Growth rate of Lipschitz constants for retractions between finite subset spaces
aa r X i v : . [ m a t h . M G ] M a y GROWTH RATE OF LIPSCHITZ CONSTANTS FOR RETRACTIONSBETWEEN FINITE SUBSET SPACES
EARNEST AKOFOR AND LEONID V. KOVALEV
Abstract.
For any metric space X , finite subset spaces of X provide a sequence ofisometric embeddings X = X (1) ⊂ X (2) ⊂ · · · . The existence of Lipschitz retractions r n : X ( n ) → X ( n −
1) depends on the geometry of X in a subtle way. Such retractions areknown to exist when X is an Hadamard space or a finite-dimensional normed space. Buteven in these cases it was unknown whether the sequence { r n } can be uniformly Lipschitz.We give a negative answer by proving that Lip( r n ) must grow with n when X is a normedspace or an Hadamard space. Introduction
Given a topological space X and a positive integer n , the nonempty subsets of X ofcardinality at most n form another topological space X ( n ) with a natural quotient topologyinduced by the map that takes each ordered tuple ( x , . . . , x n ) in the Cartesian product X n to the finite set { x , . . . , x n } in X ( n ). The space X ( n ) is called the n th finite subset space of X (the terms symmetric product or symmetric power are sometimes used as well). Thisconcept goes back to Borsuk and Ulam [8]. When X is a metric space, X ( n ) becomes ametric space with respect to Hausdorff distance which is given by d H (cid:0) { x , . . . , x n } , { x ′ , . . . , x ′ n } (cid:1) := max (cid:26) max i min j d ( x i , x ′ j ) , max i min j d ( x j , x ′ i ) (cid:27) . (1.1)See, e.g., [2, Proposition 1.2.2] for details of this metrization. The natural embeddings X = X (1) ⊂ X (2) ⊂ · · · are isometric with respect to the Hausdorff distance.If X and Y are metric spaces, a map f : X → Y is called Lipschitz if there is a number L ≥ d ( f ( x ) , f ( x ′ )) ≤ Ld ( x, x ′ ) for all x, x ′ ∈ X . The least of such numbers L isdenoted by Lip( f ) and is called the Lipschitz constant of f . If Y ⊂ X , a map r : X → Y is a retraction if its restriction to Y is the identity map. If r is in addition Lipschitz, it iscalled a Lipschitz retraction . Mathematics Subject Classification.
Primary 54E40; Secondary 46B20, 54B20, 54C15.
Key words and phrases. normed space, metric space, Hadamard space, finite subset space, Lipschitzretraction.L.V.K. supported by the National Science Foundation grant DMS-1764266.
A Lipschitz retraction X ( n ) → X ( k ), where k < n , can be interpreted as a robust choiceof k clusters within a finite set. Indeed, given a set A ⊂ X of cardinality | A | ≤ n , we mustchoose a set r ( A ) with | r ( A ) | ≤ k subject to the conditions that r ( A ) = A if | A | ≤ k , and r ( A ) a Lipschitz function of A .Some spaces X present topological obstructions to the existence of such retractions. Forexample, if X is the circle S , then X (3) is homeomorphic to 3-sphere [9] and cannotbe retracted onto X (1) = S , being simply-connected. Section 2 presents a more generalobstruction of this type. If X is a Hilbert space of any dimension (finite or infinite), for every n there exists a Lipschitz retraction r n : X ( n ) → X ( n −
1) with Lip( r n ) ≤ max (cid:0) n / , n − (cid:1) [15, (2.5)]. Question 3.2 in [15] and Remark 3.5 in [5] asked whether Lip( r n ) can be boundedindependently of n . Our first result shows that it must grow at least linearly with respectto n , provided that dim X ≥ Theorem 1.1.
Let X be a normed space over R with dim X ≥ . Suppose that r : X ( n ) → X ( k ) is a Lipschitz retraction, where ≤ k ≤ n − . Then (1.2) Lip( r ) ≥ kn π ( n − − . Moreover, if X is a Hilbert space, then (1.3) Lip( r ) ≥ knπ ( n − − . The case dim X = 1, when X is isometric to R , is covered by our second main theorem.It concerns Hadamard spaces , which share the geometric properties of Hilbert spaces butnot necessarily their linear structure. To define them, we need the notion of a geodesicspace.A geodesic in a metric space (
X, d ) is a mapping γ : [0 , → X such that for all t, s ∈ [0 , d ( γ ( t ) , γ ( s )) = | t − s | d ( γ (0) , γ (1)). In geometric terms, a geodesic is a curveparametrized proportionally to its arclength. If for any two points p, q ∈ X there exists ageodesic with γ (0) = p and γ (1) = q , then X is called a geodesic space . Definition 1.2.
