On the inverse hull of a Markov shift
aa r X i v : . [ m a t h . R A ] F e b ON THE INVERSE HULL OF A MARKOV SHIFT
ARIA BEAUPR´E, ANTHONY DICKSON, DAVID MILAN, AND CHRISTIN SUM
Abstract.
In this paper we provide an abstract characterization ofthe inverse hulls of semigroups associated with Markov shifts. As anapplication of the characterization we give an example of Markov shiftsthat are not conjugate, but have isomorphic inverse hulls. Introduction
We study inverse semigroups associated with Markov shift spaces. In [3],Starling defines an inverse semigroup associated with a one-sided subshift.He shows that the Carlsen-Matsumoto C ∗ -algebra associated with the sub-shift is generated by the inverse semigroup. He also gives a decomposition ofthe C ∗ -algebra as a partial crossed product by a free group. Recently, Exeland Steinberg [1] defined the inverse semigroups H ( S ), associated with any0-left cancellative semigroup S , called the inverse hull of S . This collectionis significant as it contains many of the examples of inverse semigroups thatappear in the study of C ∗ -algebras generated by partial isometries. Graphinverse semigroups are included in this collection as well as semigroups asso-ciated with left cancellative categories. Also, the inverse semigroups Starlingassociates to one-sided shifts are inverse hulls.One major challenge in working with the inverse hull of a 0-left cancella-tive semigroup is to obtain a useful description of the semilattice of idem-potents. Because H ( S ) is defined to be the inverse semigroup generatedby a set of partial bijections, it can be difficult to determine the possiblesets, called constructible sets in [1], on which the idempotents of H ( S ) actas identities. In section 3, we give a thorough description of the semilatticeof the inverse semigroup associated with a Markov subshift. There is a set O of mutually orthogonal idempotents such that every idempotent is com-parable to some element in O . The idempotents strictly above O (togetherwith 0) form a subsemigroup of the semilattice, with the additional propertythat each such idempotent is determined uniquely by the idempotents in O above which it sits. We describe a number of additional properties of thissubsemigroup.In section 4 we state a set of axioms on an inverse semigroup H that areequivalent to H being isomorphic to the inverse hull of the language of a Date : March 1, 2021.2010
Mathematics Subject Classification.
Markov shift. We find that it is possible to axiomatize the set O appearing insuch an inverse semigroup, although there is not necessarily a unique choicefor O . In section 5, we take advantage of this lack of uniqueness when weuse our characterization to show that different Markov shifts can give riseto isomorphic inverse hulls. In particular we find two Markov shifts withdifferent entropies and isomorphic inverse hulls. We also give an example oftwo conjugate Markov shifts whose inverse hulls are not isomorphic. Finally,we state a conjecture that if the inverse hulls of two Markov shifts areisomorphic, then their alphabets must have the same size.2. Preliminaries An inverse semigroup is a semigroup S such that for each s in S , thereexists a unique s ∗ in S such that s = ss ∗ s and s ∗ = s ∗ ss ∗ . The elements e ∈ S satisfying e = e (and hence e ∗ = e ) are called idempo-tents . The set of all idempotents in S is denoted by E ( S ).There is a natural partial order on S defined by s ≤ t if s = te forsome idempotent e . Note that the subsemigroup E ( S ) of idempotents iscommutative, and hence forms a meet semilattice under the natural partialorder with e ∧ f = ef for e, f in E ( S ).There are a number of useful relations known as Green’s relations definedon a semigroup. For inverse semigroups, these relations take the followingform: we have s L t if and only if s ∗ s = t ∗ t , s R t if and only if ss ∗ = tt ∗ ,and H = L ∩ R . If S has the property that for all s, t in S , s H t implies s = t , then S is said to be combinatorial . One can also prove that S iscombinatorial provided s ∗ s = ss ∗ implies s is idempotent for all s in S .Finally, s D t if and only if there exists a, b ∈ S such that a ∗ a = t ∗ t , aa ∗ = s ∗ s , b ∗ b = tt ∗ , bb ∗ = ss ∗ , and t = b ∗ sa . For e, f ∈ E ( S ) we have e D f if and only if there exists a ∈ S with e = a ∗ a and f = aa ∗ .An important example of an inverse semigroup is the semigroup I ( X )of partial bijections on a set X . A partial bijection on X is a bijection g : A → B with A, B ⊆ X . If g ∈ I ( X ) with domain A and range B and f ∈ I ( X ) with domain C and range D , then the product f g is thecomposition of the functions on the largest possible domain. That is, f g isthe bijection of g − ( B ∩ C ) onto f ( B ∩ C ). The map with empty domain isdenoted by 0. The inverse of f in I ( X ) is given by f − .We will now outline the construction of the inverse hull of a left cancella-tive semigroup S with zero. For more details, see [1]. For each s in S , define θ s to be the partial bijection on S − { } with domain F s = { s ∈ S : sx = 0 } , and range E s = { y ∈ S : y = sx = 0 for some x ∈ S } . N THE INVERSE HULL OF A MARKOV SHIFT 3
Then the inverse hull of S is defined to be the inverse semigroup generatedby { θ s : s ∈ S and s = 0 } .3. The semilattice of the inverse hull of a Markov shift
In this section we consider the inverse hull of a Markov shift. Our goals areto develop a thorough understanding of the semilattice and to find propertiesthat characterize these inverse semigroups.Let A be a finite alphabet and let T = { T a,b } a,b ∈ A be a matrix such that T a,b ∈ { , } for each a, b ∈ A . We refer to T as a Markov transition matrix .The set of infinite words a a a . . . such that T a i ,a i +1 = 1 for all i in N is called the Markov subshift associated with the transition matrix T . Let L T consist of all finite words occurring as subwords of elements of the Markovsubshift of T . Notice that if there is a row in T consisting only of zeros, thenno word in L T will contain the letter associated with that row. Thereforewe will always assume that T contains no zero rows.Let S T = L T ∪ { } . For x, y in L T define x ∗ y = (cid:26) xy if xy ∈ L T xy denotes the concatenation of words x and y . Then S T is a 0-cancellative semigroup under this operation. Note that S T does not containa multiplicative identity. It will sometimes be useful to adjoin one. Wedenote by S T the set S T with an identity adjoined. Similarly, L T = L T ∪{ } .Recall that H ( S T ) is the inverse semigroup with 0 generated by the maps θ w such that w ∈ L T . It will be useful to develop some basic propertiesabout products of such maps. Lemma 3.1.
