aa r X i v : . [ m a t h . M G ] O c t LINEAR ALGEBRA AND UNIFICATION OF GEOMETRIES INALL SCALES
J. DYDAK
October 2, 2019
Contents
1. Introduction 22. Multilinear forms on sets 33. Topology induced by forms 53.1. Large scale topology 64. Boundaries of large scale topological spaces 75. Boundaries of formed sets 96. Normal forms 116.1. Special normal forms 127. Boundaries of normal forms 138. Orthogonality relations 168.1. Bounded sets 168.2. Examples of small scale orthogonality 168.3. Examples of large scale orthogonality 168.4. Hyperbolic orthogonality 178.5. Freundenthal orthogonality 178.6. Normal orthogonality relations 178.7. Proximity spaces 189. ⊥ -continuous functions 189.1. Small Scale Examples 189.2. Large Scale Examples 199.3. Quotient structures 1910. Neighborhood operators 2011. Orthogonality relations vs forms 2111.1. Extending hyperbolic orthogonality relations 2212. Form-continuous functions 2313. Forms vs large scale structures 2414. Higson corona of large scale forms 2815. Compatible forms 2915.1. Gromov compactification 3016. Dimension of formed sets 32 Date : October 2, 2019.2000
Mathematics Subject Classification.
Primary 54F45; Secondary 55M10.
Key words and phrases. asymptotic dimension, CAT(0)-spaces, coarse geometry, coarsely n-to-1 functions, dimension-raising maps, Higson corona, Gromov boundary of hyperbolic spaces,the visual boundary of CAT(0)-spaces, ˇCech-Stone compactification, Samuel-Smirnov compactifi-cation, Freudenthal compactification.
17. Metrizability of form compactifications 3518. Forms on large scale spaces 3619. Asymptotic dimension 0 3620. Group actions on formed sets 3721. Parallelism 3922. Visual forms 3922.1. CoG-spaces 41References 42
Abstract.
We present an idea of unifying small scale (topology, proximityspaces, uniform spaces) and large scale (coarse spaces, large scale spaces). Itrelies on an analog of multilinear forms from Linear Algebra. Each form has alarge scale compactification and those include all well-known compactifications:Higson corona, Gromov boundary of hyperbolic spaces, the visual boundaryof CAT(0)-spaces, ˇCech-Stone compactification, Samuel-Smirnov compactifi-cation, and Freudenthal compactification.As an application we get simple proofs of results generalizing well-knowntheorems from coarse topology. A new result (at least to the author) is thefollowing (see 16.7):
A coarse bornologous function f : X → Y of metrizable large scale spaces is alarge scale equivalence if and only if it induces a homeomorphism of Higsoncoronas. This paper is an extension of [7] and, at the same time, it overrides [7]. Introduction
A topology on a set X is the same as a projection (i.e. an idempotent linearoperator) cl : 2 X → X satisfying A ⊂ cl ( A ) for all A ⊂ X . That’s a good way tosummarize Kuratowski’s closure operator.Basic geometry on a set X is a dot product · : 2 X × X → { , ∞} . Its equivalentform is an orthogonality relation on subsets of X . The optimal case is when theorthogonality relation satisfies a variant of parallel-perpendicular decompositionfrom linear algebra. Dot products are a special case of forms which act on arbitraryvectors based on a given set X .We show that this concept unifies small scale (topology, proximity spaces, uni-form spaces) and large scale (coarse spaces, large scale spaces). Using forms wedefine large scale compactifications that generalize all well-known compactifica-tions: Higson corona, Gromov boundary, the visual boundary of CAT(0)-spaces,ˇCech-Stone compactification, Samuel-Smirnov compactification, and Freudenthalcompactification. This allows to generalize many results in coarse topology fromproper metric spaces to arbitrary metric spaces or even to arbitrary large scalespaces. Example 1.1. (see 20.4) Let X be a metric space and let G be a finite group actingisometrically on X. Then X/G has the same asymptotic dimension as X .In case of proper metric spaces X , Theorem 20.4 was proved by Daniel Kasprowski [16] . Example 1.2. (see 16.5) If n ≥ and f : X → Y is a coarsely n -to- bornologousmap of large scale spaces, then asdim ( Y ) ≤ asdim ( Y ) + n − .Theorem 16.5 was proved by Austin-Virk in [1] for proper metric spaces X and Y . INEAR ALGEBRA AND UNIFICATION OF GEOMETRIES IN ALL SCALES 3 Multilinear forms on sets
Definition 2.1.
The semi-group { , ∞} has the following binary operation:1. 0 + 0 = 0,2. 0 + ∞ = ∞ + 0 = ∞ + ∞ = ∞ .Recall that a bornology B on a set X is any family of subsets closed underfinite unions so that B ⊂ B ′ ∈ B implies B ∈ B . Remark . Be aware that Wikipedia [21] assumes additionally that each point of X belongs to a bornology.Notice that bornologies B on a set X are identical with kernels of basic linearoperators ω : 2 X → { , ∞} , i.e. functions satisfying1. ω ( ∅ ) = 0,2. ω ( C ∪ D ) = ω ( C ) + ω ( D ) for all C, D ∈ X .That observation leads to the following generalization: Definition 2.3. A k -vector in X ( k ≥
1) is a k -tuple ( C , . . . , C k ) of subsets in X . The set of all vectors in X will be denoted by V ect ( X ).The concatenation V ∗ V of a k -vector V = ( C , . . . , C k ) and an m -vector V = ( D , . . . , D m ) is the ( k + m )-vector ( C , . . . , C k , D , . . . , D m ).A basic multilinear form ω on X is a symmetric function on all k -vectors of X ( k ≥
1) to { , ∞} satisfying the following properties:1. ω (( C ∪ C ) ∗ V ) = ω ( C ∗ V ) + ω ( C ∗ V ) for any k -vector V and any two1-vectors C , C ,2. ω ( ∅ ) = 0,3. ω ( C ∗ V ) = ω ( V ) if one of the coordinates of V is contained in C . Remark . One can consider forms with values in any join-semilattice , i.e. apartially ordered set ( L, ≤ ) in which each two-element subset { a, b } ⊂ L has a join(i.e. least upper bound). However, it is not clear what new applications would arisefrom that concept.Notice that { , ∞} has a natural order: namely 0 < ∞ (in fact, it is the smallestnon-trivial lattice). Lemma 2.5. If V and V ′ are two k -vectors such that V ( i ) ⊂ V ′ ( i ) for each ≤ i ≤ k , then ω ( V ) ≤ ω ( V ′ ) .Proof. Apply Axiom 1 repeatedly to see that ω ( V ) is a summand of ω ( V ′ ). (cid:3) Corollary 2.6. ω ( V ) ≤ ω ( C ) if some coordinate of V is contained in C .Proof. If V is a k vector, we may assume V ( k ) ⊂ C . Now, ω ( V ) ≤ ω ( X ∗ . . . ∗ X ∗ C )by 2.5 and the latter equals ω ( C ) by Axiom 3. (cid:3) Proposition 2.7. ω ( V ∗ V ′ ) ≤ ω ( V ) for all vectors V and V ′ .Proof. By 2.5, ω ( V ∗ V ′ ) ≤ ω ( V ∗ X ∗ . . . ∗ X ) and the latter equals ω ( V ) by Axiom3. (cid:3) Definition 2.8. A bornological set ( X, B ) is a set X equipped with a bornology B . A formed set ( X, ω ) is a set X equipped with a basic multilinear form ω . J. DYDAK
Proposition 2.9.
Each formed set ( X, ω ) induces a bornology on X via B ( ω ) := { C ⊂ X | ω ( C ) = 0 } . Conversely, given a bornology B on X , ω ( C , . . . , C k ) = 0 ⇐⇒ C ∩ . . . ∩ C k ∈ B defines a basic multilinear form on X that induces B .Proof. Left to the reader. (cid:3)
Observation 2.10.
Notice that, given a bornology B on X , the form defined aboveis the largest form that induces B . The smallest form inducing B is given by theformula ω ( C , . . . , C k ) = 0 ⇐⇒ C i ∈ B for some i. Example 2.11.
Given a topological space ( X, T ) , the basic topological form ω ( X, T ) is defined as follows: ω ( X, T )( C , . . . , C k ) = 0 ⇐⇒ k \ i =1 cl ( C i ) = ∅ . Example 2.12.
Given a topological space ( X, T ) , the basic functional form ω f ( X, T ) is defined as follows: ω f ( X, T )( C , . . . , C k ) = 0 ⇐⇒ k \ i =1 D i = ∅ for some zero-sets D i ⊃ C i . Recall D is a zero-set in a topological space X if there is a continuous function f : X → [0 , such that D = f − (0) . Example 2.13.
Given a metric space ( X, d ) , the basic small scale form ω s ( X, d ) is defined as follows: ω s ( X, d )( C , . . . , C k ) = 0 ⇐⇒ k \ i =1 B ( C i , r ) = ∅ for some r > . Example 2.14.
Given a metric space ( X, d ) , the basic large scale form ω l ( X, d ) is defined as follows: ω l ( X, d )( C , . . . , C k ) = 0 ⇐⇒ k \ i =1 B ( C i , r ) is boundedfor each r > . Example 2.15.
Given a metric space ( X, d ) , the C -form ω ( X, d ) is defined asfollows: ω ( X, d )( C , . . . , C k ) = 0 ⇐⇒ k \ i =1 B ( C i , r ) is boundedfor some r > . INEAR ALGEBRA AND UNIFICATION OF GEOMETRIES IN ALL SCALES 5
Recall that the star st ( B, U ) of a subset B of X with respect to a family U of subsets of X is the union of those elements of U that intersect B . Given twofamilies B and U of subsets of X , st ( B , U ) is the family { st ( B, U ) } , B ∈ B , of allstars of elements of B with respect to U . Example 2.16.
Given a uniform space ( X, U ) , the form ω ( U ) is defined as follows: ω ( U )( C , . . . , C k ) = 0 ⇐⇒ k \ i =1 st ( C i , V ) = ∅ for some uniform cover V of X . Example 2.17.
Given a topological space ( X, T ) and a subset A of X , the form ω ( X, A, T ) on X \ A is defined as follows: ω ( X, A, T )( C , . . . , C k ) = 0 ⇐⇒ k \ i =1 cl ( C i ) ⊂ X \ A. Here the closures are taken in X via T . Topology induced by forms
Proposition 3.1.
Given a formed set ( X, ω ) , the topology induced by ω on X consists of sets U with the property that ω ( x, X \ U ) = 0 for all x ∈ U .Proof. If x ∈ U ∩ W and ω ( x, X \ U ) = 0 = ω ( x, X \ W ) = 0, then ω ( x, ( X \ U ) ∪ ( X \ W )) = 0 and ( X \ U ) ∪ ( X \ W ) = X \ U ∩ W .If each U s , s ∈ S , is open and x ∈ W := S s ∈ S U s , then x ∈ U t for some t ∈ S , so ω ( x, X \ U t ) = 0. As X \ W ⊂ X \ U t , ω ( x, X \ W ) = 0 by 2.5. (cid:3) Here is a typical construction of open sets in the topology induced by ω : Proposition 3.2. If { C n } ∞ n =1 is an increasing sequence of subsets of a formed set ( X, ω ) such that ω ( C k , X \ C k +1 ) = 0 for each k ≥ , then U := ∞ S i =1 C k is an openset in the topology induced by ω .Proof. Given x ∈ U , there is k ≥ x ∈ C k . Therefore ω ( x, X \ C k +1 ) = 0.Since X \ U ⊂ X \ C k +1 , ω ( x, X \ U ) = 0 by 2.5. (cid:3) Proposition 3.3.
The topology induced by ω is T if and only if ω ( x, y ) = 0 whenever x = y .Proof. If ω ( x, y ) = 0 whenever x = y , then X \ { y } is open. If X \ { y } is open and x = y , then ω ( x, y ) = 0. (cid:3) Definition 3.4.
A form ω on a set X is T if ω ( x, y ) = 0 for all different points x, y ∈ X .Forms cannot induce topologies that are T but not T : Corollary 3.5.
If the topology T induced by a form ω on X is T , then ( X, T ) is T . J. DYDAK
Proof.
Given x = y in X , there is an open set U containing exactly one of the points x , y . Say y ∈ U and x / ∈ U . Therefore ω ( y, X \ U ) = 0 implying ω ( y, x ) = 0. (cid:3) Proposition 3.6. If ( X, T ) is a T topological space, then its basic topological forminduces T .Proof. Let ω be the basic topological form of X . ω ( x, X \ U ) = 0 means cl ( x ) ∩ cl ( X \ U ) = ∅ . Consequently, if U ∈ T ∈ T , then U is open in the topologyinduced by ω . Conversely, if C is closed in the topology induced by ω and x / ∈ C ,then ω ( x, C ) = 0, hence x / ∈ cl ( C ). Thus C = cl ( C ) in T . (cid:3) Of special interest are cases where the topology induced by a form has propertiesrelated to compactness.3.1.
Large scale topology.Definition 3.7. A large scale topological space ( X, T , B ) is a topological space( X, T ) in which a bornology B of open-closed subspaces is selected. Definition 3.8.
A large scale topological space ( X, T , B ) is large scale compact if and only if, for any family { U s } s ∈ S of open subsets of X , X = S s ∈ S U s impliesexistence of a finite subset F of S such that X \ S s ∈ F U s belongs to B . Proposition 3.9.
1) If ( X, T , B ) is large scale compact and topologically Hausdorff,then it is topologically regular.2) If ( X, T , B ) is large scale compact and topologically regular, then it is topologicallynormal.Proof. A is a closed subset of X not containing x . If x is open-closed,then A ⊂ X \ { x } is disjoint from x and we are done. Assume x is not open.For each point x ∈ A choose disjoint open sets U x containing x and V x containing x . Notice X = ( X \ A ) ∪ S x ∈ A U x , so there is an open-closed set B ∈ B and a finitesubset F of A such that X = B ∪ ( X \ A ) ∪ S x ∈ A U x and B does not contain x .Therefore, A ⊂ B ∪ S x ∈ A U x is disjoint from T x ∈ F V x \ B which contains x .The proof of 2) is similar or apply 1) to X/A . (cid:3) Proposition 3.10.
A continuous map f : ( X, T X , B X ) → ( Y, T Y , B Y ) of large scaletopological spaces is closed if the following conditions are satisfied:1. ( X, T X , B X ) is large scale compact.2. ( Y, T Y , B Y ) is Hausdorff.3. f ( B X ) ⊂ B Y .Proof. It suffices to show f ( X ) is closed and then apply it to f | A , A any closedsubset of X .Given y ∈ Y \ f ( X ) that is not open choose, for any z ∈ f ( X ), disjoint openneighborhoods U ( z ) of y and W ( z ) of z . Since { f − ( W ( z )) } z ∈ f ( X ) covers X , thereexist points z , . . . , z k of f ( X ) such that X \ k S i =1 f − ( W ( z i )) ∈ B X . Hence, D := f ( X ) \ k S i =1 W ( z i ) ∈ B Y is open-closed in Y , it does not contain y , and U := k T i =1 U ( z i ) \ D is a neighborhood of y missing f ( X ). (cid:3) INEAR ALGEBRA AND UNIFICATION OF GEOMETRIES IN ALL SCALES 7 Boundaries of large scale topological spaces
In this section we generalize ˇCech-Stone compactification to large scale compact-ifications of large scale topological spaces.
Definition 4.1.
The boundary ∂ ( X, T , B ) of a large scale topological space( X, T , B ) is the collection of all maximal families P in 2 X consisting of zero-setssatisfying the following condition: intersection of any finite subfamily of P is notin B .If { x } / ∈ B , then we can identify x with the family consisting of all zero subsetsof X containing x . This way we can talk about X ∪ ∂ ( X, T , B ). Observation 4.2.
Technically speaking, points x , x ∈ X such that there is nocontinuous function from X to [0 , separating them, are identified in ∂ ( X, T , B ) .Namely, they generate the same element of ∂ ( X, T , B ) consisting of all zero subsetsof X containing x . Thus, to consider X ∪ ∂ ( X, T , B ) in set-theoretic sense, weneed X to be functionally Hausdorff (i.e. every pair of different points of X can beseparated by a real-valued continuous function). We will continue to use notation X ∪ ∂ ( X, T , B ) for the union of a quotient of X and ∂ ( X, T , B ) Remark . In case of a bornological set ( X, B ) we can define its boundary as ∂ ( X, T , B ), where T is the discrete topology on X . Observation 4.4.
Suppose A and A are zero-subsets of ( X, T , B ) . If A ∪ A ∈P ∈ ∂ ( X, T , B ) , then A ∈ P or A ∈ P .Proof. Suppose A ∪ A ∈ P but A / ∈ P and A / ∈ P . In that case there exist C, D ∈ P such that A ∩ C, A ∩ D ∈ B . Consequently, ( A ∪ A ) ∩ C ∩ D ∈ B , acontradiction. (cid:3) Corollary 4.5. If A ∈ P and B ∈ B , then A \ B ∈ P . Definition 4.6. X ∪ ∂ ( X, T , B ) has a natural topology whose basis is formed (inview of 4.7) by sets N ( U ) := U ∪ {Q ∈ ∂ ( X, T , B ) | X \ U / ∈ Q} , where U is a co-zero-subset of X (i.e. the complement of a zero-set in X ). Lemma 4.7. N ( U ) ∩ N ( V ) = N ( U ∩ V ) and N ( U ) ∪ N ( V ) = N ( U ∪ V ) .Proof. Reformulate 4.4 as follows: A ∪ A / ∈ P ∈ ∂ ( X, B ) if and only if A / ∈ P and A / ∈ P . The equality N ( U ) ∩ N ( V ) = N ( U ∩ V ) is now obvious as well as N ( U ) ∪ N ( V ) = N ( U ∪ V ). (cid:3) Corollary 4.8.
Each B ∈ B is open-closed in X ∪ ∂ ( X, T , B ) .Proof. By 4.5, N ( B ) = B and N ( X \ B ) = X \ B . As N ( X ) = N ( B ) ∪ N ( X \ B ),both N ( B ) and N ( X \ B ) are closed. (cid:3) Proposition 4.9.
