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 , 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 over an algebraically closed field is presented as a certain quotient of the path algebra of a quiver , where 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 and , say and commute if there are cyclic permutations of and of such that and 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 of the quiver the simplest version of hammock, denoted , is the collection of all strings starting at the vertex. It can be equipped with an order such that in guarantees the existence of a canonical graph map between the corresponding string modules. Schröer [7, § 2.5] showed that 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 for an arbitrary string , which is the collection of all strings which admit as a left substring–this too is a bounded discrete linear order with respect to the induced order , say with and as the maximal and minimal elements respectively. For technical reasons it is necessary to write the hammock as a union , where is the interval and . Given in , the goal of this paper is to compute the order type of the interval via the assignment of a label, which we call a ‘term’, to the path from to . 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 [1, § 5] is the smallest class of linear orders containing that is closed under , and operations. Its subclass consisting of bounded discrete linear orders is denoted by . The main results of this paper (Corollary 10.14 and Proposition 10.18) show that is precisely the class of order types of hammocks for domestic string algebras.
Fix a domestic string algebra , a string and . 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 , that lies between the extended bridge quiver and the extended weak bridge quiver. For , the paths in generate all strings in in the precise sense described in [4, Lemma 3.3.4].
In this paper, we introduce the notion of a ‘decorated tree’, denoted , that linearizes the extended arch bridge quiver from the viewpoint of the string 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 so that the term associated with the root, denoted , labels . The following flowchart concisely depicts the steps of the algorithm.
To further explain a rough idea behind the proof we need some terminology and notation. Recall from [9] that for a string of positive length, we say if its first syllable is an inverse syllable, otherwise we say .
Definition 1.1.
Given , we say that is the standard partition of if
(1)
we have where for each ;
(2)
for each the string forks;
(3)
the only proper left substrings of in which fork are the ones described above.
We say that the signature type of , denoted , is . In case , we denote by .
Remark 1.2.
If fork then .
For , say that if it is of the form . Say that if it is of the form , where may not appear; here and 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 -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 , 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 and parity , we linearize/separate all paths in the extended arch bridge quiver with the help of the concept of decorated trees introduced in §4. The results in §3, §5 and §6 help to classify strings in 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 consisting of bounded discrete linear orders is of particular interest because of the next result.
Lemma 2.1.
(cf. [8, §2.5])
If is a string then is a toset. Its maximal element is and the minimal element is , where for each is the maximal length string in such that .
Consider any . Recall from [4, §2.4] that is the immediate successor of with respect to , if exists, that satisfies . We can inductively define , if exists. If exists for all then the limit of the increasing sequence is a left -string. On the other hand, if there is a maximum such that exists but does not exist then we define . Note that but .
Recall that any left -string in a domestic string algebra is an almost periodic string, i.e., is of the form for some primitive cyclic word and some finite string such that the composition is defined. Let denote the completion of by all left -strings such that . The definition of the ordering can be easily extended to include infinite left -strings so that is also a bounded total order with same extremal elements as that of .
Note that any left -string in is “far away” from its endpoints. To be more precise if exists then the interval in contains infinitely many elements of , where , where is a finite string. Thus the left -string can also be approached from the right, with respect to , as a limit of some decreasing sequence in .
Now we recall the concept of fundamental solution of the following equation from [4, §2.4]:
(1)
If one solution to the above equation exists then infinitely many solutions exist for if is a solution then so is and the fundamental solution is a solution of minimal length. If is a left substring of , then we rewrite Equation (1) as , otherwise we write . If , then we also say that for any .
Similarly if does not exist but exists then we may ask if the equation
(2)
has a solution. If one solution exists then finitely many such solutions exist for if is a solution and then is also a solution. As explained above we can define a fundamental solution of Equation (2) and introduce the notations or depending on the comparison of lengths.
In the above two paragraphs we introduced new expressions using two kinds of bracket operations, namely and . We say that such expressions are ‘terms’–the precise definition is given below.
If then the term labels the path from to in . Let us describe the path from to any element in the interval . For there exists some such that either or . In the former case, we know the label of the required path, namely , whereas in the latter case, we use concatenation operation on terms, written juxtaposition, to label the path as . Here we acknowledge that the only path to reach such from has to pass through which is captured using the bracket term. In other words, the bracket term describes a ‘ghost path’ which passes through each to reach the limit , and then directly jumps to avoiding all elements of the interval –such elements can only be accessed through reverse paths from .
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 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 . Further we can also ask if there are solutions to equations of the form
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 presented as
Figure 1. with
Figure 2 shows all connected components of the A-R quiver of along with some infinite dimensional string modules. The exceptional component is a copy of -component with a hole in the middle in and rest of the components are tubes.
Figure 2. AR components for string modules in
Note that red and arrows denote the operators and respectively whereas black and arrows denote the operators and respectively. The module represents the simple modules at vertex .
Now . Furthermore it is easy to verify that , and hence . In Examples 10.15 and 10.17 we will compute the terms, i.e., labels for the paths from to and 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 we get Figure 3. Here the vertical red colored lines (resp. the horizontal black colored lines) represent hammocks (resp. ) for different vertices . In fact we also show the infinite dimensional string modules, and thus show and . For example, the rightmost vertical line represent and the lowermost horizontal line represent .
Figure 3. Hammocks for
Now we are ready to formally define terms. Starting with symbols we simultaneously inductively define the class of simple -terms, denoted and the class of simple -terms, denoted as follows:
•
and ;
•
if and then and ;
•
if and then and .
Set and call an element of a simple term.
Remark 2.3.
Both and are in bijective correspondence with of [1, Definition 7.1] in view of [1, Theorem 7.2].
Define a map as follows:
•
and ;
•
;
•
and ;
•
.
Remark 2.4.
The map is an involution such that for each , if and only if .
Suppose and . Let be the equivalence relation on generated by
and be the equivalence relation on generated by
for each .
To make these into congruence relations, whenever and we further need the following identifications:
Remark 2.5.
The relation on coincides with of [1, Definition 6.3] on .
Say that is irreducible if either or is of the form .
Given and , say that
•
is a valid concatenation if ;
•
is a valid concatenation if .
Say that a finite concatenation of simple terms is a mixed -term (resp. mixed -term) if
•
if and only if is odd (resp. even);
•
is a valid concatenation for each .
Let and denote the classes of all mixed and -terms respectively. Set and say that a term is an element of .
The congruence relations and are naturally extended to and respectively.
Associate to each its order type, denoted , as follows:
•
;
•
;
•
.
Clearly takes values in .
Remark 2.6.
Suppose and . If is a valid concatenation then is a proper suffix of .
In view of the above remark, the map extends to by setting
Remark 2.7.
If is irreducible then , where denotes the width of 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 and . Then
•
if and only if ;
•
if and only if .
Before we close the section, let us define a natural notation, , on terms.
Suppose is an H-pair. Then if and only if there is a partition , where is a cyclic permutation of and the first syllable of is not a syllable of , such that .
Suppose is a band, is a string and . Say that the pair is an -left valid pair (-LVP for short) if is a string for some cyclic permutation of such that . Dually say that the pair is a -right valid pair (-RVP for short) if is a -LVP.
Example 3.2.
