T(A)}$ is the following quiver

${1}$${2}$$\scriptstyle{\alpha}$$\scriptstyle{\beta}$$\scriptstyle{\beta_{1}}$

Then $C=\beta_{1}\alpha^{2}\beta$ is the only elementary cycle in $KQ_{T(A)}$,
up to cyclic permutations. It follows from Definition 2.3 that the only
admissible cuts are $\\{\beta\\}$ and $\\{\beta_{1}\\}$.

Observe that if we consider the set $\\{\alpha\\}$ as a cut, then
$A/\langle\alpha\rangle$ is isomorphic to the algebra
$B=KQ^{\prime}/I^{\prime}$ where $Q^{\prime}$ is the quiver

${1}$${2}$$\scriptstyle{\beta}$$\scriptstyle{\beta_{1}}$

and $I^{\prime}=(\beta_{1}\beta)$, which is an algebra of dimension 6 and
therefore $T(B)$ is an algebra of dimension 12, while $T(A)$ is a
14-dimensional algebra. Hence $T(A)$ is not isomorphic to $T(B)$.

We note that in this example there are nonzero cycles in $T(A)$ which are not
elementary cycles, namely $\alpha$, $\alpha^{2}$ and $\beta\beta_{1}$.

######  Example 2.7.

Let $A$ be an algebra given by $Q$

${1}$${6}$${2}$${7}$${5}$${10}$${3}$${8}$${4}$${9}$$\scriptstyle{\epsilon_{1}}$$\scriptstyle{\alpha_{2}}$$\scriptstyle{\epsilon_{2}}$$\scriptstyle{\beta_{2}}$$\scriptstyle{\alpha_{1}}$$\scriptstyle{\beta_{1}}$$\scriptstyle{\gamma_{1}}$$\scriptstyle{\eta_{1}}$$\scriptstyle{\epsilon_{3}}$$\scriptstyle{\gamma_{2}}$$\scriptstyle{\epsilon_{4}}$$\scriptstyle{\eta_{2}}$

and the relations

$\alpha_{1}\alpha_{2}+\beta_{1}\beta_{2}+\gamma_{1}\gamma_{2}+\eta_{1}\eta_{2}=0$
$\epsilon_{1}\gamma_{1}\gamma_{2}=0,\,\epsilon_{1}\eta_{1}\eta_{2}=0,\,\epsilon_{2}\gamma_{1}\gamma_{2}=0,\,\epsilon_{2}\eta_{1}\eta_{2}=0,$
$\epsilon_{3}\alpha_{1}\alpha_{2}=0,\,\epsilon_{3}\beta_{1}\beta_{2}=0,\,\epsilon_{4}\alpha_{1}\alpha_{2}=0,\,\epsilon_{4}\beta_{1}\beta_{2}=0.$

Then the following elements are a $K$-basis for $\mbox{soc}_{A^{e}}A$.

$\displaystyle\epsilon_{1}\alpha_{1}\alpha_{2}=-\epsilon_{1}\beta_{1}\beta_{2},\,\epsilon_{2}\alpha_{1}\alpha_{2}=-\epsilon_{2}\beta_{1}\beta_{2},\,\epsilon_{4}\gamma_{1}\gamma_{2}=-\epsilon_{4}\eta_{1}\eta_{2}$
$\displaystyle\epsilon_{3}\gamma_{1}\gamma_{2}=-\epsilon_{3}\eta_{1}\eta_{2},\,\alpha_{1}\alpha_{2}-\beta_{1}\beta_{2},\,\gamma_{1}\gamma_{2}-\eta_{1}\eta_{2},$
$\displaystyle\epsilon_{1}\gamma_{1},\,\epsilon_{1}\eta_{1},\,\epsilon_{2}\gamma_{1},\,\epsilon_{2}\eta_{1},\,\epsilon_{3}\alpha_{1},\,\epsilon_{3}\beta_{1},\,\epsilon_{4}\alpha_{1},\textrm{\,
and\, }\epsilon_{4}\beta_{1}$

The quiver $Q_{T(A)}$ is the following.

${8}$${1}$${6}$${9}$${2}$${7}$${5}$${10}$${6}$${3}$${8}$${7}$${4}$${9}$$\scriptstyle{\delta_{7}}$$\scriptstyle{\delta_{8}}$$\scriptstyle{\epsilon_{1}}$$\scriptstyle{\alpha_{2}}$$\scriptstyle{\delta_{9}}$$\scriptstyle{\delta_{10}}$$\scriptstyle{\epsilon_{2}}$$\scriptstyle{\beta_{2}}$$\scriptstyle{\alpha_{1}}$$\scriptstyle{\beta_{1}}$$\scriptstyle{\gamma_{1}}$$\scriptstyle{\eta_{1}}$$\scriptstyle{\delta_{3}}$$\scriptstyle{\delta_{2}}$$\scriptstyle{\delta_{1}}$$\scriptstyle{\delta_{4}}$$\scriptstyle{\delta_{5}}$$\scriptstyle{\delta_{6}}$$\scriptstyle{\delta_{11}}$$\scriptstyle{\delta_{13}}$$\scriptstyle{\epsilon_{3}}$$\scriptstyle{\gamma_{2}}$$\scriptstyle{\delta_{12}}$$\scriptstyle{\delta_{14}}$$\scriptstyle{\epsilon_{4}}$$\scriptstyle{\eta_{2}}$

Where the vertices labelled by $6,7,8,9$ should be identified.

There are 18 elementary cycles, which are listed below.

$\begin{array}[]{llll}C_{1}=\epsilon_{1}\alpha_{1}\alpha_{2}\delta_{1},&C_{2}=\epsilon_{1}\beta_{1}\beta_{2}\delta_{1},&C_{3}=\epsilon_{2}\beta_{1}\beta_{2}\delta_{2},&C_{4}=\epsilon_{2}\alpha_{1}\alpha_{2}\delta_{2},\\\
C_{5}=\epsilon_{3}\gamma_{1}\gamma_{2}\delta_{3},&C_{6}=\epsilon_{3}\eta_{1}\eta_{2}\delta_{3},&C_{7}=\epsilon_{4}\gamma_{1}\gamma_{2}\delta_{4},&C_{8}=\epsilon_{4}\eta_{1}\eta_{2}\delta_{4},\\\
C_{9}=\alpha_{1}\alpha_{2}\delta_{5},&C_{10}=\gamma_{1}\gamma_{2}\delta_{6},&C_{11}=\epsilon_{1}\gamma_{1}\delta_{7},&C_{12}=\epsilon_{2}\gamma_{1}\delta_{8}\\\
C_{13}=\epsilon_{1}\eta_{1}\delta_{9},&C_{14}=\epsilon_{2}\eta_{1}\delta_{10},&C_{15}=\epsilon_{3}\alpha_{1}\delta_{11},&C_{16}=\epsilon_{3}\beta_{1}\delta_{12}\\\
C_{17}=\epsilon_{4}\alpha_{1}\delta_{13},&C_{18}=\epsilon_{4}\beta_{1}\delta_{14}&&\end{array}$

