# A Hurewicz theorem for the Assouad-Nagata dimension

**ABSTRACT** Given a function f : X → Y of metric spaces, the classical Hurewicz theorem states that dim(X) ≤ dim(f) + dim(Y). We provide analogs of this theorem for the Assouad–Nagata dimension, asymptotic Assouad–Nagata dimension, and asymptotic dimension (the latter result generalizes a theorem of Bell and Dranishnikov). As an application, we estimate the asymptotic Assouad–Nagata dimension of a finitely generated group G in terms of the asymptotic Assouad–Nagata dimensions of the groups K and H from the exact sequence 1 → K → G → H → 1.

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**ABSTRACT:**We show that a relatively hyperbolic group quasi-isometrically embeds in a product of finitely many trees if the peripheral subgroups do, and we provide an estimate on the minimal number of trees needed. Applying our result to the case of 3-manifolds, we show that fundamental groups of closed 3-manifolds have linearly controlled asymptotic dimension at most 8. To complement this result, we observe that fundamental groups of Haken 3-manifolds with non-empty boundary have asymptotic dimension 2.Algebraic & Geometric Topology 07/2012; · 0.68 Impact Factor - SourceAvailable from: Alexander N. Dranishnikov
##### Article: Asymptotic dimension

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**ABSTRACT:**The asymptotic dimension theory was founded by Gromov [M. Gromov, Asymptotic invariants of infinite groups,295] in the early 90s. In this paper we give a survey of its recent history where we emphasize two of its features: an analogy with the dimension theory of compact metric spaces and applications to the theory of discrete groups. Counting dimensions we are definitely not counting "things". Yu. Manin [71]Topology and its Applications 01/2008; 155(182). · 0.56 Impact Factor - SourceAvailable from: etd.fcla.edu[Show abstract] [Hide abstract]

**ABSTRACT:**ACKNOWLEDGMENTS I wish to thank my adviser, Alexander Dranishnikov, for his support, numerous enlightening discussions, and his patience while I was struggling. I would also like to thank the Department of Mathematics for the supportive atmosphere they have provided. I am also grateful to the College of Liberal Arts and Sciences for honoring me with the CLAS Dissertation Fellowship in my final semester. 4 TABLE OF CONTENTS page ACKNOWLEDGMENTS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .,4 ABSTRACT . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .,6 CHAPTER 1,INTRODUCTION AND BASIC DEFINITIONS . . . . . . . . . . . . . . . . . .,7 2,THE GRIGORCHUK GROUP . . . . . . . . . . . . . . . . . . . . . . . . . . . 15 3,COUNTABLE GROUPS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17 4,THE SUBLINEAR COARSE STRUCTURE . . . . . . . . . . . . . . . . . . . . 22 5,THE DIMENSION OF THE SUBLINEAR HIGSON CORONA . . . . . . . . . 29 REFERENCES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36 BIOGRAPHICAL SKETCH . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38 5 Abstract of Dissertation Presented to the Graduate School

Page 1

arXiv:math/0605416v2 [math.MG] 5 Jun 2006

HUREWICZ THEOREM FOR ASSOUAD-NAGATA

DIMENSION

N. BRODSKIY, J. DYDAK, M. LEVIN, AND A. MITRA

Abstract. Given a function f : X → Y of metric spaces, its as-

ymptotic dimension asdim(f) is the supremum of asdim(A) such

that A ⊂ X and asdim(f(A)) = 0. Our main result is

Theorem 0.1. asdim(X) ≤ asdim(f) + asdim(Y ) for any large

scale uniform function f : X → Y .

0.1 generalizes a result of Bell and Dranishnikov [3] in which f is

Lipschitz and X is geodesic. We provide analogs of 0.1 for Assouad-

Nagata dimension dimANand asymptotic Assouad-Nagata dimen-

sion asdimAN. In case of linearly controlled asymptotic dimension

l-asdim we provide counterexamples to three questions of Dranish-

nikov [14].

As an application of analogs of 0.1 we prove

Theorem 0.2. If 1 → K → G → H → 1 is an exact sequence of

groups and G is finitely generated, then

asdimAN(G,dG) ≤ asdimAN(K,dG|K) + asdimAN(H,dH)

for any word metrics metrics dGon G and dH on H.

0.2 extends a result of Bell and Dranishnikov [3] for asymptotic

dimension.

Contents

1.

2.

3.

4.

Introduction

Ostrand theorem for asymptotic dimension

Components, dimension, and coarseness

Dimension of a function

2

2

5

8

Date: May 17, 2006.

1991 Mathematics Subject Classification.

54E35, 18B30, 54D35, 54D40, 20H15.

Key words and phrases. Asymptotic dimension, coarse category, Lipschitz func-

tions, Nagata dimension.

The second-named and third-named authors were partially supported by Grant

No.2004047 from the United States-Israel Binational Science Foundation (BSF),

Jerusalem, Israel.

1

Primary: 54F45, 54C55, Secondary:

Page 2

2N. BRODSKIY, J. DYDAK, M. LEVIN, AND A. MITRA

5.

6.

7.

8.