Suppose C is a zero-subset of X . P ∈ ∂ ( X, T , B ) belongs to theclosure of C ⊂ X if and only if C ∈ P .Proof. If C ∈ P , then every base element N ( U ) containing P intersects C as C ∩ U cannot be empty. Indeed, C ∩ U = ∅ means C ⊂ X \ U resulting in X \ U ∈ P which contradicts P ∈ N ( U ). J. DYDAK
Conversely, if
C / ∈ P , then
P ∈ N ( X \ C ) and C ∩ N ( X \ C ) = ∅ . Indeed, any Q ∈ C ∩ N ( X \ C ) must be equal to all zero subsets containing some x ∈ C . Hence C ∈ Q which contradicts Q ∈ N ( X \ C ). (cid:3) Corollary 4.10.
The basis of X ∪ ∂ ( X, B ) consists of open-closed sets.Proof. In this case the topology on X is discrete and the complement of N ( U ) is N ( X \ U ). (cid:3) Theorem 4.11. X ∪ ∂ ( X, T , B ) is large scale compact and topologically normal.Proof. Suppose { N ( U ) } U ∈S is a cover of X ∪ ∂ ( X, T , B ) such that X ∪ ∂ ( X, T , B ) \ S U ∈F N ( U ) does not belong to B for all finite subfamilies F of S . Consider the family { X \ S U ∈F U } F⊂S , where F runs over all finite subfamilies of S . That family has theproperty that it is closed under finite intersections and none of its elements belongsto B . Therefore it extends to an element P of ∂ ( X, T , B ). Hence, P ∈ N ( U ) forsome U ∈ S . However, X \ U also belongs to P , a contradiction.To prove X ∪ ∂ ( X, T , B ) is topologically normal it suffices to show it is Hausdorff.Clearly, any two points x = x in X have disjoint neighborhoods in X ∪ ∂ ( X, T , B )if x ∈ B . Namely, it is { x } and its complement. If P 6 = Q are points of ∂ ( X, T , B ),then there is C ∈ P \ Q . Hence, there is D ∈ Q such that C ∩ D ∈ B . Notice E := D \ C ∩ D ∈ Q (see 4.5) and there exist disjoint co-zero subsets U and W of X containing C and E , respectively. Observe P ∈ N ( U ), Q ∈ N ( W ), and N ( U ) ∩ N ( W ) = ∅ . (cid:3) Theorem 4.12. If f : ( X, T X , B X ) → ( Y, T Y , B Y ) is a continuous function of largescale topological spaces such that f − ( B Y ) ⊂ B X , then there is a unique continuousextension ˜ f : X ∪ ∂ ( X, T X , B X ) → Y ∪ ∂ ( Y, T Y , B Y ) of f . It is given by ˜ f ( P ) ∈ \ C ∈P cl ( f ( C )) for all P ∈ ∂ ( X, T X , B X ) .Proof. First of all, let us show that T C ∈P cl ( f ( C )) = ∅ for all P , i.e. ˜ f exists. Suppose T C ∈P cl ( f ( C )) = ∅ for some P . Since Y ∪ ∂ ( Y, T Y , B Y ) is large scale compact, thereexist C i ∈ P , 1 ≤ i ≤ k , such that k T i =1 cl ( f ( C i )) ∈ B Y . Let C := k T i =1 C i . Now, C ∈ P and f ( C ) ∈ B Y . Therefore C ∈ B X , a contradiction.Our next step is to show continuity of ˜ f . Suppose ˜ f ( P ) ∈ N ( U ) for some co-zero set U in Y . By a similar argument as above, D ⊂ U for some D ∈ ˜ f ( P ),so we can choose a co-zero set W in Y and a zero-set E in Y satisfying D ⊂ W ⊂ E ⊂ U . Suppose Q ∈ N ( f − ( W )). There is C ∈ Q , C ⊂ f − ( W ). Hence˜ f ( Q ) ∈ cl ( f ( C )) ⊂ cl ( W ) ⊂ cl ( E ) which results in E ∈ ˜ f ( Q ). Now, Y \ U / ∈ ˜ f ( Q ), so ˜ f ( Q ) ∈ N ( U ) which completes the proof of continuity of ˜ f provided P ∈ N ( f − ( W )). It is so as G := X \ f − ( W ) ∈ P implies ˜ f ( P ) ∈ cl ( f ( G )) ⊂ cl ( Y \ W ),i.e. Y \ W ∈ ˜ f ( P ) which contradicts D ∈ ˜ f ( P ).The uniqueness of ˜ f follows from the fact the range is Hausdorff (see 4.11) and X is dense in X ∪ ∂ ( X, T X , B X ). (cid:3) INEAR ALGEBRA AND UNIFICATION OF GEOMETRIES IN ALL SCALES 9
Observation 4.13.
Proposition 4.14. If ( X, T X , B X ) is large scale compact, then for every P ∈ ∂ ( X, T X , B X ) there is x ∈ X \ S B X such that P equals the family of all zero-sets in X containing x . In particular, if ( X, T X , B X ) is large scale compact and Hausdorff,then ∂ ( X, T X , B X ) = X \ S B X and id : ( X, T X , B X ) → X ∪ ∂ ( X, T X , B X ) is ahomeomorphism.Proof. Suppose for each x ∈ X \ S B X there is C x ∈ P not containing x . For x ∈ S B X put C x := X \ { x } . Since X = S x ∈ X ( X \ C x ), there exists a finite subset F of X such that T x ∈ F C x ∈ B X . However, T x ∈ F C x ∈ P , a contradiction. (cid:3) Boundaries of formed sets
In this section we generalize boundaries of large scale topological spaces to bound-aries of formed sets.
Definition 5.1.
Given a set X equipped with a basic multilinear form ω , the boundary ∂ ( ω ) is defined as the set of maximal families P of subsets of X so that ω ( C , . . . , C k ) = ∞ for all sets C , . . . , C k ∈ P . Such P may be referred to as a point at infinity of X despite the possibility of P being identified with some point of X . Definition 5.2. If { x } is ω -unbounded, then we can identify x with the principalultrafilter consisting of all subsets of X containing x . This way we can talk about X ∪ ∂ ( ω ). To distinguish between X and its boundary, for each C ⊂ X we defineits non-boundary points C o as { x ∈ C | ω ( x ) = 0 } = C \ ∂ ( ω ). Remark . There may be points P of the boundary containing x ∈ X but notequal to x . That is the case if ω ( x ) = ω ( X \{ x } ) = ω ( x, X \{ x } ) = ∞ , for example. Observation 5.4. If C ∪ D ∈ Q ∈ ∂ ( ω ) , then C ∈ Q or D ∈ Q .Proof. Suppose C ∪ D ∈ Q but C / ∈ Q and
D / ∈ Q . In that case there exist V , V ∈ V ect ( Q ) such that ω ( C ∗ V ) = ω ( D ∗ V ) = 0. Consequently, ω (( C ∪ D ) ∗ V ∗ V ) = 0by 2.7, a contradiction. (cid:3) In order to extend the topology on (
X, ω ) induced by ω over X ∪ ∂ ( ω ) we needthe following concept: Definition 5.5.
Suppose (
X, ω ) is a formed set and C ⊂ X . o ( C ) is defined as o ( C ) := C o ∪ {Q ∈ ∂ ( ω ) | X \ C / ∈ Q} , Observation 5.6.
Notice C = o ( C ) ∩ X . Indeed, if x ∈ C \ C o , then X \ C doesnot belong to the principal ultrafilter generated by x . Conversely, if x ∈ X ∩ o ( C ) ,then either x ∈ C ⊂ C or x represents the principal ultrafilter not containing X \ C which implies x ∈ C . Lemma 5.7. o ( C ) ∩ o ( D ) = o ( C ∩ D ) .Proof. The equality follows from 5.4 and ( C ∩ D ) o = C o ∩ D o . (cid:3) Definition 5.8.
Suppose (
X, ω ) is a set X equipped with a basic multilinear form ω and B ( ω ) is the bornology induced by ω . X ∪ ∂ ( ω ) has a natural topology defined as follows: U ⊂ X ∪ ∂ ( ω ) is declared open if X ∩ U is open in T ( ω ) and foreach Q ∈ U ∩ ∂ ( ω ) there is an open (in T ( ω )) C ⊂ X such that Q ∈ o ( C ) ⊂ U . Remark . In view of 5.7, the above topology is well-defined and sets o ( U ), U open in X , form its basis. Lemma 5.10. Q ∈ ∂ ( ω ) belongs to cl ( C ) , C ⊂ X , if C ∈ Q .2. If C is closed in X and Q ∈ cl ( C ) , then C ∈ Q .Proof.
1. Suppose C ∈ Q and Q / ∈ cl ( C ). There is o ( D ) containing Q and disjointwith C . Hence D ⊂ X \ C resulting in C ⊂ X \ D . Consequently, X \ D ∈ Q contradicting Q ∈ o ( D ).2. If C / ∈ Q , then
Q ∈ o ( X \ C ). However, C ∩ o ( X \ C ) = ∅ contradicting Q ∈ cl ( C ). (cid:3) Corollary 5.11. C ⊂ o ( C ) ⊂ cl ( C ) for all C ⊂ X . In particular, cl ( C ) = cl ( o ( C )) .Proof. If Q ∈ o ( C ), then X \ C / ∈ Q , hence C ∈ Q . By 5.10, Q ∈ cl ( C ). (cid:3) Lemma 5.12.
1. If k T i =1 cl ( C i ) ⊂ X \ ∂ ( ω ) , then ω ( C , . . . , C k ) = 0 .2. If C i , i ≤ k , are closed in X , k T i =1 C i = ∅ , and ω ( C , . . . , C k ) = 0 , then k T i =1 cl ( o ( C i )) ⊂ X \ ∂ ( ω ) .Proof.
1. Suppose k T i =1 cl ( C i ) ⊂ X \ ∂ ( ω ) but ω ( C , . . . , C k ) = ∞ . Hence all C i are ω -unbounded and there is Q ∈ ∂ ( ω ) containing all C i , i ≤ k . By Lemma 5.10, Q ∈ k T i =1 cl ( C i ), a contradiction.2. Suppose Q ∈ k T i =1 cl ( o ( C i )), k T i =1 C i = ∅ , and ω ( C , . . . , C k ) = 0. Therefore Q cannot contain all C i ’s, so assume C / ∈ Q . Hence Q ∈ o ( X \ C ). However, o ( C ) ∩ o ( X \ C ) = ∅ , contradicting Q ∈ cl ( o ( C )). (cid:3) Proposition 5.13.
Supose D ⊂ X is closed. If { C i } ki =1 is a finite family of subsetsof X , then the following conditions are equivalent:1. k S i =1 o ( C i ) contains cl ( D ) ∩ ∂ ( ω ) .2. ω ( D, X \ C , . . . , X \ C k ) = 0 and D \ k S i =1 C i ∈ B ( ω ) .3. ω ( D, X \ C , . . . , X \ C k ) = 0 .Proof. Notice Q ∈ cl ( D ) ∩ ∂ ( ω ) \ k S i =1 o ( C i ) if and only if D, X \ C , . . . , X \ C k ∈ Q (see 5.10), so cl ( D ) ∩ ∂ ( ω ) \ k S i =1 o ( C i ) = ∅ if and only if ω ( D, X \ C , . . . , X \ C k ) = 0.Therefore 1) ⇐⇒ INEAR ALGEBRA AND UNIFICATION OF GEOMETRIES IN ALL SCALES 11 ⇐⇒
3) follows from the fact that ω ( D, X \ C , . . . , X \ C k ) = 0 implies ω ( D \ k S i =1 C i ) = 0 by 2.5. (cid:3) Corollary 5.14.
Given a covering { o ( C ) } C ∈S of cl ( D ) , D ⊂ X being closed, thereis a finite subfamily F of S covering cl ( D ) ∩ ∂ ( ω ) such that D \ S C ∈F C belongs to B ( ω ) .Proof. Suppose { o ( C ) } C ∈S is a cover of cl ( D ) that has no finite subcover of cl ( D ) ∩ ∂ ( ω ). By 5.13 it means ω ( D, X \ C , . . . , X \ C k ) = ∞ for all finite subfamilies { C i } ki =1 of S . Therefore there is Q ∈ ∂ ( ω ) containing D and all sets X \ C , C ∈ S .By 5.10, Q ∈ cl ( D ). Since Q / ∈ S { o ( C ) | C ∈ S} , we arrive at a contradiction. (cid:3) Corollary 5.15. X ∪ ∂ ( ω ) is large scale compact with respect to B ( ω ) . Proposition 5.16.
Suppose ω is a form on X . The following conditions are equiv-alent:a. X ∪ ∂ ( ω ) is large scale compact with respect to B ( ω ) and Hausdorff.b. For every two different points Q and R of ∂ ( ω ) there exist ω -closed subsets C and D of X such that C ∪ D = X , C / ∈ Q , and
D / ∈ R .Proof. a) = ⇒ b). Choose disjoint sets o ( U ) and o ( W ), where Q ∈ o ( U ), R ∈ o ( W ),and U, W are ω -open. Therefore U ∩ W = ∅ resulting in C := X \ U , D := X \ W satisfying the required conditions.b) = ⇒ a). Since every point of X \ ∂ ( ω ) is open-closed, it suffices to show thatevery two distinct points Q , R ∈ ∂ ( ω ) have disjoint neighborhoods. Choose ω -closed subsets C and D of X such that C ∪ D = X , C / ∈ Q , and
D / ∈ R . Noticethat
Q ∈ o ( X \ C ), R ∈ o ( X \ D ). Suppose S ∈ o ( X \ C ) ∩ o ( X \ D ). Therefore C / ∈ S and
D / ∈ S contradicting X = C ∪ D ∈ S . (cid:3) Proposition 5.17. If ( X, ω ) is a formed set, then ω ( C , . . . , C k ) = 0 iff k T i =1 cl ( C i ) ⊂ X \ ∂ ( ω ) .Proof. ω ( C , . . . , C k ) = ∞ if and only if there is Q ∈ ∂ ( ω ) containing all of C i .That is equivalent to Q ∈ k T i =1 cl ( C i ). (cid:3) Normal forms
In this section we seek conditions for X ∪ ∂ ( ω ) to be Hausdorff. Definition 6.1.
A formed set (
X, ω ) is normal if ω ( C , . . . , C k ) = 0 implies exis-tence of subsets D i , 1 ≤ i ≤ k , of X such that k S i =1 D i = X and ω ( C i , D i ) = 0 foreach 1 ≤ i ≤ k . Lemma 6.2. If { C i } ki =1 is a family of zero-sets in a topological space ( X, T ) whoseintersection is empty, then there exists a family of zero-sets { D i } ki =1 satisfying thefollowing conditions:a. C i ∩ D i = ∅ for each i ≤ k ,b. k S i =1 D i = X . Proof.
Choose continuous functions f i : X → [0 ,
1] such that C i = f − i (0) for each i ≤ k . Notice that g := k P i =1 f i is positive and let g i := f i /g . Define D i as g − i [1 /k, { D i } ki =1 satisfies the required conditions. (cid:3) Corollary 6.3.
The basic functional form ω f ( X, T ) of a topological space ( X, T ) is normal.Proof. Apply 6.2. (cid:3)
Corollary 6.4.
The basic topological form of a T space ( X, T ) is normal if andonly if ( X, T ) is normal.Proof. If ( X, T ) is normal, then its functional form equals the basic topologicalform, hence is normal by 6.3.Suppose ω ( T ) is normal and C, D are two disjoint closed subsets of X . Thus ω ( T )( C, D ) = 0 and there exist subsets C ′ , D ′ of X such that C ′ ∪ D ′ = X , ω ( T )( C, C ′ ) = 0, and ω ( T )( D, D ′ ) = 0. That implies C ⊂ U := X \ cl ( C ′ ), D ⊂ W := X \ cl ( D ′ ) and U ∩ W = ∅ . Since ( X, T ) is T , it is normal. (cid:3) Proposition 6.5.
If a large scale space ( X, L ) is metrizable, then the form ω ( L ) induced by L is normal.Proof. Suppose ω ( C , . . . , C k ) = 0 for some subsets of X and d is a metric on X inducing L . Choose x ∈ X and an increasing function f : N → N such that foreach n ≥ k T i =1 B ( C i , n ) ⊂ B ( x , f ( n )).For each i ≤ k , define D i as { x ∈ X | for all n ≥ , d ( x, x ) > f ( n ) = ⇒ d ( x, C i ) ≥ n } . If ω ( C i , D i ) = ∞ for some i ≤ k , then there is m ≥ B ( C i , m ) ∩ B ( D i , m )is unbounded. In particular, there is y ∈ B ( C i , m ) ∩ B ( D i , m ) satisfying d ( y, x ) >f ( m ). Thus, d ( y, C i ) ≥ m contradicting y ∈ B ( C i , m ).Suppose z ∈ X \ k S i =1 D i . Therefore, for each i ≤ k , there is m i such that d ( z, x ) >f ( m i ) but d ( z, C i ) < m i . Put M = max { m i | i ≤ k } . Notice z ∈ B ( x , f ( M )) as z ∈ k T i =1 B ( C i , M ) ⊂ B ( x , f ( M )) which contradicts d ( z, x ) > f ( M ). (cid:3) Special normal forms.
Suppose a set X has a function N : 2 X × S → X , S being a directed set such that the following conditions are satisfied:1. N ( C, s ) ⊂ N ( C, t ) if t ≤ s ,2. For all C, D ⊂ X and all t ∈ S , there is s ≥ t such that N ( C ∪ D, s ) ⊂ N ( C, t ) ∪ N ( D, t ).Notice the following defines a form: ω ( N )( C , . . . , C k ) = 0 ⇐⇒ k \ i =1 N ( C i , s ) = ∅ for some s ∈ S. Proposition 6.6.
If for each D ⊂ X \ N ( C, s ) there is t ∈ S , t ≥ s , such that N ( D, t ) ⊂ X \ N ( C, t ) , then ω ( N ) is normal. INEAR ALGEBRA AND UNIFICATION OF GEOMETRIES IN ALL SCALES 13
Proof.