Observe that the set
$\mathcal{D}=\\{\alpha_{1},\beta_{1},\delta_{3},\delta_{4},\delta_{6},\delta_{7},\delta_{8},\delta_{9},\delta_{10}\\}$
is an admissible cut of $T(A)$, however $B=T(A)/\langle\mathcal{D}\rangle$ is
not a subalgebra of $T(A)$ since $\gamma_{1}\gamma_{2}+\eta_{1}\eta_{2}=0$ is
the induced relation from the minimal relation
$\alpha_{1}\alpha_{2}+\beta_{1}\beta_{2}+\gamma_{1}\gamma_{2}+\eta_{1}\eta_{2}=0$
in $A$ and therefore $T(A)$ is not a split-by-nilpotent extension of $B$ by
[2, Theorem 2.5].

##  3\. Characterisation of trivial extensions

In this section we give a complete characterisation of the algebras that are
isomorphic to the trivial extension of some finite dimensional $K$-algebra.

Before proving the main result of this section, Theorem 3.6, we introduce the
necessary terminology and prove some preliminary results.

######  Definition 3.1.

Let $A=KQ/I$ be an algebra and let $\mathcal{E}=\\{C_{1},\dots,C_{n}\\}$ be a
set of non-zero cycles of length at least two. Then an _allowable cut
$\mathcal{D}$ of $A$ with respect to $\mathcal{E}$_ is a subset of arrows of
$Q_{A}$ containing exactly one arrow in each cycle of $\mathcal{E}$ in such a
way that if an arrow in $\mathcal{D}$ appears in a cycle $C$ of $\mathcal{E}$
then it appears only once in $C$.

######  Remark 3.2.

If $A$ is a trivial extension of an algebra $A^{\prime}$, then any admissible
cut $\mathcal{D}$ of $A$ is an allowable cut with respect the set of
elementary cycles.

In what follows, whenever the set $\mathcal{E}$ of cycles is clear from the
context, we will simply say that $\mathcal{D}$ is an allowable cut of $A$,
omitting the reference to $\mathcal{E}$.

######  Definition 3.3.

Let $A=KQ_{A}/I_{A}$ be an algebra and consider a set $\mathcal{E}$ of non-
zero cycles of length at least two and a weight function
$\omega:\mathcal{E}\to K$. Then for every vertex $x\in(Q_{A})_{0}$ we define
$I_{x}^{\mathcal{E}}$ to be the two-sided ideal in $KQ$ generated by

  1. (i)

Oriented cycles starting and ending at $x$ that do not belong to
$\mathcal{E}$.

  2. (ii)

The elements of the form $\omega(C^{\prime})C-\omega(C)C^{\prime}$, where
$C,C^{\prime}\in\mathcal{E}$.

######  Definition 3.4.

Let $A=KQ/I$ be an algebra. A finite set $\mathcal{E}$ of non-zero cycles of
length at least two is said to be a set of distinguished cycles if it admits
an allowable cut $\mathcal{D}=\\{\alpha_{1},\cdots\alpha_{r}\\}$ of $A$ and a
weight function $\omega:\mathcal{E}\to K$ such that the following holds.

  1. (i)

Every path not contained in a cycle $C$ of $\mathcal{E}$ is an element of $I$.

  2. (ii)

If $\rho$ is a linear combination of paths in $e_{x}KQ_{B}e_{y}$ such that
$q\rho\in I$ (or $\rho q\in I$) for every supplement path $q$ in a cycle
$C\in\mathcal{E}$, then $\rho$ is an element of $I$.

  3. (iii)

If $\rho$ is a linear combination of paths in $e_{x}KQ_{A}e_{y}$ such that
each summand of $\rho$ is an element of $\langle\mathcal{D}\rangle$ and $\rho
q\in I^{\mathcal{E}}_{y}$ or $q\rho\in I^{\mathcal{E}}_{x}$ for every
supplement $q$ in a cycle $C\in\mathcal{E}$, then $\rho$ is an element of $I$.

######  Lemma 3.5.

Let $A$ and $A^{\prime}$ be Artin algebras such that $T(A)\cong
T(A^{\prime})$. Then $T(A)$ is a split-by-nilpotent extension of $A^{\prime}$.

###### Proof.

Since $T(A^{\prime})$ is a split-by-nilpotent extension of $A^{\prime}$, there
exists a short exact sequence of $A^{\prime}$-$A^{\prime}$-bimodules

${0}$${DA^{\prime}}$${T(A^{\prime})}$${A^{\prime}}$${0}$$\scriptstyle{i}$$\scriptstyle{\pi}$$\scriptstyle{\iota}$

such that $\pi\circ\iota=Id_{A^{\prime}}$.

Now, let $f:T(A^{\prime})\to T(A)$ be an isomorphism of $K$-algebras. Consider
$\pi\circ f^{-1}:T(A)\to A^{\prime}$ and $f\circ\iota:A^{\prime}\to T(A)$.
Then we have a short exact sequence

${0}$${\operatorname{ker}(\pi\circ
f^{-1})}$${T(A)}$${A^{\prime}}$${0}$$\scriptstyle{\pi\circ
f^{-1}}$$\scriptstyle{f\circ\iota}$

Moreover, $(\pi\circ f^{-1})\circ(f\circ\iota)=Id_{A^{\prime}}$, then $T(A)$
is a split-by-nilpotent extension of $A^{\prime}$. ∎

######  Theorem 3.6.

Let $A=KQ/I$ be an algebra. Then $A$ is isomorphic to the trivial extension of
some finite-dimensional $K$-algebra if and only if:

  1. (a)

There is a presentation of $A$ for which there exists a set $\mathcal{E}$ of
distinguished cycles in $Q_{A}$ with weight function $\omega:\mathcal{E}\to K$
and

  2. (b)

there is an allowable cut $\mathcal{D}=\\{\gamma_{1},\cdots,\gamma_{t}\\}$ of
$A$ such that the quotient $B=A/\langle\mathcal{D}\rangle$ verifies the
following:

    1. (i)

$A$ is a split-by-nilpotent extension of $B$.

