aa r X i v : . [ m a t h . OA ] J u l A QUANTUM METRIC ON THE CANTOR SPACE
KONRAD AGUILAR AND ALEJANDRA LÓPEZ
Abstract.
The first author and Latrémolière had introduced a quantum met-ric (in the sense of Rieffel) on the algebra of complex-valued continuous func-tions on the Cantor space. We show that this quantum metric is distinctfrom the quantum metric induced by a classical metric on the Cantor space.We accomplish this by showing that the seminorms induced by each quan-tum metric (Lip-norms) are distinct on a dense subalgebra of the algebra ofcomplex-valued continuous functions on the Cantor space. In the process, wedevelop formulas for each Lip-norm on this dense subalgebra and show theseLip-norms agree on a Hamel basis of this subalgebra. Then, we use theseformulas to find families of elements for which these Lip-norms disagree.
Contents
1. Introduction and Background 12. Formulas for L d C and L λ T C
73. Separating L d C and L λ T C Introduction and Background
The study of compact quantum metric spaces was introduced by Rieffel [9, 11]to establish metric convergence of certain noncommutative algebras. This metricconvergence also serves as noncommutative analogue to the Gromov-Hausdorff dis-tance [3] which provides metric convergence of sequences of compact metric spaces[12]. This was motivated by a desire to formalize convergence of certain noncom-mutative algebras introduced in the high-energy physics literature [12]. Anotheraspect of this theory produced a way to study finite-dimensional approximations ofinfinite dimensional algebras using this strong form of metric convergence. There-fore, although this theory was developed for noncommutative algebras, the pursuitof metric finite-dimensional approximations meant that this theory could be ofinterest to study finite-dimensional approximations of commutative algebras.Producing metric finite-dimensional approximations for some commutative alge-bras was one of the consequences of the work of the first author and Latrémolièrein [2]. The commutative algebra they focused on was the algebra of complex-valuedcontinuous functions on the Cantor space (where the Cantor space is viewed as se-quences of ’s and ’s), denoted C ( C ) . They accomplished these finite-dimensional Date : July 15, 2019.2010
Mathematics Subject Classification.
Primary: 46L89, 46L30, 58B34.
Key words and phrases.
Noncommutative metric geometry, Quantum Metric Spaces, Lip-norms, C*-algebras, Cantor space. approximations by placing a quantum metric on C ( C ) using the group structure ofthe Cantor space since the Cantor space is a compact group [2, Theorem 3.5 andSection 7]. However, the Cantor space is also a compact metric space [13, Theorem30.5], and therefore has a classical quantum metric on it induced by the Lipschitzconstant associated to the metric on the Cantor space. Now, each of these quantummetrics is induced by a seminorm, called a Lip-norm (Lip is short for Lipschitz),on C ( C ) . So, the natural question is whether these Lip-norms are the same. It isnot too difficult to place distinct Lip-norms on a given commutative or noncom-mutative space, but it was also shown in [2] that these two quantum metrics arestrongly related to each other in [2, Corollary 7.6] (we also state this relation inTheorem 1.17). This relation is strong enough to provide an equivalence to whenthese Lip-norms are same on a dense subspace [10, Theorem 8.1]. Thus, it is notentirely trivial to establish a difference in these Lip-norms, which is a main accom-plishment of this paper. Hence, our work suggests that it was important for [2]to introduce this new quantum metric to achieve their finite-dimensional approx-imations. We show that these Lip-norms are distinct by introducing formulas forthem on an infinite-dimensional dense subalgebra of C ( C ) . We also show that theseLip-norms agree on a Hamel basis for this dense subalgebra, which makes it evenmore surprising when we do find a family of elements where they disagree.For the rest of this section, we provide sufficient background for the other twosections. In Section 2, we show that these Lip-norms agree on a Hamel basis fora dense subalgebra (Theorem 2.3) and also provide formulas for both Lip-normsbuilt from the structure of this dense subalgebra (Theorem 3.1 and Theorem 2.7).In Section 3, using the results from the previous section, we provide more explicitformulas for these Lip-norms on a certain finite-dimensional subspace (Theorem 3.1and Theorem 3.2) and use these formulas to separate these Lip-norms. Finally, wefind a comparison of these Lip-norms on this finite-dimensional subspace.Now, we begin the background. We start with necessary definitions to define acompact quantum metric space. A compact quantum metric space is a certain kindof algebra called a C*-algebra with a special type of seminorm defined on it. Thus,we define algebras now. Definitions (1.1—1.7) are contained in [5, Chapter I].
Definition 1.1. An associative algebra over the complex numbers C is a vectorspace A over C with an associative multiplication, denoted by concatenation, suchthat: a ( b + c ) = ab + ac and ( b + c ) a = ba + ca for all a, b, c ∈ Aλ ( ab ) = ( λa ) b = a ( λb ) for all a, b ∈ A, λ ∈ C . In other words, the associative multiplication is a bilinear map from A × A to A .We denote the zero of a vector space by A . We say that A is unital if there exists a multiplicative identity, denoted by A .That is: A a = a = a A for all a ∈ A. Convention 1.2.
All algebras are associative algebras over the complex number C . Notation 1.3.
When E is a normed vector space, then its norm will be denotedby k · k E by default. QUANTUM METRIC ON THE CANTOR SPACE 3
Definition 1.4. A normed algebra is an algebra A with a norm k · k A such that: k ab k A k a k A k b k A for all a, b ∈ A.A is a
Banach Algebra when A is complete with respect to the norm k · k A . Definition 1.5. A C*-algebra , A , is a Banach algebra such that there exists ananti-multiplicative conjugate linear involution ∗ : A −→ A , called the adjoint . Thatis, * satisfies:(1) (conjugate linear): ( λ ( a + b )) ∗ = λ ( a ∗ + b ∗ ) for all λ ∈ C , a, b ∈ A ; (2) (involution): ( a ∗ ) ∗ = a for all a ∈ A ; (3) (anti-multiplicative): ( ab ) ∗ = b ∗ a ∗ for all a, b ∈ A. Furthermore, the norm, multiplication, and adjoint together satisfy the identity:(1.1) k aa ∗ k A = k a k A for all a ∈ A called the C*-identity .We say that B ⊆ A is a C*-subalgebra of A if B is a norm closed subalgebra thatis also self-adjoint, i.e. a ∈ B ⇐⇒ a ∗ ∈ B .Our main example will be the C*-algebra of complex-valued continuous functionson a compact metric space, which we define now. Example . Let ( X, d X ) be a compact metric space. Define C ( X ) = { f : X → C | f is continuous } This is a C*-algebra under pointwise algebraic operations including pointwise com-plex conjugation as the involution. That is, if f ∈ C ( X ) , then f ∗ = f , which isdefined for all x ∈ X , by f ( x ) = f ( x ) . The C*-norm is given for all f ∈ C ( X ) by k f k C ( X ) = k f k ∞ = sup x ∈ X {| f ( x ) |} . The unit is the constant function denoted C ( X ) defind for all x ∈ X by C ( X ) ( x ) = 1 . In order to define compact quantum metric spaces we need to define anotherstructure associated to C*-algebras.
Definition 1.7.
Let A be a unital C*-algebra. Let A ′ denote the set of continuousand linear complex-valued functions on A . The state space of A is the set S ( A ) = { ϕ ∈ A ′ | ϕ (1 A ) = k ϕ k A ′ } , where k ϕ k A ′ = sup {| ϕ ( a ) | : a ∈ A, k a k A = 1 } is the operator norm.We also need the notion of a seminorm, which we will allow to take value ∞ inthis article. Definition 1.8.