A complete geodesic space is called an
Hadamard space if for every point z and every geodesic γ we have d ( γ ( t ) , z ) ≤ (1 − t ) d ( γ (0) , z ) + td ( γ (1) , z ) − t (1 − t ) d ( γ (0) , γ (1)) for all t ∈ [0 , X and ev-ery n ≥ r n : X ( n ) → X ( n −
1) with Lip( r n ) ≤ ROWTH RATE OF LIPSCHITZ CONSTANTS 3 max (cid:0) n + √ n, n / + 1 (cid:1) . The following theorem gives a lower bound for Lip( r n ), an-swering a question posed in [5, Remark 3.5]. Theorem 1.3.
Let X be either a normed space over R of dimension dim X ≥ , or anHadamard space containing more than one point. If r : X ( n ) → X ( n − is a Lipschitzretraction, then Lip( r ) ≥ n − . The paper is organized as follows. In Section 2 we collect the necessary results from thealgebraic topology of finite subset spaces. Section 3 contains preliminary results on theproperties of Lipschitz retractions. Theorems 1.1 and 1.3 are proved in sections 4 and 5,respectively. Corollary 5.2 provides a more general version of Theorem 1.3. Note thatTheorem 1.3 gives a slightly better lower bound than Theorem 1.1. On the other hand,Theorem 1.1 applies to retractions onto X ( k ) for any k < n , not only k = n − Topology of finite subset spaces
Let H n ( X ) denote the n th singular homology group of a topological space X [11, p. 108].The homology groups of finite subset spaces S ( n ) were computed by Wu [18] and theirhomotopy type was determined by Tuffley [17], see also [10]. Theorem 2.1. ( [18, Theorem III] , [17, Theorems 4–5] ). Given n ∈ N , let m be the largestodd integer not exceeding n . The homology groups H ( S ( n )) and H m ( S ( n )) are isomorphicto Z , and all other homology groups of S ( n ) are trivial. Moreover, when n is odd, theinclusion S ( n ) ⊂ S ( n + 1) induces the doubling map k k on the homology group H n ( S ( n )) . The homology presents an obstruction to the existence of continuous retractions betweenthe finite subset spaces of the circle.
Proposition 2.2. If ≤ k ≤ n − , there is no continuous retraction S ( n ) → S ( k ) .Proof. Suppose there exists a continuous retraction r : S ( n ) → S ( k ). The map inducedby r on the homology groups of S ( n ) is a left inverse of the map induced by the inclusionof S ( k ) into S ( n ) [11, p. 111]. In particular, the latter map is injective.Let m be the greatest odd integer not exceeding k . Since H m ( S ( k )) is isomorphic to Z ,the group H m ( S ( n )) must be nontrivial. Theorem 2.1 implies that m is the greatest oddinteger not exceeding n . This is only possible if n is even and k = m = n −
1. However, in thiscase the inclusion of S ( n −
1) into S ( n ) induces the doubling map on the homology groups H n − , and this map does not have a left inverse, contradicting the previous paragraph. (cid:3) EARNEST AKOFOR AND LEONID V. KOVALEV
The obstruction presented by Proposition 2.2 is the basis of our proof of Theorem 1.1.To prove Theorem 1.3 we need to find some topological obstruction within the subsets of R . It is provided by pinned finite subset spaces , which are defined as follows. Definition 2.3.
Given a set U ∈ X ( n ), the U -pinned finite subset space of X is X ( n, U ) := { A ∈ X ( n ) : U ⊂ A } . The space X ( n, U ) is empty when n < | U | . Notable examples of pinned finite subsetspaces include(2.1) D n := I ( n + 2 , { , } )where I is the interval [0 ,
1] and n = 0 , , . . . . The studies of D n go back to Schori [16]who proved that I ( n ) is a double cone of D n − when n ≥
2. Andersen, Marjanovi´c andSchori [3] called the spaces D n with even n “higher-dimensional dunce hats” because D ishomeomorphic to the “dunce hat” space introduced by Zeeman [19] and D n shares somefeatures of D when n is even. Theorem 2.4. [3, Theorem 3.4] . When n is even, D n is contractible. When n is odd, D n has the homotopy type of S n . As with the ordinary finite subset spaces, we have natural inclusions X ( k, U ) ⊂ X ( n, U )when | U | ≤ k ≤ n . For example, D ⊂ D ⊂ · · · . Corollary 2.5.
