On the conjugacy problem for finite-state automorphisms of regular rooted trees
Ievgen V. Bondarenko, Natalia V. Bondarenko, Said N. Sidki, Flavia R. Zapata
aa r X i v : . [ m a t h . G R ] A ug On the conjugacy problem for finite-stateautomorphisms of regular rooted trees with an appendix by Rapha¨el M. Jungers
Ievgen V. Bondarenko, Natalia V. Bondarenko,Said N. Sidki ∗ , Flavia R. ZapataNovember 1, 2018 Abstract
We study the conjugacy problem in the automorphism group Aut( T ) of a regular rootedtree T and in its subgroup FAut( T ) of finite-state automorphisms. We show that under thecontracting condition and the finiteness of what we call the orbit-signalizer, two finite-stateautomorphisms are conjugate in Aut( T ) if and only if they are conjugate in FAut( T ), andthat this problem is decidable. We prove that both these conditions are satisfied by boundedautomorphisms and establish that the (simultaneous) conjugacy problem in the group ofbounded automata is decidable. Mathematics Subject Classification 2010 : 20E08, 20F10
Keywords : automorphism of a rooted tree, conjugacy problem, finite-state automorphism,finite automaton, bounded automaton
The interconnection between automata theory and algebra produced in the last three decadesmany important constructions such as self-similar groups and semigroups, branch groups, iter-ated monodromy groups, self-similar (self-iterating) Lie algebras, branch algebras, permutationalbimodules, etc. (see [17, 24, 9, 3, 1, 18] and the references therein).The connection between groups and automata occurs via a natural correspondence betweeninvertible input-output automata over the alphabet X = { , , . . . , d } and automorphisms of aregular one-rooted d -ary tree T . To present this correspondence let us index the vertices ofthe tree T by the elements of the free monoid X ∗ , freely generated by the set X and orderedby v ≤ u provided u is a prefix of v . The group Aut( T ) of all automorphisms of the tree T decomposes as the permutational wreath product Aut( T ) ∼ = Aut( T ) ≀ Sym( X ), where Sym( X ) isthe symmetric group on the set X . This decomposition allows us to represent automorphisms inthe form g = ( g | , g | , . . . , g | d ) π g , where π g ∈ Sym( X ) is the permutation induced by the action of g on the first level of the tree T . Iteratively, we can define the automorphism g | v = g | x | x . . . | x n forevery vertex v = x x . . . x n of the tree T , where x i ∈ X . Then every automorphism g ∈ Aut( T )corresponds to an input-output automaton A ( g ) over the alphabet X and with the set of statesQ( g ) = { g | v | v ∈ X ∗ } . The automaton A ( g ) transforms the letters as follows: if the automatonis in state g | v and reads a letter x ∈ X then it outputs the letter y = x g | v and the state changesto g | vx ; these operations can be best described by the labeled edge g | v x | y −→ g | vx . Following theterminology of the automata theory every automorphism g | v is called the state of g at v . ∗ The author acknowledges support from the Brazilian Conselho Nacional de Pesquisa and from FAPDF. T ). A subgroup G <
Aut( T ) is called state-closed or self-similar if all states ofevery element of G are again elements of G . Self-similar groups play an important role in moderngeometric group theory, and have applications to diverse areas of mathematics. In particular, self-similar groups are connected with fractal geometry through limit spaces and also with dynamicalsystems through iterated monodromy groups as developed by V. Nekrashevych [17]. The settheoretical union of all finitely generated self-similar subgroups in Aut( T ) is a countable groupdenoted by RAut( T ) called the group of functionally recursive automorphisms [5].Automorphisms of the tree T which correspond to finite-state automata are called finite-state .More precisely, an automorphism g ∈ Aut( T ) is finite-state if the set of its states Q( g ) is finite.The set of all finite-states automorphisms forms a countable group denoted by FAut( T ). Everyfinite-state automorphism is functionally recursive, and hence the group FAut( T ) is a subgroup ofRAut( T ).Other natural subgroups of Aut( T ) are the groups Pol( n ) of polynomial automata of degree n for every n ≥ − ∞ ) = ∪ n Pol( n ). These groups were introduced byS. Sidki in [22], who tried to classify subgroups of FAut( T ) by the cyclic structure of the associatedautomata and by the growth of the number of paths in the automata avoiding the trivial state.Especially important is the group Pol(0) of bounded automata whose elements are called boundedautomorphisms. A finite-state automorphism g is bounded if the number of paths of length m in the automaton A ( g ) avoiding the trivial state is bounded independently of m . It is to benoted that most of the studied self-similar groups are subgroups of Pol(0). In particular, theGrigorchuk group [8], the Gupta-Sidki group [12], the Basilica [11] and BSV groups [6], the finite-state spinal groups [3], the iterated monodromy groups of post-critically finite polynomials [17],and many others, are generated by bounded automorphisms. Moreover, it is shown in [4] thatfinitely generated self-similar subgroups of Pol(0) are precisely those finitely generated self-similargroups whose limit space is a post-critically finite self-similar set which play an important role inthe development of analysis on fractals (see [13]).In this paper we consider the conjugacy problem and the order problem in the groups Aut( T ),RAut( T ), FAut( T ), Pol(0). It is well known that the word problem is solvable in the group FAut( T )and hence in all its subgroups, while it is an open problem in the group RAut( T ). Furthermore,the order and conjugacy problems are open in FAut( T ) and RAut( T ). The conjugacy classesof the group Aut( T ) were described in [23, 7]. It is not difficult to construct two finite-stateautomorphisms which are conjugate in Aut( T ) but not conjugate in FAut( T ) (see [9]). At thesame time, two finite-state automorphisms of finite order are conjugate in Aut( T ) if and only ifthey are conjugate in FAut( T ) (see [21]). The conjugacy classes of the group Pol( −
1) of finitaryautomorphisms were determined for the binary tree in [5] and for the general case in [19].The conjugacy problem was solved for some well-known finitely generated subgroups of Pol(0).In particular, the solution of the conjugacy problem in the Grigorchuk group was given in [14, 20],and it was generalized in [26, 10] to certain classes of branch groups and their subgroups offinite index. Moreover, it was shown in [15] that the conjugacy problem in the Grigorchuk groupis decidable in polynomial time. The conjugacy problem for the Basilica and BSV groups wastreated in [11]. A finitely generated self-similar subgroup of FAut( T ) with unsolvable conjugacyproblem was constructed in a recent preprint [25].The general approach in considering any algorithmic problem dealing with automorphisms ofthe tree T is to reduce the problem to some property of their states. The order and the conjugacyproblems lead us to the following definition. For an automorphism a ∈ Aut( T ) consider the orbits Orb a ( v ) of its action on the vertices v of the tree and define the setOS( a ) = { a m | v | v ∈ X ∗ , m = | Orb a ( v ) |} which we call the orbit-signalizer of a . It is not difficult to see that the order problem is decid-able for finite-state automorphisms with finite orbit-signalizers. We prove that every boundedautomorphism has finite orbit-signalizer and hence the order problem is decidable for boundedautomorphisms. 2 roposition 1. The order problem for bounded automorphisms is decidable.
We treat the conjugacy problem firstly in the group Aut( T ). Given two automorphisms a, b ∈ Aut( T ) we construct a conjugator graph Ψ( a, b ) based on the sets OS( a ) , OS( b ), which portraysthe inter-dependence among the different conjugacy subproblems encountered in trying to find aconjugator for the pair a, b , and which leads to the construction of a conjugator if it exists. Theorem 2.
Two finite-state automorphisms a, b with finite orbit-signalizers are conjugate in
Aut( T ) if and only if they are conjugate in RAut( T ) if and only if the conjugator graph Ψ( a, b ) isnonempty. An important class of self-similar groups are contracting groups. This property for groupscorresponds to the expanding property in a dynamical system. A finitely generated self-similargroup is contracting if the length of its elements asymptotically contracts when applied to theirstates. A finite-state automorphism is called contracting if the self-similar group generated byits states is contracting. Bounded automorphisms are contracting (see [4]), however in contrastto bounded automorphisms, contracting automorphisms do not form a group. For contractingautomorphisms with finite orbit-signalizers, we prove that conjugation is controlled by the groupof finite-state automorphisms.
Theorem 3.
Two contracting automorphisms with finite orbit-signalizers are conjugate in
Aut( T ) if and only if they are conjugate in FAut( T ) . We prove a number of results for the conjugacy problem for bounded automorphisms in Sec-tion 4, which we collect in the following theorem.
Theorem 4.
1. The (simultaneous) conjugacy problem for bounded automorphisms in
Aut( T ) is decidable.2. Two bounded automorphisms are conjugate in the group Aut( T ) if and only if they areconjugate in the group FAut( T ) .3. The (simultaneous) conjugacy problem in Pol(0) is decidable.4. Two bounded automorphisms are conjugate in the group
Pol( ∞ ) if and only if they areconjugate in the group Pol(0) . We develop two algorithms for the solution of the conjugacy problem in the group Pol(0). Thefirst one exploits the cyclic structure of bounded automorphisms. While the second exploits thenumber of active states of bounded automorphisms. This last counting argument translates toa bounded trajectory problem for nonnegative matrices which is shown to be decidable in theappendix by Rapha¨el M. Jurgens. The methods developed in this study provide a constructionfor possible conjugators whenever the associated conjugacy problems are solved.The last section presents some examples, which illustrate the solution of the conjugacy prob-lems, and describes the connection between the property of having finite orbit-signalizers andother properties of automorphisms.