Let V be a vector space over C . A seminorm s on A is a function s : A → [0 , ∞ ] such that(1) s (0 V ) = 0 ,(2) (homogeneity) s ( λa ) = | λ | s ( a ) for all λ ∈ C , a ∈ A, KONRAD AGUILAR AND ALEJANDRA LÓPEZ (3) (triangle inequality) s ( a + b ) s ( a ) + s ( b ) for all a, b ∈ A. Now, we define compact quantum metric spaces.
Definition 1.9 ([9, 10, 11, 7]) . A compact quantum metric space ( A, L ) is an orderedpair where A is a unital C*-algebra with unit A and L is a seminorm on A suchthat dom( L ) = { a ∈ A | L ( a ) < ∞} is dense in A , and:(1) L ( a ) = L ( a ∗ ) for all a ∈ A, (2) { a ∈ A | L ( a ) = 0 } = C A ,(3) the seminorm L is lower semi-continuous on A with respect to k · k A , and(4) the Monge-Kantorovich metric defined, for all two states ϕ, ψ ∈ S ( A ) , by mk L ( ϕ, ψ ) = sup {| ϕ ( a ) − ψ ( a ) | | a ∈ A, L ( a ) } is a metric on S ( A ) such that there exists D > such that(a) diam( S ( A ) , mk L ) D , and(b) the set { a ∈ A | L ( a ) , k a k A D } is compact in A .If ( A, L ) is a compact quantum metric space, then we call the seminorm L a Lip-norm .Furthermore, if there exists C > such that L ( ab ) C ( L ( a ) k b k A + L ( b ) k a k A ) for all a, b ∈ A , then we call L , C -quasi-Leibniz and we call ( A, L ) a C -quasi-Leibniz compact quantum metric space. If C = 1 , then We call L , Leibniz .One of the main examples of a quantum metric is the following and is due toKantorovich, although Kantorovich did not call such an object a quantum metric.First, some definitions, the first of which allows us to show that this quantum metricof Kantorovich recaptures the classical metric.
Definition 1.10.
Let ( X, d X ) is a compact metric space. Let x ∈ X. We definethe
Dirac point mass at x to be the function δ x : f ∈ C ( X ) f ( x ) ∈ C , where δ x ∈ S ( C ( X )) [4, Theorem VII.8.7].We note that the set of all Dirac point masses on C ( X ) is the set of certain kindsof states called pure states (see [4, Theorem VII.8.7] and [8, Theorem 5.1.6]), butwe do not need to study this fact deeper in this article. Now, we define a mainLip-norm for this paper, which shows that quantum metrics can recover classicalmetrics. Definition 1.11.
Let ( X, d X ) be a compact metric space. The Lipschitz seminorm on C ( X ) is defined for all f ∈ C ( X ) by L d X ( f ) = sup x,y ∈ X,x = y (cid:26) | f ( x ) − f ( y ) | d X ( x, y ) (cid:27) . Theorem 1.12 ([6, 9]) . If ( X, d X ) be a compact metric space, then ( C ( X ) , L d X ) is a Leibniz compact quantum metric space.Moreover, for all x, y ∈ X, it holds that d X ( x, y ) = mk L d X ( δ x , δ y ) . In this paper we will consider the particular compact metric space given by theCantor space, which we define now.
QUANTUM METRIC ON THE CANTOR SPACE 5
Convention 1.13.
The natural numbers N contain throughout this article. Definition 1.14 ([13, Corollary 30.5]) . The
Cantor space is the set of sequencesof ’s and ’s denoted by C = { ( x n ) n ∈ N ∈ N N | ∀ n ∈ N , x n ∈ { , }} . The Cantor space is a compact metric space when equipped with the metric definedfor all x = ( x n ) n ∈ N , y = ( y n ) n ∈ N ∈ C by d C ( x, y ) = ( : if x = y − min { m ∈ N | x m = y m } : otherwise . The main subset that we will compare the Lipschitz seminorm L d C on C ( C ) andthe Lip-norm from [2] is a certain dense subalgebra of C ( C ) . Thus, we now introducenotation for this subalgebra and list many facts from [2] that are important for ourwork, and are needed for defining the Lip-norm on C ( C ) from [2]. Notation 1.15 ([2, Section 7]) . Let n ∈ N . We denote the n th coordinate evalua-tion map on C by η n : ( z m ) m ∈ N ∈ C 7−→ z n ∈ C Also, define u n = 2 η n − C ( C ) where C ( C ) is the constant function on C .We note that η n , u n ∈ C ( C ) η n = η n and that the complex conjugate functions η n = η n and u n = u n , and u n = 1 C ( C ) . Furthermore, k η n k ∞ = k u n k ∞ = 1 , and η n ( C ) = { , } and u n ( C ) = {− , } .Next, set B = ∅ , and for each n ∈ N \ { } set B n = Y j ∈ F u j | ∅ 6 = F ⊆ { , . . . , n − } . Next, for each n ∈ N , set B ′ n = { C ( C ) } ∪ B n and note that | B ′ n | = 2 n since C ( C ) iscommutative and by the relations satisfied by the u n ’s where | B ′ n | is the cardinalityof B ′ n (see [1, Notation 2.1.77] and [2, Lemma 7.4]). Now, set A n = span B ′ n , which is a unital C*-subalgebra of C ( C ) by [2, Lemma 7.4], where in [2, Section7] the notation A n was used instead of A n . Also, note ∪ n ∈ N A n is an dense unitalinfinite-dimensional *-subalgebra of C ( C ) and the set Y j ∈ F u j | ∅ 6 = F ⊂ N , F is finite is a Hamel basis of ∪ n ∈ N A n by [2, Lemma 7.4] and its proof.Now, that we have introduced the appropriate algebraic properties of C ( C ) forour work, we introduce the analytical properties needed to build the Lip-norm from[2]. In particular, we use a state λ ∈ S ( C ( C )) to build this Lip-norm. However, wenote that λ is built from the algebraic structure of C viewed as a compact groupsince λ is induced by the Haar measure on this compact group (see [1, Lemma3.1.14] and [2, Section 7]). KONRAD AGUILAR AND ALEJANDRA LÓPEZ
Lemma 1.16 ([2, Section 7]) . The state λ ∈ S ( C ( C )) of [2, Notation 7.3] satisfies(1) λ (cid:16)Q j ∈ F η j (cid:17) = 2 −| F | where | F | represents the cardinality(2) λ (cid:16)Q j ∈ F u j (cid:17) = 0 for all ∅ 6 = F ⊂ N finite by [2, Lemma 7.4] and its proof.Furthermore, for each n ∈ N , the continuous linear function E n : C ( C ) → A n of [2, Theorem 3.5] associated to λ satisfies(1) E n ( f ) = P a ∈ B ′ n λ ( f a ) a for all f ∈ C ( C ) by [2, Expression (4.1) and Lemma7.4] ,(2) k E n ( f ) k ∞ k f k ∞ for all f ∈ C ( C ) by [2, Definition 3.1] (3) E n ( C ( C )) = A n by [2, Theorem 3.5] (4) E n ( a ) = a, for all a ∈ A l such that l ∈ { , ..., n } (5) E n (1 C ( C ) ) = 1 C ( C ) by (4)(6) E n ( abc ) = aE n ( b ) c for all a, c ∈ A l , where l ∈ { , ..., n } by [2, Definition3.1] (7) If k ∈ { , ..., n } , then(a) If n = 0 , then E ( u k ) = 0 , ∀ k ∈ N by proof of [2, Theorem 7.5] (b) If n > , then(i) If k ∈ { , ..., n − } E n ( u k ) = u k by (4)(ii) If k ∈ { n, n + 1 , ... } E n ( u k ) = 0 by proof of [2, Theorem 7.5] . Finally, we define the Lip-norm L λ T C on C ( C ) from [2] that we will compare with L d C . Theorem 1.17 ([2, Theorem 3.5 and Corollary 7.6]) . If we define L λ T C ( f ) = sup n ∈ N (cid:26) k f − E n ( f ) k ∞ − ( n +1) (cid:27) for all f ∈ C ( C ) , then ( C ( C ) , L λ T C ) is a -Leibniz compact quantum metric space.Moreover,(1) ∪ n ∈ N A n ⊆ dom( L λ T C ) ,(2) for all n ∈ N \ { } , it holds that L λ T C ( f ) = max k ∈{ ,...,n − } (cid:26) k f − E k ( f ) k ∞ − ( k +1) (cid:27) for all f ∈ A n , and(3) for all x, y ∈ C , it holds that mk L λ TC ( δ x , δ y ) = d C ( x, y ) = mk L d C ( δ x , δ y ) . QUANTUM METRIC ON THE CANTOR SPACE 7
It is the last expression that suggests that L λ T C and L d C could agree on a densesubspace of C ( C ) . Indeed, this expression serves as a main assumption of [10, The-orem 8.1] that provides an equivalence for this agreement. Thus, it is a main goalof this paper to show that L λ T C and L d C disagree and we separate them on ∪ n ∈ N A n .2. Formulas for L d C and L λ T C Now, we will provide the main tools we use to separate L λ T C and L d C on ∪ n ∈ N A n .We do this by providing formulas for L λ T C and L d C on each A n . Also, we show that L λ T C and L d C agree on a Hamel basis for ∪ n ∈ N A n , which provides further evidencethat L λ T C and L d C could agree, but they do not as seen in Section 3. Our first task isto show that L d C and L λ T C are even comparable. We already know that ∪ n ∈ N A n ⊆ dom( L λ T C ) by Theorem 1.17, so we will now show that ∪ n ∈ N A n ⊆ dom( L d C ) . Theorem 2.1.