For u, v in L T we have θ u θ v = ( θ uv if uv ∈ L T , otherwise.Also, θ − u θ v = θ v ′ if v = uv ′ for some v ′ ∈ L T ,θ − u ′ if u = vu ′ for some u ′ ∈ L T ,θ − u θ u if u = v , and otherwise.Proof. We leave the straightforward proof of the first assertion to the reader.For the second, suppose first that v = uv ′ . It follows thatdom θ − u θ v = dom θ v = dom θ v ′ , and that for any x ∈ dom θ v ′ we have θ − u θ v ( x ) = v ′ x = θ v ′ ( x ). Thus θ − u θ v = θ v ′ . The case where u = vu ′ for some u ′ in L T is similar, andthe case where u = v is vacuously true. Finally, suppose that there are no ARIA BEAUPR´E, ANTHONY DICKSON, DAVID MILAN, AND CHRISTIN SUM x, y ∈ L T such that vx = uy . Note that if a word x lies in dom θ − u θ v thenthere exists some y ∈ L T such that vx = uy . Thus dom( θ − u θ v ) = ∅ and θ − u θ v = 0. (cid:3) Remark 3.2.
One useful consequence of Lemma 3.1 is that for u, v ∈ L T , θ − u θ u θ v = ( θ v if uv ∈ L T H ( S T ).We omit most of the proof of the following theorem, since similar resultshave appeared in both [3] and [1]. We point out that these forms are notunique. Theorem 3.3.
All nonzero elements of H ( S T ) are of the form θ s θ − x θ x . . . θ − x n θ x n θ − w , for some n ≥ , x i ∈ A, and s, w ∈ L T .Proof. Note that this form is nearly identical to the one given in [1, Theorem7.11]. There are two small differences to consider. The first is that theidempotents in the middle of the product are associated with the letters of A , rather than arbitrary words. We do not lose generality, since for w ∈ L T , θ − w θ w = θ − a θ a where a is the last letter of w . Second, we do not assumethat the product of idempotents in the middle includes θ − s θ s or θ − w θ w . Ofcourse, this is not consequential since θ s θ − x θ x . . . θ − x n θ x n θ − w = θ s ( θ − s θ s ) θ − x θ x . . . θ − x n θ x n ( θ − w θ w ) θ − w . (cid:3) It follows quickly that the nonzero idempotents of H ( S T ) are of the form θ s θ − x θ x . . . θ − x n θ x n θ − s where n ≥ , x i ∈ A, and s ∈ L T . Based on this, it is natural to distinguishthe idempotents for which s = 1 from those for which s = 1. As we will see,the first group of idempotents sits above the second in the semilattice. Inbetween the two groups is the set O = { θ a θ − a : a ∈ A } which are the rangeidempotents associated with letters.Note that θ a θ − a is the identity map on the set aL T := { u ∈ L T : u = av for some v ∈ L T } . Choose a, b ∈ A with a = b . Since aL T ∩ bL T = ∅ , we have θ a θ − a θ b θ − b = 0.Thus we say that O consists of mutually orthogonal idempotents. Nowdefine O ↑ = { α ∈ E ( H ( S T )) : α ≥ θ a θ − a for some a ∈ A } and O ↓ = { α ∈ E ( H ( S T )) : α ≤ θ a θ − a for some a ∈ A } . N THE INVERSE HULL OF A MARKOV SHIFT 5
Proposition 3.4.
Let H ( S T ) be the inverse hull of the semigroup S T asso-ciated with a Markov transition matrix T . Set O = { θ a θ − a : a ∈ A } . Then O ↑ − O = { θ − x θ x . . . θ − x n θ x n : x i ∈ A } − { } and O ↓ = { θ s θ − x θ x . . . θ − x n θ x n θ − s : x i ∈ A, s = 1 } ∪ { } . Proof.
