COMBINATORICS OF EXCEPTIONAL SEQUENCES IN TYPE A

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1 COMBINATORICS OF EXCEPTIONAL SEQUENCES IN TYPE A ALEXANDER GARVER, KIYOSHI IGUSA, JACOB P. MATHERNE, AND JONAH OSTROFF Abstract. Exceptional sequences are certain ordered sequences of quiver representations. We introduce a class of objects called strand diagrams and use this model to classify exceptional sequences of representations of a quiver whose underlying graph is a type A n Dynkin diagram. We also use variations of this model to classify c-matrices of such quivers, to interpret exceptional sequences as linear extensions of posets, and to give a simple bijection between exceptional sequences and certain chains in the lattice of noncrossing partitions. This work extends a classification of exceptional sequences for the linearly-ordered quiver obtained in [GM5] by the first and third authors. Contents. Introduction. Preliminaries.. Quiver mutation.. Representations of quivers.. Quivers of type A n. Strand diagrams 5.. Exceptional sequences and strand diagrams 5.. Proof of Lemma.5 8. Mixed cobinary trees 5. Exceptional sequences and linear extensions 5 6. Applications Labeled trees Reddening sequences Noncrossing partitions and exceptional sequences 9 References 0. Introduction Exceptional sequences are certain sequences of quiver representations with strong homological properties. They were introduced in [GR87] to study exceptional vector bundles on P, and more recently, Crawley-Boevey showed that the braid group acts transitively on the set of complete exceptional sequences (exceptional sequences of maximal length) [CB9]. This result was generalized to hereditary Artin algebras by Ringel [Rin9]. Since that time, Meltzer has also studied exceptional sequences for weighted projective lines [Mel0], and Araya for Cohen- Macaulay modules over one dimensional graded Gorenstein rings with a simple singularity [Ara99]. Exceptional sequences have been shown to be related to many other areas of mathematics since their invention: chains in the lattice of noncrossing partitions [Bes0, HK, IT09], c-matrices and cluster algebras [ST], factorizations of Coxeter elements [IS0], and t-structures and derived categories [Bez0, BK89, Rud90]. Despite their ubiquity, very little work has been done to concretely describe exceptional sequences, even for path algebras of Dynkin quivers [Ara, GM5]. In this paper, we give a concrete description of exceptional sequences for type A n quivers of any orientation. This work extends a classification of exceptional sequences for the linearly-ordered quiver obtained in [GM5] by the first and third authors. The first author was supported by a Research Training Group, RTG grant DMS-86. The second author was supported by National Security Agency Grant H The third author was supported by a Graduate Assistance in Areas of National Need fellowship, GAANN grant P00A000.

2 Exceptional sequences are composed of indecomposable representations which have a particularly nice description. For a quiver Q of type A n, the indecomposable representations are completely determined by their dimension vectors, which are of the form p0,..., 0,,...,, 0,..., 0q. Let us denote such a representation by Xi,j ɛ, where ɛ is a vector that keeps track of the orientation of the quiver, and i ` and j are the positions where the string of s begins and ends, respectively. This simple description allows us to view exceptional sequences as combinatorial objects. Define a map Φ ɛ which associates to each indecomposable representation Xi,j ɛ a strand Φ ɛpxi,j ɛ q on a collection of n ` dots. X ɛ 0, 0 k Figure. An example of the indecomposable representation X ɛ 0, on a type A quiver and the corresponding strand Φ ɛ px ɛ 0,q As exceptional sequences are collections of representations, the map Φ ɛ allows one to regard them as collections of strands. The following lemma is the foundation for all of our results in this paper (it characterizes the homological data encoded by a pair of strands and thus by a pair of representations). Since exceptional sequences are sequences of representations, each pair of which satisfy certain homological properties, Lemma.5 allows us to completely classify exceptional sequences using strand diagrams. Lemma.5. Let Q ɛ be given. Fix two distinct indecomposable representations U, V P indprep k pq ɛ qq. aq The strands Φ ɛ puq and Φ ɛ pv q intersect nontrivially if and only if neither pu, V q nor pv, Uq are exceptional pairs. bq The strand Φ ɛ puq is clockwise from Φ ɛ pv q if and only if pu, V q is an exceptional pair and pv, Uq is not an exceptional pair. cq The strands Φ ɛ puq and Φ ɛ pv q do not intersect at any of their endpoints and they do not intersect nontrivially if and only if pu, V q and pv, Uq are both exceptional pairs. The paper is organized in the following way. In Section, we give the preliminaries on quivers and their representations which are needed for the rest of the paper. In Section., we decorate our strand diagrams with strand-labelings and oriented edges so that they can keep track of both the ordering of the representations in a complete exceptional sequence as well as the signs of the rows in the c-matrix it came from. While unlabeled diagrams classify complete exceptional collections (Theorem.6), we show that the new decorated diagrams classify more complicated objects called exceptional sequences (Theorem.9). Although Lemma.5 is the main tool that allows us to obtain these results, we delay its proof to Section.. The work of Speyer and Thomas (see [ST]) allows complete exceptional sequences to be obtained from c-matrices. In [ONA`], the number of complete exceptional sequences in type A n is given, and there are more of these than there are c-matrices. Thus, it is natural to ask exactly which c-matrices appear as strand diagrams. By establishing a bijection between the mixed cobinary trees of Igusa and Ostroff [IO] and a certain subcollection of strand diagrams, we give an answer to this question in Section. In Section 5, we ask how many complete exceptional sequences can be formed using the representations in a complete exceptional collection. It turns out that two complete exceptional sequences can be formed in this way if they have the same underlying chord diagram without chord labels. We interpret this number as the number of linear extensions of the poset determined by the chord diagram of the complete exceptional collection. This also gives an interpretation of complete exceptional sequences as linear extensions. In Section 6, we give several applications of the theory in type A, including combinatorial proofs that two reddening sequences produce isomorphic ice quivers (see [Kel] for a general proof in all types using deep category-theoretic techniques) and that there is a bijection between exceptional sequences and certain chains in the lattice of noncrossing partitions. Acknowledgements. A. Garver and J.P. Matherne gained helpful insight through conversations with E. Barnard, J. Geiger, M. Kulkarni, G. Muller, G. Musiker, D. Rupel, D. Speyer, and G. Todorov. A. Garver and J.P. Matherne also thank the 0 Mathematics Research Communities program for giving us an opportunity to work on this exciting problem as well as for giving us a stimulating (and beautiful) place to work.. Preliminaries In this section, we recall some basic definitions. We will be interested in the connection of exceptional sequences and the c-matrices of an acyclic quiver Q so we begin by defining these. After that we review the basic