Consider the algebra from [9, Example 4.24]. The only bands there are and . Choose . Since has two different cyclic permutations and with and such that and are strings, the pair is both -LVP and -LVP.
Remark 3.3.
[9, Proposition 4.23] guarantees the uniqueness of the cyclic permutation for any -LVP .
We drop the reference to in -LVP if either it is clear from the context or if does not play any role in the statement.
Proposition 3.4.
Suppose is an LVP and then for some .
Proof.
Since , can be written as for some cyclic permutation of . If is not a right substring of a finite power of then there exists a non-trivial cycle in the bridge quiver, i.e., a meta-band passing through by [4, Lemma 3.3.4], which contradicts the domesticity of the string algebra by [4, Proposition 3.4.2].
Definition 3.5.
Say that an LVP is a short pair if ; otherwise say that is a long pair.
Remark 3.6.
An LVP with is long if and only if is long.
Remark 3.7.
Suppose is an LVP. If is short then is an isomorphism.
For an LVP , let be the longest common right substring (possibly of length ) of and , and . Dually for an RVP , let be the longest common left substring (possibly of length ) of and , and .
Say that an LVP is a left tight valid pair (LTVP, for short) if ; otherwise say it is a left loose valid pair (LLVP, for short). Dually we define RTVP and RLVP.
Remark 3.8.
Suppose is an LLVP with then , and hence is a short LVP by [9, Proposition 5.3].
Proposition 3.9.
Suppose is an LVP such that . Then is a short LVP.
Proof.
Since , we have for some string by Proposition 3.4. Hence is a left substring of . The conclusion follows by Remark 3.8.
For a long LVP the above proposition gives that .
Definition 3.10.
For an LVP if is a proper right substring of (resp. of ) then say that is -long (resp. -long). If is an -long pair then we say that it is -long where . Similarly if is an -long pair then we say that it is -long where .
Proposition 3.11.
Suppose is an LVP.
•
The pair is -long if and only if is injective but not surjective.
•
The pair is -long if and only if is injective but not surjective.
Proof.
We only prove the first statement; the proof of the second is similar. Recall that .
Suppose is a string. Since is -long, is a proper right substring of which states that where is the syllable such that is a string. Since is a string and , is also a string. This shows that is indeed a well-defined map from . This map is clearly injective.
Let be the minimal string such that . Then clearly the word is not a string but the argument as in the above paragraph ensures that is a string. Therefore the map is not surjective.
For the converse, non-surjectivity of the map guarantees the existence of a string of minimal positive length with such that is a string whereas the word is not. Then there is some right substring of such that . The definition of ensures that . Since is a string we see that thus completing the proof that the pair is -long.
Example 3.12.
Consider the algebra from Figure 4. Here the bands are and . For the LVP is -long as well as -long.
Figure 4. with
Say that a long LVP is double long if it is both -long or -long; otherwise say that it is single long.
4. Decorated trees
Fix and . We wish to associate a rooted tree, , with some additional structure described in the definition below to the triple .
Definition 4.1.
A decorated tree is a tuple
where
•
is a finite height rooted tree with root ;
•
all children of lie in ;
•
is the set of all leaves;
•
for each , the relation is a strict partial order where only if and have the same parent;
•
for each and for each , the relation restricted to the children of in is a strict linear order.
Here and stands for ‘torsion’ and ‘non-torsion’ respectively. Recall that a string is a (left) torsion string if is not a string for any syllable .
Consider the set of all arrows appearing in a path in starting from . Recall from [9, Remark 6.2] that is a finite set.
Algorithm 4.2.
Construct a rooted tree as follows.
•
Add the root vertex labelled . Add one child of the root for each half -arch bridge or a zero -arch bridge from .
•
For each newly added vertex , if is a band then add one child of for each element satisfying .
•
Repeat the above step until all elements of are exhausted.
Since is domestic is finite. Let denote the set of vertices of the tree . Let denote those elements of 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 .
Let denote the set of all children of . For each non-root let denote the set of children of . Partition as by defining . For a non-root vertex let denote the parent of . The height of is the non-negative integer such that .
Define a function by
The tree “spreads out” the set by placing an “observer” at .
For let , and
For distinct non-root vertices , let denote the maximal common left substring of and . Note that . Let be the string satisfying
If is a band, let denote the longest left substring of (or, equivalently of ) in whose last syllable is not a syllable of ; otherwise set .
If is a band, let denote the cyclic permutation of for which is a string. Let denote the first syllable of .
Proposition 4.3.
Suppose . Then is a substring of .
Proof.
By definition of ([9, Definition 6.3]) we get . By definition, is a left substring of .
Consider the following cases:
•
If then is a left substring of .
•
If then . There are two subcases.
–
If then .
–
If for some string then .
Clearly in every cases is a left substring of .
Corollary 4.4.
Suppose . Then the strings and are comparable since both are left substrings of .
In view of the above corollary, we define a function as
Examples 4.5.
For , the strings and are comparable in all possible ways as the following examples demonstrate.
()
Choosing for the algebra from [9, Figure 4] the extended arch bridge quiver is shown in [9, Figure 12]. If and then is a proper left substring of .
()
In the same example, if and then is a proper left substring of .
()
Choosing for the algebra in [9, Figure 8], if and then .
Proposition 4.6.
Suppose and . Then is normal if and only if is a proper left substring of .
Proof.
Recall that both and are substrings of and hence they are comparable.
First assume that ; a similar argument works when .
If is normal then . Since , we get the required conclusion.
Conversely if is a proper left substring then is a syllable of , and hence not a syllable of . Therefore is normal.
Proposition 4.7.
Suppose and . Then is a substring of .
Proof.
Since is a left substring of and we get . Hence using , we get that . Then is a left substring of . Hence and are comparable.
Since but , we conclude that is a left substring of .
For if is a band, Proposition 4.3 allows us to define a string by . Let denote the cyclic permutation of for which is a string.
The following is an important observation from the proof of Proposition 4.3.
Corollary 4.8.
Suppose . Then is a left substring of .
Our next goal is to show that two different vertices of the tree have incomparable values of . Here is a supporting result.
Proposition 4.9.
If then is not a string for any syllable .
Proof.
Suppose is a string for some syllable . Since , is a maximal reverse torsion arch bridge, and hence is not a string. Hence as is not a string but .
On the other hand since is not a string, we get . Thus by the definition of , exists, a contradiction to the above paragraph. Hence our assumption is wrong.
Proposition 4.10.
Suppose are non-root vertices. If is a left substring of then .
Proof.
Suppose is a left substring of .
If the former is an -string then so is latter and hence they are equal.
On the other hand, if is a finite string then .
If then . By definition, is a maximal weak torsion zero bridge, and hence cannot be extended to the left. Thus .
If then Proposition 4.9 guarantees the same conclusion.
In each case we have that and that both and are either equal to or in or in . In each case it is clear that . Now [9, Theorem 7.5] gives that , and hence in .
Corollary 4.11.
If are distinct non-root vertices then and hence .
Now we define total ordering(s) on the set of children of a vertex of . Let . Say if . Let and denote the minimal and maximal elements of this order respectively. The immediate successor and immediate predecessor of in , if exist, are denoted by and respectively.
Let and . Say if . Let and denote the minimal and maximal elements of this order respectively. The immediate successor and immediate predecessor of in , if exist, are denoted by and respectively. Choosing leaves in as torsion while as non-torsion we have completely described the decorated tree structure of .
