# Tame symmetric algebras of period four

Karin Erdmann  Mathematical Institute, University of Oxford, UK
<EMAIL_ADDRESS>,  Adam Hajduk  Faculty of Mathematics and Computer
Science, Nicolaus Copernicus University, Chopina 12/18, 87-100 Torun, Poland
<EMAIL_ADDRESS>and  Adam Skowyrski  Faculty of Mathematics and Computer
Science, Nicolaus Copernicus University, Chopina 12/18, 87-100 Torun, Poland
<EMAIL_ADDRESS>
###### Abstract.

In this paper we are concerned with the structure of tame symmetric algebras
$A$ of period four (TSP4 algebras, for short). We will mostly focus on the
case when the Gabriel quiver of $A$ is biserial, i.e. there are at most two
arrows ending and at most two arrows starting at each vertex, but some of the
results can be easily extended to general case. Here, we serve a basis for
upcoming series of articles devoted to solve the problem of classification of
all TSP4 algebras with biserial Gabriel quiver. We present a range of
properties (with relatively short proofs) which must hold for the Gabriel
quiver of a tame symmetric algebra of period four. Amongst others we show that
triangles (and squares) appear naturally in the Gabriel quivers of such
algebras, so as for weighted surface algebras [6, 8, 9].

###### Key words and phrases:

Symmetric algebra, tame algebra, periodic algebra, quiver

###### 2020 Mathematics Subject Classification:

Primary: 16D50, 16E30, 16G20, 16G60

##  1\. Introduction

Classical examples of tame symmetric algebras of period four are 2-blocks of
finite-dimensional group algebras with quaternion defect groups. More recently
it was discovered that all weighted surface algebras [6] (see also [8] and
[9]) are tame symmetric of period four, and so are virtual mutations
investigated in [11] or so called weighted generalized triangulation algebras
[13], which generalize both mentioned classes. The main result of [7]
established the classification of tame symmetric algebras of period four whose
Gabriel quiver is 2-regular, which gives an evidence that general
classification may be in reach after some work. A full classification in the
biserial case seems to be an exciting challenge.

This paper is a contribution towards this goal. Here we present a range of
properties with short proofs, but which will be essential input for the
general classification (work in progress).

Throughout we fix an algebraically closed field, and we consider finite-
dimensional associative $K$-algebras with identity. We also assume that
algebras are basic and connected. Recall that an algebra $A$ is self-
injective, provided that $\Lambda$ is injective as a right $\Lambda$-module,
i.e. projective modules are also injective (see also [12]). In this paper, we
focus our attention on symmetric algebras, that is these self-injective
algebras, for which there is a nondegenerate symmetric $K$-bilinear form
$\Lambda\times\Lambda\to K$. There are many classical examples of symmetric
algebras, for instance, blocks of finite-dimensional group algebras [4] or
Hecke algebras associated to Coxeter groups [1]. Any algebra $\Lambda$ is a
quotient of its trivial extension $T(\Lambda)$, which is a symmetric algebra.

For an algebra $\Lambda$ we denote by $\operatorname{mod}\Lambda$ the category
of finitely generated (right) $\Lambda$-modules. For a module $M$ in
$\operatorname{mod}\Lambda$, its syzygy is a module $\Omega(M)=ker(\pi)$,
where $\pi:P\to M$ is a projective cover of $M$ in $\operatorname{mod}\Lambda$
(so syzygy is defined up to isomorphism).

We call a module $M$ in $\operatorname{mod}\Lambda$ a periodic module if
$\Omega^{d}(M)\cong M$, for some $d\geqslant 1$ (the smallest such $d$ is the
period of $M$). Recall that an algebra $\Lambda$ is called a periodic algebra
if $\Lambda$ is periodic as an $\Lambda$-bimodule, or equivalently, $\Lambda$
is a periodic module over its enveloping algebra
$\Lambda^{e}=\Lambda\otimes_{K}\Lambda$. Periodicity of an algebra implies
periodicity of all non-projective indecomposable $A$-modules (see for example
[14, Theorem IV.11.19]). In particular, if $\Lambda$ is a periodic algebra,
then all simple $\Lambda$-modules are periodic. Moreover, it is known [10, see
Theorem 1.4] that periodicity of simples in $\operatorname{mod}\Lambda$
implies $\Lambda$ is self-injective, and hence, periodic algebras form a
subclass in the class of self-injective algebras.

Here we work with bound quiver algebras $\Lambda=KQ/I$, where the Gabriel
quiver $Q$ is biserial: that is, at each vertex at most two arrows start and
at most two arrows end. We will consider algebras $\Lambda$ which are both
symmetric and tame, and we assume that $\Lambda$ is a periodic algebra of
period four. Any such algebra is said to be a TSP4 algebra. We will give an
overview of general properties of Gabriel quivers $Q$ and minimal generators
of ideals $I$ for such algebras. A full classification by quivers and
relations requires much more efforts. In particular, we shall see that
triangles (and squares) appear naturally; see Section 4. Moreover, in the last
section, we present partial results describing some distinguished types of
vertices.

For the necessary background in the representation theory we refer to books
[2, 14].

##  2\. Preliminaries

Let $\Lambda=KQ/I$ be an admissible presentation of $\Lambda$, where the
algebra is tame and symmetric, and has $\Omega$-period $4$, as an algebra. In
particular, all simple modules are $\Omega$-periodic as $A$-modules with
period dividing $4$ [14, Theorem IV.11.19]. In fact, we can assume that all
simples have period $4$ (see Remark 2.2). We also assume $Q$ is connected,
that is $\Lambda$ is indecomposable as an algebra. For a vertex $i\in Q$, we
denote by $P_{i}$ the indecomposable projective module in
$\operatorname{mod}\Lambda$ associated to vertex $i$, and by $p_{i}$ its
dimension vector $p_{i}:=\underline{\dim}(P_{i})$. Similarly, we write $S_{i}$
and $s_{i}$, for the simple module associated to vertex $i$ and its dimension
vector.

For a vertex $i$ of the quiver $Q$, we let $i^{-}$ be the set of arrows ending
at $i$, and $i^{+}$ the set of arrows starting at $i$. In this paper, we
assume the sizes $|i^{-}|$ and $i^{+}|$ are at most $2$. With this, $Q$ is
said to be 2-regular if $|i^{-}|=|i^{+}|=2$, and biserial if
$1\leq|i^{-}|,|i^{+}|\leq 2$. We say that $i\in Q_{0}$ is a regular vertex
($1$\- or $2$-regular), provided $|i^{-}|=|i^{+}|$ (and the size is equal $1$
or $2$, respectively). Otherwise, we call $i$ a non-regular vertex.

We will use the following notation and convention for arrows: we write
$\alpha,\bar{\alpha}$ for the arrows starting at vertex $i$, with the
convention that $\bar{\alpha}$ does not exist in case $|i^{+}|=1$. Similarly
we write $\gamma,\gamma^{*}$ for the arrows ending at some vertex $i$, where
again $\gamma^{*}$ may not exist.