    2. (ii)

The supplements of the cut arrows in the cycles in $\mathcal{E}$ are in one-
to-one correspondence with the elements of a basis of $\mbox{soc}_{B^{e}}B$.

In this case $A$ is isomorphic to $T(B)$ and the two-sided ideal
$\langle\mathcal{D}\rangle$ is isomorphic to $DB$ as a $B$-$B$-bimodule.

###### Proof.

Suppose first that $A$ is isomorphic to $T(B)$. Then it follows from Theorem
A.1 that for any $K$-basis $\mathbb{M}$ for $\mbox{soc}_{B^{e}}B$, the set of
elementary cycles $\mathcal{E}$ induced by $\mathbb{M}$ is a set of
distinguished cycles and the set $\\{\beta_{p}:p\in\mathbb{M}\\}$ is an
allowable cut of $A$ with respect to $\mathcal{E}$. Moreover, (b)(i) and
(b)(ii) hold by construction.

To prove the converse, we show that $\langle\mathcal{D}\rangle\cong DB$ as
$B$-$B$-bimodules and $A\cong T(B)$. By Lemma 3.5 we can assume that the
presentation of $A$ is as in the statement.

Let us see first that $\langle\mathcal{D}\rangle\subset DB$. Then by (b)(i)
$A$ is a split-by-nilpotent extension of $B$ by $\langle\mathcal{D}\rangle$
and thus $A=B\ltimes\langle\mathcal{D}\rangle\subset B\ltimes DB=T(B)$. Now in
order to show $\langle\mathcal{D}\rangle\subset DB$, consider a distinguished
cycle $C\in\mathcal{E}$. Up to cyclic permutation, we can write $C=\gamma q$
where $\gamma\in\mathcal{D}$ is an arrow from $x$ to $y$ and $q$ is a path
from $y$ to $x$. By hypothesis, the path $q$ is in one-to-one correspondence
with an element $p$ of $\mbox{soc}_{B^{e}}B$ from $y$ to $x$. Then, by
(b)(ii), $\gamma=p^{*}\in DB$ and $\langle\mathcal{D}\rangle\subset DB$.

To finish the proof it is enough to show that
$DB\subset\langle\mathcal{D}\rangle$. Let $p$ be an element in the basis of
$\mbox{soc}_{B^{e}}B$. Then $p$ induces at least one elementary cycle $C$ of
$T(B)$. Let $q_{p}$ be a path from $x$ to $y$ such that $\beta_{p}q_{p}=C$,
thus $q_{p}$ is a path in $B$. Since $q_{p}$ is a subpath of an elementary
cycle in $T(B)$, $q_{p}$ is non-zero in $T(B)$ and thus non-zero in $B$.
Hence, $q_{p}$ is non-zero in $A$ and there exists a path $q^{\prime}$ such
that $q_{p}q^{\prime}=C^{\prime}$ is a distinguished cycle of $A$.

We claim that $q^{\prime}$ is in fact an arrow. From the fact that $q_{p}$
belongs to $B$, we have that $q^{\prime}$ contains a unique arrow
$\gamma\in\mathcal{D}$ which appears only once in $q^{\prime}$, i.e.,
$q^{\prime}=u\gamma v$ where $u$ and $v$ are paths in $B$. If $q_{p}$ is a
maximal non-zero path in $B$ then $q_{p}$ is an element of the
$\mbox{soc}_{B^{e}}B$, and we get that $u$ and $v$ are idempotents, therefore
$q^{\prime}=\gamma$. It follows that $q_{p}$ is a supplement of $\gamma$ in
$C^{\prime}$ and then we can identify $p^{*}$ with $\gamma$.

Otherwise, $q_{p}$ is in one-to-one correspondence with an element $p$ in the
$\mbox{soc}_{B^{e}}B$ of the form $p=a_{p}q_{p}+\sum_{i=1}^{t}a_{i}q_{i}$ in
$\mbox{soc}_{B^{e}}B$. Suppose that there are two supplements $\alpha$ and $r$
of $q_{p}$, where $\alpha$ is an arrow. Then, by Definition 3.4, $\alpha-r$ is
in $I_{A}$, which contradicts the assumption that $I_{A}$ is an admissible
ideal. Hence, if there is a supplement $\alpha$ of $q_{p}$ which is an arrow,
then it is unique and it can be identified with $p^{*}$.

Now suppose that any supplement $r$ of $q_{p}$ in $A$ is of the form
$r=u\alpha v$ where $\alpha\in\mathcal{D}$ and at least one of $u$ or $v$ is a
path of positive length. Moreover, at least one of the arrows of every $r$ is
an arrow of $B$. Since $A$ is a split-by-nilpotent extension of $B$, we have
in particular that $B$ is a subalgebra of $A$. Moreover $p$ is in
$\mbox{soc}_{B^{e}}B$, if the length of $u$ is greater than zero, $p(u\gamma
v)=(pu)(\gamma v)=0(\gamma v)=0$ or, similarly, if the length of $v$ is
greater than zero $rp=0$. Hence $rp=0$ or $pr=0$ for every supplement $r$ of
$q_{p}$. By Definition 3.4 we get that $p\in I_{A}$. This, together with the
fact that $p$ is a linear combination of paths in $B$ implies that $p\in
I_{B}$. But this is a contradiction of our hypothesis that $p$ is an element
of a basis of $\mbox{soc}_{B^{e}}B$. This contradiction comes from the
supposition that $r$ is not an arrow. Thus we can conclude that $DB$ is
contained in $\langle\mathcal{D}\rangle$.

From the above we conclude that $T(B)$ and $A$ have the same quiver and that
the elementary cycles of $T(B)$ coincide with the distinguished cycles of $A$
with the same weight function. Thus, it follows from Definition 3.4 and
Theorem A.1 that $A$ is isomorphic to $T(B)$ and $\langle\mathcal{D}\rangle$
is isomorphic to $D(B)$ as $B$-$B$-bimodules. In particular $A$ is a symmetric
algebra. ∎

In Theorem 3.6 we have characterised algebras that are isomorphic to a trivial
extension in terms of allowable cuts. We now use this result to study when two
algebras have isomorphic trivial extensions in terms of admissible cuts.