When n is even, there is no continuous retraction of D n onto D n − .Proof. By Theorem 2.4, the homology group H n − ( D n − ) is isomorphic to Z while H n − ( D n )is trivial. Since the former group does not embed in the latter, D n − is not a retract of D n . (cid:3) Corollary 2.5 is a less complete result than Proposition 2.2. We do not know if D n retracts onto D n − when n is odd. An interesting related question is whether each higher-dimensional dunce hat D k is an absolute Lipschitz retract , meaning that it is a Lipschitzretract in any metric space that contains it. At present it is not known whether finite subsetspaces inherit the absolute Lipschitz retract property: see [1, 2, 14] for partial results.Another example of a pinned finite subset space is S ( n, { } ) which was studied byTuffley [17, p. 1131]. This space is homeomorphic to D n − . More specificially, the map A
7→ { e πit : t ∈ A } is a bi-Lipschitz homeomorphism of D n − onto S ( n, { } ): see the proofof Theorem 3.1 in [7]. ROWTH RATE OF LIPSCHITZ CONSTANTS 5 Properties of Lipschitz retractions
Let X be a metric space and n ≥
2. The minimum separation function δ n : X ( n ) → [0 , ∞ )is defined as follows:(3.1) δ n ( A ) = ( min { d X ( p, q ) : p, q ∈ A, p = q } if | A | = n | A | < n The importance of δ n stems from the following observation. Lemma 3.1.
Suppose X is a metric space and n ≥ . For each A ∈ X ( n ) there is B ∈ X ( n − such that d H ( A, B ) ≤ δ n ( A ) . If, in addition, X is a geodesic space, then theconclusion can be strengthened to d H ( A, B ) ≤ δ n ( A ) .Proof. If | A | < n , the set B = A satisfies the conclusion. Suppose | A | = n . Let p, q ∈ A betwo points such that d X ( p, q ) = δ n ( A ). Then the set B = A \ { p } has n − d H ( A, B ) ≤ δ n ( A ).If X is a geodesic space, let m be the midpoint of a geodesic from p to q and define B = ( A ∪ { m } ) \ { p, q } . This set has n − d H ( A, B ) ≤ δ n ( A ). (cid:3) Lemma 3.2.
Let X be a metric space and ≤ k < n . Suppose r : X ( n ) → X ( k ) is aLipschitz retraction. Then for every A ∈ X ( n ) we have (3.2) d H ( r ( A ) , A ) ≤ (Lip( r ) + 1) dist H ( A, X ( k )) where dist H ( A, X ( k )) = inf { d H ( A, B ) : B ∈ X ( k ) } .Proof. For every B ∈ X ( k ) we have r ( B ) = B , hence d H ( r ( A ) , B ) = d H ( r ( A ) , r ( B )) ≤ Lip( r ) d H ( A, B ) . By the triangle inequality, d H ( r ( A ) , A ) ≤ (Lip( r ) + 1) d H ( A, B ). Taking the infimum over B ∈ X ( k ) yields (3.2). (cid:3) Lemma 3.3.
Let X be a metric space such that there exists a Lipschitz retraction r : X ( n ) → X ( k ) for some integers ≤ k < n . Suppose Y is a metric space such that there exist Lip-schitz maps f : X → Y and g : Y → X with the property f ◦ g = id Y . Then there exists aLipschitz retraction s : Y ( n ) → Y ( k ) with Lip( s ) ≤ Lip( f ) Lip( g ) Lip( r ) .Proof. The map g induces a map g n : Y ( n ) → X ( n ) such that g n ( A ) is the image of set A under g . From the definition (1.1) of Hausdorff distance it is easy to see that Lip( g n ) =Lip( g ). Similarly, f induces a map f k : X ( k ) → Y ( k ). Let s = f k ◦ r ◦ g n . By construction, s maps Y ( n ) to Y ( k ) and has Lipschitz constant at most Lip( f ) Lip( g ) Lip( r ). If A ∈ Y ( k )then g n ( A ) ∈ X ( k ), hence s ( A ) = f k ( g n ( A )) = A by the property f ◦ g = id Y . Thus s is aLipschitz retraction onto Y ( k ). (cid:3) EARNEST AKOFOR AND LEONID V. KOVALEV
Two useful special cases of Lemma 3.3 are: (a) Y is a Lipschitz retract of X , with f being the inclusion map; (b) Y is bi-Lipschitz equivalent to X , with g = f − .4. Normed spaces: proof of theorem 1.1
Let X be a normed space over R of dimension at least 2. The following statement is aspecial case of [6, Proposition G.1] which summarizes the results of John [12] and Kadets-Snobar [13]. Lemma 4.1.