We start with the set O ↑ − O . As distinct idempotents of O areincomparable, note that O ↑ − O = { τ ∈ E ( H ( S T )) : τ > θ a θ − a for some a ∈ A } . Let τ = θ − x θ x . . . θ − x n θ x n be nonzero, where x i ∈ A for 1 ≤ i ≤ n . Since τ = 0, there is a letter a in A such that T x i ,a = 1 for all 1 ≤ i ≤ n . So x i a ∈ L T for each i . It follows thatdom( θ a θ − a ) = aL T ⊂ { w ∈ L T : x i w ∈ L T for all i } = dom τ. The above inclusion is proper because a ∈ dom τ while aL T does not containletters. Thus τ ∈ O ↑ − O .Now suppose that τ > θ a θ − a for some a ∈ L T . Then τ = 0 and we canwrite τ = θ s θ − x θ x . . . θ − x n θ x n θ − s where x i ∈ A and s ∈ L T . Suppose that s = 1. Then θ s θ − s > θ a θ − a . It follows that aL T ⊆ sL T . By our assumptionthat the row corresponding to a in T is nonzero, we know that aL T = ∅ .So there is some b ∈ A such that ab = sw for some word w ∈ L T and thus a = sw ′ for some w ′ ∈ L T . Thus s = a and w ′ = 1. But this contradicts theinequality θ s θ − s > θ a θ − a . Therefore s = 1 and 0 = τ = θ − x θ x . . . θ − x n θ x n .Next we consider the set O ↓ . Let τ = θ s θ − x θ x . . . θ − x n θ x n θ − s where s = 1.Note that dom α ⊆ sL ⊆ dom( θ l ( s ) θ − l ( s ) ) where l ( s ) denotes the last letterof s . Thus τ ≤ θ l ( s ) θ − l ( s ) . Conversely, suppose that 0 = τ ≤ θ a θ − a forsome a ∈ A and write τ = θ s θ − x θ x . . . θ − x n θ x n θ − s . If s = 1 we are done,so suppose s = 1. Then we have θ − x i θ x i θ a θ − a = 0 for all i . But one canquickly check that this is equivalent to θ − x i θ x i θ a θ − a = θ a θ − a . Therefore τ = τ θ a θ − a = θ a θ − a = θ a ( θ − a θ a ) θ − a . Since a = 1, this is of the requiredform. (cid:3) At this point we collect some observations that follow from the aboveproposition. These properties will feature in our characterization of theinverse hulls of Markov subshifts. The first observation is that each idempo-tent of H ( S T ) is comparable to some element in O . The second observationis that ( O ↑ − O ) ∪ { } is closed under multiplication. Finally, one can quicklyshow that O ↑ ∪ { } is also closed under multiplication. Elements of O aremutually orthogonal, which tells us that products of distinct elements of O are zero. Also, by Remark 3.2, products of the form θ − x θ x . . . θ − x n θ x n ( θ a θ − a )are either 0 or θ a θ − a .Now we will develop some properties for our characterization. The nextProposition says that an element of O ↑ − O is uniquely determined by theidempotents in O that it sits above. ARIA BEAUPR´E, ANTHONY DICKSON, DAVID MILAN, AND CHRISTIN SUM
Proposition 3.5.
Fix α, β in O ↑ − O . Suppose that O ∩ α ↓ = O ∩ β ↓ . Then α = β .Proof. Write α = θ − x θ x . . . θ − x n θ x n . Note that θ a θ − a ∈ O ∩ α ↓ if and onlyif θ − x i θ x i θ a θ − a = 0 for 1 ≤ i ≤ n , which is equivalent to x i a ∈ L T for1 ≤ i ≤ n . Alsodom α = { w ∈ L T : x i w ∈ L T for 1 ≤ i ≤ n } = [ (cid:8) aL T : a ∈ A, x i a ∈ L T for all i (cid:9) . Therefore, if
O ∩ α ↓ = O ∩ β ↓ , then dom α = dom β . Since α and β areidempotents, we conclude that α = β . (cid:3) Finally, we consider some properties of the D -classes and H -classes of H ( S T ). Proposition 3.6.
Fix α, β in O ↑ − O . If α D β then α = β . Also, for each θ a θ − a in O , there exists a unique α in O ↑ − O such that θ a θ − a D α .Proof. Consider arbitrary γ = θ s θ − x θ x . . . θ − x n θ x n θ − w in H ( S T ) and supposethat γγ ∗ and γ ∗ γ lie in O ↑ − O . By the same argument as in the proof ofProposition 3.4, γγ ∗ ∈ O ↑ − O implies s = 1. Also, γ ∗ γ ∈ O ↑ − O implies w = 1. Therefore γγ ∗ = γ ∗ γ . Now suppose that α, β in O ↑ − O with α D β .Then there exists γ ∈ H ( S T ) such that γγ ∗ = α and γ ∗ γ = β . By the aboveargument, α = β .For the second assertion, we have that θ a θ − a D θ − a θ a . Since no row ofthe Markov transition matrix is zero, we have 0 = θ − a θ a ∈ O ↑ − O , whichcompletes the proof. (cid:3) Proposition 3.7.