We shall use words like parent, grandparent, child, uncle, granduncle etc. to describe relations between two vertices of .
Example 4.12.
Choosing for the algebra from [9, Figure 4] the extended arch bridge quiver is shown in [9, Figure 12]. Figure 5 shows the decorated tree where the arrow with target in the tree is labeled if . Here satisfy .
Figure 5. for
5. Long elements of
Say that is long if ; otherwise say it is short. If we refer to as long or short then it is implicitly assumed that .
If is long then let be defined by . On the other hand if is short then we set . Note that .
Remark 5.1.
Suppose and . If is long and is short then . Moreover if then .
Remark 5.2.
Suppose . If with then is a left substring of .
Remark 5.3.
Suppose and . If then is a left substring of .
Remark 5.4.
If is normal then is a substring of which is a substring of . Hence .
Proposition 5.5.
Suppose . Then is long if and only if .
Proof.
Since is a left substring of , we only need to prove the forward implication. If then , and hence the conclusion is obvious. Thus we assume that . Let denote the first syllable of respectively.
Suppose is long. Then . For contradiction, assume that . Then there is a shortest string of positive length such that and is a left substring of . Hence is a left substring of a finite power of . In view of Remark 5.4 is abnormal. Minimality of guarantees that a right substring of of positive length is also a substring of a cyclic permutation of . Further [9, Propositions 4.2,4.3] gives that is not a syllable of . Thus is not a substring of . Therefore and together give that a cyclic permutation of is a substring of . Since is not a syllable of , minimality of ensures that is a substring of . Hence a cyclic permutation of is a substring of , a contradiction to the domesticity of the algebra by [4, Corollary 3.4.1].
Proposition 5.5 together with Remark 5.4 ensures that if is normal with then is long if and only if . We will show in the proof of Proposition 6.21 that the same conclusion of holds when is -long. However the following example shows that it fails if is abnormal and -long.
Example 5.6.
Consider the algebra in Figure 6. The only bands here are , and their inverses. Choose and . Then . Thus and , and hence . Here is -long, but .
Figure 6. with
Remark 5.7.
It is interesting to note that if we remove from in Figure 6 then in that algebra we get where is as chosen in Example 5.6.
For an arbitrary the best we can achieve is the following.
Corollary 5.8.
Suppose and for some . Then is long if and only if the LVP is long.
Proof.
If is long then by Proposition 5.5. Since is an H-reduced string by [9, Theorem 6.4], is long by Corollary 4.8.
Conversely if is long then and hence by Corollary 4.8. Therefore is long by Proposition 5.5.
In view of the above corollary, we freely transfer adjectives and qualifiers of the pair to .
We note an immediate consequence of the above result.
Corollary 5.9.
Suppose , and is a left substring of . If is long then is also long.
Suppose and is long. If the pair is -long then let denote the last syllable of , and let denote the last syllable of . If the pair is -long then let denote the last syllable of , and let denote the last syllable of if .
Remark 5.10.
If and then is not -long.
Suppose . If is -long then denotes the longest left substring of satisfying if . Let and satisfy and . Define a syllable and a string by .
On the other hand, if is -long then denotes the longest left substring of satisfying if . Let and satisfy and . Define a syllable and a string by .
Remark 5.11.
Suppose . If is double long then is the first syllable of exactly one of and , and hence but .
In view of the above remark, if is a double long pair and then we say it is -long, otherwise we say it is -long. Say that a -long pair is -long if it is -long. Similarly say that an -long pair is -long if it is -long.
Remark 5.12.
If is -long (resp. -long) then is a left substring of (resp. ). Moreover, if is double long then Remark 5.11 guarantees that .
Remark 5.13.
Suppose and .
•
If is -long then is a proper left substring of if and only if . As a consequence is a proper left substring of .
•
If is -long and then is a proper left substring of .
Example 5.14.
In the algebra in Figure 7 the only bands are , and their inverses. Choosing and , the LVP is -long with and . Here with with .
Figure 7. with
If we consider the algebra obtained from the same quiver by replacing the relation by in then keeping and same as above we get and . Here with .
6. Characterizing different values of
Proposition 6.1.
Suppose , and is abnormal. Then is a string if and only if if and only if .
Proof.
Note that the only possible syllables such that is a string are and the last syllable of , say . Moreover is a syllable of but not of while is a syllable of but not of . Hence is a string if and only if is the last syllable of . Since is a proper left substring of by Corollary 4.4, if the latter holds then .
On the other hand if then the last syllable of is a syllable of but not of . Hence for any cyclic permutation of if is a string then . Hence is a string.
Below we note some immediate consequences.
Corollary 6.2.
Suppose , , is abnormal and . If is -long then if and only if .
Corollary 6.3.
Suppose , , is abnormal and . Then
is -long if and only if for some string . If these equivalent conditions hold then
•
.
•
is -long.
Corollary 6.4.
Suppose , and with . If is long and is an uncle of with then could be either -long or -long but not both. Further if then , , is long and exactly one of the following happens:
•
is -long and ;
•
is -long and .
Example 6.5.
In the algebra in Figure 7 the only bands are , and their inverses. Choosing and we have and with . Here is -long and its uncle is -long.
Corollary 6.6.
Suppose , and . Further suppose that is abnormal with .
•
If then Proposition 6.1 guarantees that is a right substring of . Since we get is short.
•
If then is -long for otherwise we get , a contradiction. But here .
Proposition 6.7.
Suppose , and . If is abnormal with then is not double long.
If is -long then is the first syllable of . In this case and are strings, and . Thus and hence is not -long.
If is -long then from the above discussion it is clear that and is the first syllable of . But since , is not -long. Hence the result.
Proposition 6.8.
Suppose , , is abnormal with and . If then is a left substring of .
Proof.
Since is abnormal, is a left substring of . If then, in view of Remark 3.1, is short, and hence is a proper left substring of . Since , a string. Hence and are comparable strings.
If is a not left substring of then is a proper substring of and hence is a syllable of . Since is a proper left substring of , is a substring of , where is the last syllable of . But is the first syllable of and hence a common syllable of and , a contradiction to . Therefore is a left substring of .
Example 6.9.
In the algebra in Figure 6, choosing and we have and . Here , and hence the string satisfies .
Remark 6.10.
Suppose , and then since is not a substring of , Proposition 4.7 gives that .
Proposition 6.11.
Suppose and . Then if and only if is abnormal, is a string but is not a string.
Proof.
Suppose for some string of positive length. Since both and are substrings of and is a proper substring of , is a left substring of and hence is abnormal.
Since is a proper left substring of the last syllable of is a syllable of but not of . Hence for any cyclic permutation of if is a string then . Hence is a string. Moreover Proposition 6.1 gives that is not a string.
Now is a string if and only if is the last syllable of where is the last syllable of .
For the reverse implication clearly . Moreover by Proposition 6.1, since is not a string. Therefore .
Corollary 6.12.
Suppose and . If where . Then the following hold:
(1)
;
(2)
is -long with ;
(3)
is -long but not -long;
(4)
is long iff is a left substring of ;
Proof.
Proposition 6.11 gives that is abnormal and is a string. and hence (1) follows immediately. Moreover is a right substring of . Since is a string, is a proper left substring of . Therefore is -long. By definition of , we clearly get thus proving (2).