Let Z be a -dimensional subspace of a normed space X . Then there existsa linear projection P : X → Z , and a linear isomorphism T : R → Z such that k P k ≤ √ , k T k ≤ √ , and k T − k ≤ . Lemma 4.1 leads us to consider the geometry of finite subsets of R which is the subjectof the following lemma. Lemma 4.2.
Let S ⊂ R be the unit circle centered at , equipped with the arclengthmetric. For any set A ∈ S ( n ) and any k ∈ , . . . , n − there exists B ∈ S ( k ) such that d H ( A, B ) ≤ π ( n − / ( kn ) .Proof. Let A j be the result of rotating A by the angle 2 πj/k , and let R = S kj =1 A j . Since R has at most kn points, its complement in S contains an open arc of length 2 π/ ( kn ).The k -fold symmetry of R implies that it is covered by k uniformly spaced closed arcsΓ , . . . , Γ k ⊂ S of length 2 πk − πkn = 2 π ( n − kn . Therefore, A ⊂ S kj =1 Γ j . Let B be the set of midpoints of all arcs Γ j such that Γ j ∩ A isnonempty. Then every point of B is within distance at most π ( n − / ( kn ) of some pointof A , and vice versa. Since | B | ≤ k , the lemma is proved. (cid:3) The estimate in Lemma 4.2 is best possible when k = n −
1, as one can check using aset A of n equally spaced points. Lemma 4.2 also applies when S is equipped with chordalmetric, i.e. the restriction of the Euclidean metric on R , because the chordal metric ismajorized by arclength. Proof of Theorem 1.1.
Let Y = R and let Z , P , T be as in Lemma 4.1. The mappings f = T − ◦ P and g = T satisfy the assumptions of Lemma 3.3. Therefore, there exists aretraction s : Y ( n ) → Y ( k ) with Lip( s ) ≤ r ). When X is a Hilbert space, this can beimproved to Lip( s ) ≤ Lip( r ) because any two-dimensional subspace Z ⊂ X is isometric to R and P can be the orthogonal projection onto Z . Therefore, both cases of Theorem 1.1 ROWTH RATE OF LIPSCHITZ CONSTANTS 7 will be proved if we show that Lip( s ) ≥ knπ ( n − −
1. Suppose, toward a contradiction, thatthe constant c := π ( n − kn (Lip( s ) + 1) satisfies c < S ⊂ R be the unit circle centered at 0. From Lemmas 3.2 and 4.2 it follows that d H ( s ( A ) , A ) ≤ c for every A ∈ S ( n ). Hence s ( A ) is contained in the set W = { x ∈ R : k x k ≥ − c } . The radial projection f ( x ) = x/ k x k provides a Lipschitz retraction of W onto S . Therefore, the mapping A f ( s ( A )) is a Lipschitz retraction on S ( n ) onto S ( k ). This contradicts Proposition 2.2. (cid:3) Metric spaces: proof of theorem 1.3
We begin with the special case of retractions between the finite subset spaces of aninterval on the real line. The definition of the Hausdorff distance (1.1) implies that for anytwo nonempty finite sets
A, B ⊂ R we have(5.1) | max A − max B | ≤ d H ( A, B ) , | min A − min B | ≤ d H ( A, B ) . Theorem 5.1.
Let I = [ a, b ] ⊂ R where −∞ < a < b < ∞ . If r : I ( n ) → I ( n − is aLipschitz retraction, then (5.2) Lip( r ) ≥ ( n − , n is even n − , n is oddProof. The choice of the interval [ a, b ] does not matter in this theorem. Indeed, if φ : R → R is an invertible affine transformation, then A φ ( r ( φ − ( A ))) is a Lipschitz retractionbetween the finite subset spaces of φ ( I ), with the same Lipschitz constant Lip( r ). Thus, wecan choose any convenient interval I in the proof, and we use two different intervals for thetwo cases that follow. Case 1: n is even. Assume n ≥ n = 2. Let I = [0 , c := (Lip( r ) + 1) / ( n −
1) satisfies c < D n − ⊂ I ( n ) defined by (2.1). For every set A ∈ D n − the minimalseparation (3.1) satisfies δ n ( A ) ≤ / ( n − d H ( r ( A ) , A ) ≤ n −
1) (Lip( r ) + 1) ≤ c/ . By (5.1) we have min r ( A ) ≤ c/ r ( A ) ≥ − c/
2. For t ∈ [0 ,