The inverse hull H ( S T ) is combinatorial.Proof. Let γ = θ s θ − x θ x . . . θ − x n θ x n θ − w = 0 and suppose that γγ ∗ = γ ∗ γ .Now, dom γ ∗ γ = { wu ∈ L T : u ∈ L T , su ∈ L T , and x i u ∈ L T for all i } anddom γγ ∗ = { su ∈ L T : u ∈ L T , wu ∈ L T , and x i u ∈ L T for all i } .Since γγ ∗ = γ ∗ γ = 0 we have wu = sv for some u, v ∈ L T . Then w = sy or s = wy for some y ∈ L T . First suppose that w = sy . As γ ∗ γ = 0, thereexists a ∈ A such that wa ∈ dom γ ∗ γ . So sa ∈ L T and x i a ∈ L T for all i .Then sa ∈ dom γγ ∗ and hence sa ∈ dom γ ∗ γ . Then we have that sa = wz for some z ∈ L T . By removing the last letter we have s = wz ′ for some z ′ ∈ L T . Since w = sy for some y ∈ L T and s = wz ′ for some z ′ ∈ L T , weconclude that s = w . A symmetric argument works in the case that s = wy .Therefore γ is idempotent, which completes the proof that H ( S T ) is com-binatorial. (cid:3) The Characterization of Inverse Hulls of Markov Shifts
In this section we show that an inverse semigroup H is isomorphic tothe inverse hull of a Markov shift if and only if it is combinatorial and N THE INVERSE HULL OF A MARKOV SHIFT 7 it contains a set O of mutually orthogonal nonzero idempotents satisfyingcertain properties. Fix a combinatorial inverse semigroup H with 0 and aset O of nonzero idempotents such that O satisfies the following properties:(O1) the elements of O are mutually orthogonal,(O2) every idempotent in H is comparable to some element of O ,(O3) both O ↑ ∪ { } and ( O ↑ − O ) ∪ { } are closed under multiplication,(O4) elements of O ↑ −O are uniquely determined by the set of idempotentsin O that they lie above in the natural partial order, and(O5) for each e ∈ O , the D -class of e contains at most one element of O ↑ − O .Using the above assumptions, we will show that H contains a 0-left can-cellative semigroup S isomorphic to the semigroup associated with a Markovtransition matrix T . When S generates H , we show that H is isomorphicto H ( S T ). The first order of business it to define the sets that will serve asour alphabet and language respectively. Let A = { a ∈ H : a ∗ a ∈ O ↑ − O and aa ∗ ∈ O} , and L = { a a . . . a n = 0 : n ∈ N , a i ∈ A } . We refer to L as the language associated with O , a name that is justifiedby Theorem 4.6 below. We show that L behaves under multiplication muchlike the set of generators of the inverse hull of a Markov shift. In particular,compare Corollary 4.2 below with Lemma 3.1. Proposition 4.1.
Let H be a combinatorial inverse semigroup and suppose O is a set of nonzero idempotents that satisfies conditions (O1) – (O5). For a, b ∈ A we have:(1) ab = 0 if and only if bb ∗ ≤ a ∗ a , and(2) a ∗ b = 0 if and only if a = b .Proof. Note that both a and b are nonzero. Suppose bb ∗ ≤ a ∗ a . Then a ∗ abb ∗ = bb ∗ = 0. Thus ab = 0.Conversely, suppose ab = 0. Then a ∗ abb ∗ = 0. Since O ↑ ∪ { } is closedunder multiplication by (O3), a ∗ abb ∗ ∈ O ↑ . That is, a ∗ abb ∗ ≥ cc ∗ for some c ∈ A . Also a ∗ abb ∗ ≤ bb ∗ . Since elements of O are incomparable, a ∗ abb ∗ = bb ∗ . So a ∗ a ≥ bb ∗ .Next suppose a = b . Then aa ∗ b = a implies that a ∗ b = 0.Conversely, if a ∗ b = 0 then aa ∗ bb ∗ = 0. By (O1), aa ∗ = bb ∗ . Then for x = a ∗ b we have xx ∗ = a ∗ a and x ∗ x = b ∗ b . Thus aa ∗ D a ∗ a D b ∗ b . By (O5), a ∗ a = b ∗ b . Since H is combinatorial, a = b . (cid:3) As a corollary we have the following property about products involvingwords in L . ARIA BEAUPR´E, ANTHONY DICKSON, DAVID MILAN, AND CHRISTIN SUM
Corollary 4.2.
