The Post Correspondence Problem and equalisers for certain free group and monoid morphisms

02/18/2020
by   Laura Ciobanu, et al.
Heriot-Watt University
0

A marked free monoid morphism is a morphism for which the image of each generator starts with a different letter, and immersions are the analogous maps in free groups. We show that the (simultaneous) PCP is decidable for immersions of free groups, and provide an algorithm to compute bases for the sets, called equalisers, on which the immersions take the same values. We also answer a question of Stallings about the rank of the equaliser. Analogous results are proven for marked morphisms of free monoids.

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1. Introduction

In this paper we prove results about the classical Post Correspondence Problem (), which we state in terms of equalisers of free monoid morphisms, and the analogue problem for free groups ([CMV08], [MNU14]), and we describe the solutions to and for certain classes of morphisms. While the classical is famously undecidable for arbitrary maps of free monoids [Pos46] (see also the survey [HK97] and the recent result of Neary [Nea15]), for free groups is an important open question [DKLM19, Problem 5.1.4]. Additionally, for both free monoids and free groups there are only few results describing algebraically the solutions to classes of instances known to have decidable or . Our results apply to marked morphisms in the monoid case, and to their counterparts in free groups, called immersions. Marked morphisms are the key tool used in resolving the for the free monoid of rank two [EKR82], and therefore understanding the solutions to the for immersions is a significant step towards resolving the for the free group of rank two. The density of marked morphisms and immersions among all the free monoid or group maps is strictly positive (Appendix 1), so our results concern a significant proportion of instances.

An instance of the is a tuple , where are finite alphabets, are the respective free monoids, and are morphisms. The equaliser of is The is the decision problem:

Given , is the equaliser trivial?.

Analogously, an instance of the is a four-tuple with morphisms between the free groups and , and is the decision problem pertaining to the similarly defined in free groups. Beyond , in this paper we also consider the Algorithmic Equaliser Problem which, for an instance with either free monoid () or free group () morphisms, says:

Given , output either

  1. [label=()]

  2. a finite basis for , or

  3. an automaton whose language is .

For free groups these two problems are in fact the same (if is finitely generated), while for free monoids (1) implies (2). Part (1) of the is known to be soluble when and one of or is non-periodic and insoluble otherwise [Hol03] [HK97, Corollary 6].

As automata accepting intersections of regular languages are computable, if Part (2) of the is soluble for a class of maps then we have the following equivalent, but seemingly stronger, result: there exists an algorithm with input a finite set of morphisms from and output a finite automaton whose language is . Similarly, bases of interesections of subgroups of free groups are computable, and so we have: if Part (1) of the is soluble for a class of maps then there exists an algorithm with input a finite set of morphisms from and output a basis for .

A set of words is marked if each is non-empty and starts with a different letter of . A free monoid morphism is marked if the set is marked. An immersion of free groups is a morphism where the set is marked (see Section 3 for equivalent formulations). Halava, Hirvensalo and de Wolf [HHdW01] showed that is decidable for marked morphisms, and inspired by their methods we were able to obtain stronger results (Theorem A) for this kind of map, as well take this result to the world of free groups (Theorem C), where we employ ‘finite state automata’-like objects called Stallings graphs. In fact, our results resolve the simultaneous and , which take as input an arbitrary set of maps (not just two maps ) and ask the same questions about equalisers as in the classical setting.

Theorem A.

If is a set of marked morphisms then there exists an alphabet and a marked morphism such that . Moreover, if is a finite set then there exists an algorithm with input and output the marked morphism .

Corollary B.

The simultaneous is decidable for marked morphisms of free monoids.

Theorem C.

If is a set of immersions then there exists an alphabet and an immersion such that . Moreover, if is a finite set then there exists an algorithm with input and output the immersion .

Corollary D.

The simultaneous is decidable for immersions of free groups.


The Equaliser Conjecture. Our work was partially motivated by Stallings’ Equaliser Conjecture for free groups, which dates from 1984 [Sta87, Problems P1 & 5] (also [DV96, Problem 6] [Ven02, Conjecture 8.3] [BMS02, Problem F31]). Here stands for the rank, or minimum number of generators, of a subgroup :

Conjecture 1.1 (The Equalizer Conjecture, 1984).

If are injective morphisms then .

This conjecture has its roots in “fixed subgroups” of free group endomorphisms (if then ), where Bestvina and Handel proved that for an automorphism [BH92], and Imrich and Turner extended this bound to all endomorphisms [IT89]. Bergman further extended this bound to all sets of endomorphisms [Ber99]. Like Bergman’s result, our first corollary of Theorem C considers sets of immersions, which are injective, and answers Conjecture 1.1 for immersions.

