Minimal length maximal green sequences
Alexander Garver, Thomas McConville, Khrystyna Serhiyenko
TL;DR
This paper provides a formula for the length of minimal length maximal green sequences in quivers derived from surface triangulations, advancing understanding in representation theory and related fields.
Contribution
It introduces a novel combinatorial approach combining surface triangulations and scattering diagrams to determine minimal sequence lengths.
Findings
Derived a formula for minimal length maximal green sequences
Applied combinatorics of surface triangulations and scattering diagrams
Focused on quivers from annulus and punctured disk triangulations
Abstract
Maximal green sequences are important objects in representation theory, cluster algebras, and string theory. It is an open problem to determine what lengths are achieved by the maximal green sequences of a quiver. We combine the combinatorics of surface triangulations and the basics of scattering diagrams to address this problem. Our main result is a formula for the length of minimal length maximal green sequences of quivers defined by triangulations of an annulus or a punctured disk.
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Minimal Length Maximal Green Sequences
Alexander Garver
Laboratoire de Combinatoire et d’Informatique Mathématique, Université du Québec à Montréal
,
Thomas McConville
Department of Mathematics, Massachusetts Institute of Technology
and
Khrystyna Serhiyenko
Department of Mathematics, University of California, Berkeley
Abstract.
Maximal green sequences are important objects in representation theory, cluster algebras, and string theory. It is an open problem to determine what lengths are achieved by the maximal green sequences of a quiver. We combine the combinatorics of surface triangulations and the basics of scattering diagrams to address this problem. Our main result is a formula for the length of minimal length maximal green sequences of quivers defined by triangulations of an annulus or a punctured disk.
Contents
1. Introduction
A maximal green sequence is a distinguished sequence of local transformations, known as mutations, of a given quiver (i.e., directed graph). Maximal green sequences were introduced by Keller in [18] in order to obtain combinatorial formulas for the refined Donaldson-Thomas invariants of Kontsevich and Soibelman [19]. They are also important in string theory [2], representation theory [5, 7], and cluster algebras [14].
Recently, there have been many developments on the combinatorics of maximal green sequences (see [21] and references therein). In particular, in [21] it is shown that almost any quiver with a finite mutation class (i.e., the set of quivers obtained from by mutations) has a maximal green sequence. Our goal is to add to the known combinatorics by developing a numerical invariant of the set of maximal green sequences of : the length of minimal length maximal green sequences.
This invariant is natural from the perspective of cluster algebras. Fixing a quiver induces an orientation of the edges of the corresponding exchange graph. It turns out that the maximal green sequences of are in natural bijection with finite length maximal directed paths of the resulting oriented exchange graph (see [5]). Examples of oriented exchange graphs include the Hasse diagrams of Tamari lattices and of Cambrian lattices of type , , and [24]. In these examples, the minimal length maximal green sequences always have length equal to the number of vertices of , but, in general, the minimal length of a maximal green sequence may be larger than the number of vertices of . Thus understanding the length of minimal length maximal green sequences provides new information about oriented exchange graphs.
Our main results (see Theorems 6.1 and 9.4) are formulas for this minimal length when is of mutation type or mutation type (i.e., is in the mutation class of a type or of an affine type quiver). This number was calculated in mutation type in [10]. We also obtain explicit constructions of minimal length maximal green sequences of , in the process of proving Theorems 6.1 and 9.4.
We also suspect that the length of minimal length maximal green sequences may be significant from the perspective of representation theory of algebras. More specifically, given two quivers that define derived equivalent cluster-tilted algebras, it appears that they will have minimal length maximal green sequences of the same length (see Section 10 for more details).
The paper is organized as follows. In Section 2, we review the basics of quiver mutation, maximal green sequences, and oriented exchange graphs. We also recall the definitions of c- and g-vectors, and the result that maximal green sequences are in bijection with certain sequences of c-vectors. In Section 3, we present the basics of scattering diagrams. We use these to reformulate a result of Muller (see Theorem 3.3) that shows that a maximal green sequence of a quiver determines a maximal green sequence of any full subquiver by removing from the corresponding sequence of c-vectors all those that are not supported on the full subquiver.
In Section 4, we use Theorem 3.3 to show that the minimal length of maximal green sequences of the direct sum of two quivers and is the minimal length of maximal green sequences of plus the minimal length of maximal green sequences of . In Section 5, we use Theorem 3.3 to show that if a quiver is obtained from another quiver by “attaching” several mutation type quivers to , then the problem of finding the length of minimal length maximal green sequences of reduces to finding the length of minimal length maximal green sequences of (see Corollary 5.7).
In Sections 6 and 8, we apply our results to quivers of mutation type and calculate the minimal length of maximal green sequences in Theorem 6.1. Here, we also use combinatorics of triangulated surfaces in the sense of [12] and [13] which we review in Section 7. In particular, we recall how maximal green sequences can be interpreted geometrically using shear coordinates. In Section 9, we do likewise for quivers of mutation type (see Theorem 9.4). We remark that Corollary 5.7 plays in important role in these sections.
**Acknowledgements. ** Alexander Garver thanks Greg Muller and Rebecca Patrias for useful conversations. At various stages of this project, Alexander Garver received support from an RTG grant DMS-1148634, NSERC, and the Canada Research Chairs program. Khrystyna Serhiyenko was supported by the NSF Postdoctoral Fellowship MSPRF-1502881.
2. Maximal green sequences
2.1. Quivers and quivers mutation
A quiver is a directed graph. In other words, is a 4-tuple , where is a set of vertices, is a set of arrows, and two functions defined so that for every , we have . An ice quiver is a pair with a quiver and a set of frozen vertices with the restriction that any have no arrows of connecting them. By convention, we assume and We refer to elements of as mutable vertices. Any quiver is regarded as an ice quiver by setting .
If a given ice quiver is 2-acyclic (i.e., has no loops or 2-cycles), we can define a local transformation of called mutation. The mutation of an ice quiver at a mutable vertex , denoted , produces a new ice quiver by the three step process:
(1) For every -path in , adjoin a new arrow .
(2) Reverse the direction of all arrows incident to in .
(3) Remove any -cycles created, and remove any arrows created that connect two frozen vertices.
From now on, we will only work with 2-acyclic quivers. We show an example of mutation below with the mutable (resp., frozen) vertices in black (resp., blue).
[TABLE]
The information of an ice quiver can be equivalently described by its (skew-symmetric) exchange matrix. Given we define by Furthermore, ice quiver mutation can equivalently be defined as matrix mutation of the corresponding exchange matrix. Given an exchange matrix , the mutation of at , also denoted , produces a new exchange matrix with entries
[TABLE]
For example, the mutation of the ice quiver above (here and ) translates into the following matrix mutation. Note that mutation of matrices and of ice quivers is an involution (i.e., ).
[TABLE]
Let Mut() denote the collection of ice quivers obtainable from by finitely many mutations where such ice quivers are considered up to an isomorphism of quivers that fixes the frozen vertices. We will refer to Mut() as the mutation class of . Such an isomorphism is equivalent to a simultaneous permutation of the rows and first columns of the corresponding exchange matrices.
For our purposes, it will be useful to recall the following classification of mutation type quivers (i.e., quivers ) due to Buan and Vatne.
Lemma 2.1**.**
[9, Prop. 2.4] A connected quiver with vertices is of mutation type if and only if satisfies the following:
- i)
All non-trivial cycles in the underlying graph of are oriented and of length 3.
- ii)
Any vertex has degree at most 4.
- iii)
If a vertex has degree 4, then two of its adjacent arrows belong to one 3-cycle, and the other two belong to another 3-cycle.
- iv)
If a vertex has degree 3, then two of its adjacent arrows belong to a 3-cycle, and the third arrow does not belong to any 3-cycle.
At times we will also say that a quiver is of mutation type , meaning that is of mutation type .
2.2. Maximal green sequences and oriented exchange graphs
In this section, we review the notions of maximal green sequences and oriented exchange graphs.
Given a quiver , we define its framed (resp., coframed) quiver to be the ice quiver (resp., ) where (resp., ), , and (resp., ). We denote elements of by , and we will henceforth write where . We say that a mutable vertex of is green (resp., red) if there are no arrows in of the form (resp., ) for some . The celebrated theorem of Derksen, Weyman, and Zelevinsky [11, Theorem 1.7], known as sign-coherence of c-vectors and g-vectors, implies that given any mutable vertex of is either green or red.
Definition 2.2** ([18]).**
A maximal green sequence of is a sequence of mutable vertices of where
\begin{array}[]{rl}i)&\text{for all j\in[k]i_{j}\in[n]\mu_{i_{j-1}}\circ\cdots\circ\mu_{i_{1}}(\widehat{Q}), and}\\ ii)&\text{each vertex i\in[n]\mu_{i_{k}}\circ\cdots\circ\mu_{i_{1}}(\widehat{Q}) is red}.\end{array}
We let denote the set of maximal green sequences of . At times, we will use the abuse of language where we refer to also as a maximal green sequence. Additionally, we define and to be the length of the maximal green sequence i.
For a given quiver , it is an open question to determine what positive integers can be realized as lengths of maximal green sequences of . In [5, Lemma 2.20], it is shown that if is acyclic, then has a maximal green sequence of length and this is the shortest any maximal green sequence of a quiver can be. The following theorem is the first to address this question for an infinite family of quivers in which oriented cycles may appear.
Theorem 2.3**.**
[10, Theorem 6.5, Theorem 7.2] The length of a minimal length maximal green sequence of a mutation type quiver is |Q_{0}|+|\{\text{3-cycles of Q}\}|.
The maximal green sequences of are in natural bijection with the finite length maximal directed paths in the oriented exchange graph of [5, Proposition 2.13], which we now define. The exchange graph of , denoted , is the (a priori infinite) graph whose vertices are elements of Mut and two vertices are connected by an edge if the corresponding quivers differ by a single mutation. The oriented exchange graph of , denoted , is the directed graph whose vertex set is and whose edges are of the form where is green in . Oriented exchange graphs were introduced in [5] where they initiated the study of maximal green sequences. We show the oriented exchange graph of in Figure 1.
The oriented exchange graph of a quiver can be equivalently described using the c-vectors and c-matrices of . We say that is a c-matrix of if there exists such that is the submatrix of containing its last columns. That is, . We let c-mat() . A row vector of a c-matrix is known as a c-vector. Since a c-matrix is only defined up to a permutations of its rows, can be regarded simply as a set of c-vectors.
