Lectures on the Geometric Group Theory University of Utah Contents

Lectures on the Geometric Group Theory
University of Utah
Misha Kapovich
February 20, 2003
1 Preliminaries
2 Coarse topology
3 Ultralimits of Metric Spaces
4 Tits alternative
5 Growth of groups and Gromov's theorem
6 Quasiconformal mappings
7 Quasi-isometries of nonuniform lattices in H n .
8 A quasi-survey of QI rigidity
1 Preliminaries
1.1 Introduction
These lecture notes are based on the course that I was teaching at the University of
Utah in Fall of 2002. Our main goal is to describe various tools of quasi-isometric
rigidity and to give (essentially self-contained) proofs of several fundamental theorems
in this area: Gromov's theorem on groups of polynomial growth and Schwartz's QI
rigidity theorem for nonuniform lattices in the real-hyperbolic spaces. We conclude
with a survey of the QI rigidity theory.
The main idea of the geometric group theory is to treat nitely-generated groups
as geometric objects: with each nitely-generated group G we will associate a metric
space, the Cayley graph of G. One of the main issues of the geometric group theory is
to recover as much as possible algebraic information about G from the geometry of the
Cayley graph. A primary obsticle for this is the fact that the Cayley graph depends
not only on G but on a particular choice of a generating set of G. Cayley graphs
associated with dierent generating sets are not isometric but quasi-isometric. One
of the primary questions which we will try to address is: If G; G0 are quasi-isometric
groups, to which extent G and G0 share the same algebraic properies? The best
one can hope here is to recover the group G up to weak commensurability from its
geometry. The equivalence relation of weak commensurability is generated by two
1. Passing to a nite index subgroup (this leads to the commensurability equivalence
2. Taking nite kernel extensions G of a group ?:
1!F !G!?!1
is a short exact sequence so that F is nite.
Weak commensurability implies quasi-isometry but, in general, the converse is false.
One of the easiest examples is the following: Pick two matrices A; B 2 SL(2; Z) so
that An 6= B m for all n; m 2 Z n f0g. Dene two actions of Z on Z2 so that the
generator 1 2 Z acts by the automorphisms given by A and B respectively. Then
the semidirect products G := Z2 oA Z, G0 := Z2 oB Z are quasi-isometric but not
weakly commensurable. Observe that both groups G; G0 are polycyclic. The following
is unknown even for the group G above:
Problem 1. Suppose that ? is a group quasi-isometric to a polycyclic group G. Is
? commensurable to a polycyclic group?
An example when quasi-isometry implies weak commensurability is given by the
following theorem due to R. Schwartz:
Theorem 2. Suppose that G is a nonuniform lattice acting on the hyperbolic space
H n ; n 3. Then for each group ? quasi-isometric to G, the group ? is weakly commensurable with G.
We will present a proof of this theorem in chapter 7. Another example of quasiisometric rigidity is the following corollary from Gromov's theorem on groups of
polynomial growth:
Corollary 3. Suppose that G is a group quasi-isometric to a nilpotent group. Then
G itself is virtually nilpotent, i.e. contains a nilpotent subgroup of nite index.
Gromov's theorem and its corollary will be proven in chapter 5.
Proving these theorems are the main objectives of this course. Along the way we
will introduce several tools of the geometric group theory: coarse topology, ultralimits,
quasiconformal mappings.
1.2 Cayley graphs of nitely generated groups
Let ? be a nitely generated group with the generating set S = fs1; :::; sng, we shall
assume that the identity does not belong to S . Dene the Cayley graph C = C (?; S )
as follows: The vertices of C are the elements of ?. Two vertices g; h 2 ? are
connected by an edge if an only if there is a generator si 2 S such that h = gsi. Then
C is a locally nite graph. Dene the word metric d on C by assuming that each
edge has the unit length, this denes the length of nite PL-paths in C , nally the
distance between points p; q 2 C is the inmum (same as minimum) of the lengths of
PL-paths in C connecting p to q. For g 2 G the word length `(g) is just the distance
d(1; g) in C . It is clear that the left action of the group ? on the metric space (C; d)
is isometric.
Below are two simple examples of Cayley graphs.
Example 4. Let ? be free Abelian group on two generators s1 ; s2. Then S = fsi; i =
1; 2g. The Cayley graph C = C (?; S ) is the square grid in the Euclidean plane: The
vertices are points with integer coordinates, two vertices are connected by an edge if
and only if exactly only two of their coordinates are distinct and they dier by 1.
a -2
a -1
b -1
Figure 1: Free abelian group.
Example 5. Let ? be the free group on two generators s1; s2 . Take S = fsi; i = 1; 2g.
The Cayley graph C = C (?; S ) is the 4-valent tree (there are four edges incident to
each vertex).
See Figures 1, 2.
1.3 Quasi-isometries
Let X be a metric space. We will use the notation NR (A) to denote R-neighborhood
of a subset A X , i.e. NR (A) = fx 2 X : d(x; A) < Rg. Recall that Hausdor
distance between subsets A; B X is dened as
dHaus(A; B ) := inf fR : A NR (B ); B NR(A)g:
Two subsets of X are called Hausdor-close if they are within nite Hausdor distance
from each other.
Denition 6. Let X; Y be complete metric spaces. A map f : X ! Y is called
(L; A)-coarse Lipschitz if
dY (f (x); f (x0)) LdX (x; x0 ) + A
a -2
a -1
b -1
Figure 2: Free group.
for all x; x0 2 X . A map f : X ! Y is called a (L; A)-quasi-isometric embedding
L?1dX (x; x0 ) ? A dY (f (x); f (x0)) LdX (x; x0 ) + A
for all x; x0 2 X . Note that a quasi-isometric embedding does not have to be an
embedding in the usual sense, however distant points have distinct images.
An (L; A)-quasi-isometric embedding is called an (L; A)-quasi-isometry if it admits a quasi-inverse map f : Y ! X which is a (L; A)-quasi-isometric embedding
so that:
(x); x) A; dY (f f(y); y) A
dX (ff
for all x 2 X; y 2 Y .
We will abbreviate quasi-isometry, quasi-isometric and quasi-isometrically to QI.