Let u, v ∈ L and u = xa for some x in L and a in A .Then u ∗ v = w if v = uw for some w ∈ L,w ∗ if u = vw for some w ∈ L,a ∗ a if u = v, and otherwise.Proof. Let u = a a · · · a n and v = b b · · · b m where a i , b j ∈ L . If u ∗ v = 0then it follows from Proposition 4.1 that a = b . Moreover since b b = 0we have b b ∗ ≤ b ∗ b . Then u ∗ v = a ∗ n · · · a ∗ a ∗ b b · · · b m = a ∗ n · · · a ∗ ( b ∗ b )( b b ∗ ) b · · · b m = a ∗ n · · · a ∗ b · · · b m . By continuing in this way we conclude that u = v , u = vw , or v = uw forsome w ∈ L . In each case, the formula for u ∗ v can be verified quickly usingsimilar calculations. (cid:3) Just as for H ( S T ), we can now show that products in the inverse semi-group generated by L take on a special form. We leave the proof of thefollowing proposition to the reader. Proposition 4.3.
Any nonzero element of the inverse semigroup generatedby L is of the form wa ∗ a a ∗ a · · · a ∗ n a n v ∗ where n ≥ , a i ∈ A and w, v ∈ L . Since n ≥ wa ∗ a a ∗ a · · · a ∗ n a n v ∗ , which is formally definedin H , is always an element of H . It follows from Proposition 4.3 thatthe idempotents of the inverse semigroup generated by L are of the form wa ∗ a a ∗ a · · · a ∗ n a n w ∗ where n ≥ a i ∈ A and w ∈ L .Another useful consequence of the above proposition is that when L gen-erates H (as an inverse semigroup with 0) the sets O and O ↑ − O have aform that echoes their counterparts in the inverse hull of a Markov subshift. Proposition 4.4.
Suppose that H is generated by L . Then O = { aa ∗ : a ∈ A } and O ↑ − O = { a ∗ a a ∗ a · · · a ∗ n a n : a i ∈ A } − { } .Proof. By definition of A , aa ∗ ∈ O for each a ∈ A . Let e be a nonzeroidempotent in O . We will show that e = aa ∗ for some a ∈ L . Since L generates H , e = wa ∗ a a ∗ a · · · a ∗ n a n w ∗ where n ≥ a i ∈ A and w ∈ L .First suppose that w = 1. By definition of A , a ∗ i a i ∈ O ↑ − O for each i . By(O3), 0 = e = a ∗ a a ∗ a · · · a ∗ n a n ∈ O ↑ − O , a contradiction. Thus w = 1 andwe may write w = aw ′ for some a ∈ A and w ′ ∈ L . Then e ≤ aa ∗ ∈ O . Asidempotents in O are mutually orthogonal, we conclude that e = aa ∗ .We have shown in the previous paragraph that if a ∗ a a ∗ a · · · a ∗ n a n = 0then a ∗ a a ∗ a · · · a ∗ n a n ∈ O ↑ − O . Let e ∈ O ↑ − O . Write e = wa ∗ a a ∗ a · · · a ∗ n a n w ∗ N THE INVERSE HULL OF A MARKOV SHIFT 9 where n ≥ a i ∈ A and w ∈ L . If w = 1 then write w = aw ′ where a ∈ A and w ′ ∈ L and note that e ≤ aa ∗ , contradicting our assumptionthat e ∈ O ↑ − O . Thus e = a ∗ a a ∗ a · · · a ∗ n a n . (cid:3) L is the -left cancellative semigroup of a Markov subshift. Here we continue to assume that H is a combinatorial inverse semigroup andthat O is a set of nonzero idempotents satisfying (O1) – (O5). We also retainthe notation A and L . The set A ∗ denotes finite words over the alphabet A .If a i ∈ A for 1 ≤ i ≤ n , we will temporarily write a ◦ a ◦ a ◦ · · · ◦ a n for aword in A ∗ since a a · · · a n represents a product in H . Let M = { a ◦ a ◦ a ◦ · · · ◦ a n ∈ A ∗ : a a · · · a n ∈ L } . Proposition 4.5.
The map a ◦ a ◦ a ◦ · · · ◦ a n a a · · · a n from M to L is a bijection.Proof. The map is clearly surjective. Suppose that w = a a · · · a n = b b · · · b m in L where a i , b j ∈ A . Then 0 = w ∗ w = a ∗ n · · · a ∗ a ∗ b b · · · b m .By the same argument used in the proof of Corollary 4.2, we conclude that a = b and w ∗ w = a ∗ n · · · a ∗ b · · · b m . We may continue in this way un-til we exhaust all the a i or b j . Notice by Corollary 4.2 we also have that w ∗ w = a ∗ n a n . If n < m we have a ∗ n · · · a ∗ n − m +1 = w ∗ w = a ∗ n a n ∈ O ↑ − O .Then there exists a ∈ A such that a ∗ n · · · a ∗ n − m +1 ( aa ∗ ) = aa ∗ . This im-plies that a = a n − m +1 and a ∗ n · · · a ∗ n − m +1 = a ∗ n · · · a ∗ n − m +1 ( aa ∗ ) = aa ∗ ∈ O ,which contradicts the fact that a ∗ n · · · a ∗ n − m +1 ∈ O ↑ − O . If m < n , then b m − n +1 · · · b n = w ∗ w ∈ O ↑ − O . Similarly, there exists bb ∗ ∈ O suchthat ( bb ∗ ) b m − n +1 · · · b n = bb ∗ . Again this leads to the contradiction that b m − n +1 · · · b n ∈ O . Thus n = m and a i = b i for 1 ≤ i ≤ n , which shows thatthe map under consideration is injective. (cid:3) The proposition shows that we may identify L with a collection M ofwords in A ∗ . We do so for the rest of this section. Under this identification,we show that L is in fact the language of a Markov subshift. Define an A × A Markov transition matrix T by T ( a, b ) = ( ab ∈ L (i.e. bb ∗ ≤ a ∗ a )0 otherwise. Theorem 4.6.