Sign coherence of c-vectors, says that any c-vector is either a nonzero element of or of . In the former case, we say a c-vector is positive. In the latter case, we say a c-vector is negative. From our definition of the exchange matrix , we have that (the c-vector of vertex in ) is positive (resp., negative) if and only if is green (resp., red) in .
The set can be regarded as a directed graph whose vertices are c-matrices and whose directed edges are exactly those of the form where is positive. Now by regarding as a directed graph, it follows from [7] that and maximal directed paths in each are in natural bijection.111One can define a notion of c-matrix mutation (for example, see [7, page 35]). However, we omit this definition because we will not need it. Moreover, each directed edge of is labeled by the c-vector . We obtain the following lemma.
Lemma 2.4**.**
A maximal green sequence of a quiver is equivalent to a maximal directed path in where is the c-vector corresponding to vertex .
We conclude this section by presenting yet another equivalent description of the oriented exchange graph of , which uses g-vectors and g-matrices (see [15]). The definition of a g-vector (resp., g-matrix ) was originally made for each cluster variable (resp., seed) of a cluster algebra with principal coefficients where is the rank of the cluster algebra. The g-vectors of provide a -grading of the cluster variables by [15, Proposition 6.1].
In this paper, we define g-matrices and g-vectors using quiver mutation. We let the identity matrix be the g-matrix associated to . The rows vectors of are the g-vectors of where corresponds to mutable vertex in . Here is the th standard basis vector. Each will have an associated g-matrix whose row vectors are its g-vectors. As is the case with c-matrices, the g-matrices are only defined up to row permutations and thus can be regarded simply as sets of g-vectors.
We produce the g-matrix associated to by defining a mutation operation on g-matrices. Let be the g-matrix of whose g-vectors are . Then the g-matrix of , denoted , is defined to the be matrix whose row vectors are where
[TABLE]
We let denote the set of g-matrices of . In Figure 1, we show the g-matrices of . Perhaps surprisingly, the c-matrices and g-matrices of a quiver can be obtained from each other by a simple bijection, as the following theorem shows.
Theorem 2.5**.**
([23, Theorem 1.2],[7, Theorem 4.17]) The map
[TABLE]
sending a c-matrix or g-matrix to its inverse transpose is a bijection, and this map commutes with mutation of c-matrices and g-matrices of .
In particular, Theorem 2.5 shows that one can transfer the edge orientation of to . Thus, one can regard as a directed graph whose vertices are g-matrices and whose directed edges are exactly those of the form where is green in and . In Figure 1, we show with .
3. Scattering diagrams
In [17], scattering diagrams are shown to be extremely useful in modeling cluster algebras. They are used by Gross, Hacking, Keel, and Kontsevich to prove many long-standing conjectures on cluster algebras including the Positivity Conjecture and sign coherence of c-vectors and g-vectors. The first (resp., second) conjecture was only known for skew-symmetric cluster algebras (i.e., cluster algebras defined by quivers) from [20, Theorem 1.1] (resp., [11, Theorem 1.7]). We follow the treatment of scattering diagrams defined by quivers as presented in [22]. We recommend that the reader refer to [22] for more details.
In this section, we work with a fixed quiver , and we will associate to a particular scattering diagram whose existence follows from [17, Theorem 1.13]. This scattering diagram, which we will refer to as the consistent scattering diagram for and will denote by , will encode all of the maximal green sequences of . We begin by reviewing scattering diagram terminology. Our goal in this section is to obtain a restatement of Greg Muller’s theorem [22, Theorem 9] (see Theorem 3.3).
Define to be the formal powers series ring with coefficients coming from the Laurent polynomial ring in the variables . For (resp., ), we write (resp., ). For each , define a formal elementary transformation, denoted , by
[TABLE]
where , is the greatest common divisor of the coordinates of m, and is the exchange matrix of . The formal elementary transformation is an automorphism of with inverse given by
[TABLE]
As shown in [22], elementary transformations, in general, do not commute. For example, by [22, Prop. 4.1.1] if for , then
Define a wall to be a pair where
- •
is nonzero and
- •
is a convex polyhedral cone that spans
Given a wall , the fact that implies that Also, defines two half-spaces and , which we will refer to as the green side and red side of , respectively. We define a scattering diagram in , denoted , to be a multiset of walls222As is noted in [22], we allow multiple copies of the same wall. where each has and . We define a chamber of to be a path-connected component of .
Scattering diagrams can be used to understand diagrams of elementary transformations, as any wall is naturally associated with the elementary transformation . Given a scattering diagram in we say that a smooth path is finite transverse if
- •
and for any wall in ,
- •
for any wall of , may only intersect transversally, and in this case, must not intersect the boundary of ,
- •
intersects at most finitely many distinct walls of , and does not intersect two walls that span distinct hyperplanes at points where the two intersect (see path in Figure 2).
Let be the sequence of walls crossed by a finite transverse path . Then determines the path-ordered product of elementary transformations given by
[TABLE]
where (resp., ) if crosses from its green side to its red side (resp., from its red side to its green side).
We say that a scattering diagram with finitely many walls is consistent if any path-ordered product defined by finite transverse loop is the trivial automorphism of (see loop in Figure 2). We say that two scattering diagrams and with finitely many walls are equivalent if each smooth path that is finite transverse in and defines the same path-ordered product in as it does in . For brevity, we do not define consistency or equivalence for scattering diagrams with infinitely many walls.
Theorem 3.1**.**
[17, Theorems 1.13 and 1.28] For each quiver , there is a consistent scattering diagram, denoted , unique up to equivalence, such that
- •
for each , there is a wall of the form , and
- •
every other wall in satisfies .
The scattering diagram has two distinguished chambers, namely, the all-positive chamber and the all-negative chamber. We say that a chamber in is reachable if there exists a finite transverse path from to .
Theorem 3.2**.**
[17, Lemmas 2.9 and 5.12] Every reachable chamber of is of the form
[TABLE]
for some g-matrix . The inward-pointing normal vectors of the same reachable chamber are the row vectors of the c-matrix . These correspondences induce bijections between the reachable chambers of and elements of and elements of . In particular, the positive c-vectors of are exactly the vectors m that appear in a wall of that is incident to a reachable chamber.
A useful consequence of Theorem 3.2 is that there is a bijection between maximal green sequences of and finite transverse paths in from the all-positive chamber to the all-negative chamber with the property that whenever crosses a wall of it does so from the green side to the red side. The following is essentially proven in [22].
Theorem 3.3**.**
Let be any quiver and any full subquiver of . There is a map defined by sending to the maximal green sequence where is the unique longest subsequence of where each satisfies if
Proof.
The scattering diagram can be obtained from by considering the subset of walls of given by . Here is the canonical projection and is the coordinate inclusion. More precisely, by [22, Theorem 33], the scattering diagram is obtained from by removing all of its walls where has any nonzero entry with Now by Theorem 3.2, any wall of that is incident to a reachable chamber of has the property that is a positive c-vector of where if .
Next, let and let be a finite transverse path that defines i. Now consider the path . By the proof of [22, Theorem 9], we know that is a finite transverse path in . Thus defines a maximal green sequence , and only crosses walls of that are incident to reachable chambers of . Now by the previous paragraph, we conclude that is the unique longest subsequence of where each satisfies if ∎
As a corollary we obtain a result on the lengths of maximal green sequences that characterizes acyclic quivers.
Corollary 3.4**.**
A quiver is acyclic if and only if admits a maximal green sequence of length .
Proof.
The forward direction follows from [5, Lemma 2.20]. To show the backward direction we suppose that there exists a quiver with oriented cycles that admits a maximal green sequence i of length . Note that in this case every vertex must be mutated exactly once. Let be a subquiver of consisting of a single vertex, and observe that admits a unique maximal green sequence of length 1. Theorem 3.3 implies that consists of vectors of the form , where the only nonzero entry of equals 1 and occurs at position .
Since contains oriented cycles there exists a full subquiver of such that no vertex of is a source in . By Theorem 3.3 the sequence is a maximal green sequence for of length . Suppose begins with a mutation at vertex . By construction, there exists an arrow ending at in . Then in the c-vector at vertex becomes strictly positive at position . Furthermore, since we will not mutate at again while performing it follows that prior to a mutation at the corresponding c-vector will remain strictly positive at position . However, this contradicts the description of according to Theorem 3.3. Therefore, we can conclude that if contains oriented cycles then it cannot admit a maximal green sequence of length . ∎
4. Direct sums of quivers
In this section, we recall the definition of a direct sum of quivers following [16]. In [16], it was shown that, under certain restrictions, if a quiver can be written as a direct sum of quivers where each summand has a maximal green sequence, then the maximal green sequences of the summands can be concatenated in some way to give a maximal green sequence for . We show that, under those same restrictions, if is a direct sum of quivers, then the minimal length of a maximal green sequence of is the sum of the lengths of minimal length maximal green sequences of its summands. Throughout this section, we let and be finite ice quivers with and nonfrozen vertices, respectively. Furthermore, we assume and
Definition 4.1**.**
Let denote a -tuple of elements from and a -tuple of elements from . (By convention, we assume that the -tuple is ordered so that if unless stated otherwise.) Additionally, let and . We define the direct sum of and , denoted , to be the ice quiver with vertices
[TABLE]
and arrows
[TABLE]
Observe that we have the identification of ice quivers
[TABLE]
where in both cases.
Example 4.2**.**
Let denote the quiver shown in Figure 3. Define to be the full subquiver of on the vertices , to be the full subquiver of on the vertices , and to be the full subquiver of on the vertex 5. Note that and are each irreducible. Then
[TABLE]
where . On the other hand, we could write
[TABLE]
where . Additionally, note that
[TABLE]
where the last equality does not hold because is not defined as is not a vertex of . This shows that the direct sum of two quivers, in the sense of this paper, is not associative.
Theorem 4.3**.**
[16, Prop. 3.14] If and and is a direct sum of quivers where for any , then
Proposition 4.4**.**
Assume and and is a direct sum of quivers where for any . If for is the length of a minimal length maximal green sequence of , then is the length of a minimal length maximal green sequence of . Furthermore, the maximal green sequence of achieves this length.
Proof.
Assume that for is of minimal length. By Theorem 4.3, the sequence and has length .