It holds that ∪ n ∈ N A n ⊆ dom( L d C ) . In particular, for all n ∈ N , wehave L d C ( u n ) = 2 n +1 < ∞ . Proof.
Let n ∈ N . Fix x, y ∈ C such that x = y . Then, | u n ( x ) − u n ( y ) | = | η n ( x ) − − η n ( y ) + 1 | = 2 | η n ( x ) − η n ( y ) | . Since x = y , we know that there exists a smallest k ∈ N such that x k = y k .Therefore, d C ( x, y ) = 2 − k and | x k − y k | = 1 since x k , y k ∈ { , } and x k = y k .First, assume k ∈ { , ..., n } . If k = n , | u n ( x ) − u n ( y ) | d C ( x, y ) = 2 | η n ( x ) − η n ( y ) | − k = 2 | η k ( x ) − η k ( y ) | − k = 2 | x k − y k | − k = 2 · − k = 22 − n = 2 · n If k > n , then x n = y n since k is the first coordinate where x and y disagree. Hence, u n ( x ) = u n ( y ) by definition of u n , and thus | u n ( x ) − u n ( y ) | d C ( x, y ) = | u n ( x ) − u n ( y ) | − k = 02 − k = 0 . Thus, if n = 0 , then we would be done. Next, assume n > . The remaining caseis: k < n . Thus, k < n = ⇒ − k < − n and therefore | u n ( x ) − u n ( y ) | d C ( x, y ) = 2 | η n ( x ) − η n ( y ) | − k < | η n ( x ) − η n ( y ) | − n · − n since | η n ( x ) − η n ( y ) | = | x n − y n | for any n ∈ N as x n , y n ∈ { , } . Hence, | u n ( x ) − u n ( y ) | d c ( x, y ) − n for all x, y ∈ C , x = y and there exists x, y ∈ C such that | u n ( x ) − u n ( y ) | d c ( x,y ) = − n sincewe may just choose x, y ∈ C such that the first coordinate they disagree at is n . KONRAD AGUILAR AND ALEJANDRA LÓPEZ
Thus, L d C ( u n ) = sup x,y ∈C ,x = y (cid:26) | u n ( x ) − u n ( y ) | d C ( x, y ) (cid:27) = 2 · n = 2 n +1 < ∞ . Now, by the Leibniz rule and induction, we have that L d C ( f ) < ∞ , where f is anyfinite product of u n ’s. Thus, since L d C is a seminorm and by induction, we havethat L d C ( f ) < ∞ , where f is any finite linear combination of finite products of u n ’s.Therefore L d C ( f ) < ∞ for all f ∈ ∪ n ∈ N A n , by definition of the A n ’s. (cid:3) Thus, we are now free to compare L d C and L λ T C on ∪ n ∈ N A n . Now, by the proofof [2, Theorem 7.5], we have that L λ T C ( u n ) = 2 n +1 = L d C ( u n ) for all n ∈ N . It turnsout that we can do much more and show that L λ T C and L d C agree on all the elementsof the Hamel basis of ∪ n ∈ N A n given in Notation 1.15. The proof of Theorem 2.3follows a similar process as the proof of Theorem 2.1, but requires some differenttechniques that are crucial and acknowledge deeper structure. Thus, we first provea lemma about the algebraic structure of the E n ’s which extends (7) of Lemma1.16 to finite products of u n ’s. Lemma 2.2.
Let k ∈ N . If z ∈ N \ { } and j , . . . , j z ∈ N such that j < · · · < j z , then E k ( u j · · · u j z ) = ( u j · · · u j z : if j z < k C ( C ) : if j z > k. Proof.
First, if j z < k , then u j · · · u j z ∈ A k , and thus E k ( u j · · · u j z ) = u j · · · u j z by (4) of Lemma 1.16.Next, assume j z > k . Let a ∈ B ′ k , then if a = 1 C ( C ) , we have λ ( u j · · · u j z a ) = λ ( u j · · · u j z ) = 0 by the first (2) of Lemma 1.16. Next, by definition of B ′ k ,if a = u n · · · u n p , then n , . . . , n p ∈ N , n < · · · < n p < k j z . Set F = { j , . . . , j z }△{ n , . . . , n p } , where △ denotes symmetric difference. By assumption,we have that F = ∅ since j z ∈ F as j z = u n q for all q ∈ { , . . . , p } . Also, since u n = 1 C ( C ) for all n ∈ N by Notation 1.15, we have that u j · · · u j z a = Y m ∈ F u m . Therefore, as F = ∅ , we have that λ ( u j · · · u j z a ) = λ Y m ∈ F u m ! = 0 by the first (2) of Lemma 1.16. This exhausts all elements in B ′ k . Thus, we have E k ( u j · · · u j z ) = X a ∈ B ′ k λ ( u j · · · u j z a ) a = X a ∈ B ′ k · a = 0 C ( C ) by the second (1) of Lemma 1.16. (cid:3) Theorem 2.3.
For each n ∈ N , it holds that L d C ( u n ) = 2 n +1 = L λ T C ( u n ) . And, for each z ∈ N \ { } , j , . . . , j z ∈ N such that j < · · · < j z , it holds that L d C ( u j · · · u j z ) = 2 j z +1 = L λ T C ( u j · · · u j z ) . QUANTUM METRIC ON THE CANTOR SPACE 9
Proof.
The first equality is provided by Theorem 2.1 and the proof of [2, Theorem7.5].Next, let z ∈ N \ { } , j , . . . , j z ∈ N such that j < · · · < j z .We will first consider L d C ( u j · · · u j z ) . Let x, y ∈ C such that x = y. Thus, theresmallest k ∈ N such that x k = y k , and thus d C ( x, y ) = 2 − k . Case 1.