Then $Q$ has a subquiver

$\textstyle{x\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\gamma}$$\textstyle{y\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\gamma^{*}}$$\textstyle{i\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\alpha}$$\scriptstyle{\bar{\alpha}}$$\textstyle{j}$$\textstyle{k}$

Consider the simple module $S_{i}$, $i\in Q_{0}$. We will briefly discuss some
basic consequences of $\Omega$-periodicity of $S_{i}$, mainly, the associated
exact sequence. Recall that there are natural isomorphisms
$\Omega(S_{i})=\operatorname{rad}P_{i}=\alpha\Lambda+\bar{\alpha}\Lambda$ and
$\Omega^{-}(S_{i})\cong(\gamma,\gamma^{*})\Lambda\subset P_{x}\oplus P_{y}$.
In particular, it follows that the module $P_{i}^{+}=P_{j}\oplus P_{k}$ is a
projective cover of $\Omega(S_{i})$ and the module $P_{i}^{-}=P_{x}\oplus
P_{y}$ is an injective envelope of $\Omega^{-}(S_{i})$ ($\Lambda$ is
symmetric). Consequently, involving $\Omega$-periodicity (period $4$) of
$S_{i}$, we conclude that there is an exact sequence in
$\operatorname{mod}\Lambda$ of the form

$0\to S_{i}\to P_{i}\stackrel{{\scriptstyle
d_{3}}}{{\to}}P_{i}^{-}\stackrel{{\scriptstyle
d_{2}}}{{\to}}P_{i}^{+}\stackrel{{\scriptstyle d_{1}}}{{\to}}P_{i}\to S_{i}\to
0$ $None$

with $Im(d_{k})\cong\Omega^{k}(S_{i})$, for $k\in\\{1,2,3\\}$. By our
convention, $P_{y}$ or $P_{k}$ may not exist. Moreover, we denote by
$p_{i}^{+}$ (respectively, $p_{i}^{+}$) the dimension vector
$\underline{\dim}(P_{i}^{+})$ (respectively, $\underline{\dim}(P_{i}^{-})$).
Using the above sequence, one easily gets that $p_{i}^{+}=p_{i}^{-}$. We use
this fact (without mentioning) many times in the rest part of the paper.

Now, we will show a few examples of results obtained by using exact sequences
of the form $(*)$. As a first application note the following lemma.

######  Lemma 2.1.

If $\Lambda$ has infinite type then there is no arrow $\alpha:i\to j$ with
$i^{+}=\\{\alpha\\}=j^{-}$.

###### Proof.

Suppose there is such an arrow. Then
$\Omega(S_{i})=\alpha\Lambda\cong\Omega^{-1}(S_{j})$ and
$\Omega^{2}(S_{i})\cong S_{j}$. Therefore in the exact sequence for $S_{i}$
the projective $P_{i}^{-}$ is isomorphic to $P_{j}$ and this means that there
is a unique arrow ending at $i$ and it starts at $j$.

As well, in the exact sequence for $S_{j}$ we have $P_{j}^{+}\cong P_{i}$
since $\Omega^{2}(S_{j})\cong S_{i}$. Therefore there is a unique arrow
starting at $j$ and it ends at $i$. Now, $Q$ is connected and hence has only
two vertices and two arrows. Then $\Lambda$ is a Nakayama algebra of finite
representation type, hence a contradiction (see for example [14, Theorems
I.10.3 and 10.7]). ∎

######  Remark 2.2.

Actually, existence of arrow with the above described property implies that
$\Lambda$ is of finite type, as it is explained in the following note [5].
There is also proved this condition is equivalent to existence of a simple
with period $2$. Hence, when dealing with TSP4 algebras of infinite type, we
may assume that all simples have period exactly $4$.

We have also the following observation.

######  Lemma 2.3.

The quiver $Q$ does not have a subquiver of the form

$j\stackrel{{\scriptstyle\longrightarrow}}{{\longleftarrow}}i\longleftarrow t$

where all arrows to and from $i$ are shown.

###### Proof.

Assume this happens. Then in the exact sequence for $S_{i}$ we have
$P_{i}^{+}\cong P_{j}$ and $P_{i}^{-}\cong P_{j}\oplus P_{t}$. Since
$P_{t}\neq 0$ it follows that
$\underline{\dim}P_{i}^{+}\neq\underline{\dim}P_{i}^{-}$, a contradiction. ∎

To end this preliminary section we will give one simple lemma pertaining
vectors $p_{i}^{+}=p_{i}^{-}$, for $i\in Q_{0}$ (this common dimension vector
of two modules $P_{i}^{+}$ and $P_{i}^{-}$ will be denoted by $\hat{p}_{i}$).

It is clear from the exact sequence $(*)$ that $p_{i}$ is less or equal to
$\hat{p}_{i}+s_{i}$ (in the product order), since
$p_{i}-s_{i}=\underline{\dim}\Omega^{1}(S_{i})$ is less than
$\underline{\dim}(R^{+})=\hat{p}_{i}$. Moreover, $\hat{p}_{i}$ is greater up
to dimension, as the following shows (here we write $|x|$ for the sum
$|x|=x_{1}+\dots+x_{n}$, where $x=(x_{1},\dots,x_{n})\in\mathbb{N}^{n}$, which
corresponds to the $K$-dimension of $X$, if $x=\underline{\dim}(X)$, for a
module $X$ in $\operatorname{mod}\Lambda$).

######  Lemma 2.4.

$|\hat{p}_{i}|>|p_{i}|$.

###### Proof.

Of course, we have an exact sequence $0\to\Omega^{2}(S_{i})\to
P_{i}^{+}\to\Omega^{1}(S_{i})\to 0$, where
$\Omega^{1}(S_{i})=\operatorname{rad}P_{i}$ has dimension vector equal to
$p_{i}-s_{i}$. We claim that

$(\Box)\qquad\qquad|\hat{p}_{i}|-\dim_{K}\Omega^{1}(S_{i})>1.$

Indeed, if this is not the case, then the difference is $1$, and we conclude
that $\Omega^{2}(S_{i})\cong S_{i}$. On the other hand, $P_{i}^{+}$
(respectively, $P_{i}^{-}$) are injective envelope (respectively, projective
cover) of $\Omega^{2}(S_{i})$, so it would imply that both are isomorphic to
$P_{i}$. It means that there is a unique arrow in $Q$ starting at $i$ which
also ends at $i$, and dually, there is a unique arrow in $Q$ ending at $i$
which also starts at $i$. As a result $Q$ admits one vertex and two loops,
which is impossible, due to our assumptions. Therefore, $(\Box)$ holds. In
particular, we get
$|\hat{p}_{i}|-\dim_{K}\Omega^{1}(S_{i})=|\hat{p}_{i}|-|p_{i}|+1>1$, and hence
$|\hat{p}_{i}|-|p_{i}|>0$, so we are done. ∎

##  3\. Period 4 and minimal relations

In this section, we develop further consequences of the structure of the exact
sequence $(*)$ associated to the simple module $S_{i}$, as described in the
previous section. Actually, we will focus rather on maps and show their
connection with minimal relations defining algebra $\Lambda$.

