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On the computation of order types of hammocks for domestic string algebras

Shantanu Sardar and Amit Kuber Department of Mathematics and Statistics
Indian Institute of Technology, Kanpur
Uttar Pradesh, India
[email protected], [email protected]
Abstract.

For the representation-theoretic study of domestic string algebras, Schröer introduced a version of hammocks that are bounded discrete linear orders. He introduced a finite combinatorial gadget called the bridge quiver, which we modified in the prequel of this paper to get a variation called the arch bridge quiver. Here we use it as a tool to provide an algorithm to compute the order type of an arbitrary closed interval in such hammocks. Moreover, we characterize the class of order types of these hammocks as the bounded discrete ones amongst the class of finitely presented linear orders–the smallest class of linear orders containing finite linear orders as well as ω\omega, and that is closed under isomorphisms, order reversal, finite order sums and lexicographic products.

Key words and phrases:
domestic string algebra, hammocks, bridge quiver, finitely presented linear order
2020 Mathematics Subject Classification:
16G30

1. Introduction

This paper is the sequel of [9] where the authors embarked on the journey towards the computation of the order types of certain linear orders known as hammocks for domestic string algebras–that paper introduces a new finite combinatorial tool called the arch bridge quiver and investigates its properties. Recall that a string algebra Λ\Lambda over an algebraically closed field 𝒦\mathcal{K} is presented as a certain quotient 𝒦Q/ρ\mathcal{K}Q/\langle\rho\rangle of the path algebra 𝒦Q\mathcal{K}Q of a quiver QQ, where ρ\rho is a set of monomial relations. The vertices of the Auslander-Reiten (A-R) quiver were classified essentially in [3] in terms of certain walks on the quiver known as strings and bands, whereas its arrows were classified by Ringel and Butler [2].

A string algebra is domestic if it has only finitely many bands. For two bands 𝔟1\mathfrak{b}_{1} and 𝔟2\mathfrak{b}_{2}, say 𝔟1\mathfrak{b}_{1} and 𝔟2\mathfrak{b}_{2} commute if there are cyclic permutations 𝔟1,𝔟1\mathfrak{b}^{\prime}_{1},\mathfrak{b}^{\prime\prime}_{1} of 𝔟1\mathfrak{b}_{1} and 𝔟2,𝔟2\mathfrak{b}^{\prime}_{2},\mathfrak{b}^{\prime\prime}_{2} of 𝔟2\mathfrak{b}_{2} such that 𝔟1𝔟2\mathfrak{b}^{\prime}_{1}\mathfrak{b}^{\prime}_{2} and 𝔟2𝔟1\mathfrak{b}^{\prime\prime}_{2}\mathfrak{b}^{\prime\prime}_{1} are strings. If two bands commute in a string algebra then it is necessarily non-domestic by [4, Corollary 3.4.1].

For a vertex vv of the quiver QQ the simplest version of hammock, denoted Hl(v)H_{l}(v), is the collection of all strings starting at the vertex. It can be equipped with an order <l<_{l} such that 𝔵<l𝔶\mathfrak{x}<_{l}\mathfrak{y} in Hl(v)H_{l}(v) guarantees the existence of a canonical graph map M(𝔵)M(𝔶)M(\mathfrak{x})\to M(\mathfrak{y}) between the corresponding string modules. Schröer [7, § 2.5] showed that (Hl(v),<l)(H_{l}(v),<_{l}) is a bounded discrete linear order. Prest and Schröer showed that it has finite m-dimension (essentially [5, Theorem 1.3]). For more flexibility we consider the hammock Hl(𝔵0)H_{l}(\mathfrak{x}_{0}) for an arbitrary string 𝔵0\mathfrak{x}_{0}, which is the collection of all strings which admit 𝔵0\mathfrak{x}_{0} as a left substring–this too is a bounded discrete linear order with respect to the induced order <l<_{l}, say with 𝔪1(𝔵0)\mathfrak{m}_{1}(\mathfrak{x}_{0}) and 𝔪1(𝔵0)\mathfrak{m}_{-1}(\mathfrak{x}_{0}) as the maximal and minimal elements respectively. For technical reasons it is necessary to write the hammock as a union Hl(𝔵0)=Hl1(𝔵0)Hl1(𝔵0)H_{l}(\mathfrak{x}_{0})=H_{l}^{1}(\mathfrak{x}_{0})\cup H_{l}^{-1}(\mathfrak{x}_{0}), where Hl1(𝔵0)H_{l}^{1}(\mathfrak{x}_{0}) is the interval [𝔵0,𝔪1(𝔵0)][\mathfrak{x}_{0},\mathfrak{m}_{1}(\mathfrak{x}_{0})] and Hl1(𝔵0)=[𝔪1(𝔵0),𝔵0]H_{l}^{-1}(\mathfrak{x}_{0})=[\mathfrak{m}_{-1}(\mathfrak{x}_{0}),\mathfrak{x}_{0}]. Given 𝔵<l𝔶\mathfrak{x}<_{l}\mathfrak{y} in Hl(𝔵0)H_{l}(\mathfrak{x}_{0}), the goal of this paper is to compute the order type of the interval [𝔵,𝔶][\mathfrak{x},\mathfrak{y}] via the assignment of a label, which we call a ‘term’, to the path from 𝔵\mathfrak{x} to 𝔶\mathfrak{y}. What we call a term in this paper is referred to as a ‘real term’ in [4, § 2.4].

Some classes of finite Hausdorff rank linear orders were studied in [1]. The class LOfp\mathrm{LO}_{\mathrm{fp}} [1, § 5] is the smallest class of linear orders containing 11 that is closed under ++, ω×-\omega\times\mbox{-} and ω×-\omega^{*}\times\mbox{-} operations. Its subclass consisting of bounded discrete linear orders is denoted by dLOfpb\mathrm{dLO}_{\mathrm{fp}}^{\mathrm{b}}. The main results of this paper (Corollary 10.14 and Proposition 10.18) show that dLOfpb\mathrm{dLO}_{\mathrm{fp}}^{\mathrm{b}} is precisely the class of order types of hammocks for domestic string algebras.

Fix a domestic string algebra Λ\Lambda, a string 𝔵0\mathfrak{x}_{0} and i{1,1}i\in\{1,-1\}. Motivated by the notion of the bridge quiver introduced by Schröer for the study of hammocks for domestic string algebras, the authors introduced the notion of the extended arch bridge quiver in [9], denoted 𝒬Bai(𝔵0)\mathcal{HQ}^{\mathrm{Ba}}_{i}(\mathfrak{x}_{0}), that lies between the extended bridge quiver and the extended weak bridge quiver. For i{1,1}i\in\{1,-1\}, the paths in 𝒬Bai(𝔵0)\mathcal{HQ}^{\mathrm{Ba}}_{i}(\mathfrak{x}_{0}) generate all strings in Hli(𝔵0)H_{l}^{i}(\mathfrak{x}_{0}) in the precise sense described in [4, Lemma 3.3.4].

In this paper, we introduce the notion of a ‘decorated tree’, denoted 𝒯𝔣i(𝔵0)\mathcal{T}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0}), that linearizes the extended arch bridge quiver 𝒬Bai(𝔵0)\mathcal{HQ}^{\mathrm{Ba}}_{i}(\mathfrak{x}_{0}) from the viewpoint of the string 𝔵0\mathfrak{x}_{0} placed at the root; here the adjective decorated refers to the assignment of either a ++ or - sign to each non-root vertex and, for each vertex, a linear ordering of the children with the same sign. The data present as part of the decoration of the tree allows us to inductively associate terms to individual vertices of 𝒯𝔣i(𝔵0)\mathcal{T}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0}) so that the term associated with the root, denoted μi(𝔵0;𝔪i(𝔵0))\mu_{i}(\mathfrak{x}_{0};\mathfrak{m}_{i}(\mathfrak{x}_{0})), labels Hli(𝔵0)H_{l}^{i}(\mathfrak{x}_{0}). The following flowchart concisely depicts the steps of the algorithm.

(Λ,i,x0)𝒬Bai(x0)𝒯𝔣i(x0)μi(𝔵0;𝔪i(𝔵0))𝒪(μi(𝔵0;𝔪i(𝔵0)))Hli(x0).(\Lambda,i,x_{0})\mapsto\mathcal{HQ}^{\mathrm{Ba}}_{i}(x_{0})\mapsto\mathcal{T}^{\mathfrak{f}}_{i}(x_{0})\mapsto\mu_{i}(\mathfrak{x}_{0};\mathfrak{m}_{i}(\mathfrak{x}_{0}))\mapsto\mathcal{O}(\mu_{i}(\mathfrak{x}_{0};\mathfrak{m}_{i}(\mathfrak{x}_{0})))\cong H_{l}^{i}(x_{0}).

To further explain a rough idea behind the proof we need some terminology and notation. Recall from [9] that for a string 𝔵\mathfrak{x} of positive length, we say θ(𝔵)=1\theta(\mathfrak{x})=1 if its first syllable is an inverse syllable, otherwise we say θ(𝔵)=1\theta(\mathfrak{x})=-1.

Definition 1.1.

Given 𝔵(𝔵0)Hl(𝔵0)\mathfrak{x}(\neq\mathfrak{x}_{0})\in H_{l}(\mathfrak{x}_{0}), we say that P(𝔵):=(𝔵0;𝔵1,,𝔵n)P(\mathfrak{x}):=(\mathfrak{x}_{0};\mathfrak{x}_{1},\ldots,\mathfrak{x}_{n}) is the standard partition of 𝔵\mathfrak{x} if

  1. (1)

    we have 𝔵=𝔵n𝔵1𝔵0\mathfrak{x}=\mathfrak{x}_{n}\ldots\mathfrak{x}_{1}\mathfrak{x}_{0} where |𝔵k|>0|\mathfrak{x}_{k}|>0 for each 1kn1\leq k\leq n;

  2. (2)

    for each 1k<n1\leq k<n the string 𝔵k𝔵1𝔵0\mathfrak{x}_{k}\ldots\mathfrak{x}_{1}\mathfrak{x}_{0} forks;

  3. (3)

    the only proper left substrings of 𝔵\mathfrak{x} in Hl(𝔵0)H_{l}(\mathfrak{x}_{0}) which fork are the ones described above.

We say that the signature type of 𝔵\mathfrak{x}, denoted Θ(𝔵0;𝔵)\Theta(\mathfrak{x}_{0};\mathfrak{x}), is (θ(𝔵1),,θ(𝔵n))(\theta(\mathfrak{x}_{1}),\ldots,\theta(\mathfrak{x}_{n})). In case |𝔵0|=0|\mathfrak{x}_{0}|=0, we denote Θ(𝔵0;𝔵)\Theta(\mathfrak{x}_{0};\mathfrak{x}) by Θ(𝔵)\Theta(\mathfrak{x}).

Remark 1.2.

If 𝔵,𝔵Hl(𝔵0)\mathfrak{x},\mathfrak{x}^{\prime}\in H_{l}(\mathfrak{x}_{0}) fork then Θ(𝔵0;𝔵)Θ(𝔵0;𝔵)\Theta(\mathfrak{x}_{0};\mathfrak{x})\neq\Theta(\mathfrak{x}_{0};\mathfrak{x}^{\prime}).

For i{1,1},𝔵Hl(𝔵0),𝔵𝔵0i\in\{1,-1\},\mathfrak{x}\in H_{l}(\mathfrak{x}_{0}),\mathfrak{x}\neq\mathfrak{x}_{0}, say that Θ(𝔵0;𝔵)=C(i)\Theta(\mathfrak{x}_{0};\mathfrak{x})=C(i) if it is of the form (i,,i)(i,\ldots,i). Say that Θ(𝔵0;𝔵)=F(i)\Theta(\mathfrak{x}_{0};\mathfrak{x})=F(i) if it is of the form (i,i,,i)(i,-i,\ldots,-i), where i-i may not appear; here CC and FF stand for the words constant and flipped respectively.

The simplest possible terms are associated with the flipped signature types, and hence lot of the effort of the main proof goes into constructing strings whose signature types are so. We use these as starting points for building strings with more complex signature types. The interaction between signature types, left \mathbb{N}-strings and terms is an essential feature of this proof.

Continuing from [9] we provide a set of (counter)examples of algebras with peculiar properties which could be useful elsewhere. Even though there is lot of literature and explicit proofs for gentle algebras, we believe that these two papers are the first systematic attempt to study the combinatorics of strings for all domestic string algebras.

The rest of the paper is organized as follows. The concept of terms together with their associated order types is introduced in §2, the example of a particular domestic string algebra, namely Λ2\Lambda_{2}, is discussed in detail. Even though the bridge quiver of a domestic string algebra does not contain any directed cycle, its underlying undirected graph could still contain cycles. After fixing a string 𝔵0\mathfrak{x}_{0} and parity ii, we linearize/separate all paths in the extended arch bridge quiver 𝒬Bai(𝔵0)\mathcal{HQ}^{\mathrm{Ba}}_{i}(\mathfrak{x}_{0}) with the help of the concept of decorated trees introduced in §4. The results in §3, §5 and §6 help to classify strings in Hli(𝔵0)H^{l}_{i}(\mathfrak{x}_{0}) into various classes. Recall that there are only finitely many H-reduced strings in a domestic string algebra [9, Corollary 5.8]. In §7, we characterize all H-reduced forking strings using the vertices of the decorated tree and then use them to build all H-reduced strings in §8. The signature types of these H-reduced strings are computed in §9. Finally the algorithm (Algorithm 10.9) for computation of the term associated with a closed interval in the hammock, and hence that of the order type, is completed in §10.

The reader is advised to refer to [9] for any terminology/notation not explained in this paper.

Acknowledgements

The first author thanks the Council of Scientific and Industrial Research (CSIR) India - Research Grant No. 09/092(0951)/2016-EMR-I for the financial support.

Data Availability Statement

Data sharing not applicable to this article as no datasets were generated or analysed during the current study.

2. Labelling paths in discrete linear orders

The class dLOfpb\mathrm{dLO}_{\mathrm{fp}}^{\mathrm{b}} consisting of bounded discrete linear orders is of particular interest because of the next result.

Lemma 2.1.

(cf. [8, §2.5]) If 𝔵0\mathfrak{x}_{0} is a string then (Hl(𝔵0),<l)(H_{l}(\mathfrak{x}_{0}),<_{l}) is a toset. Its maximal element is 𝔪1(𝔵0)\mathfrak{m}_{1}(\mathfrak{x}_{0}) and the minimal element is 𝔪1(𝔵0)\mathfrak{m}_{-1}(\mathfrak{x}_{0}), where for each j{1,1}j\in\{1,-1\} 𝔪j(𝔵0)\mathfrak{m}_{j}(\mathfrak{x}_{0}) is the maximal length string in Hl(𝔵0)H_{l}(\mathfrak{x}_{0}) such that δ(𝔵0;𝔪j(𝔵0))=j\delta(\mathfrak{x}_{0};\mathfrak{m}_{j}(\mathfrak{x}_{0}))=j.

Consider any 𝔳Hl(𝔵0)\mathfrak{v}\in H_{l}(\mathfrak{x}_{0}). Recall from [4, §2.4] that l(𝔳)l(\mathfrak{v}) is the immediate successor of 𝔳\mathfrak{v} with respect to <l<_{l}, if exists, that satisfies |𝔳|<|l(𝔳)||\mathfrak{v}|<|l(\mathfrak{v})|. We can inductively define ln+1(𝔳):=l(ln(𝔳))l^{n+1}(\mathfrak{v}):=l(l^{n}(\mathfrak{v})), if exists. If ln(𝔳)l^{n}(\mathfrak{v}) exists for all n0n\geq 0 then the limit of the increasing sequence 1,l(𝔳):=limnln(𝔳)\langle 1,l\rangle(\mathfrak{v}):=\lim_{n\to\infty}l^{n}(\mathfrak{v}) is a left \mathbb{N}-string. On the other hand, if there is a maximum N1N\geq 1 such that lN(𝔳)l^{N}(\mathfrak{v}) exists but lN+1(𝔳)l^{N+1}(\mathfrak{v}) does not exist then we define [1,l](𝔳):=lN(𝔳)[1,l](\mathfrak{v}):=l^{N}(\mathfrak{v}). Note that 1,l(𝔳)Hl(𝔵0)\langle 1,l\rangle(\mathfrak{v})\notin H_{l}(\mathfrak{x}_{0}) but [1,l](𝔳)Hl(𝔵0)[1,l](\mathfrak{v})\in H_{l}(\mathfrak{x}_{0}).

Recall that any left \mathbb{N}-string in a domestic string algebra is an almost periodic string, i.e., is of the form 𝔟𝔲{}^{\infty}\mathfrak{b}\mathfrak{u} for some primitive cyclic word 𝔟\mathfrak{b} and some finite string 𝔲\mathfrak{u} such that the composition 𝔟𝔲\mathfrak{b}\mathfrak{u} is defined. Let H^l(𝔵0)\hat{H}_{l}(\mathfrak{x}_{0}) denote the completion of Hl(𝔵0)H_{l}(\mathfrak{x}_{0}) by all left \mathbb{N}-strings 𝔟𝔲{}^{\infty}\mathfrak{b}\mathfrak{u} such that 𝔟𝔲Hl(𝔵0)\mathfrak{b}\mathfrak{u}\in H_{l}(\mathfrak{x}_{0}). The definition of the ordering <l<_{l} can be easily extended to include infinite left \mathbb{N}-strings so that H^l(𝔵0)\hat{H}_{l}(\mathfrak{x}_{0}) is also a bounded total order with same extremal elements as that of Hl(𝔵0)H_{l}(\mathfrak{x}_{0}).

Note that any left \mathbb{N}-string in H^l(𝔵0)\hat{H}_{l}(\mathfrak{x}_{0}) is “far away” from its endpoints. To be more precise if 1,l(𝔳)\langle 1,l\rangle(\mathfrak{v}) exists then the interval [1,l(𝔳),𝔪1(𝔵0)][\langle 1,l\rangle(\mathfrak{v}),\mathfrak{m}_{1}(\mathfrak{x}_{0})] in H^l(𝔵0)\hat{H}_{l}(\mathfrak{x}_{0}) contains infinitely many elements of Hl(𝔵0)H_{l}(\mathfrak{x}_{0}), where 1,l(𝔳)<l𝔪1(𝔵0)\langle 1,l\rangle(\mathfrak{v})<_{l}\mathfrak{m}_{1}(\mathfrak{x}_{0}), where 𝔪1(𝔵0)\mathfrak{m}_{1}(\mathfrak{x}_{0}) is a finite string. Thus the left \mathbb{N}-string 1,l(𝔳)\langle 1,l\rangle(\mathfrak{v}) can also be approached from the right, with respect to <l<_{l}, as a limit of some decreasing sequence in Hl(𝔵0)H_{l}(\mathfrak{x}_{0}).

Now we recall the concept of fundamental solution of the following equation from [4, §2.4]:

(1) 1,l¯(𝔵)=1,l(𝔳).\langle 1,\bar{l}\rangle(\mathfrak{x})=\langle 1,l\rangle(\mathfrak{v}).

If one solution to the above equation exists then infinitely many solutions exist for if 𝔶\mathfrak{y}^{\prime} is a solution then so is l¯(𝔶)\bar{l}(\mathfrak{y}^{\prime}) and the fundamental solution is a solution of minimal length. If 𝔳\mathfrak{v} is a left substring of 𝔶\mathfrak{y}, then we rewrite Equation (1) as 𝔶=l¯,l(𝔳)\mathfrak{y}=\langle\bar{l},l\rangle(\mathfrak{v}), otherwise we write 𝔳=l,l¯(𝔶)\mathfrak{v}=\langle l,\bar{l}\rangle(\mathfrak{y}). If 𝔶=l¯,l(𝔳)\mathfrak{y}=\langle\bar{l},l\rangle(\mathfrak{v}), then we also say that 𝔶=l¯,l(ln(𝔳))\mathfrak{y}=\langle\bar{l},l\rangle(l^{n}(\mathfrak{v})) for any nn\in\mathbb{N}.

Similarly if 1,l(𝔳)\langle 1,l\rangle(\mathfrak{v}) does not exist but l(𝔳)l(\mathfrak{v}) exists then we may ask if the equation

(2) [1,l¯](𝔵)=[1,l](𝔳)[1,\bar{l}](\mathfrak{x})=[1,l](\mathfrak{v})

has a solution. If one solution exists then finitely many such solutions exist for if 𝔶\mathfrak{y}^{\prime} is a solution and [1,l¯](𝔳)l¯(𝔶)[1,\bar{l}](\mathfrak{v})\neq\bar{l}(\mathfrak{y}^{\prime}) then l¯(𝔶)\bar{l}(\mathfrak{y}^{\prime}) is also a solution. As explained above we can define a fundamental solution of Equation (2) and introduce the notations [l,l¯][l,\bar{l}] or [l¯,l][\bar{l},l] depending on the comparison of lengths.

In the above two paragraphs we introduced new expressions using two kinds of bracket operations, namely l¯,l\langle\bar{l},l\rangle and [l¯,l][\bar{l},l]. We say that such expressions are ‘terms’–the precise definition is given below.

If 𝔶=l¯,l(𝔳)\mathfrak{y}=\langle\bar{l},l\rangle(\mathfrak{v}) then the term l¯,l\langle\bar{l},l\rangle labels the path from 𝔳\mathfrak{v} to 𝔶\mathfrak{y} in H^l(𝔵0)\hat{H}_{l}(\mathfrak{x}_{0}). Let us describe the path from 𝔳\mathfrak{v} to any element 𝔲\mathfrak{u} in the interval [𝔳,𝔶][\mathfrak{v},\mathfrak{y}]. For 𝔲\mathfrak{u} there exists some n1n\geq 1 such that either 𝔲=ln(𝔳)\mathfrak{u}=l^{n}(\mathfrak{v}) or 𝔲=l¯n(𝔶)\mathfrak{u}=\bar{l}^{n}(\mathfrak{y}). In the former case, we know the label of the required path, namely lnl^{n}, whereas in the latter case, we use concatenation operation on terms, written juxtaposition, to label the path as l¯nl¯,l\bar{l}^{n}\langle\bar{l},l\rangle. Here we acknowledge that the only path to reach such 𝔲\mathfrak{u} from 𝔳\mathfrak{v} has to pass through 𝔶\mathfrak{y} which is captured using the bracket term. In other words, the bracket term l¯,l\langle\bar{l},l\rangle describes a ‘ghost path’ which passes through each ln(𝔳)l^{n}(\mathfrak{v}) to reach the limit 1,l(𝔳)\langle 1,l\rangle(\mathfrak{v}), and then directly jumps to 𝔶\mathfrak{y} avoiding all elements of the interval [1,l(𝔳),𝔶][\langle 1,l\rangle(\mathfrak{v}),\mathfrak{y}]–such elements can only be accessed through reverse paths from 𝔶\mathfrak{y}.

Once we have concatenation operation on terms as well as the bracket operation, it is natural to ask if more complex terms could be constructed by their combination. For example, we may ask if the expression l¯,l2\langle\bar{l},l\rangle^{2} labels a path between two points in some domestic string algebra. In case the answer is affirmative, we can also ask the same question for the expression 1,l¯,l:=limnl¯,ln\langle 1,\langle\bar{l},l\rangle\rangle:=\lim_{n\to\infty}\langle\bar{l},l\rangle^{n}. Further we can also ask if there are solutions to equations of the form

1,l(𝔵)\displaystyle\langle 1,l\rangle(\mathfrak{x}) =\displaystyle= 1,l¯,l(𝔳),\displaystyle\langle 1,\langle\bar{l},l\rangle\rangle(\mathfrak{v}),
1,l,l¯(𝔵)\displaystyle\langle 1,\langle l,\bar{l}\rangle\rangle(\mathfrak{x}) =\displaystyle= 1,l¯,l(𝔳)\displaystyle\langle 1,\langle\bar{l},l\rangle\rangle(\mathfrak{v})
1,[l,l¯](𝔵)\displaystyle\langle 1,[l,\bar{l}]\rangle(\mathfrak{x}) =\displaystyle= 1,[l¯,l](𝔳).\displaystyle\langle 1,[\bar{l},l]\rangle(\mathfrak{v}).

In summary, we need to close the collection of terms under concatenation and both bracket operations, whenever such a term labels a path between two elements in some hammock in some domestic string algebra.

Example 2.2.

Consider the algebra Λ2\Lambda_{2} presented as

v1{v_{1}}v2{v_{2}}v3{v_{3}}v4{v_{4}}a\scriptstyle{a}b\scriptstyle{b}c\scriptstyle{c}d\scriptstyle{d}e\scriptstyle{e}
Figure 1. Λ2\Lambda_{2} with ρ={cb,dc}\rho=\{cb,dc\}

Figure 2 shows all connected components of the A-R quiver of Λ2\Lambda_{2} along with some infinite dimensional string modules. The exceptional component is a copy of 𝐙A\mathbf{Z}A^{\infty}_{\infty}-component with a hole in the middle in and rest of the components are tubes.

Refer to caption
Figure 2. AR components for string modules in Λ2\Lambda_{2}

Note that red \hookrightarrow and \twoheadrightarrow arrows denote the operators ll and l¯\bar{l} respectively whereas black \hookrightarrow and \twoheadrightarrow arrows denote the operators rr and r¯\bar{r} respectively. The module SjS_{j} represents the simple modules at vertex vjv_{j}.

Now 1,l(eca)=(eD)eca=1,l¯(a)\langle 1,l\rangle(eca)=\ ^{\infty}(eD)eca=\langle 1,\bar{l}\rangle(a). Furthermore it is easy to verify that 1,l¯,l(eca)=(Ba)=1,l¯(1(v1,i))\langle 1,\langle\bar{l},l\rangle\rangle(eca)=\ ^{\infty}(Ba)=\langle 1,\bar{l}\rangle(1_{(v_{1},i)}), and hence l¯,l,l¯(1(v1,i))=eca\langle\langle\bar{l},l\rangle,\bar{l}\rangle(1_{(v_{1},i)})=eca. In Examples 10.15 and 10.17 we will compute the terms, i.e., labels for the paths from S1S_{1} to ecaeca and caca respectively without the involvement of any infinite string.

After connecting all possible rays and corays (in the sense of [6, § 5]) consisting of string modules in (a part of) the A-R quiver of Λ2\Lambda_{2} we get Figure 3. Here the vertical red colored lines (resp. the horizontal black colored lines) represent hammocks Hl(v)H_{l}(v) (resp. Hr(v)H_{r}(v)) for different vertices vv. In fact we also show the infinite dimensional string modules, and thus show H^l(v)\hat{H}_{l}(v) and H^r(v)\hat{H}_{r}(v). For example, the rightmost vertical line represent H^l(v1)\hat{H}_{l}(v_{1}) and the lowermost horizontal line represent H^r(v4)\hat{H}_{r}(v_{4}).

Refer to caption
Figure 3. Hammocks for Λ2\Lambda_{2}

Now we are ready to formally define terms. Starting with symbols l,l¯l,\bar{l} we simultaneously inductively define the class of simple ll-terms, denoted 𝐓l\mathbf{T}_{l} and the class of simple l¯\bar{l}-terms, denoted 𝐓l¯\mathbf{T}_{\bar{l}} as follows:

  • [l¯,l]𝐓l[\bar{l},l]\in\mathbf{T}_{l} and [l,l¯]𝐓l¯[l,\bar{l}]\in\mathbf{T}_{\bar{l}};

  • if τ𝐓l\tau\in\mathbf{T}_{l} and μ𝐓l¯\mu\in\mathbf{T}_{\bar{l}} then μ,τ𝐓l\langle\mu,\tau\rangle\in\mathbf{T}_{l} and τ,μ𝐓l¯\langle\tau,\mu\rangle\in\mathbf{T}_{\bar{l}};

  • if τ,τ𝐓l\tau,\tau^{\prime}\in\mathbf{T}_{l} and μ,μ𝐓l¯\mu,\mu^{\prime}\in\mathbf{T}_{\bar{l}} then ττ𝐓l\tau^{\prime}\tau\in\mathbf{T}_{l} and μμ𝐓l¯\mu^{\prime}\mu\in\mathbf{T}_{\bar{l}}.

Set 𝐓:=𝐓l𝐓l¯\mathbf{T}:=\mathbf{T}_{l}\cup\mathbf{T}_{\bar{l}} and call an element of 𝐓\mathbf{T} a simple term.

Remark 2.3.

Both 𝐓l\mathbf{T}_{l} and 𝐓l¯\mathbf{T}_{\bar{l}} are in bijective correspondence with 𝐀3𝐒𝐓\mathbf{A}3\mathbf{ST} of [1, Definition 7.1] in view of [1, Theorem 7.2].

Define a map :𝐓𝐓\mathcal{I}:\mathbf{T}\to\mathbf{T} as follows:

  • (l):=l¯\mathcal{I}(l):=\bar{l} and (l¯):=l\mathcal{I}(\bar{l}):=l;

  • (μ,τ):=(τ),(μ)\mathcal{I}(\langle\mu,\tau\rangle):=\langle\mathcal{I}(\tau),\mathcal{I}(\mu)\rangle;

  • ([l¯,l]):=[l,l¯]\mathcal{I}([\bar{l},l]):=[l,\bar{l}] and ([l,l¯]):=[l¯,l]\mathcal{I}([l,\bar{l}]):=[\bar{l},l];

  • (νν):=(ν)(ν)\mathcal{I}(\nu^{\prime}\nu):=\mathcal{I}(\nu^{\prime})\mathcal{I}(\nu).

Remark 2.4.

The map :𝐓𝐓\mathcal{I}:\mathbf{T}\to\mathbf{T} is an involution such that for each ν𝐓\nu\in\mathbf{T}, ν𝐓l\nu\in\mathbf{T}_{l} if and only if (ν)𝐓l¯\mathcal{I}(\nu)\in\mathbf{T}_{\bar{l}}.

Suppose τ,τ1,τ2𝐓l\tau,\tau_{1},\tau_{2}\in\mathbf{T}_{l} and μ,μ1,μ2𝐓l¯\mu,\mu_{1},\mu_{2}\in\mathbf{T}_{\bar{l}}. Let l\approx_{l} be the equivalence relation on 𝐓l\mathbf{T}_{l} generated by

μ,ττl(μ)μ,τlμ,τlμ,τmlμm,τ,μ,[l¯,l]lμ,l,[l,l¯],τll¯,τ\langle\mu,\tau\rangle\tau\approx_{l}\mathcal{I}(\mu)\langle\mu,\tau\rangle\approx_{l}\langle\mu,\tau\rangle\approx_{l}\langle\mu,\tau^{m}\rangle\approx_{l}\langle\mu^{m},\tau\rangle,\ \langle\mu,[\bar{l},l]\rangle\approx_{l}\langle\mu,l\rangle,\ \langle[l,\bar{l}],\tau\rangle\approx_{l}\langle\bar{l},\tau\rangle

and l¯\approx_{\bar{l}} be the equivalence relation on 𝐓l¯\mathbf{T}_{\bar{l}} generated by

τ,μμl¯(τ)τ,μl¯τ,μl¯τ,μml¯τm,μ,[l¯,l],μl¯l,μ,τ,[l,l¯]l¯τ,l¯\langle\tau,\mu\rangle\mu\approx_{\bar{l}}\mathcal{I}(\tau)\langle\tau,\mu\rangle\approx_{\bar{l}}\langle\tau,\mu\rangle\approx_{\bar{l}}\langle\tau,\mu^{m}\rangle\approx_{\bar{l}}\langle\tau^{m},\mu\rangle,\ \langle[\bar{l},l],\mu\rangle\approx_{\bar{l}}\langle l,\mu\rangle,\ \langle\tau,[l,\bar{l}]\rangle\approx_{\bar{l}}\langle\tau,\bar{l}\rangle

for each m2m\geq 2.

To make these into congruence relations, whenever τ1lτ2\tau_{1}\approx_{l}\tau_{2} and μ1l¯μ2\mu_{1}\approx_{\bar{l}}\mu_{2} we further need the following identifications:

τ2τlτ1τ,ττ2lττ1,μ2μl¯μ1μ,μμ2l¯μμ1,\tau_{2}\tau\approx_{l}\tau_{1}\tau,\ \tau\tau_{2}\approx_{l}\tau\tau_{1},\ \mu_{2}\mu\approx_{\bar{l}}\mu_{1}\mu,\ \mu\mu_{2}\approx_{\bar{l}}\mu\mu_{1},
μ1,τ1lμ2,τ2,τ1,μ1l¯τ2,μ2.\langle\mu_{1},\tau_{1}\rangle\approx_{l}\langle\mu_{2},\tau_{2}\rangle,\ \langle\tau_{1},\mu_{1}\rangle\approx_{\bar{l}}\langle\tau_{2},\mu_{2}\rangle.
Remark 2.5.

The relation l\approx_{l} on 𝐓l\mathbf{T}_{l} coincides with L\approx_{L} of [1, Definition 6.3] on 𝐀3𝐒𝐓\mathbf{A}3\mathbf{ST}.

Say that ν𝐓\nu\in\mathbf{T} is irreducible if either ν{l,l¯}\nu\in\{l,\bar{l}\} or ν\nu is of the form μ,τ\langle\mu,\tau\rangle.

Given τ𝐓l\tau\in\mathbf{T}_{l} and μ𝐓l¯\mu\in\mathbf{T}_{\bar{l}}, say that

  • μτ\mu\tau is a valid concatenation if (τ)μl¯(τ)\mathcal{I}(\tau)\mu\approx_{\bar{l}}\mathcal{I}(\tau);

  • τμ\tau\mu is a valid concatenation if (μ)τl(μ)\mathcal{I}(\mu)\tau\approx_{l}\mathcal{I}(\mu).

Say that a finite concatenation τnτ2τ1\tau_{n}\ldots\tau_{2}\tau_{1} of simple terms is a mixed ll-term (resp. mixed l¯\bar{l}-term) if

  • τk𝐓l\tau_{k}\in\mathbf{T}_{l} if and only if kk is odd (resp. even);

  • τk+1τk\tau_{k+1}\tau_{k} is a valid concatenation for each 1k<n1\leq k<n.

Let 𝐌l\mathbf{M}_{l} and 𝐌l¯\mathbf{M}_{\bar{l}} denote the classes of all mixed ll and l¯\bar{l}-terms respectively. Set 𝐌:=𝐌l𝐌l¯\mathbf{M}:=\mathbf{M}_{l}\cup\mathbf{M}_{\bar{l}} and say that a term is an element of 𝐌\mathbf{M}.

The congruence relations l\approx_{l} and l¯\approx_{\bar{l}} are naturally extended to 𝐌l\mathbf{M}_{l} and 𝐌l¯\mathbf{M}_{\bar{l}} respectively.

Associate to each ν𝐓\nu\in\mathbf{T} its order type, denoted 𝒪(ν)\mathcal{O}(\nu), as follows:

  • 𝒪(l)=𝒪(l¯)=𝒪([l¯,l])=𝒪([l,l¯]):=𝟏\mathcal{O}(l)=\mathcal{O}(\bar{l})=\mathcal{O}([\bar{l},l])=\mathcal{O}([l,\bar{l}]):=\mathbf{1};

  • 𝒪(μ,τ):=ω×𝒪(τ)+ω×𝒪(μ)\mathcal{O}(\langle\mu,\tau\rangle):=\omega\times\mathcal{O}(\tau)+\omega^{*}\times\mathcal{O}(\mu);

  • 𝒪(ττ):=𝒪(τ)+𝒪(τ)\mathcal{O}(\tau^{\prime}\tau):=\mathcal{O}(\tau)+\mathcal{O}(\tau^{\prime}).

Clearly 𝒪\mathcal{O} takes values in dLOfpb\mathrm{dLO}_{\mathrm{fp}}^{\mathrm{b}}.

Remark 2.6.

Suppose τ𝐓l\tau\in\mathbf{T}_{l} and μ𝐓l¯\mu\in\mathbf{T}_{\bar{l}}. If μτ\mu\tau is a valid concatenation then ω×𝒪(μ)\omega^{*}\times\mathcal{O}(\mu) is a proper suffix of 𝒪(τ)\mathcal{O}(\tau).

In view of the above remark, the map 𝒪\mathcal{O} extends to 𝐌\mathbf{M} by setting

𝒪(τnτ2τ1):=𝒪(τ1).\mathcal{O}(\tau_{n}\ldots\tau_{2}\tau_{1}):=\mathcal{O}(\tau_{1}).
Remark 2.7.

If ν𝐓\nu\in\mathbf{T} is irreducible then wd(𝒪(ν))2\mathrm{wd}(\mathcal{O}(\nu))\leq 2, where wd(L)\mathrm{wd}(L) denotes the width of LLOfpL\in\mathrm{LO}_{\mathrm{fp}} as defined in [1, Definition 8.1].

In view of Remarks 2.3 and 2.5 the next result follows from [1, Corollary 9.4].

Theorem 2.8.

Suppose τ,τ𝐌l\tau,\tau^{\prime}\in\mathbf{M}_{l} and μ,μ𝐌l¯\mu,\mu^{\prime}\in\mathbf{M}_{\bar{l}}. Then

  • 𝒪(τ)𝒪(τ)\mathcal{O}(\tau)\cong\mathcal{O}(\tau^{\prime}) if and only if τlτ\tau\approx_{l}\tau^{\prime};

  • 𝒪(μ)𝒪(μ)\mathcal{O}(\mu)\cong\mathcal{O}(\mu^{\prime}) if and only if μl¯μ\mu\approx_{\bar{l}}\mu^{\prime}.

Before we close the section, let us define a natural notation, \ast, on terms.

  1. (1)

    l1:=ll\ast 1:=l and l(1):=l¯l\ast(-1):=\bar{l}.

  2. (2)

    τ,η1:=τ,η\langle\tau,\eta\rangle\ast 1:=\langle\tau,\eta\rangle and τ,η(1):=η,τ\langle\tau,\eta\rangle\ast(-1):=\langle\eta,\tau\rangle.

  3. (3)

    [τ,η]1:=[τ,η][\tau,\eta]\ast 1:=[\tau,\eta] and [τ,η](1):=[η,τ][\tau,\eta]\ast(-1):=[\eta,\tau].

3. Long and short pairs

Recall the notion of an H-pair from [9, § 5].

Remark 3.1.

Suppose (𝔶1,𝔶2)(\mathfrak{y}_{1},\mathfrak{y}_{2}) is an H-pair. Then N(𝔟(𝔶1,𝔶2),𝔶2H𝔶1)=1N(\mathfrak{b}(\mathfrak{y}_{1},\mathfrak{y}_{2}),\mathfrak{y}_{2}\circ_{H}\mathfrak{y}_{1})=1 if and only if there is a partition 𝔶2𝔟𝔶1=𝔶2𝔟𝔶1{\mathfrak{y}_{2}\mathfrak{b}\mathfrak{y}_{1}}=\mathfrak{y}^{\prime}_{2}\mathfrak{b}^{\prime}\mathfrak{y}^{\prime}_{1}, where 𝔟\mathfrak{b}^{\prime} is a cyclic permutation of 𝔟\mathfrak{b} and the first syllable of 𝔶2\mathfrak{y}^{\prime}_{2} is not a syllable of 𝔟\mathfrak{b}, such that 𝔟𝔶1\nequivH𝔶1\mathfrak{b}^{\prime}\mathfrak{y}^{\prime}_{1}\nequiv_{H}\mathfrak{y}^{\prime}_{1}.

Suppose 𝔟\mathfrak{b} is a band, 𝔶\mathfrak{y} is a string and i{1,1}i\in\{1,-1\}. Say that the pair (𝔟,𝔶)(\mathfrak{b},\mathfrak{y}) is an ii-left valid pair (ii-LVP for short) if 𝔟𝔶𝔶\mathfrak{b}_{\mathfrak{y}}\mathfrak{y} is a string for some cyclic permutation 𝔟𝔶\mathfrak{b}_{\mathfrak{y}} of 𝔟\mathfrak{b} such that θ(𝔟𝔶)=i\theta(\mathfrak{b}_{\mathfrak{y}})=i. Dually say that the pair (𝔟,𝔶)(\mathfrak{b},\mathfrak{y}) is a ii-right valid pair (ii-RVP for short) if (𝔟1,𝔶1)(\mathfrak{b}^{-1},\mathfrak{y}^{-1}) is a (i)(-i)-LVP.

Example 3.2.

Consider the algebra Λ(iii)\Lambda^{(iii)} from [9, Example 4.24]. The only bands there are 𝔟:=cbaDEF\mathfrak{b}:=cbaDEF and 𝔟1\mathfrak{b}^{-1}. Choose 𝔶:=D\mathfrak{y}:=D. Since 𝔟\mathfrak{b} has two different cyclic permutations 𝔟1:=𝔟\mathfrak{b}_{1}:=\mathfrak{b} and 𝔟2:=DEFcba\mathfrak{b}_{2}:=DEFcba with θ(𝔟1)=1\theta(\mathfrak{b}_{1})=1 and θ(𝔟2)=1\theta(\mathfrak{b}_{2})=-1 such that 𝔟1𝔶\mathfrak{b}_{1}\mathfrak{y} and 𝔟2𝔶\mathfrak{b}_{2}\mathfrak{y} are strings, the pair (𝔟,𝔶)(\mathfrak{b},\mathfrak{y}) is both 11-LVP and 1-1-LVP.

Remark 3.3.

[9, Proposition 4.23] guarantees the uniqueness of the cyclic permutation 𝔟𝔶\mathfrak{b}_{\mathfrak{y}} for any ii-LVP (𝔟,𝔶)(\mathfrak{b},\mathfrak{y}).

We drop the reference to ii in ii-LVP if either it is clear from the context or if ii does not play any role in the statement.

Proposition 3.4.

Suppose (𝔟,𝔶)(\mathfrak{b},\mathfrak{y}) is an LVP and 𝔟𝔅(𝔶)\mathfrak{b}\in\mathfrak{B}(\mathfrak{y}) then 𝔶=𝔟𝔶𝔶\mathfrak{y}=\mathfrak{b}_{\mathfrak{y}}\mathfrak{y}^{\prime} for some 𝔶\mathfrak{y}^{\prime}.

Proof. Since N(𝔟,𝔶)>0N(\mathfrak{b},\mathfrak{y})>0, 𝔶\mathfrak{y} can be written as 𝔶=𝔶2𝔟𝔶1\mathfrak{y}=\mathfrak{y}_{2}\mathfrak{b}^{\prime}\mathfrak{y}_{1} for some cyclic permutation 𝔟\mathfrak{b}^{\prime} of 𝔟\mathfrak{b}. If 𝔶2\mathfrak{y}_{2} is not a right substring of a finite power of 𝔟𝔶\mathfrak{b}_{\mathfrak{y}} then there exists a non-trivial cycle in the bridge quiver, i.e., a meta-band passing through 𝔟\mathfrak{b} by [4, Lemma 3.3.4], which contradicts the domesticity of the string algebra by [4, Proposition 3.4.2].    \Box

Definition 3.5.

Say that an LVP (𝔟,𝔶)(\mathfrak{b},\mathfrak{y}) is a short pair if 𝔶H𝔟𝔶𝔶\mathfrak{y}\equiv_{H}\mathfrak{b}_{\mathfrak{y}}\mathfrak{y}; otherwise say that (𝔟,𝔶)(\mathfrak{b},\mathfrak{y}) is a long pair.

Remark 3.6.

An LVP (𝔟,𝔶)(\mathfrak{b},\mathfrak{y}) with |𝔶|>0|\mathfrak{y}|>0 is long if and only if (𝔟,𝔷r(𝔶))(\mathfrak{b},\mathfrak{z}_{r}(\mathfrak{y})) is long.

Remark 3.7.

Suppose (𝔟,𝔶)(\mathfrak{b},\mathfrak{y}) is an LVP. If (𝔟,𝔶)(\mathfrak{b},\mathfrak{y}) is short then 𝔵𝔶𝔵𝔟𝔶𝔶:Hl(𝔶)Hl(𝔟𝔶𝔶)\mathfrak{x}\mathfrak{y}\mapsto\mathfrak{x}\mathfrak{b}_{\mathfrak{y}}\mathfrak{y}:H_{l}(\mathfrak{y})\to H_{l}(\mathfrak{b}_{\mathfrak{y}}\mathfrak{y}) is an isomorphism.

For an LVP (𝔟,𝔶)(\mathfrak{b},\mathfrak{y}), let 𝔴¯l(𝔟,𝔶)\bar{\mathfrak{w}}_{l}(\mathfrak{b},\mathfrak{y}) be the longest common right substring (possibly of length 0) of 𝔶\mathfrak{y} and 𝔟𝔶𝔶\mathfrak{b}_{\mathfrak{y}}\mathfrak{y}, and 𝔴l(𝔟,𝔶):=ρr(𝔴¯l(𝔟,𝔶))\mathfrak{w}_{l}(\mathfrak{b},\mathfrak{y}):=\rho_{r}(\bar{\mathfrak{w}}_{l}(\mathfrak{b},\mathfrak{y})). Dually for an RVP (𝔟,𝔶)(\mathfrak{b},\mathfrak{y}), let 𝔴¯r(𝔟,𝔶)\bar{\mathfrak{w}}_{r}(\mathfrak{b},\mathfrak{y}) be the longest common left substring (possibly of length 0) of 𝔶\mathfrak{y} and 𝔶𝔟𝔶\mathfrak{y}\mathfrak{b}_{\mathfrak{y}}, and 𝔴r(𝔟,𝔶):=ρl(𝔴¯r(𝔟,𝔶))\mathfrak{w}_{r}(\mathfrak{b},\mathfrak{y}):=\rho_{l}(\bar{\mathfrak{w}}_{r}(\mathfrak{b},\mathfrak{y})).

Say that an LVP (𝔟,𝔶)(\mathfrak{b},\mathfrak{y}) is a left tight valid pair (LTVP, for short) if |𝔴¯l(𝔟,𝔶)|=0|\bar{\mathfrak{w}}_{l}(\mathfrak{b},\mathfrak{y})|=0; otherwise say it is a left loose valid pair (LLVP, for short). Dually we define RTVP and RLVP.

Remark 3.8.

Suppose (𝔟,𝔶)(\mathfrak{b},\mathfrak{y}) is an LLVP with δ(𝔴¯l(𝔟,𝔶))=0\delta(\bar{\mathfrak{w}}_{l}(\mathfrak{b},\mathfrak{y}))=0 then ρr(𝔶)=𝔴l(𝔟,𝔶)=ρr(𝔟𝔶𝔶)\rho_{r}(\mathfrak{y})=\mathfrak{w}_{l}(\mathfrak{b},\mathfrak{y})=\rho_{r}(\mathfrak{b}_{\mathfrak{y}}\mathfrak{y}), and hence (𝔟,𝔶)(\mathfrak{b},\mathfrak{y}) is a short LVP by [9, Proposition 5.3].

Proposition 3.9.

Suppose (𝔟,𝔶)(\mathfrak{b},\mathfrak{y}) is an LVP such that 𝔟𝔅(𝔶)\mathfrak{b}\in\mathfrak{B}(\mathfrak{y}). Then (𝔟,𝔶)(\mathfrak{b},\mathfrak{y}) is a short LVP.

Proof. Since 𝔟𝔅(𝔶)\mathfrak{b}\in\mathfrak{B}(\mathfrak{y}), we have 𝔶=𝔟𝔶𝔶\mathfrak{y}=\mathfrak{b}_{\mathfrak{y}}\mathfrak{y}^{\prime} for some string 𝔶\mathfrak{y}^{\prime} by Proposition 3.4. Hence 𝔟𝔶\mathfrak{b}_{\mathfrak{y}} is a left substring of 𝔴¯l(𝔟,𝔶)\bar{\mathfrak{w}}_{l}(\mathfrak{b},\mathfrak{y}). The conclusion follows by Remark 3.8.    \Box

For a long LVP (𝔟,𝔶)(\mathfrak{b},\mathfrak{y}) the above proposition gives that N(𝔟,𝔶)=0N(\mathfrak{b},\mathfrak{y})=0.

Definition 3.10.

For an LVP (𝔟,𝔶)(\mathfrak{b},\mathfrak{y}) if 𝔴l(𝔟,𝔶)\mathfrak{w}_{l}(\mathfrak{b},\mathfrak{y}) is a proper right substring of ρr(𝔟𝔶)\rho_{r}(\mathfrak{b}_{\mathfrak{y}}) (resp. of ρr(𝔶)\rho_{r}(\mathfrak{y})) then say that (𝔟,𝔶)(\mathfrak{b},\mathfrak{y}) is bb-long (resp. ss-long). If (𝔟,𝔶)(\mathfrak{b},\mathfrak{y}) is an ss-long pair then we say that it is ss^{\partial}-long where =δ(ρr(𝔶))\partial=\delta(\rho_{r}(\mathfrak{y})). Similarly if (𝔟,𝔶)(\mathfrak{b},\mathfrak{y}) is an bb-long pair then we say that it is bb^{\partial}-long where =δ(ρr(𝔟𝔶))\partial=\delta(\rho_{r}(\mathfrak{b}_{\mathfrak{y}})).

Proposition 3.11.

Suppose (𝔟,𝔶)(\mathfrak{b},\mathfrak{y}) is an LVP.

  • The pair (𝔟,𝔶)(\mathfrak{b},\mathfrak{y}) is sis^{i}-long if and only if 𝔵𝔶𝔵𝔟𝔶𝔶:Hil(𝔶)Hil(𝔟𝔶𝔶)\mathfrak{x}\mathfrak{y}\mapsto\mathfrak{x}\mathfrak{b}_{\mathfrak{y}}\mathfrak{y}:H^{i}_{l}(\mathfrak{y})\to H^{i}_{l}(\mathfrak{b}_{\mathfrak{y}}\mathfrak{y}) is injective but not surjective.

  • The pair (𝔟,𝔶)(\mathfrak{b},\mathfrak{y}) is bib^{i}-long if and only if 𝔵𝔟𝔶𝔶𝔵𝔶:Hil(𝔟𝔶𝔶)Hil(𝔶)\mathfrak{x}\mathfrak{b}_{\mathfrak{y}}\mathfrak{y}\mapsto\mathfrak{x}\mathfrak{y}:H^{i}_{l}(\mathfrak{b}_{\mathfrak{y}}\mathfrak{y})\to H^{i}_{l}(\mathfrak{y}) is injective but not surjective.

Proof. We only prove the first statement; the proof of the second is similar. Recall that i:=δ(ρr(𝔶))i:=\delta(\rho_{r}(\mathfrak{y})).

Suppose 𝔵𝔶\mathfrak{x}\mathfrak{y} is a string. Since (𝔟,𝔶)(\mathfrak{b},\mathfrak{y}) is ss-long, 𝔴l(𝔟,𝔶)\mathfrak{w}_{l}(\mathfrak{b},\mathfrak{y}) is a proper right substring of ρr(𝔶)\rho_{r}(\mathfrak{y}) which states that θ(γ)=i\theta(\gamma^{\prime})=-i where γ𝔟𝔶\gamma^{\prime}\in\mathfrak{b}_{\mathfrak{y}} is the syllable such that 𝔴l(𝔟,𝔶)γ\mathfrak{w}_{l}(\mathfrak{b},\mathfrak{y})\gamma^{\prime} is a string. Since 𝔵𝔴l(𝔟,𝔶)\mathfrak{x}\mathfrak{w}_{l}(\mathfrak{b},\mathfrak{y}) is a string and δ(𝔴l(𝔟,𝔶)γ)=0\delta(\mathfrak{w}_{l}(\mathfrak{b},\mathfrak{y})\gamma^{\prime})=0, 𝔵𝔟𝔶𝔶\mathfrak{x}\mathfrak{b}_{\mathfrak{y}}\mathfrak{y} is also a string. This shows that 𝔵𝔶𝔵𝔟𝔶𝔶\mathfrak{x}\mathfrak{y}\mapsto\mathfrak{x}\mathfrak{b}_{\mathfrak{y}}\mathfrak{y} is indeed a well-defined map from Hil(𝔶)Hil(𝔟𝔶𝔶)H^{i}_{l}(\mathfrak{y})\to H^{i}_{l}(\mathfrak{b}_{\mathfrak{y}}\mathfrak{y}). This map is clearly injective.

Let 𝔵\mathfrak{x} be the minimal string such that 𝔵ρr(𝔶)ρρ1\mathfrak{x}\rho_{r}(\mathfrak{y})\in\rho\cup\rho^{-1}. Then clearly the word 𝔵𝔶\mathfrak{x}\mathfrak{y} is not a string but the argument as in the above paragraph ensures that 𝔵𝔟𝔶𝔶\mathfrak{x}\mathfrak{b}_{\mathfrak{y}}\mathfrak{y} is a string. Therefore the map is not surjective.

For the converse, non-surjectivity of the map guarantees the existence of a string 𝔵\mathfrak{x} of minimal positive length with θ(𝔵)=i\theta(\mathfrak{x})=i such that 𝔵𝔟𝔶𝔶\mathfrak{x}\mathfrak{b}_{\mathfrak{y}}\mathfrak{y} is a string whereas the word 𝔵𝔶\mathfrak{x}\mathfrak{y} is not. Then there is some right substring 𝔶\mathfrak{y}^{\prime} of 𝔶\mathfrak{y} such that 𝔵𝔶ρρ1\mathfrak{x}\mathfrak{y}^{\prime}\in\rho\cup\rho^{-1}. The definition of ρr(𝔶)\rho_{r}(\mathfrak{y}) ensures that |𝔶||ρr(𝔶)||\mathfrak{y}^{\prime}|\leq|\rho_{r}(\mathfrak{y})|. Since 𝔵𝔟𝔶𝔶\mathfrak{x}\mathfrak{b}_{\mathfrak{y}}\mathfrak{y}^{\prime} is a string we see that |𝔴l(𝔟,𝔶)|<|𝔶||\mathfrak{w}_{l}(\mathfrak{b},\mathfrak{y})|<|\mathfrak{y}^{\prime}| thus completing the proof that the pair (𝔟,𝔶)(\mathfrak{b},\mathfrak{y}) is ss-long.    \Box

Example 3.12.

Consider the algebra Γ\Gamma from Figure 4. Here the bands are 𝔟:=dFCE\mathfrak{b}:=dFCE and 𝔟1\mathfrak{b}^{-1}. For 𝔶:=a\mathfrak{y}:=a the LVP (𝔟,𝔶)(\mathfrak{b},\mathfrak{y}) is ss-long as well as bb-long.

v1{v_{1}}v6{v_{6}}v2{v_{2}}v5{v_{5}}v7{v_{7}}v3{v_{3}}v4{v_{4}}a\scriptstyle{a}b\scriptstyle{b}e\scriptstyle{e}f\scriptstyle{f}d\scriptstyle{d}g\scriptstyle{g}c\scriptstyle{c}
Figure 4. Γ\Gamma with ρ={ba,ea,bc,dg,ecfg}\rho=\{ba,ea,bc,dg,ecfg\}

Say that a long LVP (𝔟,𝔶)(\mathfrak{b},\mathfrak{y}) is double long if it is both ss-long or bb-long; otherwise say that it is single long.