References

Asymptotic dimension of groups

Linearly controlled asymptotic dimension of groups

Assouad-Nagata dimension

Asymptotic Assouad-Nagata dimension

12

14

16

16

18

1. Introduction

The well-known Hurewicz Theorem for maps (also known as Dimension-

Lowering Theorem, see [18], Theorem 1.12.4 on p.109) says dim(X) ≤

dim(f) + dim(Y ) if f : X → Y is a closed map of separable met-

ric spaces and dim(f) is defined as the supremum of dim(f−1(y)),

y ∈ Y . Bell and Dranishnikov [3] proved a variant of Hurewicz The-

orem for asymptotic dimension without defining the asymptotic di-

mension of a function. However, Theorem 1 of [3] may be restated as

asdim(X) ≤ asdim(f)+asdim(Y ), where asdim(f) is the smallest inte-

ger n such that asdim(f−1(BR(y))) ≤ n uniformly for all R > 0. As an

application it is shown in [3] that asdim(G) ≤ asdim(K) + asdim(H)

for any exact sequence 1 → K → G → H → 1 of finitely generated

groups. That inequality was extended subsequently by Dranishnikov

and Smith [16] to all countable groups.

The purpose of this paper is to generalize Hurewicz Theorem to vari-

ants of asymptotic dimension: asymptotic Assouad-Nagata dimension

and Assouad-Nagata dimension. In the process we produce a much

simpler proof than that in [3] and a stronger result: the function is

only assumed to be large scale uniform instead of Lipschitz, and the

domain is not required to be geodesic.

One of the main tools is Kolmogorov’s idea used in his solution to

Hilbert’s 13th Problem. In dimension theory it is known as Ostrand

Theorem. Another tool is reformulating Gromov’s [19] definition of

asymptotic dimension in terms of r-components of spaces. That leads

to a definition of asymptotic dimension of a function in terms of double-

parameter components, a concept well-suited for Kolmogorov Trick.

2. Ostrand theorem for asymptotic dimension

The aim of this section is to prove a variant of Ostrand’s Theorem

for large scale dimensions. As an application we present a simple proof

of the Logarithmic Law for large scale dimensions.

Intuitively, a metric space is of dimension 0 at scale r if it can be

represented as a collection of r-disjoint and uniformly bounded subsets.

The following definition of Gromov [19] defines asymptotic dimension

Page 3

HUREWICZ THEOREM FOR ASSOUAD-NAGATA DIMENSION3

at most n in terms of being represented, for each r > 0, as the union of

n+1 sets of dimension 0 at scale r. That corresponds to the well-known

property of topologically n-dimensional spaces.

Definition 2.1. A metric space X is said to be of asymptotic dimension

at most n (notation: asdim(X) ≤ n) if there is a function DX: R+→

R+such that for all r > 0 there is a cover U =

n+1

?

i=1Uiof X so that

each Uiis r-disjoint (that means d(a,b) ≥ r for any two points a and b

belonging to different elements of Ui) and the diameter of elements of

U is bounded by DX(r).

We refer to the function DXas an n-dimensional control function for

X. When discussing variants of asymptotic dimension it is convenient

to allow DXto assume infinity as its value at some range of r.

Definition 2.2. A metric space X is said to be of Assouad-Nagata

dimension (see [20] and [7]) at most n (notation: dimAN(X) ≤ n) if it

has an n-dimensional control function DXthat is a dilation (DX(r) =

c · r for some c > 0).

A metric space X is said to be of asymptotic Assouad-Nagata dimen-

sion at most n (notation: asdimAN(X) ≤ n) if it has an n-dimensional

control function DXthat is linear (DX(r) = c·r+b for some b,c ≥ 0).

A metric space X is said to be of linearly controlled asymptotic di-

mension at most n (Dranishnikov [14], notation: l-asdim(X) ≤ n) if

it has an n-dimensional control function DX satisfying DX(r) = c · r

for some c > 0 and for all r belonging to some unbounded subset of

R+. Strictly speaking, the original definition of Dranishnikov [14] is

formulated in terms of Lebesque numbers. However, just as in the case

of asymptotic dimension, it is equivalent to our definition.

A metric space X is said to be of microscopic Assouad-Nagata dimen-

sion at most n if it has an n-dimensional control function DX: R+→

R+∪ ∞ that is a dilation near 0: DX(r) = c · r for some c > 0 and all

r smaller than some positive number M, DX(r) = ∞ for all r ≥ M.

Definition 2.3. Given a metric space X and k ≥ n+1 ≥ 1 an (n,k)-

dimensional control function for X is a function DX: R+→ R+such

that for any r > 0 there is a family U =

k?

i=1Uisatisfying the following

conditions:

(1) each Uiis r-disjoint,

(2) each Uiis DX(r)-bounded,

Page 4

4N. BRODSKIY, J. DYDAK, M. LEVIN, AND A. MITRA

(3) each element x ∈ X belongs to at least k − n elements of

U (equivalently,

?

i∈T

{1,... ,k} consisting of n + 1 elements).

Ui is a cover of X for every subset T of

The following result is an adaptation of a theorem by Ostrand [25].

Theorem 2.4. If D(n+1)

one defines a sequence of functions {D(i)

D(i)

control function of X for all k ≥ n + 1.

X

is an n-dimensional control function of X and

X}i≥n+1inductively by D(i+1)

Xis an (n,k)-dimensional

X

(r) =

X(3r)+2r for all i ≥ n+1, then each D(k)

Proof. The proof is by induction on k. The case of k = n + 1 is

obvious. Suppose the result holds for some k ≥ n+1. Let U =

k?

i=1Uibe

a family such that each Uiis 3r-disjoint, each Uiis D(k)

and each element x ∈ X belongs to at least k − n elements of U.