3 terminology of quiver representations and exceptional sequences. which serve as the starting point in our study of exceptional sequences. We conclude this section by explaining the notation we will use to discuss exceptional representations of quivers that are orientations of a type A n Dynkin diagram... Quiver mutation. A quiver Q is a directed graph without loops or -cycles. In other words, Q is a -tuple pq 0, Q, s, tq, where Q 0 rms : t,,..., mu is a set of vertices, Q is a set of arrows, and two functions s, t : Q Ñ Q 0 defined so that for every a P Q, we have spaq a ÝÑ tpaq. An ice quiver is a pair pq, F q with Q a quiver and F Ă Q 0 frozen vertices with the additional restriction that any i, j P F have no arrows of Q connecting them. We refer to the elements of Q 0 zf as mutable vertices. By convention, we assume Q 0 zf rns and F rn`, ms : tn`, n`,..., mu. Any quiver Q can be regarded as an ice quiver by setting Q pq, Hq. The mutation of an ice quiver pq, F q at mutable vertex k, denoted µ k, produces a new ice quiver pµ k Q, F q by the three step process: () For every -path i Ñ k Ñ j in Q, adjoin a new arrow i Ñ j. () Reverse the direction of all arrows incident to k in Q. () Delete any -cycles created during the first two steps. We show an example of mutation below depicting the mutable (resp. frozen) vertices in black (resp. blue). pq, F q = µ ÞÝÑ = pµ Q, F q The information of an ice quiver can be equivalently described by its (skew-symmetric) exchange matrix. Given pq, F q, we define B B pq,f q pb ij q P Z nˆm : tnˆm integer matricesu by b ij : #ti Ñ a j P Q u #tj Ñ a i P Q u. Furthermore, ice quiver mutation can equivalently be defined as matrix mutation of the corresponding exchange matrix. Given an exchange matrix B P Z nˆm, the mutation of B at k P rns, also denoted µ k, produces a new exchange matrix µ k pbq pb ijq with entries " b ij : b ij : if i k or j k b ij ` b ik b kj`b ik b kj : otherwise. For example, the mutation of the ice quiver above (here m and n ) translates into the following matrix mutation. Note that mutation of matrices (or of ice quivers) is an involution (i.e. µ k µ k pbq B).» B pq,f q fi» 0 0 fl ÞÝÑ µ 0 0 fi 0 0 fl B pµq,f q Given a quiver Q, we define its framed (resp. coframed) quiver to be the ice quiver Q p (resp. Q) q where pq 0 p Q q 0 q : Q 0 \ rn `, ns, F rn `, ns, and Q p : Q \ ti Ñ n ` i : i P rnsu (resp. Q q : Q \ tn ` i Ñ i : i P rnsu). Now given Q p we define the exchange tree of Q, p denoted ET pqq, p to be the (a priori infinite) graph whose vertices are quivers obtained from Q p by a finite sequence of mutations and with two vertices connected by an edge if and only if the corresponding quivers are obtained from each other by a single mutation. Similarly, define the exchange graph of Q, p denoted EGpQq, p to be the quotient of ET pqq p where two vertices are identified if and only if there is a frozen isomorphism of the corresponding quivers (i.e. an isomorphism that fixes the frozen vertices). Such an isomorphism is equivalent to a simultaneous permutation of the rows and columns of the corresponding exchange matrices. Given Q, p we define the c-matrix Cpnq C R pnq (resp. C C R ) of R P ET pqq p (resp. R P EGpQq) p to be the submatrix of B R where Cpnq : pb ij q iprns,jprn`,ns (resp. C : pb ij q iprns,jprn`,ns ). We let c-mat(q) : tc R : R P EGpQqu. p By definition, B R (resp. C) is only defined up to simultaneous permutations of its rows and first n columns (resp. up to permutations of its rows) for any R P EGpQq. p A row vector of a c-matrix, ÝÑ c, is known as a c-vector. The celebrated theorem of Derksen, Weyman, and Zelevinsky [DWZ0, Theorem.7], known as the sign-coherence of c-vectors, states that for any R P ET pqq p and i P rns the c-vector ÝÑ c i is a nonzero element of Z n ě0 or Z n ď0. Thus we say a c-vector is either positive or negative... Representations of quivers. A representation V ppv i q ipq0, pϕ a q apq q of a quiver Q is an assignment of a k-vector space V i to each vertex i and a k-linear map ϕ a : V spaq Ñ V tpaq to each arrow a where k is a field. The dimension vector of V is the vector dimpv q : pdim V i q ipq0. The support of V is the set

4 supppv q : ti P Q 0 : V i 0u. Here is an example of a representation, with dimpv q p,, q, of the mutable part of the quiver depicted in Section k» 0 k k Let V ppv i q ipq0, pϕ a q apq q and W ppw i q ipq0, pϱ a q apq q be two representations of a quiver Q. A morphism θ : V Ñ W consists of a collection of linear maps θ i : V i Ñ W i that are compatible with each of the linear maps in V and W. That is, for each arrow a P Q, we have θ tpaq ϕ a ϱ a θ spaq. An isomorphism of quiver representations is a morphism θ : V Ñ W where θ i is a k-vector space isomorphism for all i P Q 0. We define V W : ppv i W i q ipq0, pϕ a ϱ a q apq q to be the direct sum of V and W. We say that a nonzero representation V is indecomposable if it is not isomorphic to a direct sum of two nonzero representations. Note that representations of quivers along with morphisms between them form an abelian category denoted rep k pqq, with the indecomposable representations forming a full subcategory called indprep k pqqq. We remark that representations of Q can equivalently be regarded as modules over the path algebra kq. As such, one can define Ext s kqpv, W q (s ě 0) and Hom kq pv, W q for any representations V and W and Hom kq pv, W q is isomorphic to the vector space of all morphisms θ : V Ñ W. We refer the reader to [ASS06] for more details on representations of quivers. An exceptional sequence ξ pv,..., V k q (k ď n : #Q 0 ) is an ordered list of exceptional representations V j of Q (i.e. V j is indecomposable and Ext s kqpv j, V j q 0 for all s ě ) satisfying Hom kq pv j, V i q 0 and Ext s kqpv j, V i q 0 if i ă j for all s ě. We define an exceptional collection ξ tv,..., V k u to be a set of exceptional representations V j of Q that can be ordered in such a way that they define an exceptional sequence. When k n, we say ξ (resp. ξ) is a complete exceptional sequence (CES) (resp. complete exceptional collection (CEC)). For Dynkin quivers, a representation is exceptional if and only if it is indecomposable. The following result of Speyer and Thomas gives a beautiful connection between c-matrices of an acyclic quiver Q and CESs. It serves as motivation for our work. Before stating it we remark that for any R P ET p p Qq and any i P rns where Q is an acyclic quiver, the c-vector ÝÑ c i ÝÑ c i prq dimpv i q for some exceptional representation of Q (see [Cha]). In general, not all indecomposable representations are exceptional. The c-vectors are exactly the dimension vectors of the exceptional modules and their negatives. Notation.. Let ÝÑ c be a c-vector of an acyclic quiver Q. Define ÝÑ c : " ÝÑc : if ÝÑ c is positive ÝÑ c : if ÝÑ c is negative. Theorem. ([ST]). Let C P c-matpqq, let t ÝÑ c i u iprns denote the c-vectors of C, and let ÝÑ c i dimpv i q for some indecomposable representation of Q. There exists a permutation σ P S n such that pv σpq,..., V σpnq q is a CES with the property that if there exist positive c-vectors in C, then there exists k P rns such that ÝÝÑ c σpiq is positive if and only if i P rk, ns, and Hom kq pv i, V j q 0 if ÝÑ c i, ÝÑ c j have the same sign. Conversely, any set of n vectors having these properties defines a c-matrix whose rows are t ÝÑ c i u iprns... Quivers of type A n. For the purposes of this paper, we will only be concerned with quivers of type A n. We say a quiver Q is of type A n if the underlying graph of Q is a Dynkin diagram of type A n. By convention, two vertices i and j with i ă j in a type A n quiver Q are connected by an arrow if and only if j i ` and i P rn s. It will be convenient to denote a given type A n quiver Q using the notation Q ɛ, which we now define. Let ɛ pɛ 0, ɛ,..., ɛ n q P t`, u n` and for i P rn s define a ɛi i P Q by a ɛi i : " i Ð i ` : ɛi i Ñ i ` : ɛ i `. Then Q ɛ : ppq ɛ q 0 : rns, pq ɛ q : ta ɛi i u iprn sq Q. One observes that the values of ɛ 0 and ɛ n do not affect Q ɛ.