We show that has no maximal green sequences of length less than . Let By Theorem 3.3, the full subquivers and have maximal green sequences and such that (resp., ) is the unique longest subsequence of where each c-vector in (resp., ) satisfies if (resp., ). This implies that and . Now since and have no common vertices, we know that ∎
5. Quivers with branches of mutation type
In this section, we show that if a quiver consists of a quiver and mutation type subquivers branching out from a given set of vertices of , then we can essentially ignore these mutation type subquivers while computing (minimal length) maximal green sequences for . If these branches are made up of acyclic quivers, then can be realized as a direct sum of quivers, and we already discussed in Section 4 how to construct maximal green sequences for such . However, if the branches contain 3-cycles then we cannot use the same approach. Therefore, the results presented in this section enable us to decompose even further by disregarding such subquivers of mutation type and focusing only on .
We begin with the following lemma, which is a reformulation of [10, Lemma 2.14].
Lemma 5.1**.**
Let for some quiver , such that the following conditions hold.
- •
is composed of two full subquivers and the framed quiver ;
- •
and are connected by a single arrow , where and .
Let be a maximal green sequence for , then
- •
is composed of two full subquivers and ;
- •
and are connected by a single arrow , where is a unique vertex of with an arrow starting at the frozen vertex .
[TABLE]
Lemma 5.2**.**
Consider a quiver composed of an arbitrary quiver and a 3-cycle joined together at a vertex as shown below.
[TABLE]
Let be a maximal green sequence for , then
[TABLE]
is a maximal green sequence for .
Proof.
We begin by mutating at vertex . The resulting quiver is shown below and we note that remains a full subquiver of connected to the remaining vertices of by a single arrow.
[TABLE]
We can see that satisfies conditions of Lemma 5.1 with being the full subquiver of on vertices . Thus, by Lemma 5.1 the quiver consists of two full subquivers and connected by a single arrow where is the unique vertex of with an arrow in . Since, is a maximal green sequence for if follows that all vertices of are red in . The only remaining green vertex in is . Now, we perform which makes the vertex green, and then we perform . The resulting quivers are depicted below, where we let .
[TABLE]
We can see that in the final quiver every vertex is red. This shows that is a maximal green sequence for . ∎
Repeated applications of the lemma will allow us to construct minimal length maximal green sequences for quivers defined below.
Definition 5.3**.**
Given a quiver let be a quiver composed of full connected subquivers , such that all of the following conditions hold.
- •
.
- •
Q_{0}^{i}\cap Q_{0}^{j}=\Big{\{}\begin{array}[]{ll}\{x_{i}\}&\text{if }x_{i}=x_{j}\\ \varnothing&\text{otherwise}\end{array}.
- •
for every arrow in , whenever one of the endpoints belongs to then the other endpoint belongs to .
- •
for every the quiver is of mutation type .
[TABLE]
Note that the vertices are not necessarily distinct. With this notation consider the following results which show that in order to find a (minimal length) maximal green sequence for it suffices to find such sequence for its subquiver . In other words, one can disregard the attached mutation type quivers.
Theorem 5.4**.**
Let be a maximal green sequence for . Then the quiver admits a maximal green sequence of length
[TABLE]
where is the minimal length of a maximal green sequence for .
Proof.
Starting with we will gradually construct by adding either a 3-cycle or an arrow, that does not lies in any 3-cycle. At each step we will produce a maximal green sequence for the resulting quiver building on the sequence obtained in the previous step.
Let be a vertex of that does not belong to such that it is connected to a vertex in by an arrow . By definition of there exists a unique integer such that and the arrow is unique, because is of mutation type . Moreover, the endpoint of that lies in equals , where . Now we consider four possibilities depending on . If this arrow does not lie in a 3-cycle in then we have case (i) or (ii), and if lies in a 3-cycle then we have case (iii) or (iv) as depicted below.
[TABLE]
Now, let be a full subquiver of on vertices in cases (i) and (ii) or on vertices in cases (iii) and (iv). By Theorem 4.3 we have that or is a maximal green sequence for in case (i) or (ii) respectively. On the other hand, in case (iii) or (iv) the quiver satisfies conditions of Lemma 5.2, so we have that or is a maximal green sequence for respectively.
Next, we proceed in the same way with as we did with . That is, we pick a vertex that is connected to by a (unique) arrow . Analogously, we define , depending on the configuration around , and the corresponding maximal green sequence for this quiver. Continuing in this way we obtain a maximal green sequence for .
Observe, that at each step we either add a single vertex or a 3-cycle, that is two vertices that lie in the same 3-cycle. At the same time, we increase the length of the maximal green sequence by one or three respectively. Thus, the length of the resulting maximal green sequence for is
[TABLE]
where is the number of vertices in and is the number of 3-cycles in . By Theorem 2.3 we have
[TABLE]
which completes the proof of the theorem. ∎
Remark 5.5**.**
One can use the construction in the proof of Theorem 5.4 to build a concrete maximal green sequence for .
We obtain the following corollaries.
Corollary 5.6**.**
The quiver admits a maximal green sequence if and only if admits a maximal green sequence.
Proof.
The forward direction follows from Theorem 3.3 and the backward direction follows from Theorem 5.4. ∎
Corollary 5.7**.**
Let be a minimal length maximal green sequence for . Then the minimal length of a maximal green sequence for is
[TABLE]
where is the minimal length of a maximal green sequence for .
Proof.
Let be a minimal length maximal green sequence for . Note, that by assumption admits a maximal green sequence, so Coroallary 5.6 implies that exists. Since, is of minimal length we have
[TABLE]
by Theorem 5.4.
To show the reverse inequality consider a maximal green sequence for . Let be the full subquiver of on vertices . According to Theorem 3.3 the sequence induces and maximal green sequences for and respectively. By definition of and given in the theorem we have
[TABLE]
because and have a single vertex in common and together they make up .
Now we apply the same argument to the quiver and its corresponding sequence . Let be the full subquiver of on vertices . Then by the same reasoning as above we deduce that
[TABLE]
where and are the resulting maximal green sequences for and respectively.
Continuing in this way we see that
[TABLE]
where is a maximal green sequence for for and with the corresponding sequence . Since was arbitrary this shows the reverse inequality and completes the proof. ∎
6. Quivers of mutation type
We now focus on the minimal length maximal green sequences of quivers of mutation type . By definition, this is the family of quivers belonging to the mutation class of an orientation of a Dynkin diagram where . We recall the description of this mutation class, which was discovered by Vatne in [26, Theorem 3.1]. There are four families of quivers that make up this mutation class. Vatne refers to these as Type I, Type II, Type III, and Type IV quivers. Throughout this section, we will assume that is a connected quiver. Additionally, we say a vertex of some quiver is a connecting vertex if it has degree at most 2, and if it has degree 2, it belongs to a 3-cycle of the same quiver.
Type I: A quiver is of Type I (see Figure 5) if and only if has the following properties:
- •
it has a full subquiver of the form (we use the notation to indicate that there exists an arrow in connecting and and that arrow can be oriented in either direction) where vertices and each have degree 1 in ,
- •
the full subquiver of on the vertices is of mutation type , and
- •
the vertex is a connecting vertex of .
Type II: A quiver is of Type II (see Figure 5) if and only if has the following properties:
- •
it has a full subquiver on the vertices of the form shown in Figure 5,
- •
the subquiver obtained from by removing vertices and and the arrow consists of two connected components and , each of which is of mutation type , and
- •
the vertex (resp., ) is a connecting vertex of (resp., ).
Type III: A quiver is of Type III (see Figure 7) if and only if has the following properties:
- •
it has a full subquiver on the vertices that is an oriented 4-cycle as shown in Figure 7,
- •
the full subquiver of on the vertices consists of two connected components and , each of which is of mutation type , and
- •
the vertex (resp., ) is a connecting vertex of (resp., ).
Type IV: A quiver is of Type IV (see Figure 7) if and only if has the following properties:
- •
it has a full subquiver that is an oriented -cycle where , , and ,
- •
for each arrow , there may be a vertex that is in a 3-cycle , which is a full subquiver of , but there are no other vertices in that are connected to vertices of ,
- •
the full subquiver obtained from by removing the vertices and arrows of the subquiver consists of the quivers some of which may be empty quivers, and where each quiver is of mutation type and has as a connecting vertex.
Next we present the main theorem of this section that describes the minimal length of maximal green sequences for quivers of type depending on the particular type. The proof for quivers of Type IV uses triangulations of surfaces and is presented in the following sections.
Theorem 6.1**.**
Let be a quiver of mutation type . Let denote the length of a minimal length maximal green sequence of .
- i)
If is of Type I, then .
- ii)
If is of Type II, then
- iii)
If is of Type III, then
- iv)
If is of Type IV, then .
Proof.
i) Letting (resp., ) be the full subquiver of on the vertex (resp., ), it is clear that these have minimal length maximal green sequences of length 1.
Let be the full subquiver of of mutation type described in the definition of Type I quivers. By Theorem 2.3, there exists a minimal length maximal green sequence of with length . By the description of Type I quivers, we have that . Now the result follows from Corollary 5.7 (or from Proposition 4.4).
ii) One checks that the sequence is a maximal green sequence of the full subquiver of on the vertices and . Since this full subquiver of is not acyclic, Corollary 3.4 implies that it does not have maximal green sequences of length 4. Thus is a minimal length maximal green sequence.
Next, since the quivers and are of mutation type , Theorem 2.3 implies that there exist minimal length maximal green sequences and with lengths and , respectively. Now by Corollary 5.7, we know that the minimal length of a maximal green sequence of has the desired length.
iii) One checks that is a minimal length maximal green sequence of the full subquiver of on the vertices and . As in Case ii), we apply Theorem 2.3 to the quivers and , and the desired result then follows from Corollary 5.7.
iv) Let be the nonempty mutation type quivers obtained from by removing the vertices , and let denote the corresonding connecting vertices. By Theorem 8.12 and Theorem 8.13, we know that the full subquiver of on the vertices has a minimal length maximal green sequence of length By Theorem 2.3, each quiver has a minimal length maximal green sequence of length . Now observe that . The result thus follows from Corollary 5.7. ∎
7. Quivers defined by triangulated surfaces
To understand the minimal length maximal green sequences of quivers of Type IV, it will be useful to think of such quivers as signed adjacency quivers of tagged triangulations of a once-punctured disk. In this section, we review the basic notions of quivers arising from triangulated surfaces, following [12].