We start with our given presentation $\Lambda=KQ/I$ and a vertex $i\in Q_{0}$.
We will briefly write $J$ for the Jacobson radical $\operatorname{rad}\Lambda$
of $\Lambda$. Consider the associated exact sequence

$0\to S_{i}\to P_{i}\stackrel{{\scriptstyle
d_{3}}}{{\to}}P_{i}^{-}\stackrel{{\scriptstyle
d_{2}}}{{\to}}P_{i}^{+}\stackrel{{\scriptstyle d_{1}}}{{\to}}P_{i}\to S_{i}\to
0$ $None$

where $P_{i}^{+}=P_{j}\oplus P_{k}$ and $P_{i}^{-}=P_{x}\oplus P_{y}$. We may
assume that $d_{1}(x,y):=\alpha x+\bar{\alpha}y$, since the induced
epimorphism $(\alpha\ \bar{\alpha}):P_{j}\oplus
P_{k}\to\Omega(S_{i})=\alpha\Lambda+\bar{\alpha}\Lambda$ is a projective cover
of $\Omega(S_{i})$ in $\operatorname{mod}\Lambda$. Adjusting arrows $\gamma$
or $\gamma^{*}$ (including impact on presentation, i.e. on generators of $I$),
we can already say that $d_{3}(e_{i})=(\gamma,\gamma^{*})$ for some choice of
the arrows $\gamma,\gamma^{*}$ ending at $i$ (see [7, Proposition 4.3]).

The kernel of $d_{1}$ is then $\Omega^{2}(S_{i})=Im(d_{2})$, and it has at
most two minimal generators. They are images of idempotents $e_{x}\in
P_{x}=e_{x}\Lambda$ and $e_{y}\in P_{y}$ via $d_{2}:P_{i}^{-}\to P_{i}^{+}$.
We may write them as $\varphi$ and $\psi$, respectively, and they are
contained in $P_{j}\oplus P_{k}$, so we can also write

$\varphi=d_{2}(e_{x},0)=(\varphi_{jx},\ \varphi_{kx})\ \ \mbox{and}\ \
\psi=d_{2}(0,e_{y})=(\psi_{jy},\ \psi_{ky}),$

where $\varphi_{jx}$ belongs to $e_{j}\Lambda e_{x}$ and similarly for the
other components of $\varphi,\psi$.

The exact sequence gives information on minimal generators of the ideal $I$,
which we sometimes refer to as minimal relations. In the sense of the
following lemma, arrows of $Q$ induce minimal relations.

######  Lemma 3.1.

If there is an arrow $x\to i$ then there is a minimal generator $\rho\in
e_{i}\Lambda e_{x}$ for the ideal $I$ (given the presentation).

###### Proof.

Consider the generators $\varphi,\psi$ of the kernel of $d_{1}$. We have
$\alpha\varphi_{jx}+\bar{\alpha}\varphi_{kx}=0$ in $\Lambda$, equivalently the
element $\alpha\varphi_{jx}+\bar{\alpha}\varphi_{kx}\in KQ$ belongs to $I$. It
is a minimal relation since $\varphi$ is a minimal generator. Similarly the
generator $\psi$ gives a minimal relation in $e_{i}\Lambda e_{y}$. ∎

Recall that any homomorphism $d:P_{x}\oplus P_{y}\to P_{j}\oplus P_{k}$ in
$\operatorname{mod}\Lambda$ can be represented in the matrix form

$M={m_{jx}\ m_{jy}\choose m_{kx}\ m_{ky}},$

where $m_{ab}$ is a homomorphism $P_{b}\to P_{a}$ in
$\operatorname{mod}\Lambda$, identified with an element $m_{ab}\in
e_{a}\Lambda e_{b}$, for any $a\in\\{j,k\\}$ and $b\in\\{x,y\\}$. In this way,
$d$ becomes multiplication by $M$, i.e. $d(u)=M\cdot u$, for $u\in P_{i}^{-}$
(using column notation for vectors in $P_{i}^{-}$ and $P_{i}^{+}$).

Continuing with the generators of $\Omega^{2}(S_{i})$, let $M_{i}$ be the
matrix with column the components of $\varphi$ and $\psi$, that is $d_{2}$ is
given by matrix

$M_{i}={\varphi_{jx}\ \psi_{jy}\choose\varphi_{kx}\ \psi_{ky}}.$

Rewriting compositions $d_{1}d_{2}=0$ and $d_{2}d_{3}=0$ in matrix form, we
get identities

$(\alpha\ \bar{\alpha})\cdot M_{i}=0\mbox{ and
}M_{i}\cdot{\gamma\choose\gamma^{*}}=0$ $None$

for some choice of arrows $\gamma,\gamma^{*}$ ending at $i$ (cf. [7,
Proposition 4.3]).

######  Remark 3.2.

Basically, identities (1) determine generators (cogenerators) of
$\Omega^{2}(S_{i})$, which are encoded in columns (rows) of matrix $M_{i}$,
satisfying the following universal properties:

  1. (i)

if $\theta={\theta_{1}\choose\theta 2}\in P_{j}\oplus P_{k}$ is an element
$\theta\in\Lambda e_{z}\setminus J^{2}$ such that $[\alpha\
\bar{\alpha}]\cdot\theta=0$, then $z=x$ or $y$ and there is an exact sequence
isomorphic to $(*)$ with $\theta$ being one of the columns of $M_{i}$,

  2. (ii)

if $\mu\in P_{x}\oplus P_{y}$ is an element $\mu\in e_{z}\Lambda\setminus
J^{2}$ such that $\mu\cdot{\gamma\choose\gamma^{*}}=0$, then $z=j$ or $k$ and
there is an exact sequence isomorphic to $(*)$ with $\mu$ being one of the
rows of $M_{i}$.

Indeed, for $\theta$ as in (i), by definition $\theta\in
Ker(d_{1})=Im(d_{2})$, so $\theta$ can be written as $\theta=M_{i}\cdot\eta$,
for some $\eta={\eta_{1}\choose\eta 2}\in P_{x}\oplus P_{y}$, $\eta\in\Lambda
e_{z}$. Note also that all entries of $M_{i}$ are in $J$ (equivalently,
$d_{2}$ is in $\operatorname{rad}_{\Lambda}$), since otherwise equality
$(\alpha\ \bar{\alpha})\cdot M_{i}=0$ implies that $\alpha$ or $\bar{\alpha}$
in $J^{2}$ (or $\alpha\in K\bar{\alpha}$), which is impossible for an arrow.
But $\theta\notin J^{2}$, i.e. $\theta_{1}\notin e_{j}J^{2}e_{z}$ or
$\theta_{2}\notin e_{k}J^{2}e_{z}$, hence we infer that $\eta\notin J$,
because $\eta\in J$ would force $\theta=M_{i}\cdot\eta\in J^{2}$. As a result,
we get that $\eta_{1}\notin e_{x}Je_{z}$ or $\eta_{2}\notin e_{y}Je_{z}$.
Since for $a\neq b$ in $Q_{0}$, we have
$e_{a}Je_{b}\simeq\operatorname{rad}_{\Lambda}(P_{b},P_{a})=\operatorname{Hom}_{\Lambda}(P_{b},P_{a})\simeq
e_{a}\Lambda e_{b}$, we conclude that $z=x$ or $y$, and in both cases
$\eta_{\cdot}$ is a unit of the local algebra $e_{z}\Lambda e_{z}$. We may
assume that $z=x$ (the proof in case $z=y$ is similar). In particular, then
$\eta_{1}$ is a unit in $e_{x}\Lambda e_{x}$ (i.e. $\eta_{1}$ is a scalar
multiplication of $e_{x}$), so we obtain the following identity