5 a` a a` a Example.. Let n 5 and ɛ p, `,, `,, `q so that Q ɛ ÝÑ ÐÝ ÝÑ ÐÝ 5. Below we show its framed quiver Q p ɛ pq ɛ 5 Let Q ɛ be given where ɛ pɛ 0, ɛ,..., ɛ n q P t`, u n`. Let i, j P r0, ns : t0,,..., nu where i ă j and let X ɛ i,j ppv lq lppqɛq 0, pϕ i,j a q appqɛq q P rep k pq ɛ q be the indecomposable representation defined by V l : " k : i ` ď l ď j 0 : otherwise ϕ i,j a : " : a a ɛ k k where i ` ď k ď j 0 : otherwise. The objects of indprep k pq ɛ qq are those of the form Xi,j ɛ where 0 ď i ă j ď n, up to isomorphism.. Strand diagrams In this section, we define three different types of combinatorial objects: strand diagrams, labeled strand diagrams, and oriented strand diagrams. We will use these objects to classify exceptional collections, exceptional sequences, and c-matrices of a given type A n quiver Q ɛ. Throughout this section, we work with a given type A n quiver Q ɛ... Exceptional sequences and strand diagrams. Let S n,ɛ Ă R be a collection of n ` points arranged in a horizontal line. We identify these points with ɛ 0, ɛ,..., ɛ n where ɛ j appears to the right of ɛ i for any i, j P r0, ns : t0,,,..., nu where i ă j. Using this identification, we can write ɛ i px i, y i q P R. Definition.. Let i, j P r0, ns where i j. A strand cpi, jq on S n,ɛ is an isotopy class of simple curves in R where any γ P cpi, jq satisfies: aq the endpoints of γ are ɛ i and ɛ j, bq as a subset of R, γ Ă tpx, yq P R : x i ď x ď x j uztɛ i`, ɛ i`,..., ɛ j u, cq if k P r0, ns satisfies i ď k ď j and ɛ k ` (resp. ɛ k ), then γ is locally below (resp. above) ɛ k. There is a natural map Φ ɛ from indprep k pq ɛ qq to the set of strands on S n,ɛ given by Φ ɛ pxi,j ɛ q : cpi, jq. Remark.. It is clear that any strand can be represented by a monotone curve γ P cpi, jq (i.e. if t, s P r0, s and t ă s, then γ pq ptq ă γ pq psq where γ pq denotes the x-coordinate function of γ.). We say that two strands cpi, j q and cpi, j q intersect nontrivially if any two curves γ l P cpi l, j l q with l P t, u have at least one transversal crossing. Otherwise, we say that cpi, j q and cpi, j q do not intersect nontrivially. For example, cp, q, cp, q intersect nontrivially if and only if ɛ ɛ. Additionally, we say that cpi, j q is clockwise from cpi, j q (or, equivalently, cpi, j q is counterclockwise from cpi, j q) if and only if any γ P cpi, j q and γ P cpi, j q share an endpoint ɛ k and locally appear in one of the following six configurations up to isotopy. γ γ γ γ γ γ ɛ k = + ɛ k = + ɛ k = + γ γ γ γ ɛ k = ɛ k = ɛ k = γ γ Definition.. A strand diagram d tcpi l, j l qu lprks (k ď n) on S n,ɛ is a collection of strands on S n,ɛ that satisfies the following conditions: aq distinct strands do not intersect nontrivially, and bq the graph determined by d is a forest (i.e. a disjoint union of trees) Let D k,ɛ denote the set of strand diagrams on S n,ɛ with k strands and let D ɛ denote the set of strand diagrams with any positive number of strands. Then D ɛ ğ D k,ɛ. kprns a` a a` Example.. Let n and ɛ p, `,, `, `q so that Q ɛ ÝÑ ÐÝ ÝÑ. Then we have that d tcp0, q, cp0, q, cp, q, cp, qu P D,ɛ and d tcp0, q, cp, q, cp, qu P D,ɛ. We draw these strand diagrams 5.

6 below. The following technical lemma classifies when two distinct indecomposable representations of Q ɛ define 0,, or exceptional pairs. Its proof appears in Section.. Lemma.5. Let Q ɛ be given. Fix two distinct indecomposable representations U, V P indprep k pq ɛ qq. aq The strands Φ ɛ puq and Φ ɛ pv q intersect nontrivially if and only if neither pu, V q nor pv, Uq are exceptional pairs. bq The strand Φ ɛ puq is clockwise from Φ ɛ pv q if and only if pu, V q is an exceptional pair and pv, Uq is not an exceptional pair. cq The strands Φ ɛ puq and Φ ɛ pv q do not intersect at any of their endpoints and they do not intersect nontrivially if and only if pu, V q and pv, Uq are both exceptional pairs. Using Lemma.5 we obtain our first main result. The following theorem says that the data of an exceptional collection is completely encoded in the strand diagram it defines. Theorem.6. Let E ɛ : texceptional collections of Q ɛ u. There is a bijection E ɛ Ñ D ɛ defined by ξ ɛ tx ɛ i l,j l u lprks ÞÑ tφ ɛ px ɛ i l,j l qu lprks. Proof. Let ξ ɛ tx ɛ i l,j l u lprks be an exceptional collection of Q ɛ. Let ξ ɛ be an exceptional sequence gotten from ξ ɛ by reordering its representations. Without loss of generality, assume ξ ɛ px ɛ i l,j l q lprks is an exceptional sequence. Thus, px ɛ i l,j l, X ɛ i p,j p q is an exceptional pair for all l ă p. Lemma.5 a) implies that distinct strands of tφ ɛ px ɛ i l,j l qu lprks do not intersect nontrivially. Now we will show that tφ ɛ px ɛ i l,j l qu lprks has no cycles. Suppose that Φ ɛ px ɛ i l,j l q,..., Φ ɛ px ɛ i lp,j lp q is a cycle of length p ď k in Φ ɛ pξ ɛ q. Then, there exist l a, l b P rks with l b ą l a such that Φ ɛ px ɛ i lb,j lb q is clockwise from Φ ɛ px ɛ i la,j la q. Thus, by Lemma.5 b), px ɛ i la,j la, X ɛ i lb,j lb q is not an exceptional pair. This contradicts the fact that px ɛ i l,j l,..., X ɛ i lp,j lp q is an exceptional sequence. Hence, the graph determined by tφ ɛ px ɛ i l,j l qu lprks is a tree. We have shown that Φ ɛ pξ ɛ q P D k,ɛ. Now let d tcpi l, j l qu lprks P D k,ɛ. Since cpi l, j l q and cpi m, j m q do not intersect nontrivially, it follows that pφ ɛ pcpi l, j l qq, Φ ɛ pcpi m, j m qqq or pφ ɛ pcpi m, j m qq, Φ ɛ pcpi l, j l qqq is an exceptional pair for every l m. Notice that there exists cpi l, j l q P d such that pφ ɛ pcpi l, j l qq, Φ ɛ pcpi l, j l qqq is an exceptional pair for every cpi l, j l q P dztcpi l, j l qu. This is true because if such cpi l, j l q did not exist, then d must have a cycle. Set E Φ ɛ pcpi l, j l qq. Now, choose cpi lp, j lp q such that pφ ɛ pcpi lp, j lp qq, Φ ɛ pcpi l, j l qqq is an exceptional pair for every cpi l, j l q P dztcpi l, j l q,..., cpi lp, j lp qu inductively and put E p Φ ɛ pcpi lp, j