Define U′

elements of U′

U′

X(3r)-bounded,

ito be the r-neighborhoods of elements of Uifor i ≤ k. Notice

iare (D(k)

k+1as the collection of all sets of the form

X(3r) + 2r)-bounded and are r-disjoint. Define

?

s∈SAs\?

i/ ∈S

U′

i, where S is a

subset of {1,... ,k} consisting of exactly k − n elements and As∈ Us.

Notice that any element of U′

some Uj. Thus elements of each U′

Given two different sets A =

?

k+1is contained in a single element of

iare (D(k)

s∈SAs\?

i/ ∈S

f(3r) + 2r)-bounded.

U′

i, where S is a subset

of {1,... ,k} consisting of exactly k − n elements and As∈ Us, and

B =

?

t∈T

i/ ∈T

exactly k − n elements and Bt ∈ Ut, we need to show A and B are

r-disjoint. It is clearly so if S = T, so assume S ?= T. If a ∈ A, b ∈ B,

and d(a,b) < r, then there is s ∈ S \ T such that a ∈ Asprompting

b ∈ U ∈ U′

Suppose x ∈ X belongs exactly to k − n sets?U′

S = {i ≤ k | x ∈?U′

for some j / ∈ S, a contradiction. Thus each x ∈ X belongs to at least

k + 1 − n elements of {?U′

The product theorem for asymptotic dimension was proved in [15]

using maps to polyhedra. The product theorem for Nagata dimension

was proved in [20] using Lipschitz maps to polyhedra. Below we use

Theorem 2.4 to give a simplified proof for all dimension theories. Note

the metric on X ×Y is the sum of corresponding metrics on X and Y .

Bt\

?

U′

i, where T is a subset of {1,... ,k} consisting of

s, a contradiction.

i, i ≤ k, and let

i}. If x / ∈?U′

k+1, then x must belong to?U′

j

i}k+1

i=1.

?

Page 5

HUREWICZ THEOREM FOR ASSOUAD-NAGATA DIMENSION5

Theorem 2.5. If X and Y are metric spaces, then

D(X × Y ) ≤ D(X) + D(Y ),

where D stands for any of the following dimension theories: asymptotic

dimension, asymptotic Assouad-Nagata dimension, Assouad-Nagata di-

mension, or microscopic Assouad-Nagata dimension.

Proof. Let D(X) = m, D(Y ) = n and let k = m+n+1. Pick (m,k)-

dimension control function DX of X and (n,k)-dimensional control

function DY of Y , both of the correct type (arbitrary, linear, dilation,

or a dilation near 0). There are families {Ui}k

Y that are r-disjoint and bounded by DX(r) and DY(r) respectively,

that cover X and Y at least k −m times and k −n times respectively.

Then the family {Ui×Vi}k

contained in sets from at least k−m = n+1 families from {Ui}k

y is contained in sets from at least k−n = m+1 families from {Vi}k

so there is at least one index j such that x is covered by Ujand y is

covered by Vj. The family {Ui×Vi}k

DX(r) + DY(r).

i=1in X and {Vi}k

i=1in

i=1covers X ×Y , as for any point (x,y), x is

i=1and

i=1,

i=1is r-disjoint and is bounded by

?

3. Components, dimension, and coarseness

In this section we replace the language of r-disjoint families by the

language of r-components. This language has the advantage of being

portable to functions. It also allows for simple proofs of known results

3.11-3.12.

Definition 3.1. Let f : X → Y be a function of metric spaces, A is a

subset of X, and rX, rY are two positive numbers.

A is (rX,rY)-bounded if for any points x,x′∈ A we have

dX(x,x′) ≤ rX

anddY(f(x),f(x′)) ≤ rY.

An (rX,rY)-chain in A is a sequence of points x1,... ,xkin A such

that for every i < k the set {xi,xi+1} is (rX,rY)-bounded.

A is (rX,rY)-connected if for any points x,x′∈ A can be connected

in A by an (rX,rY)-chain.

Notice that any subset A of X is a union of its (rX,rY)-components

(the maximal (rX,rY)-connected subsets of A).

Definition 3.2. Let f : X → Y be a function of metric spaces. f is

called large scale uniform if there is function cf: R+→ R+such that

dX(x,y) ≤ r implies dY(f(x),f(y)) ≤ cf(r). The function cf will be

called a coarseness control function of f.

Page 6

6N. BRODSKIY, J. DYDAK, M. LEVIN, AND A. MITRA

Notice f is Lipschitz if and only if it has a coarseness control function

that is a dilation. f is asymptotically Lipschitz if and only if it has a

coarseness control function that is linear.

The following Lemma describes a useful case in which double pa-

rameter components coincide with single parameter components. The

proof of this Lemma is an easy exercise.

Lemma 3.3. Let f : X → Y be a function of metric spaces. cf: R+→

R+is a coarseness control function of f if and only if for any subset

A of X its (r,cf(r))-components coincide with its r-components.

The following Lemma describes the way we construct a subset of X

with all (rX,rY)-components being (RX,RY)-bounded. The proof of

this Lemma is an easy exercise.

Lemma 3.4. Let f : X → Y be a function of metric spaces, B be

a subset of X, and A be a subset of Y . If all rX-components of B

are RX-bounded and all rY-components of A are RY-bounded then all

(rX,rY)-components of the set B ∩ f−1(A) are (RX,RY)-bounded.