${\theta_{1}\ \psi_{jy}\choose\theta_{2}\ \psi_{ky}}=M_{i}\cdot{\eta_{1}\
0\choose\eta_{2}\ e_{y}}.$

Denote by $M^{\prime}_{i}$ the matrix on the left hand side and let
$N={\eta_{1}\ 0\choose\eta_{2}\ e_{y}}$. Consequently, the above identity
$M_{i}^{\prime}=M_{i}\cdot N$ translates into the following commutative
diagram in $\operatorname{mod}\Lambda$:

$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{S_{i}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{P_{i}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{d_{3}}$$\textstyle{P_{x}\oplus
P_{y}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{d_{2}}$$\textstyle{P_{j}\oplus
P_{k}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{d_{1}}$$\textstyle{P_{i}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{S_{i}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{0}$$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{S_{i}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{id}$$\textstyle{P_{i}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{d_{3}^{\prime}}$$\scriptstyle{id}$$\textstyle{P_{x}\oplus
P_{y}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{d_{2}^{\prime}}$$\scriptstyle{v}$$\textstyle{P_{j}\oplus
P_{k}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{d_{1}^{\prime}=d_{1}}$$\scriptstyle{id}$$\textstyle{P_{i}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{id}$$\textstyle{S_{i}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{id}$$\textstyle{0}$

where we identify $d_{1}=(\alpha\ \bar{\alpha})$, $d_{2}=M_{i}$,
$d_{3}={\gamma\choose\gamma^{*}}$, $d_{2}^{\prime}=M_{i}^{\prime}$, $v=N$
(which is an isomorphism, since $\eta_{1}$ and $e_{y}$ are units in the
corresponding local algebras), and $d_{3}^{\prime}=v^{-1}d_{3}$. It follows
that the bottom row of the above diagram is the required exact sequence.
Similarly, if $z=y$, then we can construct analogous matrix $M_{i}^{\prime}$,
but with $\theta$ as the second column.

In other words, one can swap the original sequence for the new one, in which
$M_{i}$ admits a fixed $\theta$ as the first (or second) column.

In a similar way, we can prove (ii), where we use cokernel of $d_{3}$ (instead
of kernel of $d_{1}$); indeed by the universal property of cokernels one can
factorize matrix ${\mu_{1}\ \ \mu_{2}\choose\varphi_{kx}\ \psi_{ky}}$ through
the cokernel of $d_{3}$ ($\cong Im(d_{2})$), and lift this factorization to a
map $u:P_{j}\oplus P_{k}\to P_{j}\oplus P_{k}$, given by a matrix
$N={\eta_{1}\ \eta_{2}\choose 0\ \ e_{k}}$, such that $N\cdot
M_{i}=M_{i}^{\prime}$ and $\eta_{1}$ is a unit. This means
$ud_{2}=d_{2}^{\prime}$, yielding analogous commutative diagram with (exact)
isomorphic rows.

Note that the conditions (i)-(ii) mentioned above explain how minimal
generators (relations) of $I$ give rise to generators of $\Omega^{2}(S_{i})$,
and how these two are connected via the exact sequence $(*)$, up to
isomorphism (here we mean both isomorphism of exact sequences and isomorphisms
of algebras, i.e. changing presentation of $\Lambda$).

Namely, we may start with a minimal generator $\rho$ of the ideal $I$ of $KQ$,
without loss of generality $\rho\in e_{i}\Lambda e_{j}$, where $i,j$ are
vertices of $Q$. Say $\alpha,\bar{\alpha}$ start at $i$ and $\beta,\beta^{*}$
end at vertex $j$. Then we can write $\rho$ as an element of $KQ$ in the
following way

$\rho=\alpha x_{1}\beta+\alpha
x_{2}\beta^{*}+\bar{\alpha}x_{3}\beta+\bar{\alpha}x_{4}\beta^{*}$ $None$

where the $x_{i}$ are linear combinations of monomials, and the expression is
unique if written in terms of the monomial basis of $KQ$. Consequently, we
infer from (i) that an element

$\theta=(x_{1}\beta+x_{2}\beta^{*},\ x_{3}\beta+x_{4}\beta^{*})$

is in the kernel of $d_{1}$, and it can be taken as a generator for
$\Omega^{2}(S_{i})$ (column of $M_{i}$), for example if $\theta\notin J^{2}$.
Similarly $\mu=(\alpha x_{1}+\bar{\alpha}x_{3},\alpha
x_{2}+\bar{\alpha}x_{4})$ gives a cogenerator (row of $M_{i}$), if $\mu\notin
J^{2}$.

######  Remark 3.3.

We note that not all minimal relations can be realized in this way. Indeed, if
$\Lambda$ is a weighted surface algebra
$\Lambda=\Lambda(Q,f,m_{\bullet},c_{\bullet})$ with at least one arrow
$\alpha\in Q_{1}$ such that $m_{\alpha}n_{\alpha}=2$ (virtual arrow), then
there exist a minimal zero relation of the form $\alpha\beta\gamma=0$, which
cannot be induced from an element in the second syzygy of a simple module.

##  4\. Triangles and squares

In this section we discuss some properties of triangles and squares in $Q$
with respect to minimal relations. As we will see in Proposition 4.1 below, it
is natural to investigate triangles in the quiver $Q$, which appear together
with paths of length $2$ involved in minimal relations (note that this was an
essential tool in [7, see Proposition 4.2]). Similarly, squares come with
paths of length $3$ as shown in paralell result (Lemma 4.5).

If $p$ is a monomial in $KQ$, we write $p\prec I$, provided that $p$ occurs as
a term (summand) in some minimal relation defining $I$ (i.e. $p$ is involved
in a minimal relation). Very often, paths of length two occur in this way as
shown in [7, Proposition 4.2].

###  4.1. Paths of length 2 and triangles

######  Proposition 4.1.

Assume $\alpha:i\to j$ and $\beta:j\to k$ are arrows such that
$\alpha\beta\prec I$. Then there is an arrow in $Q$ from $k$ to $i$, so that
$\alpha$ and $\beta$ are part of a triangle in $Q$.

###### Proof.

Recall we write $\bar{\alpha}$ for the other arrow starting at $i$ (if it
exists), and write $\beta^{*}$ for the other arrow ending at $k$ (if it
exists). Then $\alpha\beta\prec I$ means that

$\alpha\beta+\alpha z_{0}\beta+\alpha
z_{1}\beta^{*}+\bar{\alpha}z_{2}\beta+\bar{\alpha}z_{3}\beta^{*}=0$

in $\Lambda$ where $z_{0}\in J$, and $z_{i}\in\Lambda$. We may assume
$z_{0}=0$, otherwise we replace $\alpha$ by $\alpha(1+z_{0})$. We use the
exact sequence $(*)$. Then the identity above gives an element $\varphi$ in
the kernel of $d_{1}$, namely

$\varphi=(\beta+z_{1}\beta^{*},z_{2}\beta+z_{3}\beta^{*})$

Clearly, $\varphi\notin J^{2}$, because its first coordinate admits an arrow.
Therefore, using Remark 3.2(i) for $\theta:=\varphi$ (viewed as a column), we
conclude that $P_{k}$ is a direct summand of $P^{-}_{i}$, i.e. $k$ is a source
of an arrow ending at $i$, and the claim follows. ∎

Note that this holds for any symmetric periodic algebra of period 4 (i.e. also
for wild ones).