4. Decorated trees

Fix 𝔵0Str(Λ)\mathfrak{x}_{0}\in\mathrm{Str}(\Lambda) and i{1,1}i\in\{1,-1\}. We wish to associate a rooted tree, 𝒯𝔣i(𝔵0)\mathcal{T}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0}), with some additional structure described in the definition below to the triple (Λ,i,𝔵0)(\Lambda,i,\mathfrak{x}_{0}).

Definition 4.1.

A decorated tree is a tuple

𝔗:=(V,A,,V1,V1,<1,<1,T,NT,ι{1,1}),\mathfrak{T}:=(V,A,\star,V_{1},V_{-1},<_{1},<_{-1},T,NT,\iota\in\{1,-1\}),

where

  • (V={}V1V1,A)(V=\{\star\}\sqcup V_{1}\sqcup V_{-1},A) is a finite height rooted tree with root \star;

  • all children of \star lie in VιV_{-\iota};

  • TNTT\sqcup NT is the set of all leaves;

  • for each j{1,1}j\in\{1,-1\}, the relation <jVj×Vj<_{j}\subseteq V_{j}\times V_{j} is a strict partial order where 𝔲<j𝔲\mathfrak{u}<_{j}\mathfrak{u}^{\prime} only if 𝔲\mathfrak{u} and 𝔲\mathfrak{u}^{\prime} have the same parent;

  • for each 𝔲V\mathfrak{u}\in V and for each j{1,1}j\in\{1,-1\}, the relation <j<_{j} restricted to the children of 𝔲\mathfrak{u} in VjV_{j} is a strict linear order.

Here TT and NTNT stands for ‘torsion’ and ‘non-torsion’ respectively. Recall that a string 𝔶\mathfrak{y} is a (left) torsion string if α𝔶\alpha\mathfrak{y} is not a string for any syllable α\alpha.

Consider the set 𝒜fi(𝔵0)\mathcal{A}^{f}_{i}(\mathfrak{x}_{0}) of all arrows 𝔲\mathfrak{u} appearing in a path in 𝒬Bai(𝔵0)\mathcal{HQ}^{\mathrm{Ba}}_{i}(\mathfrak{x}_{0}) starting from 𝔵0\mathfrak{x}_{0}. Recall from [9, Remark 6.2] that 𝒜fi(𝔵0)\mathcal{A}^{f}_{i}(\mathfrak{x}_{0}) is a finite set.

Algorithm 4.2.

Construct a rooted tree 𝒯𝔣i(𝔵0)\mathcal{T}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0}) as follows.

  • Add the root vertex labelled 𝔵0\mathfrak{x}_{0}. Add one child of the root for each half ii-arch bridge or a zero ii-arch bridge from 𝔵0\mathfrak{x}_{0}.

  • For each newly added vertex 𝔲\mathfrak{u}, if 𝔱(𝔲)\mathfrak{t}(\mathfrak{u}) is a band then add one child of 𝔲\mathfrak{u} for each element 𝔲𝒜fi(𝔵0)\mathfrak{u}^{\prime}\in\mathcal{A}^{f}_{i}(\mathfrak{x}_{0}) satisfying 𝔰(𝔲)=𝔱(𝔲)\mathfrak{s}(\mathfrak{u}^{\prime})=\mathfrak{t}(\mathfrak{u}).

  • Repeat the above step until all elements of 𝒜fi(𝔵0)\mathcal{A}^{f}_{i}(\mathfrak{x}_{0}) are exhausted.

Since Λ\Lambda is domestic 𝒯𝔣i(𝔵0)\mathcal{T}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0}) is finite. Let 𝒱fi(𝔵0)\mathcal{V}^{f}_{i}(\mathfrak{x}_{0}) denote the set of vertices of the tree 𝒯𝔣i(𝔵0)\mathcal{T}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0}). Let 𝒰fi(𝔵0),fi(𝔵0),𝒵fi(𝔵0),fi(𝔵0)\mathcal{U}^{f}_{i}(\mathfrak{x}_{0}),\mathcal{H}^{f}_{i}(\mathfrak{x}_{0}),\mathcal{Z}^{f}_{i}(\mathfrak{x}_{0}),\mathcal{R}^{f}_{i}(\mathfrak{x}_{0}) denote those elements of 𝒱fi(𝔵0)\mathcal{V}^{f}_{i}(\mathfrak{x}_{0}) that are arch-bridges, half i-arch bridges, torsion zero i-arch bridges and torsion reverse arch bridges respectively. These sets together with the root give a partition of 𝒱fi(𝔵0)\mathcal{V}^{f}_{i}(\mathfrak{x}_{0}).

Let ξfi(𝔵0)\xi^{f}_{i}(\mathfrak{x}_{0}) denote the set of all children of 𝔵0\mathfrak{x}_{0}. For each non-root 𝔲𝒱fi(𝔵0)\mathfrak{u}\in\mathcal{V}^{f}_{i}(\mathfrak{x}_{0}) let ξf(𝔲)\xi^{f}(\mathfrak{u}) denote the set of children of 𝔲\mathfrak{u}. Partition ξf(𝔲)\xi^{f}(\mathfrak{u}) as ξf1(𝔲)ξf1(𝔲)\xi^{f}_{1}(\mathfrak{u})\sqcup\xi^{f}_{-1}(\mathfrak{u}) by defining ξfj(𝔲):={𝔲ξf(𝔲)θ(β(𝔲))=j}\xi^{f}_{j}(\mathfrak{u}):=\{\mathfrak{u}^{\prime}\in\xi^{f}(\mathfrak{u})\mid\theta(\beta(\mathfrak{u}^{\prime}))=j\}. For a non-root vertex 𝔲\mathfrak{u} let π(𝔲)\pi(\mathfrak{u}) denote the parent of 𝔲\mathfrak{u}. The height of 𝔲𝒱fi(𝔵0)\mathfrak{u}\in\mathcal{V}^{f}_{i}(\mathfrak{x}_{0}) is the non-negative integer h(𝔲)h(\mathfrak{u}) such that πh(𝔲)(𝔲)=𝔵0\pi^{h(\mathfrak{u})}(\mathfrak{u})=\mathfrak{x}_{0}.

Define a function ϕ:𝒱fi(𝔵0){1,0,1}\phi:\mathcal{V}^{f}_{i}(\mathfrak{x}_{0})\to\{1,0,-1\} by ϕ(𝔲):={0if h(𝔲)=0;iif h(𝔲)=1;θ(β(𝔲))if h(𝔲)>1.\phi(\mathfrak{u}):=\begin{cases}0&\mbox{if }h(\mathfrak{u})=0;\\ -i&\mbox{if }h(\mathfrak{u})=1;\\ \theta(\beta(\mathfrak{u}))&\mbox{if }h(\mathfrak{u})>1.\end{cases}

The tree 𝒯𝔣i(𝔵0)\mathcal{T}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0}) “spreads out” the set 𝒜fi(𝔵0)\mathcal{A}^{f}_{i}(\mathfrak{x}_{0}) by placing an “observer” at 𝔵0\mathfrak{x}_{0}.

For 𝔲𝒱fi(𝔵0)\mathfrak{u}\in\mathcal{V}^{f}_{i}(\mathfrak{x}_{0}) let 𝒫(𝔲):=(𝔵0,πh(𝔲)1(𝔲),πh(𝔲)2(𝔲),,π(𝔲),𝔲)\mathcal{P}(\mathfrak{u}):=(\mathfrak{x}_{0},\pi^{h(\mathfrak{u})-1}(\mathfrak{u}),\pi^{h(\mathfrak{u})-2}(\mathfrak{u}),\ldots,\pi(\mathfrak{u}),\mathfrak{u}), and

𝔣(𝔵0;𝔲):={𝔵0if 𝔲=𝔵0;𝔥(𝔵0;𝒫(𝔲))if 𝔲𝒵fi(𝔵0)fi(𝔵0);𝔱(𝔲)𝔥(𝔵0;𝒫(𝔲))if 𝔲fi(𝔵0)𝒰fi(𝔵0).\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u}):=\begin{cases}\mathfrak{x}_{0}&\mbox{if }\mathfrak{u}=\mathfrak{x}_{0};\\ \mathfrak{h}(\mathfrak{x}_{0};\mathcal{P}(\mathfrak{u}))&\mbox{if }\mathfrak{u}\in\mathcal{Z}^{f}_{i}(\mathfrak{x}_{0})\cup\mathcal{R}^{f}_{i}(\mathfrak{x}_{0});\\ {}^{\infty}\mathfrak{t}(\mathfrak{u})\mathfrak{h}(\mathfrak{x}_{0};\mathcal{P}(\mathfrak{u}))&\mbox{if }\mathfrak{u}\in\mathcal{H}^{f}_{i}(\mathfrak{x}_{0})\cup\mathcal{U}^{f}_{i}(\mathfrak{x}_{0}).\end{cases}

For distinct non-root vertices 𝔲,𝔲𝒱𝔣i(𝔵0)\mathfrak{u},\mathfrak{u}^{\prime}\in\mathcal{V}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0}), let 𝔣(𝔵0;𝔲,𝔲)\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u},\mathfrak{u}^{\prime}) denote the maximal common left substring of 𝔣(𝔵0;𝔲)\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u}) and 𝔣(𝔵0;𝔲)\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u}^{\prime}). Note that 𝔣(𝔵0;𝔲,𝔲)=𝔣(𝔵0;𝔲,𝔲)\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u},\mathfrak{u}^{\prime})=\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u}^{\prime},\mathfrak{u}). Let 𝔣(𝔵0;𝔲𝔲)\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u}\mid\mathfrak{u}^{\prime}) be the string satisfying

𝔣(𝔵0;𝔲)=𝔣(𝔵0;𝔲𝔲)𝔣(𝔵0;𝔲,𝔲).\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u})=\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u}\mid\mathfrak{u}^{\prime})\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u},\mathfrak{u}^{\prime}).

If 𝔱(𝔲)\mathfrak{t}(\mathfrak{u}) is a band, let 𝔠(𝔵0;𝔲)\mathfrak{c}(\mathfrak{x}_{0};\mathfrak{u}) denote the longest left substring of 𝔣(𝔵0;𝔲)\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u}) (or, equivalently of 𝔥(𝔵0;𝒫(𝔲))\mathfrak{h}(\mathfrak{x}_{0};\mathcal{P}(\mathfrak{u}))) in Hli(𝔵0)H_{l}^{i}(\mathfrak{x}_{0}) whose last syllable is not a syllable of 𝔱(𝔲)\mathfrak{t}(\mathfrak{u}); otherwise set 𝔠(𝔵0;𝔲):=𝔣(𝔵0;𝔲)\mathfrak{c}(\mathfrak{x}_{0};\mathfrak{u}):=\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u}).

If 𝔱(𝔲)\mathfrak{t}(\mathfrak{u}) is a band, let 𝔟𝔣α(𝔲)\mathfrak{b}^{\mathfrak{f}\alpha}(\mathfrak{u}) denote the cyclic permutation of 𝔱(𝔲)\mathfrak{t}(\mathfrak{u}) for which 𝔟𝔣α(𝔲)𝔠(𝔵0;𝔲)\mathfrak{b}^{\mathfrak{f}\alpha}(\mathfrak{u})\mathfrak{c}(\mathfrak{x}_{0};\mathfrak{u}) is a string. Let γ𝔣(𝔲)\gamma^{\mathfrak{f}}(\mathfrak{u}) denote the first syllable of 𝔟𝔣α(𝔲)\mathfrak{b}^{\mathfrak{f}\alpha}(\mathfrak{u}).

Proposition 4.3.

Suppose 𝔲𝒰fi(𝔵0)fi(𝔵0)\mathfrak{u}\in\mathcal{U}^{f}_{i}(\mathfrak{x}_{0})\cup\mathcal{R}^{f}_{i}(\mathfrak{x}_{0}). Then 𝔠(𝔵0;π(𝔲))\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u})) is a substring of 𝔥(𝔵0;𝒫(𝔲))\mathfrak{h}(\mathfrak{x}_{0};\mathcal{P}(\mathfrak{u})).

Proof. By definition of 𝔥(𝔵0;-)\mathfrak{h}(\mathfrak{x}_{0};\text{-}) ([9, Definition 6.3]) we get 𝔥(𝔵0;𝒫(𝔲))=𝔲H𝔥(𝔵0;𝒫(π(𝔲)))\mathfrak{h}(\mathfrak{x}_{0};\mathcal{P}(\mathfrak{u}))=\mathfrak{u}\circ_{H}\mathfrak{h}(\mathfrak{x}_{0};\mathcal{P}(\pi(\mathfrak{u}))). By definition, 𝔠(𝔵0;π(𝔲))\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u})) is a left substring of 𝔥(𝔵0;𝒫(π(𝔲)))\mathfrak{h}(\mathfrak{x}_{0};\mathcal{P}(\pi(\mathfrak{u}))).

Consider the following cases:

  • If N(𝔵0;𝔰(𝔲),𝔥(𝔵0;𝒫(𝔲)))=1N(\mathfrak{x}_{0};\mathfrak{s}(\mathfrak{u}),\mathfrak{h}(\mathfrak{x}_{0};\mathcal{P}(\mathfrak{u})))=1 then 𝔥(𝔵0;𝒫(π(𝔲)))\mathfrak{h}(\mathfrak{x}_{0};\mathcal{P}(\pi(\mathfrak{u}))) is a left substring of 𝔥(𝔵0;𝒫(𝔲))\mathfrak{h}(\mathfrak{x}_{0};\mathcal{P}(\mathfrak{u})).

  • If N(𝔵0;𝔰(𝔲),𝔥(𝔵0;𝒫(𝔲)))=0N(\mathfrak{x}_{0};\mathfrak{s}(\mathfrak{u}),\mathfrak{h}(\mathfrak{x}_{0};\mathcal{P}(\mathfrak{u})))=0 then 𝔥(𝔵0;𝒫(𝔲))=HR𝔰(𝔲)({𝔲𝔰(𝔲)𝔥(𝔵0;𝒫(π(𝔲)))})\mathfrak{h}(\mathfrak{x}_{0};\mathcal{P}(\mathfrak{u}))=\mathrm{HR}_{\mathfrak{s}(\mathfrak{u})}(\{\mathfrak{u}\mathfrak{s}(\mathfrak{u})\mathfrak{h}(\mathfrak{x}_{0};\mathcal{P}(\pi(\mathfrak{u})))\}). There are two subcases.

    • If {𝔲𝔰(𝔲)𝔥(𝔵0;𝒫(π(𝔲)))}=𝔲𝔰(𝔲)𝔥(𝔵0;𝒫(π(𝔲)))\{\mathfrak{u}\mathfrak{s}(\mathfrak{u})\mathfrak{h}(\mathfrak{x}_{0};\mathcal{P}(\pi(\mathfrak{u})))\}=\mathfrak{u}\mathfrak{s}(\mathfrak{u})\mathfrak{h}(\mathfrak{x}_{0};\mathcal{P}(\pi(\mathfrak{u}))) then 𝔥(𝔵0;𝒫(𝔲))=𝔲𝔥(𝔵0;𝒫(π(𝔲)))\mathfrak{h}(\mathfrak{x}_{0};\mathcal{P}(\mathfrak{u}))=\mathfrak{u}\mathfrak{h}(\mathfrak{x}_{0};\mathcal{P}(\pi(\mathfrak{u}))).

    • If {𝔲𝔰(𝔲)𝔥(𝔵0;𝒫(π(𝔲)))}=𝔲𝔥(𝔵0;𝒫(π(𝔲)))=𝔴𝔟𝔣α(π(𝔲))𝔠(𝔵0;π(𝔲))\{\mathfrak{u}\mathfrak{s}(\mathfrak{u})\mathfrak{h}(\mathfrak{x}_{0};\mathcal{P}(\pi(\mathfrak{u})))\}=\mathfrak{u}\mathfrak{h}(\mathfrak{x}_{0};\mathcal{P}(\pi(\mathfrak{u})))=\mathfrak{w}\mathfrak{b}^{\mathfrak{f}\alpha}(\pi(\mathfrak{u}))\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u})) for some string 𝔴\mathfrak{w} then 𝔥(𝔵0;𝒫(𝔲))=𝔴𝔠(𝔵0;π(𝔲))\mathfrak{h}(\mathfrak{x}_{0};\mathcal{P}(\mathfrak{u}))=\mathfrak{w}\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u})).

Clearly in every cases 𝔠(𝔵0;π(𝔲))\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u})) is a left substring of 𝔥(𝔵0;𝒫(𝔲))\mathfrak{h}(\mathfrak{x}_{0};\mathcal{P}(\mathfrak{u})).    \Box

Corollary 4.4.

Suppose 𝔲𝒰fi(𝔵0)𝔣i(𝔵0)\mathfrak{u}\in\mathcal{U}^{f}_{i}(\mathfrak{x}_{0})\cup\mathcal{R}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0}). Then the strings 𝔠(𝔵0;𝔲)\mathfrak{c}(\mathfrak{x}_{0};\mathfrak{u}) and 𝔠(𝔵0;π(𝔲))\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u})) are comparable since both are left substrings of 𝔥(𝔵0;𝒫(𝔲))\mathfrak{h}(\mathfrak{x}_{0};\mathcal{P}(\mathfrak{u})).

In view of the above corollary, we define a function υ:𝒱𝔣i(𝔵0){𝔵0}{1,0,1}\upsilon:\mathcal{V}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0})\setminus\{\mathfrak{x}_{0}\}\to\{1,0,-1\} as

υ(𝔲):={1if 𝔠(𝔵0;π(𝔲)) is a proper left substring of 𝔠(𝔵0;𝔲);1if 𝔠(𝔵0;𝔲) is a proper left substring of 𝔠(𝔵0;π(𝔲));0if 𝔠(𝔵0;𝔲)=𝔠(𝔵0;π(𝔲)).\upsilon(\mathfrak{u}):=\begin{cases}1&\mbox{if }\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u}))\mbox{ is a proper left substring of }\mathfrak{c}(\mathfrak{x}_{0};\mathfrak{u});\\ -1&\mbox{if }\mathfrak{c}(\mathfrak{x}_{0};\mathfrak{u})\mbox{ is a proper left substring of }\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u}));\\ 0&\mbox{if }\mathfrak{c}(\mathfrak{x}_{0};\mathfrak{u})=\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u})).\end{cases}
Examples 4.5.

For 𝔲𝒰fi(𝔵0)𝔣i(𝔵0)\mathfrak{u}\in\mathcal{U}^{f}_{i}(\mathfrak{x}_{0})\cup\mathcal{R}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0}), the strings 𝔠(𝔵0;𝔲)\mathfrak{c}(\mathfrak{x}_{0};\mathfrak{u}) and 𝔠(𝔵0;π(𝔲))\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u})) are comparable in all possible ways as the following examples demonstrate.

  • (υ=1\upsilon=-1)

    Choosing 𝔵0:=1(v1,i)\mathfrak{x}_{0}:=1_{(v_{1},i)} for the algebra Λ\Lambda^{\prime\prime} from [9, Figure 4] the extended arch bridge quiver is shown in [9, Figure 12]. If 𝔲=dF\mathfrak{u}=dF and π(𝔲)=edcbA\pi(\mathfrak{u})=edcbA then 𝔠(𝔵0;𝔲)=A\mathfrak{c}(\mathfrak{x}_{0};\mathfrak{u})=A is a proper left substring of 𝔠(𝔵0;π(𝔲))=cbA\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u}))=cbA.

  • (υ=1\upsilon=1)

    In the same example, if 𝔲=IbhG\mathfrak{u}=IbhG and π(𝔲)=dF\pi(\mathfrak{u})=dF then 𝔠(𝔵0;π(𝔲))=A\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u}))=A is a proper left substring of 𝔠(𝔵0;𝔲)=IbA\mathfrak{c}(\mathfrak{x}_{0};\mathfrak{u})=IbA.

  • (υ=0\upsilon=0)

    Choosing 𝔵0:=1(v1,i)\mathfrak{x}_{0}:=1_{(v_{1},i)} for the algebra Λ(iv)\Lambda^{(iv)} in [9, Figure 8], if 𝔲=dC\mathfrak{u}=dC and π(𝔲)=bA\pi(\mathfrak{u})=bA then 𝔠(𝔵0;𝔲)=A=𝔠(𝔵0;π(𝔲))\mathfrak{c}(\mathfrak{x}_{0};\mathfrak{u})=A=\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u})).

Proposition 4.6.

Suppose 𝔲𝒱𝔣i(𝔵0)\mathfrak{u}\in\mathcal{V}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0}) and h(𝔲)1h(\mathfrak{u})\geq 1. Then 𝔲\mathfrak{u} is normal if and only if 𝔣(𝔵0;𝔲,π(𝔲))\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u},\pi(\mathfrak{u})) is a proper left substring of 𝔠(𝔵0;𝔲)\mathfrak{c}(\mathfrak{x}_{0};\mathfrak{u}).

Proof. Recall that both 𝔣(𝔵0;𝔲,π(𝔲))\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u},\pi(\mathfrak{u})) and 𝔠(𝔵0;𝔲)\mathfrak{c}(\mathfrak{x}_{0};\mathfrak{u}) are substrings of 𝔣(𝔵0;𝔲)\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u}) and hence they are comparable.

First assume that h(𝔲)2h(\mathfrak{u})\geq 2; a similar argument works when h(𝔲)=1h(\mathfrak{u})=1.

If 𝔲\mathfrak{u} is normal then 𝔠(𝔵0;𝔲)=𝔲o𝔣(𝔵0;𝔲,π(𝔲))\mathfrak{c}(\mathfrak{x}_{0};\mathfrak{u})=\mathfrak{u}^{o}\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u},\pi(\mathfrak{u})). Since |𝔲o|>0|\mathfrak{u}^{o}|>0, we get the required conclusion.

Conversely if 𝔣(𝔵0;𝔲,π(𝔲))\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u},\pi(\mathfrak{u})) is a proper left substring 𝔠(𝔵0;𝔲)\mathfrak{c}(\mathfrak{x}_{0};\mathfrak{u}) then β(𝔲)\beta(\mathfrak{u}) is a syllable of 𝔠(𝔵0;𝔲)\mathfrak{c}(\mathfrak{x}_{0};\mathfrak{u}), and hence not a syllable of 𝔱(𝔲)\mathfrak{t}(\mathfrak{u}). Therefore 𝔲\mathfrak{u} is normal.    \Box

Proposition 4.7.

Suppose 𝔲𝒰𝔣i(𝔵0)𝔣i(𝔵0)\mathfrak{u}\in\mathcal{U}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0})\cup\mathcal{R}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0}) and N(𝔵0;𝔰(𝔲),𝔠(𝔵0;𝔲))=1N(\mathfrak{x}_{0};\mathfrak{s}(\mathfrak{u}),\mathfrak{c}(\mathfrak{x}_{0};\mathfrak{u}))=1. Then 𝔥(𝔵0;𝒫(π(𝔲)))\mathfrak{h}(\mathfrak{x}_{0};\mathcal{P}(\pi(\mathfrak{u}))) is a substring of 𝔠(𝔵0;𝔲)\mathfrak{c}(\mathfrak{x}_{0};\mathfrak{u}).

Proof. Since 𝔠(𝔵0;𝔲)\mathfrak{c}(\mathfrak{x}_{0};\mathfrak{u}) is a left substring of 𝔥(𝔵0;𝒫(𝔲))\mathfrak{h}(\mathfrak{x}_{0};\mathcal{P}(\mathfrak{u})) and N(𝔵0;𝔰(𝔲),𝔠(𝔵0;𝔲))=1N(\mathfrak{x}_{0};\mathfrak{s}(\mathfrak{u}),\mathfrak{c}(\mathfrak{x}_{0};\mathfrak{u}))=1 we get N(𝔵0;𝔰(𝔲),𝔥(𝔵0;𝒫(𝔲)))=1N(\mathfrak{x}_{0};\mathfrak{s}(\mathfrak{u}),\mathfrak{h}(\mathfrak{x}_{0};\mathcal{P}(\mathfrak{u})))=1. Hence using 𝔥(𝔵0;𝒫(𝔲))=𝔲H𝔥(𝔵0;𝒫(π(𝔲)))\mathfrak{h}(\mathfrak{x}_{0};\mathcal{P}(\mathfrak{u}))=\mathfrak{u}\circ_{H}\mathfrak{h}(\mathfrak{x}_{0};\mathcal{P}(\pi(\mathfrak{u}))), we get that 𝔥(𝔵0;𝒫(𝔲))={𝔲𝔰(𝔲)𝔥(𝔵0;𝒫(π(𝔲)))}\mathfrak{h}(\mathfrak{x}_{0};\mathcal{P}(\mathfrak{u}))=\{\mathfrak{u}\mathfrak{s}(\mathfrak{u})\mathfrak{h}(\mathfrak{x}_{0};\mathcal{P}(\pi(\mathfrak{u})))\}. Then 𝔥(𝔵0;𝒫(π(𝔲)))\mathfrak{h}(\mathfrak{x}_{0};\mathcal{P}(\pi(\mathfrak{u}))) is a left substring of 𝔥(𝔵0;𝒫(𝔲))\mathfrak{h}(\mathfrak{x}_{0};\mathcal{P}(\mathfrak{u})). Hence 𝔥(𝔵0;𝒫(π(𝔲)))\mathfrak{h}(\mathfrak{x}_{0};\mathcal{P}(\pi(\mathfrak{u}))) and 𝔠(𝔵0;𝔲)\mathfrak{c}(\mathfrak{x}_{0};\mathfrak{u}) are comparable.

Since N(𝔵0;𝔰(𝔲),𝔥(𝔵0;𝒫(π(𝔲))))=0N(\mathfrak{x}_{0};\mathfrak{s}(\mathfrak{u}),\mathfrak{h}(\mathfrak{x}_{0};\mathcal{P}(\pi(\mathfrak{u}))))=0 but N(𝔵0;𝔰(𝔲),𝔠(𝔵0;𝔲))=1N(\mathfrak{x}_{0};\mathfrak{s}(\mathfrak{u}),\mathfrak{c}(\mathfrak{x}_{0};\mathfrak{u}))=1, we conclude that 𝔥(𝔵0;𝒫(π(𝔲)))\mathfrak{h}(\mathfrak{x}_{0};\mathcal{P}(\pi(\mathfrak{u}))) is a left substring of 𝔠(𝔵0;𝔲)\mathfrak{c}(\mathfrak{x}_{0};\mathfrak{u}).    \Box

For 𝔲𝒱𝔣i(𝔵0)\mathfrak{u}\in\mathcal{V}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0}) if 𝔰(𝔲)\mathfrak{s}(\mathfrak{u}) is a band, Proposition 4.3 allows us to define a string 𝔴(𝔵0;𝔲)\mathfrak{w}(\mathfrak{x}_{0};\mathfrak{u}) by 𝔣(𝔵0;𝔲,π(𝔲))=𝔴(𝔵0;𝔲)𝔠(𝔵0;π(𝔲))\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u},\pi(\mathfrak{u}))=\mathfrak{w}(\mathfrak{x}_{0};\mathfrak{u})\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u})). Let 𝔟𝔣β\mathfrak{b}^{\mathfrak{f}\beta} denote the cyclic permutation of 𝔰(𝔲)\mathfrak{s}(\mathfrak{u}) for which 𝔟𝔣β𝔴(𝔵0;𝔲)\mathfrak{b}^{\mathfrak{f}\beta}\mathfrak{w}(\mathfrak{x}_{0};\mathfrak{u}) is a string.

The following is an important observation from the proof of Proposition 4.3.

Corollary 4.8.

Suppose 𝔲𝒰fi(𝔵0)𝔣i(𝔵0)\mathfrak{u}\in\mathcal{U}^{f}_{i}(\mathfrak{x}_{0})\cup\mathcal{R}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0}). Then 𝔴(𝔵0;𝔲)𝔠(𝔵0;π(𝔲))\mathfrak{w}(\mathfrak{x}_{0};\mathfrak{u})\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u})) is a left substring of 𝔥(𝔵0;𝒫(𝔲))\mathfrak{h}(\mathfrak{x}_{0};\mathcal{P}(\mathfrak{u})).

Our next goal is to show that two different vertices of the tree have incomparable values of 𝔣(𝔵0;-)\mathfrak{f}(\mathfrak{x}_{0};\mbox{-}). Here is a supporting result.

Proposition 4.9.

If 𝔲fi(𝔵0)\mathfrak{u}\in\mathcal{R}^{f}_{i}(\mathfrak{x}_{0}) then α𝔣(𝔵0;𝔲)\alpha\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u}) is not a string for any syllable α\alpha.

Proof. Suppose α𝔣(𝔵0;𝔲)\alpha\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u}) is a string for some syllable α\alpha. Since 𝔲fi(𝔵0)\mathfrak{u}\in\mathcal{R}^{f}_{i}(\mathfrak{x}_{0}), 𝔲\mathfrak{u} is a maximal reverse torsion arch bridge, and hence α𝔲𝔰(𝔲)\alpha\mathfrak{u}\mathfrak{s}(\mathfrak{u}) is not a string. Hence 𝔟𝔣β𝔴(𝔵0;𝔲)𝔠(𝔵0;𝔲)\nequivH𝔴(𝔵0;𝔲)𝔠(𝔵0;𝔲)\mathfrak{b}^{\mathfrak{f}\beta}\mathfrak{w}(\mathfrak{x}_{0};\mathfrak{u})\mathfrak{c}(\mathfrak{x}_{0};\mathfrak{u})\nequiv_{H}\mathfrak{w}(\mathfrak{x}_{0};\mathfrak{u})\mathfrak{c}(\mathfrak{x}_{0};\mathfrak{u}) as α𝔲o𝔟𝔣β𝔴(𝔵0;𝔲)𝔠(𝔵0;𝔲)\alpha\mathfrak{u}^{o}\mathfrak{b}^{\mathfrak{f}\beta}\mathfrak{w}(\mathfrak{x}_{0};\mathfrak{u})\mathfrak{c}(\mathfrak{x}_{0};\mathfrak{u}) is not a string but α𝔲o𝔴(𝔵0;𝔲)𝔠(𝔵0;𝔲)\alpha\mathfrak{u}^{o}\mathfrak{w}(\mathfrak{x}_{0};\mathfrak{u})\mathfrak{c}(\mathfrak{x}_{0};\mathfrak{u}).

On the other hand since α𝔲𝔰(𝔲)\alpha\mathfrak{u}\mathfrak{s}(\mathfrak{u}) is not a string, we get N(𝔵0;𝔰(𝔲),𝔣(𝔵0;𝔲))=0N(\mathfrak{x}_{0};\mathfrak{s}(\mathfrak{u}),\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u}))=0. Thus by the definition of 𝔥(𝔵0;𝒫)\mathfrak{h}(\mathfrak{x}_{0};\mathcal{P}), HR𝔰(𝔲)(𝔲o𝔟𝔣β𝔴(𝔵0;𝔲)𝔠(𝔵0;𝔲))\mathrm{HR}_{\mathfrak{s}(\mathfrak{u})}(\mathfrak{u}^{o}\mathfrak{b}^{\mathfrak{f}\beta}\mathfrak{w}(\mathfrak{x}_{0};\mathfrak{u})\mathfrak{c}(\mathfrak{x}_{0};\mathfrak{u})) exists, a contradiction to the above paragraph. Hence our assumption is wrong.    \Box

Proposition 4.10.

Suppose 𝔲,𝔲𝒱𝔣i(𝔵0)\mathfrak{u},\mathfrak{u}^{\prime}\in\mathcal{V}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0}) are non-root vertices. If 𝔣(𝔵0;𝔲)\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u}^{\prime}) is a left substring of 𝔣(𝔵0;𝔲)\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u}) then 𝔲=𝔲\mathfrak{u}=\mathfrak{u}^{\prime}.

Proof. Suppose 𝔣(𝔵0;𝔲)\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u}^{\prime}) is a left substring of 𝔣(𝔵0;𝔲)\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u}).

If the former is an \mathbb{N}^{*}-string then so is latter and hence they are equal.

On the other hand, if 𝔣(𝔵0;𝔲)\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u}^{\prime}) is a finite string then 𝔲fi(𝔵0)𝒵fi(𝔵0)\mathfrak{u}^{\prime}\in\mathcal{R}^{f}_{i}(\mathfrak{x}_{0})\cup\mathcal{Z}^{f}_{i}(\mathfrak{x}_{0}).

If 𝔲𝒵fi(𝔵0)\mathfrak{u}^{\prime}\in\mathcal{Z}^{f}_{i}(\mathfrak{x}_{0}) then 𝔣(𝔵0;𝔲)=𝔲𝔵0\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u}^{\prime})=\mathfrak{u}^{\prime}\mathfrak{x}_{0}. By definition, 𝔲\mathfrak{u}^{\prime} is a maximal weak torsion zero bridge, and hence 𝔣(𝔵0;𝔲)\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u}^{\prime}) cannot be extended to the left. Thus 𝔣(𝔵0;𝔲)=𝔣(𝔵0;𝔲)\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u}^{\prime})=\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u}).

If 𝔲fi(𝔵0)\mathfrak{u}^{\prime}\in\mathcal{R}^{f}_{i}(\mathfrak{x}_{0}) then Proposition 4.9 guarantees the same conclusion.

In each case we have that 𝔣(𝔵0;𝔲)=𝔣(𝔵0;𝔲)\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u})=\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u}^{\prime}) and that both 𝔲\mathfrak{u} and 𝔲\mathfrak{u}^{\prime} are either equal to 𝔵0\mathfrak{x}_{0} or in 𝒵fi(𝔵0)fi(𝔵0)\mathcal{Z}^{f}_{i}(\mathfrak{x}_{0})\cup\mathcal{R}^{f}_{i}(\mathfrak{x}_{0}) or in fi(𝔵0)𝒰fi(𝔵0)\mathcal{H}^{f}_{i}(\mathfrak{x}_{0})\cup\mathcal{U}^{f}_{i}(\mathfrak{x}_{0}). In each case it is clear that 𝔥(𝔵0;𝒫(𝔲))=𝔥(𝔵0;𝒫(𝔲))\mathfrak{h}(\mathfrak{x}_{0};\mathcal{P}(\mathfrak{u}))=\mathfrak{h}(\mathfrak{x}_{0};\mathcal{P}(\mathfrak{u}^{\prime})). Now [9, Theorem 7.5] gives that 𝒫(𝔲)=𝒫(𝔲)\mathcal{P}(\mathfrak{u})=\mathcal{P}(\mathfrak{u}^{\prime}), and hence 𝔲=𝔲\mathfrak{u}=\mathfrak{u}^{\prime} in 𝒯𝔣i(𝔵0)\mathcal{T}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0}).    \Box

Corollary 4.11.

If 𝔲,𝔲𝒱𝔣i(𝔵0)\mathfrak{u},\mathfrak{u}^{\prime}\in\mathcal{V}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0}) are distinct non-root vertices then |𝔣(𝔵0;𝔲𝔲)|>0|\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u}\mid\mathfrak{u}^{\prime})|>0 and hence θ(𝔣(𝔵0;𝔲𝔲))=θ(𝔣(𝔵0;𝔲𝔲))\theta(\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u}^{\prime}\mid\mathfrak{u}))=-\theta(\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u}\mid\mathfrak{u}^{\prime})).

Now we define total ordering(s) on the set of children of a vertex of 𝒯𝔣i(𝔵0)\mathcal{T}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0}). Let 𝔲1,𝔲2ξ𝔣i(𝔵0)\mathfrak{u}_{1},\mathfrak{u}_{2}\in\xi^{\mathfrak{f}}_{i}(\mathfrak{x}_{0}). Say 𝔲1𝔣i𝔲2\mathfrak{u}_{1}\sqsubset^{\mathfrak{f}}_{i}\mathfrak{u}_{2} if θ(𝔣(𝔵0;𝔲2𝔲1))=i\theta(\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u}_{2}\mid\mathfrak{u}_{1}))=i. Let ξfmini(𝔵0)\xi^{\mathrm{fmin}}_{i}(\mathfrak{x}_{0}) and ξfmaxi(𝔵0)\xi^{\mathrm{fmax}}_{i}(\mathfrak{x}_{0}) denote the minimal and maximal elements of this order respectively. The immediate successor and immediate predecessor of 𝔲1\mathfrak{u}_{1} in (ξ𝔣i(𝔵0),𝔣i)(\xi^{\mathfrak{f}}_{i}(\mathfrak{x}_{0}),\sqsubset^{\mathfrak{f}}_{i}), if exist, are denoted by 𝔲1f+\mathfrak{u}_{1}^{f+} and 𝔲1f\mathfrak{u}_{1}^{f-} respectively.

Let 𝔲𝒱𝔣i(𝔵0){𝔵0}\mathfrak{u}\in\mathcal{V}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0})\setminus\{\mathfrak{x}_{0}\} and 𝔲1,𝔲2ξ𝔣j(𝔲)\mathfrak{u}_{1},\mathfrak{u}_{2}\in\xi^{\mathfrak{f}}_{j}(\mathfrak{u}). Say 𝔲1𝔣j𝔲2\mathfrak{u}_{1}\sqsubset^{\mathfrak{f}}_{j}\mathfrak{u}_{2} if θ(𝔣(𝔵0;𝔲2𝔲1))=j\theta(\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u}_{2}\mid\mathfrak{u}_{1}))=-j. Let ξfminj(𝔲)\xi^{\mathrm{fmin}}_{j}(\mathfrak{u}) and ξfmaxj(𝔲)\xi^{\mathrm{fmax}}_{j}(\mathfrak{u}) denote the minimal and maximal elements of this order respectively. The immediate successor and immediate predecessor of 𝔲1ξ𝔣j(𝔲)\mathfrak{u}_{1}\in\xi^{\mathfrak{f}}_{j}(\mathfrak{u}) in (ξ𝔣j(𝔲),𝔣j)(\xi^{\mathfrak{f}}_{j}(\mathfrak{u}),\sqsubset^{\mathfrak{f}}_{j}), if exist, are denoted by 𝔲1f+\mathfrak{u}_{1}^{f+} and 𝔲1f\mathfrak{u}_{1}^{f-} respectively. Choosing leaves in 𝒵𝔣i(𝔵0)𝔣i(𝔵0)\mathcal{Z}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0})\cup\mathcal{R}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0}) as torsion while 𝔣i(𝔵0)𝒰𝔣i(𝔵0)\mathcal{H}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0})\cup\mathcal{U}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0}) as non-torsion we have completely described the decorated tree structure of 𝒯𝔣i(𝔵0)\mathcal{T}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0}).

We shall use words like parent, grandparent, child, uncle, granduncle etc. to describe relations between two vertices of 𝒯𝔣i(𝔵0)\mathcal{T}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0}).

Example 4.12.

Choosing 𝔵0:=1(v1,i)\mathfrak{x}_{0}:=1_{(v_{1},i)} for the algebra Λ\Lambda^{\prime\prime} from [9, Figure 4] the extended arch bridge quiver is shown in [9, Figure 12]. Figure 5 shows the decorated tree 𝒯𝔣1(𝔵0)\mathcal{T}^{\mathfrak{f}}_{1}(\mathfrak{x}_{0}) where the arrow with target 𝔲\mathfrak{u} in the tree is labeled jj if 𝔲ξ𝔣j(π(𝔲))\mathfrak{u}\in\xi^{\mathfrak{f}}_{j}(\pi(\mathfrak{u})). Here edcbA,IbAξ𝔣1(𝔵0)edcbA,IbA\in\xi^{\mathfrak{f}}_{1}(\mathfrak{x}_{0}) satisfy edcbA𝔣1IbAedcbA\sqsubset^{\mathfrak{f}}_{1}IbA.

1(v1,i){{{1_{(v_{1},i)}}}}edcbA{edcbA}IbA{IbA}dF{dF}IbhG{IbhG}+\scriptstyle{+}+\scriptstyle{+}+\scriptstyle{+}+\scriptstyle{+}
Figure 5. 𝒯𝔣1(1(v1,i))\mathcal{T}^{\mathfrak{f}}_{1}(1_{(v_{1},i)}) for Λ\Lambda^{\prime\prime}

5. Long elements of 𝒱𝔣i(𝔵0)\mathcal{V}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0})

Say that 𝔲𝒰fi(𝔵0)𝔣i(𝔵0)\mathfrak{u}\in\mathcal{U}^{f}_{i}(\mathfrak{x}_{0})\cup\mathcal{R}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0}) is long if N(𝔵0;𝔰(𝔲),𝔣(𝔵0;𝔲))=1N(\mathfrak{x}_{0};\mathfrak{s}(\mathfrak{u}),\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u}))=1; otherwise say it is short. If we refer to 𝔲\mathfrak{u} as long or short then it is implicitly assumed that 𝔲𝒰fi(𝔵0)fi(𝔵0)\mathfrak{u}\in\mathcal{U}^{f}_{i}(\mathfrak{x}_{0})\cup\mathcal{R}^{f}_{i}(\mathfrak{x}_{0}).

If 𝔲𝒱𝔣i(𝔵0)\mathfrak{u}\in\mathcal{V}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0}) is long then let 𝔴~(𝔵0;𝔲)\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u}) be defined by 𝔴(𝔵0;𝔲)=𝔟𝔣β(𝔲)𝔴~(𝔵0;𝔲)\mathfrak{w}(\mathfrak{x}_{0};\mathfrak{u})=\mathfrak{b}^{\mathfrak{f}\beta}(\mathfrak{u})\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u}). On the other hand if 𝔲\mathfrak{u} is short then we set 𝔴~(𝔵0;𝔲):=𝔴(𝔵0;𝔲)\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u}):=\mathfrak{w}(\mathfrak{x}_{0};\mathfrak{u}). Note that |𝔴~(𝔵0;𝔲)|<|𝔰(𝔲)||\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u})|<|\mathfrak{s}(\mathfrak{u})|.

Remark 5.1.

Suppose 𝔲𝒰𝔣i(𝔵0)𝔣i(𝔵0)\mathfrak{u}\in\mathcal{U}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0})\cup\mathcal{H}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0}) and 𝔲1,𝔲2ξ𝔣(𝔲)\mathfrak{u}_{1},\mathfrak{u}_{2}\in\xi^{\mathfrak{f}}(\mathfrak{u}). If 𝔲1\mathfrak{u}_{1} is long and 𝔲2\mathfrak{u}_{2} is short then 𝔣(𝔵0;𝔲1,𝔲2)=𝔴(𝔵0;𝔲2)𝔠(𝔵0;𝔲)\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u}_{1},\mathfrak{u}_{2})=\mathfrak{w}(\mathfrak{x}_{0};\mathfrak{u}_{2})\mathfrak{c}(\mathfrak{x}_{0};\mathfrak{u}). Moreover if θ(β(𝔲1))=θ(β(𝔲2))=\theta(\beta(\mathfrak{u}_{1}))=\theta(\beta(\mathfrak{u}_{2}))=\partial then 𝔲2𝔣𝔲1\mathfrak{u}_{2}\sqsubset^{\mathfrak{f}}_{\partial}\mathfrak{u}_{1}.

Remark 5.2.

Suppose 𝔲𝒱𝔣i(𝔵0)𝔵0\mathfrak{u}\in\mathcal{V}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0})\setminus\mathfrak{x}_{0}. If 𝔲1,𝔲2ξ𝔣(𝔲)\mathfrak{u}_{1},\mathfrak{u}_{2}\in\xi^{\mathfrak{f}}_{\partial}(\mathfrak{u}) with 𝔲1𝔣𝔲2\mathfrak{u}_{1}\sqsubset^{\mathfrak{f}}_{\partial}\mathfrak{u}_{2} then 𝔣(𝔵0;𝔲,𝔲1)\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u},\mathfrak{u}_{1}) is a left substring of 𝔣(𝔵0;𝔲,𝔲2)\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u},\mathfrak{u}_{2}).

Remark 5.3.

Suppose 𝔲𝒰𝔣i(𝔵0)𝔣i(𝔵0)\mathfrak{u}\in\mathcal{U}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0})\cup\mathcal{H}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0}) and 𝔲1,𝔲2,𝔲3ξ𝔣(𝔲)\mathfrak{u}_{1},\mathfrak{u}_{2},\mathfrak{u}_{3}\in\xi^{\mathfrak{f}}_{\partial}(\mathfrak{u}). If 𝔲1𝔣𝔲2𝔣𝔲3\mathfrak{u}_{1}\sqsubset^{\mathfrak{f}}_{\partial}\mathfrak{u}_{2}\sqsubset^{\mathfrak{f}}_{\partial}\mathfrak{u}_{3} then 𝔣(𝔵0;𝔲1,𝔲3)\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u}_{1},\mathfrak{u}_{3}) is a left substring of 𝔣(𝔵0;𝔲2,𝔲3)\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u}_{2},\mathfrak{u}_{3}).

Remark 5.4.

If 𝔲𝒰𝔣i(𝔵0)\mathfrak{u}\in\mathcal{U}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0}) is normal then 𝔲o\mathfrak{u}^{o} is a substring of 𝔠(𝔵0;𝔲)\mathfrak{c}(\mathfrak{x}_{0};\mathfrak{u}) which is a substring of 𝔥(𝔵0;𝒫(𝔲))\mathfrak{h}(\mathfrak{x}_{0};\mathcal{P}(\mathfrak{u})). Hence N(𝔵0;𝔰(𝔲),𝔣(𝔵0;𝔲))=N(𝔵0;𝔰(𝔲),𝔥(𝔵0;𝒫(𝔲)))=N(𝔵0;𝔰(𝔲),𝔠(𝔵0;𝔲))N(\mathfrak{x}_{0};\mathfrak{s}(\mathfrak{u}),\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u}))=N(\mathfrak{x}_{0};\mathfrak{s}(\mathfrak{u}),\mathfrak{h}(\mathfrak{x}_{0};\mathcal{P}(\mathfrak{u})))=N(\mathfrak{x}_{0};\mathfrak{s}(\mathfrak{u}),\mathfrak{c}(\mathfrak{x}_{0};\mathfrak{u})).

Proposition 5.5.

Suppose 𝔲𝒰fi(𝔵0)𝔣i(𝔵0)\mathfrak{u}\in\mathcal{U}^{f}_{i}(\mathfrak{x}_{0})\cup\mathcal{R}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0}). Then 𝔲\mathfrak{u} is long if and only if N(𝔵0;𝔰(𝔲),𝔥(𝔵0;𝒫(𝔲)))=1N(\mathfrak{x}_{0};\mathfrak{s}(\mathfrak{u}),\mathfrak{h}(\mathfrak{x}_{0};\mathcal{P}(\mathfrak{u})))=1.

Proof. Since 𝔥(𝔵0;𝒫(𝔲))\mathfrak{h}(\mathfrak{x}_{0};\mathcal{P}(\mathfrak{u})) is a left substring of 𝔣(𝔵0;𝔲)\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u}), we only need to prove the forward implication. If 𝔲𝔣i(𝔵0)\mathfrak{u}\in\mathcal{R}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0}) then 𝔣(𝔵0;𝔲)=𝔥(𝔵0;𝒫(𝔲))\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u})=\mathfrak{h}(\mathfrak{x}_{0};\mathcal{P}(\mathfrak{u})), and hence the conclusion is obvious. Thus we assume that 𝔲𝒰𝔣i(𝔵0)\mathfrak{u}\in\mathcal{U}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0}). Let γ\gamma denote the first syllable of 𝔱(𝔲)\mathfrak{t}(\mathfrak{u}) respectively.

Suppose 𝔲\mathfrak{u} is long. Then N(𝔵0;𝔰(𝔲),𝔣(𝔵0;𝔲))=1N(\mathfrak{x}_{0};\mathfrak{s}(\mathfrak{u}),\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u}))=1. For contradiction, assume that N(𝔵0;𝔰(𝔲),𝔥(𝔵0;𝒫(𝔲)))=0N(\mathfrak{x}_{0};\mathfrak{s}(\mathfrak{u}),\mathfrak{h}(\mathfrak{x}_{0};\mathcal{P}(\mathfrak{u})))=0. Then there is a shortest string 𝔴\mathfrak{w} of positive length such that N(𝔵0;𝔰(𝔲),𝔴𝔥(𝔵0;𝒫(𝔲)))=1N(\mathfrak{x}_{0};\mathfrak{s}(\mathfrak{u}),\mathfrak{w}\mathfrak{h}(\mathfrak{x}_{0};\mathcal{P}(\mathfrak{u})))=1 and 𝔴𝔥(𝔵0;𝒫(𝔲))\mathfrak{w}\mathfrak{h}(\mathfrak{x}_{0};\mathcal{P}(\mathfrak{u})) is a left substring of 𝔣(𝔵0;𝔲)\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u}). Hence 𝔴\mathfrak{w} is a left substring of a finite power of 𝔱(𝔲)\mathfrak{t}(\mathfrak{u}). In view of Remark 5.4 𝔲\mathfrak{u} is abnormal. Minimality of 𝔴\mathfrak{w} guarantees that a right substring of 𝔴\mathfrak{w} of positive length is also a substring of a cyclic permutation of 𝔰(𝔲)\mathfrak{s}(\mathfrak{u}). Further [9, Propositions 4.2,4.3] gives that γ\gamma is not a syllable of 𝔰(𝔲)\mathfrak{s}(\mathfrak{u}). Thus 𝔴\mathfrak{w} is not a substring of 𝔰(𝔲)\mathfrak{s}(\mathfrak{u}). Therefore N(𝔵0;𝔰(𝔲),𝔣(𝔵0;𝔲))=1N(\mathfrak{x}_{0};\mathfrak{s}(\mathfrak{u}),\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u}))=1 and N(𝔵0;𝔰(𝔲),𝔥(𝔵0;𝒫(𝔲)))=0N(\mathfrak{x}_{0};\mathfrak{s}(\mathfrak{u}),\mathfrak{h}(\mathfrak{x}_{0};\mathcal{P}(\mathfrak{u})))=0 together give that a cyclic permutation of 𝔰(𝔲)\mathfrak{s}(\mathfrak{u}) is a substring of 𝔴\mathfrak{w}. Since γ\gamma is not a syllable of 𝔰(𝔲)\mathfrak{s}(\mathfrak{u}), minimality of 𝔴\mathfrak{w} ensures that 𝔴\mathfrak{w} is a substring of 𝔱(𝔲)\mathfrak{t}(\mathfrak{u}). Hence a cyclic permutation of 𝔰(𝔲)\mathfrak{s}(\mathfrak{u}) is a substring of 𝔱(𝔲)\mathfrak{t}(\mathfrak{u}), a contradiction to the domesticity of the algebra by [4, Corollary 3.4.1].    \Box

Proposition 5.5 together with Remark 5.4 ensures that if 𝔲\mathfrak{u} is normal with h(𝔲)>1h(\mathfrak{u})>1 then 𝔲\mathfrak{u} is long if and only if N(𝔵0;𝔰(𝔲),𝔠(𝔵0;𝔲))=1N(\mathfrak{x}_{0};\mathfrak{s}(\mathfrak{u}),\mathfrak{c}(\mathfrak{x}_{0};\mathfrak{u}))=1. We will show in the proof of Proposition 6.21 that the same conclusion of holds when 𝔲\mathfrak{u} is bb-long. However the following example shows that it fails if 𝔲\mathfrak{u} is abnormal and ss-long.

Example 5.6.

Consider the algebra Γ(i)\Gamma^{(i)} in Figure 6. The only bands here are 𝔟1:=hbdC\mathfrak{b}_{1}:=hbdC, 𝔟2:=dfE\mathfrak{b}_{2}:=dfE and their inverses. Choose 𝔵0:=1(𝔳1,i)\mathfrak{x}_{0}:=1_{(\mathfrak{v}_{1},i)} and 𝔲:=dC\mathfrak{u}:=dC. Then π(𝔲)=hbA\pi(\mathfrak{u})=hbA. Thus 𝔠(𝔵0;π(𝔲))=A\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u}))=A and 𝔠(𝔵0;𝔲)=ChbA\mathfrak{c}(\mathfrak{x}_{0};\mathfrak{u})=ChbA, and hence υ(𝔲)=1\upsilon(\mathfrak{u})=1. Here 𝔲\mathfrak{u} is ss-long, N(𝔵0;𝔰(𝔲),𝔣(𝔵0;𝔲))=1N(\mathfrak{x}_{0};\mathfrak{s}(\mathfrak{u}),\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u}))=1 but N(𝔵0;𝔰(𝔲),𝔠(𝔵0;𝔲))=0N(\mathfrak{x}_{0};\mathfrak{s}(\mathfrak{u}),\mathfrak{c}(\mathfrak{x}_{0};\mathfrak{u}))=0.

v7{v_{7}}v5{v_{5}}v6{v_{6}}v1{v_{1}}v2{v_{2}}v3{v_{3}}v4{v_{4}}g\scriptstyle{g}h\scriptstyle{h}b\scriptstyle{b}a\scriptstyle{a}c\scriptstyle{c}d\scriptstyle{d}e\scriptstyle{e}f\scriptstyle{f}
Figure 6. Γ(i)\Gamma^{(i)} with ρ={ae,ad,be,hg,cf,hbdf}\rho=\{ae,ad,be,hg,cf,hbdf\}
Remark 5.7.

It is interesting to note that if we remove aeae from ρ\rho in Figure 6 then in that algebra we get υ(𝔲)=0\upsilon(\mathfrak{u})=0 where 𝔲\mathfrak{u} is as chosen in Example 5.6.

For an arbitrary 𝔲\mathfrak{u} the best we can achieve is the following.

Corollary 5.8.

Suppose 𝔲𝒰fi(𝔵0)fi(𝔵0)\mathfrak{u}\in\mathcal{U}^{f}_{i}(\mathfrak{x}_{0})\cup\mathcal{R}^{f}_{i}(\mathfrak{x}_{0}) and 𝔲ξ𝔣(π(𝔲))\mathfrak{u}\in\xi^{\mathfrak{f}}_{\partial}(\pi(\mathfrak{u})) for some {+,}\partial\in\{+,-\}. Then 𝔲\mathfrak{u} is long if and only if the LVP (𝔟,𝔶):=(𝔰(𝔲),𝔴~(𝔵0;𝔲)𝔠(𝔵0;π(𝔲)))(\mathfrak{b},\mathfrak{y}):=(\mathfrak{s}(\mathfrak{u}),\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u})\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u}))) is long.

Proof. If 𝔲\mathfrak{u} is long then N(𝔵0;𝔰(𝔲),𝔥(𝔵0;𝒫(𝔲)))=1N(\mathfrak{x}_{0};\mathfrak{s}(\mathfrak{u}),\mathfrak{h}(\mathfrak{x}_{0};\mathcal{P}(\mathfrak{u})))=1 by Proposition 5.5. Since 𝔥(𝔵0;𝒫(𝔲))\mathfrak{h}(\mathfrak{x}_{0};\mathcal{P}(\mathfrak{u})) is an H-reduced string by [9, Theorem 6.4], (𝔰(𝔲),𝔴~(𝔵0;𝔲)𝔠(𝔵0;π(𝔲)))(\mathfrak{s}(\mathfrak{u}),\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u})\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u}))) is long by Corollary 4.8.