Lemma 3.5. Let f : X → Y be a function of metric spaces and A,B be

subsets of X. Suppose that all (rA

bounded and all (rB

RB

A ∪ B are (RA

X,rA

Y)-components of A are (RA

Y)-components of B are (RB

Xand RB

X+ 2rA

Proof. Let x = x1,x2,...,xn= x′form an (rB

Notice that, if for some indices i < j we have xk∈ B for all i < k < j,

then xi+1and xj−1are in one (rB

dX(xi+1,xj−1) ≤ RB

If xi,xj∈ A and xk∈ B for all i < k < j, then

dX(xi,xj) ≤ dX(xi,xi+1)+dX(xi+1,xj−1)+dX(xj−1,xj) ≤ rB

and, similarly, dY(f(xi),f(xj)) ≤ rB

xi,xjbelong to the same (rA

all points in the chain x = x1,x2,...,xn= x′belonging to A are in one

(rA

Now let xsbe the first point in the chain belonging to A and xtbe

the last point in the chain belonging to A. Then

X,RA

Y)-

X,rB

X,RB

X,rB

Y)-bounded. If

Y)-components of

X+ 2rB

X< rA

Y+ 2rB

X,RA

Y< rA

Y)-bounded.

Ythen all (rB

Y+ 2rA

X,rB

Y)-chain in A ∪ B.

X,rB

Y)-component of B and therefore

Y.

Xand dY(f(xi+1),f(xj−1)) ≤ RB

X+RB

X+rB

X< rA

X

Y+RB

Y)-component of A. This implies that

Y+rB

Y< rA

Y. Thus the points

X,rA

X,rA

Y)-component of A.

dX(x,x′) ≤

dX(x1,xs−1)+dX(xs−1,xs)+dX(xs,xt)+dX(xt,xt+1)+dX(xt+1,xn) ≤

≤ RB

Similarly, dY(f(x),f(x′)) ≤ RB

X+ rB

X+ RA

X+ rB

Y+rB

X+ RB

Y+RA

X< RA

Y+rB

X+ 2rA

Y+RB

X.

Y< RA

Y+2rA

Y.

?

Page 7

HUREWICZ THEOREM FOR ASSOUAD-NAGATA DIMENSION7

Corollary 3.6. Let f : X → Y be a function of metric spaces and

{Bi}n

of Biare (R(i)

and R(i+1)

Y

+ 2r(i+1)

Y

< r(i)

i=1be subsets of X. Suppose that for every i all (r(i)

X,R(i)

Y, then all (r(n)

X,r(i)

+2r(i+1)

Y)-components

Y)-bounded. If, for every i < n, R(i+1)

X

X

< r(i)

X

X,r(n)

Y)-components of

n ?

i=1Biare

(R(1)

X+ 2r(1)

X,R(1)

Y+ 2r(1)

Y)-bounded.

Proof. By induction using Lemma 3.5.

?

Proposition 3.7. Suppose A is a subset of a metric space X, m ≥ 0,

and R > 0. If DA is an m-dimensional control function of A, then

DB(x) := DA(x + 2R) + 2R is an m-dimensional control function of

the R-neighborhood B = B(A,R) of A.

Proof.

Given r > 0 express A as

m+1

?

i=1

Ai such that (r + 2R)-

components of Ai are DA(r + 2R)-bounded. Given an r-component

of Bi := B(Ai,R), each point in that component is R-close to a

single (r + 2R)-component of Ai. Therefore r-components of Bi are

(DA(r + 2R) + 2R)-bounded.

?

Definition 3.8. Given a metric space X and r > 0 the r-scale di-

mension r-dim(X) is the smallest integer n ≥ 0 such that X can be

n+1

?

i=1

bounded.

expressed as X =Xiand r-components of each Xiare uniformly

Notice asdim(X) is the smallest integer n such that r-dim(X) ≤ n

for all r > 0. Also, asdim(X) is the smallest integer such that for

each r > 0 the space X can be expressed as X =

n+1

?

i=1Xi so that

r-dim(Xi) ≤ 0 for each i ≤ n + 1.

Corollary 3.9. Suppose X is a metric space. If, for every r > 0, there

is a subspace Xrof X such that asdim(Xr) ≤ n and r-dim(X\Xr) ≤ n,

then asdim(X) ≤ n.

Proof.

Express X \ Xr as

n+1

?

i=1Ai such that r-components of Ai

n+1

?

i=1Bi so that (R + 2r)-componentsare R-bounded. Express Xr as

of Bi are M-bounded. By Lemma 3.5, r-components of Ai∪ Biare

(M + 2R + 4r)-bounded.

?

Page 8

8N. BRODSKIY, J. DYDAK, M. LEVIN, AND A. MITRA

Corollary 3.10. Suppose G is a group with a left-invariant metric

and Gris the subgroup of G generated by B(1G,r). If, for every r > 0,

asdim(Gr) ≤ n, then asdim(G) ≤ n.

Proof. Consider gs∈ G, s ∈ S, so that [gs] enumerate G/Gr\ [Gr].

Notice Y =

?

s∈Sgs·Gris of r-dim(Y ) ≤ asdim(Gr) and Y ∪Gr= G. ?

Corollary 3.11 (Dranishnikov-Smith [16]). If G is a group with a

proper left-invariant metric, then asdim(G) is the supremum of asdim(H)

over all finitely generated subgroups H of G.