Assume that j z < k . Then j , . . . , j z < k . Hence, x j = y j , . . . , x j z = y j z since k is the first coordinatewhere x and y disagree. Thus, u j ( x ) = u j ( y ) , . . . , u j z ( x ) = u j z ( y ) by definition of u j , . . . , u j z . Therefore, | u j · · · u j z ( x ) − u j · · · u j z ( y ) | d C ( x, y ) = | u j ( x ) · · · u j z ( x ) − u j ( y ) · · · u j z ( y ) | d C ( x, y ) = 0 . Case 2.
Assume that k = j z . Hence, x j = y j , . . . , x j z − = y j z − , x j z = y j z since k = j z is the first coordi-nate where x and y disagree. Thus, u j ( x ) = u j ( y ) , . . . , u j z − ( x ) = u j z − ( y ) and u j z ( x ) = u j z ( y ) , and thus u j z ( x ) = − u j z ( y ) by definition of u j , . . . , u j z and thefact that the range of these functions is {− , } . Therefore, we have | u j · · · u j z ( x ) − u j · · · u j z ( y ) | d C ( x, y )= | u j ( x ) · · · u j z − ( x ) u j z ( x ) − u j ( y ) · · · u j z − ( y ) u j z ( y ) | − j z = | u j ( x ) · · · u j z − ( x ) u j z ( x ) − u j ( x ) · · · u j z − ( x )( − u j z ( x )) | − j z = | u j ( x ) · · · u j z − ( x ) u j z ( x ) + u j ( x ) · · · u j z − ( x ) u j z ( x ) | − j z = 2 | u j ( x ) · · · u j z − ( x ) u j z ( x ) | − j z = 2 · − j z = 2 j z +1 since u j ( x ) · · · u j z − ( x ) u j z ( x ) ∈ {− , } . Such x, y ∈ C exist and thus j z +1 doesexist in n | u j ··· u jz ( x ) − u j ··· u jz ( y ) | d C ( x,y ) | x, y ∈ C , x = y o . Case 3.
Assume that k < j z . Then, we have k < j z , which implies − k < − jz . So, | u j · · · u j z ( x ) − u j · · · u j z ( y ) | d C ( x, y ) = | u j · · · u j z ( x ) − u j · · · u j z ( y ) | − k < | u j · · · u j z ( x ) − u j · · · u j z ( y ) | − j z | u j · · · u j z ( x ) | + | u j · · · u j z ( y ) | − j z = 1 + 12 − j z = 2 j z +1 . Therefore, L d C ( u j · · · u j z ) = sup x,y ∈C ,x = y (cid:26) | u j · · · u j z ( x ) − u j · · · u j z ( y ) | d C ( x, y ) (cid:27) = 2 j z +1 . Next, we consider L λ T C ( u j · · · u j z ) . Now, u j · · · u j z ∈ A j z +1 , and thus by Theo-rem 1.17, we have that L λ T C ( u j · · · u j z ) = max k ∈{ ,...,j z } (cid:26) k u j · · · u j z − E k ( u j · · · u j z ) k ∞ − ( k +1) (cid:27) . By Lemma 2.2, we have E j z ( u j · · · u j z ) = 0 C ( C ) and E k ( u j · · · u j z ) = u j · · · u j z for all k ∈ { , . . . , j z − } . Hence L λ T C ( u j · · · u j z ) = k u j · · · u j z − C ( C ) k ∞ − ( j z +1) = 12 − ( j z +1) = 2 j z +1 since u j · · · u j z ( C ) = {− , } and the definition of the norm k · k ∞ . (cid:3) From this, we can see that L d C and L λ T C agree on A and A . Corollary 2.4. If f ∈ A , then f = α C ( C ) + α u for some α , α ∈ C and L λ T C ( f ) = 2 | α | = L d C ( f ) , and thus L λ T C = L d C on A and A .Proof. Let f ∈ A , then by definition of A and B ′ there exist α , α ∈ C suchthat f = α C ( C ) + α u .Now, since L λ T C ( α C ( C ) ) = 0 by Definition 1.9, we have by Theorem 2.3 andsince L λ T C is a seminorm | α | = | α | · = | α | L λ T C ( u ) = L λ T C ( α u )= L λ T C ( α C ( C ) + α u − α C ( C ) ) L λ T C ( α C ( C ) + α u ) + L λ T C ( α C ( C ) )= L λ T C ( f )= L λ T C ( α C ( C ) + α u ) L λ T C ( α C ( C ) ) + L λ T C ( α u ) = L λ T C ( α u )= 2 | α | . Thus L λ T C ( f ) = 2 | α | . The same argument shows this is true for L d C ( f ) by Definition1.9 and Theorem 2.3 and the fact that L d C is a seminorm. Thus they agree on A and on A since A ⊆ A . (cid:3) Thus far, we have been able to find formulas for the elements of the Hamelbasis of Notation 1.15, but now, we will develop formulas for L λ T C and L d C on A n for all n > that are built using the basis elements. We note that Corollary 2.4already provided a formula for L λ T C and L d C on A and A . This will use some of themachinery developed in the proof of Theorem 2.3 along with the following technicallemma that will help us better understand the behavior of the difference quotientsin the definition of L d C . Lemma 2.5.
Let z ∈ N \ { } and j , . . . , j z ∈ N such that j < · · · < j z . Let k ∈ N such that k < j z . Assume x, y ∈ C such that d C ( x, y ) = 2 − k .(1) u j z ( x ) − u j z ( y ) = 0 if and only if u k u j z ( x ) − u k u j z ( y ) ∈ {− , } , and u j z ( x ) − u j z ( y ) ∈ {− , } if and only if u k u j z ( x ) − u k u j z ( y ) = 0 . (2) If k = j m for some m ∈ { , . . . , z − } , then QUANTUM METRIC ON THE CANTOR SPACE 11 (a) ( Q zl =0 u j l ) ( x ) − ( Q zl =0 u j l ) ( y ) = 0 if and only if z Y l =0 ,l = m u j l ( x ) − z Y l =0 ,l = m u j l ( y ) ∈ {− , } , and(b) ( Q zl =0 u j l ) ( x ) − ( Q zl =0 u j l ) ( y ) ∈ {− , } if and only if z Y l =0 ,l = m u j l ( x ) − z Y l =0 ,l = m u j l ( y ) = 0 . (3) If j m < k < j m +1 j z for some m ∈ { , . . . , z − } , then(a) ( Q zl =0 u j l ) ( x ) − ( Q zl =0 u j l ) ( y ) = 0 if and only if u k z Y l =0 u j l ! ( x ) − u k z Y l =0 u j l ! ( y ) ∈ {− , } , and(b) ( Q zl =0 u j l ) ( x ) − ( Q zl =0 u j l ) ( y ) ∈ {− , } if and only if u k z Y l =0 u j l ! ( x ) − u k z Y l =0 u j l ! ( y ) = 0 . Moreover, if we set C k = { a ∈ ∪ n ∈ N B n | a = u p , p ∈ N , p > k or a = u j · · · u j z , j < · · · < j z ∈ N , j z > k } , and set C ′ k = { a ∈ C k | a ( x ) − a ( y ) ∈ {− , }} and C ′′ k = C k \ C ′ k = { a ∈ C k | a ( x ) − a ( y ) = 0 } , then the map ∆ k : a ∈ C ′ k u k a ∈ C ′′ k is a well-defined bijection.Proof. (1) By definition of the u ′ n s in Notation 1.15, we have that u j z ( x ) − u j z ( y ) ∈{− , , } and u k u j z ( x ) − u k u j z ( y ) ∈ {− , , } . Now assume that u j z ( x ) − u j z ( y ) =0 , then u j z ( x ) + u j z ( y ) = 0 , which implies that u j z ( x ) + u j z ( y ) ∈ {− , } since u j z ( x ) , u j z ( y ) ∈ {− , } . Now, u k ( x ) = − u k ( y ) by Case 2 of Theorem 2.3. Hence u k u j z ( x ) − u k u j z ( y ) = u k ( x ) u j z ( x ) + u k ( x ) u j z ( y )= u k ( x )( u j z ( x ) + u j z ( y )) ∈ {− , } . The other direction is similar. And, the other if and only if is simply the negationof the first since the values considered are only {− , , } . (2) If k = j m , then by the same argument as Case 2 of Theorem 2.3, we havethat u j ( x ) = u j ( y ) , . . . , u j m − ( x ) = u j m − ( y ) , u j m ( x ) = − u j m ( y ) . Hence r = u j · · · u j z ( x ) − u j · · · u j z ( y )= u j · · · u j m − ( x )( u j m ( x ) · · · u j z ( x ) − u j m ( y ) · · · u j z ( y ))= u j · · · u j m − ( x )( u j m ( x ) · · · u j z ( x ) + u j m ( x ) · · · u j z ( y ))= u j · · · u j m ( x )( u j m +1 ( x ) · · · u j z ( x ) + u j m +1 ( y ) · · · u j z ( y )) and r = u j · · · u j m − u j m +1 · · · u j z ( x ) − u j · · · u j m − u j m +1 · · · u j z ( y )= u j · · · u j m − ( x )( u j m +1 ( x ) · · · u j z ( x ) − u j m +1 ( y ) · · · u j z ( y )) Now, we note that u j · · · u j m ( x ) , u j · · · u j m − ( x ) ∈ {− , } , and thus not zero.Again by definition of the u n ’s, we have u j m +1 ( x ) · · · u j z ( x ) + u j m +1 ( y ) · · · u j z ( y ) ∈{− , , } and u j m +1 ( x ) · · · u j z ( x ) − u j m +1 ( y ) · · · u j z ( y ) ∈ {− , , } .Thus, if r = 0 , then we have u j m +1 ( x ) · · · u j z ( x ) + u j m +1 ( y ) · · · u j z ( y ) = 0 , andthus u j m +1 ( x ) · · · u j z ( x ) − u j m +1 ( y ) · · · u j z ( y ) = 0 since u j m +1 ( x ) · · · u j z ( x ) = 0 and u j m +1 ( y ) · · · u j z ( y ) = 0 , which implies that r = 0 since u j · · · u j m − ( x ) = 0 , andthus r ∈ {− , } . Next, assume we have r ∈ {− , } . Set v = u j m +1 ( x ) · · · u j z ( x ) ∈ {− , } and w = u j m +1 ( y ) · · · u j z ( y ) ∈ {− , } . First, consider r = − . Then, v − w ∈ {− , } since u j · · · u j m − ( x ) = 0 . If v − w = − , then v + w = − w + w = 2( w − . If w = − , then v + w = − , which is a contradiction since v, w ∈ {− , } . Hence w = 1 , and thus v + w = 0 , which implies r = 0 . Now, if v − w = 2 , then v + w = 2 + w + w = 2( w + 1) . If w = 1 , then v + w = 4 , which is a contradictionsince v, w ∈ {− , } . Hence w = − , and thus v + w = 0 , which implies that r = 0 . The case when r = 2 is the same proof and provides r = 0 as well. Hence, if r ∈ {− , } , then r = 0 . This concludes (2)(a).(2)(b) This is simply the negation of (2)(a) since the values considered are only {− , , } . (3) This is the same argument as (2).Now, we establish the bijection at the end of the theorem. Note that C ′′ k = { a ∈ C k | a ( x ) − a ( y ) = 0 } since a ( x ) − a ( y ) ∈ {− , , } by previous arguments. First,we show well-defined. Let a ∈ C ′ k . Now, a ( x ) − a ( y ) must be of the form givenin (1), (2), (3). If u k is not part of the product forming a , then a ( x ) − a ( y ) fallsunder the second line of (1) or (3)(b), and in either case, we have u k a ∈ C ′′ k . Now,if u k is part of the product forming a , then a ( x ) − a ( y ) falls under (2)(b). Hence a = u j · · · u k · · · u j z = u j · · · u j m · · · u j z . Thus, by Notation 1.15, we have(2.1) u k a = u j · · · u k · · · u j z = u j · · · C ( C ) · · · u j z = u j · · · u j m − u j m +1 · · · u j z . Thus, by (2)(b), we have that u k a ( x ) − u k a ( y ) = 0 , which implies that u k a ∈ C ′′ k . Hence, the map ∆ k is well-defined.Now, let’s establish surjectivity. Let b ∈ C ′′ k . Thus b ( x ) − b ( y ) = 0 . If u k is notpart of the product forming b , then b ( x ) − b ( y ) falls under the first line of (1) or(3)(a). Hence, u k b ∈ C ′ k in either case, and ∆ k ( u k b ) = u k ( u k b ) = u k b = b. Next,if u k is part of the product forming b , then b ( x ) − b ( y ) falls under (2)(a), and thesame argument of Expression (2.1) shows that u k b ∈ C ′ k , and thus, ∆ k ( u k b ) = b asin the previous case. Thus ∆ k is a surjection.Next, injectivity. Let a, a ′ ∈ C ′ k and ∆ k ( a ) = ∆ k ( a ′ ) . Then u k a = u k a ′ . Thus u k ( u k a ) = u k ( u k a ) implies u k a = u k a ′ implies a = a ′ . (cid:3) We note that we only consider formulas for elements in A n that are linear com-binations of elements in B n rather than B ′ n since both L λ T C and L d C are not affectedby C ( C ) by the argument in Corollary 2.4. Theorem 2.6.
