aa r X i v : . [ m a t h . M G ] A ug METRICS ON DOUBLES AS AN INVERSE SEMIGROUP
V. MANUILOV
Abstract.
For a metric space X we study metrics on the two copies of X . We definecomposition of such metrics and show that the equivalence classes of metrics are a semi-group M ( X ). Our main result is that M ( X ) is an inverse semigroup. Therefore, one candefine the C ∗ -algebra of this inverse semigroup, which is not necessarily commutative.If the Gromov–Hausdorff distance between two metric spaces, X and Y , is finite thentheir inverse semigroups M ( X ) and M ( Y ) (and hence their C ∗ -algebras) are isomorphic.We characterize the metrics that are idempotents, and give examples of metric spacesfor which the semigroup M ( X ) (and the corresponding C ∗ -algebra) is commutative. Wealso describe the class of metrics determined by subsets of X in terms of the closures ofthe subsets in the Higson corona of X and the class of invertible metrics. Introduction
Given metric spaces X and Y , a metric d on X ⊔ Y that extends the metrics on X and Y depends only on the values of d ( x, y ), x ∈ X , y ∈ Y , but it may be hard to checkwhich functions d : X × Y → (0 , ∞ ) determine a metric on X ⊔ Y : one has to checkthe triangle inequality too many times. The problem of description of all such extendedmetrics is difficult due to the lack of a nice algebraic structure on the set of metrics. Itwas a surprise for us to discover that in the case Y = X , there is a nice algebraic structureon the set M ( X ) of quasi-isometry classes of extended metrics on the double X ⊔ X : itis an inverse semigroup.Recall that a semigroup S is an inverse semigroup if for any u ∈ S there exists a unique v ∈ S such that u = uvu and v = vuv [1]. Philosophically, inverse semigroups describelocal symmetries in a similar way as groups describe global symmetries, and technically,the construction of the (reduced) group C ∗ -algebra of a group generalizes to that of the(reduced) inverse semigroup C ∗ -algebra [3].Thus, one can associate a new (noncommutative) C ∗ -algebra to any metric space. Inparticular, all quasi-isometry classes of metrics on the double of X are partial isometries.We characterize the metrics that are idempotents in M ( X ) and show that any two idem-potents commute (which proves that M ( X ) is an inverse semigroup). We show that ifthe Gromov–Hausdorff distance between two metric spaces, X and Y , is finite then theirinverse semigroups M ( X ) and M ( Y ) (and hence the corresponding C ∗ -algebras) are iso-morphic. We also describe the class of metrics determined by subsets of X in terms ofthe closures of the subsets in the Higson corona of X and the class of invertible metrics,and give examples of metric spaces for which the semigroup M ( X ) is commutative.Let X = ( X, d X ) be a metric space. Definition 0.1. A double of X is a metric space X × { , } with a metric d such that • the restriction of d on each copy of X in X × { , } equals d X ; • the distance between the two copies of X is non-zero.Let M ( X ) denote the set of all such metrics. The author acknowledges partial support by the RFBR grant No. 18-01-00398.
We identify X with X × { } , and write X ′ for X × { } . Similarly, we write x for ( x, x ′ for ( x, x ∈ X . Note that metrics on a double of X may differ only when twopoints lie in different copies of X . To define a metric d in M ( X ) it suffices to define d ( x, y ′ ) for all x, y ∈ X .Recall that two metrics, d , d , on the double of X are quasi-isometric if there exist α > , β ≥ − α + 1 β d ( x, y ′ ) ≤ d ( x, y ′ ) ≤ α + βd ( x, y ′ )for any x, y ∈ X . We call two metrics, d and d , on the double of X equivalent if theyare quasi-isometric. In this case we write d ∼ d , or [ d ] = [ d ].1. Composition of metrics
The idea to consider metrics on the disjoint union of two spaces as morphisms fromone space to another was suggested in [2].
Lemma 1.1.
Let ( X, d X ) , ( Y, d Y ) and ( Z, d Z ) be metric spaces, let d be a metric on X ⊔ Y , ρ a metric on Y ⊔ Z such that d | X = d X , d | Y = ρ | Y = d Y , ρ | Z = d Z . Then theformula b ( x, z ) = inf y ∈ Y [ d ( x, y ) + ρ ( y, z )] , x ∈ X, z ∈ Z, defines a metric on X ⊔ Z .Proof. Due to symmetry, it suffices to check the triangle inequality for the triangle( x , x , z ), x , x ∈ X , z ∈ Z . Fix ε > y , y ∈ Y satisfy d ( x , y ) + ρ ( y , z ) − b ( x , z ) < ε ; d ( x , y ) + ρ ( y , z ) − b ( x , z ) < ε. Then d X ( x , x ) ≤ d ( x , y ) + ρ ( y , z ) + ρ ( z, y ) + d ( y , x ) ≤ b ( x , z ) + b ( x , z ) + 2 ε ; b ( x , z ) ≤ d ( x , y ) + ρ ( y , z ) ≤ d X ( x , x ) + d ( x , y ) + ρ ( y , z ) ≤ d X ( x , x ) + b ( x , z ) + ε. Taking ε arbitrarily small, we obtain the triangle inequality. (cid:3) We shall denote the metric b by ρ ◦ d , or ρd . Corollary 1.2.
Let ρ, d be metrics on the double of X . Then the formula ρd ( x, z ′ ) = inf y ∈ X [ d ( x, y ′ ) + ρ ( y, z ′ )] , x, z ∈ X, defines the composition of d and ρ on the double of X . Lemma 1.3.
The composition of metrics is associative.Proof.
Obvious. (cid:3)
Lemma 1.4. If d ∼ ˜ d and d ∼ ˜ d then ˜ d ◦ ˜ d ∼ d ◦ d .Proof. Suppose that there exist α ≥ β ≥ d ( x, y ′ ) ≤ α + βd ( x, y ′ ) and˜ d ( x, y ′ ) ≤ α + βd ( x, y ′ ) for any x, y ∈ X . ETRICS ON DOUBLES AS AN INVERSE SEMIGROUP 3
Then˜ d ◦ ˜ d ( x, z ′ ) = inf y ∈ X [ ˜ d ( x, y ′ ) + ˜ d ( y, z ′ )] ≤ inf y ∈ X [ α + βd ( x, y ′ ) + α + βd ( y, z ′ )] ≤ inf y ∈ X [2 α + β ( d ( x, y ′ ) + d ( y, z ′ ))] ≤ α + βd ◦ d ( x, z ′ ) . Lower bound is similar. (cid:3)
Thus, the multiplication is well defined on equivalence classes of metrics on the doubleof X .Denote the set of all equivalence classes of metrics on the double of X by M ( X ) = M ( X ) / ∼ . Then M ( X ) is a semigroup. Example 1.5. If X is discrete of finite diameter then all metrics on the double of X areequivalent, so M ( X ) consists of a single element. Example 1.6.