lp qq. By construction, pe,..., E k q is a complete exceptional sequence, as desired. Our next step is to add distinct integer labels to each strand in a given strand diagram d. When these labels have what we call a good labeling, these labels will describe exactly the order in which to put the representations corresponding to strands of d so that the resulting sequence of representations is an exceptional sequence. Definition.7. A labeled diagram dpkq tpcpi l, j l q, s l qu lprks on S n,ɛ is a strand diagram on S n,ɛ whose strands are labeled by integers s l P rks bijectively. Let ɛ i P S n,ɛ and let ppcpi, j q, s q,..., pcpi, j r q, s r qq be the complete list of labeled strands of dpkq that involve ɛ i and ordered so that strand cpi, j k q is clockwise from cpi, j k q if k ă k. We say the strand labeling of dpkq is good if for each point ɛ i P S n,ɛ one has s ă ă s r. Let D k,ɛ pkq denote the set of labeled strand diagrams on S n,ɛ with k strands and with good strand labelings. a` a a` Example.8. Let n and ɛ p, `,, `, `q so that Q ɛ ÝÑ ÐÝ ÝÑ. Below we show the labeled diagrams d pq tpcp0, q, q, pcp0, q, q, pcp, q, q, pcp, q, qu and d pq tpcp0, q, q, pcp, q, q, pcp, q, qu. We have that d pq P D,ɛ pq, but d pq R D,ɛ pq. 6

7 Theorem.9. Let k P rns and let E ɛ pkq : texceptional sequences of Q ɛ of length ku. There is a bijection rφ ɛ : E ɛ pkq Ñ D k,ɛ pkq defined by ξ ɛ px ɛ i l,j l q lprks ÞÝÑ tpcpi l, j l q, k ` lqu lprks. Proof. Let ξ ɛ : pv,..., V k q P E ɛ pkq. By Lemma.5 a), Φ r ɛ pξ ɛ q has no strands that intersect nontrivially. Let pv, V q be an exceptional pair appearing in ξ ɛ with V i corresponding to strand c i in Φ r ɛ pξ ɛ q for i,, where c and c intersect only at one of their endpoints. Note that by the definition of Φ r ɛ, the strand label of c is larger than that of c. From Lemma.5 b), strand c is clockwise from c in Φ r ɛ pξ ɛ q. Thus the strand-labeling of Φ r ɛ pξ ɛ q is good, so Φ r ɛ pξ ɛ q P D k,ɛ pkq for any ξ ɛ P E ɛ pkq. Let Ψ r ɛ : D k,ɛ pkq Ñ E ɛ pkq be defined by tpcpi l, j l q, lqu lprks ÞÑ pxi ɛ k,j k, Xi ɛ k,j k,..., Xi ɛ,j q. We will show that rψ ɛ pdpkqq P E ɛ pkq for any dpkq P D k,ɛ pkq and that Ψ r ɛ Φ r ɛ. Let rψ ɛ ptpcpi l, j l q, lqu lprks q px ɛ i k,j k, X ɛ i k,j k,..., X ɛ i,j q. Consider the pair pxi ɛ s,j s, Xi ɛ s,j q with s ą s s. We will show that px ɛ i s,j s, Xi ɛ s,j s q is an exceptional pair and thus conclude that Ψ r ɛ ptpcpi l, j l q, lqu lprks q P E ɛ pkq for any dpkq P D k,ɛ pkq. Clearly, cpi s, j s q and cpi s, j s q do not intersect nontrivially. If cpi s, j s q and cpi s, j s q do not intersect at one of their endpoints, then by Lemma.5 c) pxi ɛ s,j s, Xi ɛ s,j q is exceptional. Now suppose cpi s s, j s q and cpi s, j s q intersect at one of their endpoints. Because the strand-labeling of tpcpi l, j l q, lqu lprks is good, cpi s, j s q is clockwise from cpi s, j s q. By Lemma.5 b), we have that pxi ɛ s,j s, Xi ɛ s,j s q is exceptional. To see that Ψ r ɛ Φ r ɛ, observe that rφ rψɛ ɛ ptpcpi l, j l q, lqu lprks q Φ r ɛ px i ɛ k,j k, Xi ɛ k,j k,..., Xi ɛ,j q tpcpi l, j l q, k ` pk ` lqqu lprks tpcpi l, j l q, lqu lprks. Thus r Φ ɛ rψ ɛ Dn,ɛpkq. Similarly, one shows that r Ψ ɛ rφ ɛ Eɛpkq. Thus r Φ ɛ is a bijection. The last combinatorial objects we discuss in this section are called oriented diagrams. These are strand diagrams whose strands have a direction. We will use these to classify c-matrices of a given type A n quiver Q ɛ. Definition.0. An oriented diagram ÝÑ d t ÝÑ c pi l, j l qu lprks on S n,ɛ is a strand diagram on S n,ɛ whose strands ÝÑ c pil, j l q are oriented from ɛ il to ɛ jl. Remark.. When it is clear from the context what the values of n and ɛ are, we will often refer to a strand diagram on S n,ɛ simply as a diagram. Similarly, we will often refer to labeled diagrams (resp. oriented diagrams) on S n,ɛ as labeled diagrams (resp. oriented diagrams). We now define a special subset of the oriented diagrams on S n,ɛ. As we will see, each element in this subset of oriented diagrams, denoted ÝÑ D n,ɛ, will correspond to a unique c-matrix C P c-matpq ɛ q and vice versa. Thus we obtain a diagrammatic classification of c-matrices (see Theorem.5). Definition.. Let ÝÑ D n,ɛ denote the set of oriented diagrams ÝÑ d t ÝÑ c pi l, j l qu lprns on S n,ɛ with the property that any oriented subdiagram ÝÑ d of ÝÑ d consisting only of oriented strands connected to ɛ k in S n,ɛ for some k P r0, ns is a subdiagram of one of the following: iq t ÝÑ c pk, i q, ÝÑ c pk, i q, ÝÑ c pj, kqu where i ă k ă i and ɛ k ` (shown in Figure (left)), iiq t ÝÑ c pi, kq, ÝÑ c pi, kq, ÝÑ c pk, jqu where i ă k ă i and ɛ k (shown in Figure (right)). ɛ k = + ɛ k = Figure 7

8 Lemma.. Let t ÝÑ c i u iprks be a collection of k c-vectors of Q ɛ where k ď n. Let ÝÑ c i dimpx ɛ i,i q where the sign is determined by ÝÑ c i. If t ÝÑ c i u iprks is a noncrossing collection of c-vectors (i.e. Φ ɛ px ɛ i,i q and Φ ɛ px ɛ i,i q do not intersect nontrivially for any i, i P rks), there is an injective map tnoncrossing collections t ÝÑ c i u iprks of Q ɛ u ÝÑ toriented diagrams ÝÑ d t ÝÑ c pi l, j l qu lprks u defined by ÝÑ ci ÞÝÑ " ÝÑc pi, i q : ÝÑ ci is positive ÝÑ c pi, i q : ÝÑ ci is negative. In particular, each c-matrix C ɛ P c-matpq ɛ q determines a unique oriented diagram denoted ÝÑ d Cɛ strands. with n oriented a` a a` Example.. Let n and ɛ p`, `,, `, q so that Q ɛ ÝÑ ÐÝ ÝÑ. After performing the mutation sequence µ µ to the corresponding framed quiver, we have the c-matrix with its oriented diagram.» fi ffi ffi fl The following theorem shows oriented diagrams belonging to ÝÑ D n,ɛ are in bijection with c-matrices of Q ɛ. We delay its proof until Section because it makes heavy use of the concept of a mixed cobinary tree. Theorem.5. The map c-matpq ɛ q Ñ ÝÑ D n,ɛ induced by the map defined in Lemma. is a bijection... Proof of Lemma.5. The proof of Lemma.5 requires some notions from