######  Example 4.2.

In [7, Section 11], there is a quiver $Q$ of an algebra which is symmetric and
periodic of period 4, but the algebra is wild (so out of our current interest;
see also [3] and [7, Corollary 2]). This is mentioned as a consequence of the
classification in [7], however we can observe it already, as the above
proposition implies that the algebra must be wild.

Namely, it follows from Proposition 4.1 that any path $\rho$ in $Q$ of length
two which does not involve a loop satisfies $\rho\nprec I$. Therefore
$B/J^{3}$ contains a wild subalgebra, given by a quiver of type
$\widetilde{\widetilde{E}}_{7}$ without any relations, as in [7, see the proof
of Proposition 4.2].

######  Lemma 4.3 (Triangle Lemma).

Assume $Q$ contains a triangle

---
$\textstyle{x\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\gamma}$$\textstyle{i\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\alpha}$$\textstyle{j\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\beta}$

If $\gamma\alpha\not\prec I$ then also $\alpha\beta\not\prec I$.

###### Proof.

Consider the exact sequence for the simple module $S_{x}$

$0\to S_{x}\to P_{x}\to P_{j}\oplus P_{j^{*}}\to P_{i}\oplus P_{\bar{i}}\to
P_{x}\to S_{x}\to 0,$

where $j^{*}=s(\beta^{*})$ and $\bar{i}=t(\bar{\gamma})$. Taking minimal
generators for $\Omega^{2}(S_{x})$ gives the columns of the matrix $M_{x}$,
that is

$M_{x}=\left(\begin{matrix}\varphi_{ij}&\psi_{ij^{*}}\cr\varphi_{\bar{i}j}&\psi_{\bar{i}j^{*}}\end{matrix}\right).$

It satisfies $(\gamma\ \bar{\gamma})\cdot M_{x}=0$ and
$M_{x}\cdot{\beta\choose\beta^{*}}=0$.

Suppose $\gamma\alpha\nprec I$, but (for a contradiction) $\alpha\beta\prec
I$. Then there is a minimal relation of the form

$\alpha\beta+\alpha z_{1}\beta+\alpha
z_{2}\beta^{*}+\bar{\alpha}z_{3}\beta+\bar{\alpha}z_{4}\beta^{*}=0$

with $z_{1}\in J$, and we may assume again $z_{1}=0$. Now, if we define

$\theta:=(\alpha+\bar{\alpha}z_{3},\alpha z_{2}+\bar{\alpha}z_{4}),$

then $\theta\cdot{\beta\choose\beta^{*}}=0$ and $\theta\notin J^{2}$ (since it
involves an arrow), so by Remark3.2(ii), we can take $\theta$ as the first row
of $M_{x}$. In particular, $\varphi_{ij}=\alpha+\bar{\alpha}z_{3}$, hence it
follows that
$\gamma(\alpha+\bar{\alpha}z_{3})+\bar{\gamma}\varphi_{\bar{i}j}=0$, and we
obtain $\gamma\alpha\prec I$, a contradiction. ∎

Consequently, for a triangle as above, either all paths
$\gamma\alpha,\alpha\beta,\beta\gamma\prec I$, or none of them. Hence
triangles in $Q$ split into two families: triangles, say _of type R_ (for
which all $\gamma\alpha,\alpha\beta,\beta\gamma\prec I$) and triangles _of
type N_. We will further see (Section 5) similar distinction between non-
regular vertices.

Let us finish this part with the following lemma.

######  Lemma 4.4.

Assume $i$ is a 1-vertex which is part of a triangle

$\textstyle{i\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\alpha}$$\textstyle{x\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\gamma}$$\textstyle{j\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\beta}$

Then both $x$ and $j$ must be 2-vertices.

###### Proof.

By Lemma 2.1, there must be another arrow, say $\bar{\gamma}$, starting at
$x$, and there must be another arrow, say $\alpha^{*}$, ending at $j$. From
the exact sequence for $S_{i}$ we know that $p_{x}=p_{j}$.

(i) Assume vertex $x$ is not a $2$-vertex, then $\beta$ is the only arrow
ending at $x$. Therefore

$e_{x}\Lambda/S_{x}\cong\beta\Lambda.$

Moreover, again by Lemma 2.1 there must be another arrow starting at $j$, call
it $\bar{\beta}$. Hence

${\rm rad}(P_{j})=\beta\Lambda+\bar{\beta}\Lambda.$

As a result, we get the following equalities of dimension vectors:

$p_{x}=s_{x}+\underline{\dim}(\beta\Lambda)\mbox{ and
}p_{j}=s_{j}+\underline{\dim}(\beta\Lambda+\bar{\beta}\Lambda)=s_{j}+\underline{\dim}\beta\Lambda+\underline{\dim}(\bar{\beta}\Lambda/\beta\Lambda\cap\bar{\beta}\Lambda).$

Comparing $p_{x}=p_{j}$, we conclude that
$\underline{\dim}(\bar{\beta}\Lambda/\beta\Lambda\cap\bar{\beta}\Lambda)=s_{x}-s_{j}$,
so this must be zero (i.e. $x=j$), since otherwise $s_{x}-s_{j}$ has a
negative coordinate, which cannot happen for a dimension vector of a
$\Lambda$-module.

In particular since $S_{j},S_{x}$ are simple and the vector space dimension of
$\beta\Lambda\cap\bar{\beta}\Lambda$ is equal to the vector space dimension of
$\bar{\beta}\Lambda$. However, we have an inclusion of these spaces, so they
are equal. Now
$\bar{\beta}\Lambda=\beta\Lambda\cap\bar{\beta}\Lambda\subseteq\beta\Lambda$,
hence $\bar{\beta}\in\beta\Lambda$. But this is not possible since
$\bar{\beta}$ is an arrow $\neq\beta$.

(ii) The proof that $j$ must be a 2-vertex is dual. ∎

###  4.2. Paths of length $\geq 3$

In this short paragraph we will consider a bit longer paths, i.e. of length
$3$ or $4$ (and in particular, induced squares). Suppose the quiver of
$\Lambda$ has a subquiver

$u\stackrel{{\scriptstyle\delta}}{{\longrightarrow}}i\stackrel{{\scriptstyle\alpha}}{{\longrightarrow}}k\stackrel{{\scriptstyle\beta}}{{\longrightarrow}}t\stackrel{{\scriptstyle\gamma}}{{\longrightarrow}}j$

We have the following counterpart of Proposition 4.1.

######  Lemma 4.5.

Suppose $\alpha\beta\not\prec I$ or $\beta\gamma\not\prec I$, and
$\alpha\beta\gamma\prec I$. Then there is an arrow $j\to i$.