Define a metric I on the double of X by I ( x, y ′ ) = d X ( x, y ) + 1. (Thetriangle inequality obviously holds.) Note that I ◦ d ∼ d ∼ d ◦ I for any metric on thedouble of X , hence [ I ] is the unit element in the semigroup M ( X ).For a metric d on the double of X define the adjoint metric in M ( X ) d ∗ by d ∗ ( x, y ′ ) = d ( y, x ′ ), x, y ∈ X . Then ∗ is an involution: ( d ∗ ) ∗ = d and ( d ◦ d ) ∗ = d ∗ ◦ d ∗ , and it passesto the equivalence classes, making M ( X ) a semigroup with involution.A metric d on the double of X is selfadjoint if d ∗ ∈ [ d ]. Note that if d is selfadjointthen there exists a metric ˜ d ∈ [ d ] such that ˜ d ∗ = ˜ d . Indeed, we can set ˜ d ( x, y ′ ) = ( d ( x, y ′ ) + d ( y, x ′ )), x, y ∈ X .The following simple statement from [2] is the key observation allowing to see metricsas partial isometries. Proposition 1.7.
The metrics d and d ◦ d ∗ ◦ d are equivalent for any metric d on thedouble of X .Proof. Let x, y ∈ X . On the one hand, taking t = y , s = x , we get( d ◦ d ∗ ◦ d )( x, y ′ ) = inf t,s ∈ X [ d ( x, t ′ ) + d ∗ ( t, s ′ ) + d ( s, y ′ )] ≤ d ( x, y ′ ) . On the other hand, passing to infimum in the triangle inequality, we get( d ◦ d ∗ ◦ d )( x, y ′ ) = inf t,s ∈ X [ d ( x, t ′ ) + d ∗ ( t, s ′ ) + d ( s, y ′ )] ≥ inf t,s ∈ X [ d ( x, t ′ ) + d ( t ′ , s ) + d ( s, y ′ )] ≥ d ( x, y ′ ) . (cid:3) Corollary 1.8. [ d ∗ d ] is a selfadjoint idempotent for any metric d on the double of X . Recall that a semigroup S is regular if for any d ∈ S there is b ∈ S such that d = dbd and b = bdb . Corollary 1.9. M ( X ) is a regular semigroup.Proof. Take b = d ∗ . (cid:3) There are typically a lot of idempotent metrics, i.e., metrics representing idempotents in M ( X ) (see Example 1.10) below. This means that, in general, M ( X ) is not a cancellativesemigroup. Indeed, if d ∼ d then cancellation would imply d ∼ I . V. MANUILOV
Example 1.10.
Let X = Z with the standard metric d ( n, m ) = | n − m | , n, m ∈ Z . Set d ( n, m ′ ) = (cid:26) n + m + 1 , if n, m ≥ | n − m | + 1 , otherwise.Then it is easy to see that d ∗ = d and [ d ◦ d ] = [ d ], while d is not quasi-isometric to I .2. Idempotents
Denote by d ( x, X ′ ) the distance from x ∈ X in the first copy of X to the second copy X ′ of X in the double of X . Theorem 2.1.
Let d ∗ = d be a metric on the double of X . Then [ d ] = [ d ] if and only ifthere exist α ≥ , β ≥ such that − α + β d ( x, x ′ ) ≤ d ( x, X ′ ) for any x ∈ X .Proof. First, suppose that [ d ] = [ d ]. Then there exist α ≥ β ≥ d ( x, x ′ ) ≥ − α + 1 β d ( x, x ′ ) . On the other hand, d ( x, x ′ ) = inf y ∈ X [ d ( x, y ′ ) + d ( y, x ′ )] = inf y ∈ X d ( x, y ′ ) ≤ d ( x, X ′ ) , hence d ( x, X ′ ) ≥ − α + β d ( x, x ′ ).Second, suppose that there exist α ≥ β ≥ d ( x, x ′ ) ≤ α + βd ( x, X ′ ) forany x ∈ X . We need to estimate d ( x, z ′ ) both from below and from above. The estimatefrom above is given by d ( x, z ′ ) = inf y ∈ X [ d ( x, y ′ ) + d ( y, z ′ )] ≤ d ( x, x ′ ) + d ( x, z ′ ) ≤ α + β ( d ( x, X ′ )) + d ( x, z ′ ) ≤ α + βd ( x, z ′ ) + d ( x, z ′ ) = α + ( β + 1) d ( x, z ′ ) . Here we took y = x and used that d ( x, X ′ ) ≤ d ( x, z ′ ) for any z ∈ X .To obtain an estimate from below, note that d ( x, z ′ ) = inf y ∈ X [ d ( x, y ′ ) + d ( y, z ′ )] = inf y ∈ X [ d ( x, y ′ ) + d ( y ′ , z )] ≥ d X ( x, z ) . (2.1)We also have d ( x, z ′ ) ≥ d ( x, X ′ ) + d ( z, X ′ ) ≥ − α + 1 β d ( x, x ′ ) − α + 1 β d ( z, z ′ ) , (2.2)It follows from (2.1) and (2.2) that d ( x, z ′ ) ≥ d X ( x, z ) − α + 12 β ( d ( x, x ′ ) + d ( z, z ′ )) ≥ − α + 12 β ( d X ( x, z ) + d ( x, x ′ ) + d ( z, z ′ )) . (2.3)On the other hand, the triangle inequality shows that d ( x, z ′ ) ≤ d ( x, x ′ ) + d X ( x ′ , z ′ ) + d ( z ′ , z ) = d ( x, x ′ ) + d X ( x, z ) + d ( z, z ′ ) . (2.4)Then (2.3) and (2.4) give d ( x, z ′ ) ≥ − α + 12 β d ( x, z ′ ) . (cid:3) ETRICS ON DOUBLES AS AN INVERSE SEMIGROUP 5
The next result shows that selfadjoint idempotents can be characterized only by thevalues d ( x, x ′ ), x ∈ X .We call two functions, ϕ, ψ : X → [0 , ∞ ), equivalent if there exist α ≥ β ≥ − α + β ψ ( x ) ≤ ϕ ( x ) ≤ α + βψ ( x ) for any x ∈ X . Proposition 2.2.