Since is a proper left substring of , is -long but not -long thus proving (3). Moreover if is a left substring of then is long. The converse part of (4) is by Remark 5.12.
Corollary 6.13.
Suppose , and . If is long then it is -long. Moreover if is a left substring of and there is an uncle of with such that then
•
and ;
•
is -long and .
Example 6.14.
In the algebra in Figure 8 the only bands are , and their inverses.
Figure 8. with
Choosing and we have and with . Here is -long and its uncle is -long.
Proposition 6.15.
Suppose and . Then if and only if is abnormal with and there is a partition with and is a right substring of .
Proof.
Since , and are strings. Let denote the last syllables of respectively. Then .
If then by the definition of a string algebra we must have . On the other hand if , then since are strings, we can again conclude . Therefore is abnormal and there exists a partition such that .
For the other direction observe that since , neither nor is a string. Recall from Corollary 4.4 that and are comparable. Hence Propositions 6.1 and 6.11 yield the necessary result.
Proposition 6.16.
If , and then is short.
Proof.
We only prove the result for and the proof for the another case is similar.
For a contradiction we assume that is long. Since , Proposition 6.15 guarantees that there is a partition with and is a right substring of . Here is a string. Corollary 5.8 gives that the LVP is long and hence . Moreover Corollary 4.8 states that is a left substring of . Now the maximal left substring of whose last syllable is satisfies that is a left substring of , a contradiction to . This completes the proof.
Corollary 6.17.
Suppose , and . If is long and there exist an uncle of with then the following hold:
•
is -long;
•
if then ;
•
is double -long only if . In this case is necessarily -long, , and .
Remark 6.18.
Suppose , and . If is long then it is either -long or -long.
Corollary 6.19.
Suppose , and is abnormal then
•
the LVP is -long;
•
is a right substring of .
Proof.
Since is abnormal Proposition 4.6 gives that for some string .
Propositions 6.1 (), 6.15 () and Corollary 6.12 (()) give that the LVP is -long.
The following cases describe about depending on the value of .
Case 1: .
Suppose for some positive length string .
If is long then Corollary 6.6 guarantees that either or with , and hence .
On the other hand if is short then is a proper left substring of .
Since is a string we get by Proposition 6.1 and hence by Corollary 6.3. In summary
Case 2: .
Proposition 6.16 gives that is short. Hence by Proposition 6.15 we get .
Case 3: .
Let for some positive length string . Corollary 6.12 guarantees that . Since is a string we get . Since is short by Proposition 6.16, we have . Therefore .
All the three cases above show that is a right substring of . This completes the proof.
We close this section with some expository discussion. A single monomial relation in can be responsible for making a child-parent pair in simultaneously long. When this happens then, for some , the parent is -long and the child is -long. The definition of makes sure that even if the same relation touches several other bands then it cannot make any other vertex of long. The next proposition characterizes such situation.
Example 6.20.
In the algebra from Figure 9 the only bands are , . Choosing and we have .
Figure 9. with
Here and . The monomial relation is responsible to make -long and -long.
Proposition 6.21.
Suppose , and is -long. Then is a substring of if and only if
(1)
if is abnormal then and ;
(2)
is -long with and
Proof.
We only need to prove the forward direction as the backward direction is obvious.
Since is -long, if it is abnormal then Proposition 6.16 gives . Further Corollary 6.6 gives that . To prove is -long it is enough to prove the following claim.
Claim: If is -long then .
Since is -long we get . If is normal then Remark 5.4 gives that . If is abnormal with and then . Hence the claim.
7. H-reduced forking strings
Recall that a string is forking if it can be extended on the left by two different syllables.
Proposition 7.1.
If for then is a H-reduced forking string.
Proof.
Clearly is a forking string. Then note that if is a left substring of and exists for some band then .
Thus if exists for some band then . As a consequence we get , and hence by Proposition 4.10, a contradiction.
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 is an H-reduced forking string then has one of the following forms for some :
•
;
•
.
In this section we guarantee the existence of certain distant relatives of 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 , and for some . Further suppose
(1)
is long;
(2)
for some string of positive length;
(3)
is a string.
Then there is an uncle of satisfying and
The proof is long and we need to set up more notation.
If then let satisfy . Note that the word is a string. Let be the shortest right substring of such that is a string.
If then let satisfy . If is a substring of then is a substring of . Let be the shortest right substring of such that is a string.
In both cases is a left substring of .
Since is long, Proposition 5.8 gives that is long. If is double long, then it is either -long or -long.
The proof will be completed in four cases depending on whether is single/double -long but the general plan in each case is as follows.
Step 1: To show that for some string the string is an H-reduced string.
Step 2: Since is a string the hypotheses of the following lemma are satisfied, and it yields such that is a left substring of . Then [9, Proposition 8.8,Theorem 8.9] yield an arch bridge and a weak arch bridge such that . Since and the LVP is long, we get .
Lemma 7.4.
(1)
If is an H-reduced string and is a band such that is a skeletal string that is not a substring of then there is a string and such that .
(2)
If is an H-reduced string such that is a skeletal string with respect to then there is a string and such that .
Proof.
We only prove the first statement; a similar proof works for the other case.
There are two possibilities:
(1)
If is torsion then taking to be a length string, [9, Proposition 8.8] gives that is a maximal reverse weak arch bridge.
(2)
If is not torsion then there is a such that is a string. There are further two subcases.
(a)
If is not H-reduced then there is a band such that . Since is not a substring of we get . Let be the shortest left substring of such that is a string. Since is not H-reduced, is a proper left substring of . Hence being a left substring of the H-reduced string [9, Proposition 5.11] gives that is also H-reduced. Thus is a weak arch bridge from to by [9, Proposition 8.8].
(b)
If is H-reduced then we can proceed as above by replacing by .
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.
In this step we also show that . We begin with an observation.
Remark 7.5.
Suppose , and for some . If is a substring of and then .
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 . If either is normal or is abnormal with then
•
;
•
is long if and only if is long.
Proof.
Since is normal, by the construction of . Moreover the construction of gives . If is abnormal with then , and hence the same conclusion follows.
For a contradiction suppose that . Then is non-trivial.
Claim:
Suppose not. Since is H-reduced by [9, Theorem 8.6], we get . Hence .
If is normal then . Hence , a contradiction. On the other hand if is abnormal then is a substring of , and hence , again a contradiction. This completes the proof of the claim.
Now the claim combined with gives that . Therefore is abnormal. As a consequence there is a partition such that and is a string. Since , is a string and hence is -long. Since there are two cases.
If then . On the other hand, if then is the first syllable of . In both cases is -long. Hence , a contradiction to the claim. Therefore our assumption is wrong and we conclude .
As , and we get . Hence from Corollary 5.9, is long if and only if is long.
Lemma 7.7.
Assuming the hypotheses of Lemma 7.3 and using the notations above, if is abnormal with then .
Proof.
We prove the result for . With slight modification we can get the proof for the other two cases.
Since is a proper left substring of , is long and , then Corollaries 6.2 and 6.3 together guarantee that is -long. Hence the latter corollary guarantees that and that is a substring of . Since is a left substring of , we see that . Hence is a left substring of thanks to [9, Proposition 4.3].