representation theory of finite dimensional algebras, which we now briefly review. For a more comprehensive treatment of the following notions, we refer the reader to [ASS06]. Definition.6. Given a quiver Q with #Q 0 n, the Euler characteristic (of Q) is the Z-bilinear (nonsymmetric) form Z n ˆ Z n Ñ Z defined by xdimpv q, dimpw qy ÿ iě0p q i dim Ext i kqpv, W q for every V, W P rep k pqq. For hereditary algebras A (e.g. path algebras of acyclic quivers), Ext i ApV, W q 0 for i ě and the formula reduces to xdimpv q, dimpw qy dim Hom kq pv, W q dim Ext kqpv, W q The following result gives a simple combinatorial formula for the Euler characteristic. We note that this formula is independent of the orientation of the arrows of Q. Lemma.7. [ASS06, Lemma VII..] Given an acyclic quiver Q with #Q 0 n and integral vectors x px, x,..., x n q, y py, y,..., y n q P Z n, the Euler characteristic of Q has the form xx, yy ÿ ÿ x i y i x spαq y tpαq ipq 0 αpq Next, we give a slight simplification of the previous formula. Recall that the support of V P rep k pqq is the set supppv q : ti P Q 0 : V i 0u. Thus for quivers of the form Q ɛ, any representation Xi,j ɛ P indprep kpq ɛ qq has supppxi,j ɛ q ri `, js. Lemma.8. Let Xk,l ɛ, Xɛ i,j P indprep kpq ɛ qq and A : ta P pq ɛ q : spaq, tpaq P supppxk,l ɛ q X supppxɛ i,jqu. Then xdimpxk,lq, ɛ dimpxi,jqy ɛ χ supppx ɛ k,l qxsupppxi,j ɛ q # `ta P pq ɛ q : spaq P supppxk,lq, ɛ tpaq P supppxi,jquza ɛ where χ supppx ɛ k,l qxsupppxi,j ɛ q if supppxk,l ɛ q X supppxɛ i,jq H and 0 otherwise. 8

9 Proof. We have that xdimpx ɛ k,l q, dimpxɛ i,j qy ÿ dimpxk,lq ɛ m dimpxi,jq ɛ m dimpxk,lq ɛ spaq dimpxi,jq ɛ tpaq mppq ɛq 0 appq ɛq # supppx q ɛk,l q X supppxɛ i,j #tα P pq ɛ q : spaq P supppxk,l ɛ q, tpaq P supppxɛ i,j qu # supppx q ɛk,l q X supppxɛ i,j #A # ta quza P pq ɛ q : spaq P supppx ɛk,l q, tpaq P supppxɛ i,j. Observe that if supppxk,l ɛ q X supppxɛ i,j q H, then #A #psupppxɛ k,l q X supppxɛ i,jqq. Otherwise #A 0. Thus xdimpx ɛ k,lq, dimpx ɛ i,jqy χ supppx ɛ k,l qxsupppx ɛ i,j q # `ta P pq ɛ q : spaq P supppx ɛ k,lq, tpaq P supppx ɛ i,jquza. In the sequel, we will use this formula for the Euler characteristic without further comment. We now present several lemmas that will be useful in the proof of Lemma.5. The proofs of the next four lemmas use very similar techniques so we only prove Lemma.9. The following four lemmas characterize when Hom kqɛ p, q and Ext kq ɛ p, q vanish for a given type A n quiver Q ɛ. The conditions describing when Hom kqɛ p, q and Ext kq ɛ p, q vanish are given in terms of inequalities satisfied by the indices that describe a pair of indecomposable representations of Q ɛ and the entries of ɛ. Lemma.9. Let Xk,l ɛ, Xɛ i,j P indprep kpq ɛ qq. Assume 0 ď i ă k ă j ă l ď n. iq Hom kqɛ pxi,j ɛ, Xɛ k,l q 0 if and only if ɛ k and ɛ j. iiq Hom kqɛ pxk,l ɛ, Xɛ i,j q 0 if and only if ɛ k ` and ɛ j `. iiiq Ext kq ɛ pxi,j ɛ, Xɛ k,l q 0 if and only if ɛ k ` and ɛ j `. ivq Ext kq ɛ pxk,l ɛ, Xɛ i,j q 0 if and only if ɛ k and ɛ j. Proof. We only prove iq and ivq as the proofs of iiq is very similar to that of iq the proof of iiiq is very similar to that of ivq. To prove iq, first assume there is a nonzero morphism θ : Xi,j ɛ Ñ Xɛ k,l. Clearly, θ s 0 if s R rk `, js. If θ s 0 for some s P rns, then θ s λ for some λ P k (i.e. θ s is a nonzero scalar transformation). As θ is a morphism of representations, it must satisfy that for any a P pq ɛ q the equality θ tpaq ϕ i,j a ϕ k,l a θ spaq holds. Thus for any a P ta ɛ k` k`,..., aɛj j u, we have θ tpaq θ spaq. As θ is nonzero, this implies that θ s λ for any s P rk `, js. If a a ɛ k k, then we have θ tpaq ϕ i,j a ϕ k,l a θ spaq θ tpaq 0. Thus ɛ k. Similarly, ɛ j. Conversely, it is easy to see that if ɛ k ɛ j, then θ : X ɛ i,j Ñ Xɛ k,l defined by θ s 0 if s R rk `, js and θ s otherwise is a nonzero morphism. Next, we prove ivq. Observe that by Lemma.8 we have dim Ext kq ɛ pxk,l ɛ, Xɛ i,j q dim Hom kq ɛ pxk,l ɛ, Xɛ i,j q xdimpxɛ k,l q, dimpxɛ i,j qy dim Hom kqɛ pxk,l ɛ, Xɛ i,j q `# tb quza P pq ɛ q : spbq P supppx ɛk,l q, tpbq P supppxɛ i,j. Note that # tb quza P pq ɛ q : spbq P supppx ɛk,l q, tpbq P supppxɛ i,j ď. Furthermore, the argument in the first paragraph of the proof shows that dim Hom kqɛ pxk,l ɛ, Xɛ i,j q ď. By iiq, we have that Ext kq ɛ pxk,l ɛ, Xɛ i,j q 0 if and only if ɛ k ɛ j. Lemma.0. Let Xk,l ɛ, Xɛ i,j P indprep kpq ɛ qq. Assume 0 ď i ă k ă l ă j ď n. iq Hom kqɛ pxi,j ɛ, Xɛ k,l q 0 if and only if ɛ k and ɛ l `. iiq Hom kqɛ pxk,l ɛ, Xɛ i,j q 0 if and only if ɛ k ` and ɛ l. iiiq Ext kq ɛ pxi,j ɛ, Xɛ k,l q 0 if and only if ɛ k ` and ɛ l. ivq Ext kq ɛ pxk,l ɛ, Xɛ i,j q 0 if and only if ɛ k and ɛ l `. 9 ÿ

10 Lemma.. Assume 0 ď i ă k ă j ď n. Then iq Hom kqɛ pxi,k ɛ, Xɛ k,j q 0 and Hom kq ɛ pxk,j ɛ, Xɛ i,k q 0. iiq Ext kq ɛ pxi,k ɛ, Xɛ k,j q 0 if and only if ɛ k `. iiiq Ext kq ɛ pxk,j ɛ, Xɛ i,k q 0 if and only if ɛ k. ivq Hom kqɛ pxi,k ɛ, Xɛ i,j q 0 if and only if ɛ k. vq Hom kqɛ pxi,j ɛ, Xɛ i,k q 0 if and only if ɛ k `. viq Ext kq ɛ pxi,k ɛ, Xɛ i,j q 0 and Ext kq ɛ pxi,j ɛ, Xɛ i,k q 0. viiq Hom kqɛ pxk,j ɛ, Xɛ i,j q 0 if and only if ɛ k `. viiiq Hom kqɛ pxi,j ɛ, Xɛ k,j q 0 if and only if ɛ k. ixq Ext kq ɛ pxk,j ɛ, Xɛ i,j q 0 and Ext kq ɛ pxi,j ɛ, Xɛ k,j q 0. Lemma.. Let Xk,l ɛ, Xɛ i,j P indprep kpq ɛ qq. Assume 0 ď i ă j ă k ă l ď n. Then iq Hom kqɛ pxi,j ɛ, Xɛ k,l q 0, Hom kq ɛ pxk,l ɛ, Xɛ i,j q 0, iiq Ext kq ɛ pxi,j ɛ, Xɛ k,l q 0, Ext kq ɛ pxk,l ɛ, Xɛ i,j q 0. Next, we present three geometric facts about pairs of distinct strands. These geometric facts will be crucial in our proof of Lemma.5. Lemma.. If two distinct strands cpi, j q and cpi, j q on S n,ɛ intersect nontrivially, then cpi, j q and cpi, j q can be represented by a pair of monotone curves that have a unique transversal crossing. Proof. Suppose cpi, j q and cpi, j q intersect nontrivially. Without loss of generality, we assume i ď i. Let γ k P cpi k, j k q with k P rs be