Proof. Since balls B(1G,r) are finite, the groups Gr in 3.10 are

finitely generated.

?

Corollary 3.12 (Bell-Dranishnikov [3]). Suppose X is a metric space

and {Xs}s∈Sis a family of subsets of X such that X =

?

s∈SXsand there

is a single n-dimensional control function D for all Xs. If, for every

r > 0, there is a subspace Xrof X such that asdim(Xr) ≤ n and the

family {Xs\ Xr}s∈Sis r-disjoint, then asdim(X) ≤ n.

Proof. Notice r-dim(X \ Xr) ≤ n.

?

4. Dimension of a function

When trying to generalize Hurewicz Theorem from covering dimen-

sion to any other dimension theory, the issue arises of how to define

the dimension of a function. Let us present an example of a Lipschitz

function demonstrating that replacing sup{dim(f−1(y)) | y ∈ Y } by

sup{asdim(f−1(B)) | B ⊂ Y is bounded} does not work.

Proposition 4.1. There is a Lipschitz function f : X → Y of metric

spaces such that

asdim(Y ) = 0 = sup{asdim(f−1(B)) | B ⊂ Y is bounded}

and asdim(X) > 0.

Proof. Let Y consist of points 2n, n ≥ 1, on the x-axis and X is the

union of vertical segments Inof length n and starting at 2n. f : X → Y

is the projection. Since f−1(B) is bounded for every bounded B ⊂ Y ,

sup{asdim(f−1(B)) | B ⊂ Y is bounded} = 0. Also, asdim(Y ) = 0.

However, asdim(X) > 0 as for each n it has n-components of arbitrarily

large size.

?

Remark 4.2. 4.1 shows the answer to a problem of Dranishnikov [14] in

negative. That problem asks if l-asdim(X) ≤ l-asdim(Y )+l-asdimf−1,

where l-asdimf−1is defined as sup{l-asdim(f−1(B)) | B ⊂ Y is bounded}

Page 9

HUREWICZ THEOREM FOR ASSOUAD-NAGATA DIMENSION9

and f is Lipschitz (see 2.2 for the definition of l-asdim). Notice l-asdim(Y ) =

0, l-asdimf−1= 0, and l-asdim(X) ≥ asdim(X) > 0 in 4.1.

In view of 4.1 we define asymptotic dimension of a function as follows:

Definition 4.3. Given a function f : X → Y of metric spaces we define

the asymptotic dimension asdim(f) of f as the supremum of asymptotic

dimensions of A ⊂ X so that f(A) ⊂ Y is of asymptotic dimension 0.

Definition 4.4. Given a function f : X → Y of metric spaces and

given m ≥ 0, an m-dimensional control function of f is a function

Df: R+×R+→ R+such that for all rX> 0 and RY > 0 any (∞,RY)-

bounded subset A of X can be expressed as the union of m + 1 sets

whose rX-components are Df(rX,RY)-bounded.

Proposition 4.5. Suppose f : X → Y is a function of metric spaces

and m ≥ 0. If asdim(f) ≤ m, then f has an m-dimensional control

function Df.

Proof.

yn ∈ Y such that An = f−1(B(yn,RY)) cannot be expressed as the

union of m + 1 sets whose rX-components are n-bounded. The set

∞ ?

Fix rX > 0 and RY > 0. Suppose for each n there is

C =

n=1B(yn,RY) cannot be bounded as asdim(f−1(C)) ≤ m for any

bounded subset C of Y . By passing to a subsequence we may arrange

yn→ ∞ and asdim(C) = 0, a contradiction.

?

Definition 4.6. Given a function f : X → Y of metric spaces and

given k ≥ m + 1 ≥ 1, an (m,k)-dimensional control function of f is

a function Df: R+× R+→ R+such that for all rX> 0 and RY > 0

any (∞,RY)-bounded subset A of X can be expressed as the union of

k sets {Ai}k

any x ∈ A belongs to at least k − m elements of {Ai}k

i=1whose rX-components are Df(rX,RY)-bounded so that

i=1.

Proposition 4.7. Let f : X → Y be a function of metric spaces and

m ≥ 0. Suppose there is an m-dimensional control function D(m+1)

R+→ R+of f. If one defines inductively functions D(k)

by D(k)

f

(3rX,RY)+2rX, then each D(k)

dimensional control function of f.

f

: R+×

f

for k > m+1

f(rX,RY) = D(k−1)

f

is an (m,k)-

Proof. The proof is by induction on k. The case of k = m + 1 is

obvious. Suppose the result holds for some k ≥ n + 1 and a subset

A of X is (∞,RY)-bounded. There are k subsets {Ai}k

3rX-components bounded by D(k)

m + 1 of those sets covers A.

i=1, each has

f(3rX,RY) such that the union of any

Page 10

10N. BRODSKIY, J. DYDAK, M. LEVIN, AND A. MITRA

Define A′

rX-components of each A′

are contained in rX-neighborhoods of 3rX-components of Ai.

Define A′

ito be the rX-neighborhood of Aiin A for i ≤ k. Notice

iare (D(k)

f(3rX,RY) + 2rX)-bounded as they

k+1as the union of all sets?

i∈SAi\?

i/ ∈S

A′

i, where S is a subset

of {1,... ,k} consisting of exactly k − m elements.