Let d , ρ be two idempotent metrics on the double of X , ρ ∗ = ρ , d ∗ = d .Then ρ ∼ d if and only if the functions x ρ ( x, x ′ ) and x d ( x, x ′ ) are equivalent.Proof. One direction is trivial, so let us prove the non-trivial one.Since d is a selfadjoint idempotent, there are α ≥ β ≥ d ( x, x ′ ) ≤ α + βd ( x, X ′ ) ≤ α + βd ( x, z ′ ) (2.5)for any x, z ∈ X .If the functions x ρ ( x, x ′ ) and x d ( x, x ′ ) are equivalent then there exist γ > δ ≥ ρ ( x, x ′ ) ≤ γ + δd ( x, x ′ ) ≤ γ + δ ( α + βd ( x, x ′ )) = α ′ + β ′ d ( x, z ′ ) , . (2.6)where α ′ = γ + δα , β ′ = δβ .Using (2.5) and the triangle inequality, we have d X ( x, z ) = d ( x ′ , z ′ ) ≤ d ( x ′ , x ) + d ( x, z ′ ) ≤ α + (1 + β ) d ( x, z ′ ) . (2.7)Using the triangle inequality again, together with (2.6) and (2.7), we have ρ ( x, z ′ ) ≤ ρ ( x, x ′ ) + ρ ( x ′ , z ′ ) = ρ ( x, x ′ ) + d X ( x, z ) ≤ α ′ + β ′ d ( x, z ′ ) + α + (1 + β ) d ( x, z ′ ) = ( α + α ′ ) + (1 + β + β ′ ) d ( x, z ′ ) , i.e. d dominates ρ . Symmetrically, ρ dominates d , hence they are equivalent. (cid:3) It would be interesting to find a characterization of functions on X which can beobtained from metrics on doubles. The next statement shows that selfadjoint idempotentsin M ( X ) commute. Proposition 2.3.
Let d , ρ be two idempotent metrics on the double of X , ρ ∗ = ρ , d ∗ = d .Then dρ ∼ ρd .Proof. By definition, for any ε > y ∈ X such that dρ ( x, z ′ ) = inf y ∈ X [ ρ ( x, y ′ ) + d ( y, z ′ )] ≥ ρ ( x, y ′ ) + d ( y , z ′ ) − ε. (2.8)By Theorem 2.1, there exist α ≥ β ≥ ρ ( x, y ′ ) ≥ − α + 1 β ρ ( y , y ′ ); d ( y , z ′ ) ≥ − α + 1 β d ( y , y ′ ) . Then dρ ( x, z ′ ) + ε ≥ β ( ρ ( y , y ′ ) + d ( y , y ′ )) − α. (2.9)The triangle inequality applied to the right hand side of (2.8) gives dρ ( x, z ′ ) + ε ≥ ρ ( x, y ) − ρ ( y , y ′ ) + d ( z, y ) − d ( y , y ′ ) . (2.10)On the other hand, ρd ( x, z ′ ) ≤ d ( x, y ′ ) + ρ ( z, y ′ ) ≤ d ( x, y ) + d ( y , y ′ ) + ρ ( y , y ′ ) + ρ ( y , z ) . (2.11) V. MANUILOV
Denote d ( x, y ) + ρ ( y , z ) = d X ( x, y ) + d X ( y , z ) by r and d ( y , y ′ ) + ρ ( y , y ′ ) by s .Then (2.9) and (2.10) can be written as dρ ( x, z ′ ) ≥ max (cid:16) β s − α, r − s (cid:17) − ε, and (2.11) can be written as ρd ( x, z ′ ) ≤ r + s. To finish the argument, we need the following statement.
Lemma 2.4.
There exists λ > such that r + s ≤ λ (max( β s − α, r − s ) + 2 α ) for any r, s ≥ .Proof. First, note that max( β s, r − s ) ≤ max( β s − α, r − s ) + 2 α . It remains to showthat r + s ≤ λ max (cid:16) β s, r − s (cid:17) (2.12)for some λ >
1. Set s = t r , t ∈ [0 , ∞ ). Then (2.12) becomes(1 + t ) r ≤ λ max (cid:16) tβ r, (1 − t ) r (cid:17) , or, simply, (1 + t ) ≤ λ max (cid:16) tβ , (1 − t ) (cid:17) (2.13)Taking λ = β (2 + β ), we can provide that (2.13) holds for any t ∈ [0 , ∞ ). (cid:3) Lemma 2.4 implies that ρd ( x, z ′ ) ≤ λdρ ( x, z ′ ) + 2 α + ε . Symmetry implies that ρd and dρ are equivalent. This finishes the proof of Proposition 2.3. (cid:3) Metrics from subsets
Example 3.1.
Let A ⊂ X be a subset. Define a metric d A on the double of X by d A ( x, y ′ ) = inf z ∈ A [ d X ( x, z ) + 1 + d X ( z, y )] . Then d A is selfadjoint, d A ( x, x ′ ) = inf z ∈ A [2 d ( x, z ) + 1] = 2 d ( x, A ) + 1, and d A ( x, X ′ ) = inf y ∈ X,z ∈ A [ d X ( x, z ) + 1 + d X ( z, y )] = inf z ∈ A [ d X ( x, z ) + 1]= d ( x, A ) + 1 , hence [ d A ] is an idempotent.A special case of the above metrics d A , A ⊂ X , is the case A = { x } for a fixed point x ∈ X . It is clear that the equivalence class of the metric e = d { x } does not depend onthe choice of the point x .Let A, B ⊂ X be closed subsets. In this section we establish when [ d A ] = [ d B ] underthe assumption that X is locally compact (and the metric d X is proper).Recall that the Higson compactification hX of a locally compact metric space X is theGelfand dual of the C ∗ -algebra C h ( X ) of bounded continuous functions f on X such thatlim x →∞ Var r ( f )( x ) = 0 for any r >
0, whereVar r ( f )( x ) = sup y ∈ X,d X ( x,y ) ≤ r | f ( x ) − f ( y ) | . The Gelfand dual of the quotient C h ( X ) /C ( X ) is the Higson corona νX = hX \ X . ETRICS ON DOUBLES AS AN INVERSE SEMIGROUP 7
Let J A = { f ∈ C h ( X ) : f | A = 0 } . This is an ideal in C h ( X ). Then the Gelfanddual of C h ( X ) /J A is the closure ¯ A of A in hX , and ¯ A \ A = ¯ B \ B if and only if J A + C ( X ) = J B + C ( X ). Proposition 3.2.