The semigroup L ∪ { } is isomorphic to the semigroup S T = L T ∪ { } associated with the Markov transition matrix T .Proof. We just need to verify that the set of words in L is equal to L T . Notethat for any letter a ∈ A , a ∗ a ∈ O ↑ − O and thus there is a letter b ∈ A suchthat bb ∗ ≤ a ∗ a . So ab ∈ L . Then any w = a a · · · a n ∈ L can be extendedto an infinite word a a · · · a n a n +1 a n +2 · · · such that T ( a i , a i +1 ) = 1 for each i ∈ N . Thus w ∈ L T . Conversely, let w be a subword of an infinite word inthe subshift associated with T . Then we may write w = a a · · · a n where T ( a i , a i +1 ) = 1 for each 1 ≤ i ≤ n −
1. It follows that a i a i +1 in L for each ≤ i ≤ n −
1. Note that the proof of the u = v case in Corollary 4.2depends only on the assumption that products of consecutive letters of u liein L . Thus if w = 0, then a ∗ n a n = w ∗ w = 0. By contradiction, w = 0 and so w ∈ L . (cid:3) It follows that L ∪ { } is a 0-cancellative subsemigroup of H .4.2. The isomorphism.
Here we show that if L generates H as an inversesemigroup with zero, then H is isomorphic to H ( L ). As a consequence, aninverse semigroup H is isomorphic to the inverse hull of a Markov subshiftif and only if H is combinatorial, H contains a set of nonzero idempotents O satisfying (O1) – (O5), and L generates H .We continue to fix a combinatorial inverse semigroup H that contains aset of nonzero idempotents O satisfying (O1) – (O5). Given α ∈ H , define D α = { x ∈ L : xx ∗ < α ∗ α } . We say that H is right reductive relative to L if for all α, β ∈ H ,(1) D α = D β and(2) αx = βx for all x ∈ D α implies α = β . Lemma 4.7.
Let = α = wa ∗ a a ∗ a · · · a ∗ n a n v ∗ where n ≥ , a i ∈ A , w, v ∈ L . Then:(1) For v = 1 : x ∈ D α if and only if x = vy for some y ∈ L and αx = wy ∈ L .(2) For v = 1 : x ∈ D α if and only if wx ∈ L and a i x ∈ L for each i .Proof. We give the proof for v = 1. The case where v = 1 is similar. Let x ∈ D α . Then xx ∗ < α ∗ α ≤ vv ∗ . Hence v ∗ x = 0. We can then employCorollary 4.2 to conclude that x = vy for some y ∈ L . Moreover, as vy = 0,we have that v ∗ vy = y by Corollary 4.2. By similar reasoning we have αx = wa ∗ a · · · a ∗ n a n ( v ∗ vy ) = wa ∗ a · · · ( a ∗ n a n y ) = · · · = wy. Since x ∈ D α we have αx = 0 and hence αx = wy ∈ L .Conversely suppose x = vy ∈ L for some y ∈ L and αx = wy ∈ L . Then αx = 0 and hence 0 = α ∗ αxx ∗ = va ∗ a · · · a ∗ n a n w ∗ wv ∗ vx ∗ = va ∗ a · · · a ∗ n a n ( w ∗ wy ) x ∗ = va ∗ a · · · ( a ∗ n a n y ) x ∗ = vyx ∗ = xx ∗ . So xx ∗ ≤ α ∗ α . Suppose xx ∗ = α ∗ α . As vy ∈ L we have yy ∗ ≤ v ∗ v so0 = yy ∗ = v ∗ vyy ∗ v ∗ v = v ∗ xx ∗ v = v ∗ va ∗ a · · · a ∗ n a n w ∗ w. N THE INVERSE HULL OF A MARKOV SHIFT 11
Thus yy ∗ ∈ O ↑ − O , a contradiction. Therefore xx ∗ < α ∗ α . (cid:3) Proposition 4.8. If L generates H , then H is right reductive relative to L .Proof. First we show that α = 0 implies D α = ∅ . If α = 0 then α = wa ∗ a a ∗ a · · · a ∗ n a n v ∗ for some n ≥ a i ∈ A and w, v ∈ L . As 0 = α ∗ α = va ∗ a · · · a ∗ n a n w ∗ wv ∗ we have v ∗ va ∗ a · · · a ∗ n a n w ∗ w = 0. Thus v ∗ va ∗ a · · · a ∗ n a n w ∗ w ∈ O ↑ − O and there exists a ∈ A such that aa ∗ < v ∗ va ∗ a · · · a ∗ n a n w ∗ w = v ∗ α ∗ αv .Then αva = 0, since ( αva ) ∗ αva = a ∗ a . For v = 1 we have αa = 0 impliesthat wa = 0 and a i a = 0 for all i . Thus by (2) of Lemma 4.7, a ∈ D α . For v = 1 we have αva = wa ∗ a · · · a ∗ n a n ( v ∗ va ) = wa ∗ a · · · ( a ∗ n a n a ) = · · · = wa. Thus va ∈ D α by Lemma 4.7.Now for α, β in H suppose that D α = D β and αx = βx for all x ∈ D α .By the argument above, if D α = D β = ∅ , then α = β = 0. So we mayassume D α = ∅ . Write α = wa ∗ a a ∗ a · · · a ∗ n a n v ∗ , β = tb ∗ b b ∗ b · · · b ∗ m b m z ∗ where