Suppose k T i =1 N ( C i , s ) = ∅ . Define D i := X \ N ( C i , s ) for i ≤ k . Obviously, k S i =1 D i = X . Also, for each i ≤ k there is t i > s such that N ( D i , t i ) ⊂ X \ N ( C i , t i )which means ω ( D i , C i ) = 0. (cid:3) Corollary 6.7. If ( X, U ) is a uniform space, then the form ω ( U ) induced by U isnormal.Proof. Let S be the family of uniform covers of X . Define W ≥ V as W refines V . The function N : 2 X × S → X is defined as N ( C, V ) = st ( C, V ). Notice ω ( U ) = ω ( N ) (see 2.16).Let W be a uniform cover of X such that st ( W , W ) refines V . If D ⊂ X \ st ( C, V ),then N ( D, W ) ∩ N ( C, W ) = ∅ . Indeed, suppose x ∈ st ( C, W ) ∩ st ( D, W ). Pick y ∈ C ∩ st ( x, W ) and pick z ∈ D ∩ st ( x, W ). There is W ∈ W containing x, y and there is W ∈ W containing x, z . Pick V ∈ V containing st ( W , W ). Notice W ∪ W ⊂ V and y ∈ V ∩ C . Since D ⊂ X \ V we arrive at a contradiction: z ∈ D ∩ V . (cid:3) Definition 6.8.
Suppose X is a locally compact Hausdorff space that is locallycompact. Given C ⊂ X and a compact subset K of X define N ( C, K ) as the unionof components of X \ K that intersect C .We define the order on compact subsets of X as follows: K ≤ K ′ if K ⊂ K ′ .The form ω ( N ) is called the Freundenthal form of X . Proposition 6.9.
The Freundenthal form is normal.Proof.
Observe N ( C ∪ D, K ) = N ( C, K ) ∪ N ( D, K ) for all
C, D ⊂ X and all K ⊂ X being compact. Given C ⊂ X and K ⊂ X consider D := X \ N ( C, K ). Observethat N ( D, K ) ∩ N ( C, K ) = ∅ . Apply 6.6. (cid:3) Boundaries of normal forms
In this section we show X ∪ ∂ ( ω ) is Hausdorff if and only if ω is normal and T . Lemma 7.1.
Suppose ω is a normal form on X . If D ∩ C = ∅ and ω ( D, C ) =0 , then there is an ω -open set U containing D , disjoint with C , and satisfying ω ( C, U ) = 0 .Proof.
Claim:
There exist disjoint sets C ′ and D ′ such that1. C ′ ∪ D ′ = X ,2. C ⊂ C ′ ,3. D ⊂ D ′ ,4. ω ( C, D ′ ) = 0 and ω ( D, C ′ ) = 0. Proof of Claim:
Choose sets E and F such that ω ( C, E ) = 0 = ω ( D, F ) = 0 and E ∪ F = X . Notice C ∩ E, D ∩ F ∈ B ( ω ). Put C ′ := C ∪ ( F \ F ∩ D ) and D ′ := D ∪ ( E \ E ∩ C ). (cid:4) By induction construct an increasing sequence of sets D n containing D and disjointwith C ′ so that ω ( D n , X \ D n +1 ) = 0. By 3.2, U := ∞ S i =1 D n is ω -open, D ⊂ U , and U ∩ C ′ = ∅ . Therefore U ⊂ D ′ and ω ( C, U ) = 0. (cid:3)
Corollary 7.2. If ω is a normal form on X , then x belongs to ω -closure of C ⊂ X if and only if x ∈ C or ω ( x, C ) = ∞ . Proof.
Suppose x / ∈ C and ω ( x, C ) = 0. By 7.1, x / ∈ cl ( C ).If x / ∈ cl ( C ), then x ∈ U := X \ cl ( C ) and ω ( x, X \ U ) = 0 as U is ω -open. Since C ⊂ X \ U , ω ( x, C ) = 0. (cid:3) Corollary 7.3. If ω is a normal form on X and ω ( C , . . . , C k ) = 0 , then ω ( cl ( C ) , . . . , cl ( C k )) = 0 . Proof.
Claim: If ω ( C, D ) = 0, then ω ( cl ( C ) , cl ( D )) = 0. Proof of Claim:
It is sufficient to show ω ( cl ( C ) , D ) = 0. Choose sets C ′ and D ′ such that C ′ ∪ D ′ = X , ω ( C, C ′ ) = 0, and ω ( D, D ′ ) = 0. If x ∈ D ′ \ C , then x / ∈ cl ( C ) by 7.2. Thus, cl ( C ) ⊂ C ∪ D ′ . Since ω ( C ∪ D ′ , D ) = 0, the proof ofClaim is completed. (cid:4) Suppose ω ( cl ( C ) , . . . , cl ( C k )) = ∞ and choose D i ⊂ X , i ≤ k , such that ω ( C i , D i ) =0 for all i ≤ k , and k S i =1 D i = X . There is Q ∈ ∂ ( ω ) containing each cl ( C i ), i ≤ k .As ω ( cl ( C i ) , D i ) = 0 for each i , none of D i ’s belong to Q , a contradiction as k S i =1 D i = X . (cid:3) Theorem 7.4. X ∪ ∂ ( ω ) is large scale compact with respect to B ( ω ) and Hausdorffif and only if ω is normal and T .Proof. Suppose ω is normal. Given Q 6 = R there is D ∈ Q and C , . . . , C k ∈ R such that ω ( D, C , . . . , C k ) = 0. Choose D ′ , D , . . . , D k ⊂ X such that ω ( D, D ′ ) = ω ( C i , D i ) = 0 for i ≤ k and D ′ ∪ k S i =1 D i = X . By 7.3 we may assume D ′ is ω -closedand all D i ’s are ω -closed. Notice D ′ / ∈ Q . Also, k S i =1 D i / ∈ R (as otherwise D j ∈ R for some j contradicting ω ( C j , D j ) = 0). By 5.15 and 5.16, X ∪ ∂ ( ω ) is Hausdorffand large scale compact with respect to B ( ω ).Suppose X ∪ ∂ ( ω ) is Hausdorff, large scale compact with respect to B ( ω ), and ω ( C (1) , . . . , C ( k )) = 0 for some C ( i ) ⊂ X . That means the intersection of closuresof C ( i ) (in X ∪ ∂ ( ω )) is an element B of B ( ω ). Choose, for each Q , an index i ( Q )such that Q / ∈ cl ( C ( i ( Q ))). Then choose an ω -open D ( Q ) such that cl ( o ( D ( Q ))) ∩ cl ( C ( i ( Q ))) = ∅ and Q ∈ o ( D ( Q )). Notice ω ( D ( Q ) , cl ( C ( i ( Q ))) = 0 as otherwisethere is R ∈ cl ( o ( D ( Q ))) ∩ cl ( C ( i ( Q ))).There are finitely many points Q ( j ) so that B ′ := ( X ∪ ∂ ( ω )) \ n S j =1 o ( D ( Q ( j ))) ∈ B ( ω ). For each index s ≤ k define E s as the union of all D ( Q ( j )) so that i ( Q ( j )) = s . Notice ω ( B ′ ∪ B ∪ E s , C s ) = 0 for each s ≤ k and k S s =1 ( B ′ ∪ B ∪ E s ) = X . (cid:3) Lemma 7.5.
Suppose ω , ω are normal forms on X . If ω ( C , C ) = ω ( C , C ) for all -vectors ( C , C ) in X , then ω = ω .Proof. Notice ω ( C ) = ω ( C, C ) = ω ( C, C ) = ω ( C ) for all C ⊂ X . Now, itsuffices to show ω ( C , . . . , C k ) = 0 implies ω ( C , . . . , C k ) = 0 for all k ≥ k -vectors V = ( C , . . . , C k ) of X . Choose subsets D i , 1 ≤ i ≤ k , of X such that k S i =1 D i = X and ω ( C i , D i ) = 0 for each 1 ≤ i ≤ k which implies ω ( C i , D i ) = 0 INEAR ALGEBRA AND UNIFICATION OF GEOMETRIES IN ALL SCALES 15 for each 1 ≤ i ≤ k . Therefore ω ( V ∗ D i ) = 0 for each 1 ≤ i ≤ k . Finally, ω ( V ) = ω ( V ∗ X ) = k P i =1 ω ( V ∗ D i ) = 0. (cid:3) Theorem 7.6.
Suppose ω , ω are normal forms on X and ( X, T ( ω ) , B ( ω )) islarge scale compact and Hausdorff. If T ( ω ) = T ( ω ) and B ( ω ) = B ( ω ) , then ω = ω .Proof. Notice B ( ω ) = B ( ω ) means exactly that ω ( C ) = ω ( C ) for all C ⊂ X , soour next goal is to prove ω ( C , C ) = ω ( C , C ) for all 2-vectors ( C , C ) in X . Itsuffices to show ω ( C , C ) = 0 implies ω ( C , C ) = 0 for disjoint subsets C , C of X (remove their intersection if it is not empty). Enlarge C to an open subset U of X so that ω ( C , U ) = 0 (see 7.1) and put D := cl ( C ). Now, ω ( D, U ) = 0 by 7.3.Given x ∈ D \ U , there is a neighborhood W ( x ) of x satisfying ω ( W ( x ) , U ) = 0 as x / ∈ U and U ∈ T ( ω ) (use 7.1). There is a finite subset F of D \ U for which thecomplement E of S x ∈ F W ( x ) ∪ ( X \ D ) ∪ U belongs to B ( ω ). Put W := S x ∈ F W ( x )and notice Notice D ⊂ E ∪ W ∪ ( D \ U ). Since ω ( Y, U ) = 0 for each term Y of E ∪ W ∪ ( D \ U ), ω ( D, U ) = 0 which implies ω ( C , C ) = 0.To complete the proof of 7.6 invoke 7.5. (cid:3) Theorem 7.7. If ( X, T , B ) is a large scale compact Hausdorff topological space,then there is a unique normal form ω inducing T and having ω -bounded sets iden-tical with the bornology B . That form is given by the formula ω ( C , . . . , C k ) = 0 ⇐⇒ X ∩ cl ( C ) ∩ . . . ∩ cl ( C k ) = ∅ where X := X \ S B .Proof. ω is normal by an argument similar to the proof of 6.4. First, let’s show B = B ( ω ). If B ∈ B , then B is closed and ω ( B ) = 0 as X ∩ B = ∅ . Conversely,suppose X ∩ cl ( C ) = ∅ . For each x ∈ cl ( C ) consider its neighborhood W ( x ) = { x } .Pick a finite subset F of cl ( C ) so that the complement B of S x ∈ F { x } ∪ ( X \ cl ( C ))belongs to B . Put W := S x ∈ F { x } ∈ B and notice cl ( C ) ⊂ W ∪ B ∈ B .Now, let’s show T = T ( ω ). If x ∈ U and U ∈ T , then ω ( x, X \ U ) = 0 as X ∩ { x } ∩ ( X \ U ) = ∅ . Thus U ∈ T ( ω ). Conversely, if W ∈ T ( ω ), then foreach x ∈ W , ω ( x, X \ W ) = 0 which means ∅ = X ∩ { x } ∩ cl ( X \ W ). Now, x ∈ cl ( X \ W ) is possible only if x / ∈ X , in which case { x } is open-closed, so cl ( X \ W ) ⊂ ( X \ { x } ), a contradiction. Thus, W ∩ cl ( X \ W ) = ∅ and W ∈ T .To conclude the proof of 7.7 apply 7.6. (cid:3) Proposition 7.8. If ( X, ω ) is a formed set and A ⊂ X is closed in the topology T induced by ω , then the topology on A induced by T coincides with the topologyinduced by ω A := ω | A . Moreover, if ω is normal and T , then so is ω A and thenatural function i : A ∪ ∂ ( ω A ) → X ∪ ∂ ( ω ) is a topological embedding.Proof. If U ⊂ X is open, then for each x ∈ A ∩ U one has ω A ( x, A \ U ) = 0, so U ∩ A is open in the topology induced by ω A . Conversely, if U ⊂ A has the property that ω A ( x, A \ U ) = 0 for all x ∈ U , then (as A is closed in X ) ω ( x, X \ ( U ∪ ( X \ A )) = ω ( x, A \ U ) = 0 for all x ∈ U ∪ ( X \ A ), so U ∪ ( X \ A ) is open in X and U is openin the relative topology on A . i is the identity on A and i ( P ), P ∈ ∂ ( A, ω A ), consists of all subsets C of X suchthat C ∩ A ∈ P . The complement of the image of i is exactly o ( X \ A ). Given anopen subset U of X , i − ( o ( U )) = o ( U ∩ A ), so i is continuous. Also, given an opensubset U of A , i ( o ( U )) = im ( i ) ∩ o ( U ∪ ( X \ A )), so i is an embedding. (cid:3) Orthogonality relations
In this section we describe a geometrical relation between subsets of a set X thatcan be used to generate a form on X . Definition 8.1. An orthogonality relation on subsets of a set X is a symmetricrelation ⊥ satisfying the following properties:1. ∅ ⊥ X ,2. A ⊥ ( C ∪ C ′ ) ⇐⇒ A ⊥ C and A ⊥ C ′ . Observation 8.2.
One can reduce the number of axioms by dropping symmetryand replacing Axiom 2 by2’. A ⊥ ( C ∪ C ′ ) ⇐⇒ C ⊥ A and C ′ ⊥ A . Example 8.3.
For every bornology B on a set X the relation A ⊥ C defined as A ∩ C ∈ B is an orthogonality relation. Bounded sets.Definition 8.4.
Given an orthogonality relation ⊥ on subsets of X , a boundedsubset B of X is one that is orthogonal to the whole set: B ⊥ X. Definition 8.5.
An orthogonality relation ⊥ on subsets of X is small scale if theempty set is the only subset of X that is orthogonal to itself. In particular, theonly bounded subset of X is the empty set. Definition 8.6.
An orthogonality relation ⊥ on subsets of X is large scale if eachpoint is a bounded subset of X .8.2. Examples of small scale orthogonality.Example 8.7. Set-theoretic orthogonality : Disjointness,2.
Topological orthogonality : Disjointness of closures,3.
Metric orthogonality : Disjointness of r -balls for some r > ,4. Uniform orthogonality : Disjointness of U -neighborhoods for some uniformcover U . Examples of large scale orthogonality.Example 8.8. Set-theoretic large scale orthogonality : Finiteness of inter-section,2.
Metric large scale orthogonality : Boundedness of intersection of r -balls forall r > ,3. Group large scale orthogonality : Finiteness of ( A · F ) ∩ ( C · F ) for all finitesubsets F of a group G .Same as metric ls-orthogonality for word metrics if G is finitely generated.4. Topological ls-orthogonality : Disjointness of coronas of closures in a fixedcompactification ¯ X of X . INEAR ALGEBRA AND UNIFICATION OF GEOMETRIES IN ALL SCALES 17
Hyperbolic orthogonality.
Given a metric space (
X, d ), the
Gromov prod-uct of x and y with respect to a ∈ X is defined by h x, y i a = 12 (cid:0) d ( x, a ) + d ( y, a ) − d ( x, y ) (cid:1) . Recall that metric space (
X, d ) is (Gromov) δ − hyperbolic if it satisfies the δ/ h x, y i a ≥ min {h x, z i a , h z, y i a } − δ/ , ∀ x, y, z, a ∈ X. ( X, d ) is
Gromov hyperbolic if it is δ − hyperbolic for some δ > Definition 8.9.
Two subsets A and C of a hyperbolic space X are hyperbolicallyorthogonal if there is r > h a, c i p < r for some fixed p and all ( a, c ) ∈ A × C .8.5. Freundenthal orthogonality.Definition 8.10.
Suppose X is a locally compact and locally connect topologicalspace. Two subsets A and C of X are Freundenthal orthogonal if there is acompact subset K of X such that the union of all components of X \ K intersecting A is disjoint from the union of all components of X \ K intersecting C .8.6. Normal orthogonality relations.Definition 8.11.
Given an orthogonality relation ⊥ on subsets of X , two subsets C and D ⊥ -span X if the following conditions are satisfied:1. C ⊥ D ,2. X can be decomposed as X = C ′ ∪ D ′ , where C ′ ⊥ D and D ′ ⊥ C . Remark . Obviously, we may interpret the word ”decompose” in the definitionabove as C ′ ∩ D ′ = ∅ since D ′ can be replaced by X \ C ′ . The other extreme iswhen C ⊂ C ′ and D ⊂ D ′ which can be accomplished by replacing C ′ by C ∪ C ′ and replacing D ′ by D ∪ D ′ . In that case we may think of C being parallel to C ′ , D being parallel to D ′ and interpret Definition 8.11 as an analog of parallel-perpendicular decomposition in Linear Algebra. Proposition 8.13. If C and D ⊥ -span X , then C ∩ D is ⊥ -bounded.Proof. X = C ′ ∪ D ′ , where C ′ ⊥ D and D ′ ⊥ C . Therefore C ′ ⊥ ( C ∩ D ) and D ′ ⊥ ( C ∩ D ) resulting in ( C ∩ D ) ⊥ X . (cid:3) Definition 8.14.
An orthogonality relation ⊥ on subsets of X is normal if C and D ⊥ -span X whenever C ⊥ D . Example 8.15.
The topological orthogonality relation on a topological space isnormal if X is topologically normal. Definition 8.16.
The functional orthogonality relation ⊥ on a topologicalspace X is defined as follows: C ⊥ D if there is a continuous function f : X → [0 , f ( C ) ⊂ { } and f ( D ) ⊂ { } . Proposition 8.17.
The functional orthogonality relation ⊥ on a topological space X is always normal. Proof.
For any continuous f : X → [0 ,
1] satisfying f ( C ) ⊂ { } , f ( D ) ⊂ { } , oneputs C ′ = f − [0 . , D ′ = f − [0 , .
5] and observe X = C ′ ∪ D ′ , C ′ ⊥ C , and D ′ ⊥ D . (cid:3) Proposition 8.18.
The hyperbolic orthogonality relation ⊥ on a Gromov hyperbolicspace is normal.Proof. Suppose A and C are subsets of a pointed Gromov hyperbolic space ( X, p )such that h a, c i p < r for all ( a, c ) ∈ A × C . Define A ′ as { x ∈ X | h a, x i p < r + δ ∀ a ∈ A } . Define C ′ as { x ∈ X | h c, x i p < r + δ ∀ c ∈ C ′ } . Clearly, A ′ ⊥ A and C ′ ⊥ C ,so it remains to show A ′ ∪ C ′ = X . Suppose there is x ∈ X \ ( A ′ ∪ C ′ ). There is a ∈ A and c ∈ C such that h a, x i p ≥ r + δ and h c, x i p ≥ r + δ . Therefore h a, c i p ≥ min {h x, a i p , h c, x i p } − δ/ > r a contradiction. zzz (cid:3) Proximity spaces.