The following are equivalent: (1) [ d A ] = [ d B ] ; (2) there exists C > such that A lies in the C -neighborhood of B and B lies in the C -neighborhood of A ; (3) ¯ A \ A = ¯ B \ B in the Higson corona.Proof. We begin with (1) ⇐⇒ (2). Suppose first that [ d A ] = [ d B ]. Note that d A ( x, x ′ ) =2 d X ( x, A ) + 1. In particular, d A ( x, x ′ ) = 1 when x ∈ A . Then there exists C > d B ( x, x ′ ) = 2 d B ( x, B ) < C for any x ∈ A , in other words, A lies in the C -neighborhoodof B . Similarly, B lies in the C -neighborhood of A (maybe with another C ).Assume now that (2) holds. Then d X ( x, A ) − C ≤ d X ( x, B ) ≤ d X ( x, A ) + C , hencethe functions d A ( x, x ′ ) = 2 d X ( x, A ) + 1 and d B ( x, x ′ ) = 2 d X ( x, B ) + 1 are equivalent. ByProposition 2.2 we are done.Now let us show that (2) ⇐⇒ (3). Let (2) hold, and let f ∈ J A . Let x ∈ X . Let B ⊂ N C ( A ). Let r : X → [0 , ∞ ) and µ : X → [0 , ∞ ) be defined by r ( x ) = d X ( x, x )and by µ ( x ) = d X ( x, A ) respectively. Define the map γ : X → [0 , ∞ ) × [0 , ∞ ) by γ ( x ) = ( r ( x ) , µ ( x )). Let F = γ − ([0 , ∞ ) × [0 , C ]); F = γ − ([0 , ∞ ) × [2 C, ∞ )); D k = γ − ([ k − , k ] × [ C, C ]) . Then X = F ∪ F ∪ ( S ∞ k =1 D k ).For the function f ∈ J A , set f n = sup {| f ( x ) | : x ∈ S ∞ k = n +1 D k } . As S ∞ k =1 D k ⊂ N C ( A )and as f ∈ C h ( X ) satisfies f | A = 0, one has lim n →∞ f n = 0.Let us construct a function g ∈ C h ( X ). Set g | F = 0 and g | F = f | F . Our aim is toextend g to the whole X with the following properties: k g | D n k ≤ f n − ; (3.1) k g | E n k ≤ f n , where E n = γ − ( { n } × [ C, C ]). We construct such g inductively. Suppose that we havealready extended g to F ∪ F ∪ ( S nk =1 D k ). By the Tietze Extension Theorem, extend g to D n +1 , and denote this extension on D n +1 by ˜ g . As k g | E n k ≤ f n and | f ( x ) | ≤ f n forany x ∈ γ − ([ n, n + 1] × { C } ), we have k ˜ g | D n +1 k ≤ f n .As | g ( x ) | ≤ f n +1 for any x ∈ γ − ( { n + 1 } × { C } ), there exists C ∈ [ C, C ] such that | ˜ g ( x ) | ≤ f n +1 for any x ∈ γ − ( { n + 1 } × [ C , C ]). Let ϕ : [ n, n + 1] × [ C, C ] → [0 ,
1] bea continuous function such that ϕ ( r, C ) = ϕ ( n, µ ) = 1 for any r ∈ [ n, n + 1], µ ∈ [ C, C ],and ϕ ( n + 1 , µ ) = 0 for µ ∈ [ C, C ]. Then set g ( x ) = ˜ g ( x ) ϕ ( γ ( x )) for x ∈ D n +1 . Then g is continuous on F ∪ F ∪ D ∪ · · · ∪ D n +1 , k g | D n +1 k ≤ f n and k g | E n +1 k ≤ f n +1 .By construction, g | B = 0, and it follows from (3.1) that f − g ∈ C ( X ), therefore g ∈ J B + C ( X ), i.e., J A ⊂ J B + C ( X ). Symmetrically, J B ⊂ J A + C ( X ).Now, suppose that (3) holds, i.e., J A + C ( X ) = J B + C ( X ). If A does not lie in a C -neighborhood of B for any C then there exists a sequence x n ∈ A , n ∈ N , such thatlim n →∞ d X ( x n , B ) = ∞ . Note that, necessarily, lim n →∞ x n = ∞ .Passing to a subsequence of ( x n ) n ∈ N , it may be arranged that d ( x n , B ) ≥ n and d ( x n , x j ) ≥ n + j for all n, j ∈ N . Let h n ( x ) = (1 − d ( x, x n ) /n ) + be the positive part of1 − d ( x, x n ) /n . This function is Lipschitz with constant 1 /n and supported in the ballof radius n around x n . By assumption, these balls are all disjoint and disjoint from B V. MANUILOV as well. The sum f = P n ∈ N h n belongs to C h ( X ). As f | B = 0, we have f ∈ J B , hence f ∈ J A + C ( X ). On the other hand, as f ( x n ) = 1 for any n ∈ N , hence f / ∈ J A + C ( X )(recall that all x n , n ∈ N , lie in A , and lim n →∞ x n = ∞ ). This contradiction finishes theproof. (cid:3) Note that an arbitrary idempotent metric need not be equivalent to d A for any A .4. Order structure
Let ρ and d be metrics on the double of X . We say that [ ρ ] (cid:22) [ d ] if there exists ametric d ′ ∈ [ d ] such that d ′ ( x, z ′ ) ≤ ρ ( x, z ′ ) for any x, z ∈ X . This gives a partial orderon M ( X ). Lemma 4.1.
Let d , ρ be selfadjoint idempotent metrics on the double of X . Then d (cid:22) ρ if and only if [ ρ ][ d ] = [ d ] .Proof. First, suppose that [ ρd ] = [ d ]. Since both ρ and d are selfadjoint idempotents,there exist α ≥ , β ≥ ρ ( x, X ′ ) ≥ β ρ ( x, x ′ ) − α and d ( x, X ′ ) ≥ β d ( x, x ′ ) − α for any x ∈ X . Then ρ ◦ d ( x, x ′ ) ≥ inf y ∈ X [ d ( x, y ′ ) + ρ ( y, x ′ )] ≥ d ( x, X ′ ) + ρ ( x, X ′ ) ≥ ρ ( x, X ′ ) ≥ β ρ ( x, x ′ ) − α. It follows from [ ρ ◦ d ] = [ d ] that there exist α ′ > , β ′ > ρ ◦ d ( x, x ′ ) ≤ β ′ d ( x, x ′ ) + α ′ . Combining the last two inequalities, we get β ′ d ( x, x ′ ) + α ′ ≥ β ρ ( x, x ′ ) − α ,or d ( x, x ′ ) ≥ ββ ′ ρ ( x, x ′ ) − α ′′ for some α ′′ >
0. Using the proof of Proposition 2.2, weconclude that d (cid:22) ρ .Second, suppose that d (cid:22) ρ . Without loss of generality, we may assume that d ( x, z ′ ) ≥ ρ ( x, z ′ ) for any x, z ∈ X . Then ρ ◦ d ( x, x ′ ) = inf y ∈ X [ d ( x, y ′ ) + ρ ( y, x ′ )] ≤ d ( x, x ′ ) + ρ ( x, x ′ ) ≤ d ( x, x ′ ) . This implies that [ d ] (cid:22) [ ρd ]. On the other hand, ρ ◦ d ( x, x ′ ) ≥ d ( x, X ′ ) + ρ ( x, X ′ ) ≥ d ( x, X ′ ) ≥ β ( d ( x, x ′ ) − α ) , which implies [ ρd ] (cid:22) [ d ]. Thus, [ ρd ] = [ d ]. (cid:3) C ∗ -algebra of M ( X ) Proposition 5.1.