m, n ≥ a i , b j ∈ A , and w, v, t, z ∈ L . Thus vy ∈ D α for some y ∈ L . In fact for any prefix y of y we have vy ∈ D α . In particular, va ∈ D α for some a ∈ A . Then there exists x ∈ L such that va = zx . Byright cancellativity we conclude that v = zx ′ for some x ′ ∈ L . Similarly z = vy ′ for some y ′ ∈ L . Then v = vy ′ x ′ implies that y ′ = x ′ = 1 and weconclude that v = z . It follows that wa = αva = αza = βza = ta and hence w = t .To finish the proof we will show that v ∗ α ∗ αv = w ∗ wa ∗ a · · · a ∗ n a n v ∗ v = t ∗ tb ∗ b · · · b ∗ n b n z ∗ z = z ∗ β ∗ βz. By (O4), it suffices to prove that ( v ∗ α ∗ αv ) ↓ ∩ O = ( z ∗ β ∗ βz ) ↓ ∩ O . To thatend, suppose bb ∗ ∈ ( v ∗ α ∗ αv ) ↓ ∩ O . Then 0 = vbb ∗ v ∗ ≤ α ∗ α . If vbb ∗ v ∗ = α ∗ α then bb ∗ = v ∗ α ∗ αv ∈ O ↑ − O , a contradiction. Thus zb = vb ∈ D α = D β .It follows that bb ∗ ∈ ( z ∗ β ∗ βz ) ↓ ∩ O . Thus ( v ∗ α ∗ αv ) ↓ ∩ O ⊆ ( z ∗ β ∗ βz ) ↓ ∩ O .The other inclusion follows by a symmetric argument.By (O4) we have that v ∗ α ∗ αv = z ∗ β ∗ βz and thus α = wv ∗ α ∗ αvv ∗ = tz ∗ α ∗ αzz ∗ = β. (cid:3) We now show that the set D α is the domain of the map in H ( L ) naturallyassociated with α . Proposition 4.9.
Let α = wa ∗ a · · · a ∗ n a n v ∗ in H and consider ϕ = θ w θ − a θ a · · · θ − a n θ a n θ − v in H ( L ) . Then D α = dom ϕ . Proof.
Note that x ∈ dom ϕ if and only if y ∈ L with x = vy for some y ∈ L for which a i y = 0 for 1 ≤ i ≤ n and wy = 0. By Lemma 4.7, D α = dom ϕ . (cid:3) We can now prove the main theorem of the section which characterizes theinverse semigroups that are isomorphic to inverse hulls of Markov subshifts.
Theorem 4.10.
Let H be an inverse semigroup with . Then H is isomor-phic to the inverse hull of a Markov subshift if and only if(1) H is combinatorial,(2) there is a set O of nonzero idempotents in H satisfying ( O − ( O ,and(3) the language, L , associated with O generates H .Proof. First, it was proved in section 3 that the inverse hull of a Markovsubshift satisfies properties (1), (2), and (3). Next let H satisfy properties(1), (2), and (3). Then L is the language of a Markov subshift by Theorem4.6. We would like to define a map Γ : H → H ( L ) by sending 0 to 0 and α = wa ∗ a · · · a ∗ n a n v ∗ to θ w θ − a θ a · · · θ − a n θ a n θ − v . First observe that this mapis well-defined. Indeed, if α = wa ∗ a · · · a ∗ n a n v ∗ = sb ∗ b · · · b ∗ n b n t ∗ thendom θ w θ − a θ a · · · θ − a n θ a n θ − v = D α = dom θ s θ − b θ b · · · θ − b n θ b n θ − t by Proposition 4.9. Moreover, by Lemma 4.7, for each x ∈ D α ,θ w θ − a θ a · · · θ − a n θ a n θ − v ( x ) = wx = αx = sx = θ s θ − b θ b · · · θ − b n θ b n θ − t ( x ) . Thus the map Γ is well-defined. Suppose that Γ( α ) = Γ( β ). Then D α =dom Γ( α ) = dom Γ( β ) = D β and αx = βx for all x ∈ D α . Since H is rightreductive on L , α = β . Thus Γ is injective. It is clear from the definition of H ( L ) that Γ is surjective.Finally we need to show that Γ is a homomorphism. Let α = wa ∗ a · · · a ∗ n a n v ∗ and β = sb ∗ b · · · b ∗ n b n t ∗ be nonzero. As in Corollary 4.2, we have four cases to consider. We examineone such case; the others are similar. Suppose that v = su for some u ∈ L .Then either αβ = 0 or αβ = wa ∗ a · · · a ∗ n a n ( v ∗ s ) b ∗ b · · · b ∗ n b n t ∗ = wa ∗ a · · · a ∗ n a n ( u ∗ b ∗ b ) · · · b ∗ n b n t ∗ = wa ∗ a · · · a ∗ n a n ( u ∗ b ∗ b ) · · · b ∗ n b n t ∗ ...= wa ∗ a · · · a ∗ n a n ( tu ) ∗ By Lemma 3.1 we the computation of Γ( α )Γ( β ) is the same in H ( L ), andso Γ( α )Γ( β ) = θ w θ a · · · θ − a n θ a n θ − tu N THE INVERSE HULL OF A MARKOV SHIFT 13
In the other 3 cases described in Corollary 4.2, we can similarly show thatΓ is a homomorphism. Thus H ∼ = H ( L ). (cid:3) An application