###### Proof.

We work in $KQ$, using the basis consisting of paths. We deal with the case
$\alpha\beta\not\prec I$ (the other case is dual, working with the opposite
algebra).

After possibly adjusting arrow $\beta$, there is a minimal relation of the
form

$\alpha\beta\gamma+\alpha
z_{1}\gamma^{*}+\bar{\alpha}z_{2}\gamma+\bar{\alpha}z_{3}\gamma^{*}\in I$

Consider the exact sequence for the simple module $S_{i}$,

$0\to S_{i}\to P_{i}\to P_{u}\oplus P_{u^{\prime}}\to P_{k}\oplus P_{l}\to
P_{i}\to S_{i}\to 0$

Here $\alpha:i\to k,\bar{\alpha}:i\to l$ start at $i$, and $\delta:u\to i$ and
$\delta^{*}:u^{\prime}\to i$ are the arrows ending at $i$, where by
convention, $\bar{\alpha}$ or $\delta^{*}$ may not exist (then we omit $P_{l}$
and $P_{u^{\prime}}$).

We take $\Omega(S_{i})=\alpha\Lambda+\bar{\alpha}\Lambda$ and
$\Omega^{2}(S_{i})=\\{(x,y)\in P_{k}\oplus P_{l}\mid\alpha
x+\bar{\alpha}y=0\\}$. From the exact sequence, this is equal to
$\varphi\Lambda+\psi\Lambda$ where $\varphi=\varphi e_{u}$ and $\psi=\psi
e_{u^{\prime}}$. Let $M_{i}$ be the matrix with columns $\varphi$ and $\psi$.
Then (for some choice of arrows $\delta,\delta^{*}$) we have

$(\alpha\ \bar{\alpha})\cdot M_{i}=0\ \
M_{i}\cdot{\delta\choose\delta^{*}}=0.$

The minimal relation above gives rise to the following element $\theta$ which
belongs to $\Omega^{2}(S_{i})$,

$\theta=(\beta\gamma+z_{1}\gamma^{*},\ z_{2}\gamma+z_{3}\gamma^{*})$

If $\theta\notin J^{2}$, then we may take $\varphi:=\theta$ as the first
column of $M_{i}$, since $\beta\gamma\prec\theta_{1}$ must be in $e_{k}\Lambda
e_{u}$ (see also Remark 3.2(i)). It follows that $j=u$, and hence $\delta$ is
an arrow from $j=u$ to $i$.

Suppose now $\theta\in J^{2}$. We will show that this leads to a
contradiction. The radical of $\Omega^{2}(S_{i})$ is equal to
$\Omega^{2}(S_{i})J=\varphi J+\psi J$. So we can write $\theta=\varphi v+\psi
w$ and $v,w\in J$ and we can take them in $Je_{j}$. Then

$\beta\gamma+z_{1}\gamma^{*}=\varphi_{ku}v+\psi_{ku^{\prime}}w$

Say $\beta\gamma$ occurs in $\varphi_{ku}v$. We can write
$v=ve_{j}=v_{1}\gamma+v_{2}\gamma^{*}$ with $v_{i}\in\Lambda$ (which need not
be in the radical). We can write $\varphi_{ku}=\beta y_{1}+\bar{\beta}y_{2}$
with $y_{1},y_{2}\in KQ$. Then

$\varphi_{ku}v=\beta y_{1}v_{1}\gamma+\beta
y_{1}v_{2}\gamma^{*}+\bar{\beta}y_{2}v$

Then $\beta\gamma$ is a term of $\beta y_{1}v_{1}\gamma$. Therefore
$y_{1}v_{1}$ (which we can take equal to $y_{1}v_{1}e_{t}$) is equal to
$e_{t}$ modulo the radical. It follows that $y_{1}v_{1}$ is a unit in
$e_{t}\Lambda e_{t}$.

However, it factors through vertex $u$, and it follows that $u=t$. So we have
a triangle $(\alpha,\beta,\delta)$ and $\alpha\beta\not\prec I$. It follows
(from our triangle lemma) that also $\beta\delta\not\prec I$.

On the other hand, we exploit the identity for $\varphi_{ku}v$ a bit further.
Since $y_{1}v_{1}$ is a unit, we may assume $\beta=\beta y_{1}$. Then
$\varphi_{ku}=\beta+\bar{\beta}y_{2}$ and recall this is the top left entry of
the Matrix $M_{i}$ above. We have $M_{i}{\delta\choose\delta^{*}}=0$ which
gives

$\beta\delta+\bar{\beta}y_{2}\delta+\psi_{ku^{\prime}}\delta^{*}=0,$

hence $\beta\delta\prec I$, a contradiction. ∎

The above proposition shows that paths of length $3$ involved in minimal
relations induce squares in $Q$. We have a result similar to previous Triangle
Lemma, stated as follows.

######  Lemma 4.6 (Square Lemma).

Assume $Q$ contains a square

$\textstyle{1\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\delta}$$\textstyle{2\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\alpha}$$\textstyle{4\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\gamma}$$\textstyle{3\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\beta}$

If $\alpha\beta\gamma\prec I$, then $\beta\gamma\delta\prec I$.

###### Proof.

Suppose that $\alpha\beta\gamma\prec I$, but $\beta\gamma\delta\nprec I$. In
particular, we have also $\beta\gamma\nprec I$. Consider the exact sequence
for $S_{2}$. Then $\Omega^{2}(S_{2})$ has generators being the columns of the
matrix

$M_{2}=\left(\begin{matrix}\varphi_{31}&\psi_{3,1^{*}}\cr\varphi_{\bar{3}1}&\psi_{\bar{3}1^{*}}\end{matrix}\right).$

and $(\alpha\ \bar{\alpha})M_{2}=0$ and $M_{2}{\delta\choose\delta^{*}}=0$. By
our assumption we have minimal relation
$\alpha(\beta\gamma+x_{1}\gamma+x_{2}\bar{\gamma})+\bar{\alpha}(x_{3}\gamma+x_{4}\bar{\gamma})=0$
with $x_{1}\in J^{2}$. This gives an element

$\varphi=(\beta\gamma+x_{1}\gamma+x_{2}\bar{\gamma},\
x_{3}\gamma+x_{4}\bar{\gamma})^{t}\in\Omega^{2}(S_{2})$

which cannot be in the radical of $\Omega^{2}(S_{2})$, since
$\beta\gamma\not\prec I$. So we can take this as the first column of $M_{2}$.
It follows that
$(\beta\gamma+x_{1}\gamma+x_{2}\bar{\gamma})\delta+\psi_{31^{*}}\delta^{*}=0$,
so $\beta\gamma\delta\prec I$, and we get a contradiction. ∎

The following lemma shows that sometimes one can relate paths of length 3 and
4.

######  Lemma 4.7.

Suppose we have a path in $Q$ of the form

$u\stackrel{{\scriptstyle\delta}}{{\longrightarrow}}i\stackrel{{\scriptstyle\alpha}}{{\longrightarrow}}k\stackrel{{\scriptstyle\beta}}{{\longrightarrow}}t\stackrel{{\scriptstyle\gamma}}{{\longrightarrow}}j$

with $|k^{+}|=1$ and $\delta\alpha,\alpha\beta,\beta\gamma\nprec I$. If
$\delta\alpha\beta\gamma\prec I$, then $\alpha\beta\gamma\prec I$.