Let a ∈ M ( X ) be an idempotent. Then it is selfadjoint.Proof. Note that a ∗ also must be an idempotent. Then use commutativity of selfadjointidempotents to show that a ∗ = a ∗ aa ∗ = ( a ∗ a )( aa ∗ ) = ( aa ∗ )( a ∗ a ) = aa ∗ a = a. (cid:3) Corollary 5.2.
Any two idempotents in M ( X ) commute. Recall that a semigroup S is an inverse semigroup if for any a ∈ S there exists a unique b ∈ S such that a = aba and b = bab ([1], p. 6). ETRICS ON DOUBLES AS AN INVERSE SEMIGROUP 9
Theorem 5.3. M ( X ) is an inverse semigroup.Proof. In a regular semigroup, commutativity of idempotents is equivalent to being aninverse semigroup ([1], Theorem 3). (cid:3)
By Theorem 5.3, we can define the (reduced) semigroup C ∗ -algebra C ∗ r ( M ( X )) of theinverse semigroup M ( X ) ([3], Section 4.4).Recall that if a ∈ M ( X ), V a = { b ∈ M ( X ) : bb ∗ (cid:22) a ∗ a } , then the map b ab isinjective on V a .Let l ( M ( X )) denote the Hilbert space of square-summable functions on M ( X ) (as adiscrete space) with the orthonormal basis (often uncountable) of delta-functions δ b ( c ) = (cid:26) , if c = b ;0 , if c = b, b, c ∈ M ( X ). For a ∈ M ( X ), set λ a ( δ b ) = (cid:26) δ ab , if b ∈ V a ;0 , if b / ∈ V a . Then λ a is a partial isometry for any a ∈ M ( X ), and the C ∗ -algebra C ∗ r ( M ( X )) generatedby all λ a , a ∈ M ( X ), is the reduced C ∗ -algebra of M ( X ).There is a special projection [ e ] in C ∗ r ( M ( X )), given by the metric d { x } for some x ∈ X . By definition, e ( x, y ′ ) = d X ( x, x ) + 1 + d X ( x , y ), and it is easy to see that theequivalence class [ e ] does not depend on x . Lemma 5.4. d ◦ e and e ◦ d are equivalent to e for any metric d on the double of X .Proof. Take y = x , and then use the triangle inequality to obtain e d ( x, z ′ ) = inf y ∈ X [ d ( x, y ′ ) + d X ( y, x ) + d X ( z, x ) + 1] ≤ d ( x, x ′ ) + d X ( z, x ) + 1 ≤ d X ( x, x ) + d ( x , x ′ ) + d X ( z, x ) + 1 ≤ e ( x, z ′ ) + d ( x , x ′ ) . On the other hand, by the triangle inequality, d ( x, y ′ ) + d X ( y, x ) + d X ( z, x ) + 1 = d ( x, y ′ ) + d X ( y ′ , x ′ ) + d X ( z, x ) + 1 ≥ d ( x, x ′ ) + d X ( z, x ) + 1 . (5.1)Passing to the infimum in (5.1) with respect to y ∈ X , we obtain e d ( x, z ′ ) ≥ d ( x, x ′ ) + d X ( z, x ) + 1. Another triangle inequality gives e d ( x, z ′ ) ≥ d ( x, x ′ ) + d X ( z, x ) + 1 ≥ d X ( x, x ) − d ( x , x ′ ) + d X ( z, x ) + 1= e ( x, z ′ ) − d ( x , x ′ ) . Thus, e ( x, z ′ ) − α ≤ e d ( x, z ′ ) ≤ e ( x, z ′ ) + α for any x, z ∈ X , where α = d ( x , x ′ ),hence e d ∼ e . Similarly, de ∼ e . (cid:3) Thus, [ e ] is the zero element in M ( X ). Proposition 5.5.
The set V e consists of a single element [ e ] .Proof. Let s ∈ V [ e ] . Then ss ∗ (cid:22) [ e ], hence ss ∗ = ss ∗ [ e ] = [ e ]. Then s = ss ∗ s = [ e ] s =[ e ]. (cid:3) Corollary 5.6. λ [ e ] is a rank one projection in C ∗ r ( M ( X )) . Corollary 5.7.
There is a direct sum decomposition of C ∗ -algebras C ∗ r ( M ( X )) = C ∗ ( M ( X )) ⊕ λ e C , where C ∗ ( M ( X )) = { λ a : a ∈ C ∗ r ( M ( X )) , a [ e ] = [ e ] a = 0 } . Let
A, B ⊂ X , let d X ( A, B ) = inf x ∈ A ; y ∈ B d X ( x, y ), and let B R ( x ) denote the ball ofradius R centered at x ∈ X . Proposition 5.8.