Suppose . Then is non-trivial, and is abnormal with and . But in this case is -long which guarantees that , a contradiction. Hence .
Step 3: The construction of and guarantees that if and only if is short but is long.
Example 7.8.
Continuing from Example 6.5 we get is short whereas uncle of is -long. Here indeed , and .
If and are both long or short then . We then show that .
If is short but is long, and hence , then in view of Remark 5.1 we have . The three subcases, namely , and are described in Corollaries 6.4, 6.17 and 6.13 respectively.
If is long but is short, and hence , then Corollary 6.6 gives that , , and . Hence in view of Remark 5.1.
Now we can analyse the four cases to complete the proof of Lemma 7.3.
Case I: is -long or -long.
In this case, . Remark 6.18 gives that . Further Corollary 6.3 guarantees that is normal. Moreover, is a substring of , and hence is an H-reduced string by [9, Proposition 5.11].
Step I.1: the string is H-reduced.
Suppose not. Then there is a cyclic permutation of a band such that and . Moreover there is a weak bridge . Since and is H-reduced, contains at least one syllable of and each.
If is not a substring of then and commute contradicting the domesticity of the algebra. Hence is a right substring of and . But then there is an abnormal weak bridge such that .
Let denote the cyclic permutations of and respectively such that is a right substring of and is a string. If then clearly , which yields a factorization of , a contradiction to . Hence .
In this case let so that . If then clearly . On the other hand if then [9, Proposition 5.2] together with gives that . Therefore in each case we conclude that . This contradiction proves the claim.
Step I.2:
Since is H-reduced from the claim above, Lemma 7.4 yields a string and such that .
Now [9, Proposition 8.8, Theorem 8.9] yields an arch bridge and a possibly trivial weak arch bridge such that . Lemma 7.6 guarantees that and that is long if and only if is long. Since it readily follows that .
Step I.3: We want to show that .
Suppose not. Then is non-trivial. Let . If for any cyclic permutation of , is not a word then is normal and is a substring of and hence the required identity holds. Let be its cyclic permutation such that is a word.
There are two cases.
If for some string with then there are further two subcases.
•
with .
If then can be used to show that for some , which is a contradiction to .
If then since , we obtain that is not a string. Then there is a right substring of and a left substring of such that . Since is a string, we get that . Similarly since is a string, we get that is not a right substring of . As a consequence . This is a contradiction to and .
•
for .
Here . Then using we get that and are strings, a contradiction to domesticity.
If for some string with then is also a syllable of , where is a string, and thus . Suppose is the syllable of such that is a substring of a cyclic permutation of .
If then and share with , a contradiction to domesticity.
On the other hand if then let be the cyclic permutation of such that is a string. To ensure domesticity there is a right substring of and a left substring of a cyclic permutation of such that and . Since is a string, is not a right substring of . As a consequence, , a contradiction to and .
Example 7.9.
In the algebra in Figure 10 the only bands are , . Choose and so that and .
Figure 10. with
Clearly , is -long with , and . It is readily verified that is H-reduced and is the uncle of such that .
Case II: is -long.
In view of Remark 6.18 we have three subcases depending on the value of . Here we show the proof only for the case ; the proofs for the remaining two cases are obvious modifications.
In this case . If then , and hence is short, a contradiction to our assumption. Therefore . As a consequence . Moreover, . Since is a left substring of , it is an H-reduced string by [9, Proposition 5.11].
Step II.1: is H-reduced.
If not then there is a cyclic permutation of a band such that . Suppose and is the last syllable of that is not in . Then clearly , and hence it is readily verified that and 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 such that , and that for some and possibly trivial .
Step II.3: Assuming that and are either both long or both short, we need to show that .
Suppose not. Then is non-trivial. Let . First we argue that there is a cyclic permutation of such that is a string. If not then there is a substring of such that is a string for some cyclic permutation of . Therefore is a string and the required identity follows.
There are two cases.
If is a proper left substring of then is necessarily a proper left substring of for otherwise commutes with . However if is a proper left substring of 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 is a proper left substring of . Then the first syllable of after is also a syllable of . To ensure domesticity, is a left substring of . There are two subcases:
•
If is not a substring of then for the existence of an abnormal weak bridge , it is necessary to have and that is a left substring of . Then it is readily verified that , a contradiction to .
•
If is a substring of then consider the cyclic permutation of such that is a string. To ensure there is a right substring of and a substring of such that . But in this case is long and hence which is a contradiction to .
Example 7.10.
Consider the algebra that differs in its presentation from from Figure 7 in that is removed from . Keeping and same as in Example 6.5 we get , is short and is -long with , , and . It is readily verified that is H-reduced and is the short uncle of such that .
Case III: is -long or -long. In this case .
Step III.1: is H-reduced.
First we argue that is band-free. Suppose not. Since is a left substring of , is H-reduced. Thus and hence , and there is a cyclic permutation of a band such that , the first syllable of is not a syllable of and is a proper left substring of . Since is a common syllable of both the bands, in view of [9, Proposition 4.3] there is an abnormal weak bridge from to and . On the other hand since is a string and is a left substring of there is a weak bridge from to , a contradiction to the domesticity of the algebra. This completes the proof that is band-free.
If is not band-free then in view of the above paragraph there is a cyclic permutation of a band such that . Moreover since we conclude that is a substring . Since it is readily verified that and commute, a contradiction to domesticity that proves the claim.
Step III.2:
We can argue as in Case I that there is such that , and that for some and possibly trivial .
Step III.3: Assuming that and are either both long or both short, we need to show that .
Suppose not. Then is non-trivial. Let . As in Case II, we can show that there is a cyclic permutation of such that is a string.
There are two cases.
If is a proper left substring of then there are two subcases:
•
for a string . If then our assumption gives that . Then is a string since . Hence there is a weak bridge . Since , we get that . Hence . But then it is readily verified that , a contradiction to the fact that is band-free.
•
for a string with . Then is the first syllable of . Let be the syllable of such that is a string. Clearly which implies that there is an abnormal weak arch bridge from to with . Since , we get and hence , a contradiction to .
If is a proper left substring of then to ensure domesticity no syllable of is a syllable of . On the other hand since , we conclude that is a string with at least one syllable common between and . Then it is readily verified that for an abnormal weak bridge , a contradiction to .
Example 7.11.
In the algebra in Figure 11, , and their inverses are the only bands. Choose and so that and . Here , is short and is -long with , and . It is readily verified that is H-reduced and is the short uncle of such that .
Figure 11. with
Case IV: is -long.
An argument similar to Case II guarantees that .
Step IV.1: is H-reduced.
First we show that is band-free. Recall that is a left substring of and hence H-reduced by [9, Remark 5.6]. If is not band-free then . In this case there is a cyclic permutation of a band such that . Since and there is a weak bridge from to . Moreover since is a string and is a left substring of there is a weak bridge from to , a contradiction to the domesticity of the algebra. In fact the same argument also proves that is also band-free. Hence it is H-reduced.
Step IV.2: Lemma 7.4 ensures that there is a such that is either a maximal reverse arch bridge or a weak arch-bridge from . Without loss we assume that is weak and for some and possibly trivial weak arch bridge .
Step IV.3: Assuming that and are either both long or both short, we need to show that .