In this section we consider whether there are two languages associatedwith different Markov shifts that generate isomorphic inverse hulls. Wefind languages L and L such that the associated Markov shifts are notconjugate, yet H ( L ) ∼ = H ( L ). The proof relies on the characterizationgiven in the last section which allows us to choose distinct sets O and O of idempotents satisfying axioms (O1)–(O5) from a single inverse semigroup H . The resulting Markov shifts are easily distinguished by their entropies.Consider the Markov subshift generated by the following transition matrixwith alphabet A = { a, b, c } : T = a b ca b c Let H = H ( L ) be the associated inverse hull, where L is the languageof the shift. Below is the top of the semilattice of H , with O indicated inorange text. θ − a θ a θ − b θ b θ − c θ c θ a θ − a θ c θ − c θ b θ − b θ a θ − b θ b θ − a θ a θ − c θ c θ − a θ aa θ − aa θ ac θ − ac θ ab θ − ab θ cb θ − cb θ ba θ − ba θ bc θ − bc Next, let O = { θ a θ − a , θ cb θ − cb , θ − c θ c } , the set of underlined idempotentsin the above figure. We can verify that ( H, O ) satisfies the axioms ofTheorem 4.10. The elements of O are mutually orthogonal. For example, θ a θ − a θ − c θ c ≤ θ a θ − a θ b θ − b = 0. For (O2), the fact that every idempotentin H is comparable to at least one element in O can be used to verifythat the same property holds for O . For (O3), we must show that both( O ↑ − O ) ∪ { } and O ↑ ∪ { } are closed under multiplication. For( O ↑ − O ) ∪ { } = { θ − b θ b , θ − a θ a , θ c θ − c , } it suffices to check the three products: θ − a θ a θ − b θ b = θ − b θ b , θ − a θ a θ c θ − c = θ c θ − c , and θ − b θ b θ c θ − c = θ c θ − c . Similarly, O ↑ ∪{ } is closed under multiplication. Next, we check property(O4). Notice that θ − a θ a lies above the set { θ a θ − a , θ cb θ − cb , θ − c θ c } , θ − b θ b liesabove { θ a θ − a , θ cb θ − cb } , and θ c θ − c lies above { θ cb θ − cb } . Since these three setsare distinct, (O4) is satisfied. We know that θ − a θ a , θ − b θ b and θ − c θ c are indistinct D -classes by Proposition 3.6. Since θ c θ − c D θ − c θ c , we see that (O5)is satisfied. The last property we must check is that the language L definedby O generates H as an inverse semigroup. For this it suffices to recoverthe generators θ a , θ b , θ c as products involving letters in A = { θ a , θ cb , θ − c } and their inverses. Notice that θ b = θ − c θ cb and θ c = ( θ − c ) − .Thus by Thereom 4.10 we have H = H ( L ) ∼ = H ( L ). We may recoverthe Markov transition matrix for L using the method given just beforeTheorem 4.6. For convenience let, x = θ a , y = θ cb , and z = θ − c . We findthe matrix to be T = x y zx y z The dominant eigenvalues of T and T are 2 and approximately 2 . a b ca b c (cid:20) x yx y (cid:21) which are shown to define conjugate shifts in Example 7.2.2 of [2].One can verify by inspection that the semilattices of the associated inversehulls are distinct: θ − c θ c θ − x θ x = θ − y θ y θ − a θ a θ − b θ b θ x θ − x θ y θ − y θ a θ − a θ b θ − b θ c θ − c θ − xx θ xx θ − xy θ xy θ − yx θ yx θ − yy θ yy In fact, after examining many such examples, we conjecture that isomor-phic inverse hulls of Markov shifts must have associated alphabets that arethe same size.
References
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Representations of the inverse hull of a -left cancellativesemigroup , arXiv:1802.06281, February 2018. N THE INVERSE HULL OF A MARKOV SHIFT 15
2. Douglas Lind and Brian Marcus,
An introduction to symbolic dynamics and coding ,Cambridge University Press, Cambridge, 1995.3. Charles Starling,
Inverse semigroups associated to subshifts , Journal of Algebra (2016), 211–233.
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