Conversely if (𝔰(𝔲),𝔴~(𝔵0;𝔲)𝔠(𝔵0;π(𝔲)))(\mathfrak{s}(\mathfrak{u}),\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u})\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u}))) is long then N(𝔵0;𝔰(𝔲),𝔴(𝔵0;𝔲)𝔠(𝔵0;π(𝔲)))=1N(\mathfrak{x}_{0};\mathfrak{s}(\mathfrak{u}),\mathfrak{w}(\mathfrak{x}_{0};\mathfrak{u})\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u})))=1 and hence N(𝔵0;𝔰(𝔲),𝔥(𝔵0;𝒫(𝔲)))=1N(\mathfrak{x}_{0};\mathfrak{s}(\mathfrak{u}),\mathfrak{h}(\mathfrak{x}_{0};\mathcal{P}(\mathfrak{u})))=1 by Corollary 4.8. Therefore 𝔲\mathfrak{u} is long by Proposition 5.5.    \Box

In view of the above corollary, we freely transfer adjectives and qualifiers of the pair (𝔰(𝔲),𝔴~(𝔵0;𝔲)𝔠(𝔵0;π(𝔲)))(\mathfrak{s}(\mathfrak{u}),\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u})\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u}))) to 𝔲\mathfrak{u}.

We note an immediate consequence of the above result.

Corollary 5.9.

Suppose 𝔲𝔣i(𝔵0)𝒰𝔣i(𝔵0)\mathfrak{u}\in\mathcal{H}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0})\cup\mathcal{U}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0}), 𝔲1,𝔲2ξ𝔣(𝔲)\mathfrak{u}_{1},\mathfrak{u}_{2}\in\xi^{\mathfrak{f}}(\mathfrak{u}) and 𝔴~(𝔵0;𝔲1)\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u}_{1}) is a left substring of 𝔴~(𝔵0;𝔲2)\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u}_{2}). If 𝔲2\mathfrak{u}_{2} is long then 𝔲1\mathfrak{u}_{1} is also long.

Suppose 𝔲𝔣i(𝔵0)𝒰𝔣i(𝔵0)\mathfrak{u}\in\mathcal{H}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0})\cup\mathcal{U}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0}) and (𝔱(𝔲),𝔠(𝔵0;𝔲))(\mathfrak{t}(\mathfrak{u}),\mathfrak{c}(\mathfrak{x}_{0};\mathfrak{u})) is long. If the pair is ss-long then let γ¯s(𝔲)\bar{\gamma}^{s}(\mathfrak{u}) denote the last syllable of 𝔠(𝔵0;𝔲)\mathfrak{c}(\mathfrak{x}_{0};\mathfrak{u}), and let γ¯¯s(𝔲)\bar{\bar{\gamma}}^{s}(\mathfrak{u}) denote the last syllable of 𝔟𝔣α(𝔲)\mathfrak{b}^{\mathfrak{f}\alpha}(\mathfrak{u}). If the pair is bb-long then let γ¯b(𝔲)\bar{\gamma}^{b}(\mathfrak{u}) denote the last syllable of 𝔟𝔣α(𝔲)\mathfrak{b}^{\mathfrak{f}\alpha}(\mathfrak{u}), and let γ¯¯b(𝔲)\bar{\bar{\gamma}}^{b}(\mathfrak{u}) denote the last syllable of 𝔠(𝔵0;𝔲)\mathfrak{c}(\mathfrak{x}_{0};\mathfrak{u}) if |𝔠(𝔵0;𝔲)|>0|\mathfrak{c}(\mathfrak{x}_{0};\mathfrak{u})|>0.

Remark 5.10.

If 𝔲𝔣i(𝔵0)\mathfrak{u}\in\mathcal{H}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0}) and |𝔠(𝔵0;𝔲)|=0|\mathfrak{c}(\mathfrak{x}_{0};\mathfrak{u})|=0 then (𝔱(𝔲),𝔠(𝔵0;𝔲))(\mathfrak{t}(\mathfrak{u}),\mathfrak{c}(\mathfrak{x}_{0};\mathfrak{u})) is not ss-long.

Suppose 𝔲𝒰𝔣i(𝔵0)𝔣i(𝔵0)\mathfrak{u}\in\mathcal{U}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0})\cup\mathcal{R}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0}). If 𝔲\mathfrak{u} is ss-long then 𝔯s1(π(𝔲))\mathfrak{r}^{s}_{1}(\pi(\mathfrak{u})) denotes the longest left substring of 𝔟𝔣α(π(𝔲))\mathfrak{b}^{\mathfrak{f}\alpha}(\pi(\mathfrak{u})) satisfying δ(𝔯s1(π(𝔲)))=θ(γ¯s(π(𝔲)))\delta(\mathfrak{r}^{s}_{1}(\pi(\mathfrak{u})))=\theta(\bar{\gamma}^{s}(\pi(\mathfrak{u}))) if |𝔯s1(π(𝔲))|>0|\mathfrak{r}^{s}_{1}(\pi(\mathfrak{u}))|>0. Let 𝔯~s(π(𝔲))\tilde{\mathfrak{r}}^{s}(\pi(\mathfrak{u})) and 𝔯s2(π(𝔲))\mathfrak{r}^{s}_{2}(\pi(\mathfrak{u})) satisfy 𝔯s1(π(𝔲))𝔯~s(π(𝔲))=ρr(𝔯s1(π(𝔲))𝔠(𝔵0;π(𝔲)))\mathfrak{r}^{s}_{1}(\pi(\mathfrak{u}))\tilde{\mathfrak{r}}^{s}(\pi(\mathfrak{u}))=\rho_{r}(\mathfrak{r}^{s}_{1}(\pi(\mathfrak{u}))\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u}))) and 𝔯s2(π(𝔲))𝔯s1(π(𝔲))𝔯~s(π(𝔲))ρρ1\mathfrak{r}^{s}_{2}(\pi(\mathfrak{u}))\mathfrak{r}^{s}_{1}(\pi(\mathfrak{u}))\tilde{\mathfrak{r}}^{s}(\pi(\mathfrak{u}))\in\rho\cup\rho^{-1}. Define a syllable β~s(π(𝔲))\tilde{\beta}^{s}(\pi(\mathfrak{u})) and a string 𝔯s(π(𝔲))\mathfrak{r}^{s}(\pi(\mathfrak{u})) by 𝔯s2(π(𝔲))𝔯s1(π(𝔲))=β~s(π(𝔲))𝔯s(π(𝔲))\mathfrak{r}^{s}_{2}(\pi(\mathfrak{u}))\mathfrak{r}^{s}_{1}(\pi(\mathfrak{u}))=\tilde{\beta}^{s}(\pi(\mathfrak{u}))\mathfrak{r}^{s}(\pi(\mathfrak{u})).

On the other hand, if 𝔲\mathfrak{u} is bb-long then 𝔯b1(π(𝔲))\mathfrak{r}^{b}_{1}(\pi(\mathfrak{u})) denotes the longest left substring of 𝔟𝔣α(π(𝔲))\mathfrak{b}^{\mathfrak{f}\alpha}(\pi(\mathfrak{u})) satisfying δ(𝔯b1(π(𝔲)))=θ(γ¯b(π(𝔲)))\delta(\mathfrak{r}^{b}_{1}(\pi(\mathfrak{u})))=\theta(\bar{\gamma}^{b}(\pi(\mathfrak{u}))) if |𝔯b1(π(𝔲))|>0|\mathfrak{r}^{b}_{1}(\pi(\mathfrak{u}))|>0. Let 𝔯~b(π(𝔲))\tilde{\mathfrak{r}}^{b}(\pi(\mathfrak{u})) and 𝔯b2(π(𝔲))\mathfrak{r}^{b}_{2}(\pi(\mathfrak{u})) satisfy 𝔯b1(π(𝔲))𝔯~b(π(𝔲))=ρr(𝔯b1(π(𝔲))𝔟𝔣α(π(𝔲)))\mathfrak{r}^{b}_{1}(\pi(\mathfrak{u}))\tilde{\mathfrak{r}}^{b}(\pi(\mathfrak{u}))=\rho_{r}(\mathfrak{r}^{b}_{1}(\pi(\mathfrak{u}))\mathfrak{b}^{\mathfrak{f}\alpha}(\pi(\mathfrak{u}))) and 𝔯b2(π(𝔲))𝔯b1(π(𝔲))𝔯~b(π(𝔲))ρρ1\mathfrak{r}^{b}_{2}(\pi(\mathfrak{u}))\mathfrak{r}^{b}_{1}(\pi(\mathfrak{u}))\tilde{\mathfrak{r}}^{b}(\pi(\mathfrak{u}))\in\rho\cup\rho^{-1}. Define a syllable β~b(π(𝔲))\tilde{\beta}^{b}(\pi(\mathfrak{u})) and a string 𝔯b(π(𝔲))\mathfrak{r}^{b}(\pi(\mathfrak{u})) by 𝔯b2(π(𝔲))𝔯b1(π(𝔲))=β~b(π(𝔲))𝔯b(π(𝔲))\mathfrak{r}^{b}_{2}(\pi(\mathfrak{u}))\mathfrak{r}^{b}_{1}(\pi(\mathfrak{u}))=\tilde{\beta}^{b}(\pi(\mathfrak{u}))\mathfrak{r}^{b}(\pi(\mathfrak{u})).

Remark 5.11.

Suppose 𝔲𝔣i(𝔵0)𝒰𝔣i(𝔵0)\mathfrak{u}\in\mathcal{R}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0})\cup\mathcal{U}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0}). If (𝔰(𝔲),𝔴~(𝔵0;𝔲)𝔠(𝔵0;π(𝔲)))(\mathfrak{s}(\mathfrak{u}),\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u})\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u}))) is double long then γ𝔣(π(𝔲))\gamma^{\mathfrak{f}}(\pi(\mathfrak{u})) is the first syllable of exactly one of 𝔯s1(π(𝔲))\mathfrak{r}^{s}_{1}(\pi(\mathfrak{u})) and 𝔯b1(π(𝔲))\mathfrak{r}^{b}_{1}(\pi(\mathfrak{u})), and hence |𝔯s1(π(𝔲))||𝔯b1(π(𝔲))|=0|\mathfrak{r}^{s}_{1}(\pi(\mathfrak{u}))||\mathfrak{r}^{b}_{1}(\pi(\mathfrak{u}))|=0 but |𝔯s1(π(𝔲))|+|𝔯b1(π(𝔲))|>0|\mathfrak{r}^{s}_{1}(\pi(\mathfrak{u}))|+|\mathfrak{r}^{b}_{1}(\pi(\mathfrak{u}))|>0.

In view of the above remark, if (𝔰(𝔲),𝔴~(𝔵0;𝔲)𝔠(𝔵0;π(𝔲)))(\mathfrak{s}(\mathfrak{u}),\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u})\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u}))) is a double long pair and |𝔯s1(π(𝔲))|>0|\mathfrak{r}^{s}_{1}(\pi(\mathfrak{u}))|>0 then we say it is (s,b)(s,b)-long, otherwise we say it is (b,s)(b,s)-long. Say that a (b,s)(b,s)-long pair is (b,s)(b,s)^{\partial}-long if it is bb^{\partial}-long. Similarly say that an (s,b)(s,b)-long pair is (s,b)(s,b)^{\partial}-long if it is ss^{\partial}-long.

Remark 5.12.

If 𝔲\mathfrak{u} is bb-long (resp. ss-long) then 𝔴~(𝔵0;𝔲)\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u}) is a left substring of 𝔯b1(π(𝔲))\mathfrak{r}^{b}_{1}(\pi(\mathfrak{u})) (resp. 𝔯s1(π(𝔲))\mathfrak{r}^{s}_{1}(\pi(\mathfrak{u}))). Moreover, if 𝔲\mathfrak{u} is double long then Remark 5.11 guarantees that |𝔴~(𝔵0;𝔲)|=0|\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u})|=0.

Remark 5.13.

Suppose 𝔲𝔣i(𝔵0)𝒰𝔣i(𝔵0)\mathfrak{u}\in\mathcal{H}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0})\cup\mathcal{U}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0}) and ξ𝔣(𝔲)\xi^{\mathfrak{f}}_{\partial}(\mathfrak{u})\neq\emptyset.

  • If (𝔱(𝔲),𝔠(𝔵0;𝔲))(\mathfrak{t}(\mathfrak{u}),\mathfrak{c}(\mathfrak{x}_{0};\mathfrak{u})) is bb^{\partial}-long then 𝔯b1(𝔲)\mathfrak{r}^{b}_{1}(\mathfrak{u}) is a proper left substring of 𝔴~(𝔵0;ξfmin(𝔲))\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\xi^{\mathrm{fmin}}_{\partial}(\mathfrak{u})) if and only if |𝔯b2(𝔲)|=1|\mathfrak{r}^{b}_{2}(\mathfrak{u})|=1. As a consequence 𝔯b1(𝔲)\mathfrak{r}^{b}_{1}(\mathfrak{u}) is a proper left substring of 𝔴(𝔵0;ξfmin(𝔲))\mathfrak{w}(\mathfrak{x}_{0};\xi^{\mathrm{fmin}}_{\partial}(\mathfrak{u})).

  • If (𝔱(𝔲),𝔠(𝔵0;𝔲))(\mathfrak{t}(\mathfrak{u}),\mathfrak{c}(\mathfrak{x}_{0};\mathfrak{u})) is ss^{\partial}-long and ξ𝔣(𝔲)\xi^{\mathfrak{f}}_{\partial}(\mathfrak{u})\neq\emptyset then 𝔯s1(𝔲)\mathfrak{r}^{s}_{1}(\mathfrak{u}) is a proper left substring of 𝔴(𝔵0;ξfmin(𝔲))\mathfrak{w}(\mathfrak{x}_{0};\xi^{\mathrm{fmin}}_{\partial}(\mathfrak{u})).

Example 5.14.

In the algebra Γ(ii)\Gamma^{(ii)} in Figure 7 the only bands are 𝔟1:=icBD\mathfrak{b}_{1}:=icBD, 𝔟2:=cfE\mathfrak{b}_{2}:=cfE and their inverses. Choosing 𝔵0:=1(𝔳1,1)\mathfrak{x}_{0}:=1_{(\mathfrak{v}_{1},1)} and 𝔲:=icBa\mathfrak{u}:=icBa, the LVP (𝔱(𝔲),𝔠(𝔵0;𝔲))=(𝔟1,a)(\mathfrak{t}(\mathfrak{u}),\mathfrak{c}(\mathfrak{x}_{0};\mathfrak{u}))=(\mathfrak{b}_{1},a) is b1b^{1}-long with 𝔯b1(𝔲)=B\mathfrak{r}^{b}_{1}(\mathfrak{u})=B and 𝔯b2(𝔲)=HG\mathfrak{r}^{b}_{2}(\mathfrak{u})=HG. Here GBD,cBDξ𝔣1(𝔲)GBD,cBD\in\xi^{\mathfrak{f}}_{1}(\mathfrak{u}) with ξfmin1(𝔲)=GBD\xi^{\mathrm{fmin}}_{1}(\mathfrak{u})=GBD with 𝔴~(𝔵0;ξfmin1(𝔲))=𝔯b1(𝔲)\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\xi^{\mathrm{fmin}}_{1}(\mathfrak{u}))=\mathfrak{r}^{b}_{1}(\mathfrak{u}).

v5{v_{5}}v2{v_{2}}v1{v_{1}}v4{v_{4}}v3{v_{3}}v6{v_{6}}v7{v_{7}}v8{v_{8}}d\scriptstyle{d}a\scriptstyle{a}i\scriptstyle{i}b\scriptstyle{b}c\scriptstyle{c}e\scriptstyle{e}f\scriptstyle{f}g\scriptstyle{g}h\scriptstyle{h}
Figure 7. Γ(ii)\Gamma^{(ii)} with ρ={da,bf,cg,ie,icf,dbgh}\rho=\{da,bf,cg,ie,icf,dbgh\}

If we consider the algebra obtained from the same quiver by replacing the relation dbghdbgh by dbgdbg in ρ\rho then keeping 𝔵0\mathfrak{x}_{0} and 𝔲\mathfrak{u} same as above we get 𝔯b1(𝔲)=B\mathfrak{r}^{b}_{1}(\mathfrak{u})=B and 𝔯b2(𝔲)=G\mathfrak{r}^{b}_{2}(\mathfrak{u})=G. Here cBD=ξfmin1(𝔲)cBD=\xi^{\mathrm{fmin}}_{1}(\mathfrak{u}) with 𝔴~(𝔵0;ξfmin1(𝔲))=cB\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\xi^{\mathrm{fmin}}_{1}(\mathfrak{u}))=cB.

6. Characterizing different values of υ\upsilon

Proposition 6.1.

Suppose 𝔲𝒱𝔣i(𝔵0)\mathfrak{u}\in\mathcal{V}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0}), h(𝔲)>2h(\mathfrak{u})>2 and π(𝔲)\pi(\mathfrak{u}) is abnormal. Then π(𝔲)c𝔠(𝔵0;π(𝔲))\pi(\mathfrak{u})^{c}\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u})) is a string if and only if 𝔟𝔣α(π(𝔲))=𝔟α(π(𝔲))\mathfrak{b}^{\mathfrak{f}\alpha}(\pi(\mathfrak{u}))=\mathfrak{b}^{\alpha}(\pi(\mathfrak{u})) if and only if υ(π(𝔲))=1\upsilon(\pi(\mathfrak{u}))=1.

Proof. Note that the only possible syllables α\alpha such that π(𝔲)cα\pi(\mathfrak{u})^{c}\alpha is a string are α(π(𝔲))\alpha(\pi(\mathfrak{u})) and the last syllable of 𝔟α(π(𝔲))\mathfrak{b}^{\alpha}(\pi(\mathfrak{u})), say γ\gamma^{\prime}. Moreover γ\gamma^{\prime} is a syllable of 𝔱(π(𝔲))\mathfrak{t}(\pi(\mathfrak{u})) but not of 𝔰(π(𝔲))\mathfrak{s}(\pi(\mathfrak{u})) while α(π(𝔲))\alpha(\pi(\mathfrak{u})) is a syllable of 𝔰(π(𝔲))\mathfrak{s}(\pi(\mathfrak{u})) but not of 𝔱(π(𝔲))\mathfrak{t}(\pi(\mathfrak{u})). Hence π(𝔲)c𝔠(𝔵0;π(𝔲))\pi(\mathfrak{u})^{c}\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u})) is a string if and only if α(π(𝔲))\alpha(\pi(\mathfrak{u})) is the last syllable of 𝔠(𝔵0;π(𝔲))\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u})). Since 𝔠(𝔵0;π2(𝔲))\mathfrak{c}(\mathfrak{x}_{0};\pi^{2}(\mathfrak{u})) is a proper left substring of 𝔥(𝔵0;π(𝔲))\mathfrak{h}(\mathfrak{x}_{0};\pi(\mathfrak{u})) by Corollary 4.4, if the latter holds then υ(π(𝔲))=1\upsilon(\pi(\mathfrak{u}))=1.

On the other hand if υ(π(𝔲))=1\upsilon(\pi(\mathfrak{u}))=1 then the last syllable of 𝔠(𝔵0;π(𝔲))\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u})) is a syllable of 𝔰(π(𝔲))\mathfrak{s}(\pi(\mathfrak{u})) but not of 𝔱(π(𝔲))\mathfrak{t}(\pi(\mathfrak{u})). Hence for any cyclic permutation 𝔟\mathfrak{b}^{\prime} of 𝔱(π(𝔲))\mathfrak{t}(\pi(\mathfrak{u})) if 𝔟𝔠(𝔵0;π(𝔲))\mathfrak{b}^{\prime}\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u})) is a string then 𝔟=𝔟α(π(𝔲))\mathfrak{b}^{\prime}=\mathfrak{b}^{\alpha}(\pi(\mathfrak{u})). Hence π(𝔲)c𝔠(𝔵0;π(𝔲))\pi(\mathfrak{u})^{c}\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u})) is a string.    \Box

Below we note some immediate consequences.

Corollary 6.2.

Suppose 𝔲𝒱𝔣i(𝔵0)\mathfrak{u}\in\mathcal{V}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0}), h(𝔲)>2h(\mathfrak{u})>2, π(𝔲)\pi(\mathfrak{u}) is abnormal and υ(π(𝔲))=1\upsilon(\pi(\mathfrak{u}))=1. If 𝔲\mathfrak{u} is ss-long then |𝔴~(𝔵0;𝔲)|>0|\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u})|>0 if and only if |π(𝔲)c|=0|\pi(\mathfrak{u})^{c}|=0.

Corollary 6.3.

Suppose 𝔲𝒱𝔣i(𝔵0)\mathfrak{u}\in\mathcal{V}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0}), h(𝔲)>2h(\mathfrak{u})>2, π(𝔲)\pi(\mathfrak{u}) is abnormal and υ(π(𝔲))=1\upsilon(\pi(\mathfrak{u}))=1. Then 𝔲\mathfrak{u} is bb-long if and only if π(𝔲)c=𝔴𝔴~(𝔵0;𝔲)\pi(\mathfrak{u})^{c}=\mathfrak{w}\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u}) for some string 𝔴\mathfrak{w}. If these equivalent conditions hold then

  • 𝔯b1(π(𝔲))=π(𝔲)c\mathfrak{r}^{b}_{1}(\pi(\mathfrak{u}))=\pi(\mathfrak{u})^{c}.

  • 𝔲\mathfrak{u} is bθ(β(𝔲))b^{-\theta(\beta(\mathfrak{u}))}-long.

Corollary 6.4.

Suppose 𝔲𝒱𝔣i(𝔵0)\mathfrak{u}\in\mathcal{V}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0}), h(𝔲)>2h(\mathfrak{u})>2 and υ(π(𝔲))=1\upsilon(\pi(\mathfrak{u}))=1 with 𝔠(𝔵0;π(𝔲))=𝔷𝔠(𝔵0;π2(𝔲))\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u}))=\mathfrak{z}\mathfrak{c}(\mathfrak{x}_{0};\pi^{2}(\mathfrak{u})). If 𝔲\mathfrak{u} is long and 𝔲1\mathfrak{u}_{1} is an uncle of 𝔲\mathfrak{u} with β(𝔲1)=β(𝔲)\beta(\mathfrak{u}_{1})=\beta(\mathfrak{u}) then 𝔲1\mathfrak{u}_{1} could be either bb^{\partial}-long or ss^{\partial}-long but not both. Further if π(𝔲)𝔣𝔲1\pi(\mathfrak{u})\sqsubset^{\mathfrak{f}}_{\partial}\mathfrak{u}_{1} then |π(𝔲)c|>0|\pi(\mathfrak{u})^{c}|>0, |𝔴~(𝔵0;𝔲)|=0|\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u})|=0, 𝔲\mathfrak{u} is bb^{-\partial} long and exactly one of the following happens:

  • 𝔲1\mathfrak{u}_{1} is bb^{\partial}-long and 𝔯b1(π2(𝔲))=𝔷\mathfrak{r}^{b}_{1}(\pi^{2}(\mathfrak{u}))=\mathfrak{z};

  • 𝔲1\mathfrak{u}_{1} is ss^{\partial}-long and 𝔯s1(π2(𝔲))=𝔷\mathfrak{r}^{s}_{1}(\pi^{2}(\mathfrak{u}))=\mathfrak{z}.

Example 6.5.

In the algebra Γ(ii)\Gamma^{(ii)} in Figure 7 the only bands are 𝔟1:=icBD\mathfrak{b}_{1}:=icBD, 𝔟2:=cfE\mathfrak{b}_{2}:=cfE and their inverses. Choosing 𝔵0:=1(𝔳1,1)\mathfrak{x}_{0}:=1_{(\mathfrak{v}_{1},1)} and 𝔲:=HGfE\mathfrak{u}:=HGfE we have π(𝔲)=cBD\pi(\mathfrak{u})=cBD and π2(𝔲)=icBa\pi^{2}(\mathfrak{u})=icBa with υ(𝔲)=1\upsilon(\mathfrak{u})=1. Here 𝔲\mathfrak{u} is b1b^{-1}-long and its uncle 𝔲1:=GBD\mathfrak{u}_{1}:=GBD is b1b^{1}-long.

Corollary 6.6.

Suppose 𝔲𝒱𝔣i(𝔵0)\mathfrak{u}\in\mathcal{V}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0}), h(𝔲)>2h(\mathfrak{u})>2 and υ(π(𝔲))=1\upsilon(\pi(\mathfrak{u}))=1. Further suppose that π(𝔲)\pi(\mathfrak{u}) is abnormal with |π(𝔲)c|>0|\pi(\mathfrak{u})^{c}|>0.

  • If |𝔴~(𝔵0;π(𝔲))|>0|\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\pi(\mathfrak{u}))|>0 then Proposition 6.1 guarantees that π(𝔲)cα(π(𝔲))\pi(\mathfrak{u})^{c}\alpha(\pi(\mathfrak{u})) is a right substring of 𝔴~(𝔵0;π(𝔲))\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\pi(\mathfrak{u})). Since δ(π(𝔲)cα(π(𝔲)))=0\delta(\pi(\mathfrak{u})^{c}\alpha(\pi(\mathfrak{u})))=0 we get π(𝔲)\pi(\mathfrak{u}) is short.

  • If |𝔴~(𝔵0;π(𝔲))|=0|\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\pi(\mathfrak{u}))|=0 then π(𝔲)\pi(\mathfrak{u}) is ss-long for otherwise we get υ(π(𝔲))=0\upsilon(\pi(\mathfrak{u}))=0, a contradiction. But here 𝔴(𝔵0;π(𝔲))=𝔟𝔣α(π2(𝔲))\mathfrak{w}(\mathfrak{x}_{0};\pi(\mathfrak{u}))=\mathfrak{b}^{\mathfrak{f}\alpha}{(\pi^{2}(\mathfrak{u}))}.

Proposition 6.7.

Suppose 𝔲𝒱𝔣i(𝔵0)\mathfrak{u}\in\mathcal{V}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0}), h(𝔲)>2h(\mathfrak{u})>2 and υ(π(𝔲))=1\upsilon(\pi(\mathfrak{u}))=1. If π(𝔲)\pi(\mathfrak{u}) is abnormal with |π(𝔲)c|=0|\pi(\mathfrak{u})^{c}|=0 then 𝔲\mathfrak{u} is not double long.

Proof. Since |π(𝔲)c|=0|\pi(\mathfrak{u})^{c}|=0, Corollary 6.3 ensures that |𝔯b1(π(𝔲))|=0|\mathfrak{r}^{b}_{1}(\pi(\mathfrak{u}))|=0.

If 𝔲\mathfrak{u} is bb-long then β(𝔲)\beta(\mathfrak{u}) is the first syllable of 𝔟β(π(𝔲))\mathfrak{b}^{\beta}(\pi(\mathfrak{u})). In this case β(π(𝔲))α(π(𝔲))\beta(\pi(\mathfrak{u}))\alpha(\pi(\mathfrak{u})) and β(𝔲)α(π(𝔲))\beta(\mathfrak{u})\alpha(\pi(\mathfrak{u})) are strings, and θ(α(π(𝔲)))=θ(β(π(𝔲)))\theta(\alpha(\pi(\mathfrak{u})))=\theta(\beta(\pi(\mathfrak{u}))). Thus θ(β(𝔲))=θ(α(π(𝔲)))\theta(\beta(\mathfrak{u}))=-\theta(\alpha(\pi(\mathfrak{u}))) and hence 𝔲\mathfrak{u} is not ss-long.

If 𝔲\mathfrak{u} is ss-long then from the above discussion it is clear that |𝔴~(𝔵0;𝔲)|>0|\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u})|>0 and β(π(𝔲))\beta(\pi(\mathfrak{u})) is the first syllable of 𝔴~(𝔵0;𝔲)\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u}). But since θ(β(π(𝔲)))=θ(γ¯¯(π(𝔲)))\theta(\beta(\pi(\mathfrak{u})))=-\theta(\bar{\bar{\gamma}}(\pi(\mathfrak{u}))), 𝔲\mathfrak{u} is not bb-long. Hence the result.    \Box

Proposition 6.8.

Suppose 𝔲𝒱𝔣i(𝔵0)\mathfrak{u}\in\mathcal{V}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0}), h(𝔲)>2h(\mathfrak{u})>2, π(𝔲)\pi(\mathfrak{u}) is abnormal with |π(𝔲)c|>0|\pi(\mathfrak{u})^{c}|>0 and υ(π(𝔲))=1\upsilon(\pi(\mathfrak{u}))=1. If 𝔠(𝔵0;π(𝔲))=𝔷𝔠(𝔵0;π2(𝔲))\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u}))=\mathfrak{z}\mathfrak{c}(\mathfrak{x}_{0};\pi^{2}(\mathfrak{u})) then π(𝔲)c𝔷\pi(\mathfrak{u})^{c}\mathfrak{z} is a left substring of 𝔟𝔣α(π2(𝔲))\mathfrak{b}^{\mathfrak{f}\alpha}(\pi^{2}(\mathfrak{u})).

Proof. Since π(𝔲)\pi(\mathfrak{u}) is abnormal, 𝔷\mathfrak{z} is a left substring of (𝔟𝔣α(π2(𝔲)))2(\mathfrak{b}^{\mathfrak{f}\alpha}(\pi^{2}(\mathfrak{u})))^{2}. If |π(𝔲)c|>0|\pi(\mathfrak{u})^{c}|>0 then, in view of Remark 3.1, (𝔰(π(𝔲)),𝔴~(𝔵0;π(𝔲))𝔠(𝔵0;π2(𝔲)))(\mathfrak{s}(\pi(\mathfrak{u})),\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\pi(\mathfrak{u}))\mathfrak{c}(\mathfrak{x}_{0};\pi^{2}(\mathfrak{u}))) is short, and hence 𝔷\mathfrak{z} is a proper left substring of 𝔟𝔣α(π2(𝔲))\mathfrak{b}^{\mathfrak{f}\alpha}(\pi^{2}(\mathfrak{u})). Since |𝔷|>0|\mathfrak{z}|>0, π(𝔲)c𝔷\pi(\mathfrak{u})^{c}\mathfrak{z} a string. Hence π(𝔲)c𝔷\pi(\mathfrak{u})^{c}\mathfrak{z} and 𝔟𝔣α(π(𝔲))\mathfrak{b}^{\mathfrak{f}\alpha}(\pi(\mathfrak{u})) are comparable strings.

If π(𝔲)c𝔷\pi(\mathfrak{u})^{c}\mathfrak{z} is a not left substring of 𝔟𝔣α(π2(𝔲))\mathfrak{b}^{\mathfrak{f}\alpha}(\pi^{2}(\mathfrak{u})) then 𝔟𝔣α(π2(𝔲))\mathfrak{b}^{\mathfrak{f}\alpha}(\pi^{2}(\mathfrak{u})) is a proper substring of π(𝔲)c𝔷\pi(\mathfrak{u})^{c}\mathfrak{z} and hence γ𝔣(π2(𝔲))\gamma^{\mathfrak{f}}(\pi^{2}(\mathfrak{u})) is a syllable of π(𝔲)c\pi(\mathfrak{u})^{c}. Since 𝔷\mathfrak{z} is a proper left substring of 𝔟𝔣α(π2(𝔲))\mathfrak{b}^{\mathfrak{f}\alpha}(\pi^{2}(\mathfrak{u})), γ𝔣(π2(𝔲))γ¯\gamma^{\mathfrak{f}}(\pi^{2}(\mathfrak{u}))\bar{\gamma} is a substring of π(𝔲)c\pi(\mathfrak{u})^{c}, where γ¯\bar{\gamma} is the last syllable of 𝔟𝔣α(π2(𝔲))\mathfrak{b}^{\mathfrak{f}\alpha}(\pi^{2}(\mathfrak{u})). But γ𝔣(π2(𝔲))\gamma^{\mathfrak{f}}(\pi^{2}(\mathfrak{u})) is the first syllable of 𝔷\mathfrak{z} and hence a common syllable of 𝔷\mathfrak{z} and π(𝔲)c\pi(\mathfrak{u})^{c}, a contradiction to |𝔷|>0|\mathfrak{z}|>0. Therefore π(𝔲)c𝔷\pi(\mathfrak{u})^{c}\mathfrak{z} is a left substring of 𝔟𝔣α(π2(𝔲))\mathfrak{b}^{\mathfrak{f}\alpha}(\pi^{2}(\mathfrak{u})).    \Box

Example 6.9.

In the algebra Γ(i)\Gamma^{(i)} in Figure 6, choosing 𝔵0:=1(𝔳1,i)\mathfrak{x}_{0}:=1_{(\mathfrak{v}_{1},i)} and 𝔲:=Gb\mathfrak{u}:=Gb we have π(𝔲):=dC\pi(\mathfrak{u}):=dC and π2(𝔲):=hbA\pi^{2}(\mathfrak{u}):=hbA. Here 𝔠(𝔵0;π2(𝔲))=A\mathfrak{c}(\mathfrak{x}_{0};\pi^{2}(\mathfrak{u}))=A, 𝔠(𝔵0;π(𝔲))=ChbA\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u}))=ChbA and hence the string 𝔷:=Chb\mathfrak{z}:=Chb satisfies 𝔟𝔣α(π2(𝔲))=π(𝔲)c𝔷\mathfrak{b}^{\mathfrak{f}\alpha}(\pi^{2}(\mathfrak{u}))=\pi(\mathfrak{u})^{c}\mathfrak{z}.

Remark 6.10.

Suppose 𝔲𝒱𝔣i(𝔵0)\mathfrak{u}\in\mathcal{V}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0}), h(𝔲)>2h(\mathfrak{u})>2 and υ(π(𝔲))<1\upsilon(\pi(\mathfrak{u}))<1 then since 𝔥(𝔵0;𝒫(π(𝔲)))\mathfrak{h}(\mathfrak{x}_{0};\mathcal{P}(\pi(\mathfrak{u}))) is not a substring of 𝔠(𝔵0;𝔲)\mathfrak{c}(\mathfrak{x}_{0};\mathfrak{u}), Proposition 4.7 gives that N(𝔵0;𝔰(π(𝔲)),𝔠(𝔵0;π(𝔲)))=0N(\mathfrak{x}_{0};\mathfrak{s}(\pi(\mathfrak{u})),\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u})))=0.

Proposition 6.11.

Suppose 𝔲𝒱𝔣i(𝔵0)\mathfrak{u}\in\mathcal{V}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0}) and h(𝔲)>2h(\mathfrak{u})>2. Then υ(π(𝔲))=1\upsilon(\pi(\mathfrak{u}))=-1 if and only if π(𝔲)\pi(\mathfrak{u}) is abnormal, π(𝔲)c𝔠(𝔵0;π2(𝔲))\pi(\mathfrak{u})^{c}\mathfrak{c}(\mathfrak{x}_{0};\pi^{2}(\mathfrak{u})) is a string but π(𝔲)c𝔠(𝔵0;π(𝔲))\pi(\mathfrak{u})^{c}\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u})) is not a string.

Proof. Suppose 𝔠(𝔵0;π2(𝔲))=𝔷𝔠(𝔵0;π(𝔲))\mathfrak{c}(\mathfrak{x}_{0};\pi^{2}(\mathfrak{u}))=\mathfrak{z}\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u})) for some string 𝔷\mathfrak{z} of positive length. Since both 𝔠(𝔵0;π2(𝔲))\mathfrak{c}(\mathfrak{x}_{0};\pi^{2}(\mathfrak{u})) and 𝔠(𝔵0;π(𝔲))\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u})) are substrings of 𝔥(𝔵0;𝒫(π(𝔲)))\mathfrak{h}(\mathfrak{x}_{0};\mathcal{P}(\pi(\mathfrak{u}))) and 𝔥(𝔵0;𝒫(π(𝔲)))\mathfrak{h}(\mathfrak{x}_{0};\mathcal{P}(\pi(\mathfrak{u}))) is a proper substring of 𝔟𝔣α(π(𝔲))𝔠(𝔵0;π(𝔲))\mathfrak{b}^{\mathfrak{f}\alpha}(\pi(\mathfrak{u}))\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u})), 𝔷\mathfrak{z} is a left substring of 𝔟𝔣α(π(𝔲))\mathfrak{b}^{\mathfrak{f}\alpha}(\pi(\mathfrak{u})) and hence π(𝔲)\pi(\mathfrak{u}) is abnormal.

Since 𝔠(𝔵0;π(𝔲))\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u})) is a proper left substring of 𝔠(𝔵0;π2(𝔲))\mathfrak{c}(\mathfrak{x}_{0};\pi^{2}(\mathfrak{u})) the last syllable of 𝔠(𝔵0;π2(𝔲))\mathfrak{c}(\mathfrak{x}_{0};\pi^{2}(\mathfrak{u})) is a syllable of 𝔱(π(𝔲))\mathfrak{t}(\pi(\mathfrak{u})) but not of 𝔰(π(𝔲))\mathfrak{s}(\pi(\mathfrak{u})). Hence for any cyclic permutation 𝔟\mathfrak{b}^{\prime} of 𝔰(π(𝔲))\mathfrak{s}(\pi(\mathfrak{u})) if 𝔟𝔠(𝔵0;π2(𝔲))\mathfrak{b}^{\prime}\mathfrak{c}(\mathfrak{x}_{0};\pi^{2}(\mathfrak{u})) is a string then 𝔟=𝔟α(π(𝔲))\mathfrak{b}^{\prime}=\mathfrak{b}_{\alpha}(\pi(\mathfrak{u})). Hence π(𝔲)c𝔠(𝔵0;π2(𝔲))\pi(\mathfrak{u})^{c}\mathfrak{c}(\mathfrak{x}_{0};\pi^{2}(\mathfrak{u})) is a string. Moreover Proposition 6.1 gives that π(𝔲)c𝔠(𝔵0;π(𝔲))\pi(\mathfrak{u})^{c}\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u})) is not a string.

Now π(𝔲)c𝔠(𝔵0;π2(𝔲))\pi(\mathfrak{u})^{c}\mathfrak{c}(\mathfrak{x}_{0};\pi^{2}(\mathfrak{u})) is a string if and only if γ(π(𝔲))\gamma^{\prime}(\pi(\mathfrak{u})) is the last syllable of 𝔠(𝔵0;π2(𝔲))\mathfrak{c}(\mathfrak{x}_{0};\pi^{2}(\mathfrak{u})) where γ(π(𝔲))\gamma^{\prime}(\pi(\mathfrak{u})) is the last syllable of 𝔟α(π(𝔲))\mathfrak{b}^{\alpha}(\pi(\mathfrak{u})).

For the reverse implication clearly υ(π(𝔲))0\upsilon(\pi(\mathfrak{u}))\neq 0. Moreover υ(π(𝔲))1\upsilon(\pi(\mathfrak{u}))\neq 1 by Proposition 6.1, since π(𝔲)c𝔠(𝔵0;π(𝔲))\pi(\mathfrak{u})^{c}\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u})) is not a string. Therefore υ(π(𝔲))=1\upsilon(\pi(\mathfrak{u}))=-1.    \Box

Corollary 6.12.

Suppose 𝔲𝒱𝔣i(𝔵0)\mathfrak{u}\in\mathcal{V}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0}) and h(𝔲)>2h(\mathfrak{u})>2. If 𝔠(𝔵0;π2(𝔲))=𝔷𝔠(𝔵0;π(𝔲))\mathfrak{c}(\mathfrak{x}_{0};\pi^{2}(\mathfrak{u}))=\mathfrak{z}\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u})) where |𝔷|>0|\mathfrak{z}|>0. Then the following hold:

  1. (1)

    𝔟𝔣α(π2(𝔲))=𝔟α(π(𝔲))\mathfrak{b}^{\mathfrak{f}\alpha}(\pi^{2}(\mathfrak{u}))=\mathfrak{b}_{\alpha}(\pi(\mathfrak{u}));

  2. (2)

    (𝔱(π(𝔲)),𝔠(𝔵0,π(𝔲)))(\mathfrak{t}(\pi(\mathfrak{u})),\mathfrak{c}(\mathfrak{x}_{0},\pi(\mathfrak{u}))) is bb-long with 𝔯b1(π(𝔲))=π(𝔲)c𝔷\mathfrak{r}^{b}_{1}(\pi(\mathfrak{u}))=\pi(\mathfrak{u})^{c}\mathfrak{z};

  3. (3)

    (𝔱(π(𝔲)),𝔯b1(π(𝔲))𝔠(𝔵0,π(𝔲)))(\mathfrak{t}(\pi(\mathfrak{u})),\mathfrak{r}^{b}_{1}(\pi(\mathfrak{u}))\mathfrak{c}(\mathfrak{x}_{0},\pi(\mathfrak{u}))) is bb-long but not ss-long;

  4. (4)

    𝔲\mathfrak{u} is long iff 𝔴~(𝔵0;𝔲)\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u}) is a left substring of 𝔯b1(π(𝔲))\mathfrak{r}^{b}_{1}(\pi(\mathfrak{u}));

Proof. Proposition 6.11 gives that π(𝔲)\pi(\mathfrak{u}) is abnormal and π(𝔲)c𝔠(𝔵0;π2(𝔲))\pi(\mathfrak{u})^{c}\mathfrak{c}(\mathfrak{x}_{0};\pi^{2}(\mathfrak{u})) is a string. and hence (1) follows immediately. Moreover 𝔷\mathfrak{z} is a right substring of 𝔟α(π(𝔲))\mathfrak{b}^{\alpha}(\pi(\mathfrak{u})). Since 𝔟𝔣α(π2(𝔲))𝔠(𝔵0;π2(𝔲))\mathfrak{b}^{\mathfrak{f}\alpha}(\pi^{2}(\mathfrak{u}))\mathfrak{c}(\mathfrak{x}_{0};\pi^{2}(\mathfrak{u})) is a string, 𝔷\mathfrak{z} is a proper left substring of π(𝔲)β\pi(\mathfrak{u})_{\beta}. Therefore (𝔱(π(𝔲)),𝔠(𝔵0,π(𝔲)))(\mathfrak{t}(\pi(\mathfrak{u})),\mathfrak{c}(\mathfrak{x}_{0},\pi(\mathfrak{u}))) is bb-long. By definition of 𝔯b1(π(𝔲))\mathfrak{r}^{b}_{1}(\pi(\mathfrak{u})), we clearly get 𝔯b1(π(𝔲))=π(𝔲)c𝔷\mathfrak{r}^{b}_{1}(\pi(\mathfrak{u}))=\pi(\mathfrak{u})^{c}\mathfrak{z} thus proving (2).

Since 𝔷\mathfrak{z} is a proper left substring of π(𝔲)β\pi(\mathfrak{u})_{\beta}, (𝔱(π(𝔲)),𝔯b1(π(𝔲))𝔠(𝔵0,π(𝔲)))(\mathfrak{t}(\pi(\mathfrak{u})),\mathfrak{r}^{b}_{1}(\pi(\mathfrak{u}))\mathfrak{c}(\mathfrak{x}_{0},\pi(\mathfrak{u}))) is bb-long but not ss-long thus proving (3). Moreover if 𝔴~(𝔵0;𝔲)\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u}) is a left substring of 𝔯b1(π(𝔲))\mathfrak{r}^{b}_{1}(\pi(\mathfrak{u})) then 𝔲\mathfrak{u} is long. The converse part of (4) is by Remark 5.12.    \Box

Corollary 6.13.

Suppose 𝔲𝒱𝔣i(𝔵0)\mathfrak{u}\in\mathcal{V}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0}), h(𝔲)>2h(\mathfrak{u})>2 and υ(π(𝔲))=1\upsilon(\pi(\mathfrak{u}))=-1. If 𝔲\mathfrak{u} is long then it is bb^{-\partial}-long. Moreover if 𝔠(𝔵0;π2(𝔲))\mathfrak{c}(\mathfrak{x}_{0};\pi^{2}(\mathfrak{u})) is a left substring of 𝔴~(𝔵0;𝔲)𝔠(𝔵0;π(𝔲))\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u})\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u})) and there is an uncle 𝔲1\mathfrak{u}_{1} of 𝔲\mathfrak{u} with β(𝔲1)=β(𝔲)\beta(\mathfrak{u}_{1})=\beta(\mathfrak{u}) such that π(𝔲)𝔣𝔲1\pi(\mathfrak{u})\sqsubset^{\mathfrak{f}}_{\partial}\mathfrak{u}_{1} then

  • |π(𝔲)c|>0|\pi(\mathfrak{u})^{c}|>0 and 𝔴~(𝔵0;𝔲)=𝔷\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u})=\mathfrak{z};

  • 𝔲1\mathfrak{u}_{1} is bb^{\partial}-long and |𝔯b1(π2(𝔲))|=0|\mathfrak{r}^{b}_{1}(\pi^{2}(\mathfrak{u}))|=0.

Example 6.14.

In the algebra Γ(iii)\Gamma^{(iii)} in Figure 8 the only bands are 𝔟1:=icD\mathfrak{b}_{1}:=icD, 𝔟2:=cahG\mathfrak{b}_{2}:=cahG and their inverses.

v5{v_{5}}v6{v_{6}}v7{v_{7}}v4{v_{4}}v3{v_{3}}v9{v_{9}}v8{v_{8}}v2{v_{2}}v1{v_{1}}e\scriptstyle{e}f\scriptstyle{f}i\scriptstyle{i}c\scriptstyle{c}d\scriptstyle{d}j\scriptstyle{j}g\scriptstyle{g}h\scriptstyle{h}b\scriptstyle{b}a\scriptstyle{a}
Figure 8. Γ(iii)\Gamma^{(iii)} with ρ={bh,aj,da,ce,ig,icah,def}\rho=\{bh,aj,da,ce,ig,icah,def\}

Choosing 𝔵0:=1(𝔳1,i)\mathfrak{x}_{0}:=1_{(\mathfrak{v}_{1},i)} and 𝔲:=FEahG\mathfrak{u}:=FEahG we have π(𝔲)=cD\pi(\mathfrak{u})=cD and π2(𝔲)=icaB\pi^{2}(\mathfrak{u})=icaB with υ(𝔲)=1\upsilon(\mathfrak{u})=-1. Here 𝔲\mathfrak{u} is b1b^{-1}-long and its uncle 𝔲1:=ED\mathfrak{u}_{1}:=ED is b1b^{1}-long.

Proposition 6.15.

Suppose 𝔲𝒱𝔣i(𝔵0)\mathfrak{u}\in\mathcal{V}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0}) and h(𝔲)>2h(\mathfrak{u})>2. Then υ(π(𝔲))=0\upsilon(\pi(\mathfrak{u}))=0 if and only if π(𝔲)\pi(\mathfrak{u}) is abnormal with |π(𝔲)c|>0|\pi(\mathfrak{u})^{c}|>0 and there is a partition π(𝔲)c=𝔵2𝔵1\pi(\mathfrak{u})^{c}=\mathfrak{x}_{2}\mathfrak{x}_{1} with |𝔵1|>0|\mathfrak{x}_{1}|>0 and 𝔵1\mathfrak{x}_{1} is a right substring of 𝔟𝔣α(π(𝔲))\mathfrak{b}^{\mathfrak{f}\alpha}(\pi(\mathfrak{u})).

Proof. Since 𝔠(𝔵0;π(𝔲))=𝔠(𝔵0;π2(𝔲))\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u}))=\mathfrak{c}(\mathfrak{x}_{0};\pi^{2}(\mathfrak{u})), 𝔟𝔣α(π2(𝔲))𝔠(𝔵0;π(𝔲))\mathfrak{b}^{\mathfrak{f}\alpha}(\pi^{2}(\mathfrak{u}))\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u})) and 𝔟𝔣α(π(𝔲))𝔠(𝔵0;π(𝔲))\mathfrak{b}^{\mathfrak{f}\alpha}(\pi(\mathfrak{u}))\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u})) are strings. Let α,γ,γ\alpha,\gamma^{\prime},\gamma^{\prime\prime} denote the last syllables of 𝔠(𝔵0;π(𝔲)),𝔟𝔣α(π(𝔲)),𝔟𝔣α(π2(𝔲))\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u})),\mathfrak{b}^{\mathfrak{f}\alpha}(\pi(\mathfrak{u})),\mathfrak{b}^{\mathfrak{f}\alpha}(\pi^{2}(\mathfrak{u})) respectively. Then α{γ,γ,γ𝔣(π(𝔲)),γ𝔣(π2(𝔲))}\alpha\notin\{\gamma^{\prime},\gamma^{\prime\prime},\gamma^{\mathfrak{f}}(\pi(\mathfrak{u})),\gamma^{\mathfrak{f}}(\pi^{2}(\mathfrak{u}))\}.

If γ𝔣(π(𝔲))γ𝔣(π2(𝔲))\gamma^{\mathfrak{f}}(\pi(\mathfrak{u}))\neq\gamma^{\mathfrak{f}}(\pi^{2}(\mathfrak{u})) then by the definition of a string algebra we must have γ=γ\gamma^{\prime}=\gamma^{\prime\prime}. On the other hand if γ𝔣(π(𝔲))=γ𝔣(π2(𝔲))\gamma^{\mathfrak{f}}(\pi(\mathfrak{u}))=\gamma^{\mathfrak{f}}(\pi^{2}(\mathfrak{u})), then since γ𝔣(π(𝔲))γ,γ𝔣(π2(𝔲))γ\gamma^{\mathfrak{f}}(\pi(\mathfrak{u}))\gamma^{\prime},\gamma^{\mathfrak{f}}(\pi^{2}(\mathfrak{u}))\gamma^{\prime\prime} are strings, we can again conclude γ=γ\gamma^{\prime}=\gamma^{\prime\prime}. Therefore π(𝔲)\pi(\mathfrak{u}) is abnormal and there exists a partition π(𝔲)c=𝔵2𝔵1\pi(\mathfrak{u})^{c}=\mathfrak{x}_{2}\mathfrak{x}_{1} such that |𝔵1|>0|\mathfrak{x}_{1}|>0.

For the other direction observe that since |𝔵1|>0|\mathfrak{x}_{1}|>0, neither π(𝔲)c𝔠(𝔵0;π(𝔲))\pi(\mathfrak{u})^{c}\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u})) nor π(𝔲)c𝔠(𝔵0;π2(𝔲))\pi(\mathfrak{u})^{c}\mathfrak{c}(\mathfrak{x}_{0};\pi^{2}(\mathfrak{u})) is a string. Recall from Corollary 4.4 that 𝔠(𝔵0;π2(𝔲))\mathfrak{c}(\mathfrak{x}_{0};\pi^{2}(\mathfrak{u})) and 𝔠(𝔵0;π(𝔲))\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u})) are comparable. Hence Propositions 6.1 and 6.11 yield the necessary result.    \Box

Proposition 6.16.

If 𝔲𝒱𝔣i(𝔵0)\mathfrak{u}\in\mathcal{V}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0}), h(𝔲)>2h(\mathfrak{u})>2 and υ(π(𝔲))<1\upsilon(\pi(\mathfrak{u}))<1 then π(𝔲)\pi(\mathfrak{u}) is short.

Proof. We only prove the result for υ(π(𝔲))=0\upsilon(\pi(\mathfrak{u}))=0 and the proof for the another case is similar.

For a contradiction we assume that π(𝔲)\pi(\mathfrak{u}) is long. Since υ(π(𝔲))=0\upsilon(\pi(\mathfrak{u}))=0, Proposition 6.15 guarantees that there is a partition π(𝔲)c=𝔵2𝔵1\pi(\mathfrak{u})^{c}=\mathfrak{x}_{2}\mathfrak{x}_{1} with |𝔵1|>0|\mathfrak{x}_{1}|>0 and 𝔵1\mathfrak{x}_{1} is a right substring of 𝔟𝔣α(π(𝔲))\mathfrak{b}^{\mathfrak{f}\alpha}(\pi(\mathfrak{u})). Here β(π(𝔲))𝔵2\beta(\pi(\mathfrak{u}))\mathfrak{x}_{2} is a string. Corollary 5.8 gives that the LVP (𝔟,𝔶):=(𝔰(π(𝔲)),𝔴~(𝔵0;π(𝔲))𝔠(𝔵0;π2(𝔲)))(\mathfrak{b},\mathfrak{y}):=(\mathfrak{s}(\pi(\mathfrak{u})),\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\pi(\mathfrak{u}))\mathfrak{c}(\mathfrak{x}_{0};\pi^{2}(\mathfrak{u}))) is long and hence 𝔴(𝔵0;π(𝔲))=𝔵2𝔟𝔣α(π2(𝔲))\mathfrak{w}(\mathfrak{x}_{0};\pi(\mathfrak{u}))=\mathfrak{x}_{2}\mathfrak{b}^{\mathfrak{f}\alpha}{(\pi^{2}(\mathfrak{u}))}. Moreover Corollary 4.8 states that 𝔴(𝔵0;π(𝔲))𝔠(𝔵0;π2(𝔲))\mathfrak{w}(\mathfrak{x}_{0};\pi(\mathfrak{u}))\mathfrak{c}(\mathfrak{x}_{0};\pi^{2}(\mathfrak{u})) is a left substring of 𝔥(𝔵0;𝒫(π(𝔲)))\mathfrak{h}(\mathfrak{x}_{0};\mathcal{P}(\pi(\mathfrak{u}))). Now the maximal left substring 𝔶\mathfrak{y} of 𝔟𝔣α(π2(𝔲))\mathfrak{b}^{\mathfrak{f}\alpha}{(\pi^{2}(\mathfrak{u}))} whose last syllable is α(π(𝔲))\alpha(\pi(\mathfrak{u})) satisfies that 𝔶𝔠(𝔵0;π2(𝔲)))\mathfrak{y}\mathfrak{c}(\mathfrak{x}_{0};\pi^{2}(\mathfrak{u}))) is a left substring of 𝔠(𝔵0;π(𝔲)))\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u}))), a contradiction to υ(π(𝔲))=0\upsilon(\pi(\mathfrak{u}))=0. This completes the proof.    \Box

Corollary 6.17.

Suppose 𝔲𝒱𝔣i(𝔵0)\mathfrak{u}\in\mathcal{V}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0}), h(𝔲)>2h(\mathfrak{u})>2 and υ(π(𝔲))=0\upsilon(\pi(\mathfrak{u}))=0. If 𝔲\mathfrak{u} is long and there exist an uncle 𝔲1\mathfrak{u}_{1} of 𝔲\mathfrak{u} with β(𝔲1)=β(𝔲)\beta(\mathfrak{u}_{1})=\beta(\mathfrak{u}) then the following hold:

  • 𝔲\mathfrak{u} is bb^{-\partial}-long;

  • if |𝔴~(𝔵0;𝔲)|>0|\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u})|>0 then 𝔲1𝔣π(𝔲)\mathfrak{u}_{1}\sqsubset^{\mathfrak{f}}_{\partial}\pi(\mathfrak{u});

  • 𝔲\mathfrak{u} is double (b,s)(b,s)^{-\partial}-long only if |𝔴~(𝔵0;𝔲)|=0|\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u})|=0. In this case 𝔲1\mathfrak{u}_{1} is necessarily ss^{\partial}-long, |𝔯s1(π2(𝔲))|=0|\mathfrak{r}^{s}_{1}(\pi^{2}(\mathfrak{u}))|=0, |𝔯b1(π(𝔲))|>0|\mathfrak{r}^{b}_{1}(\pi(\mathfrak{u}))|>0 and π(𝔲)𝔣𝔲1\pi(\mathfrak{u})\sqsubset^{\mathfrak{f}}_{\partial}\mathfrak{u}_{1}.

Remark 6.18.