Suppose not. Then is non-trivial. Let . As in Case II, we can show that there is a cyclic permutation of such that is a string.
There are three subcases:
•
for a string . Then is a string since . Hence there is a weak bridge . Since , we get that . Hence . But then it is readily verified that , a contradiction to the fact that is band-free.
•
for a string with and . Then is the first syllable of . Let be the syllable of such that is a string. Clearly which implies that there is an abnormal weak arch bridge from to with . Since , we get and hence , a contradiction to .
•
for a string with . To ensure domesticity we must have equality. Let be the abnormal weak bridge between and . The dual of [9, Proposition 4.3] gives that and that is a proper right substring of . It is readily verified that , a contradiction to .
Example 7.12.
Consider the algebra in Figure 11. Choose and so that and . Here , is short and is -long with , , and . It is readily verified that is H-reduced and is the short uncle of such that .
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 fails then we cannot obtain the weak arch bridge in Step 2 with . On the other hand, if only condition fails then we can obtain but it factors uniquely as . 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 .
Remark 7.13.
Suppose and . Then if and only if is an abnormal half -arch bridge. When these equivalent conditions hold then and the LVP could be simultaneously -long as well as -long. If the LVP is -long (resp. -long) then (resp. ), and the pair is -long (resp. -long).
Examples 7.14.
In the algebra from Figure 9 if and then is an abnormal half -arch bridge with and , and hence .
If we remove the relation from while keeping and the same we get , and hence .
Proposition 7.15.
Suppose , , is long, and if is -long then . Then there is an uncle of satisfying
Furthermore,
•
if is abnormal then ;
•
if is normal then iff .
Recall from Corollary 5.8 that is long if and only if the LVP is long. If these equivalent conditions hold then the LVP is long. In case the latter LVP is -long with then the proof of the uncle forking lemma still goes through where is an “invisible” long child of with and thus explains this forking string.
Proposition 7.16.
Suppose . If the LVP is -long with then
•
if and if whenever is abnormal then there exists a sibling of satisfying iff is -long and
•
if and there is a string such that then there exists a sibling of satisfying , iff is -long and
Example 7.17.
Suppose is obtained by replacing the relation by in in Figure 10. The only bands here are , and their inverses. Choosing and we have , and . Moreover the LVP is -long due to with and . Here is a sibling of such that and .
The next result is essentially a mini-version of Lemma 7.3.
Proposition 7.18.
Suppose and is normal if . Let be a proper left substring of such that and forks. Then the following hold:
If then there is a sibling of such that ;
If then there is a sibling of such that .
In both cases we have
Proof.
We only prove the first statement; a similar proof works for the other case.
Let . Let be the left substring of such that is a string. Let be the syllable with such that is a string.
Since is a string, an argument as in Step 1 of the proof of the uncle forking lemma (Lemma 7.3), we can show that is a band-free string. Then as in Step 2 there, we extend to for some using Lemma 7.4. Then an argument similar to Step 3 shows that , where is the canonical factorization such that is a possibly trivial weak arch bridge and . Since we get by Corollary 5.9 and hence . In both Step 1 and 3 the proof uses the fact that if then for some .
Example 7.19.
Consider and in the algebra from Figure 10 so that and . Clearly , is normal with and forks. Here is a sibling of with and .
Below we note without proof the consequences of the failure of condition of the hypotheses of Lemma 7.3.
Proposition 7.20.
Suppose all hypotheses of Lemma 7.3 hold except for . Then the following are equivalent:
(1)
is not -long;
(2)
is a forking string;
(3)
is abnormal with , the LVP is -long with and one of the following holds:
(a)
and ;
(b)
and .
When these equivalent conditions hold then is short by Corollary 6.6 and Proposition 6.16.
It is worth noting that either condition or 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 for arbitrary . Since abnormality is the source of anomalies, the following result states that only are possible cases.
Proposition 7.21.
Suppose for with that all hypotheses of Lemma 7.3 fold except that exactly one of or fails.
If then is a proper left substring of . As a consequence is a proper substring of , and hence . Moreover and are normal.
If then is a proper left substring of .
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 with . Assume all the hypotheses of Lemma 7.3 hold but exactly one of or fails. Furthermore assume that is a forking string. Then is short and there is a normal granduncle of satisfying
Moreover
if and only if ;
and if and only if .
Proof.
If condition in the hypothesis of Lemma 7.3 fails then is a proper substring of . In particular, . Then Proposition 7.21 guarantees that is a string for some string of positive length. Let be the shortest right substring of such that is a string. Clearly is a left substring of . In fact is also a left substring of . Then the hypotheses imply that is a band-free string, which can be extended using Lemma 7.4 to a weak arch bridge with . 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 from the hypothesis of Lemma 7.3 fails then Proposition 7.20 gives a characterization of when is a forking string. Then Corollary 6.6 and Proposition 6.16 guarantee that 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 is not a string. However satisfies the hypotheses of the latter, and this explains the forking string .
Examples 7.23.
Suppose the algebra (resp. ) is obtained by replacing by (resp. by ) in Figure 7 (resp. Figure 8). Choose and (resp. ) so that (resp. ) and (resp. ). Clearly (resp. ), is -long but not -long and all the hypotheses of Lemma 7.3 except hold. It is readily verified that satisfies the equivalent conclusions of Proposition 7.20. Moreover there is a grand-uncle (resp. ) of such that (resp. ).
In the algebra in Figure 8 choosing and we have and with . Here is -long and all the hypotheses of Lemma 7.3 except hold. It is easy to check that there is a grand-uncle of such that .
There is yet another variation of Lemma 7.3 where . 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 , , for some and is long.
If is abnormal with then is -long, is the first syllable of and .
If is abnormal with then one of the following happens:
•
If is -long, is the first syllable of and ;
•
If is -long then .
Irrespective of whether is normal or abnormal, for some string and is a string.
Lemma 7.25.
(Uncle forking lemma–opposite parity)
Suppose , , for some and is long. Further suppose that if is abnormal and is -long then is not a substring of . Then there is an uncle of constructed as in Lemma 7.3 that satisfies . Such an uncle is short if . Moreover one of the following holds:
(1)
If either of the following sets of conditions hold:
•
is normal;
•
is abnormal with , is -long and is not the first syllable of ,
then , and hence and
(2)
If is abnormal and is the first syllable of then the hypothesis ensures that there is a partition with such that is a string and is not the first syllable of . Moreover
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)
In the algebra in Figure 12 the only bands are , and their inverses. Choosing and we get and . Clearly , , and is -long. Moreover both and the uncle of are short, and .
Figure 12. with
However if we add to for the same quiver then choosing the same and we still have that is short but its uncle is -long, and .
(2)
Consider the algebra in Figure 6. Choosing and we get and . Clearly , , and is -long. Moreover is -long, the uncle of is short, and .
If we remove from for the same quiver then choosing the same and we have , both and are short, and .
(3)
In the algebra in Figure 13 the only bands are , and their inverses. Choosing and we get and . Clearly , and is -long. Moreover both and the uncle of are short, and .
Figure 13. with
However if we add to for the same quiver then choosing the same and we get that , is -long and the uncle of is short, and .
(4)
In the algebra in Figure 11 choosing , same as in Example 7.12 we get , and both and the uncle of are short. Moreover .