Suppose that there exists β ≥ such that d X ( A \ B R ( x ) , B \ B R ( x )) > β R. (5.2) Then [ d A d B ] = [ e ] .Proof. By Proposition 2.2, it suffices to compare d A d B ( x, x ′ ) and e ( x, x ′ ), x ∈ X . Recallthat d A d B ( x, x ′ ) = inf u ∈ A,v ∈ B,y ∈ X [ d X ( x, u ) + d X ( u, y ) + d X ( y, v ) + d X ( v, x ) + 2] ,e ( x, x ′ ) = 2 d X ( x, x ) + 1 . By Proposition 5.5, [ e ] (cid:22) [ d A d B ], so it remains to show that [ d A d B ] (cid:22) [ e ].Set L = β R . Take x ∈ X , and let R satisfy x ∈ B R ( x ) and x / ∈ B R + L ( x ). Then e ( x, x ) = 2 d ( x, x ) + 1 ≤ R + 1 . (5.3)Now let us estimate d A d B ( x, x ′ ). By definition, there exist u ∈ A , v ∈ B , y ∈ X suchthat d A d B ( x, x ′ ) ≥ d X ( x, u ) + d X ( u , y ) + d X ( y , v ) + d X ( v , x ) . Consider the two cases:(a) either u ∈ B R ( x ) or v ∈ B R ( x );(b) u , v / ∈ B R ( x ).In the case (a), if u ∈ B R ( x ) and x / ∈ B R + L ( x ) then d X ( x, u ) ≥ L . Otherwise, if v ∈ B R ( x ) then d X ( v , x ) ≥ L . Thus, d A d B ( x, x ′ ) ≥ L .In the case (b), by the triangle inequality and by (5.2), d A d B ( x, x ′ ) ≥ d X ( x, u ) + d X ( u , y ) + d X ( y , v ) + d X ( v , x ) ≥ d X ( x, u ) + d X ( u , v ) + d X ( v , x ) ≥ d X ( u , v ) ≥ L. Thus, in both cases we have d A d B ( x, x ′ ) ≥ L = 1 β R. (5.4)Combining (5.3) and (5.4), we get e ( x, x ′ ) ≤ βd A d B ( x, x ′ ) + 1, hence [ d A d B ] (cid:22) [ e ]. (cid:3) Corollary 5.9.
Under the assumption of Proposition 5.8, ( λ [ d A ] − λ [ e ] )( λ [ d B ] − λ [ e ] ) = 0 ,i.e., the projections λ [ d A ] − λ [ e ] and λ [ d B ] − λ [ e ] are mutually orthogonal. Example 5.10.
Let X = R with the standard metric. For ϕ ∈ [0 , π ) let A ϕ be theray from the origin with the angle ϕ to the polar axis. If ψ ∈ [0 , π ), ψ = ϕ , thenthe two rays A ϕ and A ψ satisfy the assumption of Proposition 5.8, hence the projections λ [ d Aϕ ] − λ [ e ] and λ [ d Aψ ] − λ [ e ] are mutually orthogonal. Thus, the C ∗ -algebra C ∗ ( M ( X ))has uncountably many mutually orthogonal projections. ETRICS ON DOUBLES AS AN INVERSE SEMIGROUP 11 Examples
Proposition 6.1.
Let X be a closed subset of [0 , ∞ ) with the induced metric. Then any a ∈ M ( X ) is a selfadjoint idempotent. Hence M ( X ) is commutative.Proof. First, let us show that any element of M ( X ) is selfadjoint. Suppose the contrary.Then there exists a metric d on the double of X such that d ∗ is not equivalent to d , andfor any n ∈ N we can find points y n , z n ∈ X such that n · d ( y n , z ′ n ) < d ( y ′ n , z n ) . (6.1)Since d ( X, X ′ ) >
0, the sequence d ( y ′ n , z n ) is unbounded.Passing to a subsequence, if necessary, we may assume without loss of generality that y n < z n for any n ∈ N . Then d X ( x , z n ) = d X ( x , y n ) + d X ( y n , z n ).By the triangle inequality, we have d ( y ′ n , z n ) ≤ d X ( y n , z n ) + d ( y n , z ′ n ) , (6.2)so, (6.1) and (6.2) imply that n · d ( y n , z ′ n ) < d X ( y n , z n ) + d ( y n , z ′ n ) , or, equivalently, d ( y n , z ′ n ) < n − d X ( y n , z n ) . (6.3)Another application of the triangle inequality gives d ( y n , z ′ n ) ≥ d X ( x , z n ) − ( d X ( y n , x ) + d ( x , x ′ )) . Combining this with (6.3), we get d X ( x , z n ) − ( d X ( y n , x ) + d ( x , x ′ )) < n − d X ( y n , z n ) . (6.4)By assumption, d X ( x , z n ) = d X ( x , y n ) + d X ( y n , z n ), so (6.4) implies that d X ( y n , z n ) − d ( x , x ′ ) < n − d X ( y n , z n )holds, hence the values d X ( y n , z n ) are uniformly bounded. Let C satisfy d X ( y n , z n ) < C for any n ∈ N .By the triangle inequality and (6.1), we have n ( d ( y n , y ′ n ) − C ) < n ( d ( y n , y ′ n ) − d X ( y ′ n , z ′ n )) ≤ nd ( y n , z ′ n ) < d ( y ′ n , z n ) ≤ d ( y n , y ′ n ) + d X ( y n , z n ) < d ( y n , y ′ n ) + C, hence d ( y n , y ′ n ) < n +1 n − C , and the values d ( y n , y ′ n ) are uniformly bounded.Thus we get a contradiction: the left-hand side of the triangle inequality d ( y ′ n , z n ) ≤ d ( y n , y ′ n ) + d X ( y n , z n )is unbounded, while both summands in the right-hand side are uniformly bounded.Second, we have to show that any selfadjoint metric d on the double of X is an idem-potent. Suppose the contrary: for any n ∈ N there exist points y n , z n ∈ X such that d ( y n , z ′ n ) < n d ( y n , y ′ n ) . (6.5)Once again, we have two possibilities: either d X ( x , z n ) = d X ( x , y n ) + d X ( y n , z n ) or d X ( x , y n ) = d X ( x , z n ) + d X ( z n , y n ) for infinitely many numbers n ’s, and let us assumethat the first opportunity holds true. Then, by the triangle inequality, we have d X ( z n , x ) − ( d X ( y n , x ) + d ( x , x ′ )) ≤ d ( y n , z ′ n ) , or, equivalently, d X ( y n , z n ) − d ( x , x ′ ) ≤ d ( y n , z ′ n ) , which, together with (6.5), implies that d X ( y n , z n ) ≤ d ( y n , z ′ n ) + d ( x , x ′ ) < n d ( y n , y ′ n ) + d ( x , x ′ ) . (6.6)Another triangle inequality combined with (6.5) and (6.6) gives d ( y n , y ′ n ) ≤ d ( y n , z ′ n ) + d X ( y ′ n , z ′ n ) = d ( y n , z ′ n ) + d X ( y n , z n ) < n d ( y n , y ′ n ) + d ( x , x ′ ) , which holds for infinitely many n ’s. The latter may be true only if d ( y n , y ′ n ) is boundedfor these n ’s, but this contradicts d ( X, X ′ ) >
0. Indeed, if d ( y n , y ′ n ) < C for some C > n ’s then the sequence d ( y n , z ′ n ) is not separated from 0. (cid:3) Example 6.2.