The set X ∗ is considered as the set of vertices of the tree T as described in Introduction. Thelength of a word v = x x . . . x n ∈ X ∗ for x i ∈ X is denoted by | v | = n . The set X n of wordsof length n forms the n -th level of the tree T . The vertices X ∗ are ordered by the lexicographicorder on words induced by the order on the set X .We are using right actions, so the image of a vertex v ∈ X ∗ under the action of an automorphism g ∈ Aut( T ) is written as v g or ( v ) g , and hence v g · h = ( v g ) h .3he state g | v of g at v , which was defined in Introduction, is the unique automorphism of thetree T such that the equality ( vw ) g = v g ( w ) g | v holds for all words w ∈ X ∗ . Computation of statesof automorphisms is done as follows:( g · h ) | v = g | v · h | ( v ) g , g − | v = ( g | ( v ) g − ) − , g n | v = g | v · g | ( v ) g · . . . · g | ( v ) g n − for all g, h ∈ Aut( T ) and v ∈ X ∗ . Therefore, conjugation is computed by the rules (cid:0) h − gh (cid:1) | v = (cid:0) h − (cid:1) | v ( gh ) | ( v ) h − = (cid:0) h | ( v ) h − (cid:1) − g | ( v ) h − h | ( v ) h − g ; (cid:0) h − gh (cid:1) | ( v ) h = ( h | v ) − g | v h | ( v ) g , and if ( v ) g = v then (cid:0) h − gh (cid:1) | ( v ) h = ( h | v ) − g | v h | v . The multiplication of two automorphisms expressed as g = ( g | , g | , . . . , g | d ) π g , h =( h | , h | , . . . , h | d ) π h is performed by the rule g · h = ( g | h | (1) g , g | h | (2) g , . . . , g | d h | ( d ) g ) π g π h . Every permutation π ∈ Sym( X ) can be identified with the automorphism ( e, e, . . . , e ) π of the tree T acting on the vertices by the rule ( xv ) π = x π v for x ∈ X and v ∈ X ∗ .The group RAut( T ) of functionally recursive automorphisms consists of automorphisms whichcan be constructed as follows. A finite set of automorphisms g , g , . . . , g m is called functionallyrecursive if there exist words w ij over { g ± , g ± , . . . , g ± m } and permutations π i ∈ Sym( X ) suchthat g = ( w , w , . . . , w d ) π g = ( w , w , . . . , w d ) π ... g m = ( w m , w m , . . . , w md ) π m . This system has a unique solution in the group Aut( T ), here the action of each element g i on thefirst level of the tree T is given by the permutation π i , and the action of the state g i | j is uniquelydefined by the word w ij . An automorphism of the tree is called functionally recursive provided itis an element of some functionally recursive set of automorphisms.For an automorphism g ∈ Aut( T ) define the numerical sequence θ k ( g ) = |{ v ∈ X k : g | v acts non-trivially on X }| for k ≥ , which describes the activity growth of g . Looking at the asymptotic behavior of the sequence θ k ( · )we can define different classes of automorphisms of the tree T .The elements g ∈ Aut( T ), whose sequence θ k ( g ) is eventually zero, are called finitary auto-morphisms . In other words, an automorphism g is finitary if there exists k such that g | v = 1 forall v ∈ X k , and the smallest k with this property is called the depth of g . The set of all finitaryautomorphisms forms a group denoted by Pol( − g ∈ FAut( T ) the sequence θ k ( g ) can grow either exponentiallyor polynomially (see [22, Corollary 7]). The set of all finite-state automorphisms g ∈ FAut( T ),whose sequence θ k ( g ) is bounded by a polynomial of degree n , is the group Pol( n ) of polynomialautomata of degree n . In the case n = 0, when the sequence θ k ( g ) is bounded, then the automor-phism g is called bounded and the group Pol(0) is called the group of bounded automata . We getan ascending chain of subgroups Pol( n ) < Pol( n + 1) for n ≥ −
1. The union Pol( ∞ ) = ∪ n Pol( n )is called the group of polynomial automata . If we replace the condition “ g | v acts non-trivially on X ” by “ g | v is non-trivial” in the definition of the sequence θ k ( · ) then we still get the same groupsPol( n ). 4he bounded and polynomial automorphisms can be characterized by the cyclic structure oftheir automata as described in [22]. A cycle in an automaton is called trivial if it is a loop atthe state corresponding to the trivial automorphism. Then an automorphism g ∈ FAut( T ) ispolynomial if and only if any two different non-trivial cycles in the automaton A ( g ) are disjoint.Moreover, g ∈ Pol( n ), when n is the largest number of non-trivial cycles connected by a directedpath. In particular, an automorphism g ∈ FAut( T ) is bounded if and only if any two differentnon-trivial cycles in the automaton A ( g ) are disjoint and not connected by a directed path. Wesay that g is circuit if there exists a non-empty word v ∈ X ∗ such that g = g | v , i.e. g lies on acycle in the automaton A ( g ). If g is a circuit bounded automorphism then the state g | v is finitaryfor every word v , which is not read along the circuit. Let us recall the description of the conjugacy classes in the group Aut( T ). Conjugacy classes in
Aut( T ). First, recall that every conjugacy class of the symmetricgroup Sym( X ) has a unique (left-oriented) representative of the form(1 , , . . . , n )( n + 1 , n + 2 , . . . , n ) . . . ( n k − + 1 , n k − + 2 , . . . , n k ) , (1)where 1 ≤ n ≤ n − n ≤ . . . ≤ n k − n k − and n k = d = | X | . This observation can be generalizedto the group Aut( T ) (see [23, Section 4.1]). Given an automorphism a = ( a | , a | , ..., a | d ) π a inAut( T ) we can conjugate it to a unique (left-oriented) representative of its conjugacy class usingthe following basic steps.1. Conjugate the permutation π a ∈ Sym( X ) to its unique left-oriented conjugacy representative(1).2. Consider every cycle τ i = ( n i + 1 , n i + 2 , . . . , n i +1 ) in the representative (1) of π a and define h i +1 = (cid:0) a | n i +1 , e, ( a | n i +2 ) − , ( a | n i +2 a | n i +3 ) − , ..., ( a | n i +2 a | n i +3 . . . a | n i +1 − ) − (cid:1) . Conjugate a by the the automorphism h = ( h , h , . . . , h k ) to obtain h − ah = ( a , a , . . . , a k ),where a i = (cid:0) e, ..., e, a | n i +2 . . . a | n i +1 a | n i +1 (cid:1) τ i .
3. Apply the steps 1 and 2 to the automorphisms a | n i +2 . . . a | n i +1 a | n i +1 .It is direct to see that an infinite iteration of this procedure produces a well-defined automor-phism of the tree which conjugates a into a representative and that two different representativesare not conjugate in Aut( T ).Another approach is based on the fact that two permutations are conjugate if and only if theyhave the same cycle type. The orbit type of an automorphism a ∈ Aut( T ) is the labeled graph,whose vertices are the orbits of a on X ∗ , every orbit is labeled by its cardinality, and we connecttwo orbits O and O by an edge if there exist vertices v ∈ O and v ∈ O , which are adjacentin the tree T . Then two automorphisms of the tree T are conjugate if and only if their orbit typesare isomorphic as labeled graphs (see [7, Theorem 3.1]). In particular it follows, that the groupAut( T ) is ambivalent (that is, every element is conjugate with its inverse). More generally, everyautomorphism a ∈ Aut( T ) is conjugate with a ξ for every unit ξ of the ring Z m of m -adic integers,where m is the exponent of the group Sym( X ) (see [23, Section 4.3]). Conjugation lemma . We say that an element h is a conjugator for the pair ( a, b ) if h − ah = b ,and we use the notation h : a → b . For a, b ∈ Aut( T ) and the permutations π a , π b ∈ Sym( X )induced by the action of a and b on X , the set of permutational conjugators for the pair ( π a , π b )is denoted by CΠ( a, b ) = { π ∈ Sym( X ) : π − π a π = π b } (this set can be empty).The study of the conjugacy problem in the automorphism groups of the tree T is based on thefollowing standard lemma. 5 emma 1. Let a, b, h ∈ Aut( T ) .1. If h − ah = b then | Orb a ( v ) | = | Orb b ( v h ) | for every v ∈ X ∗ and ( h | v ) − a m | v h | v = b m | v h , where m = | Orb a ( v ) | .2. Let O , O , . . . , O k be the orbits of the action of a on X . If there exists π ∈ CΠ( a, b ) suchthat a |O i | | v and b |O i | | v π are conjugate in Aut( T ) for every i = 1 , , . . . , k , where v ∈ O i isan arbitrary point, then a and b are conjugate in Aut( T ) .Proof. The first statement follows from the equalities h − a m h = b m , ( v ) a m = v .Let Orb a ( v ) = { v = v, v , . . . , v m − } , where v i = ( v ) a i . Put u = v h , then Orb b ( u ) = { u = u, u , . . . , u m − } , where u i = ( u ) b i and u i = v hi . Then b | u = ( h − ah ) | u = ( h | v ) − a | v h | v b | u = ( h − ah ) | u = ( h | v ) − a | v h | v . . . . . .b | u m − = ( h − ah ) | u m − = ( h | v m − ) − a | v m − h | v Multiplying these equations, we get( h | v ) − a m | v h | v = ( h | v ) − (cid:0) a | v a | v . . . a | v m − (cid:1) h | v == b | u b | u . . . b | u m − = b m | u . In particular ( h | v i ) − a m | v i h | v i = ( h | v i ) − (cid:0) a | v i a | v i +1 . . . a | v i − (cid:1) h | v i == b | u i b | u i +1 . . . b | u i − = b m | u i ,h | v i = (cid:0) a | v . . . a | v i − a | v i − (cid:1) − h | v (cid:0) b | u . . . b | u i − b | u i − (cid:1) = ( a i | v ) − h | v b i | u . (2)If a and b are finite-state automorphisms (we need this only for the word problem) , Lemma 1suggests a branching decision procedure for the conjugacy problem in Aut( T ). We call this pro-cedure by CP and remark that it may not stop in general. The order problem in
Aut( T ). The problem of finding the order of a given element ofAut( T ) can be handled in a manner similar to the above. The next observation gives a simplecondition used in many papers to prove that an automorphism has infinite order. Lemma 2.
Let a ∈ Aut( T ) .1. Let O , O , . . . , O k be the orbits of the action of a on X . Define a i = a m i | x i for every i = 1 , , . . . , k , where m i = |O i | and x i ∈ O i is an arbitrary point. The automorphism a hasfinite order if and only if all the states a i have finite order. Moreover, in this case, the orderof a is equal to | a | = lcm( m | a | , m | a | , . . . , m k | a k | ) .
2. Suppose a i = a for some choice of x i ∈ O i . If m i > then a has infinite order. If m i = 1 then a has finite order if and only if a j has finite order for all j = i , in which case we canremove the term m i | a i | from the right hand side of the above equality. a is a finite-state automorphism, then the word problem a i = a can be effectively solvedand Lemma 2 suggests a branching procedure to find the order of a. We call this procedure byOP and remark that it may not stop in general. Such a procedure is implemented in the programpackages [2, 16]. Orbit-signalizer.