In the most cases the quasi-isometry constants L; A do not matter, so we shall
use the words quasi-isometries and quasi-isometric embeddings without specifying
constants. If X; Y are spaces such that there exists a quasi-isometry f : X ! Y
then X and Y are called quasi-isometric. In applications X and Y will be nonempty,
however, by working with relations instead of maps one can modify this denition so
that the empty set is quasi-isometric to any bounded metric space.
Exercise 10. If f : X ! Y is a quasi-isometry and g is within nite distance from
f (i.e. sup d(f (x); g(x)) < 1) then g is also a quasi-isometry.
Exercise 11. A subset S of a metric space X is said to be r-dense in X if the
Hausdor distance between S and X is at most r. Show that if f : X ! Y is a
quasi-isometric embedding such that f (X ) is r-dense in X for some r < 1 then f is
a quasi-isometry. Hint: Construct a quasi-inverse f to the map f by mapping point
y 2 Y to x 2 X such that
dY (f (x); y) dY (f (X ); y) + 1:
For instance, the cylinder X = Sn R is quasi-isometric to Y = R ; the quasiisometry is the projection to the second factor.
Exercise 12. Show that quasi-isometry is an equivalence relation between (nonempty)
metric spaces.
A separated net in a metric space X is a subset Z X which is r-dense for some
r < 1 and such that there exists > 0 for which d(z; z0 ) , 8z 6= z0 2 Z .
Alternatively, one can describe quasi-isometric spaces as follows.
Lemma 13. Metric spaces X and Y are quasi-isometric i there are separated nets
Z X; W Y , constants L and C , and L-Lipschitz maps
f : Z ! Y; f : W ! X;
so that d(f f; id) C; d(f f; id) C .
Proof. Observe that if a map f : X ! Y is coarse Lipschitz then its restriction to
each separated net in X is Lipschitz. Conversely, if f : Z ! Y is a Lipschitz map
from a separated net in X then f admits a coarse Lipschitz extension to X .
In some cases it suces to check a weaker version of (9) to show that f is a quasiisometry.
Let X; Y be proper metric spaces. Recall that a (continuous) map f : X ! Y is
called proper if the inverse image f ?1(K ) of each compact in Y is a compact in X .
Denition 14. A map f : X ! Y is called uniformly proper if f is coarse Lipschitz
and there exists a distortion function (R) such that diam(f ?1(B (y; R))) (R) for
each y 2 Y; R 2 R + . In other words, there exists a proper function : R + ! R + such
that whenever d(x; x0 ) r, we have d(f (x); f (x0)) (r).
To see an example of a map which is proper but not uniformly proper consider the
biinnite curve ? embedded in R 2 (Figure 3):
Figure 3:
Lemma 15. Suppose that Y is a geodesic metric space, f : X ! Y is a uniformly
proper map whose image is r-dense in Y for some r < 1. Then f is a quasi-isometry.
Proof. Let's construct a quasi-inverse to the map f . Given a point y 2 Y pick a point
f(y) := x 2 X such that d(f (x); y) r. Let's check that f is coarse Lipschitz. Since
Y is a geodesic metric space it suces to verify that there is a constant A such that
for all y; y0 2 Y with d(y; y0) 1, one has:
d(f(y); f(y0)) A:
Pick t > 1 which is in the image of the distortion funcion . Then take A 2 ?1(t).
It is also clear that f; f are quasi-inverse to each other.
Lemma 16. Let X be a proper geodesic metric space. Let G be a group acting
isometrically properly discontinuously cocompactly on X . Pick a point x0 2 X . Then
the group G is nitely generated; for some choice of nite generating set S of the
group G the map f : G ! X , given by f (g) = g(x0), is a quasi-isometry. Here G is
given the word metric induced from C (G; S ).
Proof. Our proof follows [24, Proposition 10.9]. Let B = BR(x0 ) be the closed ball
of radius R in X with the center at x0 such that BR?1 (x0 ) projects onto X=G. Since
the action of G is properly discontinuous, there are only nitely many elements si 2
G ? f1g such that B \ siB 6= ;. Let S be the subset of G which consists of the above
elements si (it is clear that s?i 1 belongs to S i si does). Let
r := inf fd(B; g(B )); g 2 G ? (S [ f1g)g:
Clearly r > 0. We claim that S is a generating set of G and that for each g 2 G
`(g) d(x0; g(x0))=r + 1
where ` is the word length on G (with respect to the generating set S ). Let g 2 G,
connect x0 to g(x0) by the shortest geodesic . Let m be the smallest integer so that
d(x0; g(x0)) mr + R. Choose points x1 ; :::; xm+1 = g(x0) 2 , so that x1 2 B ,
d(xj ; xj+1) < r, 1 j m. Then each xj belongs to gj (B ) for some gj 2 G. Let
1 j m, then gj?1(xj ) 2 B and d(gj?1(gj+1(B )); B ) d(gj?1(xj ); gj?1(xj+1)) < r.
Thus the balls B; gj?1(gj+1(B )) intersect, which means that gj+1 = gj si(j) for some
si(j) 2 S [ f1g. Therefore
g = si(1) si(2) ::::si(m) :
We conclude that S is indeed a generating set for the group G. Moreover,
`(g) m (d(x0 ; g(x0)) ? R)=r + 1 d(x0 ; g(x0))=r + 1:
The word metric on the Cayley graph C = C (G; S ) of the group G is left-invariant,
thus for each h 2 G we have:
d(h; hg) = d(1; g) d(x0 ; g(x0))=r + 1 = d(h(x0 ); hg(x0))=r + 1:
Hence for any g1; g2 2 G
d(g1; g2) d(f (g1); f (g2))=r + 1:
On the other hand, the triangle inequality implies that
d(x0 ; g(x0)) t`(g)
where d(x0; s(x0 )) t 2R for all s 2 S . Thus
d(f (g1); f (g2))=t d(g1; g2):
We conclude that the map f : G ! X is a quasi-isometric embedding. Since f (G) is
R-dense in X , it follows that f is a quasi-isometry.