Corollary E.

If is a set of immersions then .

In free monoids, although equalisers of injections are free [HK97, Corollary 4] they are not necessarily regular languages (and hence not necessarily finitely generated) [HK97, Example 6]. In order to understand equalisers of sets of maps we need to understand intersections in free monoids, and although the intersection of two finitely generated free submonoids is free [Til72] and one can find a regular expression that represents a basis of [BH77], the intersection is not necessarily finitely generated [Kar84]. The following result is therefore surprising because we have finite generation, even in the intersection .

Corollary F.

If is a set of marked morphisms then is a free monoid with .


The Algorithmic Equaliser Problem. The is insoluble in general, as equalisers are not necessarily finitely generated [Ven02, Section 3], and is an open problem of Stallings’ if both maps are injective [Sta87, Problems P3 & 5]. Our next corollary of Theorem C therefore resolves this open problem for immersions.

Corollary G.

The is soluble for immersions of free groups.

The is insoluble in general, primarily as equalisers are not necessarily regular languages [ER78, Example 4.6]. Even for maps whose equalisers form regular languages, the problem remains insoluble [KS10]. Another corollary of Theorem A is the following.

Corollary H.

The is soluble for marked morphisms of free monoids.

In fact, this is a special case of Corollary 2.7, which algorithmically obtains a free basis for (and not just ).


Outline of the article. In Section 2 we prove Theorem A and its corollaries. The remainder of the paper focuses on free groups, where the central result is Theorem 6.2, which is Theorem C for . In Section 3 we reformulate immersions in terms of Stallings’ graphs. In Section 4 we define the “reduction” of an instance of the for immersions. Repeatedly computing reductions is the key process in our algorithm. In Section 5 we prove the process of reduction reduces the “prefix complexity” of an instance (so the word “reduction” makes sense). In Section 6 we prove Theorem 6.2, mentioned above. In Section 7 we prove Theorem C and its corollaries. In Section 9 we give a complexity analysis for both our free monoid and free group algorithms.

2. Marked morphisms in free monoids

In this section we prove Theorem A and its corollaries. We use the following immediate fact, which shows that , and in particular, we may assume .

Lemma 2.1.

Marked morphisms of free monoids are injective.

Consider morphisms and . The set of non-empty words over an alphabet is denoted . For , a pair is an -block if (i) starts with , and (ii) and are minimal, that is, the length is minimal among all such pairs. If the pair has blocks , , then let be the alphabet consisting of these blocks and define by and by . These maps are computable and, by an identical logic to [HHdW01, Section 2], are seen to be marked. The map , which we call (so ), is the composition of marked morphisms and hence is itself marked. We therefore have the following.

Lemma 2.2.

If and are marked morphisms then the corresponding maps , and are marked and are computable.

We require in the proof of Lemma 2.6. Meanwhile, we take .

The reduction of an instance of the marked , as defined in [HHdW01], is the instance where is defined as above, and where and are as above, but with codomain (which we may do as ). We additionally assume that ; we can do this as by Lemma 2.2.

The following relies on [HHdW01, Lemma 1], which we strengthen by replacing the notion of “equivalence” with that of “strong equivalence”: Two instances and of the are strongly equivalent if their equalisers are isomorphic, which we write as .

Lemma 2.3.

Let be the reduction of where and are marked. Then and are strongly equivalent, and .

Proof.

Firstly, note that [HHdW01, Lemma 1, paragraph 2]. From [HHdW01, Lemma 1, paragraph 1] it follows that , so . As is injective, the map is an isomorphism. Hence, and are strongly equivalent, and, by symmetry for the map, as required. ∎

We can now improve the existing result on the marked . We store a morphism as a list .

Theorem 2.4.

If is an instance of the marked then there exists an alphabet and a marked morphism such that . Moreover, there exists an algorithm with input and output the marked morphism .

Proof.

We explain the algorithm, and note at the end that the output is a marked morphism with the required properties, and so the result follows.

Begin by making reductions , starting with , the input instance. Then by [HHdW01, Section 5, paragraph 1] we will obtain an instance such that one of the following will occur:

  1. .

  2. for all .

  3. There exists some with (sequence starts cycling).

Keeping in mind the fact that reductions preserve equalisers (Lemma 2.3), we obtain in each case a subset (possibly empty) which forms a basis for : For Case (1), writing , the result holds as if then and so as roots are unique in a free monoid. For Case (2), suppose . Then and agree on the first letter of because the image of each letter has length one, and inductively we see that they agree on every letter of . Hence, a subset of forms a basis for .