Lemmas 1 and 2 lead us to define the orbit-signalizer of an automorphism a ∈ Aut( T ) as the set OS( a ) = { a m | v | v ∈ X ∗ , m = | Orb a ( v ) |} , which contains all automorphisms that may appear in the procedures OP and CP. Notice that if m = | Orb a ( v ) | , l = | Orb a ( vx ) | , and k = | Orb a m | v ( x ) | then l = mk and a l | vx = ( a m | v ) k | x . (3)This observation implies the recursive procedure to find the set OS( a ). We start from the setOS ( a ) = { a } and compute consecutivelyOS n +1 ( a ) = { b m | x | b ∈ OS n ( a ) , x ∈ X, m = | Orb b ( x ) |} . Then OS( a ) = ∪ n ≥ OS n ( a ). It follows from construction that if OS n +1 ( a ) does not containnew elements, i.e., OS n +1 ( a ) ⊂ ∪ ni =0 OS i ( n ), then we can stop and OS( a ) = ∪ ni =0 OS i ( n ). Inparticular, if the set OS( a ) is finite, then this procedure stops in finite time and we can find OS( a )algorithmically. For automorphisms with finite orbit-signalizers one can model the procedures OPand CP by finite graphs. Order graph.
Consider an automorphism a ∈ Aut( T ) which has finite orbit-signalizer. Weconstruct a finite graph Φ( a ) with the set of vertices OS( a ), called the order graph of a , whichmodels the branching procedure OP. The edges of this graph are constructed as follows. For every b ∈ OS( a ) consider all orbits O , O , . . . , O k of the action of b on X and let x i ∈ O i be the leastelement in O i . It is easy to see that b m i | x i ∈ OS( a ) for m i = |O i | , and we introduce the labelededge b m i −→ b m i | x i in the graph Φ( a ) for every i = 1 , . . . , k . Then Lemma 2 can be reformulated asfollows. Proposition 5.
Let a ∈ Aut( T ) have finite orbit-signalizer. Then a has finite order if and onlyif all edges in the directed cycles in the order graph Φ( a ) are labeled by . Moreover, in this case we can compute the order of a using the graph Φ( a ). Remove all theedges of every directed cycle in Φ( a ). Then the only dead vertex of Φ( a ), i.e. the vertex withoutoutgoing edges, is the trivial automorphism, which has order 1. Then inductively, for b ∈ OS( a )consider all outgoing edges from b , and let m , m , . . . , m k be the edge labels and b , b , . . . , b k bethe corresponding end vertices, whose order we already know. Then by Lemma 2 the order of b is equal to the least common multiple of m i | b i | . We illustrate the construction of the order graphand the solution of the order problem in Example 2 of Section 5. Conjugator graph.
Consider automorphisms a, b ∈ Aut( T ) both of which have finite orbit-signalizers. We construct a finite graph Ψ( a, b ), called the conjugator graph of the pair ( a, b ),modeled after the branching procedure CP of Lemma 1. The vertices of the graph Ψ( a, b ) are thetriples ( c, d, π ) for c ∈ OS( a ), d ∈ OS( b ), and π ∈ CΠ( c, d ) whenever this last set is nonempty.The edges are constructed as follows.Let O i ( c ) for 1 ≤ i ≤ k be the orbits of c in its action on X and let x i ( c ) denote the leastelement in each O i ( c ). We will simplify the notation by writing instead O i and x i with theunderstanding that these refer to c ∈ OS( a ) under consideration.For any vertex ( c, d, π ), if one of the sets CΠ( c m | x i , d m | x iπ ) with m = |O i | is empty, then thetriple ( c, d, π ) is a dead vertex. Otherwise we introduce in the graph the edge( c, d, π ) x i −→ ( c m | x i , d m | x iπ , τ ) with m = |O i | for every τ ∈ CΠ( c m | x i , d m | x iπ ) and i = 1 , . . . , k . Notice that c m | x i ∈ OS( a ), d m | x πi ∈ OS( b ), andhence the triple ( c m | x i , d m | x iπ , τ ) is indeed a vertex of the graph.7e simplify the graph obtained above using the following reductions. Remove the vertex( c, d, π ) which does not have an outgoing edge labeled by x i for some i . Also, remove all edgesleading to these deleted vertices. We repeat the reductions as long as possible to reach the graphΨ( a, b ).If the graph Ψ( a, b ) is empty, then the automorphisms a and b are not conjugate. Otherwisethey are conjugate and every conjugator h : a → b can be constructed level by level as follows.Choose any vertex ( a, b, π ) in Ψ( a, b ) and define the action of h on the first level by x h = x π for x ∈ X . There is an outgoing edge from ( a, b, π ) labeled by x i = x i ( a ), as explained previously.Choose an edge for every x i and let ( c i , d i , π i ) be the corresponding end vertex. We define theaction of the state h | x i by the rule ( x ) h | xi = x π i for x ∈ X . All the other states of h on the verticesof the first level are uniquely defined by Equation (2) at the end of the proof of Lemma 1, andthus we get the action of h on the second level. Similarly, we proceed further with the vertices( c i , d i , π i ) and construct the action of h on the third level, and so on. Notice that even if the samevertex ( c, d, π ) appears at different stages of the definition of h we still have a freedom to choosedifferent outgoing edges from ( c, d, π ) in each stage of the construction. Basic conjugators.
Let us construct certain conjugators, called basic conjugators for the pair( a, b ), by making as few choices as possible, in the sense that if we arrive at a triple ( c, d, π ) atsome stage of the construction then we choose the same permutation π ∈ CΠ( c, d ) whenever thepair ( c, d ) reappears further down. That is, for every two vertices ( c, d, π ) and ( c, d, π ) obtainedunder construction we insist to have π = π . More precisely every basic conjugator can be definedusing special subgraphs of the conjugator graph Ψ( a, b ). Consider the subgraph Γ of Ψ( a, b ), whichsatisfies the following properties:1. The subgraph Γ contains some vertex ( a, b, π ) for π ∈ CΠ( a, b ).2. For every vertex ( c, d, π ) of Γ and every letter x i , there exists precisely one outgoing edgefrom ( c, d, π ) labeled by x i . In particular, the graph is deterministic, and the number ofoutgoing edges at the vertex ( c, d, π ) of the graph Γ is equal to the number of orbits of c on X .3. For every c ∈ OS( a ) and d ∈ OS( b ) there is at most one vertex of the form ( c, d, ∗ ) in thegraph Γ. In other words, if ( c, d, π ) and ( c, d, π ) are vertices of Γ then π = π .If the graph Ψ( a, b ) is nonempty, there always exist subgraphs of Ψ( a, b ), which satisfy theproperties 1-3. For every such a subgraph Γ we construct the basic conjugator h = h (Γ) asfollows. We construct a functionally recursive system involving every conjugator h ( c,d ) : c → d ,where ( c, d, π ) is a vertex of Γ and thus in particular, we construct h = h ( a,b ) . First, we definethe action of the conjugator h ( c,d ) on the first level by the rule x h ( c,d ) = x π for x ∈ X , where thepermutation π ∈ CΠ( c, d ) is uniquely defined such that the triple ( c, d, π ) is a vertex of Γ. Forevery edge ( c, d, π ) x −→ ( c ′ , d ′ , π ′ ) we define the states of the conjugator h ( c,d ) on the letters fromthe orbit O = { x, ( x ) c, ( x ) c , . . . , ( x ) c m − } of x under c recursively by the rule h ( c,d ) | x = h ( c ′ ,d ′ ) and h ( c,d ) | ( x ) c i = (cid:0) c i | x (cid:1) − · h ( c ′ ,d ′ ) · d i | x π , for i = 1 , . . . , m −
1. These rules completely define the automorphisms h ( c,d ) . By Lemma 1 everyconstructed automorphism h ( c,d ) is indeed a conjugator for the pair ( c, d ). Since the graph Γ isfinite, and the automorphisms a, b are finite-state, we get a functionally recursive system whichuniquely defines the basic conjugator h = h ( a,b ) given by the subgraph Γ.We have proved the following theorem. Theorem 6.
Let a, b ∈ FAut( T ) have finite orbit-signalizers, and let Ψ( a, b ) be the correspondingconjugator graph. Then a and b are conjugate in Aut( T ) if and only if they are conjugate in RAut( T ) if and only if the graph Ψ( a, b ) is nonempty.
8n particular, the conjugacy problem for finite-state automorphisms with finite orbit-signalizersis decidable in the groups Aut( T ) and RAut( T ). We present examples of the construction of theconjugator graph and basic conjugators in Example 3 of Section 5.The same method works for the simultaneous conjugacy problem, which given automorphisms a , a , . . . , a k and b , b , . . . , b k asks for the existence of an automorphism h such that h − a i h = b i for all i . We again consider the permutations π such that π − π a i π = π b i for all i , take an orbit O i of a i on X , let x i ∈ O i be the least element and m i = |O i | . Then the problem reduces to thesimultaneous conjugacy problem for a m | x , a m | x , . . . , a m k | x k and b m | x π , b m | x π , . . . , b m k | x πk . Ifautomorphisms a i and b i are finite-state and have finite orbit-signalizers, then we can similarlyconstruct the associated conjugator graph so that Theorem 6 holds. Conjugation of contracting automorphisms.
A self-similar subgroup
G <
Aut( T ) iscalled contracting if there exists a finite set N ⊂ G with the property that for every g ∈ G thereexists n ∈ N such that g | v ∈ N for all words v of length ≥ n . The smallest set N with thisproperty is called the nucleus of the group. An automorphisms f ∈ Aut( T ) is called contracting if the self-similar group generated by all states of f is contracting. It follows from the definitionthat contracting automorphisms are finite-state. Theorem 7.
Two contracting automorphisms a, b ∈ Aut( T ) with finite orbit-signalizers are con-jugate in the group Aut( T ) if and only if they are conjugate in the group FAut( T ) .Proof. We will prove that all the basic conjugators for the pair ( a, b ) are finite-state. We need afew lemmata, which are interesting in themselves.
Lemma 3.
Let G be a contracting self-similar group, and let H be a finite subset of G . Then theset of all possible elements of the form ( . . . (( h | x · h ) | x · h ) | x · . . . · h n ) | x n , ( h n · . . . · ( h · ( h · h | x ) | x ) | x . . . ) | x n , where h i ∈ H and x i ∈ X , is finite.Proof. The statement is a reformulation of Proposition 2.11.5 in [17]. We sketch the proof forcompleteness.We can assume that the set H is self-similar, i.e. h | v ∈ H for all h ∈ H and v ∈ X ∗ (all theelements are finite-state), and contains the nucleus N of the group G . There exists a number k such that H | v ⊂ N ⊂ H for all words v of length ≥ k . Then H n | v ⊂ H n for all v ∈ X k and n ≥
1. It is sufficient to prove that there are finitely many elements of the form( . . . (( h | v · h ) | v · h ) | v · . . . · h n ) | v n for h i ∈ H k and v i ∈ X k . Then h | v ∈ H k and ( h | v · h ) | v ∈ H k | v ⊂ H k . Inductively we getthat all the above elements belong to H k .The next lemma is similar to Corollary 2.11.7 in [17], which states that different self-similarcontracting actions with the same virtual endomorphism are conjugate via a finite-state automor-phism. Lemma 4.