Corollary 18. Let S1; S2 be nite generating sets for a nitely generated group G
and d1 ; d2 be the word metrics on G corresponding to S1 ; S2 . Then the identity map
(G; d1) ! (G; d2) is a quasi-isometry.
Proof. The group G acts isometrically cocompactly on the proper metric space
(C (G; S2); d2):
Therefore the map id : G ! C (G; S2) is a quasi-isometry.
Lemma 19. Let X be a locally compact path-connected topological space, let G be
a group acting properly discontinuously cocompactly on X . Let d1 ; d2 be two proper
geodesic metrics on X (consistent with the topology of X ) both invariant under the
action of G. Then the group G is nitely generated and the identity map id : (X; d1) !
(X; d2) is a quasi-isometry.
Proof. The group G is nitely generated by Lemma 16, choose a word metric d on
G corresponding to any nite generating set (according to the previous corollary it
does not matter which one). Pick a point x0 2 X , then the maps
fi : (G; d) ! (X; di); fi(g) = g(x0)
are quasi-isometries, let fi denote their quasi-inverses. Then the map id : (X; d1) !
(X; d2) is within nite distance from the quasi-isometry f2 f1.
A (k; c)-quasigeodesic segment in a metric space X is a (k; c)-quasi-isometric embedding f : [a; b] ! X ; similarly, a complete (k; c)-quasigeodesic is a (k; c)-quasiisometric embedding f : R ! X . By abusing notation we will refer to the image of a
(k; c)-quasigeodesic as a quasigeodesic.
Corollary 20. Let d1; d2 be as in Lemma 19. Then any (complete) geodesic with
respect to the metric d1 is also a quasigeodesic with respect to the metric d2 .
1.4 Gromov-hyperbolic spaces
Roughly speaking, Gromov-hyperbolic spaces are the ones which exhibit \tree-like
behavior", at least if we restrict to nite subsets.
Let Z be a geodesic metric space. A geodesic triangle Z is called R-thin if
every side of is contained in the R-neighborhood of the union of two other sides.
An R-fat triangle is a geodesic triangle which is not R-thin. A geodesic metric space
Z is called -hyperbolic in the sense of Rips (Rips was the rst to introduce this
denition) if each geodesic triangle in Z is -thin. A nitely generated group is said
to be Gromov-hyperbolic if its Cayley graph is Gromov-hyperbolic.
Notation 21. For a subset S in a metric space X we will use the notation NR(S )
for the metric R-neighborhood of S in X .
Below is an alternative denition of -hyperbolicty due to Gromov.
Let X be a metric space (which is no longer required to be geodesic). Pick a
base-point p 2 X . For each x 2 X set jxjp := d(x; p) and dene the Gromov product
(x; y)p := 21 (jxjp + jyjp ? d(x; y)):
Note that the triangle inequality implies that (x; y)p 0 for all x; y; p; the Gromov
product measures how far the triangle inequality if from being an equality.
Exercise 22. Suppose that X is a metric tree. Then (x; y)p is the distance d(p; )
from p to the segment = xy.
In general we observe that for each point z 2 = xy
(p; x)z + (p; y)z = jzjp ? (x; y)p:
In particular, d(p; ) (x; y)p.
Suppose now that X is -hyperbolic in the sense of Rips. Then the Gromov product
is \comparable" with d(p; ):
Lemma 24.
(x; y)p d(p; ) (x; y)p + 2:
Proof. The inequality (x; y)p d(p; ) was proven above; so we have to establish
the other inequality. Note that since the triangle (pxy) is -thin, for each point
z 2 = xy we have
minf(x; p)z ; (y; p)z g minfd(z; px); d(z; py)g :
By continuity, there exists a point z 2 such that (x; p)z ; (y; p)z . By applying
the equality (23) we get:
jzjp ? (x; y)p = (p; x)z + (p; y)z 2:
Since jzjp d(p; ), we conclude that d(p; ) (x; y)p + 2.
Now dene a number p 2 [0; 1] as follows:
p := 2inf
fj8x; y; z 2 X; (x; y)p min((x; z)p ; (y; z)p) ? g:
Exercise 25. Suppose that X is a geodesic metric space. Show that X is zero-
hyperbolic (in the sense of Rips or Gromov) i X is a metric tree.
Exercise 26. If p for some p then q 2 for all q 2 X
X is said to be -hyperbolic in the sense of Gromov, if 1 > p for all p 2 X .
The advantage of this denition is that it does not require X to be geodesic and this
notion is manifestly QI-invariant:
If X; X 0 are quasi-isometric and X is -hyperbolic in the sense of Gromov then X 0 is
0{hyperbolic in the sense of Gromov. In contrast, QI invariance of Rips-hyperbolicity
is not a priori obvious. We will prove QI invariance of Rips-hyperbolicity in the
corollary 70 as a corollary of Morse lemma.
Lemma 27. (See [28, 6.3C])If X a geodesic metric space which is -hyperbolic in
Gromov's sense then X is 4-hyperbolic in the sense of Rips and vice-versa.
In what follows, we will refer to -hyperbolic spaces in the sense of Rips as being
Here are some examples of Gromov-hyperbolic spaces.
1. Let X = H n be the hyperbolic n-space. Then X is -hyperbolic for appropriate
. The reason for this is that the \largest" triangle in X is an ideal triangle, i.e. a
triangle all whose three vertices are on the boundary sphere of H n . All such triangles
are congruent to each other since Isom(H n ) acts transitively on triples of distinct
points in S n?1. Thus it suces to verify thinness of a single ideal triangle in H 2 ,
the triangle with the ideal vertices 0; 2; 1. I claim that for each point x on the arc
between 0 and m the distance to the side is < 1. Indeed, since dilations with center
at zero are hyperbolic isometries, the maximal distance from x to is realized at the
point m = 1 + i. Computing the hyperbolic length of the horizontal segment between
m and i 2 we conclude that it equals 1. Hence d(x; ) d(m; ) < 1. See Figure 4.
Remark 28. By making more careful computation with the hyperbolic distances one
can conclude that sinh(d(m; )) = 1.