1] let(5.3) f A ( t ) = t − min r ( A )max r ( A ) − min r ( A ) . Since the denominator in (5.3) is bounded below by 1 − c , the function f A is (1 − c ) − -Lipschitz with respect to t . It is also Lipschitz continuous with respect to A by virtueof (5.1). Therefore, the mapping s ( A ) := f A ( r ( A )) is Lipschitz continuous on D n − . Byconstruction, the set s ( A ) consists of at most n − s ( A ) = 0, and max s ( A ) = 1. EARNEST AKOFOR AND LEONID V. KOVALEV
Thus s ( A ) ∈ D n − . If | A | < n then r ( A ) = A , hence s ( A ) = A . We have proved that s is aLipschitz retraction of D n − onto D n − , which contradicts Corollary 2.5. Case 2: n is odd. We may assume n ≥ n ≤
3. Let I =[0 , c := (Lip( r ) + 1) / ( n −
2) satisfies c <
1. Let Y = [0 , ∪ { } and consider the pinned finite subset space E = Y ( n, { , , } ).For every set A ∈ E , all but one of its points lie in [0 , δ n ( A ) ≤ / ( n − d H ( r ( A ) , A ) ≤ n −
2) (Lip( r ) + 1) ≤ c/ . Since { , , } ⊂ A , it follows that r ( A ) meets each of the intervals [0 , c/ − c/ , c/ − c/ , r ( A ) is disjoint from the interval (1 + c/ , − c/ s ( A ) = r ( A ) ∩ [0 , c/
2] = r ( A ) ∩ [0 , − c/ . Note that | s ( A ) | ≤ r ( A ) − ≤ n − r ( A ) meets [2 − c/ , A, B ∈ E are such that ∆ := d H ( r ( A ) , r ( B )) < − c . By the definition of d H , for every a ∈ s ( A )there exists b ∈ r ( B ) such that | a − b | ≤ ∆. Then b ≤ (1 + c/
2) + ∆ < − c/
2, which implies b ∈ s ( B ). In conclusion,(5.4) d H ( s ( A ) , s ( B )) ≤ d H ( r ( A ) , r ( B ))whenever the right hand side is less than 1 − c . Since r is Lipschitz continuous, (5.4) showsthat s is also Lipschitz continuous.For t ∈ [0 ,
1] let f A ( t ) = t − min s ( A )max s ( A ) − min s ( A )and note that the denominator is bounded below by 1 − c . As in Case 1, it follows that f A ( t ) is Lipschitz with respect to both t and A . Therefore, the mapping σ ( A ) := f A ( s ( A ))is Lipschitz continuous on E . By construction, the set σ ( A ) consists of at most n − σ ( A ) = 0, and max σ ( A ) = 1. Thus, σ ( A ) ∈ D n − . If A ∈ E has fewer than n points,then r ( A ) = A , hence σ ( A ) = A ∩ [0 , D n − is isometric to E via the map ι ( A ) = A ∪ { } . The previous paragraphshows that the composition σ ◦ ι is a Lipschitz retraction of D n − onto D n − . Since n − (cid:3) The following statement is an immediate consequence of Theorem 5.1 and Lemma 3.3.
Corollary 5.2.
Suppose X is a metric space. Fix an integer n such that there exists aLipschitz retraction r : X ( n ) → X ( n − . Suppose I ⊂ R is a nondegenerate compact ROWTH RATE OF LIPSCHITZ CONSTANTS 9 interval and there exist Lipschitz maps f : X → I and g : I → X such that f ◦ g = id I .Then Lip( r ) ≥ n − f ) Lip( g ) . Proof of Theorem 1.3.
Suppose r : X ( n ) → X ( n −
1) is a Lipschitz retraction.
Case 1: X is a normed space. Let I = [0 , u ∈ X and itsnorming functional ϕ ∈ X ∗ , that is, a linear functional ϕ : X → R such that k ϕ k X ∗ = 1and ϕ ( u ) = 1. The existence of such ϕ follows from the Hahn-Banach theorem. Define g : I → X by g ( t ) = tu , and f : X → I by f ( x ) = min(max( ϕ ( x ) , , f and g are 1-Lipschitz and f ◦ g = id I . By Corollary 5.2 we have Lip( r ) ≥ n − Case 2: X is an Hadamard space. Pick any two distinct points p, q ∈ X and let I = [0 , d X ( p, q )]. Since Hadamard spaces are geodesic, there exists an isometric embedding g : I → X , namely a reparametrized geodesic connecting p to q . Since g ( I ) is a closedconvex subset of X , the nearest-point projection onto X is a 1-Lipschitz map [4, Theorem2.1.12]. Let f be the composition of this projection with g − . Since Lip( f ) = 1 = Lip( g ),Corollary 5.2 yields Lip( r ) ≥ n − (cid:3) References [1] Earnest Akofor. On Lipschitz retraction of finite subsets of normed spaces.
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