Let G be a contracting self-similar group and H ⊂ G be a finite subset. Suppose thatthe automorphism g ∈ Aut( T ) satisfies the condition that for every x ∈ X there exist h, h ′ ∈ H such that g | x = hgh ′ . Then g is finite-state.Proof. For an arbitrary word x x x . . . x n ∈ X ∗ we have g | x x x ...x n = ( h gh ′ ) | x x ...x n = (cid:0) ((( h | x · h ) | x · h ) | x · . . . · h n ) | x n · h n +1 (cid:1) · g ·· (cid:0) h ′ n +1 · ( h ′ n · . . . · ( h ′ · ( h ′ · h ′ | y ) | y ) | y . . . ) | y n (cid:1) , (4)where h i , h ′ i are some elements in H , and y i ∈ X . By Lemma 3 the above products assume a finitenumber of values, and hence g is finite-state. 9otice that in the previous lemma we only need that the groups generated by all states of h i andseparately by all states of h ′ i be contracting, while together they may generate a non-contractinggroup. Lemma 5.
Let F ⊂ Aut( T ) be a finite collection of automorphisms. Suppose that there exist twocontracting self-similar groups G , G and finite subsets H ⊂ G , H ⊂ G with the conditionthat for every g ∈ F and every letter x ∈ X there exist h ∈ H , h ∈ H and g ′ ∈ F such that g | x = h g ′ h . Then all the automorphisms in F are finite-state.Proof. The proof is the same as in Lemma 4. The only difference is that on the right hand side ofEquation (4) instead of g may appear any element of the finite set F .To finish the proof of Theorem 7 it is sufficient to notice that all the basic conjugators for thepair ( a, b ) satisfy Lemma 5, and hence all of them are finite-state.Example 4 of Section 5 shows that we cannot drop the assumption about orbit-signalizers inthe theorem.The finiteness of orbit-signalizers can be used to prove that certain automorphisms are notconjugate in the group FAut( T ). Proposition 8.
Let f, g ∈ Aut( T ) be conjugate in FAut( T ) . Then f has finite orbit-signalizer ifand only if g does.Proof. Let h − f h = g for a finite-state automorphism h , and suppose f has finite orbit-signalizer.Then m = | Orb f ( v ) | = | Orb g ( v h ) | for every v ∈ X ∗ and g m | v h = ( h | v ) − f m | v h | v ∈ ( h | v ) − OS( f ) h | v . It follows that OS( g ) ⊂ Q( h ) − OS( f )Q( h ), where Q( h ) is the set of states of h , and hence the setOS( g ) is finite. How to check that a given finite-state automorphism has finite orbit-signalizer is yet anotheralgorithmic problem. Let us show that some classes of automorphisms have finite orbit-signalizers.Every finite-state automorphism a of finite order has finite orbit-signalizer. Here the set OS( a ) isbounded by the number of all states of all powers of a , which is finite. In particular, if a finite-stateautomorphism has infinite orbit-signalizer, then it has infinite order. Proposition 9.
Every bounded automorphism has finite orbit-signalizer.Proof.
Let a be a bounded automorphism, and choose a constant C so that the number of non-trivial states a | v for v ∈ X n is not greater than C for every n ≥
0. Then for every vertex v ∈ X ∗ the state a m | v with m = | Orb a ( v ) | is a product of no more than C states of a , which is a finiteset.In particular, the orbit-signalizer of a bounded automorphism can be computed algorithmically,the procedure CP solves the conjugacy problem for bounded automorphism in Aut( T ), and theprocedure OP finds the order of a bounded automorphism. Corollary 10. (1). The order problem for bounded automorphisms is decidable.(2). The (simultaneous) conjugacy problem for bounded automorphisms in
Aut( T ) is decidable. Theorem 11.
Two bounded automorphisms are conjugate in the group
Aut( T ) if and only if theyare conjugate in the group FAut( T ) .Proof. The bounded automorphisms are contracting by [4] and have finite orbit-signalizers byProposition 9, hence we can apply Theorem 7. 10 orollary 12.
Let a be a bounded automorphism. Then a and a − are conjugate in FAut( T ) . Corollary 13.
Let a bounded automorphism f and a contracting automorphism g be conjugate in Aut( T ) . Then f and g are conjugate in FAut( T ) if and only if g has finite orbit-signalizer. Consider the conjugacy problem for bounded automorphisms in the group Pol( −
1) of finitaryautomorphisms. One of the approaches is to restrict the depth of a possible finitary conjugator.Let a, b be two bounded automorphisms. If a and b are conjugate in Pol( −
1) and h : a → b is a finitary conjugator then every state h | x for x ∈ X is a finitary conjugator for the pair( a m | x , b m | x h ) with m = | Orb a ( x ) | , and h | x has less depth than h . However, it is possible thatevery pair ( a m | x , b m | x h ) for x ∈ X with m = | Orb a ( x ) | is conjugate via a finitary conjugator ofdepth ≤ d , while ( a, b ) is not conjugate via a finitary conjugator of depth ≤ d + 1. Hence westill do not get a bound on the depth of h even if we know the bound on the depth of a finitaryconjugator for every pair ( a m | x , b m | x h ). The problem is that we need to find a finitary conjugator h | x for the pair ( a m | x , b m | x h ) so that all elements h | ( x ) a i = ( a i | x ) − · h | x · b i | x h for i = 0 , . . . , m − Configurations of orbits.
Every configuration will be of the form C = { ( α, β ) , DP C } , where( α, β ) is a pair of automorphism called the main pair of C , and DP C is the set of pairs of auto-morphism called dependent pairs . Configurations for the pair ( a, b ) are constructed recursively asfollows. At zero level we have just one configuration Λ = {C} , where C = { ( a, b ) , DP C = { ( e, e ) }} .Further we construct recursively the set Λ n +1 from the set Λ n . Take a configuration C ∈ Λ n , C = { ( α, β ) , DP C } . Consider every orbit O of the action of α on X , let x be the least element in O and m = |O| . For every π ∈ CΠ( α, β ) define new configuration C ′ = C ′ x,π = { ( α m | x , β m | x π ) , DP C ′ } ,where DP C ′ = { (( α i c ) | x , ( β i d ) | x π ) | ( c, d ) ∈ DP C and i = 0 , . . . , m − } . (5)The set Λ n +1 consists of all configurations C ′ constructed in this way. Then Λ = ∪ n ≥ Λ n is theset of configurations for ( a, b ). It follows from construction that if Λ n +1 does not contain newconfigurations, i.e., Λ n +1 ⊂ ∪ ni =0 Λ i , then we can stop and Λ = ∪ ni =0 Λ i . In particular, if the set Λis finite then it can be computed algorithmically. Lemma 6.
The set of configurations for a pair of bounded automorphisms is finite and can becomputed algorithmically.Proof.
Let C = { ( α, β ) , DP C } be a configuration for ( a, b ) and denote A C = { c | ( c, d ) ∈ DP C } . Weprove by induction that there exists a word v ∈ X ∗ such that α = a l | v and A C = { a j | v | j = 0 , , . . . , l − } , (6)where l = | Orb a ( v ) | . The basis of induction is the initial configuration C = { ( a, b ) , DP C = { ( e, e ) }} that satisfies this condition for the empty word v = ∅ . Suppose inductively that a configuration C satisfies the condition for a word v and proceed with the construction of C ′ . Let x be the leastelement in an orbit of the action of α on X and put m = | Orb α ( x ) | . Then | Orb a ( vx ) | = lm andwe get α m | x = ( a l | v ) m | x = a lm | vx and A C ′ = { ( α i c ) | x | c ∈ A C and i = 0 , , . . . , m − } == { (( a l | v ) i a j | v ) | x | i = 0 , , . . . , m − j = 0 , , . . . , l − } == { a k | vx | k = 0 , , . . . , lm − } (here k = li + j ) . Hence C ′ satisfies condition (6) for the word vx .11n particular, α ∈ OS( a ) and can assume only a finite number of values. The number ofdifferent sets { a j | v | ≤ j < | Orb a ( v ) |} , v ∈ X ∗ , for a bounded automorphism a , is finite (theproof is the same as of Proposition 9). Hence there are finitely many possibilities for the set A C .The same observation holds for β and the set B C = { d | ( c, d ) ∈ DP C } . It follows that the numberof configurations for a pair of bounded automorphisms is finite. Remark 1.
Let h − ah = b . Let O be an orbit of the action of a on X ∗ and v ∈ O be the leastelement in the orbit and m = |O| . Then C = { ( a m | v , b m | v h ) , DP C } , where DP C = { ( a i | v , b i | v h ) , for i = 0 , . . . , m − } , is a configuration for ( a, b ). Configurations satisfied by a finitary automorphism.
We say that a finitary automor-phism h satisfies a configuration C if h is a conjugator for the main pair of C and all elements c − hd for ( c, d ) ∈ DP C are finitary. We say that a configuration C has depth ≤ d if C is satisfiedby a finitary automorphism h of depth ≤ d and all elements c − hd for ( c, d ) ∈ DP C have depth ≤ d . In particular, the automorphisms a, b are conjugate in Pol( −
1) if and only if the configura-tion C = { ( a, b ) , DP C = { (1 , }} is satisfied by a finitary automorphism. Let us show that it isdecidable whether a given configuration C can be satisfied by a finitary automorphism. Lemma 7.