There is a more general structure than uniform spaces,namely a proximity (see [18]). In this section we show that those structures corre-spond to normal small scale orthogonal relations.
Definition 8.19. A proximity space ( X, δ ) is a set X with a relation δ betweensubsets of X satisfying the following properties:For all subsets A, B and C of X AδB = ⇒ BδA AδB = ⇒ A = ∅ A ∩ B = ∅ = ⇒ AδB Aδ ( B ∪ C ) ⇐⇒ ( AδB or AδC )5. ∀ E, AδE or Bδ ( X \ E ) = ⇒ AδB . Proposition 8.20.
Normal small scale orthogonality relations are in one-to-onecorrespondence with proximity relations.Proof.
Given a small scale orthogonal relation ⊥ we define AδC as ¬ ( A ⊥ C ).Conversely, given a proximity relation δ we define A ⊥ C as ¬ ( AδC ).The proof amounts to negating implications, so let’s show only the implication A ∩ B = ∅ = ⇒ AδB . If it fails, then we have two orthogonal sets A and B withnon-empty intersection A ∩ B . However, in this case A ∩ B is self-orthogonal, acontradiction. (cid:3) ⊥ -continuous functions Definition 9.1.
Given two sets X and Y equipped with orthogonality relations ⊥ X and ⊥ Y , a function f : X → Y is ⊥ -continuous if f ( A ) ⊥ Y f ( C ) = ⇒ A ⊥ X C for all subsets A, C of X .9.1. Small Scale Examples.
In the small scale ⊥ -continuous functions are ex-actly neighborhood-continuous functions with respect to the induced neighborhoodoperator. Therefore both examples below follow from [11] in view of 10.7. Example 9.2.
If both X and Y are normal spaces equipped with topological orthog-onality relations, then ⊥ -continuity is ordinary topological continuity . INEAR ALGEBRA AND UNIFICATION OF GEOMETRIES IN ALL SCALES 19
Example 9.3.
If both X and Y are uniform spaces equipped with uniform orthog-onality relations, then ⊥ -continuity is ordinary uniform continuity . Large Scale Examples.Example 9.4.
If both X and Y are metric spaces equipped with metric ls -orthogonalityrelations and f : X → Y preserves bounded sets, then ⊥ -continuity is the same as f being coarse and bornologous .Proof. Recall that f : X → Y is bornologous if, for each r >
0, there is s > f ( A )) < s if diam( A ) < r .Notice that every ⊥ -continuous function co-preserves bounded sets, i.e. it iscoarse. Suppose f is ⊥ -continuous but not bornologous. Hence, there is a se-quence B n of uniformly bounded subsets of X whose images f ( B n ) have diametersdiverging to infinity. We may reduce it to the case of each B n consisting of ex-actly two points x n and y n so that both f ( x n ) and f ( y n ) diverge to infinity. Notice A := { f ( x n ) } n ≥ and C := { f ( y n ) } n ≥ are orthogonal in Y but their point-inversesare not orthogonal in X , a contradiction.Suppose f is coarse and bornologous but not ⊥ -continuous. Choose two or-thogonal subsets A and C of Y whose point-inverses are not orthogonal. There-fore the intersection of B ( f − ( A ) , r ) and B ( f − ( C ) , r ) is unbounded for some r > s > f ( B ( Z, r )) ⊂ B ( f ( Z ) , s ) for all subsets Z of X . Therefore, the intersection of B ( A, s ) and B ( C, s ) is unbounded, a contradiction. (cid:3)
Example 9.5. If X is a metric space equipped with metric ls-orthogonality relationand Y is a compact metric space equipped with small scale metric orthogonality, then ⊥ -continuity is the same as f being slowly oscillating .Proof. Recall that f : X → Y is slowly oscillating if, for every pair of sequences { x n } n ≥ , { y n } n ≥ in X , lim n →∞ d Y ( f ( x n ) , f ( y n )) = 0 if { d X ( x n , y n ) } n ≥ is uniformlybounded and d ( x n , x ) → ∞ as n → ∞ .Suppose f is ⊥ -continuous but not slowly oscillating. Hence, there is pair ofsequences { x n } n ≥ , { y n } n ≥ in X , and ǫ > d Y ( f ( x n ) , f ( y n )) > ǫ foreach n ≥ { d X ( x n , y n ) } n ≥ is uniformly bounded. We may assume that thelimit of f ( x n ) is z , the limit of f ( y n ) is z . In particular d Y ( z , z ) ≥ ǫ . The sets B ( z , ǫ/
3) and B ( z , ǫ/
3) are orthogonal in Y but their point-inverses in X are not,a contradiction.Suppose f is slowly oscillating but not ⊥ -continuous. Choose two orthogo-nal subsets A and C of Y whose point-inverses are not orthogonal. Thereforethe intersection of B ( f − ( A ) , r ) and B ( f − ( C ) , r ) is unbounded for some r > X : { x n } n ≥ in f − ( A )and { y n } n ≥ in f − ( C ) such that d X ( x n , y n ) < r for each n . Consequently,lim n →∞ d Y ( f ( x n ) , f ( y n )) = 0 contradicting orthogonality of A and C . (cid:3) Quotient structures.
It is well-known that defining quotient maps in boththe uniform category and in the coarse category is tricky. In contrast, in setsequipped with orthogonality relations it is quite easy.
Definition 9.6.
Suppose ⊥ X is an orthogonality relation on a set X . Given asurjective function f : X → Y define C ⊥ Y D to mean f − ( C ) ⊥ X f − ( D ). It is easy to check that ⊥ Y is an orthogonality relation on Y , called the quotientorthogonality relation . Also, it is clear that the following holds: Proposition 9.7.
Suppose ⊥ X is an orthogonality relation on a set X , f : X → Y is a surjective function, and Y is equipped with the quotient orthogonality relation ⊥ Y . Given any ⊥ -continuous h : X → Z that is constant on fibers of f , there isunique ⊥ -continuous g : Y → Z such that h = g ◦ f . Neighborhood operators
This section is devoted to exploring the relation between orthogonality relationsand neighborhood operators.
Definition 10.1. [11] A neighborhood operator ≺ on a set X is a relationbetween its subsets satisfying the following conditions:( N0 ) A ≺ X for all A ⊆ X .( N1 ) if A ≺ B then X \ B ≺ X \ A .( N2 ) if A ≺ B ⊆ C , then A ≺ C .( N3 ) if A ≺ N and A ′ ≺ N ′ then A ∪ A ′ ≺ N ∪ N ′ . Observation 10.2.
Note that ( N0 ) is implied by ( N1 ) and the condition X ≺ X .Also, it is easy to see that, together, axioms ( N0 ) − ( N3 ) imply: ( N0 ′ ) ∅ ≺ A for all A ⊆ X . ( N2 ′ ) if A ⊆ B ≺ C then A ≺ C . ( N3 ′ ) if A ≺ N and A ′ ≺ N ′ then A ∩ A ′ ≺ N ∩ N ′ . Definition 10.3. A normal neighborhood operator ≺ satisfies the followingcondition:( N4 ) for every pair of subsets A ≺ C , there is a subset B with A ≺ B ≺ C . Proposition 10.4.
Each orthogonality relation ⊥ on X induces a neighborhoodoperator ≺ defined as follows: A ≺ U if A ⊥ X \ U and A ⊂ U .It is normal if and only if ⊥ is normal.Proof. Left to the reader. (cid:3)
Proposition 10.5.
Each neighborhood operator ≺ on X induces a small scaleorthogonality relation ⊥ defined as follows: A ⊥ U if A ≺ X \ U . ⊥ is normal if and only if ≺ is normal.Proof. Left to the reader. (cid:3)
Definition 10.6. [11] Let X be a set and ≺ a neighborhood operator. If A is asubset of X , then the induced neighbourhood operator ≺ A on subsets of A isdefined as follows: S ≺ A T precisely when there exists a subset T ′ of X such that S ≺ T ′ as subsets of X and T = T ′ ∩ A . Proposition 10.7.
Suppose X is a set equipped with an orthogonality relation ⊥ X and Y is a set equipped with a small scale orthogonality relation ⊥ Y . A function f : A ⊂ X → Y is neighborhood continuous (with respect to the induced neighborhoodoperators) if and only if it is ⊥ -continuous. INEAR ALGEBRA AND UNIFICATION OF GEOMETRIES IN ALL SCALES 21
Proof.
Suppose f : A ⊂ X → Y is neighborhood continuous and C ⊥ Y D . There-fore C ≺ Y Y \ D and f − ( C ) ≺ A f − ( Y \ D ). That means existence of S ⊂ X suchthat S ∩ A = f − ( Y \ D ) and f − ( C ) ≺ X S . Consequently, f − ( C ) ⊥ X ( X \ S ).Since f − ( D ) ⊂ X \ S , f − ( D ) ⊥ X f − ( C ).Suppose f : A ⊂ X → Y is ⊥ -continuous and C ≺ Y D . Hence C ⊥ Y ( Y \ D ) and f − ( C ) ⊥ X f − ( Y \ D ). That implies f − ( C ) ≺ X S , where S := X \ f − ( Y \ D ).Since S ∩ A = f − ( D ), f is neighborhood continuous. (cid:3) Corollary 10.8.
Suppose X is a set equipped with a normal orthogonality relation ⊥ X and [ a, b ] ⊂ R is equipped with the topological orthogonality relation ⊥ . If f : A ⊂ X → [ a, b ] is ⊥ -continuous, then it extends to a ⊥ -continuous ¯ f : X → [ a, b ] .Proof. In view of 10.7, it suffices to switch to neighborhood continuity and thatcase is done in [11] (Theorem 8.5). (cid:3)
Corollary 10.9.
Suppose X is a set equipped with a normal orthogonality relation ⊥ X and the set of complex numbers C is equipped with the topological orthogonalityrelation ⊥ . If f : A ⊂ X → C is ⊥ -continuous with metrically bounded image, thenit extends to a ⊥ -continuous ¯ f : X → C with metrically bounded image.Proof. To apply 10.8 it suffices to show that g, h : A → [ a, b ] are ⊥ -continuous ifand only g ∆ h : A → [ a, b ] × [ a, b ], ( g ∆ h )( x ) := ( g ( x ) , h ( x )), is ⊥ -continuous.In one direction it is obvious, so assume C, D ⊂ [ a, b ] × [ a, b ] are metricallyseparated. That means there is ǫ > | z − z | ≥ ǫ if z ∈ C and z ∈ D .Cover [ a, b ] × [ a, b ] by finitely many sets of the form B × B , where B and B areintervals of length ǫ/
4. Notice ( g ∆ h ) − ( C ∩ ( B × B )) ⊥ ( g ∆ h ) − ( D ∩ ( B ′ × B ′ ))for any choice of B , B , B ′ , B ′ . Therefore, ( g ∆ h ) − ( C ∩ ( B × B )) ⊥ ( g ∆ h ) − ( D )for any choice of B , B . Finally, ( g ∆ h ) − ( C ) ⊥ ( g ∆ h ) − ( D ). (cid:3) Observation 10.10.
Observe that the proof of 10.9 can be used to prove that, giventwo functions f, g : X → [0 , from a set equipped with an orthogonality relation ⊥ , the function h : X → [0 , × [0 , is ⊥ -continuous if and only if both f and g are ⊥ -continuous. Orthogonality relations vs forms
The purpose of this section is to show that normal orthogonality relations on aset are in one-to-one correspondence with normal forms on X . Proposition 11.1.
Every form ω on X induces a natural orthogonality relation ⊥ ω on X defined by C ⊥ ω D if and only if ω ( C, D ) = 0 . If ω is normal, then ⊥ ω is normal.Proof. Left to the reader. (cid:3)
Theorem 11.2.
Every normal orthogonality relation ⊥ on a set X extends uniquelyto a normal multilinear form ω ( ⊥ ) .Proof. Define ω ( ⊥ ) as follows: ω ( ⊥ )( C , . . . , C k ) = 0 if and only if there exist sets D i , i ≤ k such that D i ⊥ C i for each i ≤ k and k S i =1 D i = X .Suppose ω ( ⊥ )( E , C , . . . , C k ) = 0 and ω ( ⊥ )( E ′ , C ′ , . . . , C ′ k ) = 0. Choose sets D i , 0 ≤ k such that D i ⊥ C i for each 1 ≤ i ≤ k , D ⊥ E , and k S i =0 D i = X . Choose sets D ′ i , 0 ≤ k such that D ′ i ⊥ C i for each 1 ≤ i ≤ k , D ′ ⊥ E ′ , and k S i =0 D ′ i = X . Observe ( D i ∪ D ′ i ) ⊥ C i for each 1 ≤ i ≤ k , ( D ∩ D ′ ) ⊥ ( E ∪ E ′ ),and ( D ∩ D ′ ) ∪ k S i =1 ( D i ∪ D ′ i ) = X . Thus, ω ( ⊥ )( E ∪ E ′ , C , . . . , C k ) = 0.Suppose a normal form ω induces ⊥ . That means ω ( C , C ) = ω ( ⊥ )( C , C ) forall subsets C , C of X . By 7.5, ω = ω ( ⊥ ). (cid:3) Extending hyperbolic orthogonality relations.Definition 11.3.
Given a Gromov hyperbolic space and C ⊂ X , for every r > N ( C, r ) := { x ∈ X | h x, c i p > r for some c ∈ C } . Lemma 11.4. If ( X, d ) is a Gromov hyperbolic space, then1. C ∩ N ( C , r ) ⊂ N ( C , r ) ∩ N ( C , r ) for all r > .2. If C ∩ N ( C , r ) ⊂ B ( p, s ) , then N ( C , r + s + δ ) ∩ N ( C , r + s + δ ) = ∅ .Proof.
1. If x ∈ C ∩ N ( C , r ), then h x, c i p > r for some c ∈ C . Hence d ( x, p ) > r ,so x ∈ N ( C , r ) ∩ N ( C , r ).2. If x ∈ N ( C , r + s + δ ) ∩ N ( C , r + s + δ ), then h x, c i p > r + s + δ and h x, c i p >r + s + δ for some ( c , c ) ∈ C × C . Therefore h c , c i p ≥ min {h x, c i p , h c , x i p } − δ/ > r + s resulting in c ∈ C ∩ N ( C , r ) \ B ( p, s ), a contradiction. (cid:3) Corollary 11.5. If ( X, d ) is a Gromov hyperbolic space with the hyperbolic orthog-onality relation ⊥ , then the following conditions are equivalent for C , C ⊂ X :1. C ⊥ C ,2. There is r > such that C ∩ N ( C , r ) = ∅ ,3. There is r > such that C ∩ N ( C , r ) is d -bounded,4. There is r > such that N ( C , r ) ∩ N ( C , r ) = ∅ .Proof. ⇒
2. If C ⊥ C , then there is s > h c , c i p < r for all( c , c ) ∈ C × C . Therefore C ∩ N ( C , r ) = ∅ .2 = ⇒ ⇒ ⇒
1. Suppose N ( C , s ) ∩ N ( C , s ) = ∅ for some s >
0. If h c , c i p > s + δ for some ( c , c ) ∈ C × C , then consider the point x on a geodesic from p to c at the distance h c , c i p from p . Notice h c , x i p = h c , c i p > s + δ and h c , x i p ≥ min {h x, c i p , h c , c i p } − δ/ > s , a contradiction. (cid:3) Corollary 11.6. If ( X, d ) is a Gromov hyperbolic space with the hyperbolic orthog-onality relation ⊥ , then the following conditions are equivalent for C , C ⊂ X :1. C ⊥ C ,2. C ⊥ N ( C , r ) for some r > .Proof. ⇒
2. If C ⊥ C , then there is r > N ( C , r ) ∩ N ( C , r ) = ∅ .By 11.5, C ⊥ N ( C , r ).2 = ⇒
1. If C ⊥ N ( C , r ) for some r >
0, then C ∩ N ( C , r ) is d -bounded, so C ⊥ C by 11.5. (cid:3) Corollary 11.7. If ( X, d ) is a Gromov hyperbolic space then, the normal extension ω of the hyperbolic orthogonality relation ⊥ is described as follows: ω ( C , . . . , C k ) = 0 if and only if k T i =1 N ( C i , r ) = ∅ for some r > . INEAR ALGEBRA AND UNIFICATION OF GEOMETRIES IN ALL SCALES 23
Proof.
Since N ( C ∪ C , r ) = N ( C , r ) ∪ N ( C , r ), ω is indeed a form. By 11.5,the orthogonality relation induced by ω agrees with the hyperbolic orthogonalityrelation.It remains to show ω is normal. Use 6.6 and 11.4. (cid:3) Observation 11.8.
Similarly to forms, every orthogonality relation ⊥ on X in-duces a topology on X as follows: U is open if and only if x ⊥ ( X \ U ) for all x ∈ U . Form-continuous functions
Definition 12.1.
A function f : ( X, ω X ) → ( Y, ω Y ) of formed sets is form-continuous if ω Y ( f ( C ) , . . . , f ( C k )) = 0 implies ω X ( C , . . . , C k ) = 0 for all subsets C i , i ≤ k , of X . Proposition 12.2. If f : ( X, ω X ) → ( Y, ω Y ) is a function of formed sets suchthat ω Y is normal, then f is form-continuous if and only if it is ⊥ -continuous withrespect to orthogonality relations induced by ω X and ω Y , respectively.Proof. We need to show that f is form-continuous if and only if for all subsets C, D of X the equality ω Y ( f ( C ) , f ( D )) = 0 implies ω X ( C, D ) = 0.The implication in one direction is obvious, so assume ω Y ( f ( C ) , . . . , f ( C k )) = 0but ω X ( C , . . . , C k ) = ∞ for some subsets C i of X . Choose Q ∈ ∂ ( ω X ) containingall C i ’s and choose subsets D i of Y such that ω Y ( D i , f ( C i )) = 0 for each i ≤ k and k S i =1 D i = Y . There is j so that Q contains E := f − ( D j ). Now, ω Y ( f ( E ) , f ( C j ) = 0as f ( E ) ⊂ D j which implies ω X ( E, C j ) = 0, a contradiction to E, C j ∈ Q . (cid:3) Theorem 12.3.
