aa r X i v : . [ m a t h . M G ] A p r Geodesic spaces of low Nagata dimension
Martina Jørgensen & Urs LangApril 22, 2020
Abstract
We show that every geodesic metric space admitting an injective contin-uous map into the plane as well as every planar graph has Nagata dimensionat most two, hence asymptotic dimension at most two. This relies on and an-swers a question in a very recent work by Fujiwara and Papasoglu. We con-clude that all three-dimensional Hadamard manifolds have Nagata dimensionthree. As a consequence, all such manifolds are absolute Lipschitz retracts.
In [7], Fujiwara and Papasoglu show that all planar geodesic metric spaces andplanar graphs have asymptotic dimension at most three. Here, a geodesic metricspace is said to be planar if it admits an injective continuous map into R , and anot necessarily locally finite (but connected) graph, viewed as a geodesic metricspace with edges of length one, is called planar if it admits an injective map into R whose restriction to every edge is continuous. In fact, Fujiwara and Papasogluprove the stronger result that the (Assouad–)Nagata dimension, which is greaterthan or equal to the asymptotic dimension, is at most 3 (see below for the definitionsof these notions). This is achieved by showing that for some universal constant C ,every metric annulus of width comparable to s > C s such that every s -ball meets no more than two of them. Foran appropriate sequence of annuli covering the underlying space X , the union ofthe individual covers has s -multiplicity at most 4, which gives the bound on thedimension but exceeds the expected value by one.In this note, we first observe that the bound on the Nagata dimension can beimproved to 2 (see Theorem 2). This answers in particular Question 5.1 in [7] forthe asymptotic dimension. For the proof, it su ffi ces to combine the above result formetric annuli with a simplified version of the Hurewicz-type theorem from [5] forthe case of a Lipschitz function f : X → R (see Theorem 1). The latter holds inall dimensions, and we provide a streamlined proof. We then take a further stepand apply Theorem 2 and Theorem 1 to prove that all 3-dimensional Hadamardmanifolds have Nagata dimension 3 (see Theorem 3). In dimension two, this wasshown in Theorem 5.7.3 in [12], but in higher dimensions, previous results depend1n additional curvature bounds or homogeneity assumptions (compare [8, 11]).Finiteness of the asymptotic or Nagata dimension has a number of important con-sequences (see [3] for a survey). It now follows from [11] that all 3-dimensionalHadamard manifolds are absolute Lipschitz retracts (see Theorem 4). Again, thissubsumes a number of earlier results with extra conditions (compare [6, 10]).We now state the relevant definitions. Let ( X , d ) be a metric space. A collection C of subsets of X is called D-bounded , for some constant D , if every set C ∈ C has diameter diam( C ) : = sup { d ( x , y ) : x , y ∈ C } ≤ D . Given s >
0, the s-multiplicity of C is the infimum of all integers m ≥ s -ball in X meets at most m members of the collection. The asymptotic dimension asdim( X ) of X is the infimum of all integers n ≥ D : (0 , ∞ ) → (0 , ∞ ) such that for every s > X possesses a D ( s )-bounded covering of s -multiplicity at most n + Nagata dimension or Assouad–Nagata dimension dim N ( X ) of X is defined analogously with a linearcontrol function D ( s ) = cs [1, 11]. To make the constant c > X has Nagata dimension n or at most n with constant c . It is an essential feature ofthe results in [7] and those obtained here that they hold uniformly for all membersof the class of spaces considered, that is, with the same constant c .The following equivalent formulation is often useful. Let C be a collection ofsubsets of X . For s > C is called s-disjoint if d ( C , C ′ ) : = inf { d ( x , x ′ ) : x ∈ C , x ′ ∈ C ′ } ≥ s whenever C , C ′ ∈ C are distinct. More generally, we say that C is ( n + , s ) -disjoint if C = S n + i = C i for subcollections C i that are individually s -disjoint. Wethink of the indices 1 , . . . , n + colours of C . Evidently, if C is ( n + , s )-disjoint, then C has λ s -multiplicity at most n + λ ∈ (0 , ). Hence, ametric space X with an ( n + , s )-disjoint and c ′ s -bounded cover for every s > N ( X ) ≤ n with any constant c > c ′ . Conversely, the following holds(see the proof of Proposition 2.5 in [11]). Proposition 1.
If X is a metric space with a cs-bounded cover of s-multiplicity atmost n + for some c , s > , then X also admits an ( n + , λ s ) -disjoint and c ′ λ s-bounded cover, where λ, c ′ > depend only on c and n. In particular, if X hasNagata dimension at most n with constant c, then X possesses an ( n + , s ) -disjointand c ′ s-bounded cover for every s > . We now proceed to the aforementioned version of the Hurewicz theorem. Weneed the following lemma extracted from Theorem 2.4 in [5].2 emma 1.