Let C = { ( α, β ) , DP C } be a configuration. Consider all orbits O , O , . . . , O k of theaction of α on X and let x j ∈ O j be the least element in O j . The configuration C has depth ≤ d ifand only if there exists π ∈ CΠ( α, β ) such that every configuration C ′ x j ,π , j = 1 , . . . , k , has depth ≤ d − .Proof. Suppose h − αh = β and all automorphisms c − hd , ( c, d ) ∈ DP C , are finitary of depth ≤ d .For the permutation π in CΠ( α, β ) we take π h . Then( h | x j ) − α m j | x j h | x j = β m j | x πj and for every ( c, d ) ∈ DP C and i = 0 , , . . . , m − α i c ) | x j ) − · h | x j · ( β i d ) | x jπ = ( c | ( x j ) α i ) − h | ( x j ) α i d | ( x j ) α i h = ( c − hd ) | y , (7)where y = ( x j ) α i c . All automorphisms ( c − hd ) | y are finitary of depth ≤ d −
1. Hence everyconfiguration C ′ x j ,π has depth ≤ d − π ∈ CΠ( α, β ) such that every C ′ x j ,π is satisfied by a finitaryautomorphism h j . Define automorphism h by the rules π h = π and h | x j = h j and h | ( x j ) α i = ( α i | x j ) − h | j β i | x πj . Note that ( α i | x j , β i | x πj ) = (( α i c ) | x j , ( β i d ) | x πj ) for ( c, d ) = ( e, e ) ∈ DP C and hence every pair( α i | x j , β i | x πj ) belongs to DP C ′ . Therefore the automorphism h is finitary. Also h is a conjugatorfor ( α, β ) by construction and satisfies the configuration C by Equation (7). Corollary 14. If a and b are conjugate in Pol( − then there exists a finitary conjugator of depth ≤ | Λ | . In particular, the conjugacy problem for bounded automorphisms in the group Pol( − isdecidable. Instead of just running through all finitary automorphisms with a given bound on the depth,the algorithm can be realized as follows. Construct the set Λ of all configuration for a given pair( a, b ). We will consecutively label configurations by numbers which correspond to their depths.First, we identify configurations of depth 0, which are precisely configurations C = { ( α, β ) , DP C } such that α = β and c = d for all ( c, d ) ∈ DP C . Then iteratively we label a configuration C = { ( α, β ) , DP C } by number d if there exists π ∈ CΠ( α, β ) such that each C ′ x j ,π is already labeledby a number ≤ d −
1. After this process finishes, the configurations labeled by d can be satisfiedby a finitary automorphisms of depth ≤ d , while the unlabeled configurations cannot be satisfiedby finitary automorphisms. 12 .2 The conjugacy problem in the group of bounded automata In this subsection we prove that the conjugacy problem in the group of bounded automata isdecidable. We will show two approaches.
First approach: by using cyclic structure of bounded automata.
Let a and b bebounded automorphisms. We apply the following algorithm to check whether a and b are conjugatein Pol(0). The algorithm will consecutively determine the pairs from OS( a ) × OS( b ) that areconjugate in Pol(0). Further we prove that the algorithm is correct. Step 1 . Take ( c, d ) ∈ OS( a ) × OS( b ) and compute the set Λ( c, d ) of all configurations for ( c, d ).For every configuration C and every ( γ , δ ) , ( γ , δ ) ∈ DP C check whether ( γ − γ ) − c ( γ − γ ) and( δ − δ ) − d ( δ − δ ) are conjugate in Pol( − c, d are conjugate in Pol(0). Applythis step to every pair ( c, d ) ∈ OS( a ) × OS( b ). Note that since C = { ( c, d ) , DP C = { ( e, e ) }} is aconfiguration for ( c, d ) we also detect every pair ( c, d ) conjugated in Pol( −
1) (just take γ = γ = δ = δ = e ). Step 2 . Take ( c, d ) ∈ OS( a ) × OS( b ). Consider all words u ∈ X ∗ such that u c = u and | u | ≤ | OS( c ) | · | OS( d ) | . Consider all circuit automorphisms h such that h | u = h and every finitarystate of h has depth ≤ | Λ( c, d ) | . Note that there are only finitely many bounded automorphismswith these properties. For every such h check whether h − ch = d . We apply this step to everypair ( c, d ) ∈ OS( a ) × OS( b ) not detected in Step 1. Step 3 . For every pair ( c, d ) ∈ OS( a ) × OS( b ) for which we still do not know whether it isconjugate in Pol(0) proceed as follows. Consider orbits O , . . . , O k of the action of c on X , let x i ∈ O i be the least element in the orbit and m i = |O i | . Check whether there exists π ∈ CΠ( c, d )such that for every pair ( c m i | x i , d m i | x πi ) (it belongs to OS( a ) × OS( b )) we already know that itis conjugate in Pol(0). If yes then c, d are conjugate in Pol(0). We repeat this step as long aspossible until no new pairs are detected. The other pairs from OS( a ) × OS( b ) are not conjugatein Pol(0). Proof of correctness of the algorithm.
First, every pair detected in one of the steps is indeed con-jugate in Pol(0). We need to prove the converse that if a, b are conjugate in Pol(0) then the pair( a, b ) will be detected. Let h − ah = b for a bounded automorphism h . There exists a level l suchthat for every v ∈ X l the state h | v is either circuit or finitary. Consider the orbits of the action of a on X l . Let v be the least element in an orbit O and m = |O| . Then h | v is a conjugator for thepair ( c, d ) = ( a m | v , b m | v h ) ∈ OS( a ) × OS( b ).If h | v is finitary then the pair ( c, d ) is detected in Step 1.If h | v is circuit then we take a circuit conjugator g for ( c, d ) having a circuit of the shortestlength. Let u be the word which is read along the circuit, so here g | u = g . Now consider two cases.Case 1: u c = u . Then g | u c = ( c | u ) − g | u d | u g is finitary. Since g | u = g we get ( g | u ) − c ( g | u ) = d and ( g | u c ) − (cid:0) ( c | u ) − cc | u (cid:1) g | u c = ( d | u g ) − dd | u g . Hence ( c | u ) − cc | u and ( d | u g ) − dd | u g are conjugate in Pol( − w be the least element in theorbit O ′ = Orb c ( u ) and u = ( w ) c i . Let C be the configuration for ( c, d ) associated to the orbit O ′ and the conjugator g (see Remark 1). Put γ = c i | w , γ = c i +1 | w , δ = d i | w g , δ = d i +1 | w g andnote that ( γ , δ ) , ( γ , δ ) ∈ DP C . Then c | u = γ − γ and d | u g = δ − δ . Therefore the pair ( c, d )is detected in Step 1.Case 2: u c = u . In this case the states of g along the circuit do not have dependencies, and wehave a freedom to change these states without changing other states of the same level. Supposethere are two states g | v and g | v along the circuit (let | v | < | v | so v is a prefix of v ), whichsolve the same conjugacy problem ( c | v , d | v g ) = ( c | v , d | v g ). Define the automorphism f by therules f | v = g | v , f | v = g | v for v ∈ X | v | , v = v , and the action of f on X | v | is the same as theaction of g . Then f is a circuit bounded conjugator for the pair ( c, d ) and it has smaller circuitlength; we arrive at contradiction. Hence, the states of g along the circuit solve different conjugacyproblems and therefore | u | ≤ | OS( c ) | · | OS( d ) | . Now consider finitary states of g . We can assumethat g is chosen is such a way that the value of max(depth of f ), where f ranges over all finitarystates of g , is the smallest possible over all conjugators for ( c, d ) with g | u = g . Let g | w be a finitary13tate. Then every state of g along the orbit of w under c is finitary. Hence the configuration C for ( c, d ) that corresponds to the orbit of w is satisfied by a finitary automorphism. However thedepth of C is not greater than | Λ( c, d ) | by Lemma 7. Hence the depth of every state of g along theorbit of w is not greater than | Λ( c, d ) | . Therefore g satisfies the conditions in Step 2 and the pair( c, d ) is detected in this step.We have proved that every pair from the action of a on X l is detected in Steps 1 ,
2. The otherpairs coming from the orbits of the action of a on smaller levels, in particular the pair ( a, b ), aredetected in Step 3. Second approach: by calculation of active states.