Suppose 𝔲𝒱𝔣i(𝔵0)\mathfrak{u}\in\mathcal{V}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0}), h(𝔲)>2h(\mathfrak{u})>2 and υ(π(𝔲))<1\upsilon(\pi(\mathfrak{u}))<1. If 𝔲\mathfrak{u} is long then it is either bb^{-\partial}-long or (b,s)(b,s)^{-\partial}-long.

Corollary 6.19.

Suppose 𝔲𝒱𝔣i(𝔵0)\mathfrak{u}\in\mathcal{V}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0}), h(𝔲)>2h(\mathfrak{u})>2 and π(𝔲)\pi(\mathfrak{u}) is abnormal then

  • the LVP (𝔱(π(𝔲)),𝔠(𝔵0,π(𝔲)))(\mathfrak{t}(\pi(\mathfrak{u})),\mathfrak{c}(\mathfrak{x}_{0},\pi(\mathfrak{u}))) is bθ(β(π(𝔲)))b^{-\theta(\beta(\pi(\mathfrak{u})))}-long;

  • 𝔯b1(π(𝔲))\mathfrak{r}^{b}_{1}(\pi(\mathfrak{u})) is a right substring of 𝔣(𝔵0;π(𝔲),π2(𝔲))\mathfrak{f}(\mathfrak{x}_{0};\pi(\mathfrak{u}),\pi^{2}(\mathfrak{u})).

Proof. Since π(𝔲)\pi(\mathfrak{u}) is abnormal Proposition 4.6 gives that 𝔣(𝔵0;π(𝔲),π2(𝔲))=𝔷𝔠(𝔵0;π(𝔲))\mathfrak{f}(\mathfrak{x}_{0};\pi(\mathfrak{u}),\pi^{2}(\mathfrak{u}))=\mathfrak{z}\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u})) for some string 𝔷\mathfrak{z}.

Propositions 6.1 (υ(π(𝔲))=1\upsilon(\pi(\mathfrak{u}))=1), 6.15 (υ(π(𝔲))=0\upsilon(\pi(\mathfrak{u}))=0) and Corollary 6.12 ((υ(π(𝔲))=1\upsilon(\pi(\mathfrak{u}))=-1)) give that the LVP (𝔱(π(𝔲)),𝔠(𝔵0,π(𝔲)))(\mathfrak{t}(\pi(\mathfrak{u})),\mathfrak{c}(\mathfrak{x}_{0},\pi(\mathfrak{u}))) is bθ(β(π(𝔲)))b^{-\theta(\beta(\pi(\mathfrak{u})))}-long.

The following cases describe about 𝔴(𝔵0;π(𝔲))\mathfrak{w}(\mathfrak{x}_{0};\pi(\mathfrak{u})) depending on the value of υ(π(𝔲))\upsilon(\pi(\mathfrak{u})).

Case 1: υ(π(𝔲))=1\upsilon(\pi(\mathfrak{u}))=1.

Suppose 𝔠(𝔵0;π(𝔲))=𝔷𝔠(𝔵0;π2(𝔲))\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u}))=\mathfrak{z}^{\prime}\mathfrak{c}(\mathfrak{x}_{0};\pi^{2}(\mathfrak{u})) for some positive length string 𝔷\mathfrak{z}^{\prime}.

If π(𝔲)\pi(\mathfrak{u}) is long then Corollary 6.6 guarantees that either |π(𝔲)c|=0|\pi(\mathfrak{u})^{c}|=0 or |π(𝔲)c|>0|\pi(\mathfrak{u})^{c}|>0 with |𝔴~(𝔵0;π(𝔲))|=0|\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\pi(\mathfrak{u}))|=0, and hence 𝔷=𝔟𝔣α(π2(𝔲))\mathfrak{z}^{\prime}=\mathfrak{b}^{\mathfrak{f}\alpha}(\pi^{2}(\mathfrak{u})).

On the other hand if π(𝔲)\pi(\mathfrak{u}) is short then 𝔷\mathfrak{z}^{\prime} is a proper left substring of 𝔟𝔣α(π2(𝔲))\mathfrak{b}^{\mathfrak{f}\alpha}(\pi^{2}(\mathfrak{u})).

Since β(π(𝔲))π(𝔲)c\beta(\pi(\mathfrak{u}))\pi(\mathfrak{u})^{c} is a string we get π(𝔲)c=𝔷\pi(\mathfrak{u})^{c}=\mathfrak{z} by Proposition 6.1 and hence 𝔴(𝔵0;π(𝔲))=π(𝔲)c𝔷=𝔯b1(π(𝔲))𝔷\mathfrak{w}(\mathfrak{x}_{0};\pi(\mathfrak{u}))=\pi(\mathfrak{u})^{c}\mathfrak{z}^{\prime}=\mathfrak{r}^{b}_{1}(\pi(\mathfrak{u}))\mathfrak{z}^{\prime} by Corollary 6.3. In summary

𝔴~(𝔵0;π(𝔲)):={𝔯b1(π(𝔲))if π(𝔲) is long;𝔯b1(π(𝔲))𝔷otherwise.\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\pi(\mathfrak{u})):=\begin{cases}\mathfrak{r}^{b}_{1}(\pi(\mathfrak{u}))&\mbox{if $\pi(\mathfrak{u})$ is long;}\\ \mathfrak{r}^{b}_{1}(\pi(\mathfrak{u}))\mathfrak{z}^{\prime}&\mbox{otherwise}.\end{cases}

Case 2: υ(π(𝔲))=0\upsilon(\pi(\mathfrak{u}))=0.

Proposition 6.16 gives that π(𝔲)\pi(\mathfrak{u}) is short. Hence by Proposition 6.15 we get 𝔴(𝔵0;π(𝔲))=𝔴~(𝔵0;π(𝔲))=𝔯b1(π(𝔲))\mathfrak{w}(\mathfrak{x}_{0};\pi(\mathfrak{u}))=\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\pi(\mathfrak{u}))=\mathfrak{r}^{b}_{1}(\pi(\mathfrak{u})).

Case 3: υ(π(𝔲))=1\upsilon(\pi(\mathfrak{u}))=-1.

Let 𝔠(𝔵0;π2(𝔲))=𝔷𝔠(𝔵0;π(𝔲))\mathfrak{c}(\mathfrak{x}_{0};\pi^{2}(\mathfrak{u}))=\mathfrak{z}^{\prime\prime\prime}\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u})) for some positive length string 𝔷\mathfrak{z}^{\prime\prime\prime}. Corollary 6.12 guarantees that 𝔯b1(π(𝔲))=π(𝔲)c𝔷\mathfrak{r}^{b}_{1}(\pi(\mathfrak{u}))=\pi(\mathfrak{u})^{c}\mathfrak{z}^{\prime\prime\prime}. Since β(π(𝔲))π(𝔲)c\beta(\pi(\mathfrak{u}))\pi(\mathfrak{u})^{c} is a string we get 𝔴~(𝔵0;π(𝔲))=π(𝔲)c\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\pi(\mathfrak{u}))=\pi(\mathfrak{u})^{c}. Since π(𝔲)\pi(\mathfrak{u}) is short by Proposition 6.16, we have 𝔴(𝔵0;π(𝔲))=𝔴~(𝔵0;π(𝔲))=π(𝔲)c\mathfrak{w}(\mathfrak{x}_{0};\pi(\mathfrak{u}))=\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\pi(\mathfrak{u}))=\pi(\mathfrak{u})^{c}. Therefore 𝔯b1(π(𝔲))=𝔴(𝔵0;π(𝔲))𝔷\mathfrak{r}^{b}_{1}(\pi(\mathfrak{u}))=\mathfrak{w}(\mathfrak{x}_{0};\pi(\mathfrak{u}))\mathfrak{z}^{\prime\prime\prime}.

All the three cases above show that 𝔯b1(π(𝔲))\mathfrak{r}^{b}_{1}(\pi(\mathfrak{u})) is a right substring of 𝔣(𝔵0;π(𝔲),π2(𝔲))\mathfrak{f}(\mathfrak{x}_{0};\pi(\mathfrak{u}),\pi^{2}(\mathfrak{u})). This completes the proof.    \Box

We close this section with some expository discussion. A single monomial relation in ρ\rho can be responsible for making a child-parent pair in 𝒱𝔣i(𝔵0)\mathcal{V}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0}) simultaneously long. When this happens then, for some {+,}\partial\in\{+,-\}, the parent is bb^{\partial}-long and the child is ss^{\partial}-long. The definition of H\circ_{H} makes sure that even if the same relation touches several other bands then it cannot make any other vertex of 𝔲\mathfrak{u} long. The next proposition characterizes such situation.

Example 6.20.

In the algebra Γ(iv)\Gamma^{(iv)} from Figure 9 the only bands are 𝔟1:=acD\mathfrak{b}_{1}:=acD, 𝔟2:=eHG\mathfrak{b}_{2}:=eHG. Choosing 𝔵0:=1(𝔳1,i)\mathfrak{x}_{0}:=1_{(\mathfrak{v}_{1},i)} and 𝔲:=f\mathfrak{u}:=f we have π(𝔲):=e\pi(\mathfrak{u}):=e.

v1{v_{1}}v7{v_{7}}v2{v_{2}}v4{v_{4}}v5{v_{5}}v3{v_{3}}v6{v_{6}}a\scriptstyle{a}b\scriptstyle{b}e\scriptstyle{e}h\scriptstyle{h}f\scriptstyle{f}c\scriptstyle{c}d\scriptstyle{d}g\scriptstyle{g}
Figure 9. Γ(iv)\Gamma^{(iv)} with ρ={bc,ha,ed,fg,hd,feac}\rho=\{bc,ha,ed,fg,hd,feac\}

Here 𝔠(𝔵0;π(𝔲))=B\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u}))=B and 𝔠(𝔵0;𝔲)=acDaB\mathfrak{c}(\mathfrak{x}_{0};\mathfrak{u})=acDaB. The monomial relation feacρfeac\in\rho is responsible to make π(𝔲)\pi(\mathfrak{u}) bb-long and 𝔲\mathfrak{u} ss-long.

Proposition 6.21.

Suppose 𝔲𝒱𝔣i(𝔵0)\mathfrak{u}\in\mathcal{V}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0}), h(𝔲)>2h(\mathfrak{u})>2 and π(𝔲)\pi(\mathfrak{u}) is bb-long. Then β(𝔲)𝔴~(𝔵0;𝔲)\beta(\mathfrak{u})\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u}) is a substring of 𝔯b2(π(𝔲))\mathfrak{r}^{b}_{2}(\pi(\mathfrak{u})) if and only if

  1. (1)

    if π(𝔲)\pi(\mathfrak{u}) is abnormal then υ(π(𝔲))=1\upsilon(\pi(\mathfrak{u}))=1 and |π(𝔲)c|=0|\pi(\mathfrak{u})^{c}|=0;

  2. (2)

    𝔲\mathfrak{u} is ss-long with 𝔯s1(π(𝔲))=𝔴~(𝔵0;𝔲)\mathfrak{r}^{s}_{1}(\pi(\mathfrak{u}))=\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u}) and

    𝔯~s(π(𝔲)):={π(𝔲)o𝔴~(𝔵0;π(𝔲))𝔯~b(π2(𝔲))if π(𝔲) is normal;𝔴~(𝔵0;π(𝔲))𝔯~b(π2(𝔲))otherwise.\tilde{\mathfrak{r}}^{s}(\pi(\mathfrak{u})):=\begin{cases}\pi(\mathfrak{u})^{o}\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\pi(\mathfrak{u}))\tilde{\mathfrak{r}}^{b}(\pi^{2}(\mathfrak{u}))&\mbox{if $\pi(\mathfrak{u})$ is normal;}\\ \tilde{\mathfrak{w}}(\mathfrak{x}_{0};\pi(\mathfrak{u}))\tilde{\mathfrak{r}}^{b}(\pi^{2}(\mathfrak{u}))&\mbox{otherwise}.\end{cases}

Proof. We only need to prove the forward direction as the backward direction is obvious.

Since π(𝔲)\pi(\mathfrak{u}) is bb-long, if it is abnormal then Proposition 6.16 gives υ(π(𝔲))=1\upsilon(\pi(\mathfrak{u}))=1. Further Corollary 6.6 gives that |π(𝔲)c|=0|\pi(\mathfrak{u})^{c}|=0. To prove 𝔲\mathfrak{u} is ss-long it is enough to prove the following claim.

Claim: If π(𝔲)\pi(\mathfrak{u}) is bb-long then N(𝔰(π(𝔲)),𝔠(𝔵0;π(𝔲)))=1N(\mathfrak{s}(\pi(\mathfrak{u})),\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u})))=1.

Since π(𝔲)\pi(\mathfrak{u}) is bb-long we get N(𝔵0;𝔰(𝔲),𝔣(𝔵0;𝔲))=1N(\mathfrak{x}_{0};\mathfrak{s}(\mathfrak{u}),\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u}))=1. If π(𝔲)\pi(\mathfrak{u}) is normal then Remark 5.4 gives that N(𝔵0;𝔰(𝔲),𝔠(𝔵0;𝔲))=N(𝔵0;𝔰(𝔲),𝔣(𝔵0;𝔲))=1N(\mathfrak{x}_{0};\mathfrak{s}(\mathfrak{u}),\mathfrak{c}(\mathfrak{x}_{0};\mathfrak{u}))=N(\mathfrak{x}_{0};\mathfrak{s}(\mathfrak{u}),\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u}))=1. If π(𝔲)\pi(\mathfrak{u}) is abnormal with υ(π(𝔲))=1\upsilon(\pi(\mathfrak{u}))=1 and |π(𝔲)c|=0|\pi(\mathfrak{u})^{c}|=0 then 𝔠(𝔵0;π(𝔲))=𝔴~(𝔵0;π(𝔲))𝔟𝔣α(π2(𝔲))𝔠(𝔵0;π2(𝔲))\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u}))=\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\pi(\mathfrak{u}))\mathfrak{b}^{\mathfrak{f}\alpha}{(\pi^{2}(\mathfrak{u}))}\mathfrak{c}(\mathfrak{x}_{0};\pi^{2}(\mathfrak{u})). Hence the claim.    \Box

7. H-reduced forking strings

Recall that a string 𝔵Hl(𝔵0)\mathfrak{x}\in H_{l}(\mathfrak{x}_{0}) is forking if it can be extended on the left by two different syllables.

Proposition 7.1.

If 𝔲1𝔲2\mathfrak{u}_{1}\neq\mathfrak{u}_{2} for 𝔲1,𝔲2𝒱𝔣i(𝔵0){𝔵0}\mathfrak{u}_{1},\mathfrak{u}_{2}\in\mathcal{V}^{\mathfrak{f}}_{i}{(\mathfrak{x}_{0})}\setminus\{\mathfrak{x}_{0}\} then 𝔣(𝔵0;𝔲1,𝔲2)\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u}_{1},\mathfrak{u}_{2}) is a H-reduced forking string.

Proof. Clearly 𝔣(𝔵0;𝔲1,𝔲2)\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u}_{1},\mathfrak{u}_{2}) is a forking string. Then note that if 𝔶\mathfrak{y} is a left substring of 𝔣(𝔵0;𝔲1)\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u}_{1}) and HR𝔟(𝔶)\mathrm{HR}_{\mathfrak{b}}(\mathfrak{y}) exists for some band 𝔟\mathfrak{b} then 𝔟=𝔱(𝔲1)\mathfrak{b}=\mathfrak{t}(\mathfrak{u}_{1}).

Thus if HR𝔟(𝔣(𝔵0;𝔲1,𝔲2))\mathrm{HR}_{\mathfrak{b}}(\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u}_{1},\mathfrak{u}_{2})) exists for some band 𝔟\mathfrak{b} then 𝔟=𝔱(𝔲1)=𝔱(𝔲2)\mathfrak{b}=\mathfrak{t}(\mathfrak{u}_{1})=\mathfrak{t}(\mathfrak{u}_{2}). As a consequence we get 𝔣(𝔵0;𝔲1)=𝔣(𝔵0;𝔲2)\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u}_{1})=\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u}_{2}), and hence 𝔲1=𝔲2\mathfrak{u}_{1}=\mathfrak{u}_{2} by Proposition 4.10, a contradiction.    \Box

Despite the combinatorial complexity of the collection of strings, by the end of this section, we will “essentially prove” the following surprising statement.

Remark 7.2.

If 𝔶Hli(𝔵0)\mathfrak{y}\in H_{l}^{i}(\mathfrak{x}_{0}) is an H-reduced forking string then 𝔶\mathfrak{y} has one of the following forms for some 𝔲𝒱𝔣i(𝔵0)\mathfrak{u}\in\mathcal{V}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0}):

  • 𝔣(𝔵0;𝔲,𝔲f+)\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u},\mathfrak{u}^{f+});

  • 𝔣(𝔵0;𝔲,π(𝔲))\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u},\pi(\mathfrak{u})).

In this section we guarantee the existence of certain distant relatives of 𝔲𝒱𝔣i(𝔵0)\mathfrak{u}\in\mathcal{V}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0}) under certain hypotheses and thus describe a particular forking string as a forking string of two immediate siblings. There will be several different versions of this result under different hypotheses; we will prove the uncle-nephew interaction in detail and only indicate the changes in the proofs of the remaining cases.

Lemma 7.3.

(Uncle forking lemma–same parity) Suppose 𝔲𝒱𝔣i(𝔵0)\mathfrak{u}\in\mathcal{V}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0}), h(𝔲)>2h(\mathfrak{u})>2 and θ(β(𝔲))=θ(β(π(𝔲)))=\theta(\beta(\mathfrak{u}))=\theta(\beta(\pi(\mathfrak{u})))=\partial for some {+,}\partial\in\{+,-\}. Further suppose

  1. (1)

    𝔲\mathfrak{u} is long;

  2. (2)

    𝔴~(𝔵0;𝔲)𝔠(𝔵0;π(𝔲))=𝔷~𝔠(𝔵0;π2(𝔲))\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u})\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u}))=\tilde{\mathfrak{z}}\mathfrak{c}(\mathfrak{x}_{0};\pi^{2}(\mathfrak{u})) for some string 𝔷~\tilde{\mathfrak{z}} of positive length;

  3. (3)

    β(𝔲)𝔷~𝔟𝔣α(π2(𝔲))\beta(\mathfrak{u})\tilde{\mathfrak{z}}\mathfrak{b}^{\mathfrak{f}\alpha}(\pi^{2}(\mathfrak{u})) is a string.

Then there is an uncle 𝔲1\mathfrak{u}_{1} of 𝔲\mathfrak{u} satisfying β(𝔲1){β(𝔲),β(π(𝔲))}\beta(\mathfrak{u}_{1})\in\{\beta(\mathfrak{u}),\beta(\pi(\mathfrak{u}))\} and

𝔣(𝔵0;π(𝔲),𝔲1)={𝔴~(𝔵0;𝔲)𝔠(𝔵0;π(𝔲))if 𝔲1𝔣π(𝔲);𝔴~(𝔵0;π(𝔲))𝔠(𝔵0;π2(𝔲))if π(𝔲)𝔣𝔲1.\mathfrak{f}(\mathfrak{x}_{0};\pi(\mathfrak{u}),\mathfrak{u}_{1})=\begin{cases}\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u})\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u}))&\mbox{if }\mathfrak{u}_{1}\sqsubset^{\mathfrak{f}}_{\partial}\pi(\mathfrak{u});\\ \tilde{\mathfrak{w}}(\mathfrak{x}_{0};\pi(\mathfrak{u}))\mathfrak{c}(\mathfrak{x}_{0};\pi^{2}(\mathfrak{u}))&\mbox{if }\pi(\mathfrak{u})\sqsubset^{\mathfrak{f}}_{\partial}\mathfrak{u}_{1}.\end{cases}

The proof is long and we need to set up more notation.

If υ(π(𝔲))>1\upsilon(\pi(\mathfrak{u}))>-1 then let 𝔷\mathfrak{z} satisfy 𝔠(𝔵0;π(𝔲))=𝔷𝔠(𝔠0;π2(𝔲))\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u}))=\mathfrak{z}\mathfrak{c}(\mathfrak{c}_{0};\pi^{2}(\mathfrak{u})). Note that the word 𝔟𝔣α(π(𝔲))𝔷𝔟𝔣α(π2(𝔲))𝔠(𝔵0;π2(𝔲))\mathfrak{b}^{\mathfrak{f}\alpha}(\pi(\mathfrak{u}))\mathfrak{z}\mathfrak{b}^{\mathfrak{f}\alpha}(\pi^{2}(\mathfrak{u}))\mathfrak{c}(\mathfrak{x}_{0};\pi^{2}(\mathfrak{u})) is a string. Let 𝔴~\tilde{\mathfrak{w}} be the shortest right substring of 𝔷𝔟𝔣α(π2(𝔲))\mathfrak{z}\mathfrak{b}^{\mathfrak{f}\alpha}(\pi^{2}(\mathfrak{u})) such that 𝔟𝔣α(π(𝔲))𝔴~𝔰(π(𝔲))\mathfrak{b}^{\mathfrak{f}\alpha}(\pi(\mathfrak{u}))\tilde{\mathfrak{w}}\mathfrak{s}(\pi(\mathfrak{u})) is a string.

If υ(π(𝔲))=1\upsilon(\pi(\mathfrak{u}))=-1 then let 𝔷\mathfrak{z} satisfy 𝔠(𝔵0;π2(𝔲))=𝔷𝔠(𝔠0;π(𝔲))\mathfrak{c}(\mathfrak{x}_{0};\pi^{2}(\mathfrak{u}))=\mathfrak{z}\mathfrak{c}(\mathfrak{c}_{0};\pi(\mathfrak{u})). If 𝔠(𝔵0;π2(𝔲))\mathfrak{c}(\mathfrak{x}_{0};\pi^{2}(\mathfrak{u})) is a substring of 𝔴~(𝔵0;𝔲)𝔠(𝔵0;π(𝔲))\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u})\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u})) then 𝔷\mathfrak{z} is a substring of 𝔴~(𝔵0;𝔲)\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u}). Let 𝔴~\tilde{\mathfrak{w}} be the shortest right substring of 𝔟𝔣α(π2(𝔲))\mathfrak{b}^{\mathfrak{f}\alpha}(\pi^{2}(\mathfrak{u})) such that 𝔴~𝔰(π(𝔲))\tilde{\mathfrak{w}}\mathfrak{s}(\pi(\mathfrak{u})) is a string.

In both cases 𝔴~\tilde{\mathfrak{w}} is a left substring of π(𝔲)\pi(\mathfrak{u}).

Since 𝔲\mathfrak{u} is long, Proposition 5.8 gives that (𝔟,𝔶):=(𝔰(𝔲),𝔴~(𝔵0;𝔲)𝔠(𝔵0;π(𝔲)))(\mathfrak{b},\mathfrak{y}):=(\mathfrak{s}(\mathfrak{u}),\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u})\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u}))) is long. If 𝔲\mathfrak{u} is double long, then it is either (b,s)(b,s)^{-\partial}-long or (s,b)(s,b)^{-\partial}-long.

The proof will be completed in four cases depending on whether 𝔲\mathfrak{u} is single/double b/s±b/s^{\pm\partial}-long but the general plan in each case is as follows.

Step 1: To show that for some string 𝔶\mathfrak{y}^{\prime\prime} the string 𝔶β(𝔲)𝔴~(𝔵0;𝔲)𝔴~\mathfrak{y}^{\prime\prime}\beta(\mathfrak{u})\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u})\tilde{\mathfrak{w}} is an H-reduced string.

Step 2: Since β(𝔲)𝔷~𝔟𝔣α(π2(𝔲))\beta(\mathfrak{u})\tilde{\mathfrak{z}}\mathfrak{b}^{\mathfrak{f}\alpha}(\pi^{2}(\mathfrak{u})) is a string the hypotheses of the following lemma are satisfied, and it yields 𝔲¯𝒬¯Ba\bar{\mathfrak{u}}\in\overline{\mathcal{HQ}}^{\mathrm{Ba}} such that 𝔶β(𝔲)𝔴~(𝔵0;𝔲)𝔴~\mathfrak{y}^{\prime\prime}\beta(\mathfrak{u})\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u})\tilde{\mathfrak{w}} is a left substring of 𝔱(𝔲¯)𝔲¯\mathfrak{t}(\bar{\mathfrak{u}})\bar{\mathfrak{u}}. Then [9, Proposition 8.8,Theorem 8.9] yield an arch bridge 𝔲1\mathfrak{u}_{1} and a weak arch bridge 𝔲2\mathfrak{u}_{2} such that 𝔲¯=𝔲2H𝔲1\bar{\mathfrak{u}}=\mathfrak{u}_{2}\circ_{H}\mathfrak{u}_{1}. Since N(𝔰(𝔲),𝔲¯)=0N(\mathfrak{s}(\mathfrak{u}),\bar{\mathfrak{u}})=0 and the LVP (𝔰(𝔲),𝔴~(𝔵0;𝔲)𝔠(𝔵0;π(𝔲)))(\mathfrak{s}(\mathfrak{u}),\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u})\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u}))) is long, we get 𝔲1π(𝔲)\mathfrak{u}_{1}\neq\pi(\mathfrak{u}).

Lemma 7.4.
  1. (1)

    If 𝔴\mathfrak{w} is an H-reduced string and 𝔟\mathfrak{b} is a band such that 𝔴𝔟\mathfrak{w}\mathfrak{b} is a skeletal string that is not a substring of 𝔟2\mathfrak{b}^{2} then there is a string 𝔴\mathfrak{w}^{\prime} and 𝔲𝒬¯Ba\mathfrak{u}\in\overline{\mathcal{HQ}}^{\mathrm{Ba}} such that 𝔴𝔴𝔟=𝔱(𝔲)𝔲𝔟\mathfrak{w}^{\prime}\mathfrak{w}\mathfrak{b}=\mathfrak{t}(\mathfrak{u})\mathfrak{u}\mathfrak{b}.

  2. (2)

    If 𝔴\mathfrak{w} is an H-reduced string such that 𝔴𝔵0\mathfrak{w}\mathfrak{x}_{0} is a skeletal string with respect to (i,𝔵0)(i,\mathfrak{x}_{0}) then there is a string 𝔴\mathfrak{w}^{\prime} and 𝔲𝒬¯Ba\mathfrak{u}\in\overline{\mathcal{HQ}}^{\mathrm{Ba}} such that 𝔴𝔴𝔵0=𝔱(𝔲)𝔲𝔵0\mathfrak{w}^{\prime}\mathfrak{w}\mathfrak{x}_{0}=\mathfrak{t}(\mathfrak{u})\mathfrak{u}\mathfrak{x}_{0}.

Proof. We only prove the first statement; a similar proof works for the other case.

There are two possibilities:

  1. (1)

    If 𝔴𝔟\mathfrak{w}\mathfrak{b} is torsion then taking 𝔴\mathfrak{w}^{\prime} to be a length 0 string, [9, Proposition 8.8] gives that 𝔴\mathfrak{w} is a maximal reverse weak arch bridge.

  2. (2)

    If 𝔴𝔟\mathfrak{w}\mathfrak{b} is not torsion then there is a αQ1\alpha\in Q_{1} such that α𝔴𝔟\alpha\mathfrak{w}\mathfrak{b} is a string. There are further two subcases.

    1. (a)

      If α𝔴\alpha\mathfrak{w} is not H-reduced then there is a band 𝔟1\mathfrak{b}_{1} such that α𝔟1\alpha\in\mathfrak{b}_{1}. Since 𝔴𝔟\mathfrak{w}\mathfrak{b} is not a substring of 𝔟2\mathfrak{b}^{2} we get 𝔟𝔟1\mathfrak{b}\neq\mathfrak{b}_{1}. Let 𝔲\mathfrak{u} be the shortest left substring of α𝔴𝔟\alpha\mathfrak{w}\mathfrak{b} such that 𝔟1𝔲𝔟\mathfrak{b}_{1}\mathfrak{u}\mathfrak{b} is a string. Since α𝔴𝔟\alpha\mathfrak{w}\mathfrak{b} is not H-reduced, 𝔲\mathfrak{u} is a proper left substring of α𝔴𝔟\alpha\mathfrak{w}\mathfrak{b}. Hence being a left substring of the H-reduced string 𝔴\mathfrak{w} [9, Proposition 5.11] gives that 𝔲\mathfrak{u} is also H-reduced. Thus 𝔲\mathfrak{u} is a weak arch bridge from 𝔟\mathfrak{b} to 𝔟1\mathfrak{b}_{1} by [9, Proposition 8.8].

    2. (b)

      If α𝔴𝔟\alpha\mathfrak{w}\mathfrak{b} is H-reduced then we can proceed as above by replacing 𝔴𝔟\mathfrak{w}\mathfrak{b} by α𝔴𝔟\alpha\mathfrak{w}\mathfrak{b}.

Since there are only finitely many H-reduced strings by [9, Corollary 5.8], after finitely many iterations of case (2b) we must land in either case (1) or case (2a), and thus the result.    \Box

In this step we also show that β(𝔲1){β(𝔲),β(π(𝔲))}\beta(\mathfrak{u}_{1})\in\{\beta(\mathfrak{u}),\beta(\pi(\mathfrak{u}))\}. We begin with an observation.

Remark 7.5.

Suppose 𝔲𝒱𝔣i(𝔵0)\mathfrak{u}\in\mathcal{V}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0}), h(𝔲)>2h(\mathfrak{u})>2 and θ(β(𝔲))=θ(β(π(𝔲)))=\theta(\beta(\mathfrak{u}))=\theta(\beta(\pi(\mathfrak{u})))=\partial for some {+,}\partial\in\{+,-\}. If 𝔠(𝔵0;π2(𝔲))\mathfrak{c}(\mathfrak{x}_{0};\pi^{2}(\mathfrak{u})) is a substring of 𝔴~(𝔵0;𝔲)𝔠(𝔵0;π(𝔲))\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u})\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u})) and |π(𝔲)c|=0|\pi(\mathfrak{u})^{c}|=0 then υ(π(𝔲))=1\upsilon(\pi(\mathfrak{u}))=1.

In view of the above remark the following two lemmas are sufficient to complete the proof of this step in all four cases.

Lemma 7.6.

Assuming the hypotheses of Lemma 7.3 and using the notations above, suppose υ(π(𝔲))=1\upsilon(\pi(\mathfrak{u}))=1. If either π(𝔲)\pi(\mathfrak{u}) is normal or π(𝔲)\pi(\mathfrak{u}) is abnormal with |π(𝔲)|c=0|\pi(\mathfrak{u})|^{c}=0 then

  • β(𝔲1)=β(π(𝔲))=β(𝔲¯)\beta(\mathfrak{u}_{1})=\beta(\pi(\mathfrak{u}))=\beta(\bar{\mathfrak{u}});

  • 𝔲1\mathfrak{u}_{1} is long if and only if π(𝔲)\pi(\mathfrak{u}) is long.

Proof. Since π(𝔲)\pi(\mathfrak{u}) is normal, β(π(𝔲))𝔴~\beta(\pi(\mathfrak{u}))\in\tilde{\mathfrak{w}} by the construction of 𝔴~\tilde{\mathfrak{w}}. Moreover the construction of 𝔲¯\bar{\mathfrak{u}} gives β(π(𝔲))=β(𝔲¯)\beta(\pi(\mathfrak{u}))=\beta(\bar{\mathfrak{u}}). If π(𝔲)\pi(\mathfrak{u}) is abnormal with |π(𝔲)|c=0|\pi(\mathfrak{u})|^{c}=0 then |𝔴~(𝔵0;𝔲)|>0|\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u})|>0, and hence the same conclusion follows.

For a contradiction suppose that β(𝔲1)β(𝔲¯)\beta(\mathfrak{u}_{1})\neq\beta(\bar{\mathfrak{u}}). Then 𝔲2\mathfrak{u}_{2} is non-trivial.

Claim: N(𝔱(𝔲1),𝔲¯)=0N(\mathfrak{t}(\mathfrak{u}_{1}),\bar{\mathfrak{u}})=0

Suppose not. Since 𝔲¯\bar{\mathfrak{u}} is H-reduced by [9, Theorem 8.6], we get N(𝔱(𝔲1),𝔲¯)=1N(\mathfrak{t}(\mathfrak{u}_{1}),\bar{\mathfrak{u}})=1. Hence 𝔲¯={𝔲2𝔱(𝔲1)𝔲1}\bar{\mathfrak{u}}=\{\mathfrak{u}_{2}\mathfrak{t}(\mathfrak{u}_{1})\mathfrak{u}_{1}\}.

If 𝔲1\mathfrak{u}_{1} is normal then β(𝔲1)𝔲1\beta(\mathfrak{u}_{1})\in\mathfrak{u}_{1}. Hence β(𝔲¯)=β(𝔲1)\beta(\bar{\mathfrak{u}})=\beta(\mathfrak{u}_{1}), a contradiction. On the other hand if 𝔲1\mathfrak{u}_{1} is abnormal then β(𝔲1)𝔲1c\beta(\mathfrak{u}_{1})\mathfrak{u}_{1}^{c} is a substring of 𝔲¯\bar{\mathfrak{u}}, and hence β(𝔲¯)=β(𝔲1)\beta(\bar{\mathfrak{u}})=\beta(\mathfrak{u}_{1}), again a contradiction. This completes the proof of the claim.

Now the claim combined with β(𝔲1)β(𝔲¯)\beta(\mathfrak{u}_{1})\neq\beta(\bar{\mathfrak{u}}) gives that β(𝔲1)𝔱(𝔲1)\beta(\mathfrak{u}_{1})\in\mathfrak{t}(\mathfrak{u}_{1}). Therefore 𝔲1\mathfrak{u}_{1} is abnormal. As a consequence there is a partition 𝔲1e=𝔵2𝔵1\mathfrak{u}_{1}^{e}=\mathfrak{x}_{2}\mathfrak{x}_{1} such that |𝔵2|>0,|𝔵1||𝔲1c||\mathfrak{x}_{2}|>0,|\mathfrak{x}_{1}|\neq|\mathfrak{u}_{1}^{c}| and β(𝔲¯)𝔵1\beta(\bar{\mathfrak{u}})\mathfrak{x}_{1} is a string. Since β(π(𝔲))=β(𝔲¯)\beta(\pi(\mathfrak{u}))=\beta(\bar{\mathfrak{u}}), 𝔵2𝔴~(𝔵0;π(𝔲))\mathfrak{x}_{2}\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\pi(\mathfrak{u})) is a string and hence (𝔰(π(𝔲)),𝔴~(𝔵0;π(𝔲))𝔠(𝔵0;π2(𝔲)))(\mathfrak{s}(\pi(\mathfrak{u})),\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\pi(\mathfrak{u}))\mathfrak{c}(\mathfrak{x}_{0};\pi^{2}(\mathfrak{u}))) is bb-long. Since β(𝔲¯)=β(π(𝔲))\beta(\bar{\mathfrak{u}})=\beta(\pi(\mathfrak{u})) there are two cases.

If |𝔵1|<|𝔲1c||\mathfrak{x}_{1}|<|\mathfrak{u}_{1}^{c}| then β(𝔲2)=β(𝔲¯)\beta(\mathfrak{u}_{2})=\beta(\bar{\mathfrak{u}}). On the other hand, if |𝔵1|>|𝔲1c||\mathfrak{x}_{1}|>|\mathfrak{u}_{1}^{c}| then β(𝔲2)\beta(\mathfrak{u}_{2}) is the first syllable of 𝔲1β\mathfrak{u}_{1}^{\beta}. In both cases (𝔰(𝔲2),𝔴~(𝔵0;𝔲2)𝔠(𝔵0;π(𝔲2)))(\mathfrak{s}(\mathfrak{u}_{2}),\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u}_{2})\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u}_{2}))) is bb-long. Hence N(𝔱(𝔲1),𝔲¯)=1N(\mathfrak{t}(\mathfrak{u}_{1}),\bar{\mathfrak{u}})=1, a contradiction to the claim. Therefore our assumption is wrong and we conclude β(𝔲1)=β(𝔲¯)\beta(\mathfrak{u}_{1})=\beta(\bar{\mathfrak{u}}).

As π2(𝔲)=π(𝔲1)\pi^{2}(\mathfrak{u})=\pi(\mathfrak{u}_{1}), 𝔠(𝔵0;π2(𝔲))=𝔠(𝔵0;π(𝔲1))\mathfrak{c}(\mathfrak{x}_{0};\pi^{2}(\mathfrak{u}))=\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u}_{1})) and β(𝔲1)=β(π(𝔲))\beta(\mathfrak{u}_{1})=\beta(\pi(\mathfrak{u})) we get 𝔴~(𝔵0;π(𝔲))=𝔴~(𝔵0;𝔲1)\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\pi(\mathfrak{u}))=\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u}_{1}). Hence from Corollary 5.9, 𝔲1\mathfrak{u}_{1} is long if and only if π(𝔲)\pi(\mathfrak{u}) is long.    \Box

Lemma 7.7.

Assuming the hypotheses of Lemma 7.3 and using the notations above, if π(𝔲)\pi(\mathfrak{u}) is abnormal with |π(𝔲)c|>0|\pi(\mathfrak{u})^{c}|>0 then β(𝔲¯)=β(𝔲1)=β(𝔲)\beta(\bar{\mathfrak{u}})=\beta(\mathfrak{u}_{1})=\beta(\mathfrak{u}).

Proof. We prove the result for υ(π(𝔲))=1\upsilon(\pi(\mathfrak{u}))=1. With slight modification we can get the proof for the other two cases.

Since 𝔠(𝔵0;π2(𝔲))\mathfrak{c}(\mathfrak{x}_{0};\pi^{2}(\mathfrak{u})) is a proper left substring of 𝔠(𝔵0;π(𝔲))\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u})), 𝔲\mathfrak{u} is long and |π(𝔲)c|>0|\pi(\mathfrak{u})^{c}|>0, then Corollaries 6.2 and 6.3 together guarantee that 𝔲\mathfrak{u} is bb-long. Hence the latter corollary guarantees that 𝔯b1(π(𝔲))=π(𝔲)c\mathfrak{r}^{b}_{1}(\pi(\mathfrak{u}))=\pi(\mathfrak{u})^{c} and that 𝔴~(𝔵0;𝔲)\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u}) is a substring of π(𝔲)c\pi(\mathfrak{u})^{c}. Since 𝔴~(𝔵0;𝔲)𝔴~\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u})\tilde{\mathfrak{w}} is a left substring of 𝔰(π(𝔲))\mathfrak{s}(\pi(\mathfrak{u})), we see that β(𝔲¯)=β(𝔲)\beta(\bar{\mathfrak{u}})=\beta(\mathfrak{u}). Hence 𝔴~\tilde{\mathfrak{w}} is a left substring of 𝔲1\mathfrak{u}_{1} thanks to [9, Proposition 4.3].

Suppose β(𝔲¯)β(𝔲1)\beta(\bar{\mathfrak{u}})\neq\beta(\mathfrak{u}_{1}). Then 𝔲2\mathfrak{u}_{2} is non-trivial, and 𝔲1\mathfrak{u}_{1} is abnormal with |𝔴~(𝔵0;𝔲)|<|𝔲1e||\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u})|<|\mathfrak{u}_{1}^{e}| and N(𝔱(𝔲1),𝔲¯)=0N(\mathfrak{t}(\mathfrak{u}_{1}),\bar{\mathfrak{u}})=0. But in this case (𝔰(𝔲2),𝔴~(𝔵0;𝔲2)𝔠(𝔵0;π(𝔲2)))(\mathfrak{s}(\mathfrak{u}_{2}),\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u}_{2})\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u}_{2}))) is bb-long which guarantees that N(𝔱(𝔲1),𝔲¯)=1N(\mathfrak{t}(\mathfrak{u}_{1}),\bar{\mathfrak{u}})=1, a contradiction. Hence β(𝔲¯)=β(𝔲1)\beta(\bar{\mathfrak{u}})=\beta(\mathfrak{u}_{1}).    \Box

Step 3: The construction of 𝔲¯\bar{\mathfrak{u}} and 𝔲1\mathfrak{u}_{1} guarantees that π(𝔲)𝔣𝔲1\pi(\mathfrak{u})\sqsubset^{\mathfrak{f}}_{\partial}\mathfrak{u}_{1} if and only if π(𝔲)\pi(\mathfrak{u}) is short but 𝔲1\mathfrak{u}_{1} is long.

Example 7.8.

Continuing from Example 6.5 we get π(𝔲)\pi(\mathfrak{u}) is short whereas uncle 𝔲1\mathfrak{u}_{1} of 𝔲\mathfrak{u} is b1b^{-1}-long. Here indeed 𝔣(𝔵0;𝔲1)=GBDicBa\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u}_{1})=GBDicBa, 𝔣(𝔵0;π(𝔲),𝔲1)=cBa=𝔴~(𝔵0;π(𝔲))𝔠(𝔵0;π2(𝔲))\mathfrak{f}(\mathfrak{x}_{0};\pi(\mathfrak{u}),\mathfrak{u}_{1})=cBa=\newline \tilde{\mathfrak{w}}(\mathfrak{x}_{0};\pi(\mathfrak{u}))\mathfrak{c}(\mathfrak{x}_{0};\pi^{2}(\mathfrak{u})) and π(𝔲)𝔣1𝔲1\pi(\mathfrak{u})\sqsubset^{\mathfrak{f}}_{1}\mathfrak{u}_{1}.

If 𝔲1\mathfrak{u}_{1} and π(𝔲)\pi(\mathfrak{u}) are both long or short then 𝔲1𝔣π(𝔲)\mathfrak{u}_{1}\sqsubset^{\mathfrak{f}}_{\partial}\pi(\mathfrak{u}). We then show that 𝔣(𝔵0;π(𝔲),𝔲1)=𝔴~(𝔵0;𝔲)𝔠(𝔵0;π(𝔲))\mathfrak{f}(\mathfrak{x}_{0};\pi(\mathfrak{u}),\mathfrak{u}_{1})=\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u})\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u})).

If π(𝔲)\pi(\mathfrak{u}) is short but 𝔲1\mathfrak{u}_{1} is long, and hence π(𝔲)𝔣𝔲1\pi(\mathfrak{u})\sqsubset^{\mathfrak{f}}_{\partial}\mathfrak{u}_{1}, then in view of Remark 5.1 we have 𝔣(𝔵0;π(𝔲),𝔲1)=𝔴~(𝔵0;π(𝔲))𝔠(𝔵0;π2(𝔲))\mathfrak{f}(\mathfrak{x}_{0};\pi(\mathfrak{u}),\mathfrak{u}_{1})=\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\pi(\mathfrak{u}))\mathfrak{c}(\mathfrak{x}_{0};\pi^{2}(\mathfrak{u})). The three subcases, namely υ(π(𝔲))=1\upsilon(\pi(\mathfrak{u}))=1, υ(π(𝔲))=0\upsilon(\pi(\mathfrak{u}))=0 and υ(π(𝔲))=1\upsilon(\pi(\mathfrak{u}))=-1 are described in Corollaries 6.4, 6.17 and 6.13 respectively.

If π(𝔲)\pi(\mathfrak{u}) is long but 𝔲1\mathfrak{u}_{1} is short, and hence 𝔲1𝔣π(𝔲)\mathfrak{u}_{1}\sqsubset^{\mathfrak{f}}_{\partial}\pi(\mathfrak{u}), then Corollary 6.6 gives that υ(π(𝔲))=1\upsilon(\pi(\mathfrak{u}))=1, |π(𝔲)c|>0|\pi(\mathfrak{u})^{c}|>0, 𝔴~(𝔵0;𝔲1)=𝔷\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u}_{1})=\mathfrak{z} and |𝔴~(𝔵0;𝔲)|=0|\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u})|=0. Hence 𝔣(𝔵0;π(𝔲),𝔲1)=𝔴~(𝔵0;𝔲)𝔠(𝔵0;π(𝔲))\mathfrak{f}(\mathfrak{x}_{0};\pi(\mathfrak{u}),\mathfrak{u}_{1})=\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u})\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u})) in view of Remark 5.1.

Now we can analyse the four cases to complete the proof of Lemma 7.3.

Case I: 𝔲\mathfrak{u} is bb^{\partial}-long or (s,b)(s,b)^{-\partial}-long.

In this case, 𝔯1b(π(𝔲))=𝔴~(𝔵0;𝔲)\mathfrak{r}^{1}_{b}(\pi(\mathfrak{u}))=\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u}). Remark 6.18 gives that υ(π(𝔲))=1\upsilon(\pi(\mathfrak{u}))=1. Further Corollary 6.3 guarantees that π(𝔲)\pi(\mathfrak{u}) is normal. Moreover, 𝔯b1(π(𝔲))𝔴~\mathfrak{r}^{b}_{1}(\pi(\mathfrak{u}))\tilde{\mathfrak{w}} is a substring of π(𝔲)\pi(\mathfrak{u}), and hence 𝔯b1(π(𝔲))𝔴~\mathfrak{r}^{b}_{1}(\pi(\mathfrak{u}))\tilde{\mathfrak{w}} is an H-reduced string by [9, Proposition 5.11].

Step I.1: the string 𝔯b2(π(𝔲))𝔯b1(π(𝔲))𝔴~\mathfrak{r}^{b}_{2}(\pi(\mathfrak{u}))\mathfrak{r}^{b}_{1}(\pi(\mathfrak{u}))\tilde{\mathfrak{w}} is H-reduced.

Suppose not. Then there is a cyclic permutation 𝔟1\mathfrak{b}^{\prime}_{1} of a band 𝔟1\mathfrak{b}_{1} such that 𝔯b2(π(𝔲))𝔯b1(π(𝔲))𝔴~=𝔶2𝔟1𝔶1\mathfrak{r}^{b}_{2}(\pi(\mathfrak{u}))\mathfrak{r}^{b}_{1}(\pi(\mathfrak{u}))\tilde{\mathfrak{w}}=\mathfrak{y}^{\prime}_{2}\mathfrak{b}^{\prime}_{1}\mathfrak{y}^{\prime}_{1} and 𝔟1𝔶1H𝔶1\mathfrak{b}^{\prime}_{1}\mathfrak{y}^{\prime}_{1}\equiv_{H}\mathfrak{y}^{\prime}_{1}. Moreover there is a weak bridge 𝔰(π(𝔲))𝔲𝔟1\mathfrak{s}(\pi(\mathfrak{u}))\xrightarrow{\mathfrak{u}^{\prime\prime}}\mathfrak{b}_{1}. Since δ(𝔯b2(π(𝔲))𝔯b1(π(𝔲)))0\delta(\mathfrak{r}^{b}_{2}(\pi(\mathfrak{u}))\mathfrak{r}^{b}_{1}(\pi(\mathfrak{u})))\neq 0 and 𝔯b1(π(𝔲))𝔴~\mathfrak{r}^{b}_{1}(\pi(\mathfrak{u}))\tilde{\mathfrak{w}} is H-reduced, 𝔟1\mathfrak{b}^{\prime}_{1} contains at least one syllable of 𝔴~\tilde{\mathfrak{w}} and 𝔯b2(π(𝔲))\mathfrak{r}^{b}_{2}(\pi(\mathfrak{u})) each.

If 𝔯b2(π(𝔲))\mathfrak{r}^{b}_{2}(\pi(\mathfrak{u})) is not a substring of 𝔟1\mathfrak{b}^{\prime}_{1} then 𝔟\mathfrak{b} and 𝔟1\mathfrak{b}_{1} commute contradicting the domesticity of the algebra. Hence 𝔯b2(π(𝔲))\mathfrak{r}^{b}_{2}(\pi(\mathfrak{u})) is a right substring of 𝔟1\mathfrak{b}^{\prime}_{1} and |𝔶2|=0|\mathfrak{y}^{\prime}_{2}|=0. But then there is an abnormal weak bridge 𝔟1𝔲𝔰(𝔲)\mathfrak{b}_{1}\xrightarrow{\mathfrak{u}^{\prime}}\mathfrak{s}(\mathfrak{u}) such that 𝔲c=𝔯b1(π(𝔲)){\mathfrak{u}^{\prime}}^{c}=\mathfrak{r}^{b}_{1}(\pi(\mathfrak{u})).

Let 𝔟1,𝔟2\mathfrak{b}^{\prime\prime}_{1},\mathfrak{b}^{\prime}_{2} denote the cyclic permutations of 𝔟1\mathfrak{b}_{1} and 𝔰(𝔲)\mathfrak{s}(\mathfrak{u}) respectively such that 𝔯b1(π(𝔲))\mathfrak{r}^{b}_{1}(\pi(\mathfrak{u})) is a right substring of 𝔟1\mathfrak{b}^{\prime\prime}_{1} and 𝔟2𝔟1𝔯b1(π(𝔲))𝔴~\mathfrak{b}^{\prime}_{2}\mathfrak{b}^{\prime\prime}_{1}\mathfrak{r}^{b}_{1}(\pi(\mathfrak{u}))\tilde{\mathfrak{w}} is a string. If |𝔯b1(π(𝔲))|>0|\mathfrak{r}^{b}_{1}(\pi(\mathfrak{u}))|>0 then clearly 𝔟1𝔯b1(π(𝔲))𝔴~H𝔯b1(π(𝔲))𝔴~\mathfrak{b}^{\prime\prime}_{1}\mathfrak{r}^{b}_{1}(\pi(\mathfrak{u}))\tilde{\mathfrak{w}}\equiv_{H}\mathfrak{r}^{b}_{1}(\pi(\mathfrak{u}))\tilde{\mathfrak{w}}, which yields a factorization of π(𝔲)=𝔲H𝔲\pi(\mathfrak{u})=\mathfrak{u}^{\prime}\circ_{H}\mathfrak{u}^{\prime\prime}, a contradiction to π(𝔲)𝒬Ba\pi(\mathfrak{u})\in\mathcal{HQ}^{\mathrm{Ba}}. Hence |𝔯b1(π(𝔲))|=0|\mathfrak{r}^{b}_{1}(\pi(\mathfrak{u}))|=0.

In this case let 𝔟1=𝔯b2(π(𝔲))𝔶3\mathfrak{b}^{\prime}_{1}=\mathfrak{r}^{b}_{2}(\pi(\mathfrak{u}))\mathfrak{y}^{\prime}_{3} so that 𝔴~=𝔶3𝔶1\tilde{\mathfrak{w}}=\mathfrak{y}^{\prime}_{3}\mathfrak{y}^{\prime}_{1}. If δ(𝔶3)=0\delta(\mathfrak{y}^{\prime}_{3})=0 then clearly 𝔟1𝔴~H𝔴~\mathfrak{b}^{\prime\prime}_{1}\tilde{\mathfrak{w}}\equiv_{H}\tilde{\mathfrak{w}}. On the other hand if δ(𝔶3)0\delta(\mathfrak{y}^{\prime}_{3})\neq 0 then [9, Proposition 5.2] together with 𝔟1𝔶1H𝔶1\mathfrak{b}^{\prime}_{1}\mathfrak{y}^{\prime}_{1}\equiv_{H}\mathfrak{y}^{\prime}_{1} gives that 𝔟1𝔴~H𝔴~\mathfrak{b}^{\prime\prime}_{1}\tilde{\mathfrak{w}}\equiv_{H}\tilde{\mathfrak{w}}. Therefore in each case we conclude that π(𝔲)𝒬Ba\pi(\mathfrak{u})\notin\mathcal{HQ}^{\mathrm{Ba}}. This contradiction proves the claim.

Step I.2: Since 𝔯b2(π(𝔲))𝔯b1(π(𝔲))𝔴~\mathfrak{r}^{b}_{2}(\pi(\mathfrak{u}))\mathfrak{r}^{b}_{1}(\pi(\mathfrak{u}))\tilde{\mathfrak{w}} is H-reduced from the claim above, Lemma 7.4 yields a string 𝔴\mathfrak{w}^{\prime} and 𝔲¯𝒬¯Bai(𝔵0)\bar{\mathfrak{u}}\in\overline{\mathcal{HQ}}^{\mathrm{Ba}}_{i}(\mathfrak{x}_{0}) such that 𝔱(𝔲¯)𝔲¯=𝔴𝔯b2(π(𝔲))𝔯b1(π(𝔲))𝔴~\mathfrak{t}(\bar{\mathfrak{u}})\bar{\mathfrak{u}}=\mathfrak{w}^{\prime}\mathfrak{r}^{b}_{2}(\pi(\mathfrak{u}))\mathfrak{r}^{b}_{1}(\pi(\mathfrak{u}))\tilde{\mathfrak{w}}.

Now [9, Proposition 8.8, Theorem 8.9] yields an arch bridge 𝔲1\mathfrak{u}_{1} and a possibly trivial weak arch bridge 𝔲2\mathfrak{u}_{2} such that 𝔲¯=𝔲2H𝔲1\bar{\mathfrak{u}}=\mathfrak{u}_{2}\circ_{H}\mathfrak{u}_{1}. Lemma 7.6 guarantees that β(𝔲1)=β(π(𝔲))\beta(\mathfrak{u}_{1})=\beta(\pi(\mathfrak{u})) and that 𝔲1\mathfrak{u}_{1} is long if and only if π(𝔲)\pi(\mathfrak{u}) is long. Since |𝔯b2(π(𝔲))|>0|\mathfrak{r}^{b}_{2}(\pi(\mathfrak{u}))|>0 it readily follows that 𝔲1𝔣π(𝔲)\mathfrak{u}_{1}\sqsubset^{\mathfrak{f}}\pi(\mathfrak{u}).

Step I.3: We want to show that 𝔣(𝔵0;π(𝔲),𝔲1)=𝔴~(𝔵0;𝔲)𝔠(𝔵0;π(𝔲))\mathfrak{f}(\mathfrak{x}_{0};\pi(\mathfrak{u}),\mathfrak{u}_{1})=\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u})\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u})).

Suppose not. Then 𝔲2\mathfrak{u}_{2} is non-trivial. Let 𝔟1:=𝔱(𝔲1)\mathfrak{b}_{1}:=\mathfrak{t}(\mathfrak{u}_{1}). If for any cyclic permutation 𝔟1\mathfrak{b}^{\prime}_{1} of 𝔱(𝔲1)\mathfrak{t}(\mathfrak{u}_{1}), 𝔟1𝔣(𝔵0;π(𝔲),𝔲1)\mathfrak{b}^{\prime}_{1}\mathfrak{f}(\mathfrak{x}_{0};\pi(\mathfrak{u}),\mathfrak{u}_{1}) is not a word then 𝔲1\mathfrak{u}_{1} is normal and β(𝔲)𝔯b1(π(𝔲))π(𝔲)o\beta(\mathfrak{u})\mathfrak{r}^{b}_{1}(\pi(\mathfrak{u}))\pi(\mathfrak{u})^{o} is a substring of 𝔲1o\mathfrak{u}_{1}^{o} and hence the required identity holds. Let 𝔟1\mathfrak{b}^{\prime}_{1} be its cyclic permutation such that 𝔟1𝔣(𝔵0;π(𝔲),𝔲1)\mathfrak{b}^{\prime}_{1}\mathfrak{f}(\mathfrak{x}_{0};\pi(\mathfrak{u}),\mathfrak{u}_{1}) is a word.

There are two cases.

If 𝔯b1(π(𝔲))𝔠(𝔵0;π(𝔲))=𝔷𝔣(𝔵0;π(𝔲),𝔲1)\mathfrak{r}^{b}_{1}(\pi(\mathfrak{u}))\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u}))=\mathfrak{z}^{\prime}\mathfrak{f}(\mathfrak{x}_{0};\pi(\mathfrak{u}),\mathfrak{u}_{1}) for some string 𝔷\mathfrak{z}^{\prime} with |𝔷|>0|\mathfrak{z}^{\prime}|>0 then there are further two subcases.