Suppose If f : ( X, ω X ) → ( Y, ω Y ) is a form-continuous functionof formed sets. If ω Y is normal and T , then the unique continuous extension ˜ f : X ∪ ∂ ( ω X ) → Y ∪ ∂ ( ω Y ) of f is given by the formula ˜ f ( Q ) := { C ⊂ Y | ω Y ( C, f ( D )) = ∞ for all D ∈ Q} . In particular, the following statements hold:1. ˜ f ( Q ) is the unique element of ∂ ( ω Y ) containing all f ( D ) , D ∈ Q ,2. ˜ f ( Q ) is the unique element of the intersection of closures of f ( D ) in Y ∪ ∂ ( ω Y ) ,where D ∈ Q .Proof. By 7.4, Y ∪ ∂ ( ω Y ) is normal, so all we have to show is ˜ f is well-defined andcontinuous.First of all, we have to make sure that ˜ f ( Q ) exists. Observe that ω Y ( f ( E ) , f ( D )) = ∞ for all E, D ∈ Q , so ˜ f ( Q ) contains f ( Q ). In particular, it shows ˜ f ( ∂ ( ω X )) ⊂ ∂ ( ω Y ).Suppose ω ( C , . . . , C k ) = 0 for some C i ∈ ˜ f ( Q ). Choose D i ⊂ Y such that ω ( C i , D i ) = 0 for each i and k S i =1 D i = Y . Therefore E = f − ( D j ) ∈ Q for some j contradicting ω Y ( C j , f ( E )) = ∞ .Suppose ω Y ( C, C , . . . , C k ) = ∞ for all C i ∈ ˜ f ( Q ) but ω ( C, f ( E )) = 0 for some E ∈ Q . Notice f ( E ) ∈ ˜ f ( Q ), a contradiction (put k = 1 and C = f ( E )). Thus,˜ f ( Q ) ∈ ∂ ( ω Y ). Suppose ˜ f ( Q ) ∈ o ( C ), C being ω Y -open. Choose ω Y -open D ⊂ Y such that˜ f ( Q ) ∈ o ( D ) ⊂ cl ( o ( D )) ⊂ o ( C ). In particular, ω Y ( D, Y \ C ) = 0, as otherwisethere is P ∈ ∂ ( ω Y ) containing both D and Y \ C which belongs to cl ( D ) by 5.10but not to o ( C ).Hence Q ∈ o ( f − ( D )) (otherwise f − ( D ) ∈ Q and D ∈ ˜ f ( Q ) which contradicts˜ f ( Q ) ∈ o ( D )) and for any R ∈ o ( f − ( D )) we have ˜ f ( R ) ∈ o ( C ). Indeed, D ∈ ˜ f ( R )(as f − ( D ) ∈ R ), so Y \ C / ∈ ˜ f ( R ) which means ˜ f ( R ) ∈ o ( C ).Since f ( D ) ∈ P implies P ∈ cl ( f ( D )) (see 5.10), it suffices to show the validityof Statement 2. Suppose P 6 = ˜ f ( Q ) and pick C ∈ P that does not belong to ˜ f ( Q ).Hence, ω Y ( C, f ( D )) = 0 for some D ∈ Q . By 7.3, ω Y ( C, cl Y ( f ( D ))) = 0. Therefore P does not belong to the closure of f ( D ) in Y ∪ ∂ ( ω Y ). (cid:3) Proposition 12.4.
Suppose If f : ( X, ω X ) → ( Y, ω Y ) is a function of formed setssuch that ω Y ( f ( C )) = 0 implies ω X ( C ) = 0 for all subsets C of X . If f has acontinuous extension ˜ f : X ∪ ∂ ( ω X ) → Y ∪ ∂ ( ω Y ) then it is form-continuous.Proof. Suppose ω Y ( f ( C ) , . . . , f ( C k )) = 0 but ω X ( C , . . . , C k ) = ∞ for some sub-sets C i , i ≤ k , of X . Choose Q ∈ ∂ ( ω X ) containing all C i , i ≤ k . There is j suchthat f ( C j ) / ∈ ˜ f ( Q ). Hence ˜ f ( Q ) / ∈ cl ( f ( C j )) contradicting Q ∈ cl ( C j ). (cid:3) Corollary 12.5. If ( X, T ) is a completely regular topological space and ω f ( T ) is thebasic functional form of T , then X ∪ ∂ ( ω f ( T )) is the ˇCech-Stone compactificationof ( X, T ) .Proof. Since ω f is small scale, i.e. the only bounded set is ∅ , X ∪ ∂ ( ω f ( T )) iscompact. It is also Hausdorff as ω f is normal (see 6.3) and T .Let Y be any compact Hausdorff space equipped with the basic topological form ω Y . By 9.2 and 12.2 continuous maps f : X → Y are form-continuous. By 12.3every continuous function f : X → Y extends over X ∪ ∂ ( ω f ( T )) and by 12.4only continuous functions f : X → Y extend over X ∪ ∂ ( ω f ( T )). That’s thecharacteristic property of ˇCech-Stone compactification of ( X, T ). (cid:3) Corollary 12.6. If ( X, U ) is a uniform space and ω ( U ) is the induced form on X ,then X ∪ ∂ ( ω ( U )) is the Samuel-Smirnov compactification of ( X, U ) .Proof. Since ω ( U ) is small scale (see 2.16), i.e. the only bounded set is ∅ , X ∪ ∂ ( ω ( U )) is compact. It is also Hausdorff as ω ( U ) is normal (see 6.7) and T .Let Y be any compact metric space equipped with the basic topological form ω Y . By 9.3 and 12.2 uniformly continuous maps f : X → Y are form-continuous.By 12.3 every uniformly continuous function f : X → Y extends over X ∪ ∂ ( ω ( U ))and by 12.4 only uniformly continuous functions f : X → Y extend over X ∪ ∂ ( X, ω ( U )). That’s the characteristic property of Samuel-Smirnov compactificationof ( X, U ). (cid:3) Forms vs large scale structures
This section is devoted to interaction between forms and large scale structures.Namely, every form ω on X induces a large scale structure LS ( ω ) and everybornological large scale structure LS on X induces a form ω ( LS ) on X . In case INEAR ALGEBRA AND UNIFICATION OF GEOMETRIES IN ALL SCALES 25 of a metrizable large structure L , its form carries all the information about L (see13.4 and ).For basic facts related to the coarse category see [20].Recall that a coarse structure C on X is a family of subsets E (called con-trolled sets ) of X × X satisfying the following properties:(1) The diagonal ∆ = { ( x, x ) } x ∈ X belongs to C .(2) E ∈ C implies E ∈ C for every E ⊂ E .(3) E ∈ C implies E − ∈ C , where E − = { ( y, x ) } ( x,y ) ∈ E .(4) E , E ∈ C implies E ∪ E ∈ C .(5) E, F ∈ C implies E ◦ F ∈ C , where E ◦ F consists of ( x, y ) such that thereis z ∈ X so that ( x, z ) ∈ E and ( z, y ) ∈ F . Definition 13.1. [9] A large scale structure
LSS X on a set X is a non-emptyset of families B of subsets of X (called uniformly LSS X -bounded or uniformlybounded once LSS X is fixed) satisfying the following conditions:(1) B ∈ LSS X implies B ∈ LSS X if each element of B consisting of morethan one point is contained in some element of B .(2) B , B ∈ LSS X implies st ( B , B ) ∈ LSS X .A subset B of X is bounded with respect to LSS X if the family { B } belongsto LSS X . Thus, each large scale structure on X induces a bornology on X if everyfinite subset of X is bounded.Roe uses the term coarsely connected if if every finite subset of X is bounded.We find it more appropriate to use the following terminology: Definition 13.2. A bornological large scale space is a large scale space X inwhich if every finite subset of X is bounded. Thus, bounded subsets of X do forma bornology.As described in [9], the transition between the two structures is as follows:1. Given a uniformly bounded family U in X , the set S B ∈U B × B is a controlled set,2. Given a controlled set E , the family { E [ x ] } x ∈ X is uniformly bounded, where E [ x ] := { y ∈ X | ( x, y ) ∈ E } . Definition 13.3.
Given a bornological large scale space ( X, L ), the form ω ( L ) isdefined as follows: ω ( L )( C , . . . , C k ) = 0 ⇐⇒ k \ i =1 st ( C i , V ) is boundedfor each uniformly bounded cover V of X . Proposition 13.4. If ω ( L ) = ω ( L ) for two metrizable large scale structures on X , then L = L .Proof. Suppose U is uniformly bounded with respect to L induced by a metric d but is not is uniformly bounded with respect to L induced by a metric d . Thatmeans existence of two sequences C := { x n } ∞ n =1 , D := { y n } ∞ n =1 and a number r > d ( x n , y n ) < r for each n ≥ d ( x n , y n ) → ∞ . Notice ω ( L )( C, D ) = ∞ but ω ( L )( C, D ) = 0, a contradiction. (cid:3)
Proposition 13.5.
Given a form ω on X , the family of all covers U of X with theproperty ω ( C , . . . , C k ) = ω ( st ( C , U ) , . . . , st ( C k , U )) for all vectors ( C , . . . , C k ) in X induces a large scale structure LS ( ω ) whose familyof bounded subsets contains all ω -bounded sets. If B is bounded with respect to LS ( ω ) and ω ( B ) = ∞ , then ω ( C ) = ∞ for all C intersecting B . In particular, ifevery point in X is ω -bounded, then LS ( ω ) is a bornological large scale structurewith the same bornology as ω .Proof. Notice one can define LS ( ω ) as consisting of all covers U satisfying ω ( C , . . . , C k ) ≥ ω ( st ( C , U ) , C , . . . , C k )for all vectors ( C , . . . , C k ) in X . Suppose B ∈ LS ( ω ) and each element of a cover B of X consisting of more than one point is contained in some element of B .For all vectors ( C , . . . , C k ) in X , ω ( C , . . . , C k ) ≤ ω ( st ( C , B ) , . . . , st ( C k , B )) ≤ ω ( st ( C , B ) , . . . , st ( C k , B )) = ω ( C , . . . , C k ). B , B ∈ LS ( ω ) implies st ( B , B ) ∈ LS ( ω ) follows from the fact st ( C, st ( B , B )) ⊂ st ( st ( C, B ) , B )) ∪ st ( st ( C, B ) , B ).Indeed, for all vectors ( C , . . . , C k ) in X , ω ( st ( C , st ( B , B )) , C , . . . , C k ) ≤ ω ( st ( st ( C , B ) , B )) , C , . . . , C k ) + ω ( st ( st ( C , B ) , B )) , C , . . . , C k ) = ω ( C , C , . . . , C k ) + ω ( C , C , . . . , C k ) = ω ( C , C , . . . , C k ) . Thus, LS ( ω ) is indeed a large scale structure.If ω ( B ) = 0 and U = { B } ∪ S x ∈ X { x } , then for all vectors ( C , . . . , C k ) in X , ω ( C , . . . , C k ) ≤ ω ( st ( C , U ) , . . . , st ( C k , U )) ≤ ω ( C ∪ B, . . . , C k ∪ B ) = ω ( C , . . . , C k ).That means B is a bounded subset of X with respect to LS ( ω ). Suppose B = ∅ is abounded subset of X with respect to LS ( ω ) and ω ( B ) = ∞ . Let U := { B }∪ S x ∈ X { x } and suppose C intersects B . Now, ω ( C ) = ω ( st ( C, U )) = ω ( C ∪ B ) = ∞ . (cid:3) Proposition 13.6. If ω is a normal T form, then the following conditions areequivalent:a. U ∈ LS ( ω ) ,b. For each neighborhood W of P ∈ ∂ ( ω ) in X ∪ ∂ ( ω ) there is a neighborhood W ′ of P such that st ( W ′ , U ) ⊂ W ,c. The coronas of C and st ( C, U ) coincide for each C ⊂ X .Proof. a) = ⇒ b). Since X ∪ ∂ ( ω ) is compact Hausdorff, there is a neighborhood U of P in X ∪ ∂ ( ω ) whose closure is contained in W . In particular, ω ( U ∩ X, X \ W ) = 0.Now, ω ( st ( U ∩ X, U ) , st ( X \ W, U )) = 0 and B := st ( U ∩ X, U ) ∩ st ( X \ W, U ) is ω -bounded. Put W ′ := U \ st ( B, U ).b) = ⇒ c). The corona of D is defined as all P ∈ ∂ ( ω ) contained in cl ( D ).Equivalently, all P containing D .Suppose P contains st ( C, U ) but not C . There exists a vector ( C , . . . , C k )consisting of elements of P such that ω ( C, C , . . . , C k ) = 0. Consequently, ω ( st ( C, U ) , st ( C , U ) , . . . , st ( C k , U )) = 0 , a contradiction as P contains all those sets.c) = ⇒ a). Suppose U / ∈ LS ( ω ). There is a vector ( C , . . . , C k ) satisfying ω ( C , . . . , C k ) = 0 and ω ( st ( C , U ) , . . . , st ( C k , U )) = ∞ . Choose sets D , . . . , D k INEAR ALGEBRA AND UNIFICATION OF GEOMETRIES IN ALL SCALES 27 such that ω ( C i , D i ) = 0 for each i ≤ k and k S i =1 D i = X . Thus, coronas of D i and C i are disjoint for each i resulting in ω ( st ( C i , U ) , st ( D i , U )) = 0 for each i ≤ k . Thatcontradicts ω ( st ( C , U ) , . . . , st ( C k , U )) = ∞ . (cid:3) Proposition 13.7.
Given an infinite set X consider the maximal bounded geom-etry large scale structure LS on X . It consists of all covers U of X satisfying thefollowing properties:a. There is a natural number n ≥ such that each element U of U has at most n points.b. There is a natural number m ≥ such that each point x of X belongs to at most m elements U of U .The form ω ( LS ) induced by LS is characterized by the following property: ω ( LS )( V ) =0 if and only if at least one coordinate of V is finite.Proof. Suppose all C i are infinite and ω ( LS )( C , . . . , C k ) = 0. We may assume k is the smallest number with that property. Then we can maximize the number ofpairs ( C i , C j ) such that C i ∩ C j = ∅ . We claim all such pairs are disjoint if i = j .Indeed, assume C ∩ C = ∅ by reordering, if necessary. If C ∩ C is finite, we canreplace each C i by C i \ C ∩ C and increase the number of disjoint pairs. If C ∩ C is infinite, we can remove C and replace C by C ∩ C . Finally, we can pick aninfinite countable subset D i of C i for each i ≤ k . Pick bijections f i : D → D i andconsider the trivial extension U of the family U x := { x, f ( x ) , . . . , f k ( x ) } , x ∈ D .Notice that D ⊂ st ( D i , U ) for all i ≤ k , a contradiction. (cid:3) Proposition 13.8.
Suppose X is an infinite set. Let ω be defined as follows: ω ( V ) = 0 if and only if at least one coordinate of V is finite. The large scalestructure LS ( ω ) induced by ω consists of all covers U of X with the property st ( F, U ) is finite for all finite subsets F of X .Proof. Suppose
U ∈ LS ( ω ) and F ⊂ X is finite. Now, 0 = ω ( F ) = ω ( st ( F, U )), so st ( F, U ) is finite.The remainder of the proof is obvious. (cid:3) Question 13.9. Is LS ( ω ) normal if ω is the form defined in 13.8? Proposition 13.10. If λ is a normal T form on X , then ω ( LS ( λ )) ≤ λ .Proof. See 13.3 for the definition of the form induced by a large scale structure. If λ ( C , . . . , C k ) = 0, then for any U ∈ LS ( λ ) one has λ ( st ( C , U ) , . . . , st ( C k , U )) = 0.In particular, k T i =1 st ( C i , U ) must be LS ( λ )-bounded (see 13.5), hence ω ( LS ( λ ))( C , . . . , C k ) = 0. (cid:3) Question 13.11.
Suppose λ is a normal T form on a set X . Does ω ( LS ( λ )) = λ hold? The answer to the above question is yes in case of forms induced by metrics:
Proposition 13.12.
Given a metric space ( X, d ) , the basic large scale form ω l ( X, d ) satisfies LS ( ω l ( X, d )) = LS ( X, d ) .Proof. Put λ = ω l ( X, d ) (see 2.14). Observe that LS ( λ ) consists of all covers U of X that have a finite upper bound on diameters. Indeed, if U n ∈ U and diam ( U n ) → ∞ , then we can find points x n , y n ∈ U n such that d ( x n , y n ) → ∞ .In that case λ ( X , Y ) = 0, where X = { x n } ∞ n =1 and Y = { y n } ∞ n =1 . However, λ ( st ( X , U ) , st ( Y , U )) = ∞ . (cid:3) Observation 13.13.
The large scale structure LS described in 13.7 has the prop-erty that LS ( ω ( LS )) = LS . Problem 13.14.
Characterize normality of LS ( ω ) in terms of slowly oscillatingfunctions. Question 13.15. Is LS ( ω ) normal if ω is normal? Question 13.16. Is ω ( LS ) normal if LS is normal? Higson corona of large scale forms
Given a large scale normal formed set (
X, ω ) we show that its boundary ∂ ( ω ) isan analog of Higson corona if it is Hausdorff. Notice that Roe [20] (p.29) DefinesHigson coronas only for proper coarse spaces. We do not need that restriction.Recall ω is a large scale form if ω ( x, X ) = 0 for each x ∈ X . Thus, the topologyinduced by ω on X is discrete.Consider the complex numbers C with the basic topological form. Definition 14.1. A Higson function on (
X, ω ) is a form-continuous function f : X → C with bounded image. The C ∗ -algebra of Higson functions on ( X, ω ) isdenoted by C h ( ω ) (use 10.10 to see that Higson functions do form a C ∗ -algebra). C ( ω ) is the subalgebra of C h ( ω ) consisting of Higson functions vanishing at infinity,i.e. for each ǫ > { x ∈ X || f ( x ) | > ǫ } is bounded. Proposition 14.2.