7.1. Signed adjacency quivers
Let S denote an oriented Riemann surface that may or may not have a boundary and let be a finite subset of S where we require that for each component B of we have We call the elements of M marked points, we call the elements of punctures, and we call the pair a marked surface. We require that is not one of the following degenerate marked surfaces: a sphere with one, two, or three punctures; a disc with one, two, or three marked points on the boundary; or a punctured disc with one marked point on the boundary. For the remainder of Section 7, we will work with a fixed marked surface
We define an arc on S to be a curve in S such that
- •
its endpoints are marked points;
- •
does not intersect itself, except that its endpoints may coincide;
- •
except for the endpoints, is disjoint from M and from the boundary of S;
- •
does not cut out an unpunctured monogon or and unpunctured digon. (In other words, is not contractible into M or onto the boundary of S.)
An arc is considered up to isotopy relative to the endpoints of . We say two arcs and on S are compatible if they are isotopic relative to their endpoints to curves that are nonintersecting except possibly at their endpoints. An ideal triangulation of is defined to be a maximal collection of pairwise compatible arcs, denoted T.
One can also move between different triangulations of a given marked surface Let be an arc in a triangulation T that has no self-folded triangles (e.g., the region of S bounded by and in Figure 9 is an example of a self-folded triangle). Define the flip of arc to be the unique arc that produces a triangulation of given by (see Figure 8).
If is a marked surface where M contains punctures, there are triangulations of S that contain self-folded triangles. We refer to the arc (resp., ) shown in the self-folded triangle in Figure 9 as a loop (resp., a radius). As the flip of a radius of a self-folded triangle is not defined, Fomin, Shapiro, and Thurston introduced tagged arcs, a generalization of arcs, in order to develop such a notion.
A tagged arc is obtained from an arc that does not cut out a once-punctured monogon and “tagging” its ends either as plain or notched so that:
- •
an end of lying on the boundary of S is tagged plain; and
- •
both ends of a loop have the same tagging.
We use the symbol to indicate that an end of an arc is notched. We say two tagged arcs and are compatible if the following hold:
- •
Their underlying arcs and are the same, and the tagged arcs and have the same tagging at exactly one endpoint.
- •
Their underlying arcs and are distinct and compatible, and any common endpoints of and have the same tagging.
A tagged triangulation of is a maximal collection of pairwise compatible tagged arcs. It follows from the construction that any arc in a tagged triangulation can be flipped. In particular, the flip of in Figure 9 results in a new arc with one notched endpoint at the puncture and the other endpoint on the boundary where the arcs and meet. This enables us to define the exchange graph of a tagged triangulation T whose vertices are tagged triangulations and two vertices are connected by an edge if the corresponding triangulations are related by a flip.
It will be useful to keep track of the region of a triangulation where a flip of a tagged arc takes place. We refer to this region as an ideal quadrilateral. Define the ideal quadrilateral of a tagged arc in a tagged triangulation T to be the triangulated surface contained in ) that is
- •
an unpunctured disk with four marked points that consists of two triangles of T that have as a side if is not a nontrivial tagged arc in a once-punctured digon (see Figure 8) or
- •
the once-punctured digon containing and the triangulation of it determined by T (see Figure 9).
Each triangulation T of S defines a signed adjacency quiver by associating vertices to arcs and arrows based on oriented adjacencies (see Figure 10). More precisely, given an ideal triangulation T consider a map on the set of arcs defined as follows. If is a radius of a self-folded triangle then let be the corresponding loop, otherwise let . Then the quiver consists of vertices for every and arrows for every non self-folded triangle with sides and such that follows in the clockwise order. Finally, we remove a maximal collection of oriented 2-cycles from to produce a 2-acyclic quiver. Note that we defined for an ideal triangulation T, however every tagged triangulation can be transformed into an ideal triangulation by forgetting the tagging of each arc and introducing self-folded triangles around puncture whenever two underlying arcs coincide and have different tagging at . More precisely, if there are two arcs attached to with different tagging, then replace the notched arc in by a loop of a self-folded triangle enclosing in . In this way, the definition of can be extended to a tagged triangulation T by setting .
The following theorem shows that flips are compatible with mutations of the associated quiver so that .
Theorem 7.1**.**
Given a tagged triangulation and an arc , let be obtained from after the flip of , then
[TABLE]
7.2. Shear coordinates
Shear coordinates provide a way to orient and establish a bijection with . This allows one to translate the notion of maximal green sequences into the geometric setting.
A fixed orientation of a surface S induces an orientation on each component of such that the surface S lies to the right of every component. Let be a tagged arc in S. The elementary lamination is a curve that runs along within a small neighborhood of it. If has an endpoint on the boundary , then begins at a point on located near in the direction of , and proceeds along . If has an endpoint at a puncture , then spirals into : clockwise if is notched at , and counterclockwise if it is tagged plain. For example see Figure 11. A set of curves is a lamination if it consists of a set of elementary laminations arising from some pairwise compatible tagged arcs.
Definition 7.2**.**
[13, Definition 13.3] Let be a lamination, and let be an ideal triangulation. For each arc , that is not a radius of a self-folded triangle, the corresponding shear coordinate of with respect to , denoted by , is defined as a sum of contributions from all intersections of curves in with . Specifically, such an intersection contributes +1 (resp., ) to if the corresponding segment of a curve in cuts through the quadrilateral surrounding as shown in Figure 12 on the left (resp., right).
Next, we describe how to extend the definition of shear coordinates to tagged triangulations.
Let be a tagged triangulation. The shear coordinates for and some lamination are uniquely defined by the following rules:
- (i)
Suppose that tagged triangulations and coincide except that at a particular puncture , the tags of the arcs in are all different from the tags of their counterparts in . Suppose that laminations and coincide except that each curve in that spirals into has been replaced in by a curve that spirals in the opposite direction. Then for each arc and its counterpart .
- (ii)
By performing tag-changing transformations as above, we can convert any tagged triangulation into a tagged triangulation that does not contain any notched arcs except possibly inside once-punctured digons. Let denote the ideal triangulation that is represented by such as defined earlier. Let be an arc in that is not a radius of some self-folded triangle, and let be the corresponding arc in . Then, for a lamination , define by applying Definition 7.2 to the arc viewed inside the triangulation .
Note that if is a radius of a self-folded triangle in enveloping a puncture , then we can first apply the tag-changing transformation (i) to at , and then use the rule (ii) to determine the shear coordinate in question.
The following theorem establishes a relationship between shear coordinates and c-vectors.
Theorem 7.3**.**
Consider a triangulation with its associated quiver . Let and let be the corresponding triangulation. Then the c-matrix has entries, indexed by the arcs and , as follows
[TABLE]
where is the elementary lamination corresponding to .
With the above notation we say that an arc is green (resp., red) if (resp., ) for all . Note that the color of every arc in agrees with the color of the associated vertex in . This enables us to define an oriented exchange graph of T such that .
Next, we define an automorphism on the set of arcs in a given .
Definition 7.4**.**
Recall, that an orientation of a surface S induces an orientation on each component of such that the surface S lies to the right of every component. Given a marked point let be the next marked point in along , and let be the corresponding boundary segment connecting and . If is a puncture, then let and the segment equal a single point . Given an arc in S with endpoints let be the arc with endpoints that follows along the segments , and then in order. Furthermore, if is a puncture for then the tagging of is different from that of at . For example see Figure 13.
The following theorem is an consequence of [6, Theorem 3.8] and the remarks in the introduction of [6]. It will be useful for our construction of minimal length maximal green sequences.
Theorem 7.5**.**
Let T be a triangulation of a marked surface . If and is the corresponding sequence of flips, then .
8. Type IV quivers
Let be a quiver of Type IV where each of its subquivers with either consists of a single vertex or is the empty quiver. Any such quiver is the signed adjacency quiver of a triangulation of the once-punctured disk where and where denotes the unique puncture in the marked surface. A triangulation T giving rise to such a quiver is defined by the following properties (see Figure 14):
- •
if is not connected to , it belongs to a triangle whose other two sides are boundary arcs and
- •
if is connected to , it is plain at .
The triangulation T is the unique triangulation satisfying , up to the action of the universal tagged rotation on the once-punctured disk and up to the action of simultaneously changing the tagging of all ends of arcs connected to . It immediately follows, that the vertices of the quivers with are in bijection with the triangles of T containing two boundary arcs. We will construct a minimal length maximal green sequence of in terms of the triangulation T satisfying just described.
8.1. Lower bound on lengths of maximal green sequences
In this section we establish a lower bound on the lengths of maximal green sequences for quivers as defined above. Without loss of generality we give the same labels to the vertices of and the corresponding arcs of T. Given a vertex of we write , where denotes the triangulation corresponding to the flip of the arc . Throughout this section we fix a maximal green sequence for . It follows from Theorem 7.5 that . Given an arc let denote the same arc with a different tagging if possible, otherwise let .
Let , then we define a triangulation for and two sets
[TABLE]
In particular, we see that for a maximal green sequence . Also, the sets and differ only by a single arc, hence the length of is equal to the size of .
Definition 8.1**.**
Consider a map on the set of arcs
[TABLE]
defined as follows. Given an arc , there exists some such that . Let be the largest integer such that . Observe that exists because . Then is defined as the unique arc belonging to . In other words, is the arc obtained as a result of flipping when performing the mutation sequence . To simplify the notation, we omit the subscript in .
Similarly, given a nonempty set of arcs such that for some . We define to be where is the first arc in to be flipped along . More precisely, is the unique arc in where is the largest integer such that . We will often consider sets of arcs that define a triangle in , and we will refer to these sets as triangles.
The following lemma says that once an arc that appears along is flipped then it cannot appear again.
Lemma 8.2**.**
If , then .
Proof.
Given a once punctured polygon there exists the corresponding cluster category of type such that indecomposable objects of are in bijection with tagged arcs in . Furthermore, triangulations of the surface are in bijection with cluster-tilting objects in , [25]. Thus, given a triangulation of the surface we obtain the associated cluster-tilted algebra , and it follows from [1] that cluster-tilting objects in are in bijection with support -tilting modules of . Combining the two results we obtain a correspondence between triangulations of and support -tilting modules of . Moreover, this correspondence behaves well under mutations, so that a maximal green sequence for yields a maximal chain of support -tilting modules in .
Now to prove the lemma, it suffices to show that an indecomposable -rigid module over an algebra does not appear twice along a maximal chain of support -tilting modules
[TABLE]
If belongs to for some , and there is a negative mutation of at resulting in the new support -tilting module , then by definition of mutation it follows that does not belong to . Since in a maximal green sequence we have that for all , we deduce that does not belong to for any . Therefore, cannot appear again, so the corresponding arc of cannot appear more then once along a maximal green sequence. ∎
Recall that we are interested in quivers of Type IV, where each is either empty or a single vertex. The corresponding triangulation T consists of a set of arcs that are attached to the puncture , and a set of arcs that are not attached to and lie in a triangle that contains two boundary segments. Here we consider the subscripts in the notation modulo . Moreover, we assume that all arcs in are tagged plain.