2. Suppose that X is a complete Riemannian manifold of sectional curvature <
0. Then X is Gromov-hyperbolic. This follows from Rauch-Toponogov comparison
theorem. Namely, let Y be the hyperbolic plane with the curvature normalized to
be = < 0. Then Y is -hyperbolic. Let = (xyz) be a geodesic triangle
in X . Construct the comparison triangle 0 := (x0 y0z0 ) Y whose sides have
the same length as for the triangle . Then the triangle 0 is -thin. Pick a pair
of points p 2 xy; q 2 yz and the corresponding points p0 2 x0y0; q0 2 y0z0 so that
d(x; p) = d(x0; p0); d(y; q) = d(y0; q0). Then Rauch-Toponogov comparison theorem
implies that d(p; q) d(p0; q0). It immediately follows that the triangle is -thin.
Figure 4: Ideal triangle (0 2 1) in the hyperbolic plane: d(x; ) d(m; ) < 1.
1.5 Ideal boundaries
Suppose that X is a proper geodesic metric space. Introduce an equivalence relation
on the set of geodesic rays in X by declaring 0 i they are asymptotic i.e. are
within nite distance from each other. Given a geodesic ray we will denote by
(1) its equivalence class. Dene the ideal boundary of X as the collection @1X of
equivalence classes of geodesic rays in X . Our next goal is to topologize @1 X . Note
that the space of geodesic rays (parameterized by arc-length) in X has a natural
compact-open topology (we regard geodesic rays as maps from [0; 1) into X ). Thus
we topologize @1X by giving it the quotient topology .
We now restrict our attention to the case when X is -hyperbolic.
Then for each geodesic ray and a point p 2 X there exists a geodesic ray 0
with the initial point p such that (1) = 0 (1): Consider the sequence of geodesic
segments p(n) as n ! 1. Then the thin triangles property implies that these
segments are contained in a -neighborhood of [ p(0). Properness of X implies
that this sequence subconverges to a geodesic ray 0 as required.
Lemma 29. (Asymptotic rays are uniformly close). Let 1; 2 be asymptotic geodesic
rays in X such that 1 (0) = 2 (0) = p. Then for each t,
d(1(t); 2 (t)) 2:
Proof. Suppose that the raus 1 ; 2 are within distance C from each other. Take
T t. Then (since the rays are asymptotic) there is 2 R+ such that
d(1 (T ); 2( )) C:
By -thinness of the triangle (p1 (T )2( )), the point 1 (t) is within distance from a point either on p2 ( ) or on 1 (T )2( ). Since the length of 1 (T )2( ) is C
and T t, it follows that there exists t0 such that
d(1(t); 2 (t0)) :
By the triangle inequality, jt ? t0j . It follows that d(1(t); 2 (t)) 2.
Pick a base-point p 2 X . Given a number k > 2 dene a topology k on @1X
with the basis of neighborhoods of a point (1) given by
Uk;n() := f0 : d(0(t); (t)) < k; t 2 [0; n]g; n 2 R+
where the rays 0 satisfy 0 (0) = p = (0).
Lemma 30. Topologies and k coincide.
Proof. 1. Suppose that j is a sequence of rays emanating from p such that j 2=
Uk;n() for some n. If limj j = 0 then 0 2= Uk;n and by the previous lemma,
0 (1) 6= (1).
2. Conversely, if for each n, j 2 Uk;n() (provided that j is large enough), then the
sequence j subconverges to a ray 0 which belongs to each Uk;n(). Hence 0(1) =
Example 31. Suppose that X = H n is the hyperbolic n-space realized in the unit
ball model. Then the ideal boundary of X is S n?1.
Lemma 32. Let X be a proper geodesic Gromov-hyperbolic space. Then for each pair
of distinct points ; 2 @1X there exists a geodesic in X which is asymptotic to
both and .
Proof. Consider geodesic rays ; 0 emanating from the same point p 2 X and asymptotic to ; respectively. Since 6= , for each R < 1 the set
K (R) := fx 2 X : d(x; ) R; d(x; 0) Rg
is compact. Consider the sequences xn := (n); x0n := 0 (n) on ; 0 respectively.
Since the triangles pxn x0n are -thin, each segment n := xn x0n contains a point
within distance from both pxn; px0n, i.e. n \ K () 6= ;. Therefore the sequence of
geodesic segments n subconverges to a complete geodesic in X . Since N ([0 )
it follows that is asymptotic to and .
Denition 33. We say that a sequence xn 2 X converges to a point = (1) 2 @1X
in the cone topology if there is a constant C such that xn 2 NC () and the geodesic
segments x1 xn converge to a geodesic ray asymptotic to .
For instance, suppose that X = H m in the upper half-space model, = 0 2 R m?1 ,
L is the vertical geodesic from the origin. Then a sequence xn 2 X converges in
the cone topology i all the points xn belong to the Euclidean cone with the axis L
and the Euclidean distance from xn to 0 tends to zero. See Figure 5. This explains
the name cone topology.
Figure 5: Convergence in the cone topology.
Theorem 34. 1. Suppose that G is a hyperbolic group. Then @1G consists of 0, 2
or continuum of points.
2. The group G acts by homeomorphisms on @1 G as a uniform convergence
group, i.e. the action of G on Trip(@1G) is properly discontinuous and cocompact,
where Trip(@1G) consists of triples of distinct points in @1 G.
2 Coarse topology
The goal of this section is to provide tools of algebraic topology for studying quasiisometries and other concepts of the geometric group theory. The class of bounded
geometry metric cell complexes provides a class of spaces for which application of
algebraic topology is possible.
A metric space X has bounded geometry if there is a function (r) such that each
ball B (x; r) X contains at most (r) points. For instance, if G is a nitely generated
group with word metric then G has bounded geometry.
A metric cell complex is a cell complex X together with a metric.
A metric cell complex X 0 is said to have bounded geometry if:
(a) Each ball B (x; r) X intersects at most (r; k) cells of dimension k.
(b) Diameter of each k-cell is at most ck , k = 1; 2; 3; ::::.
Example 35. Let M be a compact simplicial complex. Metrize each simplex to
be isometric to the standard simplex with unit edges in the Euclidean space. Note
that for each m-simplex m and its face k , the inclusion k ! m is an isometric
embedding. This allows us to dene a path-metric on M so that each simplex is
isometrically embedded in M . Lift this metric to a cover X of M gives X structure
of a metric cell complex of bounded geometry.