######  Corollary 3.7.

Let $A=KQ_{A}/I_{A}$ be a finite-dimensional algebra with trivial extension
$T(A)$. The following are equivalent

  1. (a)

$T(A)\cong T(A^{\prime})$

  2. (b)

There exists a presentation of $T(A)$, and an admissible cut $\mathcal{D}$ of
that presentation such that

    1. (i)

$A^{\prime}\cong T(A)/\langle\mathcal{D}\rangle$,

    2. (ii)

$T(A)$ is a split-by-nilpotent extension of $A^{\prime}$,

    3. (iii)

the supplements of each arrow of $\mathcal{D}$ in the elementary cycles of
$T(A)$ are in one-to-one correspondence with the elements of
$\mbox{soc}_{A^{\prime e}}A^{\prime}$.

###### Proof.

Suppose that $T(A)\cong T(A^{\prime})$. Then the quiver of $T(A)$ is equal to
the quiver of $T(A^{\prime})$. Hence $T(A^{\prime})$ is a different
presentation of $T(A)$. Then the result follows from Theorem A.1. For the
converse it suffices to apply Theorem 3.6 to the algebra $T(A)$. ∎

In the following example, we illustrate an algorithm to decide whether an
algebra is a trivial extension.

######  Example 3.8.

Consider the algebra $A$ given by the quiver $Q_{A}$

${2}$${1}$${3}$$\scriptstyle{\alpha_{4}}$$\scriptstyle{\alpha_{1}}$$\scriptstyle{\alpha_{2}}$$\scriptstyle{\alpha_{3}}$$\scriptstyle{\alpha_{5}}$

with ideal of relations

$I_{A}=\langle\alpha_{1}^{2},\alpha_{4}\alpha_{2},\alpha_{5}\alpha_{3},\alpha_{5}\alpha_{2},\alpha_{4}\alpha_{3},\alpha_{4}\alpha_{1}\alpha_{3},\alpha_{5}\alpha_{1}\alpha_{2},\alpha_{2}\alpha_{4}-\alpha_{3}\alpha_{5}\rangle.$

We claim that the cycles $C_{1}=\alpha_{1}\alpha_{2}\alpha_{4}$ and
$C_{2}=\alpha_{1}\alpha_{3}\alpha_{5}$ and all its permutations are the
distinguished cycles of $Q_{A}$ with weights $w(C_{1})=w(C_{2})=1$. In order
to show that the claim holds we need to verify that $I_{A}$ is determined by
$C_{1}$, $C_{2}$ and its cyclic permutations, as indicated in Definition 3.4.
Indeed, all the monomial relations correspond to paths that are not subpaths
of a distinguished cycle and the relation
$\alpha_{2}\alpha_{4}-\alpha_{3}\alpha_{5}$ is obtained from Definition
3.4.(iii). Moreover the set $\mathcal{D}=\\{\alpha_{4},\alpha_{5}\\}$ form an
allowable cut of $Q_{A}$ with respect to this set of cycles. Finally, the
algebra $B=A/\langle\alpha_{4},\alpha_{5}\rangle$ is isomorphic to the path
algebra of the quiver $Q_{B}$

${2}$${1}$${3}$$\scriptstyle{\alpha_{1}}$$\scriptstyle{\alpha_{2}}$$\scriptstyle{\alpha_{3}}$

modulo $I_{B}=\langle\alpha^{2}_{1}\rangle$. A basis of the two-sided socle
soc$B_{B^{e}}$ of $B$ is given by the paths $\alpha_{1}\alpha_{2}$ and
$\alpha_{1}\alpha_{3}$. Hence there is a correspondence between a basis of
soc$B_{B^{e}}$ and the set $\mathcal{D}$. Finally, it follows from [2, Theorem
2.5] that $A$ is a split-by-nilpotent extension of $B$. Then, we can apply
Theorem 3.6 to conclude that $A$ is the trivial extension of $B$.

##  4\. Wakamatsu’s Theorem for bound path algebras

One of the motivating questions for this paper was to determine the explicit
relationship between two finite dimensional algebras that have isomorphic
trivial extensions. This has been abstractly described by Wakamatsu in [20],
where he also gives necessary and sufficient conditions to decide when two
trivial extensions of Artin algebras are isomorphic. In this section, we give
an explicit description of Wakamatu’s result in terms of quivers and relations
by providing an independent proof.

######  Theorem 4.1.

Let $A=KQ_{A}/I_{A}$ be a finite-dimensional algebra with trivial extension
$T(A)$ and set $(Q_{A})_{1}=\\{\alpha_{1},\ldots,\alpha_{n}\\}$ and
$\\{\beta_{1},\dots,\beta_{t}\\}=(Q_{T(A)})_{1}\setminus(Q_{A})_{1}$. The
following are equivalent

  1. (a)

$T(A)\cong T(A^{\prime})$.

  2. (b)

There exists an admisible cut $\mathcal{D}$ of the form

$\mathcal{D}\;=\;\\{\alpha_{1},\ldots,\alpha_{r},\beta_{1},\ldots,\beta_{s}:\alpha_{i}\in(Q_{A})_{1}\text{
and }\beta_{j}\not\in(Q_{A})_{1}\\}$

with $A^{\prime}\cong T(A)/\langle\mathcal{D}\rangle$, $T(A)$ is a split-by-
nilpotent extension of $A^{\prime}$ and the supplements in the elementary
cycles of the cut arrows are in one-to-one correspondence with the elements of
$\mbox{soc}_{A^{\prime e}}A^{\prime}$.

  3. (c)

$A\cong S\ltimes M$ and $A^{\prime}=S\ltimes D(M)$, where

    1. (i)

$S$ is the subalgebra of $T(A)$ generated by $\sum_{x\in(Q_{T_{A}})_{0}}e_{x}$
the identity of $T(A)$ and $\alpha_{r+1},\ldots,\alpha_{n}$,

    2. (ii)

$M=S\left\langle\alpha_{1},\ldots,\alpha_{r}\right\rangle S$,

    3. (iii)

and $D(M)=S\left\langle\beta_{s+1},\ldots,\beta_{t}\right\rangle S$.

###### Proof.

By Corollary 3.7 $(a)$ and $(b)$ are equivalent.

We now show that $(b)$ implies $(c)$. Suppose that $A$ and $A^{\prime}$ are as
in $(b)$ and let $S$ be the subalgebra of $T(A)$, and $M$ and $D(M)$ be the
$S$-$S$-bimodules as defined in $(c)$.