Thus, and let . Hence, and . Let
[TABLE]
and .
Definition 8.3**.**
Let be a collection of arcs such that
- •
- •
the arcs of are not attached to
- •
whenever is attached to and tagged plain.
Lemma 8.4**.**
.
Proof.
Observe that for any maximal green sequence for there exists an arc such that is the first tagged arc to appear along the mutation sequence . That is, and the set does not contain any tagged arcs. Moreover, there are precisely two plain tagged arcs attached to the puncture in the triangulation . One of these arcs is , and call the other arc . This implies that all arcs in need to be flipped prior to a flip in .
Given we make the following definition.
[TABLE]
We claim that when defined . First, observe that by construction whenever then is not attached to the puncture. Suppose such that . By definition is tagged plain and attached to while is not attached to , which implies that .
Now, suppose that such that and and we want to show that . Since is not attached to the puncture and we conclude that . Hence, intersects some arc in . However, the particular structure of the triangulation T implies that there exists a unique arc such that intersects , see Figure 15. Therefore, we have that , meaning is flipped to . This implies, that there exists a triangulation where such that .
Next we show that if then . Note that is tagged plain, attached to , and does not belong to T as it intersects . Moreover, since then cannot be a notched arc attached to the puncture. Hence, is not attached to the puncture. Finally, since does not intersect any arc in , it follows that its flip does not belong to . This implies that . It also shows the claim that if is defined. Moreover, it is easy to see that if and only if .
Let
[TABLE]
and by above we know that . Since, and is defined for every with at most two exceptions, we have .
We claim that which in turn implies that and completes the proof of the lemma. First, we observe that is not defined only if and . From Figure 15 we can see that intersects , so in particular does not belong to .
If is defined for all arcs of , then the claim holds. Suppose is the unique arc such that is not defined. Then since equals or and it does not belong to . By assumption, is defined for every arc in and , since . This shows the claim in the case when is not defined for exactly one arc in . A similar argument shows that the claim holds whenever there are precisely two arcs such that are not defined. Therefore, we obtain . ∎
Remark 8.5**.**
It follows from the definition of that for every such that , the cardinality of increases by 1 unless and for some .
Definition 8.6**.**
Define to be any subset of of cardinality . In particular, we have the following inclusions.
[TABLE]
Remark 8.7**.**
By Remark 8.5, the set is nonempty if there exists such that unless and for some .
We also define
[TABLE]
With this notation we can write as a disjoint union of three sets.
[TABLE]
To find a lower bound on the cardinality of we need to find a lower bound on the cardinality of , since we have that and . We achieve this by first constructing a map in Lemma 8.8, and then by modifying it in Lemma 8.11 to make it injective.
Lemma 8.8**.**
For every there exists an arc .
Proof.
For every there exists such that intersects . Based on the properties of we proceed to define the corresponding . The flowchart in Figure 21 illustrates various cases explained in detail below.
First we observe that is either a notched arc attached to the puncture or it is not attached to the puncture. Let be the unique integer such that .
Suppose is not attached to the puncture. Then there exists a triangle consisting of and as shown in Figure 16(a). If neither nor are boundary segments, then , and we define . In other words , where is the next arc of that is flipped along . Observe that is indeed an element of , because by construction , and since neither nor are boundary segments, it is easy to see that .
Now we consider what happens if or are boundary segments. Since is not attached to the puncture, there exists another triangle in consisting of arcs and as shown in Figure 16(a). Suppose is a boundary segment then does not belong to the original triangulation T. Indeed, by assumption which implies that , and in this case and cross. Moreover, since is tagged plain it does not belong to and since it is attached to the puncture it does not belong to . Hence, and we define . On the other hand, if is not a boundary segment but is a boundary segment then does not belong to the original triangulation T, because it crosses . So by similar reasoning as above , and we define .
Suppose is a notched arc attached to the puncture. If , then let . Again, in this case , because does not contain arcs attached to the puncture, so .
If , then in there are either two notched arcs attached to the puncture, or there is one notched arc and one tagged plain arc attached to the puncture, see Figure 16 parts (b) and (c) respectively. In the first case, since intersects it follows that . Also, since is attached to the puncture it does not belong to , so it must belong to . Therefore, we let . In the second case, we let . If then we are in the situation of case one, see Figure 16(b) and replace by . Hence, by the same reasoning as in case one . If equals or , then these arcs are not attached to the puncture, so they cannot belong to . It is also easy to see that they cannot lie in as they enclose the puncture. Hence, in this situation .
Thus, in all scenarios we provided a definition of and showed that it is well defined, which finishes the proof of the lemma. ∎
Remark 8.9**.**
Observe that no as defined in the lemma above belongs to . Recall that consists of arcs that are not attached to the puncture, but come from flips of plain tagged arcs attached to the puncture. Moreover, the arcs of appear in the surface prior to the appearance of any notched arcs in the surface.
Example 8.10**.**
Consider a triangulation T depicted in Figure 17 on the left. By definition
[TABLE]
The corresponding quiver admits a maximal green sequence
[TABLE]
and we see that
[TABLE]
Moreover,
[TABLE]
and it is depicted in Figure 17 on the right. We begin by computing . The triangulation is shown in Figure 17 in the middle, and we see that . Hence, . Observe that and , which means is not defined. However, the function is defined for the rest of the arcs in , and
[TABLE]
Let us compute and . To do so we consider the corresponding and . Observe that and will be flipped first among the rest of the arcs of along . This implies that . On the other hand consists of only two arcs, as is a boundary segment. Therefore, by definition , and in this example .
Lemma 8.11**.**
For every there exists a unique arc .
Proof.
The proof of Lemma 8.8 provides a construction of for every . Recall that the definition of can also be seen in the flowchart in Figure 21. However, it can happen that for some pair of distinct arcs and . There are three possible cases, which we discuss in detail below.
Case i. If , where arrises in Case 2 and arrises in Case 3 in the flowchart in Figure 21. Then from the flowchart, we can see that are boundary segments while is not a boundary segment. Moreover, share the same endpoint, which means , see Figure 20(a). Let , and it remains to define .
First, we claim that if is a notched arc attached to the puncture then is nonempty. Let be defined as in the proof of Lemma 8.4, that is there exists a triangulation containing plain tagged arcs that are attached to the puncture such that is the first notched arc to appear along . First we show that .
By Figure 20(a) there exists some such that , because and by construction neither of them can be flipped prior to the appearance of in the surface. Since these arcs must also belong to every triangulation for by Lemma 8.2. This shows, that there cannot be notched arcs in . Therefore, if equals , which by definition is the first notched arc to appear along , then . On the other hand, if is flipped after appears in the surface then there exists a triangulation where containing and . Therefore, because prior to a flip in , there are exactly two plain tagged arcs , and . This shows that .
Next, recall that , so by Remark 8.7 the set is nonempty unless and for some . Suppose and for some . By construction we know that , see Figure 20(a). Hence we see that and . However, implies that is a boundary segment, but this contradicts the assumption we made in the beginning of Case i. Therefore, we see that is nonempty and the claim holds.
Recall that we let , and we want to define . If is a notched arc attached to the puncture, by the claim above the set is nonempty, so we define as some arc in . Otherwise, we let provided that . To see that this definition makes sense, we need to justify that whenever is not attached to the puncture, then , or equivalently .
Note that because is not attached to the puncture and it does not intersect any arc in we conclude that . If then since , by definition of , we must have that . As before, we conclude that the corresponding is a boundary segment, which is a contradiction to the assumption we made in the beginning of Case i. Hence, . This shows that if is not attached to the puncture, then it belongs to .
Now it remains to justify that defined in this way does not coincide with some other as defined in Lemma 8.8 and the corresponding flowchart in Figure 21. Suppose , meaning is not attached to the puncture. Then cannot equal some arising in Cases 2-5, because is not attached to the puncture. Also, we note that whenever then , hence it is easy to see that if arises in Case 1 or Case 6 and is not attached to the puncture, then it does not equal a flip of a plain tagged arc. This implies that does not equal some other arising in any of the cases. If on the other hand then by Remark 8.11 it follows that does not equal some for any .
This completes Case i, and let us sum up that here we define and if is not attached to the puncture, and otherwise is some arc in .
Case ii(a). Suppose , where both and arise in Case 1 in the flowchart in Figure 21. Then from the flowchart we can see that are two triangles that do not contain boundary segments. For example see Figure 20(b). To solve this case we consider a more general situation where there are a number of such triangles that share arcs with other triangles.
Consider a triangle corresponding to an arc such that no arc of is a boundary segment. By definition we have and . If we also have that for some different triangle where again no arc of is a boundary segment, then let and
[TABLE]
Since , there exists some such that the boundary of lies in , where
[TABLE]
Observe that by construction , as none of the arcs in are attached to the puncture. Also, from Figure 20(b) it is easy to see that if is a notched arc attached to the puncture, then the corresponding plain tagged arc is an arc in . This implies that . On the other hand, if is not attached to the puncture then it cannot belong to , again see Figure 20(b). Hence, we conclude that , so .
If we also have that
[TABLE]
for some new triangle where no arc of is a boundary segment, then let
[TABLE]
Because there exists some such that , where
[TABLE]
By the same argument as before, we conclude that . Continue in this way to obtain a sequence of distinct triangles
[TABLE]
and the corresponding sequence of distinct arcs in
[TABLE]
where does not coincide with for some new triangle . For an example of such a construction when see Figure 18. Now, we define
[TABLE]
Next we justify what defined in Case ii(a) do not coincide with ’s defined in Case i or with other ’s defined in the flowchart coming from Cases 1-5. Recall that in Case ii(a) we already considered what happens if equals some other coming from Case 1. Note that defined in this way cannot be a plain tagged arc attached to the puncture, because it would equal some arc in , which would contradict Lemma 8.2. Hence, cannot equal some where arrises in Cases 2, 3. By construction cannot equal some that arrises in Cases 4. It also cannot equal in Case 5, because here the corresponding is a flip of some plain tagged arc attached to the puncture. Finally, does not equal other constructed in Case i, as these arcs are either plain tagged arcs attached to the puncture or flips of such arcs. Therefore, it remains to investigate the case when coincides with some arising in Case 6. We consider this situation next.