Recall that quasi-isometries are not necessarily continuous. We therefore have to
approximate quasi-isometries by continuous maps.
Lemma 36. Suppose that X; Y are bounded geometry metric cell complexes, Y is
uniformly contractible, and f : X ! Y is a coarse (L; A)-Lipschitz map. Then there
exists a (continuous) cellular map g : X ! Y such that d(f; g) Const, where Const
depends only on (L; A) and the geometric bounds on X and Y .
Proof. The proof of this lemma is a prototype of most of the proofs presented in this
section. We construct g by induction on skeleta of X . First, of all, for each vertex
x 2 X (0) we let g(x) denote a point in Y (0) which is nearest to f (x). It is clear
that d(f (x); g(x)) const0 , where const0 is an upper bound on the diameter of the
top-dimensional cells in Y . Note that if x; x0 belong to the boundary of a 1-cell in
X then d(g(x); g(x0)) LConst1 + A + 2const0, where Const1 is an upper bound on
the diameter of 1-cells in X .
Inductively, assume that g was constructed on X (k) . Let denote a k + 1-cell
in X . Then, inductively, diam(g(@)) Ck and d(f; gjX (k)) Ck0 . Then, using
uniform contractibility of Y , we extend g to so that diam(g()) Ck0 +1. Then
d(f; gjX (k+1)) Ck0 +1 + LConstk + A. Since X is nite-dimensional the induction
terminates after nitely many steps.
2.1 Ends of spaces
In this section we review the (historically the rst) coarse topological notion. Let
X be a locally compact connected topological space (e.g. a proper geodesic metric
space). Given a compact subset K X we consider its complement K c. Then the
system of sets 0 (K c) is an inverse system:
K L ) 0(Lc) ! 0 (K c):
Then the set of ends (X ) is dened as the inverse limit
lim (K
K X 0
c ):
The elements of (X ) are called ends of X . Analogously, one can dene \higher homotopy groups" i1(X; x ) at innity of X by considering inverse systems of higher
homotopy groups: This requires a choice of a system of base-points xk 2 K c representing a single element of (X ). The inverse limit of this sequence of base-points,
x 2 (X ), serves as a \base-point" for the homotopy group i1(X; x ).
Here is a more down-to-earth description of the ends of X . Consider a nested
sequence of compacts Ki X; i 2 N (for instance, if X is a proper metric space take
KR := BR (p) for xed p 2 X ). For each i pick a connected component Ui Kic
so that Ui Ui+1 . Then the nested sequence (Ui ) represents a single point in (X ).
Even more concretely, pick a point xi 2 Ui for each i and connect xi ; xi+1 by a curve
i Ui . The concatenation of the curves i denes a proper map : [0; 1) ! X .
Call two proper curves ; 0 : R + ! X equivalent if for each compact K X there
are points x 2 (R + ); x0 2 0 (R + ) which belong to the same connected component of
K c. The equivalence classes of such curves are in bijective correspondence with the
ends of X , the map (Ui) 7! was described above.
See Figure 6 as an example. The space X in this picture has 5 visibly dierent
ends: 1 ; :::; 5. We have K1 K2 K3. The compact K1 separates the ends 1; 2 .
The next compact K2 separates 3 from 4. Finally, the compact K3 separates 4 from
5 .
Topology on (X ). Let 2 (X ) be represented by a nested sequence (Ui). Each Ui
denes a neighborhood Ni() of consisting of all 0 2 (X ) which are represented
by nested sequences (Uj0 ) such that Uj0 Ui for all but nitely many j 2 N .
Lemma 37. If f : X ! Y is an (L; A)-quasi-isometry of proper geodesic metric
spaces then f induces a homeomorphism (X ) ! (Y ).
Figure 6: Ends of X .
Proof. Note that for each bounded subset B Y the inverse image f ?1 (B ) is again
bounded. Although for a connected subset C X the preimage f (C ) is not necessarily connected, the R := L + A-neighborhood NR (f (C )) is connected. Thus we
dene a map f : (X ) ! (Y ) as follows. Suppose that 2 (X ) is represented by
a nested sequence (Ui). Without loss of generality we may assume that for each i,
NR (Ui) Ui?1 . Thus we get a nested sequence of connected subsets NR (f (Ui)) Y
each of which is contained in a connected component Vi of the complement to the
bounded subset f (Ki?1) Y . Thus we send to f() represented by (Vi). It follows
from the construction that By considering the quasi-inverse f to f it is clear that f
has inverse map (f) . It is also clear that both f and (f) are continuous.
If G is a nitely generated group then the space of ends (G) is dened to be the set
of ends of its Cayley graph. The previous lemma implies that (G) does not depend
on the choice of a nite generating set.
Theorem 38. Properties of (X ):
1. (X ) is compact, Hausdor and totally disconnected.
2. Suppose that G is a nitely-generated group. Then (G) consists of 0, 1, 2 points
or of continuum of points. In the latter case the set (G) is perfect: Each point is a
limit point.
3. (G) is empty i G is nite. (G) consists of 2-points i G is virtually (innite)
4. j(G)j > 1 i G splits nontrivially over a nite subgroup.
All the properties listed above are relatively trivial except for the last one: if
j(G)j > 1 then G splits nontrivially over a nite subgroup, which is a theorem of
Stallings [51]. For the proof of the rest see for instance [5, Theorem 8.32].
Corollary 39. 1. Suppose that G is quasi-isometric to Z then G contains Z as a
nite index subgroup.
2. Suppose that G splits nontrivially as A B and G0 is quasi-isometric to G. Then
G0 splits nontrivially as H F E (amalgamated product) or as H F (HNN splitting)
where F is a nite group.
Theorem 40. Suppose that G is a hyperbolic group. Then there exists a continuous
equivariant surjection
: @1G ! (G)
such that the preimages ?1 ( ) are connected components of @1 G.