Remark 7.27.
The last conclusion of Proposition 7.24 guarantees that there is no variation of the granduncle forking lemma when .
8. Building H-reduced strings
Associate to each and the maximal path
in such that and does not exist for each . In particular, if then .
Set and .
Remark 8.1.
Suppose and . If and then , and hence .
Proposition 8.2.
Suppose and . Then is a proper left substring of .
Proof.
Suppose . Then and hence . Let be the syllable with such that is a string. Since , is a syllable of . Moreover is a weak half -arch bridge and hence is also a syllable of and hence the conclusion.
Suppose . Both and are left substrings of and hence they are comparable. If is a proper substring of then the conclusion is obvious. Thus we assume that for some string . Then Proposition 4.6 gives that is abnormal. Then Corollary 6.19 gives that the LVP is -long and hence is a right substring of .
If is short then is a proper left substring of . On the other hand if is long then contains a cyclic permutation of as a left substring. Therefore the conclusion follows in both the cases.
Propositions 8.2 and 4.10 together guarantee, for each with for some , the existence of a string of positive length satisfying
Remark 8.3.
Suppose , and is abnormal. In view of Proposition 8.2 there is a positive length string such that . Since and , we get .
Remark 8.4.
Suppose with satisfies the hypotheses of the granduncle forking lemma (Lemma 7.22). If and then is a proper left substring of .
Remark 8.5.
Suppose and . If and then .
Proposition 8.6.
Suppose and . If , and exists then is a proper left substring of .
Proof.
Both and are left substring of and hence are comparable. Corollary 4.4 states that is a left substring of and hence is a substring of . The conclusion is clear if is a proper left substring of . Hence we assume otherwise.
If the LVP is short then it can be readily seen that for some we have , a contradiction to . Hence the LVP is long. Since , the LVP can not be -long and -long.
Now if then is a proper left substring of . If is long then and hence the conclusion follows. Thus it remains to consider the case when is short.
There are four cases:
•
is -long. Here but since is short, is a proper left substring of .
•
is -long or -long. Since is short, is a proper left substring of while is a proper left substring of .
•
is -long. If then is a left substring of and hence by Proposition 4.10, which is a contradiction. Hence and while is a proper substring of since is short.
•
is -long or -long. The proof is similar to the second case above after replacing by .
This completes the proof.
Propositions 8.6 and 4.10 together guarantee, for each for which exists, the existence of a string of positive length satisfying
Using similar arguments as in the proofs in Propositions 8.2 and 8.6 we get the following result.
Proposition 8.7.
Suppose and . If and exists then is a proper left substring of .
Propositions 8.7 and 4.10 together guarantee, for each for which exists, the existence of a string of positive length satisfying
Combining Propositions 8.2, 8.6 and 8.7 we get the following.
Corollary 8.8.
Suppose , and . Then
•
;
•
if exists;
•
if exists.
For each , let . Note that one can identify with the path .
Definition 8.9.
Let
Define by the identity
Define .
Given distinct with and , say that if either or but . Thus we have completed the description of a decorated tree whose vertex set is .
There is a natural embedding defined by . Note that exists for each .
Equip with inorder, denoted , where for we have
•
;
•
if in ;
•
if ;
•
if ;
•
if in for some ;
•
if in for some .
Then is a total order with as the minimal element, its successor and as the maximal element.
Remark 8.10.
Suppose . Then if and only if one of the following happens:
•
and in ;
•
and in .
In fact the set can also be equipped with inorder in a similar manner.
9. Signature types of H-reduced strings
Proposition 9.1.
Suppose , and . Then
Proof.
We only prove the result when , and indicate the changes for the other case. Recall from Proposition 8.2 that and . Consider the following cases:
Case 1: is normal.
Then Proposition 4.6 gives that for some string of positive length. Note that and .
If all elements of are short then using Proposition 7.16 gives that if the LVP is -long then , and hence the conclusion is readily verified. Therefore suppose that some element of is long. Since is normal, any long satisfies condition of the hypotheses of Lemma 7.3.
Claim: is not a forking string for any .
If not then since is normal, condition of Proposition 7.20 fails, and hence satisfies condition of the hypotheses of Lemma 7.3. Now that all the hypotheses of that lemma are true, it ensures the existence of a sibling of . Since we must have but this is impossible when is normal (see Step 3 of the proof of that lemma). Thus the claim.
It follows from the claim that all long share the same exit syllable. Then it is readily verified that the conclusion follows.
When we use Proposition 7.16 but not Lemma 7.3 in the argument.
Case 2: is abnormal.
Then Proposition 4.6 gives that for some string . Further Remark 8.3 gives that . Since is the first syllable of , we have .
If is short then the conclusion is readily verified (one may need to use Proposition 7.16 when ).
On the other hand, if is long then all elements of are long. Let . An argument similar to the above paragraph shows that . Since and is a right substring of , we can easily conclude that .
With minor modifications in the proof above, we can also show the following two results.
Proposition 9.2.
.
Proposition 9.3.
Suppose then the following hold.
•
if ;
•
if .
Proposition 9.4.
Suppose and . If and then
.
Proof.
We only prove the result when ; the proof of the other case is omitted as it is very similar with only minor modifications. Recall from Proposition 8.6 that and . Clearly and . If then there is a string of positive length such that is a proper left forking substring of and . Let be the syllable such that and is a string.
There are two main cases and several subcases:
Case 1: is normal.
Here Proposition 4.6 gives that for some string of positive length. We also have .
Case 1A: for some string of positive length.
Suppose is a proper left substring of . Since we have . Moreover since is a string we conclude that . Now [4, Proposition 8.5] guarantees that there exists a semi-bridge or a torsion reverse semi-bridge, say , such that and while [4, Proposition 8.7] ensures that . Then we have , a contradiction to the definition of .
Proposition 7.18 gives that if is a proper left substring of and the latter is a substring of then there is a sibling of with and . As a consequence we get , a contradiction to the definition of .
Finally an argument similar to the proof of Case 1 of Proposition 9.1 using Lemma 7.3 and Proposition 7.16 shows that is not a substring of .
Case 1B: is a proper left substring of for some string of positive length.
The proof of this subcase is similar to the above two paragraphs.
Case 1C: for some string .
Under this assumption we showed in the proof of Proposition 8.6 that the LVP is long. Let . Then .
Claim: There is no satisfying .
Suppose not. Then the LVP is necessarily either -long or -long. Then the uncle forking lemma yields a sibling of satisfying . The existence of such will contradict the definition of , and hence the claim.
Irrespective of whether is short or long, the conclusion follows using the claim.
Since is a substring of , it immediately follows that . Corollary 6.19 gives that the LVP is -long. Furthermore Corollary 6.3 gives that . Let , where .
As in the first paragraph of Case 1A above we can show that . Since we clearly get . It remains to show that . Let . We show that is not a string.
If is long then Corollary 6.6 states that is necessarily -long and . Since , we can conclude that is always short. Hence .
If then the uncle forking lemma produces a sibling of such that . Given the assumption of Case 2A, we get , a contradiction to the definition of .
On the other hand if then . If is a string then there is some satisfying and , a contradiction. Thus is not a string. As a consequence is not a forking string, and hence we have completed the proof that .
As in the first paragraph of Case 1A above we can show that .