Given a large scale normal formed set ( X, ω ) , the C ∗ -algebra C ( ∂ ( ω )) of continuous functions on ∂ ( ω ) is isomorphic to C h ( ω ) /C ( ω ) .Proof. Given a continuous function f : ∂ ( ω ) → [ − r, r ] × [ − r, r ] we can extend itover X ∪ ∂ ( ω ) due to topological normality of X ∪ ∂ ( ω ). Any such extension isform-continuous by 12.4. The difference f − f of two extensions equals 0 on ∂ ( ω ),hence, for each ǫ >
0, the set { x ∈ X || f ( x ) | > ǫ } is bounded due to X ∪ ∂ ( ω ) beinglarge scale compact.Conversely, any form-continuous f : X → [ − r, r ] × [ − r, r ] extends to a continuousfunction on X ∪ ∂ ( ω ) by 12.3. (cid:3) Theorem 14.3.
Suppose ( X, LS ) is a large scale space of finite asymptotic dimen-sion asdim ( X, LS ) . If ( X, LS ) is metrizable and ω is the form induced by LS , then asdim ( X, LS ) = dim( ∂ ( ω )) .Proof. Pick a discrete metric d on X inducing LS . Let n = dim( ∂ ( ω )). Accordingto [10], asdim ( X, d ) is the smallest integer k ≥ − f : A ⊂ X → S k extends over X to a slowly oscillating function. If k = n it is so. Indeed, f extends over A ∪ ∂ ( ω A ) by 12.3, then over X ∪ ∂ ( ω A ) due todim( ∂ ( ω )) = n . Next, it extends over a closed naighborhood D of X ∪ ∂ ( ω A ).Since the complement of int ( D ) is discrete, we can extend over the whole X ∪ ∂ ( ω ).Notice the restriction of the last extension to X is a slowly oscillating extension of f . If k < n , then there is a continuous function g : C → S k − , C a closed subset of ∂ ( ω ) that does not extend over ∂ ( ω ). We can extend g over a closed neighborhood A INEAR ALGEBRA AND UNIFICATION OF GEOMETRIES IN ALL SCALES 29 of C in X ∪ ∂ ( ω ). The restriction h of that extension to A ∩ X is slowly oscillatingbut it does not extend over X . Inddeed, any such extension would generate anextension of g over ∂ ( ω ). (cid:3) Compatible forms
In practical applications, any large scale formed set (
X, ω l ) comes with a topology T and we want to change the topology on X ∪ ∂ ( ω l ) so that the new topology iscompact Hausdorff, is the same on ∂ ( ω l ), and equals T when restricted to X . Oneway to achieve it is to consider a small scale form ω ss on X inducing T and lookat conditions when X ∪ ∂ ( ω l + ω ss ) is the space combining both topologies.The following is obvious. Proposition 15.1.
Given two forms ω and ω on a set X , ω + ω is a form. Proposition 15.2.
Given two forms ω and ω on a set X , the topology T ( ω ) induced by ω = ω + ω equals T ( ω ) if ω is large scale.Proof. Since ω ( x, X ) = 0 for all x ∈ X , ω ( x, X \ U ) = 0 is equivalent to ω ( x, X \ U ) = 0. Thus, T ( ω ) = T ( ω ). (cid:3) Definition 15.3.
Suppose (
X, ω l ) is a large scale normal formed set and ( X, ω ss )is the small scale normal formed set induced by a normal topology T on X . ω l and ω ss are compatible if the following conditions are satisfied:1. ω := ω l + ω ss is normal and T .2. The identity function i : X → X induces a continuous extension ˜ i : X ∪ ∂ ( ω l ) → X ∪ ∂ ( ω l + ω ss ) that is a homeomorphism from ∂ ( ω l ) onto ∂ ( ω l + ω ss ) \ X . Observation 15.4.
Compatibility of ω l and ω ss implies that ( X, T ) is locally com-pact as it is an open subset of a compact Hausdorff space ∂ ( ω l + ω ss ) . Theorem 15.5.
Suppose ( X, ω l ) is a large scale normal formed set and ( X, ω ss ) is the small scale normal formed set induced by a normal topology T on X . ω := ω l + ω ss is normal and T if the following conditions are satisfied:1. For every ω l -bounded subset B of X there is an ω ss -open subset U containing B and being ω l -bounded.2. ω l ( C , . . . , C k ) = 0 implies ω l ( cl ( C ) , . . . , C k ) = 0 , where the closure is takenwith respect to T .Proof. Suppose x = y ∈ X . Since ω l is large scale, both x and y are ω l -bounded, so ω l ( x, y ) = 0. Since T is a normal topology, ω ss ( x, y ) = 0 resulting in ω ( x, y ) = 0.Thus, ω is T .Suppose ω ( C , . . . , C k ) = 0. Replacing each C i by cl ( C i ) we may assume each C i is T -closed. Now, ω l ( C , . . . , C k ) = 0 and k T i =1 C i = ∅ . Due to normality of ω l there exist sets D , . . . , D k forming a cover of X such that ω l ( C i , D i ) = 0 for each i ≥
1. Pick a T -open set U i containing C i ∩ D i such that cl ( U i ) is ω l -bounded. Let Y := k S i =1 cl ( U i ). It is a normal space, so (see 6.2) there exist T -closed sets E i ⊂ Y such that E i ∩ C i = ∅ for each i ≤ k , Y = k S i =1 E i , and k T i =1 E i = ∅ . Let D ′ i := D i ∪ E i for each i ≤ k . Observe that X = k S i =1 D ′ i , ω l ( C i , D ′ i ) = 0 and ω ss ( C i , D ′ i ) = 0 foreach i ≤ k . Therefore ω ( C i , D ′ i ) = 0 for each i ≤ k and ω l + ω ss is normal. (cid:3) Theorem 15.6.
Suppose ( X, ω l ) is a large scale normal formed set, ( X, ω ss ) isa small scale normal formed set induced by a normal topology T on X , and each ω l -bounded set is pre-compact in T . ω l and ω ss are compatible if the followingconditions are satisfied:1. For every ω l -bounded subset B of X there is an ω ss -open subset U containing B and being ω l -bounded.2. ω l ( C , . . . , C k ) = 0 implies ω l ( cl ( C ) , . . . , C k ) = 0 , where the closure is takenwith respect to T .Proof. Put ω := ∂ ( ω l + ω ss ). Since i : ( X, ω l ) → ( X, ω ) is form-continuous, itextends to a continuous map ˜ i : ˜ i : X ∪ ∂ ( ω l ) → X ∪ ∂ ( ω ). Given P ∈ ∂ ( ω l ),˜ i ( P ) consists of all C ⊂ X such that ω ( C, D ) = ∞ for all D ∈ P (see 12.3).Suppose C ∈ ˜ i ( P ) \ P . That means existence of E ∈ P such that ω l ( C, E ) = 0.Hence, cl ( C ) ∩ cl ( E ) is ω l -bounded and is contained in an ω l -bounded U ∈ T . Now, E \ U ∈ P , ω ss ( C, E \ U ) = 0 and ω l ( C, E \ U ) = 0 resulting in ω ( C, E \ U ) = 0, acontradiction.So far we have established ˜ i is an embedding on ∂ ( ω l ) and the remaining taskis to show ˜ i ( ∂ ( ω l )) = ∂ ( ω ) \ X . Suppose Q ∈ ( ∂ ( ω ) \ X ) \ ∂ ( ω l ). That meansexistence of C , . . . , C k ∈ Q satisfying ω l ( C , . . . , C k ) = 0. Notice B := k T i =1 cl ( C i ) iscompact in T as it is ω l -bounded. Since X ∪ ∂ ( ω ) is compact Hausdorff, there is aneighborhood o ( U ) of Q whose closure is disjoint with B . Now, ω l ( U, C , . . . , C k ) =0 and ω ss ( U, C , . . . , C k ) = 0, so ω ( U, C , . . . , C k ) = 0 which contradicts U ∈ Q . (cid:3) Corollary 15.7. If ( X, d ) is a proper metric space, then X ∪ ∂ ( ω ) is the Higsoncompactification of X , where ω is the sum of the large scale form ω l of ( X, d ) andthe basic topological form ω ss of ( X, d ) .Proof. ( X, d ) being proper means every d -bounded set is pre-compact in the topol-ogy T induced by d . Therefore, every ω l -bounded subset of X is pre-compact in T . Also, for each r > B ( C, r ) = B ( cl ( C ) , r ) which implies ω l and ω ss arecompatible.To show X ∪ ∂ ( ω ) is the Higson compactification of X we need to show that everycontinuous, bounded, and slowly oscillating f : X → C extends over X ∪ ∂ ( ω ) andevery continuous and bounded function f : X → C that does admit continuousextension is slowly oscillating (see [20]).Suppose f : X → C is continuous, bounded, and slowly oscillating. By 9.5and 12.2, f : X → cl ( f ( X )) is form continuous, so it extends over X ∪ ∂ ( ω )to a continuous function by 12.3. Conversely, any continuous bounded function f : X → C that extends over X ∪ ∂ ( ω ) to a continuous function must be form-continuous with respect to ω l , hence slowly oscillating. (cid:3) Gromov compactification.
In this part we consider proper geodesic δ -hyperbolic spaces ( X, d ) that are visual , i.e. there is p ∈ X such that the union ofall geodesic rays emanating from p equals X . If ω ss is the basic topological formof ( X, d ) and ω h is the hyperbolic form of X , then ω h and ω ss are compatible and X ∪ ∂ ( ω h + ω ss ) is the well-known Gromov compactification of ( X, d ). INEAR ALGEBRA AND UNIFICATION OF GEOMETRIES IN ALL SCALES 31
Proposition 15.8. If ω ss is the basic topological form of ( X, d ) and ω h is thehyperbolic form of X , then ω h and ω ss are compatible.Proof. By 11.5, C ⊥ X means C ⊂ X \ N ( X, r ) for some r >
0. Notice N ( X, r ) ⊃ X \ B ( p, r ), so C is ω h -bounded if and only if C is d -bounded, so C clearly can beenlarged to an open d -bounded set. Also, N ( cl ( C ) , r ) = N ( C, r ) for each r > (cid:3) Lemma 15.9.
For each r > there are geodesic rays l , . . . , l k emanating from p such that X = B ( p, r + δ ) ∪ k [ i =1 N ( l i , r ) . Proof.
On the r -sphere S centered at p choose points x , . . . , x k such that S = k S i =1 B ( x i , δ ). For each i ≤ k choose a geodesic ray l i emanating from p and containing x i . Suppose d ( p, x ) ≥ r + δ . Choose a geodesic ray l emanating from p andcontaining x . Let y ∈ S ∩ l and pick j ≤ k so that d ( y, x j ) < δ . Now, 2 h x, x j i p = d ( p, x ) + r + δ − d ( x, x j ) > r , so x ∈ N ( l j , r ). (cid:3) Corollary 15.10.
For each
P ∈ ∂ ( ω h ) there is a geodesic ray l emanating from p such that l ∈ P .Proof. For each n ≥ l n emanating from p such that N ( l n , n ) ∈ P (use 15.9). Let α n : [0 , ∞ ) → l n be a parametrization of l n foreach n ≥
1. By a diagonal argument choose an increasing sequence n ( k ) with theproperty α n ( k ) ( q ) is convergent to α ( q ) for each rational q . Notice α extends to aparametrization α of a ray l and α n ( k ) ( t ) is convergent to α ( t ) for each t ≥
0. Il l / ∈ P , then there is C ∈ P , C ⊥ l . Therefore C ⊥ N ( l, r ) for some r > k ≥ m = n ( k ) > r + δ , d ( α m ( r + δ ) , α ( r + δ )) < δ .Suppose x ∈ N ( l m , m ) and choose y ∈ l m such that h x, y i p > m . Put z := α ( r + δ ), t := α m ( r + δ ), and notice h y, z i p = 0 . · ( d ( y, t ) + r + δ + r + δ − d ( y, z )) > . · ( d ( y, t ) + r + δ + r + δ − ( d ( y, t ) + δ )) = 0 . · (2 r + δ ) = r + δ/
2. Therefore, h x, z i p ≥ min( h x, y i p , h z, y i p ) − δ/ > min( m, r + δ/ − δ/ > r . Consequently, N ( l m , m ) ⊂ N ( l, r ) and C ⊥ N ( l m , m ) which contradicts N ( l m , m ) ∈ P . (cid:3) Proposition 15.11.
The relation l ∼ l defined as ω h ( l , l ) = ∞ among geodesicrays emanating from p is an equivalence relation and each P ∈ ∂ ( ω h ) containsexactly one equivalence class.Proof. Pick parametrizations α i : [0 , ∞ ) → l i for i ≤
2. Notice that h α ( t ) , α ( t ) i p is an increasing function of t and it approaches ∞ if and only if ω h ( l , l ) = ∞ .Given a third ray l with parametrization α and ω h ( l , l ) = ∞ , the inequality h α ( t ) , α ( t ) i p ≥ min( h α ( t ) , α ( t ) i p , h α ( t ) , α ( t ) i p − δ/ ω h ( l , l ) = ∞ .If ω h ( l , l ) = ∞ , then for each r > M > h α ( t ) , α ( t ) i p >r + δ . If h α ( t ) , x i p > r + M + δ , then t > M + δ and h x, α ( M ) i p ≥ min( h α ( t ) , x i p , h α ( t ) , α ( t ) i p ) − δ/ > r .Thus, N ( l , r + M + δ ) ⊂ N ( l , r ). In particular, ω h ( l , C , . . . , C k ) = ∞ implies ω h ( l , l , C , . . . , C k ) = ∞ , so any P containing l contains l as well. (cid:3) Observation 15.12.
One can show existence of a universal constant K dependingon δ ony such that given parametrizations α i : [0 , ∞ ) → l i for i ≤ of two equivalent geodesic rays l and l , one has d ( α ( t ) , α ( t )) < K for all t (see Lemma 9.51 onp.229 in [6] ) which is obviously stronger than l ∼ l . Proposition 15.13.
Given a geodesic ray l emanating from p , the sets o ( N ( l, n )) , n ≥ , form a basis of open sets of P ∈ ∂ ( ω h + ω ss ) containing l .Proof. Let ω := ω h + ω ss . Since N ( l, n ) is d -open in X , it is ω -open. If P ∈ o ( U ), U being ω -open, then X \ U / ∈ P , so ( X \ U ) ∩ N ( l, k ) = ∅ for some k ≥ N ( l, k ) ⊂ U . Since X \ N ( l, k ) / ∈ P , P ∈ o ( N ( l, k )) ⊂ o ( U ). (cid:3) Theorem 15.14. If ω ss is the basic topological form of a proper visual geodesic δ -hyperbolic space ( X, d ) and ω h is the hyperbolic form of X , then X ∪ ∂ ( ω h + ω ss ) is the compactification of ( X, d ) characterized by extensions of bounded Gromovfunctions X → R .Proof. A bounded Gromov functions X → R is a continuous function such that foreach ǫ > r > | f ( x − f ( y ) | > ǫ implies h x, y i p < r . Notice Gromovfunctions are identical with bounded form-continuous functions ( X, ω ) → R , where ω = ω h + ω ss . Apply 12.3 and 12.4. (cid:3) Remark . Theorem 15.14 shows that X ∪ ∂ ( ω h + ω ss ) is the classical Gromovcompactification (see 6.4 in [20] on p.93).16. Dimension of formed sets
The goal of this section is to give simple proofs of results that generalize thework of Austin-Virk [1] on Dimension Raising maps in coarse category.
Definition 16.1.
A form-continuous function f : ( X, ω X ) → ( Y, ω Y ) of formedsets is proper if images of ω X -bounded sets are ω Y -bounded. Theorem 16.2.
Suppose f : ( X, ω X ) → ( Y, ω Y ) is a surjective and proper form-continuous function of normal T formed sets and n ≥ . The induced map ∂f : ∂ ( ω X ) → ∂ ( ω Y ) is always surjective and it is n -to- if and only if for each sequence { A i } n +1 i =1 of sub-sets of X , the condition ω X ( A i , A j ) = 0 for all i = j implies ω Y ( f ( A ) , . . . , f ( A n +1 )) =0 .Proof. See 12.3 for a description of ∂f . Using 3.10 observe that the continuousextension ˜ f of f is closed, hence the extension ˜ f of f is surjective ( Y is dense in Y ∪ ∂ ( ω Y )). In particular ∂f is surjective as no point in X \ ∂ ( ω X ) can be mappedto ∂ ( ω Y ).Suppose ∂f is not n -to-1, i.e. there exists Q ∈ ∂ ( ω Y ) and Q i ∈ ∂ ( ω X )), 1 ≤ i ≤ n + 1, such that ( ∂f )( Q i ) = Q and Q i = Q j if i = j . Since X ∪ ∂ ( ω X ) is Hausdorff,there exist mutually disjoint closed neighborhoods D i of Q i in X ∪ ∂ ( ω X ). Put A i := D i ∩ X . Notice A i ∈ Q i for each i ≤ n + 1. Also, ω X ( A i , A j ) = 0 for all i = j . However, ω Y ( f ( A ) , . . . , f ( A n +1 )) = ∞ as f ( A i ) ∈ Q for each i ≤ n + 1.Suppose there exist sets A i , i ≤ n + 1, such that ω X ( A i , A j ) = 0 whenever i = j but ω Y ( f ( A ) , . . . , f ( A n +1 )) = ∞ . We may assume each A i is closed in X by replacing it with cl ( A i ) (see 7.3). There is Q ∈ ∂ ( ω Y ) containing f ( A i ) for each i . In particular, Q ∈ ∂ ( f ( A i ) , ω Y ) for each i . Therefore, for each i ≤ n + 1, thereis Q i ∈ ∂ ( A i , ω X ) so that ( ∂f )( Q i ) = Q (use 7.8). Since Q i = Q j in ∂ ( ω X ), weconclude that ∂f is not n -to-1. (cid:3) INEAR ALGEBRA AND UNIFICATION OF GEOMETRIES IN ALL SCALES 33
Theorem 16.3.