Let a, b be two bounded automor-phisms. We check whether a, b are conjugate in Aut( T ) and if not then they are not conjugate inPol(0). So further we assume that a, b are conjugate in Aut( T ). Every conjugator for the pair ( a, b )can be constructed level by level as described in Section 3 on page 10, by choosing the conjugatingpermutation for every orbit of a . The number of orbits may grow when we pass from one level tothe next, and consequently the number of choices grows. However the number of configurationsof orbits is finite, and it is easy to see (and also follows from the previous method) that if a and b are conjugate in Pol(0) then there exists a bounded conjugator h such that for all orbits of thesame level and of the same configuration the corresponding states of h are the same. Hence it issufficient to choose a conjugating permutation only for configurations. We will show how to countthe number of active states depending on our choice.Suppose we have constructed a conjugator h up to the n -th level. Consider an orbit O of theaction of a on X n and let C = { ( α, β ); DP C } be its configuration (see Remark 1) and v be theleast element in O . The set DP C remembers only the pairs ( c, d ) which appear in the formula h | ( v ) a i = c − h | v d , here c = a i | v and d = b i | v h for i = 0 , . . . , |O| −
1; we will call h | ( v ) a i a stateof type ( c, d ). However the number of states of type ( c, d ) is lost in this way. To preserve thisinformation we introduce the nonnegative integer column-vector u C of dimension | DP C | , where u C ( c, d ) for ( c, d ) ∈ DP C is equal to the number of i such that c = a i | v and d = b i | v h . When wepass to the next level, we choose some permutation π ∈ CΠ( α, β ) and define x h | v = x π for x ∈ X .Then we check which states of h on the vertices from the orbit O are active and which are not:the state h | ( v ) a i of type ( c, d ) is active if the permutation π c − ππ d is non-trivial. We store thisinformation in the row-vector θ C ,π of dimension | DP C | by making θ C ,π ( c, d ) = 1 if π c − ππ d = e and θ C ,π ( c, d ) = 0 otherwise. Hence, when we choose the permutation π then the number of activestates of h along the orbit O is equal to θ C ,π · u C = P θ C ,π ( c, d ) u C ( c, d ).Let Λ be the set of all configurations for the pair ( a, b ). Let Π = × C∈ Λ CΠ C , where CΠ C =CΠ( α, β ) and ( α, β ) is the main pair of C . We view Π as the set of choices, so that when wechoose π ∈ Π we have chosen a conjugating permutation for every configuration. The sets Λand Π are finite. For π = ( π C ) C∈ Λ define θ π := ( θ C ,π C ) C∈ Λ and construct the square nonnegativeinteger matrix A π of dimension P C∈ Λ | DP C | , where the rows and columns of A π are indexed bypairs [ C , ( c, d )] with ( c, d ) ∈ DP C . The entry of A π in the intersection of [ C , ( c , d )]-row and[ C , ( c , d )]-column is calculated as follows. Recall the construction of configurations C ′ inducedby C and π C , and given in Equation (5). Let ( α, β ) be the main pair of C , consider orbits O , . . . , O k of the action of α on X , let x j be the least element in O j and m j = |O j | . Let C ′ j = C ′ x j ,π C be the induced configurations. Define the [ C , ( c , d )] × [ C , ( c , d )]-entry of A π as X j (cid:12)(cid:12)(cid:12)(cid:12)(cid:26) ≤ i < m j | C ′ j = C and (( α i c ) | x j , ( β i d ) | x π C j ) = ( c , d ) (cid:27)(cid:12)(cid:12)(cid:12)(cid:12) . In other words, the [ C , ( c , d )] × [ C , ( c , d )]-entry of A π is equal to the number of pairs of type( c , d ) and of configuration C induced by one pair ( c , d ) from configuration C . Now if wehave a column-vector u = ( u C ) C∈ Λ , where u C ( c, d ) is the number of states of type ( c, d ) and ofconfiguration C that we have at certain level, and we choose π ∈ Π, then the number of pairs ofeach configuration on the next level is given by the vector A π u . Put M = { A π : π ∈ Π } .Now consider all orbits of a on X n , take their configurations with respect to h defined up tothe n -th level, and define the column-vector u = ( u C ) C∈ Λ as above. To define the action of h
14n the ( n + 1)-st level we choose π = ( π C ) C∈ Λ ∈ Π, and for every orbit with configuration C wedefine the action of the states of h along this orbit using permutation π C as usual (we choose apermutation for every configuration even if not all configurations appear on the n -th level). Inthis way the conjugator h is defined up to the ( n + 1)-st level. Then the number of active statesof h on the n -th level is equal to θ π · u . The vector v = ( v C ) C∈ Λ , where v C ( c, d ) is equal to thenumber of all states of type ( c, d ) over all orbits of a on X n +1 with configuration C , is equal to v = A π u .The process starts at the zero level, where we have the vector u = ( u C ) such that u C ( e, e ) = 1for the configuration C = { ( a, b ); DP C = { ( e, e ) }} , which corresponds to the pair ( a, b ), and u C ′ ( c, d ) = 0 for all other pairs and configurations. Then we make choices π , π , . . . , π n , . . . fromΠ and construct the conjugator h . It follows from the above discussion that the activity of h canbe calculated by the following rules: θ n ( h ) = θ π n u n and u n +1 = A π n u n for all n ≥
0. If there is a choice such that the sequence θ n ( h ) is bounded, then there will bean eventually periodic choice, and hence the constructed conjugator h will be finite-state andbounded.Hence the automorphisms a and b are conjugate in the group Pol(0) if and only if there existsa sequence A n ∈ M such that the corresponding sequence θ A n u n is bounded. The last problem issolvable and can be deduced from the result presented in Appendix.In this method we don’t need to solve the auxiliary conjugacy problems in Pol( −
1) as in theprevious method, but the problem reduces to certain matrix problem which should also be solved,while the previous method was direct. We demonstrate both approaches in Examples 5 and 6 ofSection 5.We note that both approaches also solve the respective simultaneous conjugacy problems. Wehave proved the following theorem.
Theorem 15.
The (simultaneous) conjugacy problem in the group of bounded automata is decid-able.
The above methods not only solve the studied conjugacy problem but also provide a construc-tion for a possible conjugator.Similarly, one can solve the conjugacy problem for bounded automorphisms in every groupPol( n ). However, we have a stronger statement. Proposition 16.
Two bounded automorphisms are conjugate in the group
Pol( ∞ ) if and only ifthey are conjugate in the group Pol(0) .Proof.
Let a and b be two bounded automorphisms, which are conjugate in Pol( n ) for n ≥
1. Weproceed as in the first method above. Again the problem reduces to the case when a conjugator h ∈ Pol( n ) lies on a circuit. Let u be the word which is read along the circuit so that h | u = h .We consider the two cases as in the proof of correctness of the first approach.If u a = v = u then the state h | v should be in Pol( n − h = h | u = a | u h | v ( b | u h ) − ∈ Pol( n − a and b are conjugate in Pol( n − w a = w forsome word w of the form uu . . . u .Suppose w a = w for every word w of the form uu . . . u . Then h − a | w h = b w h . If a | w = e (andhence b | w h = e ) then define the automorphism g by the rules g | w = e , g | v = h | v for all v ∈ X | w | , v = w , and the action of g on X | w | is the same as that of h . Then g belongs to Pol( n −
1) and itis a conjugator for ( a, b ).If a | w = e for every word w = uu . . . u , then some state a | w is a circuit automorphism and a | w | v = a | w for some word v of the form uu . . . u . Without loss of generality we can suppose that a | u = a and b | u h = b . Then the states a | v and b | v h are finitary for all v ∈ X | u | , v = u . Considerevery orbit O of the action of a on X | u | \ u , let v ∈ O be the least element in O and m = |O| . Thenthe finitary automorphisms a m | v and b m | v h are conjugate in Aut( T ), and hence they are conjugatein Pol( − g by the rules: the action of g on X | u | is the same as that15f h , g | u = g , g | v is a finitary conjugator for the pair ( a m | v , b m | v h ), and g | ( v ) a i = (cid:0) a i | v (cid:1) − g | v b i | v h (also finitary) for every i = 1 , . . . , m − O . Then g is a bounded conjugator forthe pair ( a, b ).Inductively we get that a and b are conjugate in Pol(0). All examples will be over the binary alphabet X = { , } .We will frequently use the automorphism a ∈ Aut( T ) given by the recursion a = ( e, a ) σ ,where σ = (0 , ∈ Sym( X ) is a transposition, which is called the (binary) adding machine .The automorphism a has infinite order, and acts transitively on each level X n of the tree T . Inparticular, every automorphism which acts transitively on X n for all n , is conjugate with a in thegroup Aut( T ).In the next example we investigate the interplay between such properties as being finite-state,contracting, bounded, polynomial, having or not a finite orbit-signalizer. Example 1.
The adding machine a is a bounded automorphism, hence it is contracting and hasfinite orbit-signalizer, here OS( a ) = { a } .The automorphism b given by the recursion b = ( a, b ) is finite-state, Q( b ) = { e, a, b } , b belongsto Pol(1) \ Pol(0), and OS( b ) = { a, b } . However b is not contracting, because all elements b n =( a n , b n ) for n ≥ b = ( a, b ) σ , b = ( a, b ) belong to Pol(1) \ Pol(0), but they have infiniteorbit-signalizers. All elements a n b for n ≥ b ). At thesame time, b and b are contracting, for the self-similar group generated by a, b , b has nucleus N = { e, a ± , b ± , b ± , ( a − b ) ± , ( a − b ) ± , ( b − b ) ± } .The automorphism c = ( c, c ) σ is non-polynomial, contracting, and has finite orbit-signalizer,here OS( c ) = { e, c } .The automorphism d = ( d, d − ) σ is contracting, the nucleus of the group h d i is N = { e, d ± , d ± , d ± } . At the same time, the group h a, d i is not contracting; for da = ( da, d − )and its powers ( da ) n are different and would be in the nucleus.The automorphism g = ( a, g ) is functionally recursive but not finite-state. Hence the auto-morphism f = ( g, g − ) σ is functionally recursive, not finite-state, and has finite orbit-signalizer,here OS( f ) = { e, f } .In the next example we illustrate the solution of the order problem. Example 2.
Consider the automorphisms b = ( a, b ) and a = (1 , a ) σ . The order graph Φ( a ) isa subgraph of Φ( b ) shown in Figure 1. There is a cycle labeled by 2, hence a and b have infiniteorder.The order graph Φ( c ) for the automorphism c = ( c, σ ) is shown in Figure 1. There are nocycles with labels >
1, hence c has finite order, here | c | = 2.Let us illustrate the construction of the conjugator graph and basic conjugators. Example 3.
Consider the conjugacy problem for the trivial automorphism e with itself. HereOS( e ) = { e } and CΠ( e, e ) = Sym( X ) = { ε, σ } . The conjugator graph Ψ( e, e ) is shown in Figure 2.There are two defining subgraphs of the graph Ψ( e, e ), each consists of the one vertex ( e, e, π )with loops in it for π ∈ { ε, σ } . The corresponding basic conjugators are h = ( h , h ) = e and h = ( h , h ) σ .Consider the conjugacy problem for the adding machine a = ( e, a ) σ and its inverse a − =( a − , e ) σ . Here OS( a ) = { a } , OS( a − ) = { a − } , and CΠ( a, a − ) = { ε, σ } . There is one orbitof the action of a on { , } , a | = a and a − | π = a − for every π ∈ { ε, σ } . The conjugatorgraph Ψ( a, a − ) is shown in Figure 2. There are two defining subgraphs of the graph Ψ( a, a − ),each consists of the one vertex ( a, a − , π ) with loop in it for π ∈ { ε, σ } . The corresponding basicconjugators are h = ( h , h a − ) and h = ( h , h ) σ .16Sfrag replacements 111 11 1 22 ab c σ e Figure 1: The order graphs Φ( b ) (on the left) and Φ( c ) (on the right)PSfrag replacements ( e, e, ε ) ( e, e, σ )( a, a − , ε ) ( a, a − , σ ) ( a, b, ε ) ( a, b − , σ )( a, b − , ε ) ( a, b, σ )0000 00000000 0 , , , , Figure 2: The conjugator graph Ψ( e, e ) (on the top left), Ψ( a, a − ) (on the bottom left), andΨ( a, b ) (on the right)Consider the conjugacy problem for the adding machine a = ( e, a ) σ and the automorphism b = ( e, b − ) σ . Here OS( a ) = { a } , OS( b ) = { b, b − } , and CΠ( a, b ) = CΠ( a, b − ) = { ε, σ } . There isone orbit of the action of a on { , } , a | = a , b | π = b − , and b − | π = b for every π ∈ { ε, σ } .The conjugator graph Ψ( a, b ) is shown in Figure 2. There are four defining subgraphs of thegraph Ψ( a, b ), each consists of the two vertices ( a, b, π ) and ( a, b − , π ) with the induced edgesfor π , π ∈ { ε, σ } . The corresponding basic conjugators h , h , h , h are defined as follows h = ( g , g ) h = ( g , g ) h = ( g , ag ) σ h = ( g , ag ) σg = ( h , h b ) g = ( h , h ) σ g = ( h , h b ) g = ( h , h ) σ, where g , g , g , g are actually the basic conjugators for the pair ( a, b − ).The next example shows that the condition of having finite orbit-signalizers cannot be droppedin Theorem 7, and that Theorem 11 does not hold for polynomial automorphisms. Example 4.