monotone curves. There are two cases: aq i ď i ă j ď j bq i ď i ă j ď j. Suppose that case aq holds. Let px, y q P tpx, yq P R : x i ď x ď x j u denote a point where γ crosses γ transversally. If ɛ i (resp. ɛ i `), isotope γ relative to ɛ i and px, y q in such a way that the monotonocity of γ is preserved and so that γ lies strictly above (resp. strictly below) γ on tpx, yq P R : x i ď x ă x u. Next, if ɛ j (resp. ɛ j `), isotope γ relative to px, y q and ɛ j in such a way that the monotonicity of γ is preserved and so that γ lies strictly above (resp. strictly below) γ on tpx, yq P R : x ă x ď x j u. This process produces two monotone curves γ P cpi, j q and γ P cpi, j q that have a unique transversal crossing. The proof in case bq is very similar. Lemma.. Let cpi, j q and cpi, j q be distinct strands on S n,ɛ that intersect nontrivially. Then cpi, j q and cpi, j q do not share an endpoint. Proof. Suppose cpi, j q and cpi, j q share an endpoint. Since cpi, j q and cpi, j q intersect nontrivially, then there exist curves γ k P cpi k, j k q with k P t, u that have a unique transversal crossing. However, since cpi, j q and cpi, j q share an endpoint, γ and γ are isotopic relative to their endpoints to curves with no transversal crossing. This contradicts that cpi, j q and cpi, j q share an endpoint. Remark.5. If cpi, j q and cpi, j q are two distinct strands on S n,ɛ that do not intersect nontrivially, then cpi, j q and cpi, j q can be represented by a pair of monotone curves γ l P cpi l, j l q where l P rs that are nonintersecting, except possibly at their endpoints. We now arrive at the proof of Lemma.5. The proof is a case by case analysis where the cases are given in terms of inequalities satisfied by the indices that describe a pair of indecomposable representations of Q ɛ and the entries of ɛ. Proof of Lemma.5 a). Let X ɛ i,j : U and Xɛ k,l : V. Assume that the strands Φ ɛpx ɛ i,j q and Φ ɛpx ɛ k,l q intersect nontrivially. By Lemma., we can assume without loss of generality that either 0 ď i ă k ă j ă l ď n or 0 ď i ă k ă l ă j ď n. By Lemma., we can represent Φ ɛ px ɛ i,j q and Φ ɛpx ɛ k,l q by monotone curves γ i,j and γ k,l that have a unique transversal crossing. Furthermore, we can assume that this unique crossing occurs between ɛ k and ɛ k`. There are four possible cases: iq ɛ k ɛ k`, iiq ɛ k and ɛ k` `, iiiq ɛ k ɛ k` `, ivq ɛ k ` and ɛ k`. We illustrate these cases up to isotopy in Figure. We see that in cases iq and iiq (resp. iiiq and ivq) γ k,l lies above (resp. below) γ i,j inside of tpx, yq P R : x k` ď x ď x mintl,ju u. 0

11 ɛ k+ = + ɛ k = + ɛ k+ = + ɛ k = + ɛ k = ɛ k+ = ɛ k = ɛ k+ = Figure. The four types of crossings Suppose γ k,l lies above γ i,j inside tpx, yq P R : x k` ď x ď x mintl,ju u. Then ɛ mintl,ju " ` : mintl, ju l : mintl, ju j otherwise γ k,l and γ i,j would have a nonunique transversal crossing. If mintl, ju l, we have 0 ď i ă k ă l ă j ď n, ɛ k, and ɛ l `. Now by Lemma.0, we have that Hom kqɛ pxi,j ɛ, Xɛ k,l q 0 and Ext kq ɛ pxk,l ɛ, Xɛ i,j q 0. If mintl, ju j, then 0 ď i ă k ă j ă l ď n, ɛ k, and ɛ j. Thus, by Lemma.9, we have that Hom kqɛ pxi,j ɛ, Xɛ k,l q 0 and Ext kq ɛ pxk,l ɛ, Xɛ i,j q 0. Similarly, if γ i,j lies above γ k,l inside tpx, yq P R : x k` ď x ď x mintl,ju u, it follows that ɛ mintl,ju " : mintl, ju l ` : mintl, ju j. If mintl, ju l, then Lemma.0 implies that Hom kqɛ pxk,l ɛ, Xɛ i,j q 0 and Ext kq ɛ pxi,j ɛ, Xɛ k,lq 0. If mintl, ju j, then Lemma.9 implies that Hom kqɛ pxk,l ɛ, Xɛ i,j q 0 and Ext kq ɛ pxi,j ɛ, Xɛ k,lq 0. Thus we conclude that neither pxi,j ɛ, Xɛ k,l q nor pxɛ k,l, Xɛ i,jq are exceptional pairs. Conversely, assume that neither pu, V q nor pv, Uq are exceptional pairs where Xi,j ɛ : U and Xɛ k,l : V. Then at least one of the following is true: aq Hom kqɛ pxi,j ɛ, Xɛ k,l q 0 and Hom kq ɛ pxk,l ɛ, Xɛ i,j q 0, bq Hom kqɛ pxi,j ɛ, Xɛ k,l q 0 and Ext kq ɛ pxk,l ɛ, Xɛ i,j q 0, cq Ext kq ɛ pxi,j ɛ, Xɛ k,l q 0 and Hom kq ɛ pxk,l ɛ, Xɛ i,j q 0, dq Ext kq ɛ pxi,j ɛ, Xɛ k,l q 0 and Ext kq ɛ pxk,l ɛ, Xɛ i,j q 0. As Xi,j ɛ and Xɛ k,l are indecomposable and distinct, we have that Hom kq ɛ pxi,j ɛ, Xɛ k,l q 0 or Hom kq ɛ pxk,l ɛ, Xɛ i,j q 0. Without loss of generality, assume that Hom kqɛ pxk,l ɛ, Xɛ i,j q 0. Thus bq or dq hold so Ext kq ɛ pxk,l ɛ, Xɛ i,j q 0. Then Lemma. and Lemma. imply that 0 ď i ă k ă j ă l ď n or 0 ď i ă k ă l ă j ď n. If 0 ď i ă k ă j ă l ă n, we have ɛ k ɛ j by Lemma.9 as Hom kqɛ pxi,j ɛ, Xɛ k,l q 0 and Ext kq ɛ pxk,l ɛ, Xɛ i,j q 0. Let γ i,j P Φ ɛ pxi,j ɛ q and γ k,l P Φ ɛ pxk,l ɛ q. We can assume that there exists δpkq ą 0 such that γ i,j and γ k,l have no transversal crossing inside tpx, yq P R : x k ď x ď x k ` δpkqu. This implies that γ i,j lies above γ k,l inside tpx, yq P R : x k ď x ď x k ` δpkqu. Similarly, we can assume there exists δpjq ą 0 such that γ i,j and γ k,l have no transversal crossing inside tpx, yq P R : x j δpjq ď x ď x j u. Thus γ i,j lies below γ k,l inside tpx, yq P R : x j δpjq ď x ď x j u. This means γ i,j and γ k,l must have at least one transversal crossing. Thus Φ ɛ pxi,j ɛ q and Φ ɛpxk,l ɛ q intersect nontrivially. An analogous argument shows that if 0 ď i ă k ă l ă j ď n, then Φ ɛ pxi,j ɛ q and Φ ɛpxk,l ɛ q intersect nontrivially. Proof of Lemma.5 b). Assume that Φ ɛ puq is clockwise from Φ ɛ pv q. Then we have that one of the following holds: aq X ɛ k,j U and Xi,k ɛ V for some 0 ď i ă k ă j ď n, bq Xi,k ɛ U and Xɛ k,j V for some 0 ď i ă k ă j ď n, cq Xi,j ɛ U and Xɛ i,k V for some 0 ď i ă j ď n and 0 ď i ă k ď n, dq Xi,j ɛ U and Xɛ k,j V for some 0 ď i ă j ď n and 0 ď k ă j ď n. In Case aq, we have that ɛ k since Φ ɛ pxk,j ɛ q is clockwise from Φ ɛpxi,k ɛ q. By Lemma. iq and iiq, we have that Hom kqɛ pxi,k ɛ, Xɛ k,j q 0 and Ext kq ɛ pxi,k ɛ, Xɛ k,j q 0. Thus pxɛ k,j, Xɛ i,kq is an exceptional pair. By Lemma. iiiq, we have that Ext kq ɛ pxk,j ɛ, Xɛ i,k q 0. Thus pxɛ i,k, Xɛ k,jq is not an exceptional pair.