Let n ∈ N , n > . Let f ∈ A n such that f = P a ∈ B n α a a , where α a ∈ C for all a ∈ B n . QUANTUM METRIC ON THE CANTOR SPACE 13
Next, define C n = { x ∈ C | ∀ k > n, x k = 0 } , which has n elements. Let x, y ∈ C n and denote k x,y = − log d C ( x, y ) . Define σ x,y = { a ∈ B n | a ( x ) − a ( y ) = 0 } . We then have L d C ( f ) = max x,y ∈C n ,x = y k x,y +1 (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) X a ∈ σ x,y ± a,x,y α a (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) , where ± a,x,y is the sign of a ( x ) − a ( y ) and we note that { k x,y | x, y ∈ C n } = { , . . . , n − } and the cardinality of σ x,y is | σ x,y | = 2 n − .Proof. By definition of A n , we have that f ( x ) = f ( y ) for all x, y ∈ C such that d C ( x, y ) > n. Hence, we need only consider x, y ∈ C n with x = y . Also, thecardinality |C n | = | Powerset( { , . . . , n − } ) | = 2 n . For ease of notation in the restof the proof, set k = k x,y , and we note that k ∈ { , . . . , n − } by definition of C n . If k = 0 , define Z = ∅ and if k > , define Z k = { a ∈ B n | a = u p , p ∈ { , . . . , k − } or a = u j · · · u j z , j < · · · < j z ∈ N , j z < k } . Note by Case 1 of Theorem 2.3, we have that a ( x ) − a ( y ) = 0 for all a ∈ Z k . Since C ( C ) is commutative, we have that the cardinality | Z k | = | Powerset( { , . . . , k − } ) | − k − . Next, define S k = { a ∈ B n | a = u p , p = k or a = u j · · · u j z , j < · · · < j z ∈ N , j z = k } We note Z k ∩ S k = ∅ . Also, S k = { u k } ∪ { a ∈ B n | a = u j · · · u j z , j < · · · < j z ∈ N , j z = k } , where the two sets are disjoint. Thus, similarly to Z k , the cardinality | S k | = |{ u k }| + Powerset( { , . . . , k − } ) | − k − k By Case 2 of Theorem 2.3, we have that a ( x ) − a ( y ) = 0 and thus a ( x ) − a ( y ) ∈{− , } since a ( x ) , a ( y ) ∈ {− , } by definition of the u n ’s.Next, set C k = B n \ ( Z k ∪ S k ) . By disjoint sets, we have the cardinality, | C k | = | B n | − ( | Z k | + | S k | ) = 2 n − − (2 k − k ) = 2 n − k +1 . Also, note C k = { a ∈ B n | a = u p , p ∈ N , p > k or a = u j · · · u j z , j < · · · < j z ∈ N , j z > k } . Now, define C ′ k = { a ∈ C k | a ( x ) − a ( y ) ∈ {− , }} . and C ′′ k = C k \ C ′ k = { a ∈ C k | a ( x ) − a ( y ) = 0 } of Lemma 2.5. Also, by Lemma2.5, we have a bijection between C ′ k and C ′′ k . Hence, the cardinality | C ′ k | = | C ′′ k | . Therefore, since all sets considered are finite as B n is finite, we have | C ′ k | = | C k | −| C ′′ k | = | C k | − | C ′ k | , which implies | C ′ k | = | C k | which implies | C ′ k | = 12 | C k | = 12 (2 n − k +1 ) = 2 n − − k . Now, we have that S k ∪ C ′ k = { a ∈ B n | a ( x ) − a ( y ) = 0 } = { a ∈ B n | a ( x ) − a ( y ) ∈ {− , }} and are disjoint by construction. So, we set σ x,y = S k ∪ C ′ k and thus have, the cardinality | σ x,y | = | S k | + | C ′ k | = 2 k + 2 n − − k = 2 n − . Now f ( x ) − f ( y ) = X a ∈ B n α a ( a ( x ) − a ( y )) = X a ∈ σ x,y α a ( a ( x ) − a ( y ))= X a ∈ σ x,y α a ( a ( x ) − a ( y )) = X a ∈ σ x,y α a ( ± a,x,y X a ∈ σ x,y ± a,x,y α a , where ± a,x,y is the sign of a ( x ) − a ( y ) ∈ {− , } . Hence, | f ( x ) − f ( y ) | d C ( x, y ) = 2 (cid:12)(cid:12)(cid:12)P a ∈ σ x,y ± a,x,y α a (cid:12)(cid:12)(cid:12) − k x,y = 2 k x,y +1 (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) X a ∈ σ x,y ± a,x,y α a (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) . Therefore, as C n is finite, the proof is complete. (cid:3) Next, we find a similar formula for L λ T C , which will reveal some important andcrucial differences between the behavior of L λ T C and L d C . Theorem 2.7.
Let n ∈ N , n > . Let f ∈ A n such that f = P a ∈ B n α a a , where α a ∈ C for all a ∈ B n .Next, define C n = { x ∈ C | ∀ k > n, x k = 0 } , which has n elements. For each k ∈ { , . . . , n − } , set ρ k = { a ∈ B n | a − E k ( a ) = 0 C } = { a ∈ B n | E k ( a ) = 0 C ( C ) } . We then have L λ T C ( f ) = max k ∈{ ,...,n − } ( max x ∈C n ( k +1 (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) X a ∈ ρ k ± a,x α a (cid:12)(cid:12)(cid:12)(cid:12)(cid:12))) , where ± a,x is the sign of a ( x ) and we note that the cardinality of ρ k is | ρ k | = 2 n − k .Proof. The cardinality of C n was already determined in Theorem 2.6. Let k ∈{ , . . . , n − } . Since f ∈ A n and E k ( f ) ∈ A k ⊆ A n for all k ∈ { , . . . , n − } , wehave that ( f − E k ( f ))( x ) for x ∈ C is only determined by the values x , . . . , x n − by definition of the u n ’s. Hence, k f − E k ( f ) k ∞ = sup x ∈C | ( f − E k ( f ))( x ) | = max x ∈C n | ( f − E k ( f ))( x ) | . QUANTUM METRIC ON THE CANTOR SPACE 15
Set Z k = { a ∈ B n | a − E k ( a ) = 0 C ( C ) } . Note that C ( C ) B n and thus if a ∈ B n , then a = u j · · · u j z for j < · · · < j z n − ∈ N or a = u p for some p ∈ { , . . . , n − } . Thus E ( u p ) = 0 C ( C ) by (7)(a)of Lemma 1.16 and E ( u j · · · u j z ) = 0 C ( C ) by Lemma 2.2. In either case we havethat a − E ( a ) = a = 0 C ( C ) } . Therefore Z = ∅ . Next, assume that k > . By a similar argument, we have by Lemma 2.2 and(7)(b)(i) of Lemma 1.16 that Z k = { a ∈ B n | a = u p , p ∈ { , . . . , k − } or a = u j · · · u j z , j < · · · < j z ∈ N , j z < k } . By the proof of Theorem 2.6, we have that the cardinality | Z k | = 2 k − . Now, set ρ k = B n \ Z k = { a ∈ B n | a − E k ( a ) = 0 C ( C ) } = { a ∈ B n | E k ( a ) = 0 C ( C ) } since E k ( a ) ∈ { a, C ( C ) } by Lemma 2.2 and (7) of Lemma 1.16.Next, the cardinality | ρ k | = | B n | − | Z k | = 2 n − − (2 k −
1) = 2 n − k . Next, if x ∈ C n , we have by linearity of E k that f − E k ( f ) = X a ∈ B n α a ( a − E k ( a )) = X a ∈ ρ k α a ( a − E k ( a )) = X a ∈ ρ k α a a. Thus, by the beginning of the proof we have k f − E k ( f ) k ∞ = max x ∈C n | ( f − E k ( f ))( x ) | = max x ∈C n (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) X a ∈ ρ k α a a ! ( x ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = max x ∈C n (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) X a ∈ ρ k α a a ( x ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = max x ∈C n (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) X a ∈ ρ k ± a,x α a (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) since a ( x ) ∈ {− , } by Notation 1.15, and ± a,x is the sign of a ( x ) ∈ {− , } .Hence by Theorem 1.17, we have L λ T C ( f ) = max k ∈{ ,...,n − } (cid:26) k f − E k ( f ) k ∞ − ( k +1) (cid:27) = max k ∈{ ,...,n − } ( max x ∈C n ( k +1 (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) X a ∈ ρ k ± a,x α a (cid:12)(cid:12)(cid:12)(cid:12)(cid:12))) , which completes the proof. (cid:3) Separating L d C and L λ T C Theorems 2.6 and Theorem 2.7 provide us with a general idea of the structureof these Lip-norms with respect to the structure of the dense subalgebra ∪ n ∈ N A n . However, these formulas also gift insight into the differences between L d C and L λ T C .In particular, the cardinality between σ x,y and ρ k . The cardinality of σ x,y ondepends on the dimension of the algebra A n even though the set σ x,y is built fromthe first coordinate x and y disagree. However, the cardinality of ρ k depends on the dimension of A n and the dimension of the space E k projects onto, A k . Therefore, L λ T C captures more information from the coefficients of the element f being enteredinto the Lip-norm than L d C . We will see that this happens at A in comparingTheorem 3.1 and Theorem 3.2, where only pairs of coefficients are consider in theformula for L d C whereas pairs and triples of coefficients are consider in L λ T C . So, thehope is that we can separate L d C and L λ T C already on A without having to go tohigher dimension. In Theorem 3.3, we accomplish this and provide many elementsthat separate L λ T C and L d C on A , but first, we take a closer look at our formulas on A . As seen in the proof of Corollary 2.4, we do not need to consider C ( C ) in ourcalculations since the seminorms L d C and L λ T C vanish on C ( C ) . Theorem 3.1.