Consider the automorphisms b = ( a, b ) σ , b = ( a, b ) defined in Example 1.Inductively one can prove that the state b n | n is active for every n , and hence the automorphism b acts transitively on X n for every n . Thus a and b have the same orbit types (see page 5)and therefore they are conjugate in the group Aut( T ). Both a and b are contracting, however, b has infinite orbit-signalizer, and hence it is not conjugate with a in the group FAut( T ), byProposition 8.Finally, we illustrate the solution of the conjugacy problem in the group of bounded automata.17 xample 5. Consider the conjugacy problem for the adding machine a = ( e, a ) σ and its inverse a − = ( a − , e ) σ in the group of bounded automata.There are two configurations for the pair ( a, a − ): C = { ( a, a − ) , DP = { ( e, e ) }} , C = { ( a, a − ) , DP = { ( e, e ) , ( e, a − ) }} . Neither of them is satisfied by the trivial automorphism, and hence by a finitary automorphism.In particular, a and a − are not conjugate in the group Pol( − a, a − ) is not detectedin Step 1 of the first approach, basically, because ( a, a − ) is not conjugate in Pol( − u that satisfy Step 2, because a has no fixed vertices. Hence, a and a − are not conjugate inthe group Pol( ∞ ).For the second method we get the choice set Π = { ( ε, ε ) , ( ε, σ ) , ( σ, ε ) , ( σ, σ ) } . The configuration C induces the configuration C on the next level when we choose the conjugating permutation ε ;here the pair ( e, e ) induces one pair ( e, e ) and one pair ( e, a − ). For the choice σ , the configuration C induces C , and the pair ( e, e ) gives two pairs ( e, e ). For the choice ε , the configuration C induces C , here the pair ( e, e ) induces one pair ( e, e ) and one pair ( e, a − ), and the pair ( e, a − ) givestwo pairs ( e, a − ). For the choice σ , the configuration C induces C , here the pair ( e, e ) gives twopairs ( e, e ), and the pair ( e, a − ) gives one pair ( e, e ) and one pair ( e, a − ). We get the followingset of matrices A π and vectors θ π : A ( ε,ε ) = , A ( ε,σ ) = ,A ( σ,ε ) = , A ( σ,σ ) = .θ ( ε,ε ) = (0 , , , θ ( ε,σ ) = (0 , , , θ ( σ,ε ) = (1 , , , θ ( σ,σ ) = (1 , , . The initial vector is u = (1 , , t and on n -th step we get u n +1 = A π n u n and θ n = θ π n u n whenwe choose π n ∈ Π. For any choice { π n } n ≥ ⊂ Π the sequence θ n has exponential growth, andhence a and a − are not conjugate in the group Pol( ∞ ) of polynomial automata. Example 6.
Consider the conjugacy problem for the bounded automorphisms b = ( σ, b ) and c = ( c, σ ). Notice that the pairs σ , c and b , σ are not conjugate in Aut( T ). Hence, only σ may appear as the action on X of a possible conjugator, and we take CΠ( b, c ) = { σ } . HereOS( b ) = { e, σ, b } and OS( c ) = { e, σ, c } , CΠ( σ, σ ) = { ε, σ } . The configurations for the pair ( b, c )are the following: C = { ( b, c ) , DP = { ( e, e ) }} , C = { ( σ, σ ) , DP = { ( e, e ) }} , C = { ( e, e ) , DP = { ( e, e ) }} . Let us check what configurations are satisfied by a finitary automorphism as described afterCorollary 14. The configurations C and C are satisfied by the trivial automorphism and havedepth 0. For π ∈ CΠ( b, c ) we get that the configuration C ′ ,π induced by C is equal to C . Therefore C is not satisfied by a finitary automorphism, and hence b and c are not conjugate in Pol( − e, e ) and ( σ, σ ). In Step 2 if we take u = 1 and h = ( e, h ) σ then h − bh = c . Hence ( b, c ) is detected in Step 2 and b, c are conjugate in the groupPol(0).For the second method, we take for the choice set Π = { ( σ, ε, ε ) , ( σ, σ, ε ) , ( σ, ε, σ ) , ( σ, σ, σ ) } .All matrices A π are the same for π ∈ Π. The vectors θ π are as follows A π = , θ ( σ,ε,ε ) = (1 , , , θ ( σ,σ,ε ) = (1 , , ,θ ( σ,ε,σ ) = (1 , , , θ ( σ,σ,σ ) = (1 , , . u = (1 , , t and u n = A n u = (1 , , n −
2) independently of our choice. Ifwe choose π n = ( σ, ε, ε ) for all n ≥ θ n = (1 , , · u n = 1 is bounded. Hence b and c are conjugate in the group Pol(0). The conjugator corresponding to our choice is the addingmachine a . References [1] L. Bartholdi, Branch rings, thinned rings, tree enveloping rings.
Israel J. Math. (2006),93–139.[2] L. Bartholdi,
FR – GAP package for computations with functionally recursive groups, Version1.1.2 , 2010.[3] L. Bartholdi, R.I. Grigorchuk, Z. Sunik, Branch groups. In
Handbook of algebra , vol. 3,North-Holland, Amsterdam 2003, 989–1112.[4] E.V. Bondarenko and V.V. Nekrashevych, Post-critically finite self-similar groups.
AlgebraDiscrete Math. (2003), 21–32.[5] A. Brunner and S. Sidki, On the automorphism group of one-rooted binary trees. J. Algebra (1997), 465–486.[6] A. Brunner, S. Sidki, A. Vieira, A just-nonsolvable torsion-free group defined on the binarytree.
J. Algebra (1999), 99–114.[7] P.W. Gawron, V.V. Nekrashevych, V.I. Sushchansky, Conjugation in tree automorphismgroups.
Int. J. Algebra Comput. (2001), no. 5, 529–547.[8] R.I. Grigorchuk, On the Burnside problem on periodic groups. Functional Anal. Appl. (1980), 41–43.[9] R.I. Grigorchuk, V.V. Nekrashevych, V.I. Suschanskii, Automata, dynamical systems, andgroups. Proc. Steklov Institute (2000), 128–203.[10] R.I. Grigorchuk and J.S. Wilson, The conjugacy problem for certain branch groups.
Proc.Steklov Inst. Math. (2000), no. 4, 204–219.[11] R.I. Grigorchuk and A. Zuk. Spectral properties of a torsion-free weakly branch group definedby a three state automaton.
Contemp. Math. (2002), 57–82.[12] N. Gupta and S. Sidki, Some infinite p -groups. Algebra i Logica (1983), p.584-598.[13] Jun Kigami,
Analysis on fractals , Volume 143 of Cambridge Tracts in Mathematics, Univer-sity Press, Cambridge 2001.[14] Y.G. Leonov, The conjugacy problem in a class of 2-groups.
Mat. Zametki (1998), 573–583.[15] I. Lysenok, A. Myasnikov, A. Ushakov, The conjugacy problem in the Grigorchuk Group ispolynomial time decidable. Groups, Geometry, and Dynamics (2010), 813–833.[16] Y. Muntyan and D. Savchuk, AutomGrp – GAP package for computations in self-similargroups and semigroups, Version 1.1.2 , 2008.[17] V. Nekrashevych,
Self-similar groups.
Volume 117 of
Mathematical Surveys and Monographs ,American Mathematical Society, Providence, RI 2005.[18] V.M. Petrogradsky, I.P. Shestakov, E. Zelmanov, Nil graded self-similar algebras.
Groups,Geometry, and Dynamics (2010), 873–900.1919] M.R. Ribeiro, O grupo Finitario de Isometrias da Arvore n -´aria (The finitary group ofisometries of the n -ary tree). Doctoral Thesis, Universidade de Bras´ılia 2008.[20] A.V. Rozhkov, Conjugacy problem in an automorphism group of an infinite tree. Mat.Zametki (1998), 592–597 .[21] A.V. Russev, On conjugacy in groups of finite-state automorphisms of rooted trees. Ukr.Mat. Zh. (2008), no. 10, 1357–1366.[22] S. Sidki, Automorphisms of one-rooted trees: growth, circuit structure, and acyclicity. J.Math. Sci. (New York). (2000), no. 1, 1925–1943.[23] S. Sidki,
Regular trees and their automorphisms.
Monografias de Matematica 56, IMPA, Riode Janeiro 1998.[24] S. Sidki, Functionally recursive rings of matrices – two examples.
J. of Algebra (2009),4408–4429.[25] Z. Sunic and E. Ventura, The conjugacy problem is not solvable in automaton groups.Preprint, avilable at http://arxiv.org/abs/1010.1993.[26] J.S. Wilson and P.A. Zalesskii, Conjugacy seperability of certain torsion groups.
Arch. Math. (1997), 441–449. A On the existence of a bounded trajectory for nonnegativeinteger systems
Rapha¨el M. Jungers
The purpose of this note is to prove the following theorem.
Theorem 17.
The following bounded trajectory problem is decidable.
INSTANCE:
A finite set of nonnegative integer matrices M = { A , . . . , A m } ⊂ Z n × n and afinite set of nonnegative integer vectors V = { u , . . . , u p } ⊂ Z n . PROBLEM:
Determine whether there exists a sequence ( i t ) ∞ t =1 , i t ∈ { , . . . , m } and an initialvector v ∈ V such that the sequence of vectors determined by the recurrence v t = A i t v t − , t = 1 , , . . . (8) is bounded. In the following, M ∗ , M t denote respectively the set of all products of matrices in M , and theset of all products of length t of matrices in M . This problem is closely related to the so called joint spectral subradius of a set of matrices,which is the smallest asymptotic rate of growth of any long product of matrices in the set, when thelength of the product increases. For a survey on the joint spectral subradius and similar quantities,see [2]. While the joint spectral subradius is notoriously Turing-uncomputable in general, we willsee that in our precise situation, we are able to provide an algorithmic solution to the problem.The next lemma states that if there is a bounded trajectory, then it can be obtained with aneventually periodic sequence of matrices.
Lemma 8.