For Case (3), suppose there is a sequence of reductions beginning and ending at :

and write . By Lemma 2.3, ; thus restricts to an automorphism of , so . The automorphism is necessarily length-preserving ( for all ). Consider . Then maps the letters occurring in to letters and so and map the letters occuring in to letters, and it follows that every letter occuring in is a solution to . Hence, a subset of forms a basis for as required.

Therefore, in all three cases a subset of forms a basis for , and since is computable, this basis is as well. In order to prove the theorem, it is sufficient to prove that there is a computable immersion . Consider the map (and the analogous ). Now, each is marked, by Lemma 2.2, and so is the composition of marked morphisms and hence is marked itself. Define . This map is computable from , and as , the map is marked. As , by Lemma 2.3 and the above, the result follows. ∎

Theorem 2.4 combines with the following general result to give the non-algorithmic part of Theorem A. A subsemigroup of a free monoid is marked is it is the image of a marked morphism.

Lemma 2.5.

If is a set of marked subsemigroups of then the intersection is marked.

Proof.

Firstly, suppose for some . Then there exist two words and , with , such that and , where is a marked morphism. If and have a nontrivial common prefix, then because is marked we get , and is a prefix of both and , and in particular . By continuing this argument, if is a maximal common prefix of and , then .

Now, suppose , and suppose they both begin with some letter . By the above, their maximal common prefix is contained in each and so is contained in . Therefore, is a prefix of every element of beginning with an . It follows that is immersed, as required. ∎

We now prove the algorithmic part of Theorem A (this is independent of Lemma 2.5).

Lemma 2.6.

There exists an algorithm with input a finite set of marked morphisms and output a marked morphism such that .

Proof.

We use induction on . By Theorem 2.4, the result holds if . Suppose the result holds for all sets of marked morphisms, , and let be a set of marked morphisms. Take elements , and write . By hypothesis, we can algorithmically obtain marked morphisms and such that and .

By Lemma 2.2, there exists a (computable) marked morphism such that (the map corresponds to the map in Lemma 2.2, and to ). Then, as required:

We now prove Theorem A, which states that the equaliser is the image of a marked map.

Proof of Theorem A.

By applying Lemma 2.5 to Theorem 2.4, there exists an alphabet and a marked morphism such that , while by Lemma 2.6 if is finite then such an immersion can be algorithmically found. ∎

We now prove Corollary F, which says that is free of rank .

Proof of Corollary F.

Consider the marked morphism given by Theorem A. By Lemma 2.1, is injective so is free. As is marked the map taking each to the initial letter of is an injection, so as required. ∎

We now prove a strong form of the for marked morphisms.

Corollary 2.7.

There exists an algorithm with input a finite set of marked morphisms and output a basis for .

Proof.

To algorithmically obtain a basis for , first use the algorithm of Theorem A to obtain the marked morphism such that . Then, recalling that we store as a list , the required basis is the set of elements in this list, so the set . ∎

Corollary H, the for marked morphisms, follows from Corollary 2.7 by taking , while Corollary B, the simultaneous , also follows as is trivial if and only if its basis is empty.

3. Immersions of free groups

We denote the free group with finite generating set by , and view it as the set of all freely reduced words over , that is, words not containing , , together with the operations of concatenation and free reduction (that is, the removal of any that might occur when concatenating two words).

We now embark on our study of immersions of free groups, as defined in the introduction. We given two alternative characterisations now. The explanation of all the terms involved follows the statement of the lemma.

Lemma 3.1.

Let be a free group morphism. The following are equivalent.

  1. The map is an immersion of free groups.

  2. Every word in the language is freely reduced.

  3. For all such that , the length identity holds.

Characterisation (3) is the established definition of Kapovich [Kap00]. Characterisation (2) is the one we shall work with in this article. It uses “Stallings graphs”, which are essentially finite state automata that recognise the elements of finitely generated subgroups of free groups. We define these now.

The (unfolded) Stallings graph of the free group morphism is the directed graph formed by taking a bouquet with petals attached at a central vertex we call , where each petal consists of a path labeled by ; the elements of occur as edges traversed backwards and we denote by the edge in opposite direction, and by the sets of edges in both directions. A path , edges, is reduced if it has no backtracking, that is, for all . We denote by the initial vertex of a path and for the terminal vertex, and call a reduced path with a closed reduced circuit.