Let X be a metric space with an ( n + , s ) -disjoint and D-bounded coverfor some n ≥ and s , D > . Then there exists an ( n + , s ) -disjoint and ( D + s ) -bounded cover of X with each point x ∈ X belonging to at least two sets of di ff erentcolours.Proof. By assumption, X has an ( n + , s )-disjoint and D -bounded cover C = S n + k = C k . For C ∈ C , let C ′ denote the closed s -neighbourhood of C . Definea new cover S n + k = C ′ k , where C ′ k : = { C ′ : C ∈ C k } . Notice that this cover is( n + , s )-disjoint and ( D + s )-bounded. Now define sets of an additional colouras follows. For every B ∈ C of colour j , let B be the set B minus the union of all C ′ ∈ S k , j C ′ k , and put C ′ n + : = { B : B ∈ C } . The cover C ′ : = S n + k = C ′ k will havethe required properties.Clearly, C ′ is still ( D + s )-bounded, as diam( B ) ≤ diam( B ) ≤ D for all B ∈ C . To verify that C ′ is ( n + , s )-disjoint, it remains to check that d ( A , B ) ≥ s whenever A ∈ C i , B ∈ C j , and A , B . If i = j , then d ( A , B ) ≥ d ( A , B ) ≥ s ,because C i is s -disjoint. If i , j , then A ⊂ A and B ⊂ B \ A ′ by constructionof C ′ n + , thus d ( A , B ) ≥ d ( A , B \ A ′ ) ≥ s . Now let x ∈ X . There exist j ∈{ , . . . , n + } and B ∈ C j such that x ∈ B . If x belongs to some set C ′ ∈ C ′ k with k , j , then x ∈ B ′ ∩ C ′ . If there is no such set C ′ containing x , then x ∈ B ′ ∩ B .Thus in either case, x belongs to at least two sets of di ff erent colours. (cid:3) The following result corresponds to Theorem 7.2 in [5] for the case of a Lip-schitz function f : X → R . The argument can also easily be adapted to the asymp-totic dimension (see Theorem 1 in [2] for an earlier result). Theorem 1.
Let X be a metric space, and let f : X → R be a -Lipschitz function.Suppose that there exist n ≥ and c > such that for all r ∈ R and s > , the setf − ([ r , r + s )) possesses an ( n + , s ) -disjoint and cs-bounded cover. Then X hasNagata dimension at most n + with a constant depending only on n and c.Proof. Fix s > t : = ( n + s . For every k ∈ Z , put I k : = [ kt , ( k + t ).For every odd integer k , proceed with the following construction. By assump-tion, f − ( I k ) admits an ( n + , t )-disjoint and ct -bounded cover. Lemma 1 nowprovides an ( n + , t )-disjoint and ( c + t -bounded cover C k = S n + i = C ik with eachpoint x ∈ f − ( I k ) belonging to two sets of di ff erent colours. Let B k = S n + i = B ik bethe ( n + , s )-disjoint cover of I k such that B ik consists of the connected componentsof I k \ [ kt + ( i − s , kt + is ). Every point of I k is in n + ff erentcolours. For i = , . . . , n +
2, define D ik : = (cid:8) C ∩ f − ( B ) : C ∈ C ik , B ∈ B ik (cid:9) . Note that if x ∈ f − ( I k ), then x belongs to two sets of di ff erent colours of C k , and f ( x ) is in n + ff erent colours of B k ; thus there exists an index3 ∈ { , . . . , n + } such that x belongs to some set C ∩ f − ( B ) ∈ D ik . Hence, S n + i = D ik is a covering of f − ( I k ). Since C k is ( c + t -bounded, this cover is c ′ s -boundedfor c ′ : = ( n + c + s ≤ t and f is 1-Lipschitz, D ik is s -disjoint. For i ∈ { , . . . , n + } , let D i : = S k odd D ik . Note that D i is still s -disjoint.Now put s ′ : = ( c ′ + s and note that s ′ ≥ ( n + s = t . For every even integer k , there exists by assumption an ( n + , s ′ )-disjoint and cs ′ -bounded cover S n + i = E ik of f − ( I k ) by subsets of f − ( I k ). For i ∈ { , . . . , n + } , let E i : = S k even E ik . Theunion n + [ i = (cid:0) D i ∪ E i (cid:1) ∪ D n + is an ( n + X which we shall modify to satisfy the requiredproperties. For every E ∈ E i , let E ∗ be the union of E with all sets of D i at distance < s from E . Clearly diam( E ∗ ) ≤ cs ′ + c ′ + s = c ′′ s for some c ′′ = c ′′ ( n , c ).We claim that d ( E ∗ , F ∗ ) ≥ s whenever E , F ∈ E i are distinct. If E and F belong tothe same family E ik , this holds since d ( E , F ) ≥ s ′ ≥ s + diam( D ) for all D ∈ D i and D i is s -disjoint. In the other case, d ( E , F ) ≥ t ≥ s , and by construction no set D ∈ D i is at distance < s from both E and F , so the claim follows again since D i is s -disjoint. In the final covering, the collection of sets of colour i consists of all E ∗ with E ∈ E i and the remaining elements of D i not belonging to such an E ∗ . Thisgives an ( n + , s )-disjoint and c ′′ s -bounded cover of X . Since s > N ( X ) ≤ n + (cid:3) We now turn to planar geodesic spaces. The proof of Theorem 2 below relieson the following result from [7].