  • 𝔷=𝔯b1(π(𝔲))𝔷1\mathfrak{z}^{\prime}=\mathfrak{r}^{b}_{1}(\pi(\mathfrak{u}))\mathfrak{z}^{\prime}_{1} with |𝔷1|>0|\mathfrak{z}^{\prime}_{1}|>0.

    If |𝔯b1(π(𝔲))|>0|\mathfrak{r}^{b}_{1}(\pi(\mathfrak{u}))|>0 then δ(γ𝔣(π(𝔲))γ¯¯b(π(𝔲)))=0\delta(\gamma^{\mathfrak{f}}(\pi(\mathfrak{u}))\bar{\bar{\gamma}}^{b}(\pi(\mathfrak{u})))=0 can be used to show that π(𝔲)=𝔲2H𝔲1\pi(\mathfrak{u})=\mathfrak{u}^{\prime}_{2}\circ_{H}\mathfrak{u}_{1} for some 𝔲2\mathfrak{u}^{\prime}_{2}, which is a contradiction to π(𝔲)𝒬Bai(𝔵0)\pi(\mathfrak{u})\in\mathcal{HQ}^{\mathrm{Ba}}_{i}(\mathfrak{x}_{0}).

    If |𝔯b1(π(𝔲))|=0|\mathfrak{r}^{b}_{1}(\pi(\mathfrak{u}))|=0 then since π(𝔲)𝒬Ba\pi(\mathfrak{u})\in\mathcal{HQ}^{\mathrm{Ba}}, we obtain that 𝔟𝔣α(π(𝔲))𝔷1𝔟1\mathfrak{b}^{\mathfrak{f}\alpha}(\pi(\mathfrak{u}))\mathfrak{z}^{\prime}_{1}\mathfrak{b}^{\prime}_{1} is not a string. Then there is a right substring 𝔴1\mathfrak{w}_{1} of 𝔟1\mathfrak{b}^{\prime}_{1} and a left substring 𝔴2\mathfrak{w}_{2} of 𝔟𝔣α(π(𝔲))𝔷1\mathfrak{b}^{\mathfrak{f}\alpha}(\pi(\mathfrak{u}))\mathfrak{z}^{\prime}_{1} such that 𝔴2𝔴1ρρ1\mathfrak{w}_{2}\mathfrak{w}_{1}\in\rho\cup\rho^{-1}. Since 𝔯b2(π(𝔲))𝔷1𝔟1\mathfrak{r}^{b}_{2}(\pi(\mathfrak{u}))\mathfrak{z}^{\prime}_{1}\mathfrak{b}^{\prime}_{1} is a string, we get that |𝔷1|<|𝔴2||\mathfrak{z}^{\prime}_{1}|<|\mathfrak{w}_{2}|. Similarly since 𝔟𝔣α(π(𝔲))𝔠(𝔵0;𝔲)\mathfrak{b}^{\mathfrak{f}\alpha}(\pi(\mathfrak{u}))\mathfrak{c}(\mathfrak{x}_{0};\mathfrak{u}) is a string, we get that 𝔴1\mathfrak{w}_{1} is not a right substring of 𝔣(𝔵0;π(𝔲),𝔲1)\mathfrak{f}(\mathfrak{x}_{0};\pi(\mathfrak{u}),\mathfrak{u}_{1}). As a consequence 𝔣(𝔵0;π(𝔲),𝔲1)\nequivH𝔟1𝔣(𝔵0;π(𝔲),𝔲1)\mathfrak{f}(\mathfrak{x}_{0};\pi(\mathfrak{u}),\mathfrak{u}_{1})\nequiv_{H}\mathfrak{b}^{\prime}_{1}\mathfrak{f}(\mathfrak{x}_{0};\pi(\mathfrak{u}),\mathfrak{u}_{1}). This is a contradiction to 𝔲¯=𝔲2H𝔲1\bar{\mathfrak{u}}=\mathfrak{u}_{2}\circ_{H}\mathfrak{u}_{1} and N(𝔟1,𝔲¯)=0N(\mathfrak{b}_{1},\bar{\mathfrak{u}})=0.

  • 𝔣(𝔵0;𝔲1,π(𝔲))=𝔷1𝔠(𝔵0;π(𝔲))\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u}_{1},\pi(\mathfrak{u}))=\mathfrak{z}^{\prime}_{1}\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u})) for |𝔷1|0|\mathfrak{z}^{\prime}_{1}|\geq 0.

    Here |𝔯b1(π(𝔲))|>|𝔷1||\mathfrak{r}^{b}_{1}(\pi(\mathfrak{u}))|>|\mathfrak{z}^{\prime}_{1}|. Then using 𝔲¯=𝔲2H𝔲1\bar{\mathfrak{u}}=\mathfrak{u}_{2}\circ_{H}\mathfrak{u}_{1} we get that 𝔟𝔣α(π(𝔲))𝔟1\mathfrak{b}^{\mathfrak{f}\alpha}(\pi(\mathfrak{u}))\mathfrak{b}^{\prime}_{1} and 𝔟1𝔟𝔣α(π(𝔲))\mathfrak{b}^{\prime}_{1}\mathfrak{b}^{\mathfrak{f}\alpha}(\pi(\mathfrak{u})) are strings, a contradiction to domesticity.

If 𝔷𝔯b1(π(𝔲))𝔠(𝔵0;π(𝔲))=𝔣(𝔵0;π(𝔲),𝔲1)\mathfrak{z}^{\prime}\mathfrak{r}^{b}_{1}(\pi(\mathfrak{u}))\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u}))=\mathfrak{f}(\mathfrak{x}_{0};\pi(\mathfrak{u}),\mathfrak{u}_{1}) for some string 𝔷\mathfrak{z}^{\prime} with |𝔷|>0|\mathfrak{z}^{\prime}|>0 then γ𝔰(𝔲)\gamma\in\mathfrak{s}(\mathfrak{u}) is also a syllable of 𝔟1\mathfrak{b}_{1}, where γ𝔯b1(π(𝔲))\gamma\mathfrak{r}^{b}_{1}(\pi(\mathfrak{u})) is a string, and thus θ(γ)=\theta(\gamma)=-\partial. Suppose γ\gamma^{\prime} is the syllable of 𝔟1\mathfrak{b}_{1} such that γγ\gamma\gamma^{\prime} is a substring of a cyclic permutation of 𝔟1\mathfrak{b}_{1}.

If θ(γ)=\theta(\gamma^{\prime})=\partial then 𝔟1\mathfrak{b}^{\prime}_{1} and 𝔟𝔣α(π(𝔲))\mathfrak{b}^{\mathfrak{f}\alpha}(\pi(\mathfrak{u})) share γγ\gamma\gamma^{\prime} with δ(γγ)=0\delta(\gamma\gamma^{\prime})=0, a contradiction to domesticity.

On the other hand if θ(γ)=\theta(\gamma^{\prime})=\partial then let 𝔟1\mathfrak{b}^{\prime\prime}_{1} be the cyclic permutation of 𝔟1\mathfrak{b}_{1} such that γ𝔟1\gamma\mathfrak{b}^{\prime\prime}_{1} is a string. To ensure domesticity there is a right substring 𝔴1\mathfrak{w}_{1} of 𝔟1\mathfrak{b}^{\prime\prime}_{1} and a left substring 𝔴2\mathfrak{w}_{2} of a cyclic permutation 𝔟\mathfrak{b}^{\prime\prime} of 𝔰(𝔲)\mathfrak{s}(\mathfrak{u}) such that |𝔴2||𝔴1|>0|\mathfrak{w}_{2}||\mathfrak{w}_{1}|>0 and 𝔴2𝔴1ρρ1\mathfrak{w}_{2}\mathfrak{w}_{1}\in\rho\cup\rho^{-1}. Since 𝔟𝔷𝔯b1(π(𝔲))𝔠(𝔵0;π(𝔲))\mathfrak{b}^{\prime\prime}\mathfrak{z}^{\prime}\mathfrak{r}^{b}_{1}(\pi(\mathfrak{u}))\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u})) is a string, 𝔴1\mathfrak{w}_{1} is not a right substring of 𝔷𝔯b1(π(𝔲))𝔠(𝔵0;π(𝔲))\mathfrak{z}^{\prime}\mathfrak{r}^{b}_{1}(\pi(\mathfrak{u}))\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u})). As a consequence, 𝔟1𝔷𝔯b1(π(𝔲))𝔠(𝔵0;π(𝔲))\nequivH𝔷𝔯b1(π(𝔲))𝔠(𝔵0;π(𝔲))\mathfrak{b}^{\prime\prime}_{1}\mathfrak{z}^{\prime}\mathfrak{r}^{b}_{1}(\pi(\mathfrak{u}))\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u}))\nequiv_{H}\mathfrak{z}^{\prime}\mathfrak{r}^{b}_{1}(\pi(\mathfrak{u}))\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u})), a contradiction to 𝔲¯=𝔲2H𝔲1\bar{\mathfrak{u}}=\mathfrak{u}_{2}\circ_{H}\mathfrak{u}_{1} and N(𝔟1,𝔲¯)=0N(\mathfrak{b}_{1},\bar{\mathfrak{u}})=0.

Example 7.9.

In the algebra Γ(v)\Gamma^{(v)} in Figure 10 the only bands are 𝔟1:=aCB\mathfrak{b}_{1}:=aCB, 𝔟2:=ihG\mathfrak{b}_{2}:=ihG. Choose 𝔵0:=1(𝔳1,i)\mathfrak{x}_{0}:=1_{(\mathfrak{v}_{1},i)} and 𝔲:=j\mathfrak{u}:=j so that π(𝔲)=iDe\pi(\mathfrak{u})=iDe and π2(𝔲)=a\pi^{2}(\mathfrak{u})=a.

v1{v_{1}}v2{v_{2}}v4{v_{4}}v6{v_{6}}v8{v_{8}}v10{v_{10}}v3{v_{3}}v5{v_{5}}v7{v_{7}}v9{v_{9}}a\scriptstyle{a}c\scriptstyle{c}e\scriptstyle{e}f\scriptstyle{f}d\scriptstyle{d}i\scriptstyle{i}h\scriptstyle{h}g\scriptstyle{g}b\scriptstyle{b}j\scriptstyle{j}k\scriptstyle{k}
Figure 10. Γ(v)\Gamma^{(v)} with ρ={eb,fd,dh,jg,kjih}\rho=\{eb,fd,dh,jg,kjih\}

Clearly θ(β(𝔲))=1\theta(\beta(\mathfrak{u}))=-1, 𝔲\mathfrak{u} is b1b^{-1}-long with 𝔯1b(π(𝔲))=𝔴~(𝔵0;𝔲)=i\mathfrak{r}^{1}_{b}(\pi(\mathfrak{u}))=\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u})=i, 𝔯2b(π(𝔲))=kj\mathfrak{r}^{2}_{b}(\pi(\mathfrak{u}))=kj and 𝔴~=De\tilde{\mathfrak{w}}=De. It is readily verified that 𝔲¯:=kjiDe\bar{\mathfrak{u}}:=kjiDe is H-reduced and 𝔲1:=𝔲¯\mathfrak{u}_{1}:=\bar{\mathfrak{u}} is the uncle of 𝔲\mathfrak{u} such that 𝔣(𝔵0;π(𝔲),𝔲1)=iDea\mathfrak{f}(\mathfrak{x}_{0};\pi(\mathfrak{u}),\mathfrak{u}_{1})=iDea.

Case II: 𝔲\mathfrak{u} is bb^{-\partial}-long.

In view of Remark 6.18 we have three subcases depending on the value of υ(π(𝔲))\upsilon(\pi(\mathfrak{u})). Here we show the proof only for the case υ(π(𝔲))=1\upsilon(\pi(\mathfrak{u}))=1; the proofs for the remaining two cases are obvious modifications.

In this case |𝔴~(𝔵0;𝔲)||𝔯b1(π(𝔲))||\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u})|\neq|\mathfrak{r}^{b}_{1}(\pi(\mathfrak{u}))|. If |𝔴~(𝔵0;𝔲)|>|𝔯b1(π(𝔲))||\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u})|>|\mathfrak{r}^{b}_{1}(\pi(\mathfrak{u}))| then 𝔶H𝔟𝔶𝔶\mathfrak{y}\equiv_{H}\mathfrak{b}_{\mathfrak{y}}\mathfrak{y}, and hence (𝔟,𝔶)(\mathfrak{b},\mathfrak{y}) is short, a contradiction to our assumption. Therefore |𝔴~(𝔵0;𝔲)|<|𝔯b1(π(𝔲))||\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u})|<|\mathfrak{r}^{b}_{1}(\pi(\mathfrak{u}))|. As a consequence |𝔯b1(π(𝔲))|>0|\mathfrak{r}^{b}_{1}(\pi(\mathfrak{u}))|>0. Moreover, δ(γ𝔣(π(𝔲))γ¯¯b(π(𝔲)))=0\delta(\gamma^{\mathfrak{f}}(\pi(\mathfrak{u}))\bar{\bar{\gamma}}^{b}(\pi(\mathfrak{u})))=0. Since 𝔴~(𝔵0;𝔲)𝔴~\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u})\tilde{\mathfrak{w}} is a left substring of π(𝔲)\pi(\mathfrak{u}), it is an H-reduced string by [9, Proposition 5.11].

Step II.1: β(𝔲)𝔴~(𝔵0;𝔲)𝔴~\beta(\mathfrak{u})\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u})\tilde{\mathfrak{w}} is H-reduced.

If not then there is a cyclic permutation 𝔟1\mathfrak{b}^{\prime}_{1} of a band 𝔟1\mathfrak{b}_{1} such that β(𝔲)𝔴~(𝔵0;𝔲)𝔴~=𝔟1𝔶1\beta(\mathfrak{u})\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u})\tilde{\mathfrak{w}}=\mathfrak{b}^{\prime}_{1}\mathfrak{y}^{\prime}_{1}. Suppose 𝔟1β(𝔲)=β(𝔲)𝔟1\mathfrak{b}^{\prime}_{1}\beta(\mathfrak{u})=\beta(\mathfrak{u})\mathfrak{b}^{\prime\prime}_{1} and β\beta^{\prime} is the last syllable of 𝔟1\mathfrak{b}^{\prime\prime}_{1} that is not in 𝔴~(𝔵0;𝔲)\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u}). Then clearly θ(β)=θ(β(𝔲))=\theta(\beta^{\prime})=\theta(\beta(\mathfrak{u}))=\partial, and hence it is readily verified that 𝔟\mathfrak{b} and 𝔟1\mathfrak{b}_{1} commute, a contradiction to the domesticity of the algebra. This proves the claim.

Step II.2: Then we can argue as in Case I that there is 𝔴\mathfrak{w}^{\prime} such that 𝔱(𝔲¯)𝔲¯:=𝔴β(𝔲)𝔴~(𝔵0;𝔲)𝔴~𝒬¯Ba\mathfrak{t}(\bar{\mathfrak{u}})\bar{\mathfrak{u}}:=\mathfrak{w}^{\prime}\beta(\mathfrak{u})\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u})\tilde{\mathfrak{w}}\in\overline{\mathcal{HQ}}^{\mathrm{Ba}}, and that 𝔲¯=𝔲2H𝔲1\bar{\mathfrak{u}}=\mathfrak{u}_{2}\circ_{H}\mathfrak{u}_{1} for some 𝔲1𝒬Bai(𝔵0)\mathfrak{u}_{1}\in\mathcal{HQ}^{\mathrm{Ba}}_{i}(\mathfrak{x}_{0}) and possibly trivial 𝔲2\mathfrak{u}_{2}.

Step II.3: Assuming that 𝔲1\mathfrak{u}_{1} and π(𝔲)\pi(\mathfrak{u}) are either both long or both short, we need to show that 𝔣(𝔵0;π(𝔲),𝔲1)=𝔴~(𝔵0;𝔲)𝔠(𝔵0;π(𝔲))\mathfrak{f}(\mathfrak{x}_{0};\pi(\mathfrak{u}),\mathfrak{u}_{1})=\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u})\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u})).

Suppose not. Then 𝔲2\mathfrak{u}_{2} is non-trivial. Let 𝔟1:=𝔱(𝔲1)\mathfrak{b}_{1}:=\mathfrak{t}(\mathfrak{u}_{1}). First we argue that there is a cyclic permutation 𝔟1\mathfrak{b}^{\prime}_{1} of 𝔱(𝔲1)\mathfrak{t}(\mathfrak{u}_{1}) such that 𝔟1𝔣(𝔵0;π(𝔲),𝔲1)\mathfrak{b}^{\prime}_{1}\mathfrak{f}(\mathfrak{x}_{0};\pi(\mathfrak{u}),\mathfrak{u}_{1}) is a string. If not then there is a substring 𝔶¯\bar{\mathfrak{y}} of 𝔲¯\bar{\mathfrak{u}} such that 𝔟1𝔶¯𝔣(𝔵0;π(𝔲),𝔲1)\mathfrak{b}^{\prime\prime}_{1}\bar{\mathfrak{y}}\mathfrak{f}(\mathfrak{x}_{0};\pi(\mathfrak{u}),\mathfrak{u}_{1}) is a string for some cyclic permutation 𝔟1\mathfrak{b}^{\prime\prime}_{1} of 𝔱(𝔲1)\mathfrak{t}(\mathfrak{u}_{1}). Therefore 𝔶¯𝔴~(𝔵0;𝔲)𝔴~\bar{\mathfrak{y}}\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u})\tilde{\mathfrak{w}} is a string and the required identity follows.

There are two cases.

If 𝔣(𝔵0;π(𝔲),𝔲1)\mathfrak{f}(\mathfrak{x}_{0};\pi(\mathfrak{u}),\mathfrak{u}_{1}) is a proper left substring of 𝔴~(𝔵0;𝔲)𝔠(𝔵0;π(𝔲))\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u})\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u})) then 𝔣(𝔵0;π(𝔲),𝔲1)\mathfrak{f}(\mathfrak{x}_{0};\pi(\mathfrak{u}),\mathfrak{u}_{1}) is necessarily a proper left substring of 𝔠(𝔵0;π(𝔲))\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u})) for otherwise 𝔟1\mathfrak{b}_{1} commutes with 𝔱(π(𝔲))\mathfrak{t}(\pi(\mathfrak{u})). However if 𝔣(𝔵0;π(𝔲),𝔲1)\mathfrak{f}(\mathfrak{x}_{0};\pi(\mathfrak{u}),\mathfrak{u}_{1}) is a proper left substring of 𝔠(𝔵0;π(𝔲))\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u})) then the proof of impossibility is similar to the proof of the first subcase of the first case of the proof of the same identity from Case I.

If 𝔴~(𝔵0;𝔲)𝔠(𝔵0;π(𝔲))\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u})\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u})) is a proper left substring of 𝔣(𝔵0;π(𝔲),𝔲1)\mathfrak{f}(\mathfrak{x}_{0};\pi(\mathfrak{u}),\mathfrak{u}_{1}). Then the first syllable of 𝔱(π(𝔲))\mathfrak{t}(\pi(\mathfrak{u})) after 𝔴~(𝔵0;𝔲)\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u}) is also a syllable of 𝔟1\mathfrak{b}_{1}. To ensure domesticity, 𝔣(𝔵0;π(𝔲),𝔲1)\mathfrak{f}(\mathfrak{x}_{0};\pi(\mathfrak{u}),\mathfrak{u}_{1}) is a left substring of 𝔯b1(π(𝔲))𝔠(𝔵0;π(𝔲))\mathfrak{r}^{b}_{1}(\pi(\mathfrak{u}))\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u})). There are two subcases:

  • If 𝔯~b(π(𝔲))\tilde{\mathfrak{r}}^{b}(\pi(\mathfrak{u})) is not a substring of 𝔟1\mathfrak{b}_{1} then for the existence of an abnormal weak bridge 𝔟1𝔲π(𝔲)\mathfrak{b}_{1}\xrightarrow{\mathfrak{u}^{\prime\prime}}\pi(\mathfrak{u}), it is necessary to have 𝔣(𝔵0;π(𝔲),𝔲1)=𝔯b1(π(𝔲))𝔠(𝔵0;π(𝔲))\mathfrak{f}(\mathfrak{x}_{0};\pi(\mathfrak{u}),\mathfrak{u}_{1})=\mathfrak{r}^{b}_{1}(\pi(\mathfrak{u}))\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u})) and that 𝔯b2(π(𝔲))\mathfrak{r}^{b}_{2}(\pi(\mathfrak{u})) is a left substring of 𝔟1\mathfrak{b}^{\prime}_{1}. Then it is readily verified that π(𝔲)=𝔲H𝔲1\pi(\mathfrak{u})=\mathfrak{u}^{\prime\prime}\circ_{H}\mathfrak{u}_{1}, a contradiction to π(𝔲)𝒬Ba\pi(\mathfrak{u})\in\mathcal{HQ}^{\mathrm{Ba}}.

  • If 𝔯~b(π(𝔲))\tilde{\mathfrak{r}}^{b}(\pi(\mathfrak{u})) is a substring of 𝔟1\mathfrak{b}_{1} then consider the cyclic permutation 𝔟1\mathfrak{b}^{\prime\prime\prime}_{1} of 𝔟1\mathfrak{b}_{1} such that β(𝔲2)𝔟1\beta(\mathfrak{u}_{2})\mathfrak{b}^{\prime\prime\prime}_{1} is a string. To ensure π(𝔲)𝒬Ba\pi(\mathfrak{u})\in\mathcal{HQ}^{\mathrm{Ba}} there is a right substring 𝔶1\mathfrak{y}_{1} of 𝔟1\mathfrak{b}^{\prime\prime\prime}_{1} and a substring 𝔶2\mathfrak{y}_{2} of 𝔱(π(𝔲))\mathfrak{t}(\pi(\mathfrak{u})) such that 𝔶2𝔶1ρρ1\mathfrak{y}_{2}\mathfrak{y}_{1}\in\rho\cup\rho^{-1}. But in this case 𝔲2\mathfrak{u}_{2} is long and hence N(𝔱(𝔲1),𝔲¯)=1N(\mathfrak{t}(\mathfrak{u}_{1}),\bar{\mathfrak{u}})=1 which is a contradiction to N(𝔱(𝔲1),𝔴~(𝔵0;𝔲)𝔴~)=0N(\mathfrak{t}(\mathfrak{u}_{1}),\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u})\tilde{\mathfrak{w}})=0.

Example 7.10.

Consider the algebra Γ~(ii)\tilde{\Gamma}^{(ii)} that differs in its presentation from Γ(ii)\Gamma^{(ii)} from Figure 7 in that dbghdbgh is removed from ρ\rho. Keeping 𝔵0\mathfrak{x}_{0} and 𝔲\mathfrak{u} same as in Example 6.5 we get θ(β(𝔲))=1\theta(\beta(\mathfrak{u}))=1, π(𝔲)\pi(\mathfrak{u}) is short and 𝔲\mathfrak{u} is b1b^{-1}-long with 𝔴~(𝔵0;𝔲)=1(𝔳3,i)\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u})=1_{(\mathfrak{v}_{3},i)}, 𝔯1b(π(𝔲))=c\mathfrak{r}^{1}_{b}(\pi(\mathfrak{u}))=c, 𝔯2b(π(𝔲))=i\mathfrak{r}^{2}_{b}(\pi(\mathfrak{u}))=i and 𝔴~=BD\tilde{\mathfrak{w}}=BD. It is readily verified that 𝔲¯:=HGBD\bar{\mathfrak{u}}:=HGBD is H-reduced and 𝔲1:=𝔲¯\mathfrak{u}_{1}:=\bar{\mathfrak{u}} is the short uncle of 𝔲\mathfrak{u} such that 𝔣(𝔵0;π(𝔲),𝔲1)=Ba=𝔴~(𝔵0;𝔲)𝔠(𝔵0;π(𝔲))\mathfrak{f}(\mathfrak{x}_{0};\pi(\mathfrak{u}),\mathfrak{u}_{1})=Ba=\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u})\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u})).

Case III: 𝔲\mathfrak{u} is ss^{\partial}-long or (b,s)(b,s)^{-\partial}-long. In this case 𝔴~(𝔵0;𝔲)=𝔯s1(π(𝔲))\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u})=\mathfrak{r}^{s}_{1}(\pi(\mathfrak{u})).

Step III.1: β(𝔲)𝔴~(𝔵0;𝔲)𝔴~\beta(\mathfrak{u})\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u})\tilde{\mathfrak{w}} is H-reduced.

First we argue that 𝔴~(𝔵0;𝔲)𝔴~\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u})\tilde{\mathfrak{w}} is band-free. Suppose not. Since 𝔴~\tilde{\mathfrak{w}} is a left substring of π(𝔲)\pi(\mathfrak{u}), 𝔴~\tilde{\mathfrak{w}} is H-reduced. Thus |𝔴~||𝔴~(𝔵0;𝔲)|>0|\tilde{\mathfrak{w}}||\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u})|>0 and hence γ𝔣(π(𝔲))𝔴~(𝔵0;𝔲)\gamma^{\mathfrak{f}}(\pi(\mathfrak{u}))\in\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u}), and there is a cyclic permutation 𝔟¯1\bar{\mathfrak{b}}^{\prime}_{1} of a band 𝔟¯1\bar{\mathfrak{b}}_{1} such that 𝔴~(𝔵0;𝔲)𝔴~=𝔶2𝔟¯1𝔶1\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u})\tilde{\mathfrak{w}}=\mathfrak{y}^{\prime}_{2}\bar{\mathfrak{b}}^{\prime}_{1}\mathfrak{y}^{\prime}_{1}, the first syllable of 𝔶2\mathfrak{y}^{\prime}_{2} is not a syllable of 𝔟¯1\bar{\mathfrak{b}}_{1} and 𝔶1\mathfrak{y}^{\prime}_{1} is a proper left substring of 𝔴~\tilde{\mathfrak{w}}. Since γ𝔣(π(𝔲))\gamma^{\mathfrak{f}}(\pi(\mathfrak{u})) is a common syllable of both the bands, in view of [9, Proposition 4.3] there is an abnormal weak bridge from 𝔰(𝔲)\mathfrak{s}(\mathfrak{u}) to 𝔟¯1\bar{\mathfrak{b}}_{1} and |𝔶2|>0|\mathfrak{y}^{\prime}_{2}|>0. On the other hand since 𝔟𝔣α(π(𝔲))𝔴~\mathfrak{b}^{\mathfrak{f}\alpha}(\pi(\mathfrak{u}))\tilde{\mathfrak{w}} is a string and 𝔴~(𝔵0;𝔲)\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u}) is a left substring of 𝔟𝔣α(π(𝔲))\mathfrak{b}^{\mathfrak{f}\alpha}(\pi(\mathfrak{u})) there is a weak bridge from 𝔟¯1\bar{\mathfrak{b}}_{1} to 𝔰(𝔲)\mathfrak{s}(\mathfrak{u}), a contradiction to the domesticity of the algebra. This completes the proof that 𝔴~(𝔵0;𝔲)𝔴~\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u})\tilde{\mathfrak{w}} is band-free.

If β(𝔲)𝔴~(𝔵0;𝔲)𝔴~\beta(\mathfrak{u})\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u})\tilde{\mathfrak{w}} is not band-free then in view of the above paragraph there is a cyclic permutation 𝔟1\mathfrak{b}^{\prime}_{1} of a band 𝔟1\mathfrak{b}_{1} such that β(𝔲)𝔴~(𝔵0;𝔲)𝔴~=𝔟1𝔶1\beta(\mathfrak{u})\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u})\tilde{\mathfrak{w}}=\mathfrak{b}^{\prime}_{1}\mathfrak{y}^{\prime}_{1}. Moreover since δ(β(𝔲)𝔯s1(π(𝔲))𝔯~s(π(𝔲)))0\delta(\beta(\mathfrak{u})\mathfrak{r}^{s}_{1}(\pi(\mathfrak{u}))\tilde{\mathfrak{r}}^{s}(\pi(\mathfrak{u})))\neq 0 we conclude that β(𝔲)𝔯s1(π(𝔲))γ¯s(π(𝔲))\beta(\mathfrak{u})\mathfrak{r}^{s}_{1}(\pi(\mathfrak{u}))\bar{\gamma}^{s}(\pi(\mathfrak{u})) is a substring 𝔟1\mathfrak{b}^{\prime}_{1}. Since θ(γ¯¯s(π(𝔲)))=θ(γ𝔣(π(𝔲)))\theta(\bar{\bar{\gamma}}^{s}(\pi(\mathfrak{u})))=-\theta(\gamma^{\mathfrak{f}}(\pi(\mathfrak{u}))) it is readily verified that 𝔟1\mathfrak{b}_{1} and 𝔰(𝔲)\mathfrak{s}(\mathfrak{u}) commute, a contradiction to domesticity that proves the claim.

Step III.2: We can argue as in Case I that there is 𝔴\mathfrak{w}^{\prime} such that 𝔱(𝔲¯)𝔲¯:=𝔴β(𝔲)𝔴~(𝔵0;𝔲)𝔴~𝒬¯Ba\mathfrak{t}(\bar{\mathfrak{u}})\bar{\mathfrak{u}}:=\mathfrak{w}^{\prime}\beta(\mathfrak{u})\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u})\tilde{\mathfrak{w}}\in\overline{\mathcal{HQ}}^{\mathrm{Ba}}, and that 𝔲¯=𝔲2H𝔲1\bar{\mathfrak{u}}=\mathfrak{u}_{2}\circ_{H}\mathfrak{u}_{1} for some 𝔲1𝒬Bai(𝔵0)\mathfrak{u}_{1}\in\mathcal{HQ}^{\mathrm{Ba}}_{i}(\mathfrak{x}_{0}) and possibly trivial 𝔲2\mathfrak{u}_{2}.

Step III.3: Assuming that 𝔲1\mathfrak{u}_{1} and π(𝔲)\pi(\mathfrak{u}) are either both long or both short, we need to show that 𝔣(𝔵0;π(𝔲),𝔲1)=𝔴~(𝔵0;𝔲)𝔠(𝔵0;π(𝔲))\mathfrak{f}(\mathfrak{x}_{0};\pi(\mathfrak{u}),\mathfrak{u}_{1})=\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u})\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u})).

Suppose not. Then 𝔲2\mathfrak{u}_{2} is non-trivial. Let 𝔟1:=𝔱(𝔲1)\mathfrak{b}_{1}:=\mathfrak{t}(\mathfrak{u}_{1}). As in Case II, we can show that there is a cyclic permutation 𝔟1\mathfrak{b}^{\prime}_{1} of 𝔟1\mathfrak{b}_{1} such that 𝔟1𝔣(𝔵0;π(𝔲),𝔲1)\mathfrak{b}^{\prime}_{1}\mathfrak{f}(\mathfrak{x}_{0};\pi(\mathfrak{u}),\mathfrak{u}_{1}) is a string.

There are two cases.

If 𝔣(𝔵0;π(𝔲),𝔲1)\mathfrak{f}(\mathfrak{x}_{0};\pi(\mathfrak{u}),\mathfrak{u}_{1}) is a proper left substring of 𝔴~(𝔵0;𝔲)𝔠(𝔵0;π(𝔲))\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u})\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u})) then there are two subcases:

  • 𝔠(𝔵0;π(𝔲))=𝔷𝔣(𝔵0;π(𝔲),𝔲1)\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u}))=\mathfrak{z}^{\prime}\mathfrak{f}(\mathfrak{x}_{0};\pi(\mathfrak{u}),\mathfrak{u}_{1}) for a string 𝔷\mathfrak{z}^{\prime}. If |𝔷|=0|\mathfrak{z}^{\prime}|=0 then our assumption gives that |𝔴~(𝔵0;𝔲)|>0|\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u})|>0. Then 𝔯s1(π(𝔲))𝔷𝔟1𝔣(𝔵0;π(𝔲),𝔲1)\mathfrak{r}^{s}_{1}(\pi(\mathfrak{u}))\mathfrak{z}^{\prime}\mathfrak{b}^{\prime}_{1}\mathfrak{f}(\mathfrak{x}_{0};\pi(\mathfrak{u}),\mathfrak{u}_{1}) is a string since 𝔲¯=𝔲2H𝔲1\bar{\mathfrak{u}}=\mathfrak{u}_{2}\circ_{H}\mathfrak{u}_{1}. Hence there is a weak bridge 𝔱(𝔲1)𝔲𝔰(𝔲)\mathfrak{t}(\mathfrak{u}_{1})\xrightarrow{\mathfrak{u}^{\prime}}\mathfrak{s}(\mathfrak{u}). Since π(𝔲)𝒬Ba\pi(\mathfrak{u})\in\mathcal{HQ}^{\mathrm{Ba}}, we get that π(𝔲)𝔲H𝔲1\pi(\mathfrak{u})\neq\mathfrak{u}^{\prime}\circ_{H}\mathfrak{u}_{1}. Hence N(𝔱(𝔲1),𝔲H𝔲1)=1N(\mathfrak{t}(\mathfrak{u}_{1}),\mathfrak{u}^{\prime}\circ_{H}\mathfrak{u}_{1})=1. But then it is readily verified that N(𝔱(𝔲1),𝔲¯)=1N(\mathfrak{t}(\mathfrak{u}_{1}),\bar{\mathfrak{u}})=1, a contradiction to the fact that 𝔴~\tilde{\mathfrak{w}} is band-free.

  • 𝔣(𝔵0;π(𝔲),𝔲1)=𝔷𝔠(𝔵0;π(𝔲))\mathfrak{f}(\mathfrak{x}_{0};\pi(\mathfrak{u}),\mathfrak{u}_{1})=\mathfrak{z}^{\prime}\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u})) for a string 𝔷\mathfrak{z}^{\prime} with 0<|𝔷|<|𝔴~(𝔵0;𝔲)|0<|\mathfrak{z}^{\prime}|<|\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u})|. Then γ𝔣(π(𝔲))\gamma^{\mathfrak{f}}(\pi(\mathfrak{u})) is the first syllable of 𝔷\mathfrak{z}^{\prime}. Let β\beta be the syllable of 𝔱(𝔲1)\mathfrak{t}(\mathfrak{u}_{1}) such that β𝔣(𝔵0;π(𝔲),𝔲1)\beta\mathfrak{f}(\mathfrak{x}_{0};\pi(\mathfrak{u}),\mathfrak{u}_{1}) is a string. Clearly θ(β)=δ(𝔯s1(π(𝔲)))=θ(γ𝔣(π(𝔲)))=θ(γ¯¯s(π(𝔲)))\theta(\beta)=-\delta(\mathfrak{r}^{s}_{1}(\pi(\mathfrak{u})))=-\theta(\gamma^{\mathfrak{f}}(\pi(\mathfrak{u})))=\theta(\bar{\bar{\gamma}}^{s}(\pi(\mathfrak{u}))) which implies that there is an abnormal weak arch bridge 𝔲\mathfrak{u}^{\prime} from 𝔰(𝔲)\mathfrak{s}(\mathfrak{u}) to 𝔱(𝔲1)\mathfrak{t}(\mathfrak{u}_{1}) with |𝔲c|>0|\mathfrak{u}^{\prime c}|>0. Since θ(β(𝔲))=δ(𝔲c)\theta(\beta(\mathfrak{u}^{\prime}))=-\delta(\mathfrak{u}^{\prime c}), we get 𝔟β(𝔲)𝔲c𝔴~H𝔲c𝔴~\mathfrak{b}^{\beta}(\mathfrak{u}^{\prime})\mathfrak{u}^{\prime c}\tilde{\mathfrak{w}}\equiv_{H}\mathfrak{u}^{\prime c}\tilde{\mathfrak{w}} and hence 𝔲1=𝔲Hπ(𝔲)\mathfrak{u}_{1}=\mathfrak{u}^{\prime}\circ_{H}\pi(\mathfrak{u}), a contradiction to 𝔲1𝒬Bai(𝔵0)\mathfrak{u}_{1}\in\mathcal{HQ}^{\mathrm{Ba}}_{i}(\mathfrak{x}_{0}).

If 𝔴~(𝔵0;𝔲)𝔠(𝔵0;π(𝔲))\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u})\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u})) is a proper left substring of 𝔣(𝔵0;π(𝔲),𝔲1)\mathfrak{f}(\mathfrak{x}_{0};\pi(\mathfrak{u}),\mathfrak{u}_{1}) then to ensure domesticity no syllable of 𝔴~(𝔵0;𝔲)𝔴~\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u})\tilde{\mathfrak{w}} is a syllable of 𝔟1\mathfrak{b}^{\prime}_{1}. On the other hand since 𝔲¯=𝔲2H𝔲1\bar{\mathfrak{u}}=\mathfrak{u}_{2}\circ_{H}\mathfrak{u}_{1}, we conclude that 𝔟1𝔴~(𝔵0;𝔲)𝔴~\mathfrak{b}^{\prime}_{1}\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u})\tilde{\mathfrak{w}} is a string with at least one syllable common between 𝔱(𝔲1)\mathfrak{t}(\mathfrak{u}_{1}) and 𝔰(𝔲)\mathfrak{s}(\mathfrak{u}). Then it is readily verified that 𝔲1=𝔲Hπ(𝔲)\mathfrak{u}_{1}=\mathfrak{u}^{\prime\prime}\circ_{H}\pi(\mathfrak{u}) for an abnormal weak bridge 𝔰(𝔲)𝔲𝔱(𝔲1)\mathfrak{s}(\mathfrak{u})\xrightarrow{\mathfrak{u}^{\prime\prime}}\mathfrak{t}(\mathfrak{u}_{1}), a contradiction to 𝔲1𝒬Ba\mathfrak{u}_{1}\in\mathcal{HQ}^{\mathrm{Ba}}.

Example 7.11.

In the algebra Γ(vi)\Gamma^{(vi)} in Figure 11, 𝔟1:=bK\mathfrak{b}_{1}:=bK, 𝔟2:=feG\mathfrak{b}_{2}:=feG and their inverses are the only bands. Choose 𝔵0:=1(𝔳1,i)\mathfrak{x}_{0}:=1_{(\mathfrak{v}_{1},i)} and 𝔲:=ji\mathfrak{u}:=ji so that π(𝔲):=feca\pi(\mathfrak{u}):=feca and π2(𝔲):=b\pi^{2}(\mathfrak{u}):=b. Here θ(β(𝔲))=1\theta(\beta(\mathfrak{u}))=-1, π(𝔲)\pi(\mathfrak{u}) is short and 𝔲\mathfrak{u} is s1s^{-1}-long with 𝔴~(𝔵0;𝔲)=𝔯1s(π(𝔲))=fe\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u})=\mathfrak{r}^{1}_{s}(\pi(\mathfrak{u}))=fe, 𝔯2s(π(𝔲))=ji\mathfrak{r}^{2}_{s}(\pi(\mathfrak{u}))=ji and 𝔴~=ca\tilde{\mathfrak{w}}=ca. It is readily verified that 𝔲¯:=ifeca\bar{\mathfrak{u}}:=ifeca is H-reduced and 𝔲1:=𝔲¯\mathfrak{u}_{1}:=\bar{\mathfrak{u}} is the short uncle of 𝔲\mathfrak{u} such that 𝔣(𝔵0;π(𝔲),𝔲1)=𝔴~(𝔵0;𝔲)𝔠(𝔵0;π(𝔲))=fecab\mathfrak{f}(\mathfrak{x}_{0};\pi(\mathfrak{u}),\mathfrak{u}_{1})=\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u})\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u}))=fecab.

v2{v_{2}}v3{v_{3}}v5{v_{5}}v8{v_{8}}v9{v_{9}}v1{v_{1}}v4{v_{4}}v7{v_{7}}v6{v_{6}}v10{v_{10}}a\scriptstyle{a}c\scriptstyle{c}g\scriptstyle{g}e\scriptstyle{e}i\scriptstyle{i}j\scriptstyle{j}b\scriptstyle{b}k\scriptstyle{k}d\scriptstyle{d}h\scriptstyle{h}f\scriptstyle{f}
Figure 11. Γ(vi)\Gamma^{(vi)} with ρ={ak,cd,gc,fh,ig,jifeca}\rho=\{ak,cd,gc,fh,ig,jifeca\}

Case IV: 𝔲\mathfrak{u} is ss^{-\partial}-long.

An argument similar to Case II guarantees that |𝔴~(𝔵0;𝔲)|<|𝔯s1(π(𝔲))||\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u})|<|\mathfrak{r}^{s}_{1}(\pi(\mathfrak{u}))|.

Step IV.1: β(𝔲)𝔴~(𝔵0;𝔲)𝔴~\beta(\mathfrak{u})\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u})\tilde{\mathfrak{w}} is H-reduced.

First we show that 𝔴~(𝔵0;𝔲)𝔴~\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u})\tilde{\mathfrak{w}} is band-free. Recall that 𝔴~\tilde{\mathfrak{w}} is a left substring of π(𝔲)\pi(\mathfrak{u}) and hence H-reduced by [9, Remark 5.6]. If 𝔴~(𝔵0;𝔲)𝔴~\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u})\tilde{\mathfrak{w}} is not band-free then |𝔴~||𝔴~(𝔵0;𝔲)|>0|\tilde{\mathfrak{w}}||\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u})|>0. In this case there is a cyclic permutation 𝔟¯1\bar{\mathfrak{b}}^{\prime}_{1} of a band 𝔟¯1\bar{\mathfrak{b}}_{1} such that 𝔴~(𝔵0;𝔲)𝔴~=𝔶2𝔟¯1𝔶1\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u})\tilde{\mathfrak{w}}=\mathfrak{y}^{\prime}_{2}\bar{\mathfrak{b}}^{\prime}_{1}\mathfrak{y}^{\prime}_{1}. Since δ(𝔴~(𝔵0;𝔲)𝔯~s(π(𝔲)))0\delta(\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u})\tilde{\mathfrak{r}}^{s}(\pi(\mathfrak{u})))\neq 0 and δ(γ𝔣(π(𝔲))γ¯¯s(π(𝔲)))=0\delta(\gamma^{\mathfrak{f}}(\pi(\mathfrak{u}))\bar{\bar{\gamma}}^{s}(\pi(\mathfrak{u})))=0 there is a weak bridge from 𝔰(𝔲)\mathfrak{s}(\mathfrak{u}) to 𝔟¯1\bar{\mathfrak{b}}_{1}. Moreover since 𝔟𝔣α(π(𝔲))𝔴~\mathfrak{b}^{\mathfrak{f}\alpha}(\pi(\mathfrak{u}))\tilde{\mathfrak{w}} is a string and 𝔴~(𝔵0;𝔲)\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u}) is a left substring of 𝔟𝔣α(π(𝔲))\mathfrak{b}^{\mathfrak{f}\alpha}(\pi(\mathfrak{u})) there is a weak bridge from 𝔟¯1\bar{\mathfrak{b}}_{1} to 𝔰(𝔲)\mathfrak{s}(\mathfrak{u}), a contradiction to the domesticity of the algebra. In fact the same argument also proves that β(𝔲)𝔴~(𝔵0;𝔲)𝔴~\beta(\mathfrak{u})\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u})\tilde{\mathfrak{w}} is also band-free. Hence it is H-reduced.

Step IV.2: Lemma 7.4 ensures that there is a 𝔴\mathfrak{w}^{\prime} such that 𝔱(𝔲¯)𝔲¯:=𝔴β(𝔲)𝔴~(𝔵0;𝔲)𝔴~\mathfrak{t}(\bar{\mathfrak{u}})\bar{\mathfrak{u}}:=\mathfrak{w}^{\prime}\beta(\mathfrak{u})\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u})\tilde{\mathfrak{w}} is either a maximal reverse arch bridge or a weak arch-bridge from 𝔰(π(𝔲))\mathfrak{s}(\pi(\mathfrak{u})). Without loss we assume that 𝔲¯\bar{\mathfrak{u}} is weak and 𝔲¯=𝔲2H𝔲1\bar{\mathfrak{u}}=\mathfrak{u}_{2}\circ_{H}\mathfrak{u}_{1} for some 𝔲1𝒬Bai(𝔵0)\mathfrak{u}_{1}\in\mathcal{HQ}^{\mathrm{Ba}}_{i}(\mathfrak{x}_{0}) and possibly trivial weak arch bridge 𝔲2\mathfrak{u}_{2}.

Step IV.3: Assuming that 𝔲1\mathfrak{u}_{1} and π(𝔲)\pi(\mathfrak{u}) are either both long or both short, we need to show that 𝔣(𝔵0;π(𝔲),𝔲1)=𝔴~(𝔵0;𝔲)𝔠(𝔵0;π(𝔲))\mathfrak{f}(\mathfrak{x}_{0};\pi(\mathfrak{u}),\mathfrak{u}_{1})=\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u})\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u})).

Suppose not. Then 𝔲2\mathfrak{u}_{2} is non-trivial. Let 𝔟1:=𝔱(𝔲1)\mathfrak{b}_{1}:=\mathfrak{t}(\mathfrak{u}_{1}). As in Case II, we can show that there is a cyclic permutation 𝔟1\mathfrak{b}^{\prime}_{1} of 𝔟1\mathfrak{b}_{1} such that 𝔟1𝔣(𝔵0;π(𝔲),𝔲1)\mathfrak{b}^{\prime}_{1}\mathfrak{f}(\mathfrak{x}_{0};\pi(\mathfrak{u}),\mathfrak{u}_{1}) is a string.

There are three subcases:

  • 𝔠(𝔵0;π(𝔲))=𝔷𝔣(𝔵0;π(𝔲),𝔲1)\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u}))=\mathfrak{z}^{\prime}\mathfrak{f}(\mathfrak{x}_{0};\pi(\mathfrak{u}),\mathfrak{u}_{1}) for a string 𝔷\mathfrak{z}^{\prime}. Then 𝔯s1(π(𝔲))𝔷𝔟1𝔣(𝔵0;π(𝔲),𝔲1)\mathfrak{r}^{s}_{1}(\pi(\mathfrak{u}))\mathfrak{z}^{\prime}\mathfrak{b}^{\prime}_{1}\mathfrak{f}(\mathfrak{x}_{0};\pi(\mathfrak{u}),\mathfrak{u}_{1}) is a string since 𝔲¯=𝔲2H𝔲1\bar{\mathfrak{u}}=\mathfrak{u}_{2}\circ_{H}\mathfrak{u}_{1}. Hence there is a weak bridge 𝔱(𝔲1)𝔲𝔰(𝔲)\mathfrak{t}(\mathfrak{u}_{1})\xrightarrow{\mathfrak{u}^{\prime}}\mathfrak{s}(\mathfrak{u}). Since π(𝔲)𝒬Ba\pi(\mathfrak{u})\in\mathcal{HQ}^{\mathrm{Ba}}, we get that π(𝔲)𝔲H𝔲1\pi(\mathfrak{u})\neq\mathfrak{u}^{\prime}\circ_{H}\mathfrak{u}_{1}. Hence N(𝔱(𝔲1),𝔲H𝔲1)=1N(\mathfrak{t}(\mathfrak{u}_{1}),\mathfrak{u}^{\prime}\circ_{H}\mathfrak{u}_{1})=1. But then it is readily verified that N(𝔱(𝔲1),𝔲¯)=1N(\mathfrak{t}(\mathfrak{u}_{1}),\bar{\mathfrak{u}})=1, a contradiction to the fact that 𝔴~\tilde{\mathfrak{w}} is band-free.

  • 𝔣(𝔵0;π(𝔲),𝔲1)=𝔷𝔠(𝔵0;π(𝔲))\mathfrak{f}(\mathfrak{x}_{0};\pi(\mathfrak{u}),\mathfrak{u}_{1})=\mathfrak{z}^{\prime}\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u})) for a string 𝔷\mathfrak{z}^{\prime} with 0<|𝔷|<|𝔯s1(π(𝔲))|0<|\mathfrak{z}^{\prime}|<|\mathfrak{r}^{s}_{1}(\pi(\mathfrak{u}))| and 𝔷𝔴~(𝔵0;𝔲)\mathfrak{z}^{\prime}\neq\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u}). Then γ𝔣(π(𝔲))\gamma^{\mathfrak{f}}(\pi(\mathfrak{u})) is the first syllable of 𝔷\mathfrak{z}^{\prime}. Let β\beta be the syllable of 𝔱(𝔲1)\mathfrak{t}(\mathfrak{u}_{1}) such that β𝔣(𝔵0;π(𝔲),𝔲1)\beta\mathfrak{f}(\mathfrak{x}_{0};\pi(\mathfrak{u}),\mathfrak{u}_{1}) is a string. Clearly θ(β)=δ(𝔯s1(π(𝔲)))=θ(γ𝔣(π(𝔲)))=θ(γ¯¯s(π(𝔲)))\theta(\beta)=-\delta(\mathfrak{r}^{s}_{1}(\pi(\mathfrak{u})))=-\theta(\gamma^{\mathfrak{f}}(\pi(\mathfrak{u})))=\theta(\bar{\bar{\gamma}}^{s}(\pi(\mathfrak{u}))) which implies that there is an abnormal weak arch bridge 𝔲\mathfrak{u}^{\prime} from 𝔰(𝔲)\mathfrak{s}(\mathfrak{u}) to 𝔱(𝔲1)\mathfrak{t}(\mathfrak{u}_{1}) with |𝔲c|>0|\mathfrak{u}^{\prime c}|>0. Since θ(β(𝔲))=δ(𝔲c)\theta(\beta(\mathfrak{u}^{\prime}))=-\delta(\mathfrak{u}^{\prime c}), we get 𝔟β(𝔲)𝔲c𝔴~H𝔲c𝔴~\mathfrak{b}^{\beta}(\mathfrak{u}^{\prime})\mathfrak{u}^{\prime c}\tilde{\mathfrak{w}}\equiv_{H}\mathfrak{u}^{\prime c}\tilde{\mathfrak{w}} and hence 𝔲1=𝔲Hπ(𝔲)\mathfrak{u}_{1}=\mathfrak{u}^{\prime}\circ_{H}\pi(\mathfrak{u}), a contradiction to 𝔲1𝒬Bai(𝔵0)\mathfrak{u}_{1}\in\mathcal{HQ}^{\mathrm{Ba}}_{i}(\mathfrak{x}_{0}).

  • 𝔣(𝔵0;π(𝔲),𝔲1)=𝔷𝔠(𝔵0;π(𝔲))\mathfrak{f}(\mathfrak{x}_{0};\pi(\mathfrak{u}),\mathfrak{u}_{1})=\mathfrak{z}^{\prime}\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u})) for a string 𝔷\mathfrak{z}^{\prime} with |𝔷||𝔯s1(π(𝔲))||\mathfrak{z}^{\prime}|\geq|\mathfrak{r}^{s}_{1}(\pi(\mathfrak{u}))|. To ensure domesticity we must have equality. Let 𝔲\mathfrak{u}^{\prime\prime} be the abnormal weak bridge between 𝔰(𝔲)\mathfrak{s}(\mathfrak{u}) and 𝔱(𝔲1)\mathfrak{t}(\mathfrak{u}_{1}). The dual of [9, Proposition 4.3] gives that 𝔰(𝔲)=𝔱(𝔲1)\mathfrak{s}(\mathfrak{u}^{\prime\prime})=\mathfrak{t}(\mathfrak{u}_{1}) and that 𝔲c{\mathfrak{u}^{\prime\prime}}^{c} is a proper right substring of 𝔯s1(π(𝔲))\mathfrak{r}^{s}_{1}(\pi(\mathfrak{u})). It is readily verified that π(𝔲)=𝔲H𝔲1\pi(\mathfrak{u})=\mathfrak{u}^{\prime\prime}\circ_{H}\mathfrak{u}_{1}, a contradiction to π(𝔲)𝒬Ba\pi(\mathfrak{u})\in\mathcal{HQ}^{\mathrm{Ba}}.

Example 7.12.

Consider the algebra Γ(vi)\Gamma^{(vi)} in Figure 11. Choose 𝔵0:=1(𝔳1,i)\mathfrak{x}_{0}:=1_{(\mathfrak{v}_{1},i)} and 𝔲:=HeG\mathfrak{u}:=HeG so that π(𝔲):=feca\pi(\mathfrak{u}):=feca and π2(𝔲):=b\pi^{2}(\mathfrak{u}):=b. Here θ(β(𝔲))=1\theta(\beta(\mathfrak{u}))=1, π(𝔲)\pi(\mathfrak{u}) is short and 𝔲\mathfrak{u} is s1s^{-1}-long with 𝔴~(𝔵0;𝔲)=e\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u})=e, 𝔯1s(π(𝔲))=fe\mathfrak{r}^{1}_{s}(\pi(\mathfrak{u}))=fe, 𝔯2s(π(𝔲))=ji\mathfrak{r}^{2}_{s}(\pi(\mathfrak{u}))=ji and 𝔴~=ca\tilde{\mathfrak{w}}=ca. It is readily verified that 𝔲¯:=Heca\bar{\mathfrak{u}}:=Heca is H-reduced and 𝔲1:=𝔲¯\mathfrak{u}_{1}:=\bar{\mathfrak{u}} is the short uncle of 𝔲\mathfrak{u} such that 𝔣(𝔵0;π(𝔲),𝔲1)=𝔴~(𝔵0;𝔲)𝔠(𝔵0;π(𝔲))=ecab\mathfrak{f}(\mathfrak{x}_{0};\pi(\mathfrak{u}),\mathfrak{u}_{1})=\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u})\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u}))=ecab.

Thus we have completed the proof of all cases of Lemma 7.3.

We note the significance of the hypotheses of the above lemma. If only condition (3)(3) fails then we cannot obtain the weak arch bridge 𝔲¯\bar{\mathfrak{u}} in Step 2 with 𝔰(𝔲¯)=𝔰(π(𝔲))\mathfrak{s}(\bar{\mathfrak{u}})=\mathfrak{s}(\pi(\mathfrak{u})). On the other hand, if only condition (2)(2) fails then we can obtain 𝔲¯𝒬¯Ba\bar{\mathfrak{u}}\in\overline{\mathcal{HQ}}^{\mathrm{Ba}} but it factors uniquely as 𝔲¯=𝔲2Hπ(𝔲)\bar{\mathfrak{u}}=\mathfrak{u}_{2}\circ_{H}\pi(\mathfrak{u}). Therefore the recipe in the proof of the uncle forking lemma fails to explain such a forking string. Our search leads us to the granduncle-grandchild interaction (see Lemma 7.22).

This raises a natural question, namely “how many distant relative pairs do we need to justify each forking string?” Fortunately we will be able to justify that uncle-nephew and granduncle-grandchild are the only ones necessary as explained in Proposition 7.21.

Our next goal is to ensure the existence of an uncle for long children of elements of 𝔣i(𝔵0)\mathcal{H}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0}).

Remark 7.13.