We shall view as a finite state automaton with start and accept states both equal to . Then the extended language accepted by is the set of words labelling reduced closed circuits at in :

Immersions are precisely those maps such that every element of is freely reduced; this corresponds to the automaton and the “reversed” automaton , where edge directions are reversed, both being deterministic (map in Figure 1 is not an immersion; although the automaton is deterministic, is not). For such maps, is precisely the image of the map [KM02, Proposition 3.8].

Example 3.2.

Let be the map defined by and . Then the graph , where the double arrow represents and the single arrow , is depicted in Figure 1. The map is not an immersion since there are two edges labeled entering (violating Characterisation (2)). Similarly, and both start with (violating Characterisation (1)) and (violating Characterisation (3)).

Figure 1. The graph for the map defined by , .

Using Characterisation (2), we see that immersions are injective [KM02, Proposition 3.8]:

Lemma 3.3.

If is an immersion then it is injective.

4. The reduction of an instance in free groups

By an immersed instance of the we mean an instance where both and are immersions. In this section we define the “reduction” of an immersed instance of the , which is similar to the reduction in the free monoid case.

Let be a directed, labeled graph and a vertex of . The core graph of at , written , is the maximal subgraph of containing but no vertices of degree , except possibly itself. Note that . For , directed, labeled graphs, the product graph of and , denoted , is the subgraph of with vertex set and edge set . One may think of the standard construction of an automaton recognising the intersection of two regular languages, each given by a finite state automaton with start state , where the core of at is the automaton recognising this intersection.


Core graph of a pair of morphisms. Let , be morphisms. The core graph of the pair , denoted , is the core graph of at the vertex , so . We shall refer to as the central vertex of . Note that represents the intersection of the two images [KM02, Lemma 9.3], in the sense that

Write and for the restriction of to the and components, respectively, so , etc.

Now, let be immersions. The graph is a bouquet and every element of is freely reduced [KM02, Lemma 9.2]. We therefore have free group morphisms and induced by the maps , , where and . These maps are computable [KM02, Corollary 9.5]. Let be the alphabet whose elements consist of the petals of . Then generates the free group , so , and we see that both and are immersions with , . The map , which we shall call (so ) is the composition of immersions and hence is itself an immersion. We therefore have the following.

Lemma 4.1.

If and are immersions then the corresponding maps , and are immersions and are computable.

We require that in the proof of Lemma 7.2. Meanwhile, we take .


Reduction. The reduction of an immersed instance of the is the instance where is defined as above, and where and are as above, but with codomain (which we may do as ). We additionally assume that ; we can do this as by Lemma 4.1. As is immersed, it follows from Lemma 4.1 that is also immersed. In the next section we show that the name “reduction” makes sense, as it reduces the “prefix complexity” of instances.

Example 4.2.

Consider the maps given by and .

Then the graph is a bouquet with two petals labelled and , and . Moreover, and Then we can take , and the maps given by and are the reduction of .

We now prove that reduction preserves equalisers. Two instances and of the are strongly equivalent if the equalisers are isomorphic, which we write as .

Lemma 4.3.

Let be the reduction of where and are immersions. Then and are strongly equivalent, and .

Proof.

It is sufficient to prove that is injective and ; that follows as .

As is an immersion it is injective, by Lemma 3.3. Therefore, is injective. To see that , suppose . Writing , we have and so , as required.

To see that , suppose . Then there exists a path , , such that [KM02, Proposition 9.4], where is the canonical morphism of directed, labeled graphs from to the bouquet with petals. Hence, writing for the element of corresponding to , we have that . As and are injective, by Lemma 3.3, we have that as required. ∎

5. Prefix complexity of immersions in free groups

In this section we associate to an instance of the a certain complexity, called the “prefix complexity”. We prove that the process of reduction does not increase this complexity, and that for all there are only finitely many instances with complexity .

Let be an immersive instance of the . We define, analogously to [HHdW01, Section 4] (see also [EKR82]), the prefix complexity as:

In the maps in Example 4.2, , and

The process of reduction does not increase the prefix complexity, and we prove this by using the fact that, for any , the proper prefixes of and are in bijection with the proper initial subpaths of the petals of and , respectively.

Lemma 5.1.

Let be an instance of the with and immersions, and let be the reduction of . Then .

Proof.

We write for the set of vertices in the whose first component is the central vertex of , and similarly for . Note that .

By construction, each petal of and corresponds to a letter , and we shall denote the petal also by . Write for the set of reduced paths in a graph . Similarly to and , we map write for the petal in corresponding to . From now on, all paths are assumed to be reduced. Define

and define and analogously. Hence, and analogously .

For let be a subpath of . Denote by the shortest subpath of intersecting at only one point, their common end vertex (that is, ), such that