2.2 Rips complexes and coarse connectedness
Let X be a metric space of bounded geometry, R 2 R + . Then the R-Rips complex
RipsR(X ) is the simplicial complex whose vertices are points of X ; vertices x1 ; :::; xn
span a simplex i d(xi; xj ) R for each i; j . Note that the system of Rips complexes
of X is a direct system Rips (X ) of simplicial complexes:
For each pair 0 r R < 1 we have a natural embedding r;R : Ripsr (X ) !
RipsR(X ) and r; = R; r;R provided that r R .
One can metrize RipsR (X ) by declaring each simplex to be isometric to a regular
Euclidean simplex with unit edges. Note that the assumption that X has bounded
geometry implies that RipsR(X ) is nite-dimensional for each R. Moreover, RipsR(X )
is a metric cell complex of bounded geometry.
The following simple observation explains why Rips complexes are useful for analyzing quasi-isometries:
Lemma 41. Let f : X ! Y be an L-Lipschitz map. Then f induces a (continuous)
simplicial map Ripsd (X ) ! RipsLd(Y ) for each d 0.
Proof. Consider an (m ? 1)-simplex in Ripsd (X ), the vertices of are points
x1 ; :::; xm within distance R from each other. Since f is L-Lipschitz, the points
f (x1 ); :::; f (xm) are within distance LR from each other, hence they span a simplex
0 of dimension m ? 1 in RipsLd (Y ). The map f sends vertices of to vertices of
0, extend this map linearly to the simplex . It is clear that this extension denes a
(continuous) simplicial map of simplicial complexes Ripsd(X ) ! RipsLd (Y ).
Denition 42. A metric space X is coarsely k-connected if for each r there exists
R r so that the mapping Ripsr (X ) ! RipsR(X ) induces a trivial map of i for
0 i k.
For instance, X is coarsely 0-connected if there exists a number R such that each
pair of points x; y 2 X can be connected by an R-chain of points xi 2 X , i.e. a chain
of points where d(xi; xi+1 ) R for each i. Note that for k 1 coarse k-connectedness
of X is equivalent to the property that RipsR(X ) is k-connected for suciently large
Properties of the direct system of Rips complexes:
Lemma 43. Let r; C < 1, then each simplicial spherical cycle of diameter C
in Ripsr bounds a disk of diameter C + d within Ripsr+C .
Proof. Pick a point x 2 . Then Ripsr+C contains a simplicial cone () over with
the origin at x. Clearly ( ) r + C .
Corollary 44. Let
f; g : Ripsd1 (X ) ! Ripsd2 (Y )
be L-Lipschitz within distance C from each other. Then there exists d3 d2 such
that the maps f; g : Ripsd1 ! Ripsd3 (Y ) are homotopic via a homotopy whose tracks
have lengths C 0 = C 0 (C; d1 ; d2; L).
Proof. Construct the homotopy via induction on skeleta using the previous lemma.
We will refer to the maps f; g above as being coarsely homotopic. In the same
way one denes coarse homotopy equivalence between the direct systems of Rips
Corollary 45. Suppose that f; g : X ! Y be L-Lipschitz maps within nite distance
from each other. Then they induce coarsely homotopic maps Ripsd (X ) ! RipsLd (Y )
for each d 0.
Corollary 46. if f : X ! Y is a quasi-isometry, then f induces a coarse homotopyequivalence of the Rips complexes: Rips (X ) ! Rips (Y ).
Corollary 47. Coarse k-connectedness is a QI invariant.
Proof. Suppose that X 0 is coarsely k-connected and f : X ! X 0 is an L-Lipschitz
quasi-isometry with L-Lipschitz quasi-inverse f : X 0 ! X . Let be a spherical
i-cycle in Ripsd (X ), 0 i k. Then we have the induced spherical i-cycle f ( ) RipsLd(X 0). Since X 0 is coarsely k-connected, there exists d0 Ld such that f ( )
bounds a singular i + 1-disk within Ripsd (X 0). Consider now f( ) RipsL2 d(X ).
The boundary of this singular disk is a singular i-sphere f( ). Since f f is homotopic
to id within Ripsd (X ), d00 L2 d, there exists a singular cylinder in Ripsd (X )
which cobounds and f( ). Note that d00 does not depend on . By combining and f( ) we get a singular i + 1-disk in Ripsd (X ) whose boundary is . Hence X is
coarsely k-connected.
Our next goal is to nd a large supply of examples of metric spaces which are
coarsely k-connected.
Denition 48. A bounded geometry metric cell complex X is said to be uniformly
k-connected if there is a function (k; r) such that for each i k, each singular
i-sphere of diameter r in X (i+1) bounds a singular i + 1-disk of diameter (k; r).
For instance, if X is a nite-dimensional contractible complex which admits a cocompact cellular group action, then X is uniformly k-connected for each k.
Here is an example of a simply-connected complex which is not uniformly simplyconnected. Take S 1 R + with the product metric and attach to this complex a 2-disk
along the circle S 1 f0g.
Theorem 49. Suppose that X is a metric cell complex of bounded geometry such that
X is uniformly n-connected. Then Z := X (0) is coarsely n-connected.
Proof. Let : S k ! RipsR (Z ) be a spherical m-cycle in RipsR (Z ), 0 k n.
Without loss of generality (using simplicial approximation) we can assume that is a
simplicial cycle, i.e. the sphere S k is given a triangulation so that sends simplices
of S k to simplices in RipsR(Z ) so that the restriction of to each simplex is a linear
map. Let 1 be a k-simplex in S k . Then (1) is spanned by points x1 ; :::; xk+1 2 Z
which are within distance R from each other. Since X is uniformly k-connected,
there is a singular k-disk 1(1) containing x1 ; :::; xk+1 and having diameter R0,
where R0 depends only on R. Namely, we construct 1 by induction on skeleta: First
connect each pair of points xi ; xj by a path in X (of length bounded in terms of R),
this denes the map 1 on the 1-skeleton of 1 . Then continue inductively. This
construction ensures that if 2 is a k-simplex in S k which shares an m-face with 1
then 2 and 1 agree on 1 \ 2 . As the result, we have \approximated" by a
singular spherical k-cycle 0 : S k ! X (k) (the restriction of 0 to each i equals i).