Let X = { ( n, n,
0) : n ∈ N } ∪ { ( n, − n,
0) : n ∈ N } ⊂ R with thestandard metric, and let X ′ = { ( x, − y,
1) : ( x, y, ∈ X } . For a n = ( n, n, ∈ X wehave a ′ n = ( n, − n, ⊂ X ′ , and for b n = ( n, − n, ∈ X we have b ′ n = ( n, n, ∈ X ′ .Let the metric d on the double of X be inherited from the standard metric of R . Then d ( a n , a ′ n ) = √ n + 1, while d ( a n , X ′ ) = d ( a n , b ′ n ) = 1, hence [ d ] is selfadjoint, but notidempotent. Example 6.3.
Let X = R with the standard metric, and let A = [0 , ∞ ), B = ( −∞ , x, y ∈ X , set d ( x, y ′ ) = (cid:26) | x + y | + 1 , if x ∈ A, y ∈ B ; | x | + | y | + 1 , otherwise.Then d ∗ d ( x, y ′ ) = inf z ∈ X d ( x, z ′ ) + d ( y, z ′ ). If x, y ∈ A then one may take z = − x , inthis case d ( x, − x ′ ) + d ( y, − x ′ ) = | x − y | + 2. In other cases one may take z = 0, and d ∗ d ( x, y ′ ) = | x | + | y | + 2. Thus, d ∗ d ( x, y ′ ) = d A ( x ) + 1, hence [ d ∗ d ] = [ d A ]. Similary, wecan see that [ dd ∗ ] = [ d B ]. Thus, [ d ] is a partial isometry from [ d A ] to [ d B ].7. Examples from extended metrics
When X is non-compact, the inverse semigroup M ( X ) is infinite. Here we considerthe case when metrics are replaced by the so-called extended metrics, which are the sameas usual metrics, except that they are allowed to take infinite values. This gives a lot ofexamples with finite M ( X ).Note that setting d ( x, y ′ ) = ∞ for any x, y ∈ X gives the zero element 0 ∈ M ( X ), as d d = 0 for any metric d on the double of X . Example 7.1.
Let X be a one-point space, X = { a } . Any two finite metrics on thedouble of X are equivalent, but an infinite metric with d ( a, a ′ ) = ∞ is not equivalent to afinite metric, so M ( X ) = { I, } . We have V I = { I, } and V = { } . Then the C ∗ -algebraof X is a subalgebra in the algebra M ( C ) of 2 × C ⊕ C . Example 7.2.
Let X be the space consisting of two points, a and b , with d X ( a, b ) = ∞ .Any metric in M ( X ) is determined by the 4 values: d ( a, a ′ ), d ( a, b ′ ), d ( b, a ′ ) and d ( b, b ′ ).Metrics with any finite value are equivalent to those with this value equal to 1, so non-equivalent classes of metrics should take values 1 and ∞ . Taking into account the triangleinequality, there are 7 possible metrics in M ( X ): ETRICS ON DOUBLES AS AN INVERSE SEMIGROUP 13 (1) 0( a, a ′ ) = 0( a, b ′ ) = 0( b, a ′ ) = 0( b, b ′ ) = ∞ ;(2) I ( a, a ′ ) = I ( b, b ′ ) = 1, I ( a, b ′ ) = I ( b, a ′ ) = ∞ ;(3) p ( a, a ′ ) = 1, p ( a, b ′ ) = p ( b, b ′ ) = p ( b, a ′ ) = ∞ ;(4) q ( b, b ′ ) = 1, q ( b, a ′ ) = q ( a, a ′ ) = q ( a, b ′ ) = ∞ ;(5) u ( a, b ′ ) = 1, u ( a, a ′ ) = u ( b, a ′ ) = u ( b, b ′ ) = ∞ ;(6) u ∗ ( b, a ′ ) = 1, u ∗ ( b, b ′ ) = u ∗ ( a, b ′ ) = u ∗ ( a, a ′ ) = ∞ ;(7) s ( a, b ′ ) = s ( b, a ′ ) = 1, s ( a, a ′ ) = s ( b, b ′ ) = ∞ .Note that p, q are idempotents, u and u ∗ are partial isometries, u ∗ u = p , uu ∗ = q , and s is a symmetry. Let L = h δ i , L = h δ p , δ u ∗ i , L = h δ q , δ u i , L = h δ I , δ s i . Then V = L , V p = V u = L ⊕ L , V q = V u ∗ = L ⊕ L , V I = V s = L ⊕ L ⊕ L ⊕ L .We have λ d | L = id for any d ∈ M ( X ), λ u ( L ) = L , λ u ∗ ( L ) = L and λ s | L ⊕ L = λ u | L ⊕ L + λ u ∗ | L ⊕ L , so λ u , λ u ∗ and λ s restricted to the invariant subspace L ⊕ L generatethe algebra isomorphic to M ( C ). Taking into account the invariant subspaces L and L , where M ( X ) acts by scalars, we get C ∗ ( M ( X )) ∼ = C ⊕ C ⊕ M ( C ) ⊂ M ( C ).8. Sufficient condition for an isomorphism M ( X ) ∼ = M ( Y )Given two metric spaces, X and Y , consider all metrics d on the disjoint union X ⊔ Y such that • d | X = d X , d | Y = d Y ; • d ( X, Y ) = 0.Let M ( X, Y ) denote the set of all such metrics.Recall that, given a metric d on X ⊔ Y , the Hausdorff distance between X and Y is d H ( X, Y ) = max(sup x ∈ X d ( x, Y ) , sup y ∈ Y d ( y, X )), and the Gromov–Hausdorff distancebetween X and Y is inf d H ( X, Y ), where the infimum is over all metrics on X ⊔ Y thatequal d X and d Y on X and Y , respectively. Note that the Gromov–Hausdorff distancemay be (and often is) infinite. Lemma 8.1.
The Gromov–Hausdorff distance between X and Y equals inf d ∈M ( X,Y ) d H ( X, Y ) .Proof. If d is a metric on X ⊔ Y that equals d X and d Y on X and Y respectively, and d H ( X, Y ) = 0 then, for any ε >
0, set d ε | X = d X , d ε | Y = d Y , and d ε ( x, y ) = d ( x, y ) + ε for x ∈ X , y ∈ Y . It is clear that d ε is a metric in M ( X, Y ) and d εH ( X, Y ) ≥ ε , so it sufficesto take the infimum over metrics for which the distance between X and Y is non-zero. (cid:3) Proposition 8.2.