Suppose 𝔲𝒱𝔣i(𝔵0)\mathfrak{u}\in\mathcal{V}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0}) and h(𝔲)=2h(\mathfrak{u})=2. Then υ(π(𝔲))=0\upsilon(\pi(\mathfrak{u}))=0 if and only if π(𝔲)\pi(\mathfrak{u}) is an abnormal half ii-arch bridge. When these equivalent conditions hold then υ(𝔲)0\upsilon(\mathfrak{u})\geq 0 and the LVP (𝔰(𝔲),𝔵0)(\mathfrak{s}(\mathfrak{u}),\mathfrak{x}_{0}) could be simultaneously bb-long as well as ss-long. If the LVP (𝔰(𝔲),𝔵0)(\mathfrak{s}(\mathfrak{u}),\mathfrak{x}_{0}) is (b,s)(b,s)-long (resp. (s,b)(s,b)-long) then |𝔯b1(π(𝔲))|1|\mathfrak{r}_{b}^{1}(\pi(\mathfrak{u}))|\geq 1 (resp. |𝔯s1(π(𝔲))|1|\mathfrak{r}_{s}^{1}(\pi(\mathfrak{u}))|\geq 1), and the pair is (b,s)i(b,s)^{i}-long (resp. (s,b)i(s,b)^{i}-long).

Examples 7.14.

In the algebra Γ(iv)\Gamma^{(iv)} from Figure 9 if 𝔵0:=1(𝔳4,1)\mathfrak{x}_{0}:=1_{(\mathfrak{v}_{4},1)} and 𝔲:=e\mathfrak{u}:=e then π(𝔲)\pi(\mathfrak{u}) is an abnormal half 11-arch bridge with 𝔠(𝔵0;π(𝔲))=𝔵0\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u}))=\mathfrak{x}_{0} and 𝔠(𝔵0;𝔲)=acD\mathfrak{c}(\mathfrak{x}_{0};\mathfrak{u})=acD, and hence υ(𝔲)>0\upsilon(\mathfrak{u})>0.

If we remove the relation feacfeac from ρ\rho while keeping 𝔵0\mathfrak{x}_{0} and 𝔲\mathfrak{u} the same we get 𝔠(𝔵0;π(𝔲))=𝔵0=𝔠(𝔵0;𝔲)\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u}))=\mathfrak{x}_{0}=\mathfrak{c}(\mathfrak{x}_{0};\mathfrak{u}), and hence υ(𝔲)=0\upsilon(\mathfrak{u})=0.

Proposition 7.15.

Suppose 𝔲𝒱𝔣i(𝔵0)\mathfrak{u}\in\mathcal{V}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0}), h(𝔲)=2h(\mathfrak{u})=2, 𝔲\mathfrak{u} is long, and if 𝔲\mathfrak{u} is ss-long then |𝔯s2(π(𝔲))|>1|\mathfrak{r}^{s}_{2}(\pi(\mathfrak{u}))|>1. Then there is an uncle 𝔲1\mathfrak{u}_{1} of 𝔲\mathfrak{u} satisfying

𝔣(𝔵0;π(𝔲),𝔲1)=𝔴~(𝔵0;𝔲)𝔠(𝔵0;π(𝔲)).\mathfrak{f}(\mathfrak{x}_{0};\pi(\mathfrak{u}),\mathfrak{u}_{1})=\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u})\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u})).

Furthermore,

  • if π(𝔲)\pi(\mathfrak{u}) is abnormal then π(𝔲)𝔣i𝔲1\pi(\mathfrak{u})\sqsubset^{\mathfrak{f}}_{i}\mathfrak{u}_{1};

  • if π(𝔲)\pi(\mathfrak{u}) is normal then θ(β(𝔲))=i\theta(\beta(\mathfrak{u}))=i iff π(𝔲)𝔣i𝔲1\pi(\mathfrak{u})\sqsubset^{\mathfrak{f}}_{i}\mathfrak{u}_{1}.

Recall from Corollary 5.8 that 𝔲\mathfrak{u} is long if and only if the LVP (𝔰(𝔲),𝔴~(𝔵0;𝔲))𝔠(𝔵0;𝔲))(\mathfrak{s}(\mathfrak{u}),\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u}))\mathfrak{c}(\mathfrak{x}_{0};\mathfrak{u})) is long. If these equivalent conditions hold then the LVP (𝔰(𝔲),𝔠(𝔵0;π(𝔲)))(\mathfrak{s}(\mathfrak{u}),\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u}))) is long. In case the latter LVP is bb-long with |𝔯b2(π(𝔲))|=1|\mathfrak{r}^{b}_{2}(\pi(\mathfrak{u}))|=1 then the proof of the uncle forking lemma still goes through where 𝔲\mathfrak{u} is an “invisible” long child of π(𝔲)\pi(\mathfrak{u}) with 𝔴~(𝔵0;𝔲)=𝔯b1(π(𝔲))\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u})=\mathfrak{r}^{b}_{1}(\pi(\mathfrak{u})) and thus explains this forking string.

Proposition 7.16.

Suppose 𝔲𝔣i(𝔵0)𝒰𝔣i(𝔵0)\mathfrak{u}\in\mathcal{H}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0})\cup\mathcal{U}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0}). If the LVP (𝔟,𝔶):=(𝔱(𝔲),𝔠(𝔵0;𝔲))(\mathfrak{b},\mathfrak{y}):=(\mathfrak{t}(\mathfrak{u}),\mathfrak{c}(\mathfrak{x}_{0};\mathfrak{u})) is bb-long with |𝔯b2(𝔲)|=1|\mathfrak{r}^{b}_{2}(\mathfrak{u})|=1 then

  • if 𝔲𝔣i(𝔵0)\mathfrak{u}\in\mathcal{H}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0}) and if |𝔯b1(𝔲)|>0|\mathfrak{r}^{b}_{1}(\mathfrak{u})|>0 whenever 𝔲\mathfrak{u} is abnormal then there exists a sibling 𝔲1\mathfrak{u}_{1} of 𝔲\mathfrak{u} satisfying 𝔲𝔣i𝔲1\mathfrak{u}\sqsubset^{\mathfrak{f}}_{i}\mathfrak{u}_{1} iff (𝔟,𝔶)(\mathfrak{b},\mathfrak{y}) is bib^{i}-long and

    𝔣(𝔵0;𝔲1,𝔲)=𝔯b1(𝔲)𝔠(𝔵0;𝔲).\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u}_{1},\mathfrak{u})=\mathfrak{r}^{b}_{1}(\mathfrak{u})\mathfrak{c}(\mathfrak{x}_{0};\mathfrak{u}).
  • if 𝔲𝒰𝔣i(𝔵0)\mathfrak{u}\in\mathcal{U}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0}) and there is a string 𝔶\mathfrak{y} such that θ(𝔶𝔯b2(𝔲)𝔯b1(𝔲)𝔠(𝔵0;𝔲)𝔣(𝔵0;π(𝔲))=θ(β(𝔲))\theta(\mathfrak{y}\mathfrak{r}^{b}_{2}(\mathfrak{u})\mathfrak{r}^{b}_{1}(\mathfrak{u})\mathfrak{c}(\mathfrak{x}_{0};\mathfrak{u})\mid\mathfrak{f}(\mathfrak{x}_{0};\pi(\mathfrak{u}))=\theta(\beta(\mathfrak{u})) then there exists a sibling 𝔲1\mathfrak{u}_{1} of 𝔲\mathfrak{u} satisfying θ(β(𝔲1))=θ(β(𝔲))\theta(\beta(\mathfrak{u}_{1}))=\theta(\beta(\mathfrak{u})), 𝔲1𝔣θ(β(𝔲))𝔲\mathfrak{u}_{1}\sqsubset^{\mathfrak{f}}_{\theta(\beta(\mathfrak{u}))}\mathfrak{u} iff (𝔟,𝔶)(\mathfrak{b},\mathfrak{y}) is bθ(β(𝔲))b^{\theta(\beta(\mathfrak{u}))}-long and

    𝔣(𝔵0;𝔲1,𝔲)=𝔯b1(𝔲)𝔠(𝔵0;𝔲).\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u}_{1},\mathfrak{u})=\mathfrak{r}^{b}_{1}(\mathfrak{u})\mathfrak{c}(\mathfrak{x}_{0};\mathfrak{u}).
Example 7.17.

Suppose Γ¯(v)\bar{\Gamma}^{(v)} is obtained by replacing the relation kjihkjih by jihjih in ρ\rho in Figure 10. The only bands here are 𝔟1:=aCB\mathfrak{b}_{1}:=aCB, 𝔟2:=ihG\mathfrak{b}_{2}:=ihG and their inverses. Choosing 𝔵0:=1(𝔳1,i)\mathfrak{x}_{0}:=1_{(\mathfrak{v}_{1},i)} and 𝔲:=iDe\mathfrak{u}:=iDe we have π(𝔲)=a\pi(\mathfrak{u})=a, θ(β(𝔲))=1\theta(\beta(\mathfrak{u}))=-1 and 𝔠(𝔵0;𝔲)=Dea\mathfrak{c}(\mathfrak{x}_{0};\mathfrak{u})=Dea. Moreover the LVP (𝔱(𝔲),𝔠(𝔵0;𝔲))(\mathfrak{t}(\mathfrak{u}),\mathfrak{c}(\mathfrak{x}_{0};\mathfrak{u})) is b1b^{-1}-long due to jihρjih\in\rho with 𝔯b1(𝔲)=i\mathfrak{r}^{b}_{1}(\mathfrak{u})=i and 𝔯b2(𝔲)=j\mathfrak{r}^{b}_{2}(\mathfrak{u})=j. Here 𝔲1:=kjiDe\mathfrak{u}_{1}:=kjiDe is a sibling of 𝔲\mathfrak{u} such that 𝔣(𝔵0;𝔲1,𝔲)=iDea\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u}_{1},\mathfrak{u})=iDea and 𝔲1𝔣θ(β(𝔲))𝔲\mathfrak{u}_{1}\sqsubset^{\mathfrak{f}}_{\theta(\beta(\mathfrak{u}))}\mathfrak{u}.

The next result is essentially a mini-version of Lemma 7.3.

Proposition 7.18.

Suppose 𝔲𝒱𝔣i(𝔵0)\mathfrak{u}\in\mathcal{V}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0}) and 𝔲\mathfrak{u} is normal if 𝔲𝒰𝔣i(𝔵0)𝔣i(𝔵0)\mathfrak{u}\in\mathcal{U}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0})\cup\mathcal{H}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0}). Let 𝔶\mathfrak{y} be a proper left substring of 𝔲o\mathfrak{u}^{o} such that |𝔶|>0|\mathfrak{y}|>0 and 𝔶𝔣(𝔵0;𝔲,π(𝔲))\mathfrak{y}\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u},\pi(\mathfrak{u})) forks. Then the following hold:

  • h(𝔲)=1h(\mathfrak{u})=1

    If θ(𝔶;𝔲o)=i\theta(\mathfrak{y};\mathfrak{u}^{o})=i then there is a sibling 𝔲1\mathfrak{u}_{1} of 𝔲\mathfrak{u} such that 𝔲1𝔣i𝔲\mathfrak{u}_{1}\sqsubset^{\mathfrak{f}}_{i}\mathfrak{u};

  • h(𝔲)>1h(\mathfrak{u})>1

    If θ(𝔶;𝔲o)=θ(β(𝔲))\theta(\mathfrak{y};\mathfrak{u}^{o})=-\theta(\beta(\mathfrak{u})) then there is a sibling 𝔲1\mathfrak{u}_{1} of 𝔲\mathfrak{u} such that 𝔲1𝔣θ(β(𝔲))𝔲\mathfrak{u}_{1}\sqsubset^{\mathfrak{f}}_{\theta(\beta(\mathfrak{u}))}\mathfrak{u}.

In both cases we have

𝔣(𝔵0;𝔲,𝔲1)=𝔶𝔣(𝔵0;𝔲,π(𝔲)).\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u},\mathfrak{u}_{1})=\mathfrak{y}\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u},\pi(\mathfrak{u})).

Proof. We only prove the first statement; a similar proof works for the other case.

Let :=θ(β(𝔲))\partial:=\theta(\beta(\mathfrak{u})). Let 𝔴~\tilde{\mathfrak{w}} be the left substring of 𝔲\mathfrak{u} such that 𝔲o𝔴~\mathfrak{u}^{o}\tilde{\mathfrak{w}} is a string. Let α\alpha be the syllable with θ(α)=δ\theta(\alpha)=\delta such that α𝔶\alpha\mathfrak{y} is a string.

Since α𝔶𝔴~\alpha\mathfrak{y}\tilde{\mathfrak{w}} is a string, an argument as in Step 1 of the proof of the uncle forking lemma (Lemma 7.3), we can show that α𝔶𝔴~\alpha\mathfrak{y}\tilde{\mathfrak{w}} is a band-free string. Then as in Step 2 there, we extend α𝔶𝔴~\alpha\mathfrak{y}\tilde{\mathfrak{w}} to 𝔱(𝔲¯)𝔲¯\mathfrak{t}(\bar{\mathfrak{u}})\bar{\mathfrak{u}} for some 𝔲¯𝒬¯Ba\bar{\mathfrak{u}}\in\overline{\mathcal{HQ}}^{\mathrm{Ba}} using Lemma 7.4. Then an argument similar to Step 3 shows that α𝔶𝔴~𝔱(𝔲1)𝔲1\alpha\mathfrak{y}\tilde{\mathfrak{w}}\in\mathfrak{t}(\mathfrak{u}_{1})\mathfrak{u}_{1}, where 𝔲¯=𝔲2H𝔲1\bar{\mathfrak{u}}=\mathfrak{u}_{2}\circ_{H}\mathfrak{u}_{1} is the canonical factorization such that 𝔲2\mathfrak{u}_{2} is a possibly trivial weak arch bridge and 𝔲1𝒬Ba\mathfrak{u}_{1}\in\mathcal{HQ}^{\mathrm{Ba}}. Since β(𝔲1)=β(𝔲)\beta(\mathfrak{u}_{1})=\beta(\mathfrak{u}) we get 𝔣(𝔵0;𝔲,𝔲1)=𝔶𝔣(𝔵0;𝔲,π(𝔲))\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u},\mathfrak{u}_{1})=\mathfrak{y}\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u},\pi(\mathfrak{u})) by Corollary 5.9 and hence 𝔲1𝔣𝔲\mathfrak{u}_{1}\sqsubset^{\mathfrak{f}}_{\partial}\mathfrak{u}. In both Step 1 and 3 the proof uses the fact that if N(𝔱(𝔲1),𝔲¯)=0N(\mathfrak{t}(\mathfrak{u}_{1}),\bar{\mathfrak{u}})=0 then 𝔲=𝔲H𝔲1\mathfrak{u}=\mathfrak{u}^{\prime}\circ_{H}\mathfrak{u}_{1} for some 𝔲𝒬¯Ba\mathfrak{u}^{\prime}\in\overline{\mathcal{HQ}}^{\mathrm{Ba}}.    \Box

Example 7.19.

Consider 𝔵0:=1(𝔳1,i)\mathfrak{x}_{0}:=1_{(\mathfrak{v}_{1},i)} and 𝔲:=iDe\mathfrak{u}:=iDe in the algebra Γ(v)\Gamma^{(v)} from Figure 10 so that π(𝔲):=a\pi(\mathfrak{u}):=a and 𝔶:=e\mathfrak{y}:=e. Clearly h(𝔲)>1h(\mathfrak{u})>1, 𝔲\mathfrak{u} is normal with 𝔲o=De\mathfrak{u}^{o}=De and 𝔶𝔣(𝔵0;𝔲,π(𝔲))=ea\mathfrak{y}\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u},\pi(\mathfrak{u}))=ea forks. Here 𝔲1:=fe\mathfrak{u}_{1}:=fe is a sibling of 𝔲\mathfrak{u} with 𝔲1𝔣θ(β(𝔲))𝔲\mathfrak{u}_{1}\sqsubset^{\mathfrak{f}}_{\theta(\beta(\mathfrak{u}))}\mathfrak{u} and 𝔣(𝔵0;𝔲,𝔲1)=ea\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u},\mathfrak{u}_{1})=ea.

Below we note without proof the consequences of the failure of condition (3)(3) of the hypotheses of Lemma 7.3.

Proposition 7.20.

Suppose all hypotheses of Lemma 7.3 hold except for (3)(3). Then the following are equivalent:

  1. (1)

    𝔲\mathfrak{u} is not ss^{\partial}-long;

  2. (2)

    𝔴~(𝔵0;𝔲)𝔠(𝔵0;π(𝔲))\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u})\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u})) is a forking string;

  3. (3)

    π(𝔲)\pi(\mathfrak{u}) is abnormal with |π(𝔲)c|>0|\pi(\mathfrak{u})^{c}|>0, the LVP (𝔰(π(𝔲)),𝔠(𝔵0;π2(𝔲)))(\mathfrak{s}(\pi(\mathfrak{u})),\mathfrak{c}(\mathfrak{x}_{0};\pi^{2}(\mathfrak{u}))) is bb^{\partial}-long with 𝔯b2(π2(𝔲))=β(𝔲)\mathfrak{r}^{b}_{2}(\pi^{2}(\mathfrak{u}))=\beta(\mathfrak{u}) and one of the following holds:

    1. (a)

      υ(π(𝔲))=1\upsilon(\pi(\mathfrak{u}))=-1 and |𝔯b1(π2(𝔲))|=0|\mathfrak{r}^{b}_{1}(\pi^{2}(\mathfrak{u}))|=0;

    2. (b)

      υ(π(𝔲))=1\upsilon(\pi(\mathfrak{u}))=1 and 𝔯b1(π2(𝔲))𝔠(𝔵0;π2(𝔲))=𝔠(𝔵0;π(𝔲))\mathfrak{r}^{b}_{1}(\pi^{2}(\mathfrak{u}))\mathfrak{c}(\mathfrak{x}_{0};\pi^{2}(\mathfrak{u}))=\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u})).

When these equivalent conditions hold then π(𝔲)\pi(\mathfrak{u}) is short by Corollary 6.6 and Proposition 6.16.

It is worth noting that either condition (2)(2) or (3)(3) of the hypotheses of Lemma 7.3 fails if and only if exactly one of those conditions fails. When this happens, there are yet unexplained forking strings then we get the existence of a granduncle, which is the content of Lemma 7.22. However this raises a natural question whether we can guarantee the existence of a sibling of πn(𝔲)\pi^{n}(\mathfrak{u}) for arbitrary n>0n>0. Since abnormality is the source of anomalies, the following result states that only n=1,2n=1,2 are possible cases.

Proposition 7.21.

Suppose for 𝔲𝒱𝔣i(𝔵0)\mathfrak{u}\in\mathcal{V}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0}) with h(𝔲)3h(\mathfrak{u})\geq 3 that all hypotheses of Lemma 7.3 fold except that exactly one of (2)(2) or (3)(3) fails.

If h(𝔲)>3h(\mathfrak{u})>3 then 𝔣(𝔵0;π3(𝔲),π(𝔲))\mathfrak{f}(\mathfrak{x}_{0};\pi^{3}(\mathfrak{u}),\pi(\mathfrak{u})) is a proper left substring of 𝔠(𝔵0;π(𝔲))\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u})). As a consequence 𝔠(𝔵0;π3(𝔲))\mathfrak{c}(\mathfrak{x}_{0};\pi^{3}(\mathfrak{u})) is a proper substring of 𝔠(𝔵0;π(𝔲))\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u})), and hence υ(π2(𝔲))=1\upsilon(\pi^{2}(\mathfrak{u}))=1. Moreover π2(𝔲)\pi^{2}(\mathfrak{u}) and π(𝔲)Hπ2(𝔲)\pi(\mathfrak{u})\circ_{H}\pi^{2}(\mathfrak{u}) are normal.

If h(𝔲)=3h(\mathfrak{u})=3 then 𝔣(𝔵0;π3(𝔲),π(𝔲))\mathfrak{f}(\mathfrak{x}_{0};\pi^{3}(\mathfrak{u}),\pi(\mathfrak{u})) is a proper left substring of 𝔴~(𝔵0;𝔲)𝔠(𝔵0;π(𝔲))\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u})\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u})).

We do not include a proof but only the first statement needs one, where if we suppose that the first conclusion fails then the impossibility can be verified in several long but straightforward cases using Propositions 6.11, 6.15, Corollary 6.12 and [9, Proposition 4.3].

Lemma 7.22.

(Granduncle forking lemma) Suppose 𝔲𝒱𝔣i(𝔵0)\mathfrak{u}\in\mathcal{V}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0}) with h(𝔲)3h(\mathfrak{u})\geq 3. Assume all the hypotheses of Lemma 7.3 hold but exactly one of (2)(2) or (3)(3) fails. Furthermore assume that 𝔴~(𝔵0;𝔲)𝔠(𝔵0;π(𝔲))\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u})\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u})) is a forking string. Then π(𝔲)\pi(\mathfrak{u}) is short and there is a normal granduncle 𝔲1\mathfrak{u}_{1} of 𝔲\mathfrak{u} satisfying

𝔣(𝔵0;π(𝔲),𝔲1)=𝔣(𝔵0;π2(𝔲),𝔲1)=𝔴~(𝔵0;𝔲)𝔠(𝔵0;π(𝔲)).\mathfrak{f}(\mathfrak{x}_{0};\pi(\mathfrak{u}),\mathfrak{u}_{1})=\mathfrak{f}(\mathfrak{x}_{0};\pi^{2}(\mathfrak{u}),\mathfrak{u}_{1})=\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u})\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u})).

Moreover

  • (h(𝔲)=3)(h(\mathfrak{u})=3)

    𝔲1𝔣iπ2(𝔲)\mathfrak{u}_{1}\sqsubset^{\mathfrak{f}}_{i}\pi^{2}(\mathfrak{u}) if and only if δ(π(𝔲)e)=i\delta(\pi(\mathfrak{u})^{e})=i;

  • (h(𝔲)>3)(h(\mathfrak{u})>3)

    β(𝔲1)=β(π2(𝔲))\beta(\mathfrak{u}_{1})=\beta(\pi^{2}(\mathfrak{u})) and π2(𝔲)𝔣θ(β(π2(𝔲)))𝔲1\pi^{2}(\mathfrak{u})\sqsubset^{\mathfrak{f}}_{\theta(\beta(\pi^{2}(\mathfrak{u})))}\mathfrak{u}_{1} if and only if θ(β(π2(𝔲)))=\theta(\beta(\pi^{2}(\mathfrak{u})))=-\partial.

Proof. If condition (2)(2) in the hypothesis of Lemma 7.3 fails then 𝔴~(𝔵0;𝔲)𝔠(𝔵0;π(𝔲))\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u})\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u})) is a proper substring of 𝔠(𝔵0;π2(𝔲))\mathfrak{c}(\mathfrak{x}_{0};\pi^{2}(\mathfrak{u})). In particular, υ(π(𝔲))=1\upsilon(\pi(\mathfrak{u}))=-1. Then Proposition 7.21 guarantees that 𝔟𝔣α(π(𝔲))𝔷¯𝔟𝔣α(π3(𝔲))𝔠(𝔵0;π3(𝔲))\mathfrak{b}^{\mathfrak{f}\alpha}(\pi(\mathfrak{u}))\bar{\mathfrak{z}}\mathfrak{b}^{\mathfrak{f}\alpha}(\pi^{3}(\mathfrak{u}))\mathfrak{c}(\mathfrak{x}_{0};\pi^{3}(\mathfrak{u})) is a string for some string 𝔷¯\bar{\mathfrak{z}} of positive length. Let 𝔴~\tilde{\mathfrak{w}} be the shortest right substring of 𝔷¯𝔟𝔣α(π3(𝔲))\bar{\mathfrak{z}}\mathfrak{b}^{\mathfrak{f}\alpha}(\pi^{3}(\mathfrak{u})) such that 𝔟𝔣α(π(𝔲))𝔴~𝔰(π2(𝔲))\mathfrak{b}^{\mathfrak{f}\alpha}(\pi(\mathfrak{u}))\tilde{\mathfrak{w}}\mathfrak{s}(\pi^{2}(\mathfrak{u})) is a string. Clearly 𝔴~\tilde{\mathfrak{w}} is a left substring of π2(𝔲)\pi^{2}(\mathfrak{u}). In fact 𝔴~(𝔵0;𝔲)𝔴~\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u})\tilde{\mathfrak{w}} is also a left substring of π2(𝔲)\pi^{2}(\mathfrak{u}). Then the hypotheses imply that β(𝔲)𝔴~(𝔵0;𝔲)𝔴~\beta(\mathfrak{u})\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u})\tilde{\mathfrak{w}} is a band-free string, which can be extended using Lemma 7.4 to a weak arch bridge 𝔲¯=𝔲2H𝔲1\bar{\mathfrak{u}}=\mathfrak{u}_{2}\circ_{H}\mathfrak{u}_{1} with 𝔲1𝒬Ba\mathfrak{u}_{1}\in\mathcal{HQ}^{\mathrm{Ba}}. Using Propositions 6.11, 7.21 and Corollaries 6.12, 6.13 as tools while following the idea of the proof of uncle forking lemma the proof can be completed.

If condition (3)(3) from the hypothesis of Lemma 7.3 fails then Proposition 7.20 gives a characterization of when 𝔴~(𝔵0;𝔲)𝔠(𝔵0;π(𝔲))\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u})\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u})) is a forking string. Then Corollary 6.6 and Proposition 6.16 guarantee that π(𝔲)\pi(\mathfrak{u}) is short. The idea of the proof is a combination of Lemma 7.3 and Proposition 7.16. The proof of the former fails marginally as β(𝔲)𝔷~𝔟𝔣α(π2(𝔲))\beta(\mathfrak{u})\tilde{\mathfrak{z}}\mathfrak{b}^{\mathfrak{f}\alpha}(\pi^{2}(\mathfrak{u})) is not a string. However π2(𝔲)\pi^{2}(\mathfrak{u}) satisfies the hypotheses of the latter, and this explains the forking string 𝔴~(𝔵0;𝔲)𝔠(𝔵0;π(𝔲))\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u})\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u})).    \Box

Examples 7.23.

Suppose the algebra Γ¯(ii)\bar{\Gamma}^{(ii)} (resp. Γ¯(iii)\bar{\Gamma}^{(iii)}) is obtained by replacing dbghdbgh by dbgdbg (resp. defdef by dede) in Figure 7 (resp. Figure 8). Choose 𝔵0:=1(𝔳1,i)\mathfrak{x}_{0}:=1_{(\mathfrak{v}_{1},i)} and 𝔲:=HGfE\mathfrak{u}:=HGfE (resp. 𝔲:=FEahG\mathfrak{u}:=FEahG) so that π(𝔲):=cBD\pi(\mathfrak{u}):=cBD (resp. cDcD) and π2(𝔲):=icBa\pi^{2}(\mathfrak{u}):=icBa (resp. icaBicaB). Clearly υ(𝔲)=1\upsilon(\mathfrak{u})=1 (resp. υ(𝔲)=1\upsilon(\mathfrak{u})=-1), 𝔲\mathfrak{u} is b1b^{-1}-long but not ss-long and all the hypotheses of Lemma 7.3 except (3)(3) hold. It is readily verified that 𝔲\mathfrak{u} satisfies the equivalent conclusions of Proposition 7.20. Moreover there is a grand-uncle 𝔲1:=HGBa\mathfrak{u}_{1}:=HGBa (resp. FEaBFEaB) of 𝔲\mathfrak{u} such that 𝔣(𝔵0;𝔲,𝔲1)=𝔣(𝔵0;π(𝔲),𝔲1)=𝔣(𝔵0;π2(𝔲),𝔲1)=Ba\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u},\mathfrak{u}_{1})=\mathfrak{f}(\mathfrak{x}_{0};\pi(\mathfrak{u}),\mathfrak{u}_{1})=\mathfrak{f}(\mathfrak{x}_{0};\pi^{2}(\mathfrak{u}),\mathfrak{u}_{1})=Ba (resp. aBaB).

In the algebra Γ(iii)\Gamma^{(iii)} in Figure 8 choosing 𝔵0:=1(𝔳1,i)\mathfrak{x}_{0}:=1_{(\mathfrak{v}_{1},i)} and 𝔲:=JhG\mathfrak{u}:=JhG we have π(𝔲)=cD\pi(\mathfrak{u})=cD and π2(𝔲)=icaB\pi^{2}(\mathfrak{u})=icaB with υ(𝔲)=1\upsilon(\mathfrak{u})=-1. Here 𝔲\mathfrak{u} is b1b^{-1}-long and all the hypotheses of Lemma 7.3 except (2)(2) hold. It is easy to check that there is a grand-uncle 𝔲1:=JB\mathfrak{u}_{1}:=JB of 𝔲\mathfrak{u} such that 𝔣(𝔵0;𝔲,𝔲1)=𝔣(𝔵0;π(𝔲),𝔲1)=𝔣(𝔵0;π2(𝔲),𝔲1)=𝔴~(𝔵0;𝔲)𝔠(𝔵0;π(𝔲))=B\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u},\mathfrak{u}_{1})=\mathfrak{f}(\mathfrak{x}_{0};\pi(\mathfrak{u}),\mathfrak{u}_{1})=\mathfrak{f}(\mathfrak{x}_{0};\pi^{2}(\mathfrak{u}),\mathfrak{u}_{1})=\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u})\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u}))=B.

There is yet another variation of Lemma 7.3 where θ(β(π(𝔲)))=θ(β(𝔲))\theta(\beta(\pi(\mathfrak{u})))=-\theta(\beta(\mathfrak{u})). The proof techniques are similar, and thus the very long proof is omitted. Before we state the variation, we give a supporting result, again without proof.

Proposition 7.24.

Suppose 𝔲𝒱𝔣i(𝔵0)\mathfrak{u}\in\mathcal{V}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0}), h(𝔲)>2h(\mathfrak{u})>2, θ(β(π(𝔲)))==θ(β(𝔲))\theta(\beta(\pi(\mathfrak{u})))=\partial=-\theta(\beta(\mathfrak{u})) for some {+,}\partial\in\{+,-\} and 𝔲\mathfrak{u} is long.

If π(𝔲)\pi(\mathfrak{u}) is abnormal with |π(𝔲)c|>0|\pi(\mathfrak{u})^{c}|>0 then 𝔲\mathfrak{u} is bb^{-\partial}-long, β(𝔲)\beta(\mathfrak{u}) is the first syllable of π(𝔲)β\pi(\mathfrak{u})^{\beta} and 𝔴~(𝔵0;𝔲)=𝔯b1(π(𝔲))\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u})=\mathfrak{r}^{b}_{1}(\pi(\mathfrak{u})).

If π(𝔲)\pi(\mathfrak{u}) is abnormal with |π(𝔲)c|=0|\pi(\mathfrak{u})^{c}|=0 then one of the following happens:

  • If 𝔲\mathfrak{u} is bb^{-\partial}-long, β(𝔲)\beta(\mathfrak{u}) is the first syllable of π(𝔲)β\pi(\mathfrak{u})^{\beta} and 𝔴~(𝔵0;𝔲)=𝔯b1(π(𝔲))\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u})=\mathfrak{r}^{b}_{1}(\pi(\mathfrak{u}));

  • If 𝔲\mathfrak{u} is ss^{\partial}-long then 0<|𝔴~(𝔵0;𝔲)|<|𝔯s1(π(𝔲))|0<|\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u})|<|\mathfrak{r}^{s}_{1}(\pi(\mathfrak{u}))|.

Irrespective of whether π(𝔲)\pi(\mathfrak{u}) is normal or abnormal, 𝔴~(𝔵0;𝔲)𝔠(𝔵0;π(𝔲))=𝔷~𝔠(𝔵0;π2(𝔲))\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u})\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u}))=\tilde{\mathfrak{z}}\mathfrak{c}(\mathfrak{x}_{0};\pi^{2}(\mathfrak{u})) for some string 𝔷~\tilde{\mathfrak{z}} and β(𝔲)𝔷~𝔟𝔣α(π2(𝔲))\beta(\mathfrak{u})\tilde{\mathfrak{z}}\mathfrak{b}^{\mathfrak{f}\alpha}(\pi^{2}(\mathfrak{u})) is a string.

Lemma 7.25.

(Uncle forking lemma–opposite parity) Suppose 𝔲𝒱𝔣i(𝔵0)\mathfrak{u}\in\mathcal{V}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0}), h(𝔲)>2h(\mathfrak{u})>2, θ(β(π(𝔲)))==θ(β(𝔲))\theta(\beta(\pi(\mathfrak{u})))=\partial=-\theta(\beta(\mathfrak{u})) for some {+,}\partial\in\{+,-\} and 𝔲\mathfrak{u} is long. Further suppose that if π(𝔲)\pi(\mathfrak{u}) is abnormal and 𝔲\mathfrak{u} is bb^{-\partial}-long then 𝔱(𝔲)𝔲\mathfrak{t}(\mathfrak{u})\mathfrak{u} is not a substring of π(𝔲)β𝔟β(π(𝔲))\pi(\mathfrak{u})^{\beta}\mathfrak{b}_{\beta}(\pi(\mathfrak{u})). Then there is an uncle 𝔲1\mathfrak{u}_{1} of 𝔲\mathfrak{u} constructed as in Lemma 7.3 that satisfies θ(β(𝔲1))=θ(β(π(𝔲)))\theta(\beta(\mathfrak{u}_{1}))=\theta(\beta(\pi(\mathfrak{u}))). Such an uncle is short if υ(π(𝔲))<1\upsilon(\pi(\mathfrak{u}))<1. Moreover one of the following holds:

  1. (1)

    If either of the following sets of conditions hold:

    • π(𝔲)\pi(\mathfrak{u}) is normal;

    • π(𝔲)\pi(\mathfrak{u}) is abnormal with |π(𝔲)c|=0|\pi(\mathfrak{u})^{c}|=0, 𝔲\mathfrak{u} is ss^{\partial}-long and β(𝔲)\beta(\mathfrak{u}) is not the first syllable of π(𝔲)β\pi(\mathfrak{u})^{\beta},

    then β(𝔲1)=β(π(𝔲))\beta(\mathfrak{u}_{1})=\beta(\pi(\mathfrak{u})), and hence π(𝔲)𝔣𝔲1\pi(\mathfrak{u})\sqsubset^{\mathfrak{f}}_{\partial}\mathfrak{u}_{1} and

    𝔣(𝔵0;π(𝔲),𝔲1)=𝔴~(𝔵0;𝔲)𝔠(𝔵0;π(𝔲)).\mathfrak{f}(\mathfrak{x}_{0};\pi(\mathfrak{u}),\mathfrak{u}_{1})=\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u})\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u})).
  2. (2)

    If π(𝔲)\pi(\mathfrak{u}) is abnormal and β(𝔲)\beta(\mathfrak{u}) is the first syllable of π(𝔲)β\pi(\mathfrak{u})^{\beta} then the hypothesis ensures that there is a partition π(𝔲)β=𝔶2𝔶1\pi(\mathfrak{u})^{\beta}=\mathfrak{y}_{2}\mathfrak{y}_{1} with |𝔶1||𝔶2|>0|\mathfrak{y}_{1}||\mathfrak{y}_{2}|>0 such that β(𝔲1)𝔶1\beta(\mathfrak{u}_{1})\mathfrak{y}_{1} is a string and β(𝔲1)\beta(\mathfrak{u}_{1}) is not the first syllable of 𝔶2\mathfrak{y}_{2}. Moreover

    𝔣(𝔵0;π(𝔲),𝔲1)={HR𝔰(π(𝔲))(𝔶1𝔴~(𝔵0;𝔲)𝔠(𝔵0;π(𝔲)))if 𝔲1𝔣π(𝔲);𝔴~(𝔵0;π(𝔲))𝔠(𝔵0;π2(𝔲))if π(𝔲)𝔣𝔲1.\mathfrak{f}(\mathfrak{x}_{0};\pi(\mathfrak{u}),\mathfrak{u}_{1})=\begin{cases}\mathrm{HR}_{\mathfrak{s}(\pi(\mathfrak{u}))}(\mathfrak{y}_{1}\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u})\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u})))&\mbox{if }\mathfrak{u}_{1}\sqsubset^{\mathfrak{f}}_{\partial}\pi(\mathfrak{u});\\ \tilde{\mathfrak{w}}(\mathfrak{x}_{0};\pi(\mathfrak{u}))\mathfrak{c}(\mathfrak{x}_{0};\pi^{2}(\mathfrak{u}))&\mbox{if }\pi(\mathfrak{u})\sqsubset^{\mathfrak{f}}_{\partial}\mathfrak{u}_{1}.\end{cases}

Since we omit the proof of the above lemma we indicate some examples below to illustrate some cases of its proof.

Examples 7.26.
  1. (1)

    In the algebra Γ(vii)\Gamma^{(vii)} in Figure 12 the only bands are 𝔟1:=hbdC\mathfrak{b}_{1}:=hbdC, 𝔟2:=dfE\mathfrak{b}_{2}:=dfE and their inverses. Choosing 𝔵0:=1(𝔳1,i)\mathfrak{x}_{0}:=1_{(\mathfrak{v}_{1},i)} and 𝔲:=Gb\mathfrak{u}:=Gb we get π(𝔲)=dC\pi(\mathfrak{u})=dC and π2(𝔲)=hA\pi^{2}(\mathfrak{u})=hA. Clearly υ(𝔲)=1\upsilon(\mathfrak{u})=1, θ(β(π(𝔲)))=θ(β(𝔲))=1\theta(\beta(\pi(\mathfrak{u})))=-\theta(\beta(\mathfrak{u}))=1, 𝔶1=b\mathfrak{y}_{1}=b and 𝔲\mathfrak{u} is b1b^{-1}-long. Moreover both π(𝔲)\pi(\mathfrak{u}) and the uncle 𝔲1:=GbdC\mathfrak{u}_{1}:=GbdC of 𝔲\mathfrak{u} are short, 𝔲1𝔣1π(𝔲)\mathfrak{u}_{1}\sqsubset^{\mathfrak{f}}_{1}\pi(\mathfrak{u}) and 𝔣(𝔵0;π(𝔲),𝔲1)=HR𝔰(π(𝔲))(𝔶1𝔴~(𝔵0;𝔲)𝔠(𝔵0;π(𝔲)))=A\mathfrak{f}(\mathfrak{x}_{0};\pi(\mathfrak{u}),\mathfrak{u}_{1})=\mathrm{HR}_{\mathfrak{s}(\pi(\mathfrak{u}))}(\mathfrak{y}_{1}\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u})\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u})))=A.

    v7{v_{7}}v2{v_{2}}v3{v_{3}}v1{v_{1}}v5{v_{5}}v4{v_{4}}v6{v_{6}}g\scriptstyle{g}h\scriptstyle{h}a\scriptstyle{a}b\scriptstyle{b}c\scriptstyle{c}d\scriptstyle{d}e\scriptstyle{e}f\scriptstyle{f}
    Figure 12. Γ(vii)\Gamma^{(vii)} with ρ={be,ab,hg,cf,hbdf}\rho=\{be,ab,hg,cf,hbdf\}

    However if we add agag to ρ\rho for the same quiver then choosing the same 𝔵0\mathfrak{x}_{0} and 𝔲\mathfrak{u} we still have that π(𝔲)\pi(\mathfrak{u}) is short but its uncle 𝔲1:=GbdC\mathfrak{u}_{1}:=GbdC is ss-long, π(𝔲)𝔣1𝔲1\pi(\mathfrak{u})\sqsubset^{\mathfrak{f}}_{1}\mathfrak{u}_{1} and 𝔣(𝔵0;π(𝔲),𝔲1)=𝔴~(𝔵0;π(𝔲))𝔠(𝔵0;π2(𝔲))=dChA\mathfrak{f}(\mathfrak{x}_{0};\pi(\mathfrak{u}),\mathfrak{u}_{1})=\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\pi(\mathfrak{u}))\mathfrak{c}(\mathfrak{x}_{0};\pi^{2}(\mathfrak{u}))=dChA.

  2. (2)

    Consider the algebra Γ(i)\Gamma^{(i)} in Figure 6. Choosing 𝔵0:=1(𝔳1,i)\mathfrak{x}_{0}:=1_{(\mathfrak{v}_{1},i)} and 𝔲:=Gb\mathfrak{u}:=Gb we get π(𝔲)=dC\pi(\mathfrak{u})=dC and π2(𝔲)=hbA\pi^{2}(\mathfrak{u})=hbA. Clearly υ(𝔲)=1\upsilon(\mathfrak{u})=1, θ(β(π(𝔲)))=θ(β(𝔲))=1\theta(\beta(\pi(\mathfrak{u})))=-\theta(\beta(\mathfrak{u}))=1, 𝔶1=b\mathfrak{y}_{1}=b and 𝔲\mathfrak{u} is b1b^{-1}-long. Moreover π(𝔲)\pi(\mathfrak{u}) is ss-long, the uncle 𝔲1:=GbdC\mathfrak{u}_{1}:=GbdC of 𝔲\mathfrak{u} is short, 𝔲1𝔣1π(𝔲)\mathfrak{u}_{1}\sqsubset^{\mathfrak{f}}_{1}\pi(\mathfrak{u}) and 𝔣(𝔵0;π(𝔲),𝔲1)=HR𝔰(π(𝔲))(𝔶1𝔴~(𝔵0;𝔲)𝔠(𝔵0;π(𝔲)))=bA\mathfrak{f}(\mathfrak{x}_{0};\pi(\mathfrak{u}),\mathfrak{u}_{1})=\mathrm{HR}_{\mathfrak{s}(\pi(\mathfrak{u}))}(\mathfrak{y}_{1}\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u})\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u})))=bA.

    If we remove aeae from ρ\rho for the same quiver then choosing the same 𝔵0\mathfrak{x}_{0} and 𝔲\mathfrak{u} we have υ(𝔲)=0\upsilon(\mathfrak{u})=0, both π(𝔲)\pi(\mathfrak{u}) and 𝔲1:=GbdC\mathfrak{u}_{1}:=GbdC are short, π(𝔲)𝔣1𝔲1\pi(\mathfrak{u})\sqsubset^{\mathfrak{f}}_{1}\mathfrak{u}_{1} and 𝔣(𝔵0;π(𝔲),𝔲1)=𝔴~(𝔵0;π(𝔲))𝔠(𝔵0;π2(𝔲))=A\mathfrak{f}(\mathfrak{x}_{0};\pi(\mathfrak{u}),\mathfrak{u}_{1})=\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\pi(\mathfrak{u}))\mathfrak{c}(\mathfrak{x}_{0};\pi^{2}(\mathfrak{u}))=A.

  3. (3)

    In the algebra Γ(viii)\Gamma^{(viii)} in Figure 13 the only bands are 𝔟1:=bED\mathfrak{b}_{1}:=bED, 𝔟2:=fcH\mathfrak{b}_{2}:=fcH and their inverses. Choosing 𝔵0:=1(𝔳1,i)\mathfrak{x}_{0}:=1_{(\mathfrak{v}_{1},i)} and 𝔲:=jDH\mathfrak{u}:=jDH we get π(𝔲)=fc\pi(\mathfrak{u})=fc and π2(𝔲)=ba\pi^{2}(\mathfrak{u})=ba. Clearly υ(𝔲)=1\upsilon(\mathfrak{u})=1, θ(β(π(𝔲)))=θ(β(𝔲))=1\theta(\beta(\pi(\mathfrak{u})))=-\theta(\beta(\mathfrak{u}))=-1 and 𝔲\mathfrak{u} is b1b^{1}-long. Moreover both π(𝔲)\pi(\mathfrak{u}) and the uncle 𝔲1:=jD\mathfrak{u}_{1}:=jD of 𝔲\mathfrak{u} are short, π(𝔲)𝔣1𝔲1\pi(\mathfrak{u})\sqsubset^{\mathfrak{f}}_{1}\mathfrak{u}_{1} and 𝔣(𝔵0;π(𝔲),𝔲1)=𝔴~(𝔵0;π(𝔲))𝔠(𝔵0;π2(𝔲))=ba\mathfrak{f}(\mathfrak{x}_{0};\pi(\mathfrak{u}),\mathfrak{u}_{1})=\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\pi(\mathfrak{u}))\mathfrak{c}(\mathfrak{x}_{0};\pi^{2}(\mathfrak{u}))=ba.

    v1{v_{1}}v2{v_{2}}v3{v_{3}}v4{v_{4}}v5{v_{5}}v9{v_{9}}v8{v_{8}}v6{v_{6}}v7{v_{7}}a\scriptstyle{a}b\scriptstyle{b}e\scriptstyle{e}c\scriptstyle{c}h\scriptstyle{h}f\scriptstyle{f}g\scriptstyle{g}d\scriptstyle{d}j\scriptstyle{j}i\scriptstyle{i}
    Figure 13. Γ(viii)\Gamma^{(viii)} with ρ={ea,hb,cd,fg,ih,je,hde}\rho=\{ea,hb,cd,fg,ih,je,hde\}

    However if we add ifcbaifcba to ρ\rho for the same quiver then choosing the same 𝔵0\mathfrak{x}_{0} and 𝔲\mathfrak{u} we get that 𝔶1=D\mathfrak{y}_{1}=D, π(𝔲)\pi(\mathfrak{u}) is ss-long and the uncle 𝔲1:=jD\mathfrak{u}_{1}:=jD of 𝔲\mathfrak{u} is short, 𝔲1𝔣1π(𝔲)\mathfrak{u}_{1}\sqsubset^{\mathfrak{f}}_{1}\pi(\mathfrak{u}) and 𝔣(𝔵0;π(𝔲),𝔲1)=HR𝔰(π(𝔲))(𝔶1𝔴~(𝔵0;𝔲)𝔠(𝔵0;π(𝔲)))=Dba\mathfrak{f}(\mathfrak{x}_{0};\pi(\mathfrak{u}),\mathfrak{u}_{1})=\mathrm{HR}_{\mathfrak{s}(\pi(\mathfrak{u}))}(\mathfrak{y}_{1}\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u})\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u})))=Dba.

  4. (4)

    In the algebra Γ(vi)\Gamma^{(vi)} in Figure 11 choosing 𝔵0\mathfrak{x}_{0}, 𝔲\mathfrak{u} same as in Example 7.12 we get υ(𝔲)=1\upsilon(\mathfrak{u})=1, θ(β(π(𝔲)))=θ(β(𝔲))=1\theta(\beta(\pi(\mathfrak{u})))=-\theta(\beta(\mathfrak{u}))=-1 and both π(𝔲)\pi(\mathfrak{u}) and the uncle 𝔲1\mathfrak{u}_{1} of 𝔲\mathfrak{u} are short. Moreover 𝔣(𝔵0;π(𝔲),𝔲1)=𝔴~(𝔵0;𝔲)𝔠(𝔵0;π(𝔲))=ecab\mathfrak{f}(\mathfrak{x}_{0};\pi(\mathfrak{u}),\mathfrak{u}_{1})=\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u})\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u}))=ecab.

Remark 7.27.

The last conclusion of Proposition 7.24 guarantees that there is no variation of the granduncle forking lemma when θ(β(π(𝔲)))=θ(β(𝔲))\theta(\beta(\pi(\mathfrak{u})))=-\theta(\beta(\mathfrak{u})).

8. Building H-reduced strings

Associate to each 𝔲𝒱𝔣i(𝔵0)\mathfrak{u}\in\mathcal{V}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0}) and j{1,1}j\in\{1,-1\} the maximal path

𝒫j(𝔲):=(𝔵0,𝔲0,𝔲1,,𝔲k=𝔲,𝔲k+1,,𝔲k[j])\mathcal{P}_{j}(\mathfrak{u}):=(\mathfrak{x}_{0},\mathfrak{u}_{0},\mathfrak{u}_{1},\ldots,\mathfrak{u}_{k}=\mathfrak{u},\mathfrak{u}_{k+1},\ldots,\mathfrak{u}_{k[j]})

in 𝒯𝔣i(𝔵0)\mathcal{T}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0}) such that ϕ(𝔲p+1)=j\phi(\mathfrak{u}_{p+1})=-j and 𝔲p+1f\mathfrak{u}_{p+1}^{f-} does not exist for each kp<k[j]k\leq p<k[j]. In particular, if 𝔲kfi(𝔵0)𝒵fi(𝔵0)\mathfrak{u}_{k}\in\mathcal{R}^{f}_{i}(\mathfrak{x}_{0})\cup\mathcal{Z}^{f}_{i}(\mathfrak{x}_{0}) then 𝔲k[±1]=𝔲k\mathfrak{u}_{k[\pm 1]}=\mathfrak{u}_{k}.

Set 𝔲min:=ξfmini(𝔵0)[i]\mathfrak{u}_{\mathrm{min}}:=\xi^{\mathrm{fmin}}_{i}(\mathfrak{x}_{0})_{[i]} and 𝔲max:=ξfmaxi(𝔵0)[i]\mathfrak{u}_{\mathrm{max}}:=\xi^{\mathrm{fmax}}_{i}(\mathfrak{x}_{0})_{[-i]}.

Remark 8.1.

Suppose 𝔲𝒱𝔣i(𝔵0)\mathfrak{u}\in\mathcal{V}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0}) and j{1,1}j\in\{1,-1\}. If 𝔲𝒫j(𝔲)\mathfrak{u}^{\prime}\in\mathcal{P}_{j}(\mathfrak{u}) and h(𝔲)>h(𝔲)h(\mathfrak{u}^{\prime})>h(\mathfrak{u}) then 𝒫j(𝔲)=𝒫j(𝔲)\mathcal{P}_{j}(\mathfrak{u})=\mathcal{P}_{j}(\mathfrak{u}^{\prime}), and hence 𝔲[j]=𝔲[j]\mathfrak{u}_{[j]}=\mathfrak{u}^{\prime}_{[j]}.

Proposition 8.2.

Suppose 𝔲𝒱𝔣i(𝔵0)\mathfrak{u}\in\mathcal{V}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0}) and h(𝔲)2h(\mathfrak{u})\geq 2. Then 𝔣(𝔵0;π(𝔲),π2(𝔲))\mathfrak{f}(\mathfrak{x}_{0};\pi(\mathfrak{u}),\pi^{2}(\mathfrak{u})) is a proper left substring of 𝔣(𝔵0;𝔲,π(𝔲))\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u},\pi(\mathfrak{u})).

Proof. Suppose h(𝔲)=2h(\mathfrak{u})=2. Then π2(𝔲)=𝔵0\pi^{2}(\mathfrak{u})=\mathfrak{x}_{0} and hence 𝔣(𝔵0;π(𝔲),π2(𝔲))=𝔵0\mathfrak{f}(\mathfrak{x}_{0};\pi(\mathfrak{u}),\pi^{2}(\mathfrak{u}))=\mathfrak{x}_{0}. Let αi\alpha_{i} be the syllable with θ(αi)=i\theta(\alpha_{i})=i such that αi𝔵0\alpha_{i}\mathfrak{x}_{0} is a string. Since π(𝔲)𝔣i(𝔵0)\pi(\mathfrak{u})\in\mathcal{H}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0}), αi\alpha_{i} is a syllable of 𝔣(𝔵0;π(𝔲))\mathfrak{f}(\mathfrak{x}_{0};\pi(\mathfrak{u})). Moreover 𝔲Hπ(𝔲)\mathfrak{u}\circ_{H}\pi(\mathfrak{u}) is a weak half ii-arch bridge and hence αi\alpha_{i} is also a syllable of 𝔣(𝔵0;𝔲)\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u}) and hence the conclusion.

Suppose h(𝔲)>2h(\mathfrak{u})>2. Both 𝔣(𝔵0;π(𝔲),π2(𝔲))\mathfrak{f}(\mathfrak{x}_{0};\pi(\mathfrak{u}),\pi^{2}(\mathfrak{u})) and 𝔣(𝔵0;𝔲,π(𝔲))\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u},\pi(\mathfrak{u})) are left substrings of 𝔣(𝔵0;π(𝔲))\mathfrak{f}(\mathfrak{x}_{0};\pi(\mathfrak{u})) and hence they are comparable. If 𝔣(𝔵0;π(𝔲),π2(𝔲))\mathfrak{f}(\mathfrak{x}_{0};\pi(\mathfrak{u}),\pi^{2}(\mathfrak{u})) is a proper substring of 𝔠(𝔵0;π(𝔲))\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u})) then the conclusion is obvious. Thus we assume that 𝔣(𝔵0;π(𝔲),π2(𝔲))=𝔷𝔠(𝔵0;π(𝔲))\mathfrak{f}(\mathfrak{x}_{0};\pi(\mathfrak{u}),\pi^{2}(\mathfrak{u}))=\mathfrak{z}\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u})) for some string 𝔷\mathfrak{z}. Then Proposition 4.6 gives that π(𝔲)\pi(\mathfrak{u}) is abnormal. Then Corollary 6.19 gives that the LVP (𝔱(π(𝔲)),𝔠(𝔵0,π(𝔲)))(\mathfrak{t}(\pi(\mathfrak{u})),\mathfrak{c}(\mathfrak{x}_{0},\pi(\mathfrak{u}))) is bb-long and hence 𝔯b1(π(𝔲))\mathfrak{r}^{b}_{1}(\pi(\mathfrak{u})) is a right substring of 𝔣(𝔵0;π(𝔲),π2(𝔲))\mathfrak{f}(\mathfrak{x}_{0};\pi(\mathfrak{u}),\pi^{2}(\mathfrak{u})).

If 𝔲\mathfrak{u} is short then 𝔯b1(π(𝔲))\mathfrak{r}^{b}_{1}(\pi(\mathfrak{u})) is a proper left substring of 𝔴~(𝔵0;𝔲)\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u}). On the other hand if 𝔲\mathfrak{u} is long then 𝔴(𝔵0;𝔲)\mathfrak{w}(\mathfrak{x}_{0};\mathfrak{u}) contains a cyclic permutation of 𝔟𝔣α(π(𝔲))\mathfrak{b}^{\mathfrak{f}\alpha}(\pi(\mathfrak{u})) as a left substring. Therefore the conclusion follows in both the cases.    \Box

Propositions 8.2 and 4.10 together guarantee, for each 𝔲𝒱𝔣i(𝔵0){𝔵0}\mathfrak{u}\in\mathcal{V}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0})\setminus\{\mathfrak{x}_{0}\} with 𝔲=ξfmin(π(𝔲))\mathfrak{u}=\xi^{\mathrm{fmin}}_{\partial}(\pi(\mathfrak{u})) for some {+,}\partial\in\{+,-\}, the existence of a string 𝔮0(𝔵0;𝔲)\mathfrak{q}_{0}(\mathfrak{x}_{0};\mathfrak{u}) of positive length satisfying

𝔮0(𝔵0;𝔲)𝔣(𝔵0;𝔲,π(𝔲)):={𝔣(𝔵0;𝔲,ξfminϕ(𝔲)(𝔲))if ξ𝔣ϕ(𝔲)(𝔲);𝔣(𝔵0;𝔲)if ξ𝔣ϕ(𝔲)(𝔲)=.\mathfrak{q}_{0}(\mathfrak{x}_{0};\mathfrak{u})\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u},\pi(\mathfrak{u})):=\begin{cases}\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u},\xi^{\mathrm{fmin}}_{\phi(\mathfrak{u})}(\mathfrak{u}))&\mbox{if }\xi^{\mathfrak{f}}_{\phi(\mathfrak{u})}(\mathfrak{u})\neq\emptyset;\\ \mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u})&\mbox{if }\xi^{\mathfrak{f}}_{\phi(\mathfrak{u})}(\mathfrak{u})=\emptyset.\end{cases}
Remark 8.3.