###### Proof.

Suppose that $\delta\alpha\beta\gamma\prec I$ and let $\bar{\delta}:u\to
u^{\prime}$ and $\bar{\alpha}:i\to i^{\prime}$ denote the second arrow
starting at $u$ and $i$ (if exist). By the assumptions on $k$, we conclude
that any path $p\in e_{i}A$ starting from $\delta\alpha$ must go through
$\delta\alpha\beta$. Hence $\delta\alpha\beta\gamma$ as a minimal generator of
$I$ is involved in a minimal relation of the form

$\delta\alpha\beta\gamma+\delta\alpha\beta
p+\delta\bar{\alpha}q+\bar{\delta}r=0,$

where $p,q,r\in J$. After adjusting $\gamma:=\gamma+p$, we may change the
presentation to get $p=0$. Consequently, the element
$\rho=(\alpha\beta\gamma+\bar{\alpha}q,r)$ belongs to
$\Omega^{2}(S_{u})=ker([\delta\ \bar{\delta}])$. Finally, if
$u^{-}=\\{\sigma,\sigma^{*}\\}$, and $v=s(\sigma)$,
$v^{\prime}=s(\sigma^{*})$, then $\Omega^{2}(S_{u})\cong Im(M_{u})$, where
$M_{u}:P_{v}\oplus P_{v^{\prime}}\to P_{i}\oplus P_{i^{\prime}}$ is given by
the matrix

$\left(\begin{matrix}\varphi_{uv}&\psi_{uv^{\prime}}\cr\varphi_{u^{\prime}v}&\psi_{u^{\prime}v^{\prime}}\end{matrix}\right),$

hence $\rho=M_{u}\cdot{\kappa_{1}\choose\kappa_{2}}$, for some $\kappa_{1}\in
P_{v}$ and $\kappa_{2}\in P_{v^{\prime}}$. But then

$\alpha\beta\gamma+\bar{\alpha}q=\varphi_{uv}\kappa_{1}+\psi_{uv^{\prime}}\kappa_{2},$

and therefore, $\alpha\beta\gamma$ is generated by minimal relations. But,
$\alpha\beta\nprec I$ and $\beta\gamma\nprec I$, hence $\alpha\beta\gamma$ is
also involved in some minimal relation of $I$, as claimed. ∎

##  5\. Non-regular vertices

In this section, we give some partial results describing non-regular vertices.

Clearly, for $Q$ biserial, the non-negular vertices $i$ satisfy either
$|i^{-}|=1$ and $|i^{+}|=2$ or $|i^{-}|=2$ and $|i^{+}|=1$. In the first case
$i$ is called a $(1,2)$-vertex, whilst in the second a $(2,1)$-vertex. Let $i$
be a $(1,2)$-vertex of the form

$\textstyle{j\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\alpha}$$\textstyle{i\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\beta}$$\scriptstyle{\bar{\beta}}$$\textstyle{k}$$\textstyle{l}$

We call $i$ a vertex _of type R_ (respectively, _of type N_), provided that
both $\alpha\beta\prec I$ and $\alpha\bar{\beta}\prec I$ (respectively, both
$\alpha\beta\nprec I$ and $\alpha\bar{\beta}\nprec I$). If $k\neq l$, $i$ is
said to be _proper_. Similar notions can be defined for $(2,1)$-vertices.
Whenever we consider a $(1,2)$-vertex $i$, we keep the above notation for
arrows starting and ending at $i$.

######  Remark 5.1.

We recall that there exist infinitely many pairwise non-isomorphic TSP4
algebras $A$ containing arbitrary large number of $(1,2)$\- and
$(2,1)$-vertices of both types R or N. Indeed, one may take any weighted
surface algebra $\Lambda$ (see [8]) containing arbitrary number of ’blocks’ of
the form

$\textstyle{\circ\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{y_{i}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\alpha_{i}}$$\textstyle{x_{i}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\bullet\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\xi_{i}}$$\textstyle{\bullet\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$

with $\xi_{i}$ being a virtual arrow. Then using results of [11, see also
Section 4], we conclude that the virtual mutation $A=\Lambda(\xi)$ with
respect to the sequence $\xi=(\xi_{1},\dots,\xi_{n})$ of virtual arrows is a
TSP4 algebra and vertices $x_{1},\dots,x_{n}$ are $(1,2)$-vertices of type N
(in $Q_{A}$), whereas $y_{1},\dots,y_{n}$ are $(2,1)$-vertices of type N (in
definition of $\Lambda$, we have to pick weights
$m_{\xi_{i}}=m_{\alpha_{i}}=1$, for any $i\in\\{1,\dots,n\\}$).

For vertices of type R one has to consider so called weighted generalized
triangulation algebras [13], given by quivers which are glueings of blocks of
five types I-V. Without going into details, we only mention that for any such
algebra $A$ (it is a TSP4 algebra in most cases), its Gabriel quiver contains
two $(1,2)$-vertices and two $(2,1)$-vertices per each block of type V, and
all these vertices are of type R (this follows directly from the shape of
relations in $A$). Therefore, one can easily construct a TSP4 algebra with
arbitrary large number of non-regular vertices of type R (in this case the
number of $(1,2)$-vertices is equal to the number of $(2,1)$-vertices). But
then the Gabriel quiver of $A$ is not biserial.

It turns out that there are no non-regular vertices of type R, if the Gabriel
quiver is biserial. For proper ones, it is pretty easy to see, as the
following lemma shows.

######  Lemma 5.2.

There are no proper non-regular vertices of type R.

###### Proof.

Suppose $i$ is a $(1,2)$-vertex of type R. In particular, $\alpha\beta\prec I$
yields an arrow $\gamma:k\to j$, whereas $\alpha\bar{\beta}:l\to j$, an arrow
$\delta:l\to j$, due to Proposition 4.1. Suppose now $i$ is proper. Then
$j^{-}=\\{\gamma,\delta\\}$, since $Q$ is biserial, and hence
$p_{j}^{-}=p_{k}+p_{l}$. But $p_{i}^{-}=p_{i}^{+}$ gives $p_{j}=p_{k}+p_{l}$,
because $i$ is a $(1,2)$-vertex, and therefore, we get
$p_{j}=p_{j}^{-}=\hat{p}_{j}$, which is a contradiction with Lemma 2.4. Dual
arguments provide the proof for $(2,1)$-vertices. ∎

We complete the claim as follows.

######  Theorem 5.3.

There are no non-regular vertices of type R.

