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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:
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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
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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,
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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.
?
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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.
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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.
?
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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.
?
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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}
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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 DIMENSION 13
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 DIMENSION 15
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.
?
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