Suppose 𝔲𝒱𝔣i(𝔵0)\mathfrak{u}\in\mathcal{V}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0}), h(𝔲)>2h(\mathfrak{u})>2 and π(𝔲)\pi(\mathfrak{u}) is abnormal. In view of Proposition 8.2 there is a positive length string 𝔷¯\bar{\mathfrak{z}} such that 𝔣(𝔵0;𝔲,π(𝔲))=𝔷¯𝔣(𝔵0;π(𝔲),π2(𝔲))\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u},\pi(\mathfrak{u}))=\bar{\mathfrak{z}}\mathfrak{f}(\mathfrak{x}_{0};\pi(\mathfrak{u}),\pi^{2}(\mathfrak{u})). Since 𝔣(𝔵0;π(𝔲),π2(𝔲))=𝔯b1(π(𝔲))𝔠(𝔵0;π(𝔲))\mathfrak{f}(\mathfrak{x}_{0};\pi(\mathfrak{u}),\pi^{2}(\mathfrak{u}))=\mathfrak{r}^{b}_{1}(\pi(\mathfrak{u}))\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u})) and 𝔣(𝔵0;𝔲,π(𝔲))=𝔴(𝔵0,𝔲)𝔠(𝔵0;π(𝔲))\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u},\pi(\mathfrak{u}))=\mathfrak{w}(\mathfrak{x}_{0},\mathfrak{u})\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u})), we get 𝔴(𝔵0;𝔲)=𝔷¯𝔯b1(π(𝔲))\mathfrak{w}(\mathfrak{x}_{0};\mathfrak{u})=\bar{\mathfrak{z}}\mathfrak{r}^{b}_{1}(\pi(\mathfrak{u})).

Remark 8.4.

Suppose 𝔲𝒱𝔣i(𝔵0)\mathfrak{u}\in\mathcal{V}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0}) with h(𝔲)>2h(\mathfrak{u})>2 satisfies the hypotheses of the granduncle forking lemma (Lemma 7.22). If π(𝔲)=ξfmin(π2(𝔲))\pi(\mathfrak{u})=\xi^{\mathrm{fmin}}_{\partial}(\pi^{2}(\mathfrak{u})) and 𝔲=ξfmin(π(𝔲))\mathfrak{u}=\xi^{\mathrm{fmin}}_{\partial}(\pi(\mathfrak{u})) then 𝔴~(𝔵0;𝔲)𝔠(𝔵0;π(𝔲))\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u})\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u})) is a proper left substring of 𝔣(𝔵0;π(𝔲),π2(𝔲))\mathfrak{f}(\mathfrak{x}_{0};\pi(\mathfrak{u}),\pi^{2}(\mathfrak{u})).

Remark 8.5.

Suppose 𝔲𝒱𝔣i(𝔵0),𝔲𝔵0\mathfrak{u}\in\mathcal{V}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0}),\ \mathfrak{u}\neq\mathfrak{x}_{0} and ξ𝔣(𝔲)\xi^{\mathfrak{f}}_{\partial}(\mathfrak{u})\neq\emptyset. If 𝔲ξfmaxθ(β(𝔲))(π(𝔲))\mathfrak{u}\neq\xi^{\mathrm{fmax}}_{\theta(\beta(\mathfrak{u}))}(\pi(\mathfrak{u})) and β(𝔲)β(𝔲f+)\beta(\mathfrak{u})\neq\beta(\mathfrak{u}^{f+}) then 𝔣(𝔵0;𝔲,𝔲f+)=𝔣(𝔵0;𝔲,π(𝔲))\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u},\mathfrak{u}^{f+})=\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u},\pi(\mathfrak{u})).

Proposition 8.6.

Suppose 𝔲𝒱𝔣i(𝔵0)\mathfrak{u}\in\mathcal{V}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0}) and h(𝔲)>2h(\mathfrak{u})>2. If θ(β(π(𝔲)))=\theta(\beta(\pi(\mathfrak{u})))=\partial, 𝔲=ξfmin(π(𝔲))\mathfrak{u}=\xi^{\mathrm{fmin}}_{\partial}(\pi(\mathfrak{u})) and π(𝔲)f\pi(\mathfrak{u})^{f-} exists then 𝔣(𝔵0;π(𝔲),π(𝔲)f)\mathfrak{f}(\mathfrak{x}_{0};\pi(\mathfrak{u}),\pi(\mathfrak{u})^{f-}) is a proper left substring of 𝔣(𝔵0;π(𝔲),𝔲)\mathfrak{f}(\mathfrak{x}_{0};\pi(\mathfrak{u}),\mathfrak{u}).

Proof. Both 𝔣(𝔵0;π(𝔲),π(𝔲)f)\mathfrak{f}(\mathfrak{x}_{0};\pi(\mathfrak{u}),\pi(\mathfrak{u})^{f-}) and 𝔣(𝔵0;π(𝔲),𝔲)\mathfrak{f}(\mathfrak{x}_{0};\pi(\mathfrak{u}),\mathfrak{u}) are left substring of 𝔣(𝔵0;π(𝔲))\mathfrak{f}(\mathfrak{x}_{0};\pi(\mathfrak{u})) and hence are comparable. Corollary 4.4 states that 𝔠(𝔵0;π(𝔲))\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u})) is a left substring of 𝔥(𝔵0;𝒫(𝔲))\mathfrak{h}(\mathfrak{x}_{0};\mathcal{P}(\mathfrak{u})) and hence 𝔠(𝔵0;π(𝔲))\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u})) is a substring of 𝔣(𝔵0;π(𝔲),𝔲)\mathfrak{f}(\mathfrak{x}_{0};\pi(\mathfrak{u}),\mathfrak{u}). The conclusion is clear if 𝔣(𝔵0;π(𝔲),π(𝔲)f)\mathfrak{f}(\mathfrak{x}_{0};\pi(\mathfrak{u}),\pi(\mathfrak{u})^{f-}) is a proper left substring of 𝔠(𝔵0;π(𝔲))\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u})). Hence we assume otherwise.

If the LVP (𝔟,𝔶):=(𝔱(π(𝔲)),𝔣(𝔵0;π(𝔲),π(𝔲)f))(\mathfrak{b},\mathfrak{y}):=(\mathfrak{t}(\pi(\mathfrak{u})),\mathfrak{f}(\mathfrak{x}_{0};\pi(\mathfrak{u}),\pi(\mathfrak{u})^{f-})) is short then it can be readily seen that for some 𝔲𝒬¯Ba\mathfrak{u}^{\prime}\in\overline{\mathcal{HQ}}^{\mathrm{Ba}} we have π(𝔲)f=𝔲Hπ(𝔲)\pi(\mathfrak{u})^{f-}=\mathfrak{u}^{\prime}\circ_{H}\pi(\mathfrak{u}), a contradiction to π(𝔲)f𝒬Ba\pi(\mathfrak{u})^{f-}\in\mathcal{HQ}^{\mathrm{Ba}}. Hence the LVP (𝔟,𝔶)(\mathfrak{b},\mathfrak{y}) is long. Since θ(𝔣(𝔵0;π(𝔲)fπ(𝔲)))=\theta(\mathfrak{f}(\mathfrak{x}_{0};\pi(\mathfrak{u})^{f-}\mid\pi(\mathfrak{u})))=\partial, the LVP (𝔟,𝔶)(\mathfrak{b},\mathfrak{y}) can not be (s,b)(s,b)^{-\partial}-long and (b,s)(b,s)^{\partial}-long.

Now if 𝔣(𝔵0;π(𝔲),π(𝔲)f)=𝔷𝔠(𝔵0;π(𝔲))\mathfrak{f}(\mathfrak{x}_{0};\pi(\mathfrak{u}),\pi(\mathfrak{u})^{f-})=\mathfrak{z}\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u})) then 𝔷\mathfrak{z} is a proper left substring of 𝔟𝔣α(π(𝔲))\mathfrak{b}^{{\mathfrak{f}\alpha}}(\pi(\mathfrak{u})). If 𝔲\mathfrak{u} is long then 𝔣(𝔵0;π(𝔲),𝔲)=𝔴~(𝔵0;𝔲)(π(𝔲))𝔟𝔣α(π(𝔲))𝔠(𝔵0;π(𝔲))\mathfrak{f}(\mathfrak{x}_{0};\pi(\mathfrak{u}),\mathfrak{u})=\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u})(\pi(\mathfrak{u}))\mathfrak{b}^{\mathfrak{f}\alpha}{(\pi(\mathfrak{u}))}\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u})) and hence the conclusion follows. Thus it remains to consider the case when 𝔲\mathfrak{u} is short.

There are four cases:

  • (𝔟,𝔶)(\mathfrak{b},\mathfrak{y}) is bb^{\partial}-long. Here 𝔷=𝔯b1(π(𝔲))\mathfrak{z}=\mathfrak{r}^{b}_{1}(\pi(\mathfrak{u})) but since 𝔲\mathfrak{u} is short, 𝔯b1(π(𝔲))\mathfrak{r}^{b}_{1}(\pi(\mathfrak{u})) is a proper left substring of 𝔴~(𝔵0;𝔲)\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u}).

  • (𝔟,𝔶)(\mathfrak{b},\mathfrak{y}) is bb^{-\partial}-long or (b,s)(b,s)^{-\partial}-long. Since 𝔲\mathfrak{u} is short, 𝔯b1(π(𝔲))\mathfrak{r}^{b}_{1}(\pi(\mathfrak{u})) is a proper left substring of 𝔴~(𝔵0;𝔲)\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u}) while 𝔷\mathfrak{z} is a proper left substring of 𝔯b1(π(𝔲))\mathfrak{r}^{b}_{1}(\pi(\mathfrak{u})).

  • (𝔟,𝔶)(\mathfrak{b},\mathfrak{y}) is ss^{\partial}-long. If |𝔯s2(π(𝔲))|=1|\mathfrak{r}^{s}_{2}(\pi(\mathfrak{u}))|=1 then 𝔣(𝔵0;π(𝔲)f)\mathfrak{f}(\mathfrak{x}_{0};\pi(\mathfrak{u})^{f-}) is a left substring of 𝔣(𝔵0;π(𝔲))\mathfrak{f}(\mathfrak{x}_{0};\pi(\mathfrak{u})) and hence π(𝔲)=π(𝔲)f\pi(\mathfrak{u})=\pi(\mathfrak{u})^{f-} by Proposition 4.10, which is a contradiction. Hence |𝔯s2(π(𝔲))|>1|\mathfrak{r}^{s}_{2}(\pi(\mathfrak{u}))|>1 and 𝔷=𝔯s1(π(𝔲))\mathfrak{z}=\mathfrak{r}^{s}_{1}(\pi(\mathfrak{u})) while 𝔯s1(π(𝔲))\mathfrak{r}^{s}_{1}(\pi(\mathfrak{u})) is a proper substring of 𝔴~(𝔵0;𝔲)\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u}) since 𝔲\mathfrak{u} is short.

  • (𝔟,𝔶)(\mathfrak{b},\mathfrak{y}) is ss^{-\partial}-long or (s,b)(s,b)^{\partial}-long. The proof is similar to the second case above after replacing 𝔯b1\mathfrak{r}^{b}_{1} by 𝔯s1\mathfrak{r}^{s}_{1}.

This completes the proof.    \Box

Propositions 8.6 and 4.10 together guarantee, for each 𝔲𝒱𝔣i(𝔵0){𝔵0}\mathfrak{u}\in\mathcal{V}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0})\setminus\{\mathfrak{x}_{0}\} for which 𝔲f\mathfrak{u}^{f-} exists, the existence of a string 𝔮ϕ(𝔲)(𝔵0;𝔲)\mathfrak{q}_{\phi(\mathfrak{u})}(\mathfrak{x}_{0};\mathfrak{u}) of positive length satisfying

𝔮ϕ(𝔲)(𝔵0;𝔲)𝔣(𝔵0;𝔲,𝔲f):={𝔣(𝔵0;𝔲,ξfminϕ(𝔲)(𝔲))if ξ𝔣ϕ(𝔲)(𝔲);𝔣(𝔵0;𝔲)if ξ𝔣ϕ(𝔲)(𝔲)=.\mathfrak{q}_{\phi(\mathfrak{u})}(\mathfrak{x}_{0};\mathfrak{u})\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u},\mathfrak{u}^{f-}):=\begin{cases}\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u},\xi^{\mathrm{fmin}}_{\phi(\mathfrak{u})}(\mathfrak{u}))&\mbox{if }\xi^{\mathfrak{f}}_{\phi(\mathfrak{u})}(\mathfrak{u})\neq\emptyset;\\ \mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u})&\mbox{if }\xi^{\mathfrak{f}}_{\phi(\mathfrak{u})}(\mathfrak{u})=\emptyset.\end{cases}

Using similar arguments as in the proofs in Propositions 8.2 and 8.6 we get the following result.

Proposition 8.7.

Suppose 𝔲𝒱𝔣i(𝔵0)\mathfrak{u}\in\mathcal{V}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0}) and h(𝔲)2h(\mathfrak{u})\geq 2. If 𝔲=ξfminθ(β(π(𝔲)))(π(𝔲))\mathfrak{u}=\xi^{\mathrm{fmin}}_{-\theta(\beta(\pi(\mathfrak{u})))}(\pi(\mathfrak{u})) and π(𝔲)f+\pi(\mathfrak{u})^{f+} exists then 𝔣(𝔵0;π(𝔲),π(𝔲)f+)\mathfrak{f}(\mathfrak{x}_{0};\pi(\mathfrak{u}),\pi(\mathfrak{u})^{f+}) is a proper left substring of 𝔣(𝔵0;π(𝔲),𝔲)\mathfrak{f}(\mathfrak{x}_{0};\pi(\mathfrak{u}),\mathfrak{u}).

Propositions 8.7 and 4.10 together guarantee, for each 𝔲𝒱𝔣i(𝔵0){𝔵0}\mathfrak{u}\in\mathcal{V}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0})\setminus\{\mathfrak{x}_{0}\} for which 𝔲f+\mathfrak{u}^{f+} exists, the existence of a string 𝔮ϕ(𝔲)(𝔵0;𝔲)\mathfrak{q}_{-\phi(\mathfrak{u})}(\mathfrak{x}_{0};\mathfrak{u}) of positive length satisfying

𝔮ϕ(𝔲)(𝔵0;𝔲)𝔣(𝔵0;𝔲,𝔲f+):={𝔣(𝔵0;𝔲,ξfminϕ(𝔲)(𝔲))if ξ𝔣ϕ(𝔲)(𝔲);𝔣(𝔵0;𝔲)if ξ𝔣ϕ(𝔲)(𝔲)=.\mathfrak{q}_{-\phi(\mathfrak{u})}(\mathfrak{x}_{0};\mathfrak{u})\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u},\mathfrak{u}^{f+}):=\begin{cases}\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u},\xi^{\mathrm{fmin}}_{-\phi(\mathfrak{u})}(\mathfrak{u}))&\mbox{if }\xi^{\mathfrak{f}}_{-\phi(\mathfrak{u})}(\mathfrak{u})\neq\emptyset;\\ \mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u})&\mbox{if }\xi^{\mathfrak{f}}_{-\phi(\mathfrak{u})}(\mathfrak{u})=\emptyset.\end{cases}

Combining Propositions 8.2, 8.6 and 8.7 we get the following.

Corollary 8.8.

Suppose |𝒱𝔣i(𝔵0)|>1|\mathcal{V}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0})|>1, 𝔲𝒱𝔣i(𝔵0)\mathfrak{u}\in\mathcal{V}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0}) and h(𝔲)1h(\mathfrak{u})\geq 1. Then

  • 𝔣(𝔵0;𝔲min)=𝔮0(𝔵0;𝔲min)𝔮0(𝔵0;π(𝔲min))𝔮0(𝔵0;πh(𝔲min)1(𝔲min))𝔵0\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u}_{\mathrm{min}})=\mathfrak{q}_{0}(\mathfrak{x}_{0};\mathfrak{u}_{\mathrm{min}})\mathfrak{q}_{0}(\mathfrak{x}_{0};\pi(\mathfrak{u}_{\mathrm{min}}))\ldots\mathfrak{q}_{0}(\mathfrak{x}_{0};\pi^{h(\mathfrak{u}_{\mathrm{min}})-1}(\mathfrak{u}_{\mathrm{min}}))\mathfrak{x}_{0};

  • 𝔣(𝔵0;𝔲[ϕ(𝔲)]𝔲f)=𝔮0(𝔵0;𝔲[ϕ(𝔲)])𝔮0(𝔵0;ξfminϕ(𝔲)(𝔲))𝔮ϕ(𝔲)(𝔵0;𝔲)𝔣(𝔵0;𝔲,𝔲f)\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u}_{[-\phi(\mathfrak{u})]}\mid\mathfrak{u}^{f-})=\mathfrak{q}_{0}(\mathfrak{x}_{0};\mathfrak{u}_{[-\phi(\mathfrak{u})]})\ldots\mathfrak{q}_{0}(\mathfrak{x}_{0};\xi^{\mathrm{fmin}}_{\phi(\mathfrak{u})}(\mathfrak{u}))\mathfrak{q}_{\phi(\mathfrak{u})}(\mathfrak{x}_{0};\mathfrak{u})\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u},\mathfrak{u}^{f-})
    if 𝔲f\mathfrak{u}^{f-} exists;

  • 𝔣(𝔵0;𝔲[ϕ(𝔲)]𝔲f+)=𝔮0(𝔵0;𝔲[ϕ(𝔲)])𝔮0(𝔵0;ξfminϕ(𝔲)(𝔲))𝔮ϕ(𝔲)(𝔵0;𝔲)𝔣(𝔵0;𝔲,𝔲f+)\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u}_{[\phi(\mathfrak{u})]}\mid\mathfrak{u}^{f+})=\mathfrak{q}_{0}(\mathfrak{x}_{0};\mathfrak{u}_{[\phi(\mathfrak{u})]})\ldots\mathfrak{q}_{0}(\mathfrak{x}_{0};\xi^{\mathrm{fmin}}_{-\phi(\mathfrak{u})}(\mathfrak{u}))\mathfrak{q}_{-\phi(\mathfrak{u})}(\mathfrak{x}_{0};\mathfrak{u})\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u},\mathfrak{u}^{f+})
    if 𝔲f+\mathfrak{u}^{f+} exists.

For each n0n\geq 0, let 𝒱𝔣i(𝔵0)n:={𝔲𝒱𝔣i(𝔵0)h(𝔲)=n}\mathcal{V}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0})_{n}:=\{\mathfrak{u}\in\mathcal{V}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0})\mid h(\mathfrak{u})=n\}. Note that one can identify 𝔲𝒱𝔣i(𝔵0)n\mathfrak{u}\in\mathcal{V}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0})_{n} with the path (𝔵0,πn1(𝔲),,𝔲)(\mathfrak{x}_{0},\pi^{n-1}(\mathfrak{u}),\ldots,\mathfrak{u}).

Definition 8.9.

Let

𝒱¯𝔣i(𝔵0):={(𝔲,(s1,s2,,sh(𝔲)1))𝔲𝒱𝔣i(𝔵0),(s1,s2,,sh(𝔲)1)h(𝔲)1}.\overline{\mathcal{V}}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0}):=\{(\mathfrak{u},(s_{1},s_{2},\ldots,s_{h(\mathfrak{u})-1}))\mid\mathfrak{u}\in\mathcal{V}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0}),(s_{1},s_{2},\ldots,s_{h(\mathfrak{u})-1})\in\mathbb{N}^{h(\mathfrak{u})-1}\}.

Define 𝔣(𝔵0;(𝔲,(s1,s2,,sh(𝔲)1)))\mathfrak{f}(\mathfrak{x}_{0};(\mathfrak{u},(s_{1},s_{2},\ldots,s_{h(\mathfrak{u})-1}))) by the identity

R𝔰(𝔲)sh(𝔲)1R𝔰(πh(𝔲)2(𝔲))s1𝔣(𝔵0;(𝔲,(s1,s2,,sh(𝔲)1)))=𝔣(𝔵0;𝔲).\mathrm{R}_{\mathfrak{s}(\mathfrak{u})}^{s_{h(\mathfrak{u})-1}}\ldots\mathrm{R}_{\mathfrak{s}(\pi^{h(\mathfrak{u})-2}(\mathfrak{u}))}^{s_{1}}\mathfrak{f}(\mathfrak{x}_{0};(\mathfrak{u},(s_{1},s_{2},\ldots,s_{h(\mathfrak{u})-1})))=\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u}).

Define π((𝔲,(s1,,sh(𝔲)1))):=(π(𝔲),(s1,,sh(𝔲)2)),ϕ((𝔲,(s1,,sh(𝔲)1))):=ϕ(𝔲)\pi((\mathfrak{u},(s_{1},\ldots,s_{h(\mathfrak{u})-1}))):=(\pi(\mathfrak{u}),(s_{1},\ldots,s_{h(\mathfrak{u})-2})),\phi((\mathfrak{u},(s_{1},\ldots,s_{h(\mathfrak{u})-1}))):=\phi(\mathfrak{u}).

Given distinct (𝔲,(s1,s2,,sh(𝔲)2),s),(𝔲,(s1,s2,,sh(𝔲)2,t))(\mathfrak{u},(s_{1},s_{2},\ldots,s_{h(\mathfrak{u})-2}),s),(\mathfrak{u}^{\prime},(s_{1},s_{2},\ldots,s_{h(\mathfrak{u})-2},t)) with π(𝔲)=π(𝔲)\pi(\mathfrak{u})=\pi(\mathfrak{u}^{\prime}) and 𝔲,𝔲ξ𝔣(π(𝔲))\mathfrak{u},\mathfrak{u}^{\prime}\in\xi^{\mathfrak{f}}_{\partial}(\pi(\mathfrak{u})), say that (𝔲,(s1,s2,,sh(𝔲)2,s))𝔣(𝔲,(s1,s2,,sh(𝔲)2,t))(\mathfrak{u},(s_{1},s_{2},\ldots,s_{h(\mathfrak{u})-2},s))\sqsubset^{\mathfrak{f}}_{\partial}(\mathfrak{u}^{\prime},(s_{1},s_{2},\ldots,s_{h(\mathfrak{u})-2},t)) if either 𝔲𝔣𝔲\mathfrak{u}\sqsubset^{\mathfrak{f}}_{\partial}\mathfrak{u}^{\prime} or 𝔲=𝔲\mathfrak{u}=\mathfrak{u}^{\prime} but s<ts<t. Thus we have completed the description of a decorated tree 𝒯¯𝔣i(𝔵0)\overline{\mathcal{T}}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0}) whose vertex set is 𝒱¯𝔣i(𝔵0)\overline{\mathcal{V}}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0}).

There is a natural embedding 𝒱𝔣i(𝔵0)𝒱¯𝔣i(𝔵0)\mathcal{V}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0})\to\overline{\mathcal{V}}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0}) defined by 𝔲(𝔲,(0,0,,0))\mathfrak{u}\mapsto(\mathfrak{u},(0,0,\ldots,0)). Note that 𝔲¯f+\bar{\mathfrak{u}}^{f+} exists for each 𝔲𝒱¯𝔣i(𝔵0){𝔵0,ξfmaxi(𝔵0)}\mathfrak{u}\in\overline{\mathcal{V}}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0})\setminus\{\mathfrak{x}_{0},\xi^{\mathrm{fmax}}_{i}(\mathfrak{x}_{0})\}.

Equip 𝒱𝔣i(𝔵0)\mathcal{V}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0}) with inorder, denoted i\prec_{i}, where for 𝔲𝒱𝔣i(𝔵0){𝔵0}\mathfrak{u}\in\mathcal{V}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0})\setminus\{\mathfrak{x}_{0}\} we have

  • 𝔵0i𝔲\mathfrak{x}_{0}\prec_{i}\mathfrak{u};

  • 𝔲i𝔲\mathfrak{u}^{\prime}\prec_{i}\mathfrak{u}^{\prime\prime} if 𝔲𝔣i𝔲\mathfrak{u}^{\prime}\sqsubset^{\mathfrak{f}}_{i}\mathfrak{u}^{\prime\prime} in ξ𝔣i(𝔵0)\xi^{\mathfrak{f}}_{i}(\mathfrak{x}_{0});

  • 𝔲i𝔲\mathfrak{u}^{\prime}\prec_{i}\mathfrak{u} if 𝔲ξ𝔣i(𝔲)\mathfrak{u}^{\prime}\in\xi^{\mathfrak{f}}_{-i}(\mathfrak{u});

  • 𝔲i𝔲\mathfrak{u}\prec_{i}\mathfrak{u}^{\prime} if 𝔲ξ𝔣i(𝔲)\mathfrak{u}^{\prime}\in\xi^{\mathfrak{f}}_{i}(\mathfrak{u});

  • 𝔲i𝔲\mathfrak{u}^{\prime}\prec_{i}\mathfrak{u}^{\prime\prime} if 𝔲𝔣i𝔲\mathfrak{u}^{\prime}\sqsubset^{\mathfrak{f}}_{-i}\mathfrak{u}^{\prime\prime} in ξ𝔣i(𝔲)\xi^{\mathfrak{f}}_{-i}(\mathfrak{u}) for some 𝔲𝔵0\mathfrak{u}\neq\mathfrak{x}_{0};

  • 𝔲i𝔲\mathfrak{u}^{\prime\prime}\prec_{i}\mathfrak{u}^{\prime} if 𝔲𝔣i𝔲\mathfrak{u}^{\prime}\sqsubset^{\mathfrak{f}}_{i}\mathfrak{u}^{\prime\prime} in ξ𝔣i(𝔲)\xi^{\mathfrak{f}}_{i}(\mathfrak{u}) for some 𝔲𝔵0\mathfrak{u}\neq\mathfrak{x}_{0}.

Then (𝒱𝔣i(𝔵0),i)(\mathcal{V}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0}),\prec_{i}) is a total order with 𝔵0\mathfrak{x}_{0} as the minimal element, 𝔲min\mathfrak{u}_{\mathrm{min}} its successor and 𝔲max\mathfrak{u}_{\mathrm{max}} as the maximal element.

Remark 8.10.

Suppose 𝔲1,𝔲2𝒱𝔣i(𝔵0)\mathfrak{u}_{1},\mathfrak{u}_{2}\in\mathcal{V}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0}). Then 𝔲1i𝔲2\mathfrak{u}_{1}\prec_{i}\mathfrak{u}_{2} if and only if one of the following happens:

  • i=1i=1 and 𝔣(𝔵0;𝔲1)<l𝔣(𝔵0;𝔲2)\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u}_{1})<_{l}\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u}_{2}) in H^l(𝔵0)\hat{H}_{l}(\mathfrak{x}_{0});

  • i=1i=-1 and 𝔣(𝔵0;𝔲2)<l𝔣(𝔵0;𝔲1)\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u}_{2})<_{l}\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u}_{1}) in H^l(𝔵0)\hat{H}_{l}(\mathfrak{x}_{0}).

In fact the set 𝒱¯𝔣i(𝔵0)\overline{\mathcal{V}}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0}) can also be equipped with inorder in a similar manner.

9. Signature types of H-reduced strings

Proposition 9.1.

Suppose 𝔲𝒱𝔣i(𝔵0)\mathfrak{u}\in\mathcal{V}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0}), h(𝔲)>1h(\mathfrak{u})>1 and 𝔲=ξfmin(π(𝔲))\mathfrak{u}=\xi^{\mathrm{fmin}}_{\partial}(\pi(\mathfrak{u})). Then

Θ(𝔮0(𝔵0;𝔲))=C().\Theta(\mathfrak{q}_{0}(\mathfrak{x}_{0};\mathfrak{u}))=C(\partial).

Proof. We only prove the result when ξ𝔣(𝔲)\xi^{\mathfrak{f}}_{\partial}(\mathfrak{u})\neq\emptyset, and indicate the changes for the other case. Recall from Proposition 8.2 that |𝔮0(𝔵0;𝔲)|>0|\mathfrak{q}_{0}(\mathfrak{x}_{0};\mathfrak{u})|>0 and 𝔣(𝔵0;𝔲,ξfmin(𝔲))=𝔮0(𝔵0;𝔲)𝔣(𝔵0;𝔲,π(𝔲))\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u},\xi^{\mathrm{fmin}}_{\partial}(\mathfrak{u}))=\mathfrak{q}_{0}(\mathfrak{x}_{0};\mathfrak{u})\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u},\pi(\mathfrak{u})). Consider the following cases:

Case 1: 𝔲\mathfrak{u} is normal.

Then Proposition 4.6 gives that 𝔠(𝔵0;𝔲)=𝔷𝔣(𝔵0;𝔲,π(𝔲))\mathfrak{c}(\mathfrak{x}_{0};\mathfrak{u})=\mathfrak{z}\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u},\pi(\mathfrak{u})) for some string 𝔷\mathfrak{z} of positive length. Note that 𝔠(𝔵0;𝔲)=𝔲o𝔴(𝔵0;𝔲)𝔠(𝔵0;π(𝔲))\mathfrak{c}(\mathfrak{x}_{0};\mathfrak{u})=\mathfrak{u}^{o}\mathfrak{w}(\mathfrak{x}_{0};\mathfrak{u})\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u})) and θ(𝔮0(𝔵0;𝔲))=\theta(\mathfrak{q}_{0}(\mathfrak{x}_{0};\mathfrak{u}))=\partial.

Proposition 7.18 gives that Θ(𝔣(𝔵0;𝔲,π(𝔲));𝔲o𝔣(𝔵0;𝔲,π(𝔲)))=C()\Theta(\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u},\pi(\mathfrak{u}));\mathfrak{u}^{o}\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u},\pi(\mathfrak{u})))=C(\partial).

If all elements of ξ𝔣(𝔲)\xi^{\mathfrak{f}}_{\partial}(\mathfrak{u}) are short then using 𝔲=ξfmin(π(𝔲))\mathfrak{u}=\xi^{\mathrm{fmin}}_{\partial}(\pi(\mathfrak{u})) Proposition 7.16 gives that if the LVP (𝔱(𝔲),𝔠(𝔵0;𝔲))(\mathfrak{t}(\mathfrak{u}),\mathfrak{c}(\mathfrak{x}_{0};\mathfrak{u})) is bb^{\partial}-long then |𝔯b2(𝔲)|>1|\mathfrak{r}^{b}_{2}(\mathfrak{u})|>1, and hence the conclusion is readily verified. Therefore suppose that some element of ξ𝔣(𝔲)\xi^{\mathfrak{f}}_{\partial}(\mathfrak{u}) is long. Since 𝔲\mathfrak{u} is normal, any long 𝔲ξ𝔣(𝔲)\mathfrak{u}^{\prime}\in\xi^{\mathfrak{f}}_{\partial}(\mathfrak{u}) satisfies condition (2)(2) of the hypotheses of Lemma 7.3.

Claim: 𝔴~(𝔵0;𝔲)\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u}^{\prime}) is not a forking string for any 𝔲ξ𝔣(𝔲)\mathfrak{u}^{\prime}\in\xi^{\mathfrak{f}}_{\partial}(\mathfrak{u}).

If not then since 𝔲\mathfrak{u} is normal, condition (3)(3) of Proposition 7.20 fails, and hence 𝔲\mathfrak{u}^{\prime} satisfies condition (3)(3) of the hypotheses of Lemma 7.3. Now that all the hypotheses of that lemma are true, it ensures the existence of a sibling 𝔲1\mathfrak{u}_{1} of 𝔲\mathfrak{u}. Since 𝔲=ξfmin(π(𝔲))\mathfrak{u}=\xi^{\mathrm{fmin}}_{\partial}(\pi(\mathfrak{u})) we must have 𝔲𝔣𝔲1\mathfrak{u}\sqsubset^{\mathfrak{f}}_{\partial}\mathfrak{u}_{1} but this is impossible when 𝔲\mathfrak{u} is normal (see Step 3 of the proof of that lemma). Thus the claim.

It follows from the claim that all long 𝔲ξ𝔣(𝔲)\mathfrak{u}^{\prime}\in\xi^{\mathfrak{f}}_{\partial}(\mathfrak{u}) share the same exit syllable. Then it is readily verified that the conclusion follows.

When ξ𝔣(𝔲)=\xi^{\mathfrak{f}}_{\partial}(\mathfrak{u})=\emptyset we use Proposition 7.16 but not Lemma 7.3 in the argument.

Case 2: 𝔲\mathfrak{u} is abnormal.

Then Proposition 4.6 gives that 𝔣(𝔵0;𝔲,π(𝔲))=𝔷𝔠(𝔵0;𝔲)\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u},\pi(\mathfrak{u}))=\mathfrak{z}^{\prime}\mathfrak{c}(\mathfrak{x}_{0};\mathfrak{u}) for some string 𝔷\mathfrak{z}^{\prime}. Further Remark 8.3 gives that 𝔴(𝔵0;ξfmin(𝔲))=𝔮0(𝔵0;𝔲)𝔯b1(𝔲)\mathfrak{w}(\mathfrak{x}_{0};\xi^{\mathrm{fmin}}_{\partial}(\mathfrak{u}))=\mathfrak{q}_{0}(\mathfrak{x}_{0};\mathfrak{u})\mathfrak{r}^{b}_{1}(\mathfrak{u}). Since β(𝔲)\beta(\mathfrak{u}) is the first syllable of 𝔮0(𝔵0;𝔲)\mathfrak{q}_{0}(\mathfrak{x}_{0};\mathfrak{u}), we have θ(𝔷¯)=\theta(\bar{\mathfrak{z}})=\partial.

If ξfmin(𝔲)\xi^{\mathrm{fmin}}_{\partial}(\mathfrak{u}) is short then the conclusion is readily verified (one may need to use Proposition 7.16 when |𝔲c|=0|\mathfrak{u}^{c}|=0).

On the other hand, if ξfmin(𝔲)\xi^{\mathrm{fmin}}_{\partial}(\mathfrak{u}) is long then all elements of ξ𝔣(𝔲)\xi^{\mathfrak{f}}_{\partial}(\mathfrak{u}) are long. Let 𝔟𝔣α(𝔲)=𝔶¯𝔯b1(𝔲)\mathfrak{b}^{\mathfrak{f}\alpha}(\mathfrak{u})=\bar{\mathfrak{y}}\mathfrak{r}^{b}_{1}(\mathfrak{u}). An argument similar to the above paragraph shows that Θ(𝔶¯)=C()\Theta(\bar{\mathfrak{y}})=C(\partial). Since δ(𝔯b1(𝔲)𝔯~b(𝔲))=\delta(\mathfrak{r}^{b}_{1}(\mathfrak{u})\tilde{\mathfrak{r}}^{b}(\mathfrak{u}))=-\partial and 𝔯~b(𝔲)\tilde{\mathfrak{r}}^{b}(\mathfrak{u}) is a right substring of 𝔶¯\bar{\mathfrak{y}}, we can easily conclude that Θ(𝔮0(𝔵0;𝔲))=C()\Theta(\mathfrak{q}_{0}(\mathfrak{x}_{0};\mathfrak{u}))=C(\partial).    \Box

With minor modifications in the proof above, we can also show the following two results.

Proposition 9.2.

Θ(𝔮0(𝔵0;ξfmini(𝔵0)))=F(i)\Theta(\mathfrak{q}_{0}(\mathfrak{x}_{0};\xi^{\mathrm{fmin}}_{i}(\mathfrak{x}_{0})))=F(i).

Proposition 9.3.

Suppose 𝔲=ξfmaxi(𝔵0)\mathfrak{u}=\xi^{\mathrm{fmax}}_{i}(\mathfrak{x}_{0}) then the following hold.

  • Θ(𝔵0;𝔣(𝔵0;𝔲))=C(i)\Theta(\mathfrak{x}_{0};\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u}))=C(i) if ξ𝔣i(𝔲)=\xi^{\mathfrak{f}}_{i}(\mathfrak{u})=\emptyset;

  • Θ(𝔵0;𝔣(𝔵0;𝔲,ξfmini(𝔲)))=C(i)\Theta(\mathfrak{x}_{0};\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u},\xi^{\mathrm{fmin}}_{i}(\mathfrak{u})))=C(i) if ξ𝔣i(𝔲)\xi^{\mathfrak{f}}_{i}(\mathfrak{u})\neq\emptyset.

Proposition 9.4.

Suppose 𝔲𝒱𝔣i(𝔵0)\mathfrak{u}\in\mathcal{V}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0}) and h(𝔲)>1h(\mathfrak{u})>1. If :=θ(β(𝔲))\partial:=\theta(\beta(\mathfrak{u})) and 𝔲ξfmin(π(𝔲))\mathfrak{u}\neq\xi^{\mathrm{fmin}}_{\partial}(\pi(\mathfrak{u})) then Θ(𝔮(𝔵0;𝔲))=F()\Theta(\mathfrak{q}_{\partial}(\mathfrak{x}_{0};\mathfrak{u}))=F(-\partial).

Proof. We only prove the result when ξ𝔣(𝔲)\xi^{\mathfrak{f}}_{\partial}(\mathfrak{u})\neq\emptyset; the proof of the other case is omitted as it is very similar with only minor modifications. Recall from Proposition 8.6 that |𝔮(𝔵0;𝔲)|>0|\mathfrak{q}_{\partial}(\mathfrak{x}_{0};\mathfrak{u})|>0 and 𝔣(𝔵0;𝔲,ξfmin(𝔲))=𝔮(𝔵0;𝔲)𝔣(𝔵0;𝔲,𝔲f)\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u},\xi^{\mathrm{fmin}}_{\partial}(\mathfrak{u}))=\mathfrak{q}_{\partial}(\mathfrak{x}_{0};\mathfrak{u})\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u},\mathfrak{u}^{f-}). Clearly θ(𝔮(𝔵0;𝔲))=\theta(\mathfrak{q}_{\partial}(\mathfrak{x}_{0};\mathfrak{u}))=-\partial and 𝔣(𝔵0;𝔲,𝔲f)𝔣(𝔵0;𝔲,π(𝔲))\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u},\mathfrak{u}^{f-})\neq\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u},\pi(\mathfrak{u})). If Θ(𝔮(𝔵0;𝔲))F()\Theta(\mathfrak{q}_{\partial}(\mathfrak{x}_{0};\mathfrak{u}))\neq F(-\partial) then there is a string 𝔶\mathfrak{y} of positive length such that 𝔶𝔣(𝔵0;𝔲,𝔲f)\mathfrak{y}\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u},\mathfrak{u}^{f-}) is a proper left forking substring of 𝔣(𝔵0;𝔲,ξfmin(𝔲))\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u},\xi^{\mathrm{fmin}}_{\partial}(\mathfrak{u})) and θ(𝔶𝔣(𝔵0;𝔲,𝔲f);𝔣(𝔵0;𝔲,π(𝔲)))=\theta(\mathfrak{y}\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u},\mathfrak{u}^{f-});\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u},\pi(\mathfrak{u})))=-\partial. Let α\alpha be the syllable such that θ(α)=\theta(\alpha)=\partial and α𝔶𝔣(𝔵0;𝔲,𝔲f)\alpha\mathfrak{y}\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u},\mathfrak{u}^{f-}) is a string.

There are two main cases and several subcases:

Case 1: 𝔲\mathfrak{u} is normal.

Here Proposition 4.6 gives that 𝔠(𝔵0;𝔲)=𝔷𝔣(𝔵0;𝔲,π(𝔲))\mathfrak{c}(\mathfrak{x}_{0};\mathfrak{u})=\mathfrak{z}\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u},\pi(\mathfrak{u})) for some string 𝔷\mathfrak{z} of positive length. We also have 𝔠(𝔵0;𝔲)=𝔲o𝔴(𝔵0;𝔲)𝔠(𝔵0;π(𝔲))\mathfrak{c}(\mathfrak{x}_{0};\mathfrak{u})=\mathfrak{u}^{o}\mathfrak{w}(\mathfrak{x}_{0};\mathfrak{u})\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u})).

Case 1A: 𝔣(𝔵0;𝔲,π(𝔲))=𝔷𝔣(𝔵0;𝔲,𝔲f)\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u},\pi(\mathfrak{u}))=\mathfrak{z}\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u},\mathfrak{u}^{f-}) for some string 𝔷\mathfrak{z} of positive length.

Suppose 𝔶𝔣(𝔵0;𝔲,𝔲f)\mathfrak{y}\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u},\mathfrak{u}^{f-}) is a proper left substring of 𝔣(𝔵0;𝔲,π(𝔲))\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u},\pi(\mathfrak{u})). Since θ(𝔶)=\theta(\mathfrak{y})=-\partial we have δ(α𝔶)=0\delta(\alpha\mathfrak{y})=0. Moreover since α𝔶\alpha\mathfrak{y} is a string we conclude that α(𝔰(𝔲))\alpha\in\mathcal{E}(\mathfrak{s}(\mathfrak{u})). Now [4, Proposition 8.5] guarantees that there exists a semi-bridge or a torsion reverse semi-bridge, say 𝔲1\mathfrak{u}_{1}, such that 𝔰(𝔲1)=𝔰(𝔲)\mathfrak{s}(\mathfrak{u}_{1})=\mathfrak{s}(\mathfrak{u}) and β(𝔲1)=α\beta(\mathfrak{u}_{1})=\alpha while [4, Proposition 8.7] ensures that 𝔲1𝒬Ba\mathfrak{u}_{1}\in\mathcal{HQ}^{\mathrm{Ba}}. Then we have 𝔲f𝔣𝔲𝔣𝔲\mathfrak{u}^{f-}\sqsubset^{\mathfrak{f}}_{\partial}\mathfrak{u}^{\prime}\sqsubset^{\mathfrak{f}}_{\partial}\mathfrak{u}, a contradiction to the definition of 𝔲f\mathfrak{u}^{f-}.

Proposition 7.18 gives that if 𝔣(𝔵0;𝔲,π(𝔲))\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u},\pi(\mathfrak{u})) is a proper left substring of 𝔶𝔣(𝔵0;𝔲,𝔲f)\mathfrak{y}\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u},\mathfrak{u}^{f-}) and the latter is a substring of 𝔠(𝔵0;𝔲)\mathfrak{c}(\mathfrak{x}_{0};\mathfrak{u}) then there is a sibling 𝔲1\mathfrak{u}_{1} of 𝔲\mathfrak{u} with β(𝔲1)=β(𝔲)\beta(\mathfrak{u}_{1})=\beta(\mathfrak{u}) and 𝔣(𝔵0;𝔲,𝔲1)=𝔶𝔣(𝔵0;𝔲,𝔲f)\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u},\mathfrak{u}_{1})=\mathfrak{y}\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u},\mathfrak{u}^{f-}). As a consequence we get 𝔲f𝔣𝔲1𝔣𝔲\mathfrak{u}^{f-}\sqsubset^{\mathfrak{f}}_{\partial}\mathfrak{u}_{1}\sqsubset^{\mathfrak{f}}_{\partial}\mathfrak{u}, a contradiction to the definition of 𝔲f\mathfrak{u}^{f-}.

Finally an argument similar to the proof of Case 1 of Proposition 9.1 using Lemma 7.3 and Proposition 7.16 shows that 𝔠(𝔵0;𝔲)\mathfrak{c}(\mathfrak{x}_{0};\mathfrak{u}) is not a substring of 𝔶𝔣(𝔵0;𝔲,𝔲f)\mathfrak{y}\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u},\mathfrak{u}^{f-}).

Case 1B: 𝔣(𝔵0;𝔲,𝔲f)=𝔷𝔣(𝔵0;𝔲,π(𝔲))\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u},\mathfrak{u}^{f-})=\mathfrak{z}\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u},\pi(\mathfrak{u})) is a proper left substring of 𝔠(𝔵0;𝔲)\mathfrak{c}(\mathfrak{x}_{0};\mathfrak{u}) for some string 𝔷\mathfrak{z} of positive length.

The proof of this subcase is similar to the above two paragraphs.

Case 1C: 𝔣(𝔵0;𝔲,𝔲f)=𝔷𝔠(𝔵0;𝔲)\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u},\mathfrak{u}^{f-})=\mathfrak{z}\mathfrak{c}(\mathfrak{x}_{0};\mathfrak{u}) for some string 𝔷\mathfrak{z}.

Under this assumption we showed in the proof of Proposition 8.6 that the LVP (𝔱(𝔲),𝔣(𝔵0;𝔲,𝔲f))(\mathfrak{t}(\mathfrak{u}),\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u},\mathfrak{u}^{f-})) is long. Let k:=max{|𝔯b1(𝔲)|,|𝔯s1(𝔲)|}k:=\max\{|\mathfrak{r}^{b}_{1}(\mathfrak{u})|,|\mathfrak{r}^{s}_{1}(\mathfrak{u})|\}. Then |𝔷|k|\mathfrak{z}|\leq k.

Claim: There is no 𝔲ξ𝔣(𝔲)\mathfrak{u}^{\prime}\in\xi^{\mathfrak{f}}_{\partial}(\mathfrak{u}) satisfying |𝔷|<|𝔴~(𝔵0;𝔲)|k|\mathfrak{z}|<|\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u}^{\prime})|\leq k.

Suppose not. Then the LVP (𝔱(𝔲),𝔣(𝔵0;𝔲,𝔲f))(\mathfrak{t}(\mathfrak{u}),\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u},\mathfrak{u}^{f-})) is necessarily either bb^{-\partial}-long or ss^{-\partial}-long. Then the uncle forking lemma yields a sibling 𝔲1\mathfrak{u}_{1} of 𝔲\mathfrak{u} satisfying 𝔣(𝔵0;𝔲,𝔲1)=𝔴~(𝔵0;𝔲)𝔠(𝔵0;𝔲)\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u},\mathfrak{u}_{1})=\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u}^{\prime})\mathfrak{c}(\mathfrak{x}_{0};\mathfrak{u}). The existence of such 𝔲1\mathfrak{u}_{1} will contradict the definition of 𝔲f\mathfrak{u}^{f-}, and hence the claim.

Irrespective of whether ξfmin(𝔲)\xi^{\mathrm{fmin}}_{\partial}(\mathfrak{u}) is short or long, the conclusion follows using the claim.

Case 2: 𝔲\mathfrak{u} is abnormal.

Here Proposition 4.6 gives that 𝔣(𝔵0;𝔲,π(𝔲))=𝔷𝔠(𝔵0;𝔲)\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u},\pi(\mathfrak{u}))=\mathfrak{z}\mathfrak{c}(\mathfrak{x}_{0};\mathfrak{u}) for some string 𝔷\mathfrak{z}.

Case 2A: 𝔠(𝔵0;𝔲)=𝔷1𝔣(𝔵0;𝔲,𝔲f)\mathfrak{c}(\mathfrak{x}_{0};\mathfrak{u})=\mathfrak{z}_{1}\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u},\mathfrak{u}^{f-}) for some string 𝔷1\mathfrak{z}_{1} of positive length.

Since 𝔠(𝔵0;π(𝔲))\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u})) is a substring of 𝔣(𝔵0;𝔲,𝔲f)\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u},\mathfrak{u}^{f-}), it immediately follows that υ(𝔲)=1\upsilon(\mathfrak{u})=1. Corollary 6.19 gives that the LVP (𝔱(𝔲),𝔠(𝔵0;𝔲))(\mathfrak{t}(\mathfrak{u}),\mathfrak{c}(\mathfrak{x}_{0};\mathfrak{u})) is bb^{-\partial}-long. Furthermore Corollary 6.3 gives that 𝔯b1(𝔲)=𝔲c\mathfrak{r}^{b}_{1}(\mathfrak{u})=\mathfrak{u}^{c}. Let 𝔮(𝔵0;𝔲)=𝔷2𝔲c𝔷1\mathfrak{q}_{\partial}(\mathfrak{x}_{0};\mathfrak{u})=\mathfrak{z}_{2}\mathfrak{u}^{c}\mathfrak{z}_{1}, where |𝔷1||𝔷2|>0|\mathfrak{z}_{1}||\mathfrak{z}_{2}|>0.

  1. |𝔲c|>0|\mathfrak{u}^{c}|>0

    As in the first paragraph of Case 1A above we can show that Θ(𝔷1)=F()\Theta(\mathfrak{z}_{1})=F(-\partial). Since δ(β(𝔲)𝔲c)=0\delta(\beta(\mathfrak{u})\mathfrak{u}^{c})=0 we clearly get Θ(𝔷2)=C()\Theta(\mathfrak{z}_{2})=C(\partial). It remains to show that Θ(𝔲c𝔷1)=F()\Theta(\mathfrak{u}^{c}\mathfrak{z}_{1})=F(-\partial). Let 𝔲ξ𝔣(𝔲)\mathfrak{u}^{\prime}\in\xi^{\mathfrak{f}}_{\partial}(\mathfrak{u}). We show that β(𝔲)𝔴~(𝔵0;𝔲)𝔠(𝔵0;𝔲)\beta(\mathfrak{u}^{\prime})\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u}^{\prime})\mathfrak{c}(\mathfrak{x}_{0};\mathfrak{u}) is not a string.

    If 𝔲\mathfrak{u} is long then Corollary 6.6 states that 𝔲\mathfrak{u} is necessarily ss-long and 𝔴(𝔵0;𝔲)=𝔟𝔣α(π(𝔲))\mathfrak{w}(\mathfrak{x}_{0};\mathfrak{u})=\mathfrak{b}^{\mathfrak{f}\alpha}{(\pi(\mathfrak{u}))}. Since β(𝔲)β(𝔲f)\beta(\mathfrak{u})\neq\beta(\mathfrak{u}^{f-}), we can conclude that 𝔲f\mathfrak{u}^{f-} is always short. Hence 𝔣(𝔵0;𝔲f,π(𝔲))=𝔴~(𝔵0;𝔲f)𝔠(𝔵0;π(𝔲))\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u}^{f-},\pi(\mathfrak{u}))=\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u}^{f-})\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u})).

    If |𝔴~(𝔵0;𝔲)|>0|\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u}^{\prime})|>0 then the uncle forking lemma produces a sibling 𝔲1\mathfrak{u}_{1} of 𝔲\mathfrak{u} such that 𝔣(𝔵0;𝔲,𝔲1)=𝔴~(𝔵0;𝔲)𝔠(𝔵0;𝔲)\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u},\mathfrak{u}_{1})=\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u}^{\prime})\mathfrak{c}(\mathfrak{x}_{0};\mathfrak{u}). Given the assumption of Case 2A, we get 𝔲f𝔣𝔲1𝔣𝔲\mathfrak{u}^{f-}\sqsubset^{\mathfrak{f}}_{\partial}\mathfrak{u}_{1}\sqsubset^{\mathfrak{f}}_{\partial}\mathfrak{u}, a contradiction to the definition of 𝔲f\mathfrak{u}^{f-}.

    On the other hand if |𝔴~(𝔵0;𝔲)|=0|\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u}^{\prime})|=0 then δ(β(𝔲)𝔷1)=0\delta(\beta(\mathfrak{u}^{\prime})\mathfrak{z}_{1})=0. If β(𝔲)𝔴~(𝔵0;𝔲)𝔠(𝔵0;𝔲)\beta(\mathfrak{u})\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u}^{\prime})\mathfrak{c}(\mathfrak{x}_{0};\mathfrak{u}) is a string then there is some 𝔲ξ𝔣(π(𝔲))\mathfrak{u}^{\prime\prime}\in\xi^{\mathfrak{f}}_{\partial}(\pi(\mathfrak{u})) satisfying β(𝔲)=β(𝔲)\beta(\mathfrak{u}^{\prime\prime})=\beta(\mathfrak{u}^{\prime\prime}) and 𝔲f𝔣𝔲𝔣𝔲\mathfrak{u}^{f-}\sqsubset^{\mathfrak{f}}_{\partial}\mathfrak{u}^{\prime\prime}\sqsubset^{\mathfrak{f}}_{\partial}\mathfrak{u}, a contradiction. Thus β(𝔲)𝔴~(𝔵0;𝔲)𝔠(𝔵0;𝔲)\beta(\mathfrak{u})\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u}^{\prime})\mathfrak{c}(\mathfrak{x}_{0};\mathfrak{u}) is not a string. As a consequence 𝔴~(𝔵0;𝔲)𝔠(𝔵0;π(𝔲))\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u}^{\prime})\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u})) is not a forking string, and hence we have completed the proof that θ(𝔲c𝔷1)=F()\theta(\mathfrak{u}^{c}\mathfrak{z}_{1})=F(-\partial).

  2. |𝔲c|=0|\mathfrak{u}^{c}|=0

    As in the first paragraph of Case 1A above we can show that Θ(𝔷1)=F()\Theta(\mathfrak{z}_{1})=F(-\partial).

    If ξfmin(𝔲)\xi^{\mathrm{fmin}}_{\partial}(\mathfrak{u}) is short then Θ(𝔷2)=C()\Theta(\mathfrak{z}_{2})=C(\partial) follows from the fact δ(β(𝔲)γ¯b(𝔲))=0\delta(\beta(\mathfrak{u})\bar{\gamma}^{b}(\mathfrak{u}))=0.

    If ξfmin(𝔲)\xi^{\mathrm{fmin}}_{\partial}(\mathfrak{u}) is long then the LVP (𝔱(𝔲),𝔠(𝔵0;𝔲))(\mathfrak{t}(\mathfrak{u}),\mathfrak{c}(\mathfrak{x}_{0};\mathfrak{u})) is ss^{\partial}-long with 𝔯s1(𝔲)=𝔴~(𝔵0;ξfmin(𝔲))\mathfrak{r}^{s}_{1}(\mathfrak{u})=\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\xi^{\mathrm{fmin}}_{\partial}(\mathfrak{u})) and |𝔯s1(𝔲)|>0|\mathfrak{r}^{s}_{1}(\mathfrak{u})|>0.

    If ξfmin(𝔲)\xi^{\mathrm{fmin}}_{\partial}(\mathfrak{u}) satisfies the hypotheses of the uncle forking lemma then the lemma will produce a sibling 𝔲1\mathfrak{u}_{1} of 𝔲\mathfrak{u} satisfying 𝔲f𝔣𝔲1𝔣𝔲\mathfrak{u}^{f-}\sqsubset^{\mathfrak{f}}_{\partial}\mathfrak{u}_{1}\sqsubset^{\mathfrak{f}}_{\partial}\mathfrak{u}, a contradiction. Hence condition (3)(3) fails. Since δ(β(𝔲)𝔴~(𝔵0;ξfmin(𝔲))𝔷1)=0\delta(\beta(\mathfrak{u})\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\xi^{\mathrm{fmin}}_{\partial}(\mathfrak{u}))\mathfrak{z}_{1})=0 we conclude that |𝔯s2(𝔲)|=1|\mathfrak{r}^{s}_{2}(\mathfrak{u})|=1 and hence that 𝔴~(𝔵0;ξfmin(𝔲))𝔠(𝔵0;𝔲)\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\xi^{\mathrm{fmin}}_{\partial}(\mathfrak{u}))\mathfrak{c}(\mathfrak{x}_{0};\mathfrak{u}) is not a forking string. Then it immediately follows that Θ(𝔷2)=C()\Theta(\mathfrak{z}_{2})=C(\partial).

Case 2B: 𝔠(𝔵0;𝔲)=𝔣(𝔵0;𝔲,𝔲f)\mathfrak{c}(\mathfrak{x}_{0};\mathfrak{u})=\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u},\mathfrak{u}^{f-}).