Let M , V be an instance of the bounded trajectory problem. There exists a sequence ( v t ) as given by Equation (8) which is bounded if and only if there exist matrices A, B ∈ M ∗ anda vector v ∈ V such that the sequence u t = A t Bv is bounded. roof. The if-part is obvious. In the other direction, if the set { v t = A i t . . . A i v } is bounded itmust be finite. Thus, there actually exist A, B ∈ M ∗ such that v = Bv and Av = v .As it turns out it is possible to check in polynomial time, given a nonnegative integer matrix A and a vector v, whether the sequence u t = A t v is bounded. In fact, as we show below, thisdoes not really depend on the actual value of the entries of A and v, but only for each entry of A whether it is equal to zero, one, or larger than one, and for each entry of v whether it is equal tozero or larger than zero. For this reason we introduce two operators that get rid of the inessentialinformation. Definition 1.
Given any nonnegative matrix (or vector) M ∈ Z n × n , we denote by σ ( M ) thematrix in { , , } n × n in which all entries larger than two are set to two, while the other entriesare equal to the corresponding ones in M .Similarly, we denote by τ ( M ) the matrix in { , } n × n in which all entries larger than zeroare set to one, while the other entries are equal to zero.We can now prove the main ingredient of our algorithm. Theorem 18.
Given a nonnegative matrix A ∈ Z n × n , and two indices ≤ i, j ≤ n, the sequence ( A t ) i,j remains bounded when t grows if and only if the sequence ( σ ( A ) t ) i,j remains bounded.Moreover, the boundedness of the sequence ( A t ) i,j can be checked in polynomial time.Proof. We consider the matrix A as the adjacency matrix of a directed graph on n vertices. Theedges of this graph are given by the nonzero entries of A . The graph may have loops, i.e., edgesfrom a node to itself, which correspond to diagonal entries. We say that there is a path (of length t ) from i to j if there is a power A t of A such that ( A t ) i,j ≥ . Equivalently, there exist indices1 ≤ i , . . . , i t ≤ n, i = i, i t = j, such that for all 0 ≤ t ′ ≤ t − , A i t ′ ,i t ′ +1 ≥
1. It is obvious thatif there is a path from i to j, then there is such a path of length less than n .We recall some easy facts from graph theory (see [3] for proofs and references). For anydirected graph, there is a partition of the set V of its vertices in nonempty disjoint sets (the strongly connected components ) V , . . . , V I such that for all v, w ∈ V, v = w, there is a path from v to w and a path from w to v if and only if they belong to the same set in the partition. If thereis no path from v to itself, then { v } is said to be a trivial connected component. Moreover thereexists a (non necessarily unique) ordering of the subsets in the partition such that for any twovertices i ∈ V k , j ∈ V l , there cannot be a path from i to j whenever k > l . There is an algorithmto obtain this partition in O ( n ) operations (with n the number of vertices). In matrix terms, thismeans that one can find a permutation matrix P such that the matrix P T AP is in block upperdiagonal form, where each block on the diagonal corresponds to a strongly connected component.In the following, we suppose for the sake of clarity that A is already in block triangular shape.It is clear that entries in the blocks under the diagonal remain equal to zero in any power of A. We need a different treatment for the entries within diagonal blocks and the entries in blocks abovethe diagonal. • Diagonal blocks.
Let us consider an arbitrary diagonal block B l , which is strongly con-nected by definition. It is easy to see that either all the entries in the block remain boundedor all the entries are unbounded. This occurs if and only if the spectral radius of B l islarger than one. It is easy to see that given a nonnegative matrix with integer entries whosecorresponding graph is strongly connected, its spectral radius is larger than one if and onlyif one of these conditions is satisfied: – There is an entry in B l larger than one. – There is a row in B l with two entries larger than zero.Observe that these conditions do only depend on σ ( A ) . • Non-diagonal blocks.
Let us consider a particular ( i, j )-entry in a non-diagonal block.We will prove that this entry is unbounded if and only if one of the following conditionsholds (and these conditions can be checked in polynomial time):21. There is a path ( i = i , i , . . . , i t − , i t = j ) from i to j, and one of the entries ( i s , i s ) isunbounded for 0 ≤ s ≤ t. II. There exists t such that A ti,i , A ti,j , A tj,j ≥ . (9)Moreover, if this condition holds, there is such a t smaller than n [3, Proposition 1].III. There exist two indices i ′ = j ′ such that there is a path from i to i ′ , a path from j ′ to j, and such that the pair ( i ′ , j ′ ) satisfies condition II above.It is straightforward to check that any of these three conditions implies that the ( i, j )-entry isunbounded.We claim that if the ( i, j )-entry is unbounded yet I and II fail, then III should hold. We provethe claim by induction on the number of vertices. The claim is obvious for n = 1. Take now anarbitrary n, and suppose that the claim holds for n − . We consider an n -by- n matrix such thatthe ( i, j )-entry is unbounded, but I and II fail.First, we must have that either ( A t ) i,i = 0 for all t or ( A t ) j,j = 0 for all t . Indeed, it is notdifficult to see that if there exist t , t , t such that ( A t ) i,i ≥ , ( A t ) j,j ≥ , ( A t ) i,j ≥ , thencondition II holds (see [3, proof of Proposition 1] for a proof). We thus suppose without loss ofgenerality that ( A t ) j,j = 0 for all t , which means that { j } is a trivial connected component. (If itis not the case, then the proof is symmetrically the same replacing j with i ).Now, since ( A t ) i,j = X k A t − i,k A k,j , it comes that there is an index k = j such that ( A t ) i,k is unbounded and A k,j ≥ . Moreover k = i because Condition I does not hold. Thus, if the pair ( i, k ) satisfies Condition II the proof is done,because there is a path from k to j. If not, we now show that one can remove the row and columncorresponding to j in the matrix A and obtain a submatrix A ′ which fulfills the assumptions ofthe claim.Firstly, the entry ( i, k ) is also unbounded in the powers of A ′ . Indeed, we know that { j } isa trivial component and there is no path from j to k . In matrix terms, it means that A can beblock-upper triangularized with the entry corresponding to k before the entry corresponding to j, and k, j in different blocks. Hence, one can erase all the rows and columns of all blocks after theone corresponding to k without changing the successive values of the entry ( i, k ).Secondly, we just assumed that ( i, k ) does not satisfy Condition II, and it cannot satisfyCondition I either, because then Condition I would also hold on ( i, j ) in the matrix A , since thereis a path from k to j in A . Thus, one can apply the induction hypothesis and the claim is proved,because, for any node j ′ , if there is a path in A ′ from j ′ to k, there is a path in A from j ′ to j (obtained by appending the edge ( k, j )).Finally, remark that all the conditions here only depend on which entries are different from zero(since they amount to check the existence of paths), except for the condition on the boundednessof the ( i, i )-entry and the ( j, j )-entry in Condition I, which is treated in the first part of this proof(diagonal blocks).We are now in position to present our algorithm: Algorithm for solving the bounded trajectory problem.
I. Construct a new instance of the bounded trajectory problem: M ′ = { σ ( A ) : A ∈ M} and V ′ = { τ ( v ) : v ∈ V } . II. REPEAT • V ′ ← V ′ ∪ { τ ( Av ) : A ∈ M ′ , v ∈ V ′ }• M ′ ← M ′ ∪ { σ ( AB ) : A ∈ M ′ , B ∈ M ′ } V ′ , M ′ . III. For every pair (
A, v ) ∈ M ′ × V ′ :IF the sequence u t = A t v is bounded, RETURN YES and STOP.IV. RETURN NO. Theorem 19.
Algorithm is correct and stops in finite time.Proof.
We first show how to implement Line III in the algorithm. For any column correspondingto a nonzero entry of v, one just has to check whether all the entries of this column remain boundedin the sequence of matrices A t . Thanks to Theorem 18, it is possible to fulfill this requirementBy Lemma 8 we need to check whether there exist
A, B ∈ M ∗ and v ∈ V such that A t Bv is bounded. Note that A t Bv is bounded if and only if σ ( A ) t τ ( Bv ) is bounded. The finite sets { σ ( A ) : A ∈ M ∗ } and { τ ( Bv ) : B ∈ M ∗ , v ∈ V } are precisely the sets M ′ and V ′ obtainedafter the loop at Line II in the algorithm. Therefore the algorithm is correct and stops in finitetime.Let us show that one should not expect a polynomial time algorithm for the problem. Proposition 20.
Unless P = N P, there is no polynomial time algorithm for solving the boundedtrajectory problem.Proof.
Our proof is by reduction from the mortality problem which is known to be NP-hard, evenfor nonnegative integer matrices [1, p. 286]. In this problem, one is given a set of matrices M , andit is asked whether there exists a product of matrices in M ∗ which is equal to the zero matrix.We now construct an instance M ′ , V of the bounded trajectory problem such that there is abounded trajectory for this instance if and only if the set M is mortal: take M ′ = { A ′ = 2 A : A ∈ M} and v = e (the ”all ones vector”) as the unique vector in V. Now, it is straightforward that there exists a sequence ( i t ) ∞ t =1 , i t ∈ { , . . . , m } , such that thesequence of vectors A ′ i t . . . A ′ i e = 2 t A i t . . . A i e is bounded if and only if the set M is mortal. Indeed, the matrices in M have nonnegative integerentries, and if the vector A i t . . . A i e is different from zero, then its (say, Euclidean) norm is greateror equal to one.Also, if one relaxes the requirement that the matrices and the vectors are nonnegative, thenthe problem becomes undecidable, as shown in the next proposition. Proposition 21.
The bounded trajectory problem is undecidable if the matrices and vectors inthe instance can have negative entries.Proof. (sketch) It is known that the mortality problem with entries in Z is undecidable [2, Corollary2.1]. We reduce this problem to the bounded trajectory problem in a way similar as in Proposition20, except that we build much larger matrices: we make 2 n copies of each matrix in M and placethem in a large block-diagonal matrix. That is, our matrices in M ′ are of the shape { diag(2 A, A, . . . , A ) : A ∈ M} . Now we take V = { v } , where v ∈ {− , } n n is the concatenation of all the different n -dimensional {− , } -vectors. This vector has a bounded trajectory if and only if there existsa zero product in M ∗ . eferences [1] V. D Blondel and J. N. Tsitsiklis. When is a pair of matrices mortal? Information ProcessingLetters , (1997), 283–286.[2] R. M. Jungers. The joint spectral radius, theory and applications. In Lecture Notes in Controland Information Sciences , volume 385, Springer-Verlag, Berlin 2009.[3] R. M. Jungers, V. Protasov, and V. D. Blondel. Efficient algorithms for deciding the typeof growth of products of integer matrices.
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