Let f ∈ A such that f = α u + α u + α u u for some α , α , α ∈ C . It holds that L d C ( f ) = max (cid:26) | α − α | , | α − α | , | α + α | , | α + α | , | α − α | , | α + α | (cid:27) . Proof.
Note B = { u , u , u u } by Notation 1.15. By Theorem 2.6, we have that L d C ( f ) = max x,y ∈C ,x = y k x,y +1 (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) X a ∈ σ x,y ± a,x,y α a (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) , where we set α u = α , α u = α , α u u = α , and for all x, y ∈ C , x = y , we have σ x,y = { a ∈ B | a ( x ) − a ( y ) = 0 } and ± a,x,y is the sign of a ( x ) − a ( y ) for all a ∈ B , and k x,y = − log d C ( x, y ) . Let x, y ∈ C , x = y. First, assume that k x,y = 0 . Thus x = y . (1) If x = 0 , y = 1 , then u ( x ) = 2 η ( x ) − · x − − , u ( y ) =2 η ( y ) − · y − , and hence u ( x ) − u ( y ) = − .(a) If x = 0 , y = 0 , then u ( x ) − u ( y ) = 1 − and u ( x ) u ( x ) − u ( y ) u ( y ) = − · ( − − · ( −
1) = 2 . Thus, we have k x,y +1 (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) X a ∈ σ x,y ± a,x,y α a (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = 2 | − α + α | = 2 | α − α | . (b) If x = 0 , y = 1 , then u ( x ) − u ( y ) = − − − and u ( x ) u ( x ) − u ( y ) u ( y ) = − · ( − − · − . Thus, we have k x,y +1 (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) X a ∈ σ x,y ± a,x,y α a (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = 2 | − α − α | = 2 | α + α | . (c) If x = 1 , y = 0 , then u ( x ) − u ( y ) = 1 − ( −
1) = 2 and u ( x ) u ( x ) − u ( y ) u ( y ) = − · − · ( −
1) = − . Thus, we have k x,y +1 (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) X a ∈ σ x,y ± a,x,y α a (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = 2 | − α + α | = 2 | α − α | . QUANTUM METRIC ON THE CANTOR SPACE 17 (d) If x = 1 , y = 1 , then u ( x ) − u ( y ) = 1 − (1) = 0 and u ( x ) u ( x ) − u ( y ) u ( y ) = − · − · (1) = − − − . Thus, we have k x,y +1 (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) X a ∈ σ x,y ± a,x,y α a (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = 2 | − α − α | = 2 | α + α | . (2) If x = 1 , y = 0 , then u ( x ) = 1 , u ( y ) = − , and u ( x ) − u ( y ) =1 − ( −
1) = 2 .(a) If x = 0 , y = 0 , then u ( x ) − u ( y ) = − − ( −
1) = 0 and u ( x ) u ( x ) − u ( y ) u ( y ) = 1 · ( − − ( − · ( −
1) = − . Thus, we have k x,y +1 (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) X a ∈ σ x,y ± a,x,y α a (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = 2 | α − α | . (b) If x = 0 , y = 1 , then u ( x ) − u ( y ) = − − − and u ( x ) u ( x ) − u ( y ) u ( y ) = 1 · ( − − ( − · − . Thus, we have k x,y +1 (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) X a ∈ σ x,y ± a,x,y α a (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = 2 | α − α | . (c) If x = 1 , y = 0 , then u ( x ) − u ( y ) = 1 − ( −
1) = 2 and u ( x ) u ( x ) − u ( y ) u ( y ) = 1 · − ( − · ( −
1) = 1 − . Thus, we have k x,y +1 (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) X a ∈ σ x,y ± a,x,y α a (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = 2 | α + α | . (d) If x = 1 , y = 1 , then u ( x ) − u ( y ) = 1 − (1) = 0 and u ( x ) u ( x ) − u ( y ) u ( y ) = 1 · − ( − · (1) = 1 + 1 = 2 . Thus, we have k x,y +1 (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) X a ∈ σ x,y ± a,x,y α a (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = 2 | α + α | . Second, assume that k x,y = 1 , then x = y and thus u ( x ) − u ( y ) = 0 , and x = y . (1) If x = y = 0 , then u ( x ) = − and u ( y ) = − and(a) If x = 0 , y = 1 , then u ( x ) − u ( y ) = − − − , and u ( x ) u ( x ) − u ( y ) u ( y ) = − · ( − − ( − · . Thus, we have k x,y +1 (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) X a ∈ σ x,y ± a,x,y α a (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = 4 | − α + α | = 4 | α − α | . (b) If x = 1 , y = 0 , then u ( x ) − u ( y ) = 1 − ( −
1) = 2 , and u ( x ) u ( x ) − u ( y ) u ( y ) = − · (1) − ( − · ( −
1) = − − − . Thus, we have k x,y +1 (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) X a ∈ σ x,y ± a,x,y α a (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = 4 | α − α | . (2) If x = y = 1 , then u ( x ) = 1 and u ( y ) = 1 and (a) If x = 0 , y = 1 , then u ( x ) − u ( y ) = − − − , and u ( x ) u ( x ) − u ( y ) u ( y ) = 1 · ( − − (1) · − − − . Thus, we have k x,y +1 (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) X a ∈ σ x,y ± a,x,y α a (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = 4 | − α − α | = 4 | α + α | . (b) If x = 1 , y = 0 , then u ( x ) − u ( y ) = 1 − ( −
1) = 2 , and u ( x ) u ( x ) − u ( y ) u ( y ) = 1 · (1) − (1) · ( −
1) = 1 + 1 = 2 . Thus, we have k x,y +1 (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) X a ∈ σ x,y ± a,x,y α a (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = 4 | α + α | . Thus, all cases are finished since k x,y , and the proof is complete. (cid:3) Next, we calculate L λ T C on A . Theorem 3.2.
Let f ∈ A such that f = α u + α u + α u u for some α , α , α ∈ C . It holds that L λ T C ( f ) = max (cid:26) | α + α − α | , | α − α + α | , | α − α − α | , | α + α + α | , | α − α | , | α + α | (cid:27) . Proof.