12 In Case bq, we have that ɛ k ` since Φ ɛ pxi,k ɛ q is clockwise from Φ ɛpxk,j ɛ q. By Lemma. iq and iiiq, we have that Hom kqɛ pxk,j ɛ, Xɛ i,k q 0 and Ext kq ɛ pxk,j ɛ, Xɛ i,k q 0. Thus pxɛ i,k, Xɛ k,jq is an exceptional pair. By Lemma. iiq, we have that Ext kq ɛ pxi,k ɛ, Xɛ k,j q 0. Thus pxɛ k,j, Xɛ i,kq is not an exceptional pair. In Case cq, if j ă k, it follows that ɛ j. Indeed, Φ ɛ pxi,j ɛ q is clockwise from Φ ɛpxi,k ɛ q and so by Lemma. the two do not intersect nontrivially. Now by Remark.5, we can choose monotone curves γ i,k P Φ ɛ pxi,k ɛ q and γ i,j P Φ ɛ pxi,j ɛ q such that γ i,k lies strictly above γ i,j on tpx, yq P R : x i ă x ď x j u. Thus ɛ j. By Lemma. vq and viq, we have that Hom kqɛ pxi,k ɛ, Xɛ i,j q 0 and Ext kq ɛ pxi,k ɛ, Xɛ i,j q 0 so that pxɛ i,j, Xɛ i,kq is an exceptional pair. By Lemma. ivq, we have that Hom kqɛ pxi,j ɛ, Xɛ i,k q 0. Thus pxɛ i,k, Xɛ i,jq is not an exceptional pair. Similarly, one shows that if k ă j, then ɛ k `. By Lemma. ivq and viq, we have that Hom kqɛ pxi,k ɛ, Xɛ i,j q 0 and Ext kq ɛ pxi,k ɛ, Xɛ i,j q 0 so that pxɛ i,j, Xɛ i,kq is an exceptional pair. By Lemma. vq, we have that Hom kqɛ pxi,j ɛ, Xɛ i,k q 0. Thus pxɛ i,k, Xɛ i,jq is not an exceptional pair. The proof in Case dq is completely analogous to the proof in Case cq so we omit it. Conversely, let U Xi,j ɛ and V Xɛ k,l and assume that pxɛ i,j, Xɛ k,l q is an exceptional pair and pxɛ k,l, Xɛ i,j q is not an exceptional pair. This implies that at least one of the following holds: q Hom kqɛ pxk,l ɛ, Xɛ i,j q 0, Ext kq ɛ pxk,l ɛ, Xɛ i,j q 0, and Hom kq ɛ pxi,j ɛ, Xɛ k,l q 0, q Hom kqɛ pxk,l ɛ, Xɛ i,j q 0, Ext kq ɛ pxk,l ɛ, Xɛ i,j q 0, and Ext kq ɛ pxi,j ɛ, Xɛ k,l q 0. By Lemma., we know that ri, js X rk, ls H. This implies that either iq Φ ɛ pxi,j ɛ q and Φ ɛpxk,l ɛ q share an endpoint, iiq 0 ď i ă k ă j ă l ď n, iiiq 0 ď i ă k ă l ă j ď n, ivq 0 ď k ă i ă l ă j ď n, vq 0 ď k ă i ă j ă l ď n. We will show that Φ ɛ pxi,j ɛ q and Φ ɛpxk,l ɛ q share an endpoint. Suppose 0 ď i ă k ă j ă l ď n. Then since Hom kqɛ pxk,l ɛ, Xɛ i,j q 0, Ext kq ɛ pxk,l ɛ, Xɛ i,jq 0, we have by Lemma.9 iiq and ivq that either ɛ k and ɛ j ` or ɛ k ` and ɛ j. However, as Hom kqɛ pxi,j ɛ, Xɛ k,l q 0 or Ext kq ɛ pxi,j ɛ, Xɛ k,l q 0, Lemma.9 iq and iiiq we have that ɛ k ɛ j or ɛ k ɛ j `. This contradicts that 0 ď i ă k ă j ă l ď n. An analogous argument shows that i, j, k, l do not satisfy 0 ď k ă i ă l ă j ď n. Suppose 0 ď i ă k ă l ă j ď n. Then since Hom kqɛ pxk,l ɛ, Xɛ i,j q 0, Ext kq ɛ pxk,l ɛ, Xɛ i,jq 0, we have by Lemma.0 iiq and ivq that either ɛ k ɛ l ` or ɛ k ɛ l. However, as Hom kqɛ pxi,j ɛ, Xɛ k,l q 0 or Ext kq ɛ pxi,j ɛ, Xɛ k,l q 0, Lemma.0 iq and iiiq we have that ɛ k and ɛ l ` or ɛ k ` and ɛ l. This contradicts that 0 ď i ă k ă l ă j ď n. An analogous argmuent shows that i, j, k, l do not satisfy 0 ď k ă i ă j ă l ď n. We conclude that Φ ɛ puq and Φ ɛ pv q share an endpoint. Thus we have that one of the following holds where we forget the previous roles played by i, j, k: aq Xk,j ɛ U and Xɛ i,k V for some 0 ď i ă k ă j ď n, bq Xi,k ɛ U and Xɛ k,j V for some 0 ď i ă k ă j ď n, cq Xi,j ɛ U and Xɛ i,k V for some 0 ď i ă j ď n and 0 ď i ă k ď n, dq Xi,j ɛ U and Xɛ k,j V for some 0 ď i ă j ď n and 0 ď k ă j ď n. Suppose Case aq holds. Then since pu, V q is an exceptional pair, we have Ext kq ɛ pxi,k ɛ, Xɛ k,jq 0. By Lemma. iiq, we have that ɛ k. Thus Φ ɛ puq is clockwise from Φ ɛ pv q. Suppose Case bq holds. Then since pu, V q is an exceptional pair, we have Ext kq ɛ pxk,j ɛ, Xɛ i,kq 0. By Lemma. iiiq, we have that ɛ k `. Thus Φ ɛ puq is clockwise from Φ ɛ pv q. Suppose Case cq holds. Assume k ă j. Then Lemma. ivq and the fact that Hom kqɛ pxi,k ɛ, Xɛ i,j q 0 imply that ɛ k `. Thus we have that Φ ɛ puq Φ ɛ pxi,j ɛ q is clockwise from Φ ɛpv q Φ ɛ pxi,k ɛ q. Now suppose j ă k. Then Lemma. vq and Hom kqɛ pxi,k ɛ, Xɛ i,j q 0 imply that ɛ j. Thus we have that Φ ɛ puq Φ ɛ pxi,j ɛ q is clockwise from Φ ɛ pv q Φ ɛ pxi,k ɛ q. The proof in Case dq is very similar so we omit it. Proof of Lemma.5 cq. Observe that two strands cpi, j q and cpi, j q share and endpoint if and only if one of the two strands is clockwise from the other. Thus Lemma.5 aq and bq implies that Φ ɛ puq and Φ ɛ pv q do not intersect at any of their endpoints and they do not intersect nontrivially if and only if both pu, V q and pv, Uq are exceptional pairs.