Suppose x ∈ A belongs exactly to k −m sets Aisuch that i ≤ k and

let S = {i ≤ k | x ∈ Ai}. If x / ∈ A′

some j / ∈ S. Thus each x ∈ A belongs to at least k + 1 − m elements

of {A′

Notice that any rX-component of A′

component of some Aj resulting in rX-components of each A′

(D(k)

k+1, then x must belong to A′

jfor

i}k+1

i=1.

k+1is contained in a single rX-

ibeing

f(3rX,RY) + 2rX)-bounded.

?

Proposition 4.8. Let f : X → Y be a function of metric spaces and

k,m ≥ 0. If Df: R+×R+→ R+is an (m,k)-dimensional control func-

tion of f, then for any B ⊂ Y whose rY-components are RY-bounded,

the set f−1(B) can be covered by k sets whose (rX,rY)-components are

Df(rX,RY)-bounded and every element of f−1(B) belongs to at least

k − m elements of that covering.

Proof. Given an rY-component S of B express f−1(S) as AS

AS

element of f−1(S) belongs to at least k −m elements of that covering.

Put Ai=?

S

in an rX-component of some AS

1∪...∪

ksuch that rX-components of AS

iare Df(rX,RY)-bounded and every

AS

iand notice each (rX,rY)-component of Aiis contained

i.

?

Theorem 4.9. Let k = m+n+1, where m,n ≥ 0. Suppose f : X → Y

is a large scale uniform function of metric spaces and asdim(Y ) ≤ n.

asdim(X) ≤ m + n

if there is an (m,k)-dimensional control function Df of f.

over, if one can choose the coarseness control function cf of f, the

n-dimensional control function of Y , and Dfto be linear (respectively,

a dilation), then X has a (k − 1)-dimensional control function that is

linear (respectively, a dilation).

More-

Proof. Let cfbe a coarseness control function of f. Let DY: R+→ R+

be an (n,k)-dimensional control function of Y . Notice we may require

cf(r) > r, DY(r) > r, and Df(r,R) > r + R as we may redefine those

functions by adding r or r + R without losing their properties or type

(linear, dilation or dilation near 0).

Page 11

HUREWICZ THEOREM FOR ASSOUAD-NAGATA DIMENSION11

Given a number r > 0 we are going to construct a number DX(r)

and represent the space X as a union of k sets {Dj}k

components of Djbeing DX(r)-bounded for every j.

Define inductively a sequence of numbers r(n+1)

R(n)

Y

< ··· < r(1)

Y

moving to lower indices so that for every i we have R(i)

r(i)

Y

.

Express Y as the union of k sets {Ai}k

of Aiare R(0)

set Aias the union of k sets {Uj

Uj

least m sets.

Define inductively a sequence of numbers r(n+1)

R(n)

R(i)

X

For every i we express the set f−1(Ai) as the union of k sets {Bj

such that all (r(i)

every j and every point x ∈ f−1(Ai) belongs to at least n sets.

Put Dj

Dj’s cover X by the use of Kolmogorov’s argument: given x ∈ X there

is i so that f(x) ∈ Ai. The set of j’s such that x ∈ Bj

k − m elements, the set of j’s such that f(x) ∈ Uj

elements, so they cannot be disjoint.

Notice all (r(i)

By 3.6 all (r(n+1)

X

,r(n+1)

Y

)-components of the set Djare (3R(1)

bounded. Since (r(n+1)

X

,r(n+1)

Y

) = (r,cf(r)), by 3.3 all r-components

of Djare 3R(1)

spectively, a dilation), if the coarseness control function cf of f, the

n-dimensional control function of Y , and Df are linear (respectively,

dilations).

j=1with all r-

Y

< R(n+1)

Y

< r(n)

Y

<

Y < R(1)

Y < r(0)

Y < R(0)

starting from r(n+1)

Y

Y= DY(r(i)

= cf(r) and

Y) and

Y= 3R(i+1)

i=1such that all r(0)

Y-components

Y-bounded for every i. Since Ai⊂ Y , we can express the

i}k

Y-bounded for every j and every point y ∈ Aibelongs to at

j=1such that all r(i)

Y-components of

iare R(i)

X

< R(n+1)

X

< r(n)

X <

X··· < r(1)

X= Df(r(i)

X< R(1)

X,R(i)

Xstarting with r(n+1)

Y) and r(i)

X

= r and for every i we have

X= 3R(i+1)

.

i}k

j=1

X,r(i)

Y)-components of Bj

iare (R(i)

X,R(i)

Y)-bounded for

i= Bj

i∩ f−1(Uj

i) and let Djbe the union of all Dj

i. Notice

ihas at least

ihas at least k − n

X,r(i)

Y)-components of the set Dj

iare (R(i)

X,R(i)

Y)-bounded.

X,3R(1)

Y)-

X-bounded. Observe, 3R(1)

Xis a linear function of r (re-

?

Corollary 4.10. Suppose f : X → Y is a function of metric spaces

and m ≥ 0. asdim(f) ≤ m if and only if f has an m-dimensional

control function.

Proof. In one direction use 4.5. In the other direction apply 4.9 in

the case of n = 0.

?

Theorem 4.11.

asdim(X) ≤ asdim(f) + asdim(Y )

Page 12

12N. BRODSKIY, J. DYDAK, M. LEVIN, AND A. MITRA

for any large scale uniform function f : X → Y .

Proof. Follows from 4.9, 4.10, and 4.7.

In [3] there is a concept of a family {Xα} of subsets of X satisfying

asdim(Xα) ≤ n uniformly. Notice that in our language it means there

is one function that serves as an n-dimensional control function for all

Xα.