Suppose ( X, L X ) and ( Y, L Y ) are large scale spaces such that theinduced forms ω X , ω Y are normal and T . If f : ( X, L X ) → ( Y, L Y ) is coarsely n -to- for some n ≥ , then the induced map ∂ ( ω X ) → ∂ ( ω Y ) of form coronas is n -to- .Proof. Suppose the induced map ∂ ( ω X ) → ∂ ( ω Y ) of form coronas is not n -to-1. ByTheorem 16.2 there exist sets A i ⊂ X , i ≤ n + 1 such that ω X ( A i , A j ) = 0 if i = j but ω Y ( f ( A ) , . . . , f ( A n +1 )) = ∞ . Therefore, there exists a uniformly boundedcover U of Y such that n +1 T i =1 st ( f ( A i ) , U ) is ω Y -unbounded. Choose a uniformlybounded cover W of X such that each set f − ( U ), U ∈ st ( U , U ), can be covered byat most n -elements of W . Define C i as A i \ B i , where B i := S j = i st ( A i , W ) ∩ st ( A j , W )is ω X -bounded. If x ∈ n +1 T i =1 st ( C i , f − ( U )), then st ( x, f − ( U )) ∩ C i = ∅ for each i ,which implies existence of i = j such that st ( C i , W ) ∩ st ( C j , W ) = ∅ , a contradic-tion. That means n +1 T i =1 st ( C i , f − ( U )) = ∅ . Consequently, n +1 T i =1 st ( f ( C i ) , U ) = ∅ and n +1 T i =1 st ( f ( A i ) , U )) ⊂ n +1 S i =1 st ( f ( B i ) , U ) is ω Y -bounded, a contradiction. (cid:3) Theorem 16.4.
Suppose f : X → Y is a coarse bornologous function of large scalespaces whose induced forms are normal and T . If X is metrizable and n ≥ , then f is coarsely n -to- if and only if the induced map of their boundaries at infinity is n -to- .Proof. Suppose f is not coarsely n -to-1 and the induced map of the boundariesat infinity is n -to-1. That means the existence of a uniformly bounded cover U of Y such that for each k ≥ U k ∈ U such that f − ( U k ) cannot becovered by at most n sets of diameter at most 2 k . That implies existence, foreach k ≥
1, of points x k , . . . , x kn +1 of f − ( U k ) such that d ( x ki , x kj ) ≥ k whenever i = j . Let A i := { x ki } ∞ k =1 . Notice ω X ( A i , A j ) = 0 if i = j . By Theorem 16.2, ω Y ( f ( A ) , . . . , f ( A n +1 )) = 0. Let C i = st ( f ( A i ) , U ) for i ≤ n + 1. Therefore ω Y ( C , . . . , C n +1 ) = 0. Since W := ∞ S i =1 U n ⊂ C i for each i , ω Y ( W ) = 0 and W isbounded in Y . Hence f − ( W ) is bounded in X , a contradiction as in that case,for k > diam ( f − ( W )), f − ( U k ) can be covered by one set of diameter at most 2 k ,namely f − ( W ).Apply 16.3 to conclude the proof. (cid:3) Theorem 16.5. If n ≥ and f : X → Y is a coarsely n -to- bornologous map oflarge scale spaces, then asdim ( Y ) ≤ asdim ( X ) + n − .Proof. Case 1:
First consider the case of X and Y being metrizable. It is known(see [1]) that asdim ( Y ) is finite. By 14.3 it suffices to show dim( ∂ ( ω Y )) ≤ dim( ∂ ( ω X ))+ n −
1. It is so (see 16.4) as the induced map between boundaries of forms is n -to-1and Theorem 3.3.7 on page 196 in [13] does say that dimension of the image of an n -to-1 map between compact spaces is at most dimension of the range plus ( n − Case 2: Y is metrizable. Fix a metric d Y on Y inducing its large scale structure.By induction construct a sequence of uniformly bounded covers }V i } ∞ i =1 of X and asequence of real numbers r i diverging to infinity satisfying the following conditions:
1. Each element of f − ( { B ( y, r i ) } y ∈ Y ) is contained in a union of at most n elementsof V i for each i ≥ · r i < r i +1 for each i ≥ st ( V i , V i ) refines V i +1 for each i ≥ V i is of multiplicity at most asdim ( X ) + 1 for each i ≥ f ( V i ) refines { B ( y, r i +1 ) } y ∈ Y for each i ≥ d X : X × X → [0 , ∞ ] as follows:a. d X ( x, x ) = 0 for each x ∈ X ,b. If x = y ∈ X and there is ≥ x and y belong to an element of V k , then d X ( x, y ) is the smallest such number k ,c. d X ( x, y ) = ∞ if x, y do not qualify in a) and b).Consider the partition of X into equivalence classes determined by x ∼ y if d X ( x, y ) < ∞ . There are at most n such classes. Indeed, if S ⊂ X is a set containing n + 1 elements pairwise non-equivalent, then f ( S ) is contained in B ( y, r i ) for some y ∈ Y and some r i . Therefore S is contained in a union of at most n elements of V i , so two elements of S belong to one one element of V i , a contradiction.Notice that, for each equivalence class X ′ of X , f : X ′ → f ( X ′ ) is coarsely n -to-1. By Case 1, asdim ( f ( X ′ )) ≤ asdim ( X ) + n −
1. Therefore asdim ( Y ) ≤ asdim ( X ) + n − General Case:
Suppose there is a uniformly bounded cover U of Y which can-not be coarsened to a uniformly bounded cover of multiplicity at most asdim ( X )+ n .By induction construct a sequence of uniformly bounded covers {V i } ∞ i =1 of X anda sequence of uniformly bounded covers {U i } ∞ i =1 of Y satisfying the following con-ditions:1. Each element of f − ( U i ) is contained in a union of at most n elements of V i foreach i ≥ st ( U i , U i ) refines U i +1 for each i ≥ st ( V i , V i ) refines V i +1 for each i ≥ V i is of multiplicity at most asdim ( X ) + 1 for each i ≥ f ( V i ) refines U i +1 for each i ≥ X that refine V i for some i ≥ LS X on X . Similarly, all covers of Y that refine U i for some i ≥ LS Y on Y . Also, f : ( X, LS X ) → ( Y, LS Y ) is coarsely n -to-1. Consider the following equivalence relation on Y : y ∼ y if there is i ≥ U of U i containing both y and y . By Case 2, each equivalenceclass of Y is of asymptotic dimension at most asdim ( X ) + n −
1. Consequently, foreach equivalence class Y ′ of Y there is the smallest integer k ( Y ′ ) such that U | Y ′ can be coarsened to a cover refining U k and of multiplicity at most asdim ( X ) + n .Therefore there is an infinite sequence Y i of equivalence classes of Y so that k ( Y i )is strictly increasing. Put Z := ∞ S i =1 Y i . For each i > U i by adding i − S j =1 Y i to it. Similarly, each i > V i by adding i − S j =1 f − ( Y i ) to it. Thenew covers induce metrizable large scale structures on Z and f − ( Z ) such that f : f − ( Z ) → Z is coarsely n -to-1. Moreover, asdim ( f − ( Z )) ≤ asdim ( X ). ByCase 1, asdim ( Z ) ≤ asdim ( X ) + n −
1, so there is a uniformly bounded cover U of Z coarsening U | Z of multiplicity at most asdim ( X ) + n . That contradicts k ( Y i )being strictly increasing. (cid:3) INEAR ALGEBRA AND UNIFICATION OF GEOMETRIES IN ALL SCALES 35
Remark . Theorem 16.5 was proved by Austin-Virk in [1] for proper metricspaces X and Y . Theorem 16.7.
A coarse bornologous function f : X → Y of metrizable largescale spaces is a large scale equivalence if and only if it induces a homeomorphismof Higson coronas.Proof. Suppose f : X → Y induces a homeomorphism of Higson coronas. Case 1 : X ⊂ Y and f is the inclusion. Given a metric d on Y we need toshow existence of r > B ( X, r ) = Y . If such r does not exist, then Y i := Y \ B ( X, i ) is a sequence of unbounded sets of Y and there is P ∈ ∂ ( ω Y )containing a sequence y i ∈ Y i . There is Q ∈ ∂ ( ω X ), ˜ f ( Q ) = P . Since X ∈ Q , X ∈ P leading to a contradiction: ω Y ( X, { y i } ∞ i =1 ) = 0. General Case : By Case 1, Z := f ( X ) induces a large scale equivalence i : Z → Y .Apply 16.4 to conclude f : X → Z is coarsely 1-to-1. That means precisely that theinverse of a uniformly bounded cover of Z is uniformly bounded in X from whichit follows that f is a large scale equivalence. (cid:3) Corollary 16.8.
A coarse bornologous function f : X → Y of metrizable large scalespaces is a large scale embedding if and only if it induces a topological embedding ofHigson coronas.Proof. f : X → f ( X ) induces a homeomorphism of Higson coronas. Apply 16.7. (cid:3) Metrizability of form compactifications
Theorem 17.1.
Suppose ( X, ω ) is a normal T formed set. The following condi-tions are equivalent:1. X ∪ ∂ ( ω ) is metrizable,2. There exists a countable family S of subsets of X such that for any subsets C , . . . , C k of X satisfying ω ( C , . . . , C k ) = 0 there are D i ∈ S , i ≤ k , such that C i ⊂ D i for each i ≤ k and ω ( D , . . . , D k ) = 0 ,3. There exist subsets C n , n ≥ , of X such that for any two subsets D, E of X satisfying ω ( D, E ) = 0 there is C k such that D ⊂ C k and ω ( C k , E ) = 0 .Proof.
1) = ⇒ d on X ∪ ∂ ( ω ), notice that complements B n of allballs B ( ∂ ( ω ) , n ), n ≥
1, are ω -bounded and we start S by enlisting all of them. As( ∂ ( ω ) , d ) is compact metric, there is a countable family { D } of subsets of X suchthat sets o ( D ) form a basis at points of ∂ ( ω ). Add all finite unions of elements of { D } ∪ { B n } ∞ n =1 to S . If subsets C , . . . , C k of X satisfy ω ( C , . . . , C k ) = 0, then theintersection of sets cl ( C i ) ∩ ∂ ( ω ) is empty, so there is a finite union E i of elementsof { D } such that C i \ E i is ω -bounded and the intersection of sets cl ( E i ) ∩ ∂ ( ω ) isempty. Each set C i \ E i must be contained in some B n , so all of them are containedin a single B m . Define D i as E i ∪ B m and we are done.3) = ⇒ ω -bounded elements of the family { C n } form a basis of ω -bounded subsets of X in the following sense: if B is ω -bounded, then there is n such that C n is ω -bounded and B ⊂ C n . List those sets as { B n } . Claim : The family { int ( cl ( C n ) \ B m ) } forms a basis of points of ∂ ( ω ) in X ∪ ∂ ( ω ). Proof of Claim : Suppose
Q ∈ ∂ ( ω ) and Q ∈ o ( U ) ⊂ cl ( o ( U )) ⊂ o ( W ). Now, ω ( U, X \ W ) = 0, so there exists D ∈ { C n } containing U such that ω ( D, X \ W ) = 0.Choose B n containing D ∩ ( X \ W ) and notice cl ( D \ B n ) ∩ cl ( X \ W ) = ∅ . Therefore Q ∈ int ( cl ( D \ B n )) ⊂ o ( W ). (cid:4) Given a countable basis { U n } of points of ∂ ( ω ) in X ∪ ∂ ( ω ) choose, for everypair ( U n , U m ) such that cl ( U n ) ⊂ U m , a continuous function f m,n : X ∪ ∂ ( ω ) → [0 , − m − n ] such that f ( cl ( U n )) ⊂ { } and f ( X ∪ ∂ ( X, ω ) \ U m ) ⊂ { − m − n } . Add allDirac functions of points in X that are ω -bounded and the resulting family { f s } s ∈ S has continuous sum f := P s ∈ S f s . Now, { f s /f } s ∈ S is a partition of unity on X ∪ ∂ ( ω )whose carriers form a basis of X ∪ ∂ ( ω ). By [8], X ∪ ∂ ( ω ) is metrizable. (cid:3) Forms on large scale spaces
Proposition 18.1. If A is a subset of a large scale space ( X, L ) , then the forminduced by L| A coincides with ω ( L ) | A .Proof. Suppose C i ⊂ A , i ≤ k , and for each uniformly bounded cover U of X , theset k T i =1 st ( C i , U ) is bounded. Suppose W is a uniformly bounded cover of X . Toshow ω ( L )( C , . . . , C k ) = 0 it suffices to prove inclusion k \ i =1 st ( C i , W ) ⊂ st ( k \ i =1 st ( C i , st ( W , W ) | A ) , W ) . If x ∈ k T i =1 st ( C i , W ) \ A , then for each i there is W i ∈ W such that C i ∩ W = ∅ and x ∈ C i ∪ W i . Choose y i ∈ C i ∩ W and notice y j ∈ st ( C i , st ( W , W ) | A ) for all i, j ≤ k . Therefore x ∈ st ( k T i =1 st ( C i , st ( W , W ) | A ) , W ). (cid:3) Proposition 18.2.
Suppose f, g : ( X, L X ) → ( Y, L Y ) are coarse maps of largescale spaces.a. If the induced form ω Y := ω ( L Y ) is normal T and f is close to g , then ∂f = ∂g .b. If ( Y, L Y ) is metrizable and ∂f = ∂g , then f is close to g .Proof. a. U := { f ( x ) , g ( x ) } x ∈ X is a uniformly bounded family in Y . Suppose Q ∈ ∂ ( ω ( L X ) and ( ∂f )( Q ) = ( ∂g )( Q ). By 12.3 there is C ∈ Q such that f ( C ) / ∈ ( ∂g )( Q ). Using 12.3 again, we detect D ∈ Q so that ω Y ( f ( C ) , g ( D )) = 0. Hence ω Y ( st ( f ( C )) , U ) , g ( D )) = 0, a contradiction as g ( C ) ⊂ st ( f ( C )) , U ).b. If f is not close to g , then for each k ≥ x k ∈ X such that d ( f ( x k ) , g ( x k )) > k . Put A := { x k } ∞ k =1 and notice it is unbounded. Notice ω Y ( f ( A ) , g ( A )) = 0, so for any Q ∈ ∂ ( ω ( L X )) containing A one has ( ∂f )( Q ) =( ∂g )( Q ), a contradiction. (cid:3) Asymptotic dimension 0
It is well-known that a proper metric space X is of asymptotic dimension 0 ifand only if its Higson corona is of dimension 0. In this section we generalize thisreasult to large scale structures induced by normal T forms. Definition 19.1.
A bornological large space X is coarsely totally disconnected if U -components of every uniformly bounded cover U of X are bounded. INEAR ALGEBRA AND UNIFICATION OF GEOMETRIES IN ALL SCALES 37
Lemma 19.2. If X is coarsely totally disconnected and W is the family of all U -components of X for some uniformly bounded cover U , then st ( B, W ) is boundedfor all bounded sets B .Proof. If st ( B, W ) is unbounded, then the U ′ -component of X containing B , U ′ := U ∪ { B } , contains st ( B, W ) and is unbounded, a contradiction. (cid:3) Theorem 19.3.
Suppose ω is a normal T form on X . If dim( X ∪ ∂ ( ω )) = 0 and LS ( ω ) is coarsely totally disconnected, then the asymptotic dimension of LS ( ω ) equals .Proof. Suppose U is an element of LS ( ω ) such that W , the collection of U -componentsof X , is not uniformly bounded. That means existence of an ω -unbounded C ⊂ X such that X ∩ cl ( C ) ( X := ∂ ( ω )) is a proper subset of X ∩ cl ( st ( C, W )). Choose Q ∈ X belonging to cl ( st ( C, W )) \ cl ( C ). There is an open-closed subset Y of X ∪ ∂ ( ω ) containing cl ( C ) and missing Q . Let Y := X ∩ Y .Choose a continuous function α : X ∪ ∂ ( ω ) → S such that α ( Y ) and α ( X \ Y )are disjoint. Consider S := { U ∈ U| α ( U ) = S } . B := S S is ω -bounded asotherwise, for each U ∈ S , we can pick x U , y U ∈ U satisfying α ( x U ) = − α ( y U ) = 1. In that case both D := { x U } U ∈S and E := { y U } U ∈S are ω -unboundedwith disjoint coronas contradicting E ⊂ st ( D, U ).Consider Y \ st ( B, W ) and ( X \ Y ) \ st ( B, W ). There is no U -chain joining thesetwo sets, hence the stars of the two sets with respect to W are disjoint, contradicting Q belonging to cl ( st ( C, W )) \ cl ( C ). (cid:3) Example 19.4.
Given an infinite set X consider ω defined as follows: ω ( V ) = 0 if and only if at least one coordinate of V is finite. In that case LS ( ω ) consists ofall covers U of X by finite sets such that st ( F, U ) is finite for each finite subset F of X . Notice LS ( ω ) is not coarsely totally disconnected. Indeed, choose an infinitesequence { x n } ∞ n =1 in X and notice the trivial extension U of { x n , x n +1 } ∞ n =1 hasinfinite U -component. Conjecture 19.5.
If If dim( X ∪ ∂ ( ω )) = 0 and the asymptotic dimension of LS ( ω ) does not equal , then there is an unbounded subset C of X whose corona consistsof exactly one point. Group actions on formed sets
The goal of this section is to give sufficient conditions for finite group actions on X to produce spaces of orbits X/G of the same asymptotic dimension as asdim ( X ). Definition 20.1.
Suppose a group G acts on a formed set ( X, ω ). ω is G -invariant if ω ( V ) = ω ( g · V )for each g ∈ G and each vector V in X .Naturally, g · ( C , . . . , C k ) is defined as ( g · C , . . . , g · C k ).The space of orbits X/G consists of sets of the form G · x . There is a naturalprojection π : X → X/G ( π ( x ) := G · x ) and X/G is equipped with the form ω G defined as follows: ω G ( C , . . . , C k ) = ω ( π − ( C ) , . . . , π − ( C k ))for all vectors ( C , . . . , C k ) in X/G . Proposition 20.2.
Suppose G is a finite group acting on X . If ω is a normal, T ,and G -invariant form on X , then the induced form ω G on X/G is normal and T .Moreover, ˜ π : X ∪ ∂ ( ω ) → X/G ∪ ∂ ( ω G ) is a closed map.Proof. Given two different orbits G · x and G · y , ω ( G · x, G · y ) = 0 (use additivityof ω ). Thus ω G is T .Supposae ω ( π − ( C ) , . . . , π − ( C k )) = 0 for a vector ( C , . . . , C k ) in X/G . Choosesets D i in X so that ω ( D i , π − ( C i )) = 0 for each i ≤ k and { D i } ki =1 covers X . Foreach g ∈ G , ω ( g · D i , π − ( C i )) = ω ( g · D i , g · π − ( C i )) = ω ( D i , π − ( C i )) = 0.Therefore, ω ( G · D i , π − ( C i )) = 0 and sets E i := π ( G · D i ) form a cover of X/G such that ω G ( E i , C i ) = 0 for each i ≤ k .To show ˜ π is closed we apply 3.10 by observing π ( B ) is ω G -bounded if B is ω -bounded. Indeed, ω G ( π ( B ) , X/G ) = ω ( G · B, X ) = 0 as G · B is a finite union of ω -bounded sets g · B . (cid:3) Corollary 20.3.