If is short then follows from the fact .
If is long then the LVP is -long with and .
If satisfies the hypotheses of the uncle forking lemma then the lemma will produce a sibling of satisfying , a contradiction. Hence condition fails. Since we conclude that and hence that is not a forking string. Then it immediately follows that .
Case 2B: .
If then for some string of positive length with by Corollary 6.12. Then it is readily verified that for some , a contradiction. Hence . Let .
Note that for otherwise is a proper left substring of , a contradiction to the assumption of this case. Hence .
An easy impossibility argument using the uncle forking lemma shows that there is no such that . Hence .
As we have . Hence it is readily seen that .
Case 2C: is a proper left substring of for some string of positive length.
Since is a proper left substring of , we have .
If then is a left substring of for otherwise and hence factors through as explained in the first paragraph of Case 2B above.
Let with and . The argument that and is similar to the proof of Case 2B.
Case 2D: for some string of positive length.
Clearly , and . We also have for otherwise factors through .
As argued in the second paragraph of the proof of Proposition 8.6, we see that the LVP is long.
If then Corollary 6.19 gives that . If then . These two statements together contradict the fact that is long. Therefore .
Further since is long we get that is -long with and a proper left substring of , a contradiction by Proposition 6.16. Thus .
Now if then . This together with the identity guaranteed by Corollary 6.19 give a contradiction the fact that is long. Therefore . If then is long gives that is -long, again a contradiction by Proposition 6.16, and hence .
In this case it is readily verified that is -long with and . Then the conclusion easily follows.
The following is the half -arch bridge version of the above whose proof uses similar techniques.
Proposition 9.5.
Suppose and . If then
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 and . If , then .
Proposition 9.7.
Suppose and . If then
Combining Propositions 9.1-9.7 with Corollary 8.8 we get the following result.
Corollary 9.8.
Suppose , and . Then the following hold.
(1)
;
(2)
if exists;
(3)
if exists;
(4)
.
The last equality follows from Proposition 9.3 once we recall that and that there is no syllable with such that is a string.
In view of Definition 8.9, there is an obvious generalization of Corollary 9.8 to .
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 , , say that is a label of the path from to in , written , if one of the following holds:
•
and is isomorphic to the interval in ;
•
and is isomorphic to the interval in .
Here is an observation.
Remark 10.1.
Suppose exists for some string for some . Then does not exist. Hence is not a forking string.
Moreover if for some , forking string with then . As a consequence, if exists, otherwise .
Proposition 10.2.
Suppose is a proper left substring of and for some .
(1)
If is a left -string then .
(2)
If is a left torsion string then .
In addition, if is a forking string then is a fundamental solution of either or above.
Proof.
We only prove the first statement; the proof of the second is similar.
Without loss assume that . Note that exists since .
Suppose exists for some and is a left substring of . If is not a left substring of then let be the maximal common forking left substring of and . Remark 10.1 yields that . Thus by the definition of we have , and hence , a contradiction to . Thus we have concluded that is a left substring of . Since is infinite and is not a forking string by Remark 10.1, it is readily seen that exists. Thus the argument is complete using induction.
The next result follows by combining Proposition 10.2 and Corollary 9.8.
Corollary 10.3.
Suppose , and . Then the following hold.
(1)
(2)
if exists.
(3)
if exists.
(4)
We only justify the last identity. Using the fact that and the definition of we clearly have . The reverse inequality follows by combining Proposition 9.3 with Proposition 10.2.
Definition 10.4.
Given and , define to be the minimal height element of such that .
Remark 10.5.
Suppose , and . Then the following hold.
(1)
;
(2)
exists if and only if ;
(3)
If then exists;
(4)
If then exists.
Using Corollary 10.3 and Remark 10.5 we get the following.
Corollary 10.6.
Suppose and . Then the fundamental solution, say , of
(3)
is given by
Lemma 10.7.
Suppose . Then .
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.
The assignment of for as in Definition 8.9 has a simple consequence.
Proposition 10.8.
Suppose and . Then
(4)
and hence
(5)
Our goal is to associate a term to each non-root using induction on subtrees, such that
(6)
Algorithm 10.9.
Suppose , and are all the elements of for and . Suppose for each we already have inductively defined terms , for , satisfying
(7)
We abbreviate the above equations as follows:
(8)
Define
Proposition 10.10.
With the above hypotheses and notations, we have
(9)
Proof.
Without loss we assume that and that . The proofs in the other cases are simpler, and hence omitted.
First note by definition of that . Hence . Hence by Corollary 10.6 we obtain
(10)
If then the same corollary gives that
(11)
In fact if we extend the list by adding for and as in Definition 8.9 then also Equation (11) holds true. Hence
The required equation in the statement now follows from Equations (10), (11),(12) and (13).
There is an obvious generalization of the above when .
The string may not be the fundamental solution of Equation (9) as the following example demonstrates.
Example 10.11.
In the algebra in Figure 11 consider , and . It is readily verified that and by Algorithm 10.9 but the fundamental solution of Equation (9) is .
Lemma 10.12.
Suppose , and .
Then
(14)
Proof.
Since , we get .
If then Remark 10.5(1) ensures the existence of . Finally the proof of is similar to the proof of Equation (11) using Corollary 10.6.
Now we are ready to inductively define the term for .
(15)
Combining Corollary 10.6, Proposition 10.10 and Lemma 10.12 we get the main result of this section.
Theorem 10.13.
For each , Equation (6) indeed holds. In particular,
Corollary 10.14.
If is a domestic string algebra, is a string for and then , where is the class of finitely presented bounded discrete linear orders.
Example 10.15.
Continuing from Example 2.2, choosing the only non-root vertices are and , where and . Here . Since is a leaf we get and by Algorithm 10.9. Moreover , and hence the same algorithm gives and . Since and , Corollary 10.6 gives . The same corollary gives . Therefore we get
The definition of naturally extends to . Indeed the computation of for needs at most many iterations of Algorithm 10.9.
For the following result gives the recipe to find a term such that
Lemma 10.16.
Suppose . Then
for some
, and .
Proof.
Recall from Lemma 2.1 that is a bounded discrete linear order. If is a forking string then both and exist. Since and , we get that the sequence of lengths has a local minimum at .
Hence if for any then is not a forking string and the sequence of lengths of strings does not have a local minimum at . Hence for some , and . The conclusion then follows from Lemma 10.7.
Since , the last clause of Corollary 10.6 gives that . Since , we get by the third clause of the same corollary. Therefore we get
by Equation (6). Now using the computation in Example 10.15 together with Equation (15) we have
Since we get
.
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 and , there is a domestic string algebra , a string and such that .
References
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[3]
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[4]
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[5]
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Appl. Algebra, 170(2-3), 295–307 (2002)
[6]
Ringel, C.M.: Infinite length modules. Some examples as introduction, Birkhäuser, Basel, 1-73 (2000)
[7]
Schröer, J.: Hammocks for string algebras. PhD thesis, Universität Bielefeld (1997)
[8]
Schröer, J.: On the infinite radical of a module category. Proc. Lond. Math. Soc. 81(3), 651–674 (2000)
[9] Sardar, S., Kuber, A.: Variations of the bridge quiver for domestic string algebras, preprint (2021) available at https://arxiv.org/abs/2109.06592
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