###### Proof.

By previous lemma, it is sufficient to prove that there is no $(1,2)$-vertex
$i$ of type R with $k=l$ (i.e. non-proper one). Suppose to the contrary, that
such a vertex exists. For simplicity, we will use notation $1,2,3$ for
vertices, respectively, $i,k=l$ and $j$. Since $1=i$ is of type R, we get an
arrow $\gamma:2\to 3$ (see Proposition 4.1), and consequently, $Q$ admits the
following subuiver

$\textstyle{1\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\bar{\beta}}$$\scriptstyle{\beta}$$\textstyle{2\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\gamma}$$\textstyle{3\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\alpha}$

Of course, $1$ is a non-regular vertex, by the assumption. We claim that also
$2$ is a non-regular vertex. Indeed, if this is not the case, then there is an
arrow $\bar{\gamma}:2\to x$, $\bar{\gamma}\neq\gamma$, and moreover, $x\neq
3$, because otherwise, we would get a subquiver
$\textstyle{1\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{2\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{3}$,
hence $p_{1}=p_{3}$, because $p_{2}^{-}=p_{2}^{+}$, and so
$\hat{p}_{1}=p_{1}^{-}=p_{3}=p_{1}$, which gives a contradiction with Lemma
2.4. Consequently, we have no arrows $x\to 1$, so both
$\beta\bar{\gamma}\nprec I$ and $\bar{\beta}\bar{\gamma}\nprec I$, due to
Proposition 4.1. But then $A$ admits a wild (hereditary) factor algebra of the
form
$\textstyle{1\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{2\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\bar{\gamma}}$$\textstyle{x}$,
a contradiction. This proves that $2$ is a $(2,1)$-vertex.

In particular, using $p_{1}^{-}=p_{1}^{+}$ and $p_{2}^{-}=p_{2}^{+}$ one gets
$p_{1}=p_{2}$ and $p_{3}=2p_{1}$. It follows also that $3^{-}\cup
3^{+}\supsetneq\\{\gamma,\alpha\\}$, and hence $3$ is a $2$-vertex (note:
$p_{3}^{-}=p_{3}^{+}$). In particular, there may be a loop $\rho$ at vertex
$3$, and then $Q_{1}=\\{\alpha,\beta,\bar{\beta},\gamma,\rho\\}$). Otherwise,
instead of $\rho$ there are arrows starting and ending at $3$, and other
vertices or arrows. We start with properties which hold in both cases.

We may assume that $e_{1}J^{3}=e_{1}\beta\gamma J$ and $\bar{\beta}\gamma\in
e_{1}J^{3}$. By tameness of $A$, there must be a minimal relation involving at
least one of $\beta\gamma,\bar{\beta}\gamma$. This means that $\dim
e_{1}J^{2}/e_{1}J^{3}\leq 1$ and clearly it is non-zero. So we may assume it
is spanned by the coset of $\beta\gamma$. Then
$\bar{\beta}\gamma=c\beta\gamma+\psi$ for $c\in K$ and $\psi\in e_{1}J^{3}$.
If $c\neq 0$ then we replace the arrow $\bar{\beta}$ by $\bar{\beta}-c\beta$,
to get the claim.

We write the relation as

$\bar{\beta}\gamma=\beta\gamma\Theta_{1}\gamma+\beta\gamma\Theta_{2}\rho\ \ \
(\Theta_{1}\in J^{2},\ \ \Theta_{2}\in\Lambda).$ $None$

Note also that we may assume $\gamma\alpha=0$. Indeed, using the exact
sequence for the simple module $S_{1}$, one can see that the second syzygy
$\Omega^{2}(S_{1})$ has one minimal generator, say $(\varphi_{23},\psi_{23})$,
satisfying both $\varphi_{23}\alpha=0$ and $\psi_{23}\alpha=0$. Applying now
the relation (*), we may fix the generator of the form
$(-(\gamma\Theta_{1}\gamma+\gamma\Theta_{2}\rho),\gamma)$ (previously
adjusting $\gamma$, to get $\psi_{23}=\gamma$). In particular,
$\gamma\alpha=0$ for this choice.

Consider the exact sequence for $S_{2}$. This gives

$0\to\Omega^{-1}(S_{2})\cong(\beta,\bar{\beta})\Lambda\to P_{1}\oplus P_{1}\to
P_{3}\to\gamma\Lambda\cong\Omega(S_{2})\to 0$

Since $\gamma\alpha=0$ we have $\alpha\Lambda\subset\Omega^{2}(S_{2})$. It is
not in the radical of $\Omega^{2}(S_{2})$ since $\alpha$ is an arrow. From the
exact sequence, $\Omega^{2}(S_{2})$ has one more generator, call it
$\varphi_{31}$, and we have the minimal relation

$\alpha\beta+\varphi_{31}\bar{\beta}=0$ $None$

Assume now that there is a loop at vertex $3$ and consider the exact sequence
for $S_{3}$. This gives an exact sequence

$0\to\Omega^{-1}(S_{3})\cong(\gamma,\rho^{\prime})\Lambda\to P_{2}\oplus
P_{3}\to\ P_{1}\oplus
P_{3}\to\alpha\Lambda\oplus\rho\Lambda\cong\Omega(S_{3})\to 0$

where $\rho^{\prime}$ is a version of $\rho$, and the middle map is given by
matrix $M_{3}={\varphi_{12}\ \psi_{13}\choose\varphi_{32}\ \psi_{33}}$, which
describes the generators of $\Omega^{2}(S_{3})$. Then $(\alpha\ \rho)M_{3}=0$
and $M_{3}{\gamma\choose\rho^{\prime}}=0$. We can write the minimal relation
(**) as $0=\alpha\beta\ +\
\alpha\phi^{\prime}\bar{\beta}+\rho\phi^{\prime\prime}\bar{\beta}$ where
$\phi^{\prime}\in\Lambda$ and $\phi^{\prime\prime}\in J$, and therefore
$\Omega^{2}(S_{3})$ has a generator
$(\beta+\phi^{\prime}\bar{\beta},\phi^{\prime\prime}\bar{\beta})$. This can be
taken as the first column in $M_{3}$. The first row of $M_{3}$ gives now that
we have a minimal relation
$(\beta+\varphi^{\prime}\bar{\beta})\gamma+\psi_{13}\rho^{\prime}=0$. Now
$\bar{\beta}\gamma$ is in $\beta\gamma J$ and $\psi_{13}\rho^{\prime}$ also is
in $J^{3}$. Therefore $\beta\gamma\in J^{3}$. But we have seen that
$e_{1}J^{3}=\beta\gamma J$, so we deduce $e_{1}J^{3}\subseteq e_{1}J^{4}$ and
hence $e_{1}J^{3}=0$. As a result, $\beta\gamma J=0$ and $\beta\gamma\in{\rm
soc}(e_{1}\Lambda)\cap e_{1}\Lambda e_{3}=0$, which is not possible for a
symmetric algebra (see [4, I.3.5]).

In general, we have arrows $\bar{\alpha}:3\to x$ and $\gamma^{*}:y\to 3$. The
exaxt sequence for $S_{3}$ is now of the form

$0\to\Omega^{-1}(S_{3})\cong(\gamma,\gamma^{*})\Lambda\to P_{2}\oplus
P_{y}\to\ P_{1}\oplus
P_{x}\to\alpha\Lambda\oplus\bar{\alpha}\Lambda\cong\Omega(S_{3})\to 0$

Exactly as in the first case, we rewrite (**) which gives the first column of
the matrix $M_{3}$. Then from the first row we get the identity
$0=(\beta+\phi^{\prime}\bar{\beta})\gamma+\psi_{1y}\gamma^{*}$, and we get the
same contradiction as before. ∎

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