See gure 7 in the case k = 1.
γ ’ (Dk+1)
γ(S )
γ ’ (∆)
γ ’ (S k )
Figure 7:
Since X is k-connected, the map 0 extends to a cellular map 0 : Dk+1 ! X (k+1) .
Let D denote the maximal diameter of a k + 1-cell in X . For each simplex Dk+1
the diameter of 0() is at most D. We therefore can \push" the singular disk 0 (Dk+1)
into RipsD (Z ) by replacing each linear map 0 : ! 0() X with the linear map
00 : ! 00 () RipsD (Z ) where 00 () is the simplex spanned by the vertices of
0(). This yields a map 00 : Dk+1 ! RipsD (Z ). Observe that the map 00 is a
cellular map with respect to a subdivision 0 of the initial triangulation of S k .
Note however that and 00jS k are dierent maps. Let V denote the vertices of a
k-simplex S k ; let V 00 denote the set of vertices of 0 within the simplex . Then
the diameter of 00 (V 00 ) is at most R0. Hence (V ) 00 (V ) is contained in a simplex
in RipsR+R (Z ). Therefore, by taking = R + D + R0 we conclude that the maps
; 00 : S k ! Rips(Z ) are homotopic. See Figure 8. Thus the map is nil-homotopic
within Rips(Z ).
Corollary 50. Suppose that G is a nitely-presented group with the word metric.
Then G is coarsely simply-connected.
Corollary 51. (See for instance [5, Proposition 8.24]) Finite presentability is a QI
γ(S )
γ " (S )
Homotopy between γ and γ " .
Figure 8:
Proof. It remains to show that each coarsely 1-connected group G is nitely presentable. The Rips complex X := RipsR(G) is 1-connected for large R. The group
G acts on X properly discontinuously and cocompactly. Therefore G is nitely presentable.
Denition 52. A group G is said to be of type Fn (n 1) if its admits a cellular
action on a cell complex X such that for each k n: (1) X (k+1) =G is compact. (2)
X (k+1) is k-connected. (3) The action G y X is free.
Example 53. (See [3].) Let F 2 be free group on 2 generators a; b. Consider the group
G = F n2 which is the direct product of F 2 with itself n times. Dene a homomorphism
: G ! Z which sends each generator ai; bi of G to the same generator of Z. Let
K := Ker(). Then K is of type Fn?1 but not of type Fn.
Thus, analogously to Corollary 51 we get:
Theorem 54. (See [29, 1.C2]) Type Fn is a QI invariant.
Proof. It remains to show that each coarsely n-connected group has type Fn . The
proof below follows [34]. We build the complex X on which G would act as required
by the denition of type Fn. We build this complex and the action by induction on
(0). X (1) , is a Cayley graph of G; the action of G is cocompact, free, cellular.
(i) i+1). Suppose that X (i) has been constructed. Using i-connectedness of
Rips(G) we construct (by induction on skeleta) a G-equivariant cellular map f :
X (i) ! RipsD (G) for a suciently large D. If G were torsion-free, the action
G y RipsD (G) is free; this allows one to we construct (by induction on skeleta)
a G-equivariant \retraction" : RipsD (G)(i) ! X (i), i.e. a map such that the composition f is G-equivariantly homotopic to the identity.
However, if G contains nontrivial elements of nite order, we have to use a more
complicated construction.
Suppose that 2 i n and an i ? 1-connected complex X (i) together with a free
discrete cocompact action G y X (i) was constructed. Let x0 2 X (0) be a base-point.
Lemma 55. There are nitely many spherical i-cycles 1 ; :::; k in X (i) such that
their G-orbits normally generate 1 (X (i) ), in the sense that the normal closure of the
cycles fg^j : j = 1; :::; k; g 2 Gg is i (X (i) ), where each ^j is obtained from j by
attaching a \tail" from x0 .
Proof. Without loss of generality we can assume that X (i) is a (metric) simplicial
complex. Let f : X (i) ! Y := RipsD (Z ) be a G-equivariant continuous map as
Here is the construction of j 's:
Let : S i ! Y (i) ; 2 N , denote the attaching maps of the i + 1-cells in Y ,
these maps are just simplicial homeomorphic embeddings from the boundary S i of
the standard i +1-simplex into Y (i) . Starting with a G-equivariant projection Y (0) !
X (0) one inductively constructs a (non-equivariant!) map f : Y (i) ! X (i) so that
f f : Y (i) ! Y (i+1) is within distance Const from the identity. Hence (by
coarse connectedness of Z ) this composition is homotopic to the identity inclusion
within RipsD (Z ). The homotopy H is such that its tracks have \uniformly bounded
complexity", i.e. the compositions
H ( id) : S i I ! RipsD (Z )
are simplicial maps with a uniform upper bound on the number of simplices in a
triangulation of S i I . Let B X (i) denote a compact subset such that GB = X (i) .
We let j denote the composition g f where g 2 G are chosen so that the
image of j intersects B .
We now equivariantly attach i+1-cells along G-orbits of the cycles j : for each j and
g 2 G we attach an i + 1-cell along g(j ). Note that if j is stabilized by a subgroup
of order m = m(j ) in G, then we attach m copies of the i + 1-dimensional cell along
j . We let X (i+1) denote the resulting complex and we extend the G-action to X (i+1)
in obvious fashion. It is clear that G y X (i+1) is free, discrete and cocompact.
2.3 Coarse separation
Suppose that X is a metric cell complex and Y X is a subset. We let NR (Y )
denote the metric R-neighborhood of Y in X . Let C be a complementary component
of NR(Y ) in Y . Dene the inradius, inrad(C ), of C to be the supremum of radii of
metric balls in X contained in C . A component C is called shallow if inrad(C ) is
< 1 and deep if inrad(C ) = 1.
Example 56. Suppose that Y is compact. Then deep complementary components
of X n NR (Y ) are those components which have innite diameter.
A subcomplex Y is said to coarsely separate X if there is R such that NR(Y ) has
at least two distinct deep complementary components.
Example 57. The curve ? in R 2 does not coarsely separate R2 . A straight line in
R 2 coarsely separates R 2 .