The first step is to show that $A\cong S\ltimes M$. Let $N\subset T(A)$ be the
$S$-$S$-bimodule generated by $\\{\beta_{s+1},\ldots,\beta_{t}\\}$, hence
$N^{2}=0$. By construction $M\subset\langle\mathcal{D}\rangle$ and therefore
we also have that $M^{2}=0$. We claim that $S\cap(M+N)=0$ in $T(A)$. Suppose
that we have $0\neq s\in S\cap(M+N)$. Then $s=\sum a_{i}u_{i}=\sum b_{j}v_{j}$
where each $u_{i}$ is a path from $x$ to $y$ in $S$ and each $v_{j}$ is a path
from $x$ to $y$ containing at least an arrow in
$\\{\alpha_{1},\ldots,\alpha_{r}\\}$ or in
$\\{\beta_{s+1},\ldots,\beta_{t}\\}$. If $s$ is a non-zero element of $T(A)$,
at least one of the $u_{i}$ is non-zero, lets call it $u_{k}$. By Theorem A.1
there exists a path $q_{k}\in T(A)$ from $y$ to $x$ such that $q_{k}u_{k}$ is
an elementary cycle in $T(A)$, that is, $q_{k}$ is a supplement of $u_{k}$.
Recall that each elementary cycle of $T(A)$ contains exactly one arrow of each
admissible cut. Hence the elementary cycle $q_{k}u_{k}$ contains exactly one
arrow in $\mathcal{D}$ and one arrow in $\\{\beta_{1},\ldots,\beta_{t}\\}$.
Since $u_{k}$ is in $S$, it contains no arrow in
$\mathcal{D}\cup\\{\beta_{1},\ldots,\beta_{t}\\}$. Thus $q_{k}$ contains
exactly one arrow in $\mathcal{D}$ and one arrow in
$\\{\beta_{1},\ldots,\beta_{t}\\}$. Furthermore $q_{k}s$ is non-zero and thus
there exists $j$ such that $q_{k}v_{j}$ is non-zero. By definition $v_{j}$ has
at least one arrow of $\mathcal{D}$ or $\\{\beta_{1},\ldots,\beta_{t}\\}$.
This implies that $q_{k}v_{j}$ has two arrows in the same admissible cut. This
implies that $q_{k}v_{j}$ is not a subpath of an elementary cycle and thus
$q_{k}v_{j}=0$ by Theorem A.1 and we arrive at a contradiction. Thus
$S\cap(M+N)=0$. In particular, we have that $S\cap M=0$ and $S\cap N=0$.

Recall that $S$ is a subalgebra of $T(A)$ generated by $e_{x}$ and
$\alpha_{r+1},\ldots,\alpha_{n}$. By Theorem A.1 the relations of $T(A)$ only
depend on the relations of $A$ and the elementary cycles in $T(A)$. Thus $S$
can be seen as a subalgebra of $A$ generated by $e_{x}$ and
$\alpha_{r+1},\ldots,\alpha_{n}$. Now let $a\in A$, then $a=u+v$ where $u\in
S$ and $v\in M$. Then for $m\in M$, $am=um+vm=um$ since $vm\in M^{2}=0$. By
symmetry $M$ is a two-sided ideal of $A$. Since $S\cap M=0$, the map
$h:S\ltimes M\to A$ defined by $h(s,m)=s+m$ is an isomorphism and $S\ltimes
M\cong A$.

Now, we prove that $A^{\prime}=S\ltimes DM$. Since $N$ is a $S$-$S$-bimodule
such that $N^{2}=0$ we get that $\overline{A^{\prime}}:=S+N$ is a subalgebra
of $T(A)$. We have that $\overline{A^{\prime}}$ is isomorphic to $S\ltimes N$
using a similar argument that we used to show that $A\cong S\ltimes M$.

We now show that $\overline{A^{\prime}}\cong A^{\prime}$. Consider the
canonical map $\pi:T(A)\to A^{\prime}$ restricted to $\overline{A^{\prime}}$.
Since $\pi|_{\overline{A^{\prime}}}$ surjects the generators of
$\overline{A^{\prime}}$ onto the generators of $A^{\prime}$ and the relations
of $A^{\prime}$ are induced by the relations in $T(A)$, we have that
$\pi|_{\overline{A^{\prime}}}$ is a surjective algebra morphism.

To show that $\pi|_{\overline{A^{\prime}}}$ is an isomorphism, we need to show
that the kernel
$\ker(\pi|_{\overline{A^{\prime}}})=\langle\mathcal{D}\rangle\cap\overline{A^{\prime}}=\langle\mathcal{D}\rangle\cap(S+N)=0$.
Let $0\neq\rho\in\langle\mathcal{D}\rangle\cap(S+N)$. Then $\rho$ is in
$\langle\mathcal{D}\rangle$ and there are $s\in S$ and $n\in N$ such that
$\rho=s+n$, where $s=\sum a_{k}u_{k}$ and $n=\sum b_{k}v_{k}$. Assume $\rho$
is a linear combination of paths from $x$ to $y$. Note that
$\langle\mathcal{D}\rangle\cap N=0$ since $\langle\mathcal{D}\rangle$ is
generated by $\\{\beta_{1},\dots,\beta_{s}\\}$, $N^{\prime}$ is the ideal
generated by $\\{\beta_{s+1},\dots,\beta_{t}\\}$ and
$\\{\beta_{1},\dots,\beta_{t}\\}$ is an admissible cut of $T(A)$. If $s$ is
non-zero, there exists a $k$ such that $u_{k}$ is a non-zero path in $T(A)$.
Then there exists a path $q_{k}$ from $y$ to $x$ such that $q_{k}u_{k}$ is an
elementary cycle. Thus $q_{k}u_{k}$ contains exactly one arrow of the
admissible cut $\mathcal{D}$ and one arrow of
$\\{\beta_{1},\dots,\beta_{t}\\}$. Since $u_{k}$ is a path in $S$ we have that
$q_{k}$ must contain one arrow of the admissible cut $\mathcal{D}$ and one
arrow of $\\{\beta_{1},\dots,\beta_{t}\\}$. This implies that $q_{k}v_{k}=0$
for all $k$ because $N$ is generated by $\\{\beta_{s+1},\dots,\beta_{t}\\}$.
Thus that $q_{k}\rho=q_{k}s\neq 0$. On the other hand,
$\rho\in\langle\mathcal{D}\rangle\cap(S+N)$. Then we can write $\rho=\sum
c_{j}w_{j}\in\langle\mathcal{D}\rangle=\langle\alpha_{1},\dots,\alpha_{r},\beta_{1},\dots,\beta_{t}\rangle$.
So $w_{j}$ contains an arrow of $\mathcal{D}$, for all $j$. Since we have seen
that $q_{k}$ also contains an arrow of $\mathcal{D}$ we have that
$q_{k}w_{j}=0$, for all $j$. Hence $q_{k}\rho=0$. A contradiction that arise
from the assumption that $\rho\neq 0$. Then
$(S+N)\cap\langle\mathcal{D}\rangle=0$, as claimed.