Proposition 2.
There is a universal constant c such that the following holds. Sup-pose that X is a planar geodesic metric space or a planar graph, and let z ∈ X bea base point. Then for any r , t > , the metric annulus { x ∈ X : r ≤ d ( z , x ) ≤ r + t } admits a c t-bounded cover of t-multiplicity at most 2. See Lemma 4.4 in [7] for M = m . One can take c = . Theorem 2.
There is a universal constant c such that every planar geodesic met-ric space or planar graph has Nagata dimension at most with constant c .Proof. Let X be a planar geodesic metric space or a planar graph, let z ∈ X bea base point, and put f : = d ( z , · ). By Proposition 2, for any r , t >
0, the set f − ([ r , r + t ]) admits a c t -bounded cover of t -multiplicity at most 2. It followsfrom Proposition 1 that f − ([ r , r + t ]) possesses a (2 , λ t )-disjoint and c ′ λ t -boundedcover, where λ and c ′ depend only on c . If λ <
1, put s : = λ t and c : = c ′ . If λ ≥ s : = t and c : = c ′ λ . In either case, for any r , s >
0, the set f − ([ r , r + s ]) has a(2 , s )-disjoint and cs -bounded cover. Now the result follows from Theorem 1. (cid:3)
4e proceed to complete, simply connected Riemannian manifolds of nonposi-tive sectional curvature.
Theorem 3.
There is a universal constant c such that every 3-dimensionalHadamard manifold has Nagata dimension with constant c .Proof. Let ( X , d ) be a 3-dimensional Hadamard manifold. Fix a Busemann func-tion f : X → R . Recall that f is 1-Lipschitz, all horoballs f − (( −∞ , r ]) are convex,and the nearest point retraction from X onto any horoball is 1-Lipschitz. Fur-thermore, every horosphere f − { r } is homeomorphic to R , and the induced innermetric on f − { r } is finite and complete, hence geodesic. (In fact, Busemann func-tions and horospheres are C , see Proposition 3.1 in [9], but this is not neededhere.) By Theorem 2, every f − ( r ) has Nagata dimension at most 2 with constant c . Now fix r ∈ R and s >
0. By Proposition 1 there is a universal constant c ′ such that the horosphere H : = f − { r − s } , with the induced inner metric d H , pos-sesses a (3 , s )-disjoint and c ′ s -bounded cover B = S i = B i by subsets of H . Equip A : = f − ([ r − s , r + s ]) with the induced inner metric d A , and let π : A → H be thenearest point projection. Note that π does not increase the length of curves, thus π is 1-Lipschitz also with respect to d A and d H . The sets π − ( B ) with B ∈ B form a(3 , s )-disjoint and ( c ′ + s -bounded cover of ( A , d A ). Furthermore, by taking theintersections with A ′ : = f − ([ r , r + s ]), we get a cover of A ′ that is (3 , s )-disjointand ( c ′ + s -bounded with respect to d , because d ( x , y ) ≤ d A ( x , y ) for all x , y ∈ A ′ ,with equality if d ( x , y ) ≤ s . Since r ∈ R and s > X has Nagata dimension at most 3 with a universal constant c . Infact, dim N ( X ) =
3, as X is locally bi-Lipschitz homeomorphic to R . (cid:3) A metric space X is an absolute C-Lipschitz retract , for a constant C ≥ e of X into another metric space Y there is a C -Lipschitz retraction of Y onto e ( X ). Equivalently, for every metric space W and every λ -Lipschitz map f : V → X with V ⊂ W , there exists a C λ -Lipschitzextension ¯ f : W → X of f (see Proposition 1.2 in [4]). It was shown in [10]that Hadamard manifolds with pinched negative sectional curvature, homogeneousHadamard manifolds, and all 2-dimensional Hadamard manifolds are absolute Lip-schitz retracts (in the last case, one can take C = √ f = id on Y = Z ), we can now settle the 3-dimensional case completely. Theorem 4.
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