Suppose that the Gromov–Hausdorff distance between X and Y is finite.Then M ( X ) and M ( Y ) are isomorphic.Proof. By assumption, there exists ρ ∈ M ( X, Y ) and
C > ρ ( x, Y ) < C and ρ ( y, X ) for any x ∈ X and any y ∈ Y . Then ρ ∗ ρ ∈ M ( X ), and ρ ∗ ρ ( x, x ′ ) =inf z ∈ X ρ ( x, z ′ ) < C for any x ∈ X , hence ρ ∗ ρ ∼ I , where I ∈ M ( X ) is defined inExample 1.6. Similarly, ρρ ∗ ∼ I in M ( Y ).For d ∈ M ( X ), b ∈ M ( Y ), set ϕ ( d ) = ρdρ ∗ ∈ M ( Y ), ψ ( b ) = ρ ∗ bρ . Clearly, ϕ and ψ pass to maps ¯ ϕ : M ( X ) → M ( Y ) and ¯ ψ : M ( Y ) → M ( X ), respectively. These mapsare semigroup homomorphisms, as [ ρ ∗ ρ ] and [ ρρ ∗ ] are the unit elements in M ( X ) and in M ( Y ), respectively.Finally, ψ ◦ ϕ ( d ) = ρ ∗ ρdρ ∗ ρ ∼ d , hence ¯ ψ ◦ ¯ ϕ = id M ( X ) . Similarly, ¯ ϕ ◦ ¯ ψ = id M ( Y ) . (cid:3) Subgroup of invertibles
An element [ d ] ∈ M ( X ) is invertible if [ d ∗ d ] = [ dd ∗ ] = [ I ]. It is clear that the invertibleelements form a group. Here we describe this group. Definition 9.1.
A map f : X → X is an almost isometry if(i1) there exists C > d X ( x, ˜ x ) − C ≤ d Y ( f ( x ) , f (˜ x )) ≤ d X ( x, ˜ x ) + C for any x, ˜ x ∈ X ;(i2) there exist a map g : X → X and D > d X ( g ◦ f ( x ) , x ) < D and d X ( f ◦ g ( x ) , x ) < D for any x ∈ X .Note that if such a map g exists then it automatically satisfies (i1), possibly withdifferent C .Any isometry is patently an almost isometry. Another example of an almost isometryfor X = Γ, where Γ is a finitely generated group with the word-length metric, is providedby conjugation by a fixed element g ∈ Γ.Given an almost isometry f : X → X , set d f ( x, y ′ ) = inf z ∈ X d X ( x, z ) + C + d X ( f ( z ) , y ) , x, y ∈ X. It was shown in [2] that d f is a metric (one has to check four triangle inequalities).If f, g : X → X are almost isometries as in Definition 9.1 then (i2) implies that[ d f d g ] = [ d g d f ] = [ I ]. Proposition 9.2.
Let d ∈ M ( X ) and let [ d ] ∈ M ( X ) be invertible. Then there exists analmost isometry f of X such that [ d f ] = [ d ] .Proof. If [ d ] is invertible then [ d ∗ d ] = [ dd ∗ ] = [ I ], so there exists C > z ∈ X [ d ( x, z ′ ) + d ( z ′ , x )] < C and inf z ∈ X [ d ( x ′ , z ) + d ( z, x ′ )] < C for any x ∈ X . Thereforethere exist u, v ∈ X such that d ( x, u ′ ) < C/ d ( x ′ , v ) < C/ f ( x ) = u , g ( x ) = v . Then d ( x, f ( x ) ′ ) < C/ d ( x ′ , g ( x )) < C/ x ∈ X ,hence d ( f ( x ) ′ , g ( f ( x ))) < C/
2, and, by the triangle inequality, d X ( x, g ◦ f ( x )) ≤ d ( x, f ( x ) ′ ) + d ( f ( x ) ′ , g ( f ( x ))) < C. Similarly one gets d X ( x, f ◦ g ( x )) < C .Let x, ˜ x ∈ X . Then, by the triangle inequality, d X ( f ( x ) , f (˜ x )) ≤ d ( f ( x ) , x ′ ) + d X ( x ′ , ˜ x ′ ) + d (˜ x ′ , f (˜ x )) ≤ d X ( x, ˜ x ) + C and d X ( f ( x ) , f (˜ x )) ≥ − d ( f ( x ) , x ′ ) + d X ( x ′ , ˜ x ′ ) − d (˜ x ′ , f (˜ x )) ≥ d X ( x, ˜ x ) − C, hence f is an almost isometry.It remains to check that [ d f ] = [ d ]. Taking z = x and using the triangle inequality, weget d f ( x, y ′ ) = inf z ∈ X [ d X ( x, z ) + d X ( f ( z ) , y ) + C ] ≤ d X ( f ( x ) , y ) + C = d X ( f ( x ) ′ , y ′ ) + C ≤ d ( f ( x ) ′ , x ) + d ( x, y ′ ) + C ≤ d ( x, y ′ ) + 3 C/ . To prove the estimate from below, note that d X ( f ( z ) , y ) ≥ d X ( g ◦ f ( z ) , g ( y )) − C/ ≥ d X ( z, g ( y )) − d X ( g ◦ f ( z ) , z ) − C/ ≥ d X ( z, g ( y )) − C − C/ d X ( z, g ( y )) − C/ , ETRICS ON DOUBLES AS AN INVERSE SEMIGROUP 15 hence, by the triangle inequality, d f ( x, y ′ ) = inf z ∈ X [ d X ( x, z ) + d X ( f ( z ) , y ) + C ] ≥ inf z ∈ X [ d X ( x, z ) + d X ( z, g ( y ))] − C/ ≥ d X ( x, g ( y )) − C/ ≥ d ( x, y ′ ) − d ( y ′ , g ( y )) − C − C/ ≥ d ( x, y ′ ) − C/ . (cid:3) Coarse version
Two metrics, d , d , on X are coarsely equivalent if there exists a monotonely increasingfunction f on [0 , ∞ ) with lim t →∞ f ( t ) = ∞ such that f − ( d ( x, y )) ≤ d ( x, y ) ≤ f ( d ( x, y ))for any x, y ∈ X .All our results hold also for the coarse equivalence classes of metrics on the double of X . This gives a smaller quotient inverse semigroup M c ( X ) of coarse equivalence classes.We may also use the fact that the image of an inverse semigroup, under a semigrouphomomorphism, is an inverse semigroup. Acknowledgment.
The author expresses his gratitude to the referees for valuablecomments and for suggested nicer versions of certain proofs.
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