To finish, we need to show that $D(M)\cong N$. We recall that
$N=\langle\beta_{s+1},\dots,\beta_{t}\rangle$ where $\beta_{k}=(0,p_{k}^{*})$
and $p_{k}^{*}:A\to K$ is an element of the basis of $D(soc_{A^{e}}(A))$ which
we extend to a basis of $D(A)$. Thus $N\subset D(A)$. We define $f:N\to D(M)$
where $f(w):M\to K$ is given by $f(w)(m)=w(m)$ for all $w\in N$ and $m\in M$.

We show that $f:N\to D(M)$ is injective. Let $0\neq w\in N$ such that
$f(w)(m)=0$ for all $m\in M$. Without loss of generality, $w=\sum b_{k}v_{k}$,
where $v_{k}$ is a path from the vertex $x$ to $y$. If $w$ is not zero, there
exists a $k$ such that $v_{k}$ is not zero in $T(A)$. Then there exists a path
$q_{k}$ from $y$ to $x$ such that $q_{k}v_{k}$ is an elementary cycle. By
construction, $v_{k}$ contains exactly one arrow
$\\{\beta_{s+1},\dots,\beta_{t}\\}$. Hence $q_{k}$ does not contain any arrow
in $\\{\beta_{1},\dots,\beta_{t}\\}$, and $q_{k}$ is an element of $A$. On the
other hand, the elementary cycle $q_{k}v_{k}$ contains exactly one arrow
$\alpha$ in $\mathcal{D}$ and $v_{k}$ does not, thus $q_{k}$ contains exactly
one arrow $\alpha$ of $\mathcal{D}$. This implies that $q_{k}\in M$. Then
$f(w)(q_{k})=w(q_{k})\neq 0$, which is a contradiction. So $f$ is injective.

Since $(b)$ implies $(a)$ by Corollary 3.7, $T(A)\cong T(A^{\prime})$. Then
$\dim_{k}T(A)=\dim_{k}T(A^{\prime})$ and this implies that
$\dim_{k}T(A)=2\dim_{k}A$. In particular $\dim_{k}A=\dim_{k}A^{\prime}$. From
$A=S\ltimes M$ and $A^{\prime}=S\ltimes N^{\prime}$ it follows that
$\dim_{k}DM=\dim_{k}M=\dim_{k}N$. Then $f$ is surjective and $N$ is isomorphic
to $DM$. We conclude that $A^{\prime}\cong S\ltimes DM$.

Now we prove that $(c)$ implies $(a)$. Since $A=S\ltimes M$ and
$A^{\prime}=S\ltimes DM$ we have that $T(A)=A\ltimes DA=S\ltimes
M\ltimes(DS\ltimes DM)$ and $T(A)=A^{\prime}\ltimes DA^{\prime}=S\ltimes
DM\ltimes(DS\ltimes M)$. It is not difficult to show that $T(A)\cong
T(A^{\prime}).$ ∎

##  Appendix A The ideal of relations of the trivial extension of an algebra
by Elsa Fernandez

In this appendix we give an explicit description of the ideal of relations
$I_{T(A)}$ of the trivial extension $T(A)=A\ltimes D(A)$ for any finite-
dimensional algebra $A=KQ/I$.

For completeness, we first restate the explicit description of the ordinary
quiver of $T(A)$ based on the quiver of $A$, as given in Section 2. This
description depends on a $K$-basis $\mathbb{M}=\\{p_{1},\dots,p_{n}\\}$ of
$\mbox{soc}_{A^{e}}A$. The set of vertices $(Q_{A})_{0}$ and $(Q_{T(A)})_{0}$
coincide. The set of arrows $(Q_{T(A)})_{1}$ is the disjoint union of
$(Q_{A})_{1}$ and $\\{\beta_{p_{1}},\dots,\beta_{p_{n}}\\}$ where
$s(\beta_{p_{i}})=t(p_{i})$ and $t(\beta_{p_{i}})=s(p_{i})$ for every
$p_{i}\in\mathbb{M}$.

Before giving an explicit description of the ideal of relations $I_{T(A)}$ of
$T(A)$, we recall the following definition from [7]. For each vertex $x\in
Q_{T(A)}$ we define the two-sided ideal $I_{x}^{\prime}$ in $KQ_{T(A)}$
generated by

  1. (i)

oriented cycles from $x$ to $x$ which are not contained in an elementary
cycle;

  2. (ii)

the elements of the form $\omega(C^{\prime})C-\omega(C)C^{\prime}$, where $C$
and $C^{\prime}$ are elementary cycles starting and ending at $x$.

######  Theorem A.1.

Let $A=KQ_{A}/I_{A}$ be a finite-dimensional algebra and let
$T(A)=KQ_{T(A)}/I_{T(A)}$ be its trivial extension. Then the quiver $Q_{T(A)}$
is as above and the ideal $I_{T(A)}$ is generated by the union of the
following sets.

  1. (1)

A generating set of the ideal of relations $I_{A}$ of $A$.

  2. (2)

The paths that are not contained in an elementary cycle.

  3. (3)

For any vertices $x$ and $y$ in $Q_{T(A)}$, the linear combinations of paths
$\rho\in e_{x}KQ_{T(A)}e_{y}$ such that $q\rho\in I_{x}^{\prime}$ or $\rho
q\in I_{y}^{\prime}$ for any supplement path $q$ in an elementary cycle $C$.

In order to prove our result we need to recall and show some preliminary
results. We start with the following remark.

######  Remark A.2.

  1. (i)

Note that it follows directly from Definition 2.1 that each non-zero path $q$
in $KQ_{A}$ can be completed to an elementary cycle in $KQ_{T(A)}$. In other
words, each non-zero path in $KQ_{A}$ is contained in an elementary cycle $C$.

  2. (ii)

In Item $3.$ of Theorem A.1, if we set $x=y$, then $\rho$ is a relation
generated by the second condition in the definition of $I^{\prime}_{x}$ above.
Moreover, every relation of the form $\omega(C^{\prime})C-\omega(C)C^{\prime}$
arises in this way by taking $q=e_{x}$ the trivial path at $x$.

  3. (iii)

Also note that it follows from Item $2.$ of Theorem A.1 that if
$C=\alpha_{1}\dots\alpha_{n}$ is an elementary cycle in $T(A)$, then the paths
$C\alpha_{1}$ and $\alpha_{n}C$ are zero in $T(A)$.

Consider the morphism of $K$-algebras $\Phi:KQ_{T(A)}\to T(A)$ defined on the
stationary paths and arrows as follows.