Case ii(b). If , where arrises in Case 6, then
[TABLE]
for some triangulated polygon . Recall that by Case ii(a) the polygon is made up of triangles
[TABLE]
where no triangle contains a boundary segment. Therefore, there exists a triangulation containing both and . Observe that cannot equal , because in accordance with Case 6 the arc cannot lie on the boundary of some triangulated polygon, see Figure 16(c). Without loss of generality let , see Figure 20(c). The same argument can be made if .
Since we assume that , it follows from the structure of that belongs to a triangle for some . Hence, , but from the construction it is clear that , see Figure 20(c). Also, in accordance with Case ii(a) we have , therefore it remains to define .
First, notice that by construction of the set consists of arcs of the form and where the corresponding triangles . Moreover, two arcs in share an endpoint if and only if there exists some ending in the point and the corresponding . In particular we see that any arc having as an endpint does not belong to as it crosses .
Next, we consider the set of arcs as shown in Figure 20(c). Observe that these arcs are in the analogous situation as the arcs in were prior to a flip in . Next, we claim that . Let and indeed it is easy to see that if equals or then . If then by definition of the set we have that . But also has the endpoint in common with located on the boundary of the disk. By properties of discussed in the paragraph above we see that . This shows the claim that .
In addition, we claim that for any does not coincide with any other or coming from Case i, and Cases 2-6. We can see that cannot be a plain tagged arc attached to the puncture, so it cannot equal any arising in Case 2 or Case 3. By construction cannot equal some arising in Case 5 because a notched arc would have to be present in the triangulation prior to a flip in . The same conclusion holds in Case 6, because we already mutated in . Observe that in Case 4, if , then we must have . Recall that by assumption , so in particular does not cross any arc of . However, if such that crosses , then we would also have that crosses , because and share the same endpoint. But this is a contradiction to . This shows that cannot equal some arising in Case 4. Finally, does not equal some arising in Case i, because such arcs are never notched and come from flips of plain tagged arcs attached to the puncture.
However, it can happen that there exists some such that for some triangulated polygon as defined in Case ii(a). If no such exists, then set . Otherwise, set , but then again we need to define .
We can proceed in the same manner as before with the new set of arcs . Continuing in this way, we obtain a sequence of sets
[TABLE]
and a sequence of triangulated polygons
[TABLE]
such that for all . Also, we define for and finally for any . Indeed this holds, because the sizes of the triangulated polygons must eventually decrease to zero as the endpoints of move closer together as increases. Hence, we define . This completes Case ii(b). Moreover, we managed to define ’s in Cases ii(a) and ii(b) such that if and only if .
We already justified that ’s as defined in the three cases above do not coincide with other ’s or ’s. Therefore, it remains to show that the scenarios discussed in this lemma are the only ones when for some pair of distinct arcs . Recall that above we investigated when Case 1 overlapped with Cases 1 and 6, and Case 2 overlapped with Case 3. In Case 1 the arc cannot be tagged plain and attached to the puncture, so it cannot overlap with Cases 2 and 3. Also, by definition this cannot equal for some , hence Cases 1 and 4 cannot overlap. By construction Case 1 and Case 5 cannot overlap, because as defined in Case 5 is a flip of some plain tagged arc attached to the puncture. It is easy to see that Case 2 cannot overlap with Cases 4, 5, 6, as none of these yield a plain tagged arc attached to the puncture. Also, Case 2 cannot overlap with another Case 2, because are defined uniquely by the corresponding . Similarly, Case 3 cannot overlap with Cases 3, 4, 5, 6.
Suppose arrises in Case 4, then it cannot equal another arising in Cases 4, because whenever then . Suppose that , where arrises in Case 4 and arrises in Case 5. Then by definition, see the flowchart in Figure 21, there exists a triangulation depicted in Figure 19, such that is a notched arc attached to the puncture and belonging to . However, and share an endpoint on the boundary of the disk, and since and cross, see the flowchart, it follows that and also cross. This implies that , which is a contradiction. Hence, Case 4 cannot overlap with Case 5. Also, Case 4 cannot overlap with Case 6, because the arcs arising in Case 6 are not attached to the puncture. Moreover, if where is defined in Case 6, then we obtain the same triangulation as in Case 5. Proceeding in the same manner as above, we also obtain a contradiction.
Again, we can see that Case 5 can be though of as a special situation of Case 6 when , but Case 5 cannot overlap with Case 5, because in this scenario there cannot be two notched arcs belonging to . Finally, by the same reasoning Case 6 cannot overlap with another Case 6. Therefore, considering various scenarios we can see that Cases i, ii(a), ii(b) are the only possibilities when definitions of ’s can overlap. In all other cases we let , which completes the proof of the lemma.
∎
Theorem 8.12**.**
Suppose is a quiver of Type IV, where each is either empty or a single vertex. The length of a maximal green sequence for is at least .
Proof.
By definition . Moreover, the three sets are disjoint, and , , and by the lemma above . Therefore, . ∎
8.2. Minimal length maximal green sequences
For convenience, throughout this section, we refer to all tagged triangulations (resp., tagged arcs) simply as triangulations (resp., arcs). We construct minimal length maximal green sequences of using the notion of a fan of arcs. A fan with respect to is a sequence of arcs in a given triangulation of where
- a)
each has as one of its endpoints and
- b)
the arc is immediately clockwise from about in the triangulation of .
A fan with respect to is complete if every arc of that is incident to appears in exactly once.
Given any sequence of arcs of a triangulation , we let denote the sequence of vertices of in the same order as the arcs of . In particular, any fan defines such a sequence of vertices of . Similarly, we let to denote corresponding sequence of flips of arcs in . If is the empty sequence of arcs, then and will also be empty sequences.
Each nonempty quiver , determines a complete fan of arcs in T as follows. Define
[TABLE]
where is calculated333We will henceforth perform all such calculations mod . mod and where we abuse notation and identify the vertices of with the arcs of T.
Define the following sequence of vertices of . If at least one of the quivers is the empty quiver or if none of the quivers is the empty quiver and there is an even number of quivers , we let
[TABLE]
where is a set of complete fans of arcs of T as defined in the previous paragraph, enumerated in any order, with the property that an arc of T belongs to at most one of these fans and the set of all arcs appearing in these fans is maximal with respect to this property.
On the other hand, if none of the quivers is the empty quiver and there is an odd number of quivers , we first fix an arc . Then we let
[TABLE]
where is a set of complete fans of arcs of , enumerated in any order, with the property that an arc of belongs to at most one of these fans and the set of all arcs appearing in these fans is maximal with respect to this property.
Next, we let
[TABLE]
where is a set of fans of arcs in with respect to , enumerated in any order, such that every arc of with an endpoint at belongs to one of the fans and where the set of arcs appearing in each fan is as large as possible.
Let denote the sequence of arcs in , enumerated in any order, where each arc appears in a triangle of with the following properties:
- •
one of its other arcs is from —one of the fans defining —and
- •
the arc is immediately counterclockwise from about their common endpoint444In other words, there is an arrow in the quiver whose source (resp., target) is the vertex of corresponding to (resp., )..
In addition, we require that each arc does not have as an endpoint. Let
[TABLE]
and let denote the sequence of flips along .
Let and where is any fan of arcs of with respect to where each arc appearing in is plain at and where the underlying set of arcs of is as large as possible.
The last component of our sequence is defined inductively. Let denote the sequence of arcs of , enumerated in any order, where each satisfies the following:
- •
the arc appears in a triangle of whose other two arcs are notched at , and
- •
the arc does not appear in a triangle of whose other two arcs are boundary arcs.
We define analogously. Assume that has been defined and now let denote the sequence of arcs of , enumerated in any order, where each satisfies the following:
- •
the arc appears in a triangle of whose other two arcs are notched at , and
- •
the arc does not appear in a triangle of whose other two arcs are boundary arcs.
Define and where is the empty sequence of arcs.
Theorem 8.13**.**
Let be a Type IV quiver where each of its subquivers with either consists of a single vertex or is the empty quiver. Then the sequence is a maximal green sequence.
Corollary 8.14**.**
The length of the maximal green sequence is where k_{\text{in}}=|\{\text{arcs of \textbf{T}p}\}|, k_{\text{out}}=|\{\text{arcs of \textbf{T}p}\}|, and
[TABLE]
Example 8.15**.**
Here we give an example of the maximal green sequence from Theorem 8.13 in the context of Figure 22. We have that , , and Notice that the arc obtained by flipping also satisfies , as desired.
In the proof of Theorem 8.13, we will need to consider how a specific part of a triangulation of is affected by flips. In particular, we will need to understand how a single flip affects the colors of arcs in that are close to in . We formulate what we mean by a specific part of a triangulation as follows.
Let be a triangulation of and let . We define to be the marked surface obtained from by taking the disjoint union of all ideal quadrilaterals in of arcs appearing in and then identifying any common triangles appearing in these ideal quadrilaterals. The resulting (possibly disconnected) surface clearly has a triangulation given by the arcs of . If we are working with a specific set of arcs , for convenience, we write to denote . Similarly, when we are considering a fan of arcs in , we write to denote where is the underlying set of arcs in . Note that for such a fan the surface will be connected.
The following facts, which we will use in the proof of Theorem 8.13, are straightforward to verify.
Lemma 8.16**.**
Let and be two subsets of arcs of a triangulation of , let be the quiver corresponding to , and let and be total orderings of the arcs in and , respectively.
If the intersection of and , regarding the two as subspaces of , is empty or consists only of boundary arcs of these two surfaces, then
[TABLE]
If an arc is neither an interior arc nor a boundary arc of , then the color of as an arc of is the same as its color as an arc of
Proof of Theorem 8.13.
We assume that at least one of the quivers is the empty quiver or that none of the quivers is the empty quiver and is even. The proof when each of the quivers consists of a single vertex and is odd is similar so we omit it. We prove that is a maximal green sequence by analyzing the effect each subsequence of has on the triangulation . In the course of the proof, we also justify why in the definition of the sequence for we can enumerate the subsequences defining it in any order. Similarly, we explain why the arcs in for may be flipped in any order.
We first consider the effect of on T. Let . One observes that the intersection of the surfaces and where is either empty or consists of a boundary arc of these surfaces. Therefore, by Lemma 8.16 , we can assume . Now the sequence can be expressed as the composition of a unique family of sequences of the form where and if each of the quivers is a single vertex (and, in this case, there is exactly one sequence in this family) or and are empty quivers. Given such a sequence , we show the two possible ways that may affect the triangulation T in Figure 23. It is clear that all flips in take place at green arcs.