13 . Mixed cobinary trees We recall the definition of an ɛ-mixed cobinary tree and construct a bijection between the set of (isomorphism classes of) such trees and the set of maximal oriented strand diagrams on S n,ɛ. Definition.. [IO] Given a sign function ɛ : r0, ns Ñ t`, u, an ɛ-mixed cobinary tree (MCT) is a tree T embedded in R with vertex set tpi, y i q i P r0, nsu and edges straight line segments and satisfying the following conditions. aq None of the edges is horizontal. bq If ɛ i ` then y i ě z for any pi, zq P T. So, the tree goes under pi, y i q. cq If ɛ i then y i ď z for any pi, zq P T. So, the tree goes over pi, y i q. dq If ɛ i ` then there is at most one edge descending from pi, y i q and at most two edges ascending from pi, y i q and not on the same side. eq If ɛ i then there is at most one edge ascending from pi, y i q and at most two edges descending from pi, y i q and not on the same side. Two MCT s T, T are isomorphic as MCT s if there is a graph isomorphism T T which sends pi, y i q to pi, yi q and so that corresponding edges have the same sign of their slopes. Given a MCT T, there is a partial ordering on r0, ns given by i ă T j if the unique path from pi, y i q to pj, y j q in T is monotonically increasing. Isomorphic MCT s give the same partial ordering by definition. Conversely, the partial ordering ă T determines T uniquely up to isomorphism since T is the Hasse diagram of the partial ordering ă T. We sometimes decorate MCT s with leaves at vertices so that the result is trivalent, i.e., with three edges incident to each vertex. See, e.g., Figure 5. The ends of these leaves are not considered to be vertices. In that case, each vertex with ɛ ` forms a Y and this pattern is vertically inverted for ɛ. The position of the leaves is uniquely determined. p, q p, q p0, 0q p, 0q p, q Figure. A MCT with ɛ ɛ, ɛ ` and any value for ɛ 0, ɛ. Figure 5. This MCT (in blue) has added green leaves showing that ɛ p, `,, q. In Figure 5, the four vertices have coordinates p0, y 0 q, p, y q, p, y q, p, y q where y i can be any real numbers so that y 0 ă y ă y ă y. This inequality defines an open subset of R which is called the region of this tree T. More generally, for any MCT T, the region of T, denoted R ɛ pt q, is the set of all points y P R n` with the property that there exists a mixed cobinary tree T which is isomorphic to T so that the vertex set of T is tpi, y i q i P rnsu. Theorem.. [IO] Let n and ɛ : rns Ñ t`, u be fixed. Then, for every MCT T, the region R ɛ pt q is convex and nonempty. Furthermore, every point y py 0,, y n q in R n` with distinct coordinates lies in R ɛ pt q for a unique T (up to isomorphism). In particular these regions are disjoint and their union is dense in R n`. For a fixed n and ɛ : rns Ñ t`, u we will construct a bijection between the set T ɛ of isomorphism classes of mixed cobinary trees with sign function ɛ and the set ÝÑ D n,ɛ defined in Definition.. Lemma.. Let ÝÑ d t ÝÑ c pi l, j l qu lprns P ÝÑ D n,ɛ. Let p, q be two points on this graph so that q lies directly above p. Then each edge of ÝÑ d in the unique path γ from p to q is oriented in the same direction as γ. Proof. The proof will be by induction on the number m of internal vertices of the path γ. If m with internal vertex ɛ i then the path γ has only two edges of ÝÑ d : one going from p to ɛ i, say to the left, and the other going from ɛ i back to q. Since ÝÑ d P ÝÑ D n,ɛ, the edge coming into ɛ i from its right is below the edge going out from ɛ i to q. Therefore the orientation of these two edges in ÝÑ d matches that of γ. Now suppose that m ě and the lemma holds for smaller m. There are two cases. Case : The path γ lies entirely on one side of p and q (as in the case m ). Case : γ has internal vertices on both sides of p, q.

14 Case : Suppose by symmetry that γ lies entirely on the left side of p and q. Let j be maximal so that ɛ j is an internal vertex of γ. Then γ contains an edge connecting ɛ j to either p or q, say p. And the edge of γ ending in q contains a unique point r which lies above ɛ j. This forces the sign to be ɛ j. By induction on m, the rest of the path γ, which goes from ɛ j to r has orientation compatible with that of ÝÑ d. So, it must be oriented outward from ɛ j. Any other edge at ɛ is oriented inward. So, the edge from p to ɛ j is oriented from p to ɛ j as required. The edge coming into r from the left is oriented to the right (by induction). So, this same edge continues to be oriented to the right as it goes from r to q. The other subcases (when ɛ j is connected to q instead of p and when γ lies to the right of p and q) are analogous. Case : Suppose that γ on both sides of p and q. Then γ passes through a third point, say r, on the same vertical line containing p and q. Let γ 0 and γ denote the parts of γ going from p to r and from r to q respectively. Then γ 0, γ each have at least one internal vertex. So, the lemma holds for each of them separately. There are three subcases: either (a) r lies below p, (b) r lies above q or (c) r lies between p and q. In subcase (a), we have, by induction on m, that γ 0, γ are both oriented away from r. So, r ɛ k ` which contradicts the assumption that q lies above r. Similarly, subcase (c) is not possible. In subcase (b), we have by induction on m that the orientations of the edges of ÝÑ d are compatible with the orientations of γ 0 and γ. So, the lemma holds in subcase (b), which is the only subcase of Case which is possible. Therefore, the lemma holds in all cases. Theorem.. For each ÝÑ d t ÝÑ c pi l, j l qu lprns P ÝÑ D n,ɛ, let Rp ÝÑ d q denote the set of all y P R n` so that y i ă y j for any ÝÑ c pi, jq in ÝÑ d. Then Rp ÝÑ d q R ɛ pt q for a uniquely determined mixed cobinary tree T. Furthermore, this gives a bijection ÝÑ D n,ɛ T ɛ. Proof. We first verify the existence of a mixed cobinary tree T for every choice of y P Rp ÝÑ d q. Since the strand diagram is a tree, the vector y is uniquely determined by y 0 P R and y jl y il ą 0, l P rns, which are arbitrary. Given such a y, we need to verify that the n line segments L l in R connecting the pairs of points pi l, y il q, pj l, y jl q meet only on their endpoints. This follows from the lemma above. If two of these line segments, say L k, L l, meet then they come from two distinct points p P ÝÑ c pi k, j k q and q ÝÑ c pi l, j l q in the strand diagram which lie one the same vertical line. If q lies above p in the strand diagram then, by Lemma., the unique path γ from p to q is oriented positively. This implies that the y coordinate of the point in L k corresponding to p is less that the y coordinate of the point in L l corresponding to q. Thus, this intersection is not possible. So, T is a linearly embedded tree. The lemma also implies that the tree T lies above all negative vertices and below all positive vertices. The other parts of Definition. follow from the definition of an oriented strand diagram. Therefore T P T ɛ. Since this argument works for every y P Rp ÝÑ d q, we see that Rp ÝÑ d q R ɛ pt q as claimed. A description of the inverse mapping T ɛ Ñ ÝÑ D n,ɛ is given as follows. Take any MCT T and deform the tree by moving all vertices vertically to the subset rns ˆ 0 on the x-axis and deforming the edges in such a way that they are always embedded in the plane with no vertical tangents and so that their interiors do not meet. The result is an oriented strand diagram ÝÑ d with Rp ÝÑ d q R ɛ pt q. It is clear that these are inverse mappings giving the desired bijection ÝÑ D n,ɛ T ɛ. Example.5. The MCTs in Figures and 5 above give the oriented strand diagrams: and the oriented strand diagram in Example. gives the MCT: ðñ We now arrive at the proof of Theorem.5. This theorem follows from the fact that oriented diagrams belonging to ÝÑ D n,ɛ can be regarded as mixed cobinary trees by Theorem..