Note B = { u , u , u u } by Notation 1.15. By Theorem 2.7, we have L λ T C ( f ) = max k ∈{ , } ( max x ∈C ( k +1 (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) X a ∈ ρ k ± a,x α a (cid:12)(cid:12)(cid:12)(cid:12)(cid:12))) , where we set α u = α , α u = α , α u u = α , and ρ k = { a ∈ B | E k ( a ) = 0 C ( C ) } for all k ∈ { , } and ± a,x is the sign of a ( x ) for all a ∈ B , x ∈ C . First, let k = 0 . By Lemma 2.2 and (7) of Lemma 1.16, we have that ρ = { u , u , u u } . Let x ∈ C . (1) If x = x = 0 , then u ( x ) = − , u ( x ) = − , u ( x ) u ( x ) = 1 , and thus k +1 (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) X a ∈ ρ k ± a,x α a (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = 2 | − α − α + α | = 2 | α + α − α | . (2) If x = 0 , x = 1 , then u ( x ) = − , u ( x ) = 1 , u ( x ) u ( x ) = − , and thus k +1 (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) X a ∈ ρ k ± a,x α a (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = 2 | − α + α − α | = 2 | α − α + α | . (3) If x = 1 , x = 0 , then u ( x ) = 1 , u ( x ) = − , u ( x ) u ( x ) = − , and thus k +1 (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) X a ∈ ρ k ± a,x α a (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = 2 | α − α − α | . (4) If x = 1 , x = 1 , then u ( x ) = 1 , u ( x ) = 1 , u ( x ) u ( x ) = 1 , and thus k +1 (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) X a ∈ ρ k ± a,x α a (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = 2 | α + α + α | . Second, let k = 1 . By Lemma 2.2 and (7) of Lemma 1.16, we have ρ = { u , u u } . Let x ∈ C . QUANTUM METRIC ON THE CANTOR SPACE 19 (1) If x = x = 0 , then u ( x ) = − , u ( x ) = − , u ( x ) u ( x ) = 1 , and thus k +1 (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) X a ∈ ρ k ± a,x α a (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = 4 | − α + α | = 4 | α − α | . (2) If x = 0 , x = 1 , then u ( x ) = − , u ( x ) = 1 , u ( x ) u ( x ) = − , and thus k +1 (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) X a ∈ ρ k ± a,x α a (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = 4 | α − α | . (3) If x = 1 , x = 0 , then u ( x ) = 1 , u ( x ) = − , u ( x ) u ( x ) = − , and thus k +1 (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) X a ∈ ρ k ± a,x α a (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = 4 | − α − α | = 4 | α + α | . (4) If x = 1 , x = 1 , then u ( x ) = 1 , u ( x ) = 1 , u ( x ) u ( x ) = 1 , and thus k +1 (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) X a ∈ ρ k ± a,x α a (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = 4 | α + α | , which completes the proof. (cid:3) Therefore, the A case displays how L λ T C seems to be more sensitive to the coeffi-cients by allowing one to vary them to impact the entire quantity rather than onlybeing able to compare coefficients pairwise in L d C . Let’s now use these formulas tofind many elements in A that separate L λ T C and L d C . Theorem 3.3. If f = α u + α u + α u u such that α , α , α ∈ R with α > α + α and α = α > , then L d C ( f ) = max { α + α ) , α } and L λ T C ( f ) = 2( α + 2 α ) = 2( α + 2 α ) and L d C ( f ) < L λ T C ( f ) . In particular, L d C (4 u + u + u u ) = 10 <
12 = L λ T C (4 u + u + u u ) . Proof.
First, we consider L d C . First, α − α > α > , so | α − α | = 2( α − α ) < α + α ) = 2( α + α ) since α > . Similarly, | α − α | = 2( α − α ) < α + α ) . Also, | α + α | = 2( α + α ) and | α + α | = 2( α + α ) = 2( α + α ) . Now, | α − α | = 0 and | α + α | = 8 α . This proves the formula for L d C ( f ) . Second, we consider L λ T C . Now, α + α − α > α + α + α − α = 2 α > .Thus | α + α − α | = 2( α + α − α ) < α + α + α ) since α > . Similarly | α − α + α | = 2( α − α + α ) < α + α + α ) and | α − α − α | = 2( α − α − α ) < α + α + α ) . However α + α + α ) = 2( α + 2 α ) = 2( α + 2 α ) . Now | α − α | = 0 and | α + α | = 8 α . But α + α + α ) > α + α + α + α ) =2(4 α ) = 8 α , which proves the formula for L λ T C ( f ) . Next, α + α ) < α + 2 α ) since α > . And, we already showed that α < α + 2 α ) , which establishes L d C ( f ) < L λ T C ( f ) . Lastly, α = 4 , α = 1 , α =1 satisfy the assumption and completes the proof. (cid:3) Although we showed that L d C and L λ T C separate on A , the fact that A is finite-dimensional still allows us to compare L d C and L λ T C . Indeed, since L d C and L λ T C vanish on the same subspace and A is finite-dimensionl, we have that L d C and L λ T C are equivalent since all norms on finite-dimensional vector spaces are equivalent(see [4, Theorem III.3.1]) and L d C and L λ T C are norms on the quotient space (whichis still finite-dimensional) given by the subspace they vanish on. However, thisresult is about existence and doesn’t provide a way to find explicit constants forequivalence. So, in Theorem 3.6, we find such constants. First, we present somebasic inequaliites. Lemma 3.4. If x, y ∈ C , then | x | max {| x + y | , | x − y |} . Proof.
We have | x | = | x | = | x + x | = | x + y − y + x | | x + y | + | x − y | {| x + y | , | x − y |} , which implies | x | max {| x + y | , | x − y |} . (cid:3) Lemma 3.5. If x, y, z ∈ C , then | x + y − z | max { | x + y | , | x + z | , | x − z |} . Proof.
We have | x + y − z | | x + y | + | z | {| x + y | , | z |} {| x + y | , max {| x + z | , | x − z |}} = 2 max {| x + y | , | x + z | , | x − z |} , where the second to last line is provided by Lemma 3.4. (cid:3) Theorem 3.6. If f ∈ A , then L d C ( f ) L λ T C ( f ) L d C ( f ) . Proof.
We begin with the first inequality. Since L d C (1 C ( C ) ) = L λ T C (1 C ( C ) ) = 0 sincethey are Lip-norms, we only need to consider f ∈ A such that f = α u + α u + α u u for some α , α , α ∈ C by the argument of Corollary 2.4.By Lemma 3.4 and Theorem 3.2, we have | α − α | max { | α − α − α | , | α + α − α |} L λ T C ( f ) , | α − α | max { | α − α − α | , | α − α + α |} L λ T C ( f ) , | α + α | max { | α + α − α | , | α + α + α |} L λ T C ( f ) , | α + α | max { | α + α + α | , | α − α + α |} L λ T C ( f ) , and | α − α | L λ T C ( f ) and | α + α | L λ T C ( f ) . Hence, by Theorem 3.1, each term defining L d C ( f ) is less than or equal to L λ T C ( f ) , and thus the maximum, L d C ( f ) , is less than or equal to L λ T C ( f ) . Therefore, L d C ( f ) L λ T C ( f ) . QUANTUM METRIC ON THE CANTOR SPACE 21
Next, we consider the second inequality. By Lemma 3.5 and Theorem 3.1, wehave | α + α − α | { | α + α | , | α + α | , | α − α |} { L d C ( f ) , L d C ( f ) , L d C ( f ) } = 2 L d C ( f ) , | α − α + α | { | α − α | , | α + α | , | α − α |} L d C ( f ) , | α − α − α | { | α − α | , | α − α | , | α + α |} L d C ( f ) , | α + α + α | { | α + α | , | α + α | , | α − α |} L d C ( f ) , and | α − α | L d C ( f ) and | α + α | L d C ( f ) . Hence, by Theorem 3.1, each term defining L λ T C ( f ) is less than or equal to L d C ( f ) , and thus the maximum, L λ T C ( f ) , is less than or equal to L d C ( f ) . Therefore, L λ T C ( f ) L d C ( f ) , which completes the proof. (cid:3) Thus, we have successfully separated L λ T C and L d C , while also discovering someinteresting structural differences and similarities between the two. One route to gonext would be to see if we can continue finding equivalence constants for higherdimensional spaces than A . We note that we are not even certain if we havefound the tightest constant on the right inequality in Theorem 3.6. It may bethat the tighter number is less than . Another route is to compare the domains dom( L λ T C ) and dom( L d C ) . Yes, Theorem 1.17 and Theorem 2.1 show that ∪ n ∈ N A n ⊆ dom( L λ T C ) ∩ dom( L d C ) , but this doesn’t mean the domains are necessarily equal. Ourformulas for L λ T C and L d C (Theorem 2.7 and Theorem 2.6, respectively) may be thekey to figuring this out. The formula for L λ T C will continue to consider more andmore coefficients as the dimension approaches infinity in comparison to L d C . Thus,it may be the case that value approaches infinity while the other stays bounded,which would establish difference in domains. References [1] K. Aguilar.
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School of Mathematical and Statistical Sciences, Arizona State University, 901S. Palm Walk, Tempe, AZ 85287-1804
E-mail address : [email protected] URL : https://math.la.asu.edu/~kaguilar/ Department of Mathematics, Purdue University, 150 N. University Street, WestLafayette, IN 47907-2067
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