$\Phi(e_{x})=(e_{x},0)$, |  | for all $x\in(Q_{A})_{0}$
---|---|---
$\Phi(\alpha)=(\alpha,0)$, |  |  for all $\alpha\in(Q_{A})_{1}$,
$\Phi(\beta_{p})=(0,p^{*})$, |  |  and $p\in\mathbb{M}$.

It follows immediately from the definition of $\Phi$ that it is a surjective
morphism of algebras and $\ker\Phi\cong I_{T(A)}$. Associated with $\Phi$ we
consider two morphisms

$\phi_{1}=\pi_{1}\Phi:KQ_{T(A)}\to
A\qquad\text{and}\qquad\phi_{2}=\pi_{2}\Phi:KQ_{T(A)}\to D(A)$

where $\pi_{1}$ and $\pi_{2}$ are the natural projections induced by the
decomposition of $T(A)=A\oplus D(A)$ as $K$-vector spaces. We denote by
$I_{\mathbb{M}}=\langle\beta_{p}\rangle_{p\in\mathbb{M}}$ the ideal of $T(A)$
generated by the elements $\beta_{p}$ in $T(A)$. The following lemma was shown
in [7, Lemma 3.5] provided that each cycle in $A$ is zero, but the same proof
holds for any algebra $A=KQ/I$.

######  Lemma A.3.

Let $A=KQ_{A}/I_{A}$ be an algebra and let $q,u$ be paths in $KQ_{T(A)}$.

  1. (a)

If $v\in KQ_{T(A)}$ is such that $v=v_{1}+v_{2}$ with $v_{1}\in KQ_{A}$ and
$v_{2}\in I_{\mathbb{M}}$, then $\Phi(v)=(\phi_{1}(v_{1}),\phi_{2}(v_{2}))$.

  2. (b)

$\phi_{2}(q)\neq 0$ implies $q\in I_{\mathbb{M}}$.

  3. (c)

$\phi_{2}(q)=0$ if $q$ contains at least two arrows $\beta_{p}$,
$p\in\mathbb{M}$.

  4. (d)

$\phi_{2}(q)(u)\neq 0$ implies that $u$ is a supplement of $q$.

  5. (e)

$\phi_{2}(v)(u)=\phi_{2}(vu)(e_{y})=\phi(uv)(e_{x})$ if $u$ is a path from $x$
to $y$ in $Q_{A}$.

  6. (f)

If $v=\sum a_{i}q_{i}$, with $q_{i}$ different paths and $\phi_{2}(v)\neq 0$,
then there exists a supplement $u$ of the $q_{i}$ such that $\phi_{2}(uv)\neq
0$ and $\phi_{2}(vu)\neq 0$.

  7. (g)

Let $C$ be an elementary cycle with origin $e$. Then
$\phi_{2}(C)(e)=\omega(C)$ and $\phi_{2}(C)(u)=0$ for any path $u\neq e$.

  8. (h)

If $q$ has a supplement, then $\Phi(q)\neq 0$.

As a consequence of Lemma A.3(g) we have that any elementary cycle is nonzero
in $T(A)$. It was proven in [7, Corollary 2.8] that if any oriented cycle in
$Q_{A}$ is zero in A, then any nonzero cycle in $T(A)$ is an elementary cycle.
We illustrate with the following example that the previous statement is not
true in general, see Example 2.6.

The following result was shown in [7, Proposition 3.6] under the assumption
that $A$ is monomial. However, the proof directly generalises to every finite-
dimensional algebra $A$.

######  Proposition A.4.

Let $A$ be a finite-dimensional algebra and $\Phi$ the $K$-algebra morphism
defined as above. Then $I_{x}^{\prime}$ is a subset of
$\ker\Phi\cap(e_{x}KQ_{T(A)}e_{x}$) for any $x\in(Q_{A})_{0}$.

######  Remark A.5.

As a direct consequence of the preceding proposition we have that the classes
of elementary cycles starting at a vertex $x$ of $KQ_{T(A)}$ generate a one
dimensional subspace of $KQ_{T(A)}/I^{\prime}_{x}$.

We now show Theorem A.1

###### Proof of Theorem A.1.

Let $I^{\prime}$ be the ideal generated by 1.-3. in the statement. It is
sufficient to prove that $I^{\prime}=\ker\Phi$.

First we show that $I^{\prime}\subset\ker\Phi$. It is clear that
$I_{A}\subset\ker\Phi$. Now, Remark A.2(i) and Lemma A.3(d) implies that
$\Phi(u)=0$ for every path $u$ which is not contained in an elementary cycle,
since if $u$ is contained in an elementary cycle it is non-zero on its
supplement.

Assume now that $w=\sum_{k=1}^{l}a_{k}v_{k}\not\in\ker\Phi$, where $v_{k}$ are
different paths in $I_{\mathbb{M}}$ from $x$ to $y$. Then
$\Phi(w)=(0,\phi_{2}(w))$. We know that $I^{\prime}_{x}\subset\ker\Phi$ by
Proposition A.4 for all $x\in(Q_{T(A)})_{0}$. Then Lemma A.3.(f) states that
if $\phi_{2}(w)\neq 0$ then there are no supplements $\rho_{k}$ of $v_{k}$
such that $\sum_{k=1}^{l}a_{k}\rho_{k}v_{k}\in I^{\prime}_{x}$. Then $w$ is
not in $I^{\prime}$. This shows that $I^{\prime}\subset\ker\Phi$.

Since $\Phi:KQ_{T(A)}\to T(A)$ is surjective and $I^{\prime}\subset\ker\Phi$,
it is enough to show that
$\dim_{K}(KQ_{T(A)}/I^{\prime})=\dim_{K}T(A)=2\dim_{K}A$. Note that the
inclusion morphism $\sigma:A\to T(A)$ factors through $KQ_{T(A)}/I^{\prime}$
because $I_{A}\subset I^{\prime}$, which implies that $\iota:A\to
KQ_{T(A)}/I^{\prime}$ is an algebra monomorphism.

We have that $KQ_{T(A)}=KQ_{A}+I_{\mathbb{M}}$. Then
$e_{x}KQ_{T(A)}e_{y}=e_{x}KQ_{A}e_{y}+e_{x}I_{\mathbb{M}}e_{y}$ for each
$x,y\in(Q_{T(A)})_{0}$. Let $\pi:KQ_{T(A)}\to KQ_{T(A)}/I^{\prime}$ be the
canonical epimorphism. We define the subspaces
$\mathscr{P}_{xy}=\iota(e_{x}Ae_{y})$ and
$\mathcal{F}_{xy}=\pi(e_{x}I_{\mathbb{M}}e_{y})$. Note that
$\sum_{xy}\dim_{K}\mathscr{P}_{xy}=\dim_{K}A$.