Next, we consider the effect of on . From the description of given by Figure 23, the surfaces and with can only intersect at their boundaries. By Lemma 8.16 , we can enumerate the fans appearing in in any order.
The sequence is either defined by at least two fans and or it is defined by a single fan . Let be one of the fans defining . In the former case, the surface is an unpunctured disk, triangulated by a fan triangulation (see Figure 24 ). In the latter case, the surface is a once-punctured disk, triangulated by a fan of arcs about the puncture (see Figure 24 and )666These three cases can equivalently be described by saying Figure 24 , Figure 24 , and Figure 24 , respectively, is the case in which among the fans defining at least two fans consist of three arcs, exactly one fan consists of three arcs, and no fans consist of three arcs.. Since Figure 23 shows that the only green arcs of are those in , all flips in take place at green arcs.
One checks that if , then consists of at least two arcs. In the case when consists of exactly two arcs, the triangulation T is one of three shown in Figure 25. In these cases, we leave it to the reader to verify that is a maximal green sequence, and we focus on the generic case when .
In Figure 24, we also show the effect of performing the transformation on the triangulated surface in each of these three cases. The Figures 26, 27, and 28 justify why the resulting triangulation of is what appears in the lower image in Figure 24 and that the arcs in that triangulation have the indicated colors.
The Figures 26, 27, and 28 also show that an arc of one of whose endpoints is is green if and only if the arc is plain at . Furthermore, we conclude that in the cases of Figures 24 and an arc of none of whose endpoints is is green if and only if for some where was flipped when applying . Lastly, in the case of Figure 24 , an arc of none of whose endpoints is is green if and only if for some and the arcs and with form a triangle in or .
We now apply to By referring to Figures 26, 27, and 28, we see that applying to means that we flip exactly the green arcs from Figure 26 and Figure 27 . Observe that none of these flips turn any red arcs green. Furthermore, the surfaces and do not intersect nontrivially so the arcs defining may be flipped in any order.
Next, we apply to Note that if and only if the sequence is defined by a single fan where is a once-punctured disk with a fan triangulation given by the arcs of as in Figure 24 .
If , then every boundary arc of the surface is an internal arc of the triangulation . Indeed, if is defined by at least two fans (resp., a single fan), the boundary arcs of are exactly the arcs from Figure 24 (resp., and from Figure 24 ). In the former case, every boundary arc of is red in , and, in the latter case, the arc is the only boundary arc of that is red in .
Let and consider . We show how affects in Figure 29. In particular, the elementary laminations appearing in Figure 29 do appear because we have previously determined that the arcs (resp., the boundary arcs of ) are green (resp., red) in . Furthermore, every flip in occurs at a green arc.
Lastly, we apply to By the definition of the arcs in and Lemma 8.16 , we can enumerate the arcs appearing in is any order. We notice from Figures 28 and 29 and our description of the boundary arcs of that any arc appearing in is of the form or and each arc appearing in is green in . In fact, Figures 24, 26, 27, 28, and 29 and our description of the boundary arcs of show that each arc of is of the form or .
Next, we show that each flip in takes place at a green arc. We have already shown that each flip in takes place at a green arc. Now assume that for some the arcs in are green in for each . In Figure 30 , we show the generic configuration of arcs from around an arc appearing in . Here, either for some arc in where or for some and is an arc in one of the fans defining .
Suppose . Then Figures 24, 26, 27, 28, and 29 and our description of the boundary arcs of shows that where is the last arc to be flipped in some fan defining and is either a boundary component of or it is an arc of the form . Now assume where . Then Figures 24, 26, 27, and 28 show that the arcs and may be boundary components of or these are arcs of the form .
We can now conclude that if an arc in the left image in Figure 30 is not an boundary arc of , then it has the indicated color in . The elementary laminations (resp., ) in Figure 30 are those witnessing that (resp., ) is red (resp., green) in . These elementary laminations also show that the arcs other than in Figure 30 have the indicated color in . We will justify that is red in shortly.
The remaining cases are that where is the first arc to be flipped in some fan defining or that is an arc or . If , then Figures 24, 26, 27, 28 show that appears in one of the leftmost configurations of arcs in Figure 30 , , or . In Figures 30 and , the arcs and are either boundary arcs of or they are arcs or . If is an arc or , then plays the role of arc in the middle configuration of Figures 30 and .
Using Figures 24, 26, 27, 28, one checks that the elementary laminations shown in Figure 30 , , or are present in these configurations. Furthermore, one checks that these elementary laminations show that the arcs other than have the indicated colors in . However, each arc is realizable as an arc in a configuration of one of the forms shown in Figure 30 about some arc where , but plays the role of in the configuration. Thus will be red in .
We conclude that all flips in take place at green arcs. By induction, we see that all flips of take place at green arcs. Moreover, the rightmost configurations in Figure 30 , , and show that all arcs in are red. Thus is a maximal green sequence.∎
Proof of Corollary 8.14.
As in the proof of Theorem 8.13, we prove that has the desired length when at least one of the quivers is the empty quiver or none of the quivers is the empty quiver and is even.
Observe that applying the sequence of flips to T flips each arc of T exactly once. Thus is a green sequence of with length . It remains to show that .
We first show that . We claim that there is a bijection . Let where and where is the arc appearing in a triangle of with from the definition of From the structure of , we have that for and that .
Let where . We need to show that . Observe that for each there exists such that This implies and for . We conclude that and that this vertex has degree 4.
It remains to show that is surjective. Observe that for any degree 4 vertex , either appears in the fan –one of the fans that defines –or is an arc that gives rise to one of the arcs in . As such, the vertex is either the image under of an arc in or of an arc in .
Lastly, we show that . By Theorem 8.13 and Theorem 7.5, we know that In particular, we know that has exactly arcs connected to that are notched at . In the triangulation (see Figures 28 and 29), there are exactly two arcs connected to that are notched at . Since each flip in produces a triangulation with exactly one additional arc connected to that is notched at , we conclude that . ∎
9. Quivers of mutation type
We now focus on the minimal length maximal green sequences of quivers that we refer to as mutation type where and are positive integers. Let denote an affine type Dynkin diagram with vertices where . By definition, the mutation type quivers are those that are mutation-equivalent to an acyclic orientation of , denoted , where has edges are oriented clockwise, edges are oriented counterclockwise, and . Note that the notion of clockwise and counterclockwise arrows is only unique up to changing the roles of the clockwise and counterclockwise arrows.
9.1. The mutation class of quivers of mutation type quivers
Here we recall the description of mutation type quivers that was found in [3]. We first mention the following result, which shows that the mutation class of mutation type quivers depends only on the choice of the parameters and .
Lemma 9.1**.**
[12, Lemma 6.8] Let and be two acyclic orientations of . Then and are mutation-equivalent if and only if and .
Definition 9.2**.**
[3, Definition 3.3] Let be the class of connected quivers with vertices that satisfy the following conditions:
- i)
There exists precisely one full subquiver of which is an acyclic orientation of , denoted , for some satisfying .
- ii)
for each arrow in , there may be a vertex such that the 3-cycle is a full subquiver of , and there are no other vertices in that are connected to vertices of ;
- iii)
the full subquiver obtained from by removing the vertices and arrows of consists of the quivers , one for each , and each nonempty quiver is of mutation type .
Let be a quiver in . We refer to the quiver as the non-oriented cycle of . Now let be an arrow of its non-oriented cycle .
Example 9.3**.**
In Figure 31, we give an example of a quiver belonging to The full subquiver of on the vertices 2, 3, 4, 5, 6, 7, 8, and 9 is the acyclic orientation of described in part i) of Definition 9.2. The quiver is the full subquiver of on the vertices 1, 10, 11, 13, 12, 14, 15, and 16. It consists of a quiver of mutation type , a quiver of mutation type , and quiver of mutation type .
It is clear that any acyclic orientation of belongs to . By [3, Lemma 3.5], is closed under mutation. Thus each quiver in is mutation equivalent to an acyclic orientation of .
Theorem 9.4**.**
Let be a quiver of mutation type . The length of a minimal length maximal green sequence of is
Proof.
If there exists a clockwise arrow and counterclockwise arrow where and are both the empty quivers, then for some quivers and . It follows from Definition 9.2 that and are both of mutation type . By Theorem 2.3, there exist minimal length maximal green sequences and where and . By Theorem 4.3 and Proposition 4.4, and this is a minimal length maximal green sequence of the desired length.
Suppose there does not exist a clockwise arrow and counterclockwise arrow where and are both the empty quivers. By Definition 9.2, the quiver is an example of a quiver described by Definition 5.3 where the quivers are the nonempty quivers in from Definition 9.2 and is the full subquiver of on the vertex set where the vertices are those appearing in Definition 9.2.
The quivers are all of mutation type so by Theorem 2.3 there exist minimal length maximal green sequences where . By Theorem 9.8, there exist a minimal length maximal green sequence where . Now by Theorem 5.4, the quiver has a minimal length maximal green sequence i with ∎
9.2. Construction of minimal length maximal green sequences
In this section, we construct a particular mutation sequence of a quiver of mutation type that will turn out to be a minimal length maximal green sequence of the quiver.
Let be a quiver of mutation type . In this section, for brevity, we let be the non-oriented cycle of . Assume that cannot be expressed as a nontrivial direct sum of two mutation type quivers. This implies that there do not exist distinct arrows such that for some full subquivers and of . In particular, it follows that all clockwise arrows of belong to a 3-cycle of or all counterclockwise arrows of belong to a 3-cycle of . Without loss of generality, we will assume that all clockwise arrows of belong to a 3-cycle of .
Lemma 9.5**.**
The length of any maximal green sequence of is at least .
Proof.
We will assume that is not the quiver of mutation type where is the Kronecker quiver with arrows and both of its quivers and are single vertices. We will also assume that is not the quiver of mutation type where is the Kronecker quiver with arrows and where is a single vertex and is the empty quiver. If is either of these two quivers, one checks that its minimal length maximal green sequences have the desired length.