?

Corollary 4.12 (Bell-Dranishnikov [3]). Let f : X → Y be a Lipschitz

function of metric spaces. Suppose that, for every R > 0,

asdim{f−1(BR(y))} ≤ n

uniformly (in y ∈ Y ). If X is geodesic, then asdim(X) ≤ asdim(Y )+n.

Proof.

dimensional control function, so apply 4.9 and 4.7.

asdim{f−1(BR(y))} ≤ n uniformly means f has an n-

?

Remark 4.13. Notice Bell-Dranishnikov’s version of Hurewicz type the-

orem 4.12 (see [3]) assumes that X is geodesic. We do not use this

assumption in 4.11.

5. Asymptotic dimension of groups

J.Smith [27] showed that any two proper and left-invariant metrics

on a given countable group G are coarsely equivalent. We generalize

that result as follows.

Proposition 5.1. Suppose f : G → H is a homomorphism of groups

and dG, dH are left-invariant metrics on G and H, respectively. If

f : (G,dG) → (H,dH) is coarsely proper (i.e., it sends bounded subsets

of G to bounded subsets of H), then f : (G,dG) → (H,dH) is large scale

uniform.

Proof.

R > 0 such that dH(1H,f(g)) < R for all g ∈ B(1G,r). If x,y ∈ G

satisfy dG(x,y) < r, then x−1· y ∈ B(1G,r), so dH(f(x),f(y)) =

dH(1H,f(x−1y)) < R.

Suppose r > 0. Since f(B(1G,r)) is bounded, there is

?

Corollary 5.2. Suppose f : G → H is a homomorphism of groups and

dG, dH are left-invariant metrics on G and H, respectively. If dG is

proper, then f : (G,dG) → (H,dH) is large scale uniform.

Proof. Since bounded subsets of G are finite, f is coarsely proper.

?

Corollary 5.3 (J.Smith [27]). Any two proper left-invariant metrics

on a group G are coarsely equivalent.

Page 13

HUREWICZ THEOREM FOR ASSOUAD-NAGATA DIMENSION13

The subsequent result is derived in [16] from the corresponding the-

orem for finitely generated groups. We deduce it directly from 4.11.

Theorem 5.4 (Dranishnikov-Smith [16]). If 1 → K → G → H → 1

is a short exact sequence of countable groups, then

asdim(G) ≤ asdim(K) + asdim(H).

Proof. It suffices to show asdim(f) = k = asdim(K), where f : G →

H. Given RH > 0, the subspace f−1(B(1H,RH)) is in a bounded

neighborhood of K as B(1H,RH) is finite, so it has the same asymptotic

dimension as K. Fix the dimension control function D(rG,RH) for that

space. Notice that all spaces f−1(B(h,RH)), h ∈ H, are isometric to

f−1(B(1H,RH)), so D(rG,RH) is a dimension control function of all of

them. By 4.10, asdim(f) ≤ k and by applying 4.11 we are done.

In [16] the asymptotic dimension of a group G is defined as the supre-

mum of asdim(H), H ranging through all finitely generated subgroups

of G. We provide an alternative definition which will be applied in the

next section.

?

Definition 5.5. Given a finite subset S of a group G, an S-component

of G is an equivalence class of G of the relation x ∼ y iff x can be

connected to y by a finite chain xiso that x−1

A family of subsets {Ai}i∈Jis S-bounded if x−1·y ∈ S for each i ∈ J

and all x,y ∈ Ai.

i

· xi+1∈ S for all i.

Proposition 5.6. Let G be a group. asdim(G) ≤ n if and only if

for each finite subset S of G there is a finite subset T of G and a

decomposition G = A1∪ ... ∪ An+1such that S-components of Aiare

T-bounded for all i ≤ n + 1.

Proof. Suppose for each finite subset S of G there is a finite subset T

of G and a decomposition G = A1∪...∪An+1such that S-components

of Aiare T-bounded for all i ≤ n + 1. Given a countable subgroup

H of G and given a proper left-invariant metric dH on H, we put

S = B(1H,r) and R = max{dH(1H,t) | t ∈ T · T ∩ H} (by T · T we

mean all products t · s, where t,s ∈ T). Notice that r-components of

Ai∩ H are R-bounded.

Conversely, suppose asdim(H) ≤ n for all finitely generated sub-

groups H of G. Given a finite subset S of G let H be the subgroup of

G generated by S and let d be the word metric on H induced by S.

Since asdim(H) ≤ n, there is a decomposition of H into A1∪...∪An+1

such that 1-components of Aiare m-bounded for all i ≤ n+1 and some

m > 0. Let T = B(1H,m + 1). Pick a representative gj∈ G, j ∈ J, of

Page 14

14N. BRODSKIY, J. DYDAK, M. LEVIN, AND A. MITRA

each left coset of H in G. Put Bi=?

j∈J

gj·Aiand notice S components

of Biare T-bounded.

?

6. Linearly controlled asymptotic dimension of groups

Unlike the asymptotic dimension of countable groups, the asymptotic

Assouad-Nagata dimension may depend on the left-invariant metric.

Piotr Nowak [23] constructed finitely generated groups Gnof asymp-

totic dimension n ≥ 2 and asdimAN(Gn) = ∞. Obviously, there is

no such example for n = 0 as finitely generated groups of asymptotic

dimension 0 are finite. However, 6.4 does provide a countable group G

with a proper, left-invariant metric d such that asdim(G,d) = 0 and

asdimAN(G,d) = ∞.