Suppose G is a finite group acting on X . If ω is a normal, T ,and G -invariant form on X , then dim( X ∪ ∂ ( ω )) = dim( X/G ∪ ∂ ( ω G )) .Proof. In order to apply Proposition 9.2.16 from [19] we need to show that ˜ π : X ∪ ∂ ( ω ) → X/G ∪ ∂ ( ω G ) is open and its fibers are finite. Here is the full statementof that proposition:Let X, Y be weakly paracompact, normal spaces. Let f : X → Y be a continuous,open surjection. If for every point y ∈ Y the pre-image f − ( y ) is finite, thendim( X ) = dim( Y ).Notice we can extend the action of G over X ∪ ∂ ( ω ). Namely, g · P := { g · C | C ∈P} . Observe g · o ( U ) = o ( g · U ), so G acts by homeomorphisms on X ∪ ∂ ( X, ω ). Itsuffices to show that fibers of ˜ π coincide with the orbits of G , then ˜ π is open as itis quotient (it is actually closed) and the inverse of ˜ π ( W ) is G · W for each opensubset W of X ∪ ∂ ( X, ω ).Suppose ˜ π ( Q ) = ˜ π ( P ) but Q 6 = g · P for all g ∈ G . For each g ∈ G there areopen subsets U g , W g of X such that Q ∈ o ( U g ), g · P ∈ o ( W g ) and closures of o ( U g ) and o ( W g ) are disjoint. In particular, ω ( U g , W g ) = 0. Put U = T g ∈ G U g and W = T g ∈ G g − · W g . Q ∈ o ( U ), P ∈ o ( W ), and ω ( G · U, G · W ) = 0, a contradictionas G · U ∈ ˜ π ( Q ) and G · W ∈ ˜ π ( P ). (cid:3) Theorem 20.4.
Let X be a metric space and let G be a finite group acting isometri-cally on X . Then X/G equipped with the Hausdorff metric has the same asymptoticdimension as X if asdim ( X ) < ∞ .Proof. Let ω be the basic large scale form on X induced by its metric d . We willshow that the basic large scale form λ on X/G induced by the Hausdorff metric d H equals the form ω G on X/G obtained from ω via the projection π : X → X/G .Since both forms are normal and T , it suffices to show ω G ( C, D ) = λ ( C, D ) for all
C, D ⊂ X/G (see 7.5).Suppose ω G ( C, D ) = ∞ . That means B ( G · C, r ) ∩ B ( G · D, r ) is unbounded forsome r >
0. Pick points c i ∈ G · C and d i ∈ G · D such that d ( c i , d i ) < r foreach i ≥ { c i } ∞ i =1 is unbounded. Notice d H ( G · c i , G · d i ) < r for each i ≥ λ ( G · C, G · D ) = ∞ .Conversely, if λ ( G · C, G · D ) = ∞ , then there is r > c i ∈ G · C and d i ∈ G · D such that d H ( G · c i , G · d i ) < r for each i ≥ { G · c i } ∞ i =1INEAR ALGEBRA AND UNIFICATION OF GEOMETRIES IN ALL SCALES 39 is unbounded in d H . We may pick the sequences c i ∈ G · C so that { c i } ∞ i =1 isunbounded in X . For each i we may choose the closest point in G · d i thus adjustingthe sequence d i ∈ G · D so that d ( c i , d i ) < r for each i ≥
1. That means Thatmeans B ( G · C, r ) ∩ B ( G · D, r ) is unbounded and ω G ( C, D ) = ∞ .Notice π : ( X, d ) → ( X/G, d H ) is coarsely | G | -to-1. Therefore asdim ( X/G ) isfinite and equal to dim( ∂ ( ω G )) = dim( ω X ) = asdim ( X ) by 20.3. (cid:3) Remark . In case of proper metric spaces X , Theorem 20.4 was proved byDaniel Kasprowski [16]. 21. Parallelism
It is traditional in two-dimensional geometry to call two lines parallel if theyare either equal or disjoint. A more natural definition of being parallel is to havethe same coronas. However, on the hyperbolic plane only identical lines would beparallel under that definition.
Definition 21.1.
Suppose ω is a normal T form on a set X . C ⊂ X is parallel to D ⊂ X if their coronas are equal.See [15] for other interesting equivalence relations in coarse geometry. Proposition 21.2.
Suppose ω is a normal T form on a set X . C ⊂ X is parallelto D ⊂ X if and only if E ⊥ C ⇐⇒ E ⊥ D for all E ⊂ X , where ⊥ is theorthogonality relation induced by ω .Proof. E ⊥ C means that the corona of E is disjoint from the corona of C . There-fore assume E ⊥ C ⇐⇒ E ⊥ D for all E ⊂ X but P is in the corona of C but not in the corona of D . As D / ∈ P , there exists a vector ( C , . . . , C k ) whosecoordinates are in P but ω ( D, C , . . . , C k ) = 0. By 12.3 applied to id X , there is D ′ ⊥ D belonging to P . Hence D ′ ⊥ C , a contradiction as both D ′ and C belongto P . (cid:3) Visual forms
In this section we consider proper geodesic spaces (
X, d ) satisfying the followingtwo conditions:1. For every two points x, y ∈ X there is a unique geodesic joining them,2. For every point p ∈ X , X is the union of all infinite geodesic rays emanatingfrom p (those spaces will be called visual ).Given a proper geodesic space ( X, d ) with uniqueness of geodesics, for each twopoints x, y ∈ X we denote the geodesics from x to y by [ x, y ]. The function g ( x, y, t ),where x, y ∈ X and t ≤ d ( x, y ) is the parametrization of [ x, y ].Given a geodesic ray l emanating from p ∈ X , its parametrization is denoted by g ( p, l, t ), t ≥ Definition 22.1.
Given C ⊂ X and r > N p ( C, r ) as the closure of theset of all α ( r ), α ranging over all parametrizations α : [0 , ∞ ) → X of geodesic raysemanating from p such that α [ r, ∞ ) intersects cl ( C ).The form ω ( N p ) induced by the function N p will be called the visual form on( X, d ) with respect to p . Definition 22.2.
Given a subset D of the r -sphere S ( p, r ) centered at p , Cone p ( D )is defined as the union of all α [ r, ∞ ), α ranging over all parametrizations α :[0 , ∞ ) → X of geodesic rays emanating from p such that α ( r ) ∈ D . Proposition 22.3. If D is a subset of the r -sphere S ( p, r ) centered at p , then N p ( Cone p ( D ) , r ) = D .Proof. Clearly, N p ( Cone ( D ) , r ) ⊂ D . Conversely, if X ∈ D , there is a parametriza-tion α : [0 , ∞ ) → X of a geodesic ray such that α ( r ) = x . Consequently, D ⊂ N p ( Cone p ( D ) , r ). (cid:3) Corollary 22.4. If D is a closed subset of the r -sphere S ( p, r ) centered at p , and D ⊂ S ( p, r ) \ N ( C, r ) , then N p ( Cone p ( D ) , r ) ∩ N p ( C, r ) = ∅ .Proof. Obvious, as N p ( Cone p ( D ) , r ) = D . (cid:3) Corollary 22.5. ω ( N p ) is a normal T form that is compatible with the form ω ss induced by the topology of ( X, d ) .Proof. To show normality of ω ( N p ) suppose k T i =1 N p ( C i , r ) = ∅ . Now, k S i =1 ( S ( p, r ) \ N p ( C i , r )) = S ( p, r ) and we can choose closed subsets D k of S ( p, r ) such that D i ⊂ S ( p, r ) \ N p ( C i , r ) for each i ≤ k and k S i =1 D i = S ( p, r ). By 22.4, ω ( N p )( D i , C i ) = 0for each i ≤ k which proves ω ( N p ) is normal. N p ( C, r ) = ∅ if and only if cl ( C ) ⊂ B ( p, r ). Thus, ω ( N p )-bounded sets are iden-tical with pre-compact subsets of X . Since N p ( C, r ) = N p ( cl ( C ) , r ), compatibilityof forms follows from 15.6. (cid:3) Definition 22.6.
Fix p ∈ X . For each n ≥ π n +1 n : S ( p, n +1) → S ( p, n ) be de-fined as π n +1 n ( x ) = g ( p, x, n ). Notice π n +1 n is continuous. Given C ⊂ X let Inv p ( C )be the set of all sequences { x n } ∞ n =1 such that x n ∈ S ( p, n ) and π n +1 n ( x n +1 ) = x n for each n ≥
1. If C is unbounded, then Inv p ( C ) = ∅ as Inv p ( C ) is identical withthe inverse limit of { N p ( C ) , π n +1 n } . Proposition 22.7.
For each
P ∈ ∂ ( ω ( N p ) there is a unique geodesic ray l P at p belonging to P . Moreover, C ∈ P if and only if N p ( l P , C ) = ∞ .Proof. Given n ≥
1, any family C , . . . , C k ∈ P induces non-empty intersection of N p ( C i , n ), i ≤ k . Hence all N p ( C, n ), C ∈ P , have non-empty interesection denotedby N p ( P , n ). Observe π n +1 n ( N p ( P , n + 1)) ⊂ N p ( P , n ) for each n ≥
1, so thereis a geodesic ray l P at p satisfying g ( p, l P , n ) ∈ N p ( P , n ) for each n ≥
1. Since { g ( p, { l P , n ) } = N p ( l P , n ) ∈ N p ( C, n ) for each n ≥ C ∈ P , l P ∈ P . Anygeodesic ray l at p belonging to P must satisfy g ( p, l, n ) = g ( p, l P , n ) for all n ≥ l = l P . (cid:3) Proposition 22.8.
Given
P ∈ ∂ ( ω ( N p ) , the sets o ( B ( p, n + 1) ∪ Cone p ( B ( l P , ǫ ) ∩ S ( p, n ))) form a basis of open neighborhoods of P in ∂ ( ω ( N p ) + ω ss ) .Proof. Suppose U is open in ( X, d ) and
P ∈ o ( U ), i.e. ( X \ U ) / ∈ P . Hence N ( l P , n ) ∩ N p ( X \ U, n ) = ∅ for some n ≥ ǫ > cl ( B ( g ( p, l P , ǫ ))) ∩ N p ( X \ U, n ) = ∅ . Suppose Q ∈ o ( B ( p, n + 1) ∪ Cone p ( B ( l P , ǫ ) ∩ S ( p, n ))). Therefore Cone p ( B ( l P , ǫ ) ∩ S ( p, n )) ∈ Q , so N p ( Q , n ) ⊂ B ( l P , ǫ ) ∩ S ( p, n )(see the proof of 22.4) and N p ( Q , n ) ∩ ( X \ U ) = ∅ resulting in X \ U / ∈ Q (see22.7), i.e. Q ∈ o ( U ). (cid:3) INEAR ALGEBRA AND UNIFICATION OF GEOMETRIES IN ALL SCALES 41
CoG-spaces.
In this part we discuss a necessary and sufficient condition on(
X, d ) for ω ( N p ) = ω ( N q ) for all p, q ∈ X . That condition is satisfied by CAT(0)-spaces. Definition 22.9.
A sequence of geodesics [ p, x n ] converges to a geodesic ray l at p if lim d ( p, x n ) = ∞ and lim g ( p, x n , t ) = g ( p, l, t ) for all t ≥ Observation 22.10.
If one defines convergence of a sequence of geodesics [ p, x n ] , n ≥ , converges to a geodesics [ p, x ] analogously to 22.9, then it simply amountsto lim x n = x . In the next definition CoG is an initialism for ”convergence of geodesics”.
Definition 22.11.
A proper geodesic visual space (
X, d ) is a
CoG-space if, when-ever a sequence of geodesics [ p, x n ], n ≥
1, converges to a geodesic ray l p , then forevery point q ∈ X there is a geodesic ray l q at q such that the sequence of geodesics[ q, x n ], n ≥
1, converges to l q . Proposition 22.12.
Each CAT(0)-space ( X, d ) is a CoG-space and ∂ ( ω ( N p )) isthe visual boundary at infinity of ( X, d ) .Proof. CAT(0)-spaces have the property that for each geodesic ray l p at p ∈ X andeach q ∈ X there is K > l q at q such that d ( g ( p, l p , t ) , g ( q, l q , t ) 1, converges to a geodesic ray l p and let l q be a geodesic ray at q suchthat for some constant K > r ≥ d ( g ( p, l p , t ) , g ( q, l q , t ) < K forall t ≥ 0. Given r > d ( p, q ) choose m ≥ d ( g ( p, x n , r ) , g ( p, l p , r )) < K forall n ≥ m . Observe that d ( g ( p, x n , r ) , d ( q, x n , r )) ≤ d ( p, q ) (see the Claim below).Therefore, d ( q, x n , r ) , g ( q, l q , r )) leqd ( p, q ) + d ( p, q ) + K for all n ≥ m . That is suf-ficient to conclude [ q, x n ] converges to l q . Indeed, given k ≥ s > 1, put r = s · k · ( d ( p, q )+ d ( p, q )+ K ) and find m ≥ d ( q, x n , r ) , g ( q, l q , r )) leqd ( p, q )+ d ( p, q ) + K for all n ≥ m . By applying the comparison to an euclidean triangle, weconclude d ( g ( q, x n , s ) , g ( q, l q , s )) ≤ d ( p,q )+ d ( p,q )+ Kk · ( d ( p,q )+ d ( p,q )+ K ) = 1 /k for all n ≥ m . Claim : Given a triangle ∆( P QR ) on the euclidean plane and r ≤ min( P Q, P R ),the distance | Q ′ R ′ | between points Q ′ ∈ P Q and R ′ ∈ P R is at most | QR | if | QQ ′ | = r and | RR ′ | = r . Proof of Claim: It is so if Q ′ R ′ is parallel to QR . Otherwise, without loss ofgenerality, assume Q ′ is higher than R ′ , i.e. the line l parallel to QR and passingthrough R ′ intersects QQ ′ at a point Q ′′ . Consider the ray l parallel to the ray RP and emanating from Q . Look at the point A on l such that | AQ | = r . Notice that ∠ ( AQ ′ R ′ ) > ∠ ( R ′ AQ ′ ), so | AR ′ | > | Q ′ R ′ . (cid:3) Theorem 22.13. Let ( X, d ) be a proper geodesic visual space with uniqueness ofgeodesics. The visual form does not depend on the base point p if and only if ( X, d ) is a CoG-space.Proof. Suppose ω ( N p ) = ω ( N q ) for all p, q ∈ X and { [ p, x n ] } ∞ n =1 is a sequence ofgeodesics converging to a geodesic ray l at p . If { [ q, x n ] } ∞ n =1 does not converge to ageodesic ray at q , then (by a double diagonal process) we can find two subsequences C = { c n } ∞ n =1 and D = { d n } ∞ n =1 of { x n } ∞ n =1 such that { [ q, c n ] } ∞ n =1 converges to ageodesic ray l C at q , { [ q, d n ] } ∞ n =1 converges to a geodesic ray l D at q , and l C = l D .Notice ω ( N q )( l C , l D ) = 0 but ω ( N p )( l C , l D ) = ∞ . Suppose ( X, d ) is a CoG-space. Since both ω ( N p ) and ω ( N q ) are normal, itsuffices to show N p ( C, D ) = 0 implies N q ( C, D ) = 0 for all C, D ⊂ X .Suppose N p ( C, r ) ∩ N p ( D, r ) = ∅ but ω ( N q )( C, D ) = ∞ . Choose P ∈ ∂ ( ω ( N q ))containing both C and D and consider l P (see 22.7). There is a sequence { x n } ∞ n =1 such that [ q, x n ] converges to l q := l P and x k ∈ cl ( C ) for all k odd, x k ∈ cl ( D )for all k even. Let l p be the ray at p such that [ p, x n ] converges to l p . Notice l p ( r ) ∈ N p ( C, r ) ∩ N p ( D, r ), a contradiction. (cid:3) Proposition 22.14. Suppose ( X, d X ) and ( Y, d Y ) are two CoG-spaces. A function f : X → Y is visual form-continuous if and only if it has the following property:for any sequence of geodesics { [ p, x n ] } ∞ n =1 converging to a geodesic ray l p at p ∈ X ,the sequence of geodesics { [ q, f ( x n )] } ∞ n =1 converges to some geodesic ray l q at q = f ( p ) ∈ Y .Proof. Suppose there is a sequence of geodesics { [ p, x n ] } ∞ n =1 converging to a geodesicray l p at p ∈ X such that the sequence of geodesics { [ q, f ( x n )] } ∞ n =1 does not convergeto any geodesic ray at q = f ( p ) ∈ Y . As in the proof of 22.13 we can find twosubsequences C = { c n } ∞ n =1 and D = { d n } ∞ n =1 of { x n } ∞ n =1 such that { [ q, f ( c n )] } ∞ n =1 converges to a geodesic ray l C at q , { [ q, f ( d n )] } ∞ n =1 converges to a geodesic ray l D at q , and l C = l D . Notice ω ( N p )( C, D ) = ∞ but ω ( N q )( f ( C ) , f ( D )) = 0, so f isnot form-continuous.Suppose f is visual form-continuous and { [ p, x n ] } ∞ n =1 is a sequence of geodesicsconverging to a geodesic ray l p at p ∈ X but the sequence of geodesics { [ q, f ( x n )] } ∞ n =1 does not converge to any geodesic ray at q = f ( p ) ∈ Y . As above, we can find twosubsequences C = { c n } ∞ n =1 and D = { d n } ∞ n =1 of { x n } ∞ n =1 such that { [ q, f ( c n )] } ∞ n =1 converges to a geodesic ray l C at q , { [ q, f ( d n )] } ∞ n =1 converges to a geodesic ray l D at q , and l C = l D . Notice ω ( N p )( C, D ) = ∞ but ω ( N q )( f ( C ) , f ( D )) = 0. (cid:3) Observation 22.15. 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