Now let and let be an arrow in the 3-cycle . By Theorem 3.3, we let denote the induced maximal green sequence on the full subquiver of whose vertices are and . One checks that the sequence of c-vectors includes at least one c-vector with two nonzero entries. Thus the sequence includes at least one c-vector with two nonzero entries for each 3-cycle of . It follows that . ∎
Now assume that has the property that the family of subquivers of , using the notation of Definition 9.2, contains only empty quivers or quivers of mutation type . It immediately follows that has the property that each arrow of satisfies one of the following:
- •
belongs to a (necessarily, unique) 3-cycle of
- •
is not an arrow in an oriented cycle of .
Let denote a quiver satisfying all of the assumptions mentioned in the previous two paragraphs.
We now define two families and of full subquivers of that will be important in our construction of minimal length maximal green sequences of . We note that the elements of and will form a set partition of the arrows of . Define the clockwise components of , denoted , to be the set of full, connected subquivers of such that
- •
contains no counterclockwise arrows of ,
- •
each clockwise arrow of belongs to a 3-cycle, and
- •
is an inclusion-maximal full subquiver of with respect to these conditions.
Let and order the clockwise arrows in as so that and for each Define to be the following sequence of vertices of :
[TABLE]
Next, define the counterclockwise components of , denoted , to be the set of full, connected subquivers of such that
- •
contains no clockwise arrows of and
- •
is an inclusion-maximal full subquiver of with respect to these conditions.
Let and order the counterclockwise arrows in as so that and are vertices of some quivers belonging to and for each . Define to be the following sequence of vertices of :
[TABLE]
Let be a sequence of some distinct vertices of , first containing all of the vertices in , followed by the vertices of for which there exists an arrow where is in . We exclude from this sequence if there exists an arrow and is part of some clockwise component of .
Now let . Let be the vertices of , ordered such that there exists vertices with a 2-path in for . Let be the unique vertex in with an arrow . Then is part of some counterclockwise component . Let be the vertex of with an arrow . If this arrow is part of a 3-cycle, then we define to be the sequence . Otherwise, we let be the sequence .
Example 9.6**.**
Consider the quiver shown in Figure 32. The set (resp., ) consists of the 3 full subquivers of that have solid (resp., dashed) arrows. For the quiver , the sequences of vertices defined above are as follows.
[TABLE]
Remark 9.7**.**
There is an obvious bijection between the quivers in and those in given by sending to so that is satisfied. Thus |C(Q)|=|CC(Q)|=|\{\text{sources of \eta(Q)}\}|. We will henceforth write and so that for each as we have done in Example 9.6.
We define to be the following sequence of vertices.
[TABLE]
Theorem 9.8**.**
The sequence i is a minimal length maximal green sequence of with .
The remainder of this section is dedicated to proving Theorem 9.8. As usual, we let denote a quiver with frozen vertices whose mutable part is . Given two full subquivers and of some quiver , we define (resp., ) to be the full subquiver of whose vertex set is (resp., ). If is a clockwise or counterclockwise component of , we refer to the unique source (resp., sink) of as the first (resp., last) vertex of . Similarly, we define the first (resp., last) arrow of to be the unique arrow of incident to the first (resp., last) vertex of .
Let us first remark that the sequence is of the appropriate length. Every vertex except the first and last in a clockwise component of is in the sequence . The sequence contains all of the vertices in the main cycle that lie in a counterclockwise component of . Together with , this now includes all of the vertices in and the vertices not in that are in a clockwise component. The sequence contains the vertices in each counterclockwise component of outside the main cycle, as well as one additional vertex for each 3-cycle in a counterclockwise component unless the 3-cycle contains the last vertex of that component. Finally, for each 3-cycle in a clockwise component, there is one corresponding vertex in , as well as one vertex in the sequence for each 3-cycle in a counterclockwise component that contains the last vertex of that component. In total, there is a mutation for each vertex of and an additional mutation corresponding to each 3-cycle.
To show that this sequence is a maximal green sequence, we characterize the quiver obtained from by applying mutations at each of the sequences for . We say a quiver is of Type I if it satisfies the defining criteria for . We say a quiver is of Type II if the following holds.
- •
For each :
- –
contains at least three vertices in .
- –
The first and last arrows of are not part of a 3-cycle, and every other arrow of is part of a 3-cycle.
- –
If is a vertex in , then there is a unique arrow with source , and is a red vertex with arrows and .
- –
If is the last vertex in , then there is a unique arrow in with target , and is a green vertex with arrows .
- –
If is the second-to-last vertex in , then is red with an arrow .
- •
Every other vertex is green with an arrow .
Lemma 9.9**.**
Applying the mutation sequence to gives a quiver of Type II, up to a permutation of the mutable vertices.
Proof.
Let . Except for the endpoints, the vertices of any counterclockwise component are not adjacent to any vertex in the sequence . Consequently, they will remained unchanged in .
Let be a clockwise component of . After mutating at each of the vertices in , the resulting quiver is a directed path with a clockwise orientation. The vertices in this path alternate between and . For each vertex , there is an arrow in . For a vertex and arrow in , there are arrows and . Finally, if is the first vertex of , then it is green with .
Now let be the result of mutating at each of the vertices in except the first and last. Then is a clockwise component of . Each vertex of except the first and last is part of a 3-cycle in but lies in . Every other vertex of lies on the main cycle in . Since we assumed that every clockwise arrow in is part of a 3-cycle in , we deduce that contains at least 3 vertices. Furthermore, every arrow of except the first and last is part of a 3-cycle.
If is a vertex in besides the first and last vertex, then there is a unique arrow in with source . For each such pair we swap their labels in the mutable part of . After doing this permutation, it is straight-forward to check that the component satisfies the remaining conditions to be a clockwise component of a quiver of Type II. ∎
We say a quiver is of Type III if the following holds.
- •
For each :
- –
Every arrow in is part of a 3-cycle.
- –
If is the last vertex in , then is red with arrow .
- –
If is any other vertex in , then is green with arrow .
- –
If is in , and is the unique arrow with source , then is red with arrows and .
- •
For each :
- –
If is the first vertex in , then is green with arrow .
- –
If is any other vertex in , then is red with arrow .
- –
If is in , and is the unique arrow with target , then is green with arrows and .
Lemma 9.10**.**
Applying to a quiver of Type II produces a quiver of Type III, up to a permutation of the mutable vertices.
Proof.
Let be a quiver of Type II. Let be a counterclockwise component of . Let be the list of vertices in such that there exist arrows . Let be the unique arrow outside whose source is , and be the unique arrow outside whose target is . Since is of Type II, both of the arrows and are not part of a 3-cycle. In particular, mutation at the vertices only affects , and the vertices in .
Let . Let be the counterclockwise component of containing . Then is the directed path . Furthermore, contains a 3-cycle . In particular, any arrow in a clockwise component of is part of a 3-cycle.
There is a 3-cycle in if and only if there is a 3-cycle in .
Swap the labels on the mutable vertices and in . Once again, it is routine to check that satisfies the remaining conditions of a quiver of Type III.∎
Let be a Type III quiver. We define a mutation sequence on some of the vertices of the counterclockwise components of such that for each , the restriction of to consists of the vertices in , followed by the vertices in such that for some . If is the last vertex of , then it is removed from the list .
We say a quiver is of Type IV if the following holds.
- •
The clockwise components satisfy most of the same rules as a Type III quiver. The only exception is that if is the last vertex of some clockwise component, then either is red with an arrow or it is green with an arrow .
- •
For :
- –
The first arrow of is not part of a 3-cycle.
- –
If is the first vertex in , then it is green with arrow .
- –
If is the last vertex in , then it is either green with arrow or red with arrow . Furthermore, if is green, then the last arrow is not part of a 3-cycle, and is red with arrows and .
- –
Every other vertex in is red with arrow .
Lemma 9.11**.**
Applying the mutation sequence to a quiver of Type III produces a quiver of Type IV, up to a permutation of the mutable vertices.
Proof.
Let be a quiver of Type III. It is clear that the mutation sequence does not affect any clockwise component with the possible exception of the last vertex.
Let be a counterclockwise component of . Let be the result of mutating at each of the vertices in . Then is a directed path. If is a vertex of and is the unique vertex in with , then is red in with arrows and . Furthermore, is green in with an arrow . All other vertices of are still red.
Let be the result of mutating at the remaining vertices in the mutation sequence . Let and be the same vertices as before, and suppose is not the last vertex in . Then is red in with an arrow and is red in with an arrow . If is the last vertex, then and remain as they were in .
Up to swapping some labels in the mutable part of , the result of applying to is a quiver of Type IV. ∎
Given a quiver of Type IV, we define a mutation sequence as follows. For each clockwise component , if is then the restriction of to is if is red and if is green.
With this setup, we have the following result whose proof we omit.
Lemma 9.12**.**
Applying the mutation sequence to a quiver of Type IV produces a quiver whose vertices are all red.
Since the mutation sequences and are the same as and up to relabeling mutable vertices as done in the proofs of Lemmas 9.9, 9.10, and 9.11, we have now proved Theorem 9.8.
10. Additional Remarks
Any quiver in the mutation class of an acyclic quiver gives rise to a finite dimensional -algebra, denoted , known as a cluster-tilted algebra (as defined in [8]). The quivers of mutation types , , and , which we have considered in this paper, all define cluster-tilted algebras of the corresponding type. By referring to the known results on derived equivalence of cluster-tilted algebras of these types, we suspect that the length of minimal length maximal green sequences is constant on derived equivalence classes. We now briefly summarize these known results.
Let and be two derived equivalent cluster-tilted algebras. It follows from [9, Theorem 5.1] that and are of type if and only if the minimal length maximal green sequences of have the same length as those of . It follows from [3, Theorem 1.1] that if and are of type , then they have the same number of 3-cycles. Thus the minimal length maximal green sequences of will have the same length as those of .
In the case of cluster-tilted algebras of type , there is no complete derived equivalence classification. From [4, Theorem 2.3], any cluster-tilted algebra of type is derived equivalent to an algebra whose quiver is in one of six classes of mutation type quivers. Bastian, Holm, and Ladkani call these six classes standard forms and denote them by . Any two distinct standard forms which are not of the class are not derived equivalent.
If is not self-injective, it may be put into its standard form by mutations that the authors call good mutations and good double mutations. Using Theorem 6.1, one checks that the length of minimal length maximal green sequences is invariant under good mutations and good double mutations. If is self-injective, then [4, Lemma 4.5] and Theorem 6.1, shows that the quiver of the standard form of has minimal length maximal green sequences with the same length as those of These observations suggest the following question.
Question 10.1**.**
Suppose and are two derived equivalent cluster-tilted algebras where and are minimal length maximal green sequences. Is it true that ?
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