We start by showing that $\mathscr{P}_{xy}\neq 0$ if and only if
$\mathcal{F}_{yx}\neq 0$. Indeed, if $\mathcal{F}_{yx}\neq 0$ then there
exists a non-zero path $q\in\mathcal{F}_{yx}$ which admits a supplement $r$ in
an elementary cycle $C$ by Remark A.2(i). Moreover, $r\in KQ_{A}$ since $q$
contains an arrow $\beta\in I_{\mathbb{M}}$ and $C$ is an elementary cycle.
Hence $0\neq r\in\mathscr{P}_{xy}$ and $\mathscr{P}_{xy}\neq 0$.

Conversely, if $\mathscr{P}_{xy}\neq 0$ there exists a non-zero path
$q\in\mathscr{P}_{xy}$ which is contained in an elementary cycle $C$ with
supplement $r$. Moreover, we know that there exists a unique arrow $\beta\in
I_{\mathbb{M}}$ which is contained in the cycle $C$. Since $C=rq$ and
$q\in\mathscr{P}_{xy}$, we have that $r$ contains $\beta$. Thus
$r\in\mathcal{F}_{yx}$ and $\mathcal{F}_{yx}$ is non-zero.

We now prove that $\dim_{K}\mathscr{P}_{xy}\geq\dim_{K}\mathcal{F}_{yx}$.
Suppose to the contrary that
$n:=\dim_{K}\mathscr{P}_{xy}<\dim_{K}\mathcal{F}_{yx}$. Then there is a set of
linearly independent paths $\\{\mu_{1},\dots,\mu_{n+1}\\}$ in
$\mathcal{F}_{yx}$. Furthermore, $\mu_{t}$ does not belong to $I^{\prime}$ for
all $1\leq t\leq n+1$. This implies that $\mu_{t}$ is included in an
elementary cycle $C_{t}$ and admits a supplement $\gamma_{t}$ for all $1\leq
t\leq n+1$. Note that, $0\neq\gamma_{t}\in\mathscr{P}_{xy}$ for all $t$. Since
$n=\dim_{K}\mathscr{P}_{xy}$, there exists $1\leq s\leq n+1$ such that
$\gamma_{s}=\sum_{i=1}^{s-1}a_{i}\gamma_{i}$, where
$\\{\gamma_{1},\dots,\gamma_{s-1}\\}$ is a linearly independent set in
$\mathscr{P}_{xy}$. Then there exists $1\leq r\leq s-1$ such that
$\mu_{s}\gamma_{r}\neq 0$ since $0\neq
a_{s}\mu_{s}\gamma_{s}=\sum_{i=1}^{s-1}a_{i}\mu_{s}\gamma_{i}$.

Now, since $\mu_{s}\gamma_{r}$ is a non-zero path going from the vertex $x$ to
itself we have the existence of a cycle $\rho$ from $x$ to $x$ in $KQ_{A}$
such that $\rho\mu_{s}\gamma_{r}$ is an elementary cycle in $KQ_{A}$. By
hypothesis $A$ is a finite-dimensional algebra, implying the existence of
$m\in\mathbb{N}$ such that $\rho^{m-1}\neq 0$ and $\rho^{m}=0$. Moreover we
have that $\\{\rho^{m-1}\mu_{1},\dots,\rho^{m-1}\mu_{n+1}\\}$ is a linearly
independent set in $\mathcal{F}_{yx}$. We know from Remark A.5 that the set of
elementary cycles from $x$ to $x$ generate a subspace of dimension one in
$KQ_{T(A)}/I^{\prime}_{i}$. Thus, there exists a non-zero $b\in K$ such that
$\rho\mu_{s}\gamma_{r}=b\mu_{r}\gamma_{r}$ and then the element
$\rho\mu_{s}-b\mu_{r}$ belongs to class (3) of $I^{\prime}$. Therefore,
$\rho\mu_{s}-b\mu_{r}=0$. As a consequence $\mu_{r}=b^{-1}\rho\mu_{s}$. Then
$0\neq\rho^{m-1}\mu_{r}=b^{-1}\rho^{m}\mu_{s}=0$, a contradiction. So
$\dim_{K}\mathscr{P}_{xy}\geq\dim_{K}\mathcal{F}_{yx}$ as claimed. From this
we can conclude that $\dim_{K}T(A)\geq\dim_{K}KQ_{TA}/{I^{\prime}}$, since

$\displaystyle\dim_{K}T(A)$ $\displaystyle=$ $\displaystyle
2\dim_{K}A=\sum_{x,y\in
Q_{0}}\dim_{K}\mathscr{P}_{xy}+\dim_{K}\mathscr{P}_{yx}$ $\displaystyle\geq$
$\displaystyle\sum_{x,y\in
Q_{0}}\dim_{K}\mathscr{P}_{xy}+\dim_{K}\mathcal{F}_{yx}\geq\dim_{k}(KQ_{TA}/I^{\prime}).$

Now, using that $KQ_{TA}/I^{\prime}$ maps onto $T(A)$, we obtain that
$\dim_{K}(KQ_{TA}/I^{\prime})\geq\dim_{K}T(A)$. This completes the proof of
the theorem. ∎

### Acknowledgements

The authors would like to thank Ibrahim Assem for insightful comments on a
previous version of this manuscript. EF and ST have been partially supported
by a PICT of ANPCyT, Argentina. SS and HT were supported by the EPSRC through
the Early Career Fellowship, EP/P016294/1. HT was partially funded by the DFG
under Germany’s Excellence Strategy Programme – EXC-2047/1 – 390685813. HT is
also supported by the European Union’s Horizon 2020 research and innovation
programme under the Marie Sklodowska-Curie grant agreement No 893654. YV is
also supported by the European Union’s Horizon 2020 research and innovation
programme under the Marie Sklodowska-Curie grant agreement No H2020-MSCA-
IF-2018-838316. SS acknowledges partial support from the DFG through the
project SFB/TRR 191 Symplectic Structures in Geometry, Algebra and Dynamics
(Projektnummer 281071066– TRR 191). SS, HT and YV would like to thank the
Isaac Newton Institute for Mathematical Sciences, Cambridge, for support and
hospitality during the programme “Cluster Algebras and Representation Theory”
where part of the work on this paper was undertaken. This work was supported
by EPSRC grant no EP/K032208/1 and the Simons Foundation.

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