The purpose of this section is to solve in negative the following two

problems of Dranishnikov [14].

Problem 6.1. Does l-asdim(X) = asdimAN(X) hold for metric spaces?

Problem 6.2. Find a metric space X of minimal asdim(X) such that

asdim(X) < l-asdim(X).

Here is an answer to 6.2.

Proposition 6.3. There is a proper, left-invariant metric dGon G =

∞

?

n=2Z/n such that l-asdim(G,dG) > 0 = asdim(G,dG).

Proof. Pick a generator gnof Z/n and assign it the norm of n for

n ≥ 2. Extend the norm over all elements g of G as the minimum

∞

?

n=1|kn| · n, where g =

dG(g,h) := |g − h| is a proper and invariant metric on G. Suppose

there is an unbounded subset U of R+such that for some C > 0 all

r-components of G are (C·r)-bounded for r ∈ U. Pick r ∈ U such that

4C + 4 < r and let n be an integer such that r − 2 < 2n ≤ r. Notice

all 2n-components of G are contained in some r-components of G, so

they are (n − 1) · (n + 1)-bounded as C <r−4

However, the 2n-component of 0 contains Z/2n and |n · g2n| = 2n2>

(n − 1) · (n + 1), a contradiction.

of

∞

?

n=1kn· gnand knare integers. Notice that

4

≤n−2

2

and r < 2n + 2.

?

Proposition 6.4. There is an Abelian torsion group G with a proper

invariant metric dGsuch that l-asdim(G,dG) = 0 and asdimAN(G,dG) =

∞.

Proof.

asdim(Zn+1) > n, there is r(n) > 0 such that any decomposition of

Consider Zn+1with the standard word metric ρn. Since

Page 15

HUREWICZ THEOREM FOR ASSOUAD-NAGATA DIMENSION15

Zn+1as the union X0∪...∪Xnforces one Xito have r(n)-components

without a common upper bound. Without loss of generality we may

assume r(n) is an integer greater than n.

and consider Gn= (Zs(n))n+1with the metric dnequal to t(n) times

the standard word metric. Numbers t(n) are chosen so that t(n+1) >

diam(G1)+...+diam(Gn) for all n ≥ 1. Since diam(Gn) = 4nt(n)r(n),

we want t(n + 1) = 1 + 4 · t(1)r(1) + ... + 4n · t(n)r(n) for all n ≥ 1.

The group G is the direct sum of all Gnwith the obvious metric d:

d({xn},{yn}) =?dn(xn,yn). Notice that

of size at most t(n). Indeed, any

x + G1⊕ ... ⊕ Gn−1for some x ∈ G. Thus l-asdim(G,d) = 0.

Consider the projection πn: Zn+1→ Gnand notice it is t(n)-Lipschitz.

Moreover, if two points x and y in Zn+1satisfy ρn(x,y) ≤ 4nr(n), then

dn(x,y) = t(n) · ρn(x,y).

Suppose Gnequals Y0∪ ... ∪ Yn. Assume r(n)-components of some

Xj= π−1

x0,... ,xk in Xj such that ρn(xi,xi+1) < r(n) and (n + 2) · r(n) >

ρn(x0,xk) > n·r(n). Assume such a sequence does not exist with num-

ber of elements smaller than k + 1. Therefore dn(πn(xi),πn(xi+1)) =

t(n)·ρn(xi,xi+1) < t(n)·r(n) and (n+2)·t(n)r(n) > ρn(πn(x0),πn(xk)) >

n·t(n)r(n). Thus, one (t(n)·r(n))-component of Yiis of diameter bigger

than n · t(n)r(n).

Suppose asdimAN(G,d) = k < ∞ and DG(r) = c · r + b is a k-

dimension control function of (G,d). Pick n > max(k,|c|,|b|) + 2 and

choose a decomposition Y0∪...∪Ykof G so that (t(n)·r(n))-components

of each Yiare bounded by c·t(n)·r(n)+b < (c·t(n)+1)·r(n). Notice

there is a 1-Lipschitz projection pn: G → Gn. Therefore each set pn(Yi)

has (t(n) · r(n))-components bounded by (c · t(n) + 1) · r(n). On the

other hand, for some j, there is an (t(n) · r(n))-component of pn(Yj)

of diameter bigger than nt(n) · r(n). Thus, c · t(n) + 1 > nt(n) and

c > n − 1/t(n), a contradiction.

Put s(n) = 8 · n · r(n)

t(n)

2-components of G are

2-component of G is contained in

t(n)

n(Yj) are not uniformly bounded. Therefore there is a sequence

?

Remark 6.5. Proposition 6.4 solves Problem 6.1 in negative.

Proposition 6.6. For any countable group G there is a proper left-

invariant metric dGsuch that asdimAN(G,dG) = asdim(G).

Proof.

increasing sequence S0⊂ S1... of its finite subsets so that {1G} = S0,

each Siis symmetric, and for each i there is a decomposition of G as

Ai

dG(x,y) as the smallest i so that x−1·y ∈ Si. Notice that in the metric

dGall i-components of Ai

Let asdim(G) = n − 1.Express G as the union of an

1∪ ... ∪ Ai

nso that Si-components of Ai

kare Si+1-bounded. Define

kare (i + 1)-bounded.

?