If υ(𝔲)=1\upsilon(\mathfrak{u})=-1 then 𝔟𝔣α(𝔲)=𝔷~𝔯b1(𝔲)\mathfrak{b}^{\mathfrak{f}\alpha}(\mathfrak{u})=\tilde{\mathfrak{z}}\mathfrak{r}^{b}_{1}(\mathfrak{u}) for some string 𝔷~\tilde{\mathfrak{z}} of positive length with δ(𝔷~)=0\delta(\tilde{\mathfrak{z}})=0 by Corollary 6.12. Then it is readily verified that 𝔲f=𝔲H𝔲\mathfrak{u}^{f-}=\mathfrak{u}^{\prime\prime}\circ_{H}\mathfrak{u} for some 𝔲\mathfrak{u}^{\prime\prime}, a contradiction. Hence υ(𝔲)1\upsilon(\mathfrak{u})\neq-1. Let 𝔮(𝔵0;𝔲)=𝔷2𝔯b1(𝔲)\mathfrak{q}_{\partial}(\mathfrak{x}_{0};\mathfrak{u})=\mathfrak{z}_{2}\mathfrak{r}^{b}_{1}(\mathfrak{u}).

Note that β(𝔲)β(𝔲f)\beta(\mathfrak{u})\neq\beta(\mathfrak{u}^{f-}) for otherwise 𝔠(𝔵0;𝔲)\mathfrak{c}(\mathfrak{x}_{0};\mathfrak{u}) is a proper left substring of 𝔣(𝔵0;𝔲,𝔲f)\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u},\mathfrak{u}^{f-}), a contradiction to the assumption of this case. Hence |𝔯b1(𝔲)|>0|\mathfrak{r}^{b}_{1}(\mathfrak{u})|>0.

An easy impossibility argument using the uncle forking lemma shows that there is no 𝔲ξ𝔣(𝔲)\mathfrak{u}^{\prime}\in\xi^{\mathfrak{f}}_{\partial}(\mathfrak{u}) such that 0<|𝔴~(𝔵0;𝔲)|<|𝔯b1(𝔲)|0<|\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u}^{\prime})|<|\mathfrak{r}^{b}_{1}(\mathfrak{u})|. Hence Θ(𝔯b1(𝔲))=F()\Theta(\mathfrak{r}^{b}_{1}(\mathfrak{u}))=F(-\partial).

As |𝔯b1(𝔲)|>0|\mathfrak{r}^{b}_{1}(\mathfrak{u})|>0 we have δ(β(𝔲)𝔯b1(𝔲))=0\delta(\beta(\mathfrak{u})\mathfrak{r}^{b}_{1}(\mathfrak{u}))=0. Hence it is readily seen that Θ(𝔷2)=C()\Theta(\mathfrak{z}_{2})=C(\partial).

Case 2C: 𝔣(𝔵0;𝔲,𝔲f)=𝔷1𝔠(𝔵0;𝔲)\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u},\mathfrak{u}^{f-})=\mathfrak{z}_{1}\mathfrak{c}(\mathfrak{x}_{0};\mathfrak{u}) is a proper left substring of 𝔣(𝔵0;𝔲,π(𝔲))\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u},\pi(\mathfrak{u})) for some string 𝔷1\mathfrak{z}_{1} of positive length.

Since 𝔣(𝔵0;𝔲,𝔲f)\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u},\mathfrak{u}^{f-}) is a proper left substring of 𝔣(𝔵0;𝔲,π(𝔲))\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u},\pi(\mathfrak{u})), we have β(𝔲)β(𝔲f)\beta(\mathfrak{u})\neq\beta(\mathfrak{u}^{f-}).

If υ(𝔲)=1\upsilon(\mathfrak{u})=-1 then 𝔠(𝔵0;π(𝔲))\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u})) is a left substring of 𝔣(𝔵0;𝔲,𝔲f)\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u},\mathfrak{u}^{f-}) for otherwise β(𝔲)=β(𝔲f)\beta(\mathfrak{u})=\beta(\mathfrak{u}^{f-}) and hence 𝔲f\mathfrak{u}^{f-} factors through 𝔲\mathfrak{u} as explained in the first paragraph of Case 2B above.

Let 𝔮(𝔵0;𝔲)=𝔷2𝔷1\mathfrak{q}_{\partial}(\mathfrak{x}_{0};\mathfrak{u})=\mathfrak{z}_{2}\mathfrak{z}_{1} with |𝔷2||𝔷1|>0|\mathfrak{z}_{2}||\mathfrak{z}_{1}|>0 and 𝔷1𝔴~(𝔵0;𝔲f)𝔠(𝔵0;π(𝔲))=𝔯b1(𝔲)𝔠(𝔵0;𝔲)\mathfrak{z}_{1}\tilde{\mathfrak{w}}(\mathfrak{x}_{0};\mathfrak{u}^{f-})\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u}))=\mathfrak{r}^{b}_{1}(\mathfrak{u})\mathfrak{c}(\mathfrak{x}_{0};\mathfrak{u}). The argument that Θ(𝔷1)=F()\Theta(\mathfrak{z}_{1})=F(-\partial) and Θ(𝔷2)=C()\Theta(\mathfrak{z}_{2})=C(\partial) is similar to the proof of Case 2B.

Case 2D: 𝔣(𝔵0;𝔲,𝔲f)=𝔷3𝔣(𝔵0;𝔲,π(𝔲))\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u},\mathfrak{u}^{f-})=\mathfrak{z}_{3}\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u},\pi(\mathfrak{u})) for some string 𝔷3\mathfrak{z}_{3} of positive length.

Clearly β(𝔲)=β(𝔲f)\beta(\mathfrak{u})=\beta(\mathfrak{u}^{f-}), θ(𝔷3)=\theta(\mathfrak{z}_{3})=\partial and |𝔷3|>0|\mathfrak{z}_{3}|>0. We also have δ(𝔷3)=\delta(\mathfrak{z}_{3})=\partial for otherwise 𝔲f\mathfrak{u}^{f-} factors through 𝔲\mathfrak{u}.

As argued in the second paragraph of the proof of Proposition 8.6, we see that the LVP (𝔟,𝔶):=(𝔱(𝔲),𝔣(𝔵0;𝔲,𝔲f))(\mathfrak{b},\mathfrak{y}):=(\mathfrak{t}(\mathfrak{u}),\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u},\mathfrak{u}^{f-})) is long.

If υ(𝔲)=1\upsilon(\mathfrak{u})=-1 then Corollary 6.19 gives that 𝔣(𝔵0;𝔲,π(𝔲))=𝔲c𝔠(𝔵0;π(𝔲))=𝔯b1(𝔲)𝔠(𝔵0;𝔲)\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u},\pi(\mathfrak{u}))=\mathfrak{u}^{c}\mathfrak{c}(\mathfrak{x}_{0};\pi(\mathfrak{u}))=\mathfrak{r}^{b}_{1}(\mathfrak{u})\mathfrak{c}(\mathfrak{x}_{0};\mathfrak{u}). If |𝔲c|>0|\mathfrak{u}^{c}|>0 then δ(β(𝔲)𝔲c)=0\delta(\beta(\mathfrak{u})\mathfrak{u}^{c})=0. These two statements together contradict the fact that (𝔟,𝔶)(\mathfrak{b},\mathfrak{y}) is long. Therefore |𝔲c|=0|\mathfrak{u}^{c}|=0.

Further since (𝔟,𝔶)(\mathfrak{b},\mathfrak{y}) is long we get that 𝔲\mathfrak{u} is bb^{\partial}-long with |𝔯b1(π(𝔲))|=0|\mathfrak{r}^{b}_{1}(\pi(\mathfrak{u}))|=0 and 𝔷3\mathfrak{z}_{3} a proper left substring of 𝔯b2(π(𝔲))\mathfrak{r}^{b}_{2}(\pi(\mathfrak{u})), a contradiction by Proposition 6.16. Thus υ(𝔲)>1\upsilon(\mathfrak{u})>-1.

Now if |𝔯b1(𝔲)|>0|\mathfrak{r}^{b}_{1}(\mathfrak{u})|>0 then δ(β(𝔲)𝔯b1(𝔲))=0\delta(\beta(\mathfrak{u})\mathfrak{r}^{b}_{1}(\mathfrak{u}))=0. This together with the identity 𝔣(𝔵0;𝔲,π(𝔲))=𝔯b1(𝔲)𝔠(𝔵0;𝔲)\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u},\pi(\mathfrak{u}))=\mathfrak{r}^{b}_{1}(\mathfrak{u})\mathfrak{c}(\mathfrak{x}_{0};\mathfrak{u}) guaranteed by Corollary 6.19 give a contradiction the fact that (𝔟,𝔶)(\mathfrak{b},\mathfrak{y}) is long. Therefore |𝔯b1(𝔲)|=0|\mathfrak{r}^{b}_{1}(\mathfrak{u})|=0. If υ(𝔲)=0\upsilon(\mathfrak{u})=0 then (𝔟,𝔶)(\mathfrak{b},\mathfrak{y}) is long gives that 𝔲\mathfrak{u} is ss-long, again a contradiction by Proposition 6.16, and hence υ(𝔲)=1\upsilon(\mathfrak{u})=1.

In this case it is readily verified that (𝔟,𝔶)(\mathfrak{b},\mathfrak{y}) is ss^{\partial}-long with 𝔯s1(𝔲)𝔣(𝔵0;𝔲,π(𝔲))=𝔣(𝔵0;𝔲,𝔲f)\mathfrak{r}^{s}_{1}(\mathfrak{u})\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u},\pi(\mathfrak{u}))=\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u},\mathfrak{u}^{f-}) and |𝔯s1(𝔲)|>0|\mathfrak{r}^{s}_{1}(\mathfrak{u})|>0. Then the conclusion easily follows.    \Box

The following is the half ii-arch bridge version of the above whose proof uses similar techniques.

Proposition 9.5.

Suppose 𝔲𝒱𝔣i(𝔵0)\mathfrak{u}\in\mathcal{V}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0}) and h(𝔲)=1h(\mathfrak{u})=1. If 𝔲ξfmini(𝔵0)\mathfrak{u}\neq\xi^{\mathrm{fmin}}_{i}(\mathfrak{x}_{0}) then

Θ(𝔮i(𝔵0;𝔲))=F(i).\Theta(\mathfrak{q}_{-i}(\mathfrak{x}_{0};\mathfrak{u}))=F(i).

The following two results are the opposite parity versions of Propositions 9.4 and 9.5 whose proof uses similar techniques but Proposition 7.25 as the main tool instead of Proposition 7.3.

Proposition 9.6.

Suppose 𝔲𝒱𝔣i(𝔵0)\mathfrak{u}\in\mathcal{V}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0}) and h(𝔲)>1h(\mathfrak{u})>1. If θ(β(𝔲))=\theta(\beta(\mathfrak{u}))=\partial, 𝔲ξfmax(π(𝔲))\mathfrak{u}\neq\xi^{\mathrm{fmax}}_{\partial}(\pi(\mathfrak{u})) then Θ(𝔮(𝔵0;𝔲))=F()\Theta(\mathfrak{q}_{-\partial}(\mathfrak{x}_{0};\mathfrak{u}))=F(\partial).

Proposition 9.7.

Suppose 𝔲𝒱𝔣i(𝔵0)\mathfrak{u}\in\mathcal{V}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0}) and h(𝔲)=1h(\mathfrak{u})=1. If 𝔲ξfmaxi(𝔵0)\mathfrak{u}\neq\xi^{\mathrm{fmax}}_{i}(\mathfrak{x}_{0}) then

Θ(𝔮i(𝔵0;𝔲))=F(i).\Theta(\mathfrak{q}_{i}(\mathfrak{x}_{0};\mathfrak{u}))=F(-i).

Combining Propositions 9.1-9.7 with Corollary 8.8 we get the following result.

Corollary 9.8.

Suppose |𝒱𝔣i(𝔵0)|>1|\mathcal{V}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0})|>1, 𝔲𝒱𝔣i(𝔵0)\mathfrak{u}\in\mathcal{V}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0}) and h(𝔲)1h(\mathfrak{u})\geq 1. Then the following hold.

  1. (1)

    Θ(𝔵0;𝔣(𝔵0;𝔲min))=F(i)\Theta(\mathfrak{x}_{0};\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u}_{\mathrm{min}}))=F(i);

  2. (2)

    Θ(𝔣(𝔵0;𝔲,𝔲f);𝔣(𝔵0;𝔲[ϕ(𝔲)]))=F(ϕ(𝔲))\Theta(\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u},\mathfrak{u}^{f-});\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u}_{[-\phi(\mathfrak{u})]}))=F(-\phi(\mathfrak{u})) if 𝔲f\mathfrak{u}^{f-} exists;

  3. (3)

    Θ(𝔣(𝔵0;𝔲,𝔲f+);𝔣(𝔵0;𝔲[ϕ(𝔲)]))=F(ϕ(𝔲))\Theta(\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u},\mathfrak{u}^{f+});\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u}_{[\phi(\mathfrak{u})]}))=F(\phi(\mathfrak{u})) if 𝔲f+\mathfrak{u}^{f+} exists;

  4. (4)

    Θ(𝔪i(𝔵0);𝔣(𝔵0;𝔲max))=F(i)\Theta(\mathfrak{m}_{i}(\mathfrak{x}_{0});\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u}_{\mathrm{max}}))=F(-i).

The last equality follows from Proposition 9.3 once we recall that δ(𝔵0;𝔪i(𝔵0))=i\delta(\mathfrak{x}_{0};\mathfrak{m}_{i}(\mathfrak{x}_{0}))=i and that there is no syllable α\alpha with θ(α)=i\theta(\alpha)=i such that α𝔪i(𝔵0)\alpha\mathfrak{m}_{i}(\mathfrak{x}_{0}) is a string.

In view of Definition 8.9, there is an obvious generalization of Corollary 9.8 to 𝒱¯𝔣i(𝔵0)\overline{\mathcal{V}}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0}).

10. Terms from bridge quivers

Now we use the notation introduced in §2 to get useful consequences of the results in the above sections.

Given 𝔶Hli(𝔵0)\mathfrak{y}\in H_{l}^{i}(\mathfrak{x}_{0}), 𝔶𝔵0\mathfrak{y}\neq\mathfrak{x}_{0}, say that ν𝐌\nu\in\mathbf{M} is a label of the path from 𝔵0\mathfrak{x}_{0} to 𝔶\mathfrak{y} in Hli(𝔵0)H_{l}^{i}(\mathfrak{x}_{0}), written ν(𝔵0)=𝔶\nu(\mathfrak{x}_{0})=\mathfrak{y}, if one of the following holds:

  • i=1i=1 and 𝒪(ν)\mathcal{O}(\nu) is isomorphic to the interval (𝔵0,𝔶](\mathfrak{x}_{0},\mathfrak{y}] in Hl1(𝔵0)H_{l}^{1}(\mathfrak{x}_{0});

  • i=1i=-1 and 𝒪(ν)\mathcal{O}(\nu) is isomorphic to the interval [𝔶,𝔵0)[\mathfrak{y},\mathfrak{x}_{0}) in Hl1(𝔵0)H_{l}^{-1}(\mathfrak{x}_{0}).

Here is an observation.

Remark 10.1.

Suppose (lj)(𝔵)(l\ast j)(\mathfrak{x}) exists for some string 𝔵\mathfrak{x} for some j{1,1}j\in\{1,-1\}. Then (lj)((lj)(𝔵))(l\ast-j)((l\ast j)(\mathfrak{x})) does not exist. Hence (lj)(𝔵)(l\ast j)(\mathfrak{x}) is not a forking string.

Moreover if (lj)n(𝔵)=𝔷𝔶(l\ast j)^{n}(\mathfrak{x})=\mathfrak{z}\mathfrak{y} for some n1n\geq 1, forking string 𝔶\mathfrak{y} with (|𝔶||𝔵|)|𝔷|>0(|\mathfrak{y}|-|\mathfrak{x}|)|\mathfrak{z}|>0 then θ(𝔷)=j\theta(\mathfrak{z})=-j. As a consequence, Θ(𝔵;1,lj(𝔵))=F(j)\Theta(\mathfrak{x};\langle 1,l\ast j\rangle(\mathfrak{x}))=F(j) if 1,lj(𝔵)\langle 1,l\ast j\rangle(\mathfrak{x}) exists, otherwise Θ(𝔵;[1,lj](𝔵))=F(j)\Theta(\mathfrak{x};[1,l\ast j](\mathfrak{x}))=F(j).

Proposition 10.2.

Suppose 𝔵\mathfrak{x} is a proper left substring of 𝔶\mathfrak{y} and Θ(𝔵;𝔶)=F(j)\Theta(\mathfrak{x};\mathfrak{y})=F(j) for some j{1,1}j\in\{1,-1\}.

  1. (1)

    If 𝔶\mathfrak{y} is a left \mathbb{N}-string then 𝔶=1,lj(𝔵)\mathfrak{y}=\langle 1,l\ast j\rangle(\mathfrak{x}).

  2. (2)

    If 𝔶\mathfrak{y} is a left torsion string then 𝔶=[1,lj](𝔵)\mathfrak{y}=[1,l\ast j](\mathfrak{x}).

In addition, if 𝔵\mathfrak{x} is a forking string then 𝔵\mathfrak{x} is a fundamental solution of either (1)(1) or (2)(2) above.

Proof. We only prove the first statement; the proof of the second is similar.

Without loss assume that j=1j=1. Note that l(𝔵)l(\mathfrak{x}) exists since θ(𝔵;𝔶)=1\theta(\mathfrak{x};\mathfrak{y})=1.

Suppose ln(𝔵)l^{n}(\mathfrak{x}) exists for some n1n\geq 1 and ln1(𝔵)l^{n-1}(\mathfrak{x}) is a left substring of 𝔶\mathfrak{y}. If ln(𝔵)l^{n}(\mathfrak{x}) is not a left substring of 𝔶\mathfrak{y} then let 𝔴\mathfrak{w} be the maximal common forking left substring of ln(𝔵)l^{n}(\mathfrak{x}) and 𝔶\mathfrak{y}. Remark 10.1 yields that |𝔴||ln1(𝔵)|+1|\mathfrak{w}|\geq|l^{n-1}(\mathfrak{x})|+1. Thus by the definition of ll we have θ(𝔴;ln(𝔵))=1\theta(\mathfrak{w};l^{n}(\mathfrak{x}))=-1, and hence θ(𝔴;𝔶)=1\theta(\mathfrak{w};\mathfrak{y})=1, a contradiction to Θ(𝔵;𝔶)=F(1)\Theta(\mathfrak{x};\mathfrak{y})=F(1). Thus we have concluded that ln(𝔵)l^{n}(\mathfrak{x}) is a left substring of 𝔶\mathfrak{y}. Since 𝔶\mathfrak{y} is infinite and ln(𝔵)l^{n}(\mathfrak{x}) is not a forking string by Remark 10.1, it is readily seen that ln+1(𝔵)l^{n+1}(\mathfrak{x}) exists. Thus the argument is complete using induction.

The last statement follows from Remark 10.1.    \Box

The next result follows by combining Proposition 10.2 and Corollary 9.8.

Corollary 10.3.

Suppose |𝒱𝔣i(𝔵0)|>1|\mathcal{V}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0})|>1, 𝔲𝒱𝔣i(𝔵0)\mathfrak{u}\in\mathcal{V}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0}) and h(𝔲)1h(\mathfrak{u})\geq 1. Then the following hold.

  1. (1)

    𝔣(𝔵0;𝔲min)={1,li(𝔵0)if 𝔲min𝒰𝔣i(𝔵0)𝔣i(𝔵0);[1,li](𝔵0)if 𝔲min𝔣i(𝔵0)𝒵𝔣i(𝔵0).\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u}_{\mathrm{min}})=\begin{cases}\langle 1,l\ast i\rangle(\mathfrak{x}_{0})&\mbox{if }\mathfrak{u}_{\mathrm{min}}\in\mathcal{U}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0})\cup\mathcal{H}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0});\\ [1,l\ast i](\mathfrak{x}_{0})&\mbox{if }\mathfrak{u}_{\mathrm{min}}\in\mathcal{R}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0})\cup\mathcal{Z}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0}).\end{cases}

  2. (2)

    𝔣(𝔵0;𝔲[ϕ(𝔲)])={1,l(ϕ(𝔲))(𝔣(𝔵0;𝔲,𝔲f))if 𝔲[ϕ(𝔲)]𝒰𝔣i(𝔵0)𝔣i(𝔵0);[1,l(ϕ(𝔲))](𝔣(𝔵0;𝔲,𝔲f))if 𝔲[ϕ(𝔲)]𝔣i(𝔵0)𝒵𝔣i(𝔵0),\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u}_{[-\phi(\mathfrak{u})]})=\begin{cases}\langle 1,l\ast(-\phi(\mathfrak{u}))\rangle(\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u},\mathfrak{u}^{f-}))&\mbox{if }\mathfrak{u}_{[-\phi(\mathfrak{u})]}\in\mathcal{U}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0})\cup\mathcal{H}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0});\\ [1,l\ast(-\phi(\mathfrak{u}))](\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u},\mathfrak{u}^{f-}))&\mbox{if }\mathfrak{u}_{[-\phi(\mathfrak{u})]}\in\mathcal{R}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0})\cup\mathcal{Z}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0}),\end{cases}
    if 𝔲f\mathfrak{u}^{f-} exists.

  3. (3)

    𝔣(𝔵0;𝔲[ϕ(𝔲)])={1,lϕ(𝔲)(𝔣(𝔵0;𝔲,𝔲f+))if 𝔲[ϕ(𝔲)]𝒰𝔣i(𝔵0)𝔣i(𝔵0);[1,lϕ(𝔲)](𝔣(𝔵0;𝔲,𝔲f+))if 𝔲[ϕ(𝔲)]𝔣i(𝔵0)𝒵𝔣i(𝔵0),\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u}_{[\phi(\mathfrak{u})]})=\begin{cases}\langle 1,l\ast\phi(\mathfrak{u})\rangle(\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u},\mathfrak{u}^{f+}))&\mbox{if }\mathfrak{u}_{[\phi(\mathfrak{u})]}\in\mathcal{U}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0})\cup\mathcal{H}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0});\\ [1,l\ast\phi(\mathfrak{u})](\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u},\mathfrak{u}^{f+}))&\mbox{if }\mathfrak{u}_{[\phi(\mathfrak{u})]}\in\mathcal{R}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0})\cup\mathcal{Z}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0}),\end{cases}
    if 𝔲f+\mathfrak{u}^{f+} exists.

  4. (4)

    𝔣(𝔵0;𝔲max)={1,li(𝔪i(𝔵0))if 𝔲max𝒰𝔣i(𝔵0)𝔣i(𝔵0);[1,li](𝔪i(𝔵0))if 𝔲max𝔣i(𝔵0)𝒵𝔣i(𝔵0).\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u}_{\mathrm{max}})=\begin{cases}\langle 1,l\ast-i\rangle(\mathfrak{m}_{i}(\mathfrak{x}_{0}))&\mbox{if }\mathfrak{u}_{\mathrm{max}}\in\mathcal{U}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0})\cup\mathcal{H}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0});\\ [1,l\ast-i](\mathfrak{m}_{i}(\mathfrak{x}_{0}))&\mbox{if }\mathfrak{u}_{\mathrm{max}}\in\mathcal{R}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0})\cup\mathcal{Z}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0}).\end{cases}

We only justify the last identity. Using the fact that δ(𝔵0;𝔪i(𝔵0))=i\delta(\mathfrak{x}_{0};\mathfrak{m}_{i}(\mathfrak{x}_{0}))=i and the definition of ξfmaxi(𝔵0)li\xi^{\mathrm{fmax}}_{i}(\mathfrak{x}_{0})^{l\ast-i} we clearly have |ξfmaxi(𝔵0)li||𝔪i(𝔵0)||\xi^{\mathrm{fmax}}_{i}(\mathfrak{x}_{0})^{l\ast-i}|\geq|\mathfrak{m}_{i}(\mathfrak{x}_{0})|. The reverse inequality follows by combining Proposition 9.3 with Proposition 10.2.

Definition 10.4.

Given 𝔲𝒱𝔣i(𝔵0)\mathfrak{u}\in\mathcal{V}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0}) and j{1,1}j\in\{1,-1\}, define κj(𝔲)\kappa_{j}(\mathfrak{u}) to be the minimal height element of 𝒫j(𝔲)\mathcal{P}_{j}(\mathfrak{u}) such that 𝒫j(κj(𝔲))=𝒫j(𝔲)\mathcal{P}_{j}(\kappa_{j}(\mathfrak{u}))=\mathcal{P}_{j}(\mathfrak{u}).

Remark 10.5.

Suppose 𝔲𝒱𝔣i(𝔵0)\mathfrak{u}\in\mathcal{V}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0}), ϕ(𝔲)0\phi(\mathfrak{u})\neq 0 and j{1,1}j\in\{1,-1\}. Then the following hold.

  1. (1)

    𝔲=κϕ(𝔲)(𝔲)\mathfrak{u}=\kappa_{\phi(\mathfrak{u})}(\mathfrak{u});

  2. (2)

    𝔲f\mathfrak{u}^{f-} exists if and only if 𝔲=κϕ(𝔲)(𝔲)\mathfrak{u}=\kappa_{-\phi(\mathfrak{u})}(\mathfrak{u});

  3. (3)

    If ϕ(κj(𝔲))=j\phi(\kappa_{j}(\mathfrak{u}))=-j then κj(𝔲)f\kappa_{j}(\mathfrak{u})^{f-} exists;

  4. (4)

    If ϕ(κϕ(𝔲)(𝔲))=ϕ(𝔲)\phi(\kappa_{-\phi(\mathfrak{u})}(\mathfrak{u}))=\phi(\mathfrak{u}) then κϕ(𝔲)(𝔲)f\kappa_{-\phi(\mathfrak{u})}(\mathfrak{u})^{f-} exists.

Using Corollary 10.3 and Remark 10.5 we get the following.

Corollary 10.6.

Suppose 𝔲𝒱¯𝔣i(𝔵0)\mathfrak{u}\in\overline{\mathcal{V}}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0}) and j{1,1}j\in\{1,-1\}. Then the fundamental solution, say 𝔲lj\mathfrak{u}^{l\ast j}, of

(3) 𝔣(𝔵0;𝔲[j])={1,lj(𝔵)if 𝔲[j]𝒰𝔣i(𝔵0)𝔣i(𝔵0);[1,lj](𝔵)if 𝔲[j]𝔣i(𝔵0)𝒵𝔣i(𝔵0),\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u}_{[j]})=\begin{cases}\langle 1,l\ast j\rangle(\mathfrak{x})&\mbox{if }\mathfrak{u}_{[j]}\in\mathcal{U}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0})\cup\mathcal{H}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0});\\ [1,l\ast j](\mathfrak{x})&\mbox{if }\mathfrak{u}_{[j]}\in\mathcal{R}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0})\cup\mathcal{Z}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0}),\end{cases}

is given by 𝔲lj={𝔵0if ϕ(κj(𝔲))=0;𝔣(𝔵0;κj(𝔲),κj(𝔲)f)if ϕ(κj(𝔲))=j;𝔣(𝔵0;κj(𝔲),κj(𝔲)f+)if ϕ(κj(𝔲))=j and (j,κj(𝔲))(i,ξfmaxi(𝔵0));𝔪i(𝔵0)if (j,κj(𝔲))=(i,ξfmaxi(𝔵0)).\mathfrak{u}^{l\ast j}=\begin{cases}\mathfrak{x}_{0}&\mbox{if }\phi(\kappa_{j}(\mathfrak{u}))=0;\\ \mathfrak{f}(\mathfrak{x}_{0};\kappa_{j}(\mathfrak{u}),\kappa_{j}(\mathfrak{u})^{f-})&\mbox{if }\phi(\kappa_{j}(\mathfrak{u}))=-j;\\ \mathfrak{f}(\mathfrak{x}_{0};\kappa_{j}(\mathfrak{u}),\kappa_{j}(\mathfrak{u})^{f+})&\mbox{if }\phi(\kappa_{j}(\mathfrak{u}))=j\mbox{ and }(j,\kappa_{j}(\mathfrak{u}))\neq(-i,\xi^{\mathrm{fmax}}_{i}(\mathfrak{x}_{0}));\\ \mathfrak{m}_{i}(\mathfrak{x}_{0})&\mbox{if }(j,\kappa_{j}(\mathfrak{u}))=(-i,\xi^{\mathrm{fmax}}_{i}(\mathfrak{x}_{0})).\end{cases}

Lemma 10.7.

Suppose Fli(𝔵0):={𝔵Hli(𝔵0)𝔵 is a forking string}{𝔵0,𝔪i(𝔵0)}F_{l}^{i}(\mathfrak{x}_{0}):=\{\mathfrak{x}\in H_{l}^{i}(\mathfrak{x}_{0})\mid\mathfrak{x}\mbox{ is a forking string}\}\cup\{\mathfrak{x}_{0},\mathfrak{m}_{i}(\mathfrak{x}_{0})\}. Then Fli(𝔵0)={(𝔲,(s1,s2,,sh(𝔲)1))lϕ(𝔲)(𝔲,(s1,s2,,sh(𝔲)1))𝒱¯𝔣i(𝔵0)}F_{l}^{i}(\mathfrak{x}_{0})=\{(\mathfrak{u},(s_{1},s_{2},\ldots,s_{h(\mathfrak{u})-1}))^{l\ast\phi(\mathfrak{u})}\mid(\mathfrak{u},(s_{1},s_{2},\ldots,s_{h(\mathfrak{u})-1}))\in\overline{\mathcal{V}}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0})\}.

Proof. The inclusion of the right hand side into the left hand side follows from Corollary 10.6. The other inclusion follows by Proposition 10.2.    \Box

The assignment of 𝔣(𝔵0;𝔲¯)\mathfrak{f}(\mathfrak{x}_{0};\bar{\mathfrak{u}}) for 𝔲¯𝒱¯𝔣i(𝔵0)\bar{\mathfrak{u}}\in\overline{\mathcal{V}}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0}) as in Definition 8.9 has a simple consequence.

Proposition 10.8.

Suppose (𝔲,(s1,,sh(𝔲)1))𝒱¯𝔣i(𝔵0),h(𝔲)1(\mathfrak{u},(s_{1},\ldots,s_{h(\mathfrak{u})-1}))\in\overline{\mathcal{V}}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0}),h(\mathfrak{u})\geq 1 and 𝔲¯ξ𝔣(𝔲)\bar{\mathfrak{u}}\in\xi^{\mathfrak{f}}(\mathfrak{u}). Then

(4) R𝔰(𝔲)((𝔲¯,(s1,,sh(𝔲)1,s+1))lϕ(𝔲))=(𝔲¯,(s1,,sh(𝔲)1,s))lϕ(𝔲),\mathrm{R}_{\mathfrak{s}(\mathfrak{u})}((\bar{\mathfrak{u}},(s_{1},\ldots,s_{h(\mathfrak{u})-1},s+1))^{l\ast\phi(\mathfrak{u})})=(\bar{\mathfrak{u}},(s_{1},\ldots,s_{h(\mathfrak{u})-1},s))^{l\ast\phi(\mathfrak{u})},

and hence

(5) lims(𝔲¯,(s1,,sh(𝔲)1,s))lϕ(𝔲)=𝔣(𝔵0;(𝔲,(s1,,sh(𝔲)1))).\lim_{s\to\infty}(\bar{\mathfrak{u}},(s_{1},\ldots,s_{h(\mathfrak{u})-1},s))^{l\ast\phi(\mathfrak{u})}=\mathfrak{f}(\mathfrak{x}_{0};(\mathfrak{u},(s_{1},\ldots,s_{h(\mathfrak{u})-1}))).

Our goal is to associate a term μi(𝔵0;𝔲)\mu_{i}(\mathfrak{x}_{0};\mathfrak{u}) to each non-root 𝔲𝒱𝔣i(𝔵0)\mathfrak{u}\in\mathcal{V}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0}) using induction on subtrees, such that

(6) μi(𝔵0;𝔲)(𝔵0)=𝔲lϕ(𝔲).\mu_{i}(\mathfrak{x}_{0};\mathfrak{u})(\mathfrak{x}_{0})=\mathfrak{u}^{l\ast\phi(\mathfrak{u})}.
Algorithm 10.9.

Suppose 𝔲𝒱𝔣i(𝔵0)\mathfrak{u}\in\mathcal{V}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0}), h(𝔲)1h(\mathfrak{u})\geq 1 and 𝔲1,𝔣𝔣𝔲m,\mathfrak{u}_{1,\partial}\sqsubset^{\mathfrak{f}}_{\partial}\ldots\sqsubset^{\mathfrak{f}}_{\partial}\mathfrak{u}_{{m_{\partial}},\partial} are all the elements of ξ𝔣(𝔲)\xi^{\mathfrak{f}}_{\partial}(\mathfrak{u}) for mm_{\partial}\in\mathbb{N} and {+,}\partial\in\{+,-\}. Suppose for each 𝔲k,ξ𝔣(𝔲)\mathfrak{u}_{k,\partial}\in\xi^{\mathfrak{f}}_{\partial}(\mathfrak{u}) we already have inductively defined terms τj(𝔲k,)\tau^{j}(\mathfrak{u}_{k,\partial}), for j{+,}j\in\{+,-\}, satisfying

(7) 𝔣(𝔵0;𝔲k,)={1,τj(𝔲k,)(𝔲k,lj)if 𝔲k,𝒰𝔣i(𝔵0)𝔣i(𝔵0);[1,τj(𝔲k,)](𝔲k,lj)if 𝔲k,𝔣i(𝔵0)𝒵𝔣i(𝔵0).\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u}_{k,\partial})=\begin{cases}\langle 1,\tau^{j}(\mathfrak{u}_{k,\partial})\rangle(\mathfrak{u}_{k,\partial}^{l\ast j})&\mbox{if }\mathfrak{u}_{k,\partial}\in\mathcal{U}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0})\cup\mathcal{H}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0});\\ [1,\tau^{j}(\mathfrak{u}_{k,\partial})](\mathfrak{u}_{k,\partial}^{l\ast j})&\mbox{if }\mathfrak{u}_{k,\partial}\in\mathcal{R}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0})\cup\mathcal{Z}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0}).\end{cases}

We abbreviate the above equations as follows:

(8) 𝔣(𝔵0;𝔲k,)={1,τj(𝔲k,)}(𝔲k,lj).\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u}_{k,\partial})=\{1,\tau^{j}(\mathfrak{u}_{k,\partial})\}(\mathfrak{u}_{k,\partial}^{l\ast j}).

Define τj(𝔲):={{τj(𝔲mj,j),τj(𝔲mj,j)}{τj(𝔲1,j),τj(𝔲1,j)}if mj>0;ljotherwise.\tau^{j}(\mathfrak{u}):=\begin{cases}\{\tau^{-j}(\mathfrak{u}_{m_{-j},-j}),\tau^{j}(\mathfrak{u}_{m_{-j},-j})\}\ldots\{\tau^{-j}(\mathfrak{u}_{1,-j}),\tau^{j}(\mathfrak{u}_{1,-j})\}&\mbox{if }m_{-j}>0;\\ l\ast j&\mbox{otherwise}.\end{cases}

Proposition 10.10.

With the above hypotheses and notations, we have

(9) 𝔣(𝔵0;𝔲)={1,τj(𝔲)}(𝔲lj)\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u})=\{1,\tau^{j}(\mathfrak{u})\}(\mathfrak{u}^{l\ast j})

Proof. Without loss we assume that 𝔲𝒰𝔣i(𝔵0)𝔣i(𝔵0)\mathfrak{u}\in\mathcal{U}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0})\cup\mathcal{H}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0}) and that mj>0m_{-j}>0. The proofs in the other cases are simpler, and hence omitted.

First note by definition of 𝔲1,j\mathfrak{u}_{1,-j} that 𝒫j(𝔲1,j)=𝒫j(𝔲)\mathcal{P}_{j}(\mathfrak{u}_{1,-j})=\mathcal{P}_{j}(\mathfrak{u}). Hence κj(𝔲1,j)=κj(𝔲)\kappa_{j}(\mathfrak{u}_{1,-j})=\kappa_{j}(\mathfrak{u}). Hence by Corollary 10.6 we obtain

(10) 𝔲1,jlj=𝔲lj.\mathfrak{u}_{1,-j}^{l\ast j}=\mathfrak{u}^{l\ast j}.

If 1k<mj1\leq k<m_{-j} then the same corollary gives that

(11) 𝔲k,jlj=𝔣(𝔵0;κj(𝔲k,j),κj(𝔲k,j)f+)=𝔣(𝔵0;κj(𝔲k+1,j),κj(𝔲k+1,j)f)=𝔲k+1,jlj.\mathfrak{u}_{k,-j}^{l\ast-j}=\mathfrak{f}(\mathfrak{x}_{0};\kappa_{-j}(\mathfrak{u}_{k,-j}),\kappa_{-j}(\mathfrak{u}_{k,-j})^{f+})=\mathfrak{f}(\mathfrak{x}_{0};\kappa_{-j}(\mathfrak{u}_{k+1,-j}),\kappa_{-j}(\mathfrak{u}_{k+1,-j})^{f-})=\mathfrak{u}_{k+1,-j}^{l\ast j}.

In fact if we extend the list ξ𝔣j(𝔲)\xi^{\mathfrak{f}}_{-j}(\mathfrak{u}) by adding (𝔲k,j,(0,,0,s))(\mathfrak{u}_{k,-j},(0,\ldots,0,s)) for s0s\geq 0 and 1kmj1\leq k\leq m_{-j} as in Definition 8.9 then also Equation (11) holds true. Hence

(12) {τj(𝔲k,j),τj(𝔲k,j)}(𝔲k,j,(0,,0,s))lj=((𝔲k,j,(0,,0,s))f+)lj.\{\tau^{-j}(\mathfrak{u}_{k,-j}),\tau^{j}(\mathfrak{u}_{k,-j})\}(\mathfrak{u}_{k,-j},(0,\ldots,0,s))^{l\ast j}=((\mathfrak{u}_{k,-j},(0,\ldots,0,s))^{f+})^{l\ast j}.

Finally for each 1kmj1\leq k\leq m_{-j} Proposition 10.8 yields

(13) lims(𝔲k,j,(0,,0,s))lj=lims(𝔲k,j,(0,,0,s))lj=𝔣(𝔵0;𝔲).\lim_{s\to\infty}(\mathfrak{u}_{k,-j},(0,\ldots,0,s))^{l\ast j}=\lim_{s\to\infty}(\mathfrak{u}_{k,-j},(0,\ldots,0,s))^{l\ast-j}=\mathfrak{f}(\mathfrak{x}_{0};\mathfrak{u}).

The required equation in the statement now follows from Equations (10), (11),(12) and (13).    \Box

There is an obvious generalization of the above when 𝔲𝒱¯𝔣i(𝔵0)\mathfrak{u}\in\overline{\mathcal{V}}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0}).

The string 𝔲lj\mathfrak{u}^{l\ast j} may not be the fundamental solution of Equation (9) as the following example demonstrates.

Example 10.11.

In the algebra Γ(vi)\Gamma^{(vi)} in Figure 11 consider 𝔵0:=1(𝔳1,i)\mathfrak{x}_{0}:=1_{(\mathfrak{v}_{1},i)}, 𝔲:=feca\mathfrak{u}:=feca and j=1j=-1. It is readily verified that 𝔲lj=ab\mathfrak{u}^{l\ast j}=ab and τj(𝔲)=l,l¯\tau^{j}(\mathfrak{u})=\langle l,\bar{l}\rangle by Algorithm 10.9 but the fundamental solution of Equation (9) is bb.

Lemma 10.12.

Suppose 𝔲𝒱𝔣i(𝔵0)\mathfrak{u}\in\mathcal{V}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0}), ϕ(𝔲)0\phi(\mathfrak{u})\neq 0 and j{1,1}j\in\{1,-1\}. Then

(14) 𝔲lϕ(𝔲)={κϕ(𝔲)(𝔲)lϕ(𝔲)if ϕ(κϕ(𝔲)(𝔲))ϕ(𝔲);(κϕ(𝔲)(𝔲)f)lϕ(𝔲)=κϕ(𝔲)(𝔲)lϕ(𝔲)if ϕ(κϕ(𝔲)(𝔲))=ϕ(𝔲).\mathfrak{u}^{l\ast-\phi(\mathfrak{u})}=\begin{cases}\kappa_{-\phi(\mathfrak{u})}(\mathfrak{u})^{l\ast-\phi(\mathfrak{u})}&\mbox{if }\phi(\kappa_{-\phi(\mathfrak{u})}(\mathfrak{u}))\neq\phi(\mathfrak{u});\\ (\kappa_{-\phi(\mathfrak{u})}(\mathfrak{u})^{f-})^{l\ast\phi(\mathfrak{u})}=\kappa_{-\phi(\mathfrak{u})}(\mathfrak{u})^{l\ast-\phi(\mathfrak{u})}&\mbox{if }\phi(\kappa_{-\phi(\mathfrak{u})}(\mathfrak{u}))=\phi(\mathfrak{u}).\end{cases}

Proof. Since 𝒫ϕ(𝔲)(κϕ(𝔲)(𝔲))=𝒫ϕ(𝔲)(𝔲)\mathcal{P}_{-\phi(\mathfrak{u})}(\kappa_{-\phi(\mathfrak{u})}(\mathfrak{u}))=\mathcal{P}_{-\phi(\mathfrak{u})}(\mathfrak{u}), we get 𝔲lϕ(𝔲)=κϕ(𝔲)(𝔲)lϕ(𝔲)\mathfrak{u}^{l\ast-\phi(\mathfrak{u})}=\kappa_{-\phi(\mathfrak{u})}(\mathfrak{u})^{l\ast-\phi(\mathfrak{u})}.

If ϕ(κϕ(𝔲)(𝔲))=ϕ(𝔲)\phi(\kappa_{-\phi(\mathfrak{u})}(\mathfrak{u}))=\phi(\mathfrak{u}) then Remark 10.5(1) ensures the existence of κϕ(𝔲)(𝔲)f\kappa_{-\phi(\mathfrak{u})}(\mathfrak{u})^{f-}. Finally the proof of (κϕ(𝔲)(𝔲)f)lϕ(𝔲)=κϕ(𝔲)(𝔲)lϕ(𝔲)(\kappa_{-\phi(\mathfrak{u})}(\mathfrak{u})^{f-})^{l\ast\phi(\mathfrak{u})}=\kappa_{-\phi(\mathfrak{u})}(\mathfrak{u})^{l\ast-\phi(\mathfrak{u})} is similar to the proof of Equation (11) using Corollary 10.6.    \Box

Now we are ready to inductively define the term μi(𝔵0;𝔲)\mu_{i}(\mathfrak{x}_{0};\mathfrak{u}) for 𝔲𝒱𝔣i(𝔵0)\mathfrak{u}\in\mathcal{V}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0}).

(15) μi(𝔵0;𝔲):={1if ϕ(𝔲)=0;{τϕ(𝔲)(𝔲),τϕ(𝔲)(𝔲)}μi(𝔵0;κϕ(𝔲)(𝔲)f)if ϕ(𝔲)=ϕ(κϕ(𝔲)(𝔲));{τϕ(𝔲)(𝔲),τϕ(𝔲)(𝔲)}μi(𝔵0;κϕ(𝔲)(𝔲)) if ϕ(𝔲)ϕ(κϕ(𝔲)(𝔲)).\mu_{i}(\mathfrak{x}_{0};\mathfrak{u}):=\begin{cases}1&\mbox{if }\phi(\mathfrak{u})=0;\par\\ \{\tau^{\phi(\mathfrak{u})}(\mathfrak{u}),\tau^{-\phi(\mathfrak{u})}(\mathfrak{u})\}\mu_{i}(\mathfrak{x}_{0};\kappa_{-\phi(\mathfrak{u})}(\mathfrak{u})^{f-})&\mbox{if }\phi(\mathfrak{u})=\phi(\kappa_{-\phi(\mathfrak{u})}(\mathfrak{u}));\par\\ \{\tau^{\phi(\mathfrak{u})}(\mathfrak{u}),\tau^{-\phi(\mathfrak{u})}(\mathfrak{u})\}\mu_{i}(\mathfrak{x}_{0};\kappa_{-\phi(\mathfrak{u})}(\mathfrak{u}))&\mbox{ if }\phi(\mathfrak{u})\neq\phi(\kappa_{-\phi(\mathfrak{u})}(\mathfrak{u})).\end{cases}

Combining Corollary 10.6, Proposition 10.10 and Lemma 10.12 we get the main result of this section.

Theorem 10.13.

For each 𝔲𝒱𝔣i(𝔵0)\mathfrak{u}\in\mathcal{V}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0}), Equation (6) indeed holds. In particular,

μi(𝔵0;ξfmaxi(𝔵0))(𝔵0)=𝔪i(𝔵0).\mu_{i}(\mathfrak{x}_{0};\xi^{\mathrm{fmax}}_{i}(\mathfrak{x}_{0}))(\mathfrak{x}_{0})=\mathfrak{m}_{i}(\mathfrak{x}_{0}).
Corollary 10.14.

If Λ\Lambda is a domestic string algebra, 𝔵0\mathfrak{x}_{0} is a string for Λ\Lambda and i{1,1}i\in\{1,-1\} then (Hli(𝔵0),<l)dLOfpb(H_{l}^{i}(\mathfrak{x}_{0}),<_{l})\in\mathrm{dLO}_{\mathrm{fp}}^{\mathrm{b}}, where dLOfpb\mathrm{dLO}_{\mathrm{fp}}^{\mathrm{b}} is the class of finitely presented bounded discrete linear orders.

Example 10.15.

Continuing from Example 2.2, choosing 𝔵0:=1(𝔳1,i)\mathfrak{x}_{0}:=1_{(\mathfrak{v}_{1},i)} the only non-root vertices are 𝔲0:=a\mathfrak{u}_{0}:=a and 𝔲1:=ec\mathfrak{u}_{1}:=ec, where π(𝔲1)=𝔲0\pi(\mathfrak{u}_{1})=\mathfrak{u}_{0} and π(𝔲0)=𝔵0\pi(\mathfrak{u}_{0})=\mathfrak{x}_{0}. Here 𝔪1(𝔵0)=eca\mathfrak{m}_{-1}(\mathfrak{x}_{0})=eca. Since 𝔲1\mathfrak{u}_{1} is a leaf we get τ1(𝔲1)=l\tau^{1}(\mathfrak{u}_{1})=l and τ1(𝔲1)=l¯\tau^{-1}(\mathfrak{u}_{1})=\bar{l} by Algorithm 10.9. Moreover ξ𝔣1(𝔲0)={𝔲1}\xi^{\mathfrak{f}}_{-1}(\mathfrak{u}_{0})=\{\mathfrak{u}_{1}\}, ξ𝔣1(𝔲0)=\xi^{\mathfrak{f}}_{1}(\mathfrak{u}_{0})=\emptyset and hence the same algorithm gives τ1(𝔲0)=τ1(𝔲1),τ1(𝔲1)=l¯,l\tau^{1}(\mathfrak{u}_{0})=\langle\tau^{-1}(\mathfrak{u}_{1}),\tau^{1}(\mathfrak{u}_{1})\rangle=\langle\bar{l},l\rangle and τ1(𝔲0)=l¯\tau^{-1}(\mathfrak{u}_{0})=\bar{l}. Since κ1(𝔲0)=𝔵0\kappa_{-1}(\mathfrak{u}_{0})=\mathfrak{x}_{0} and ϕ(𝔵0)=0\phi(\mathfrak{x}_{0})=0, Corollary 10.6 gives 𝔲0l¯=𝔵0\mathfrak{u}_{0}^{\bar{l}}=\mathfrak{x}_{0}. The same corollary gives 𝔲0l=𝔪1(𝔵0)\mathfrak{u}_{0}^{l}=\mathfrak{m}_{-1}(\mathfrak{x}_{0}). Therefore we get

μ1(𝔵0;ξfmax1(𝔵0))=l¯,l,l¯.\mu_{-1}(\mathfrak{x}_{0};\xi^{\mathrm{fmax}}_{-1}(\mathfrak{x}_{0}))=\langle\langle\bar{l},l\rangle,\bar{l}\rangle.

The definition of μi(𝔵0;-)\mu_{i}(\mathfrak{x}_{0};\mbox{-}) naturally extends to 𝒱¯𝔣i(𝔵0)\overline{\mathcal{V}}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0}). Indeed the computation of μi(𝔵0;𝔲¯)\mu_{i}(\mathfrak{x}_{0};\bar{\mathfrak{u}}) for 𝔲¯:=(𝔲,(s1,s2,,sh(𝔲)1))𝒱¯𝔣i(𝔵0)\bar{\mathfrak{u}}:=(\mathfrak{u},(s_{1},s_{2},\ldots,s_{h(\mathfrak{u})-1}))\in\overline{\mathcal{V}}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0}) needs at most 2j=1sh(𝔲)1(sj+1)2\prod_{j=1}^{s_{h(\mathfrak{u})}-1}(s_{j}+1) many iterations of Algorithm 10.9.

For 𝔶Hli(𝔵0)\mathfrak{y}\in H_{l}^{i}(\mathfrak{x}_{0}) the following result gives the recipe to find a term τi(𝔵0;𝔶)\tau_{i}(\mathfrak{x}_{0};\mathfrak{y}) such that

τi(𝔵0;𝔶)(𝔵0)=𝔶.\tau_{i}(\mathfrak{x}_{0};\mathfrak{y})(\mathfrak{x}_{0})=\mathfrak{y}.
Lemma 10.16.

Suppose 𝔶Hli(𝔵0)\mathfrak{y}\in H_{l}^{i}(\mathfrak{x}_{0}). Then

τi(𝔵0;𝔶)=(lj)nμi(𝔵0;(𝔲,(s1,,sh(𝔲)1)))\tau_{i}(\mathfrak{x}_{0};\mathfrak{y})=(l\ast j)^{n}\mu_{i}(\mathfrak{x}_{0};(\mathfrak{u},(s_{1},\ldots,s_{h(\mathfrak{u})-1})))

for some (𝔲,(s1,,sh(𝔲)1))𝒱¯𝔣i(𝔵0)(\mathfrak{u},(s_{1},\ldots,s_{h(\mathfrak{u})-1}))\in\overline{\mathcal{V}}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0}), nn\in\mathbb{N} and j{1,1}j\in\{1,-1\}.

Proof. Recall from Lemma 2.1 that Hli(𝔵0)H_{l}^{i}(\mathfrak{x}_{0}) is a bounded discrete linear order. If 𝔵Hli(𝔵0)\mathfrak{x}\in H_{l}^{i}(\mathfrak{x}_{0}) is a forking string then both l(𝔵)l(\mathfrak{x}) and l¯(𝔵)\bar{l}(\mathfrak{x}) exist. Since |𝔵|<|l(𝔵)||\mathfrak{x}|<|l(\mathfrak{x})| and |𝔵|<|l¯(𝔵)||\mathfrak{x}|<|\bar{l}(\mathfrak{x})|, we get that the sequence of lengths has a local minimum at 𝔵\mathfrak{x}.

Hence if 𝔶(𝔲,(s1,,sh(𝔲)1))lϕ(𝔲)\mathfrak{y}\neq(\mathfrak{u},(s_{1},\ldots,s_{h(\mathfrak{u})-1}))^{l\ast\phi(\mathfrak{u})} for any (𝔲,(s1,,sh(𝔲)1))𝒱¯𝔣i(𝔵0)(\mathfrak{u},(s_{1},\ldots,s_{h(\mathfrak{u})-1}))\in\overline{\mathcal{V}}^{\mathfrak{f}}_{i}(\mathfrak{x}_{0}) then 𝔶\mathfrak{y} is not a forking string and the sequence of lengths of strings does not have a local minimum at 𝔶\mathfrak{y}. Hence 𝔶=(lj)n(𝔶)\mathfrak{y}=(l\ast j)^{n}(\mathfrak{y}^{\prime}) for some n1n\geq 1, j{1,1}j\in\{1,-1\} and 𝔶Fli(𝔵0)\mathfrak{y}^{\prime}\in F_{l}^{i}(\mathfrak{x}_{0}). The conclusion then follows from Lemma 10.7.    \Box

This defines a map μi(𝔵0;-):Hli(𝔵0)𝐌li\mu_{i}(\mathfrak{x}_{0};\mbox{-}):H_{l}^{i}(\mathfrak{x}_{0})\to\mathbf{M}_{l\ast i}.

Example 10.17.

Continuing from Example 10.15, we compute τ1(𝔵0;ca)\tau_{-1}(\mathfrak{x}_{0};ca).

Since κ1(𝔲1)=𝔲0\kappa_{1}(\mathfrak{u}_{1})=\mathfrak{u}_{0}, the last clause of Corollary 10.6 gives that 𝔲1l=𝔪1(𝔵0)\mathfrak{u}_{1}^{l}=\mathfrak{m}_{-1}(\mathfrak{x}_{0}). Since κ1(𝔲1)=𝔲1\kappa_{-1}(\mathfrak{u}_{1})=\mathfrak{u}_{1}, we get 𝔲1l¯=a\mathfrak{u}_{1}^{\bar{l}}=a by the third clause of the same corollary. Therefore we get

a=μ1(𝔵0;𝔲1)(𝔵0)a=\mu_{-1}(\mathfrak{x}_{0};\mathfrak{u}_{1})(\mathfrak{x}_{0})

by Equation (6). Now using the computation in Example 10.15 together with Equation (15) we have

μ1(𝔵0;𝔲1)=τ1(𝔲1),τ1(𝔲1)μ1(𝔵0;𝔲0)=l¯,ll¯,l,l¯.\mu_{-1}(\mathfrak{x}_{0};\mathfrak{u}_{1})=\langle\tau^{-1}(\mathfrak{u}_{1}),\tau^{1}(\mathfrak{u}_{1})\rangle\mu_{-1}(\mathfrak{x}_{0};\mathfrak{u}_{0})=\langle\bar{l},l\rangle\langle\langle\bar{l},l\rangle,\bar{l}\rangle.

Since ca=l¯(a)ca=\bar{l}(a) we get

τ1(𝔵0;ca)=l¯l¯,ll¯,l,l¯\tau_{-1}(\mathfrak{x}_{0};ca)=\bar{l}\langle\bar{l},l\rangle\langle\langle\bar{l},l\rangle,\bar{l}\rangle

.

The following result is neither surprising nor difficult to prove using the theory of decorated trees and terms that we have developed so far.

Proposition 10.18.

For i{1,1}i\in\{1,-1\} and ν𝐌li\nu\in\mathbf{M}_{l\ast i}, there is a domestic string algebra Λ\Lambda, a string 𝔵0\mathfrak{x}_{0} and 𝔶(𝔵0)Hli(𝔵0)\mathfrak{y}(\neq\mathfrak{x}_{0})\in H_{l}^{i}(\mathfrak{x}_{0}) such that ν=τi(𝔵0;𝔶)\nu=\tau_{i}(\mathfrak{x}_{0};\mathfrak{y}).

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Shantanu Sardar
Indian Institute of Technology Kanpur
Uttar Pradesh, India
Email: [email protected]

Corresponding Author: Amit Kuber
Indian Institute of Technology Kanpur
Uttar Pradesh, India
Email: [email protected]
Phone: (+91) 512 259 6721
Fax: (+91) 512 259 7500