Theorem 58. Suppose that Y; X be uniformly contractible metric cell complexes of
bounded geometry which are homeomorphic to R n?1 and R n respectively. Then for
each uniformly proper map f : Y ! X , the image f (Y ) coarsely separates X . Moreover, the number of deep complementary components is 2.
Proof. Actually, our proof will use the assumption on the topology of Y only weakly:
to get coarse separation it suces to assume that Hcn?1(Y; R) 6= 0.
Let W := f (Y ). Given R 2 R + we dene a retraction : NR(W ) ! Y , so that
d( f; idY ) const, where const depends only on the distortion function of f and on
the geometry of X and Y . Here NR(W ) is the smallest subcomplex in X containing
the R-neighborhood of W in X . We dene by induction on skeleta of NR(W ).
For each vertex x 2 NR(W ) we pick a vertex (x) := y 2 Y such that the distance
d(x; f (y)) is the smallest possible. If there are several such points y, we pick one of
them arbitrarily. The fact that f is a uniform proper embedding ensures that
d( f; idY 0 ) const0:
Note also that for any 1-cell in NR (W ), diam((@)) Const0 . Suppose that we
have constructed on NR(k)(W ). Inductively we assume that:
d( f; idY k ) constk ; diam((@)) Constk ;
for each k +1-cell . We extend to the k +1-skeleton by using uniform contractibility
of Y : For each k + 1-cell there exists a singular disk : Dk+1 ! Y in Y k+1 of
diameter (Constk ) whose boundary is (@). Then we extend to via . It is
clear that the extension satises the inequalities (59) with k replaced with k + 1.
Since Y is uniformly contractible we get a homotopy f = idY , whose tracks are
uniformly bounded (construct it by induction on skeleta the same way as before).
Recall that we have a system of isomorphisms
P : Hcn?1(Nr ) = H1(X; X n Nr )
given by the Poincare duality in R n . This isomorphism moves support sets of n ? 1cocycles by a uniformly bounded amount (to support sets of 1-cycles). Let ! be
a generator of Hcn?1(Y ). Given R > 0 consider \retraction" as above and the
pull-back !R := (!). If for some 0 < r < R the restriction !r of !R to Nr (W )
is zero then we get a contradiction, since f = id on the compactly supported
cohomology of Y . Thus !r is nontrivial. Applying the Poincare duality operator P
to the cohomology class !r we get a nontrivial relative homology class
P (!r ) 2 H1(X; X n Nr ) = H~ 0(X n Nr ):
We note that for each R r the class P (!r ) 2 H1(Nr ; @ Nr ) is represented by \restriction" of the class P (!R) 2 H1(NR ; @ NR ) to Nr , see Figure 9. In particular,
the images r ; R of P (!r ); P (!R) in H~ 0(X n Nr ), H~ 0(X n NR) are homologous in
H~ 0(X n Nr ). Moreover, R restricts nontrivially to 1 2 H~ 0(X n N1).
Therefore, we get sequences of points
xi ; x0i 2 @ Ni ; i 2 N ;
such that xi ; x0i belong to the support sets of i for each i, xi ; xi+1 belong to the
same component of X n Ni, x0i ; x0i+1 belong to the same component of X n Ni, but
x i+1
N i (W)
x ’i
N (W)
x ’i+1
Figure 9: Coarse separation.
the points xi; x0i belong to distinct components C; C 0 of X n N1. It follows that C; C 0
are distinct deep complementary components of W . The same argument run in the
reverse implies that there are exactly two deep complementary components (although
we will not use this fact).
I refer to [20], [33] for further discussion and generalization of coarse separation and
coarse Poincare/Alexander duality.
2.4 Other notions of coarse equivalence
Theorem 60. (Gromov, [29], see also de la Harpe [12, page 98]) Groups G and ? are
QI i they admit commuting (i.e. extending to an action of G ?) proper cocompact
topological actions on a locally compact topological space Y .
Proof. 1. Suppose that there exists an (L; A)-quasi-isometry G ! ?. Consider the
collection F of all (L; A)-quasi-isometries f from G to ?, given the compact-open
topology. By Arcela-Ascoli, the space F is locally compact. The groups G and ? act
on F by left and right multiplication:
g(f )(x) = f (g?1(x)); g 2 G;
g(f )(x) = f (x); 2 ?:
It is clear that these are commuting topological actions. Since both G; ? act on
themselves properly, both actions G; ? y F are proper. Let fj 2 F , then, since the
action of ? on itself is transitive, there exists a sequence j 2 ? such that j fj (1) = 1.
Hence, by Arcela-Ascoli theorem, the action ? y F is cocompact. (So far, everything
works if instead of QI mappings we use QI embeddings). On the other hand, since
for each fj the image fj (G) is A-dense in ?, for each j there exists xj 2 G such that
d(fj (xj ); 1) A. Hence the sequence (x?j 1 )fj is also relatively compact in F . Hence
both actions G; ? y F are cocompact.
2. Suppose that G; ? y Y are commuting actions. Pick a compact K Y which
maps onto both Y=G; Y=?. Choose a point k 2 K and consider the mapping f : G !
? which sends g 2 G to an element ?1 2 ? such that g(k) 2 (K ). I claim that f is
a quasi-isometry. Let's rst check that f is Lipschitz. Let S = fs1 ; :::; smg be a nite
generating set of G. It suces to check that f distorts each edge of the corresponding
Cayley graph by a uniformly bounded amount. Pick g 2 G; ?1 := f (g).
Since S is nite, Kb := [s2S K is compact, hence there exists a nite subset ?
such that
Kb Ke := [2 (K ):
In addition dene a nite set
0 := f 2 ? : (K ) \ Ke 6= ;g
Set L := maxfd?(; 1); 2 0 g.
Recall that the group operation on G is dened so that hg = gh. Thus d(si g; g) =
1 for each si 2 S . We have:
si g(k) = si (y) = si (y) 2 (K ); for some y 2 K; 2 :
Observe that 0 := [f (gsi)]?1 also satises si g(k) 2 0(K ). Hence ?1 0(K ) \
(K ) 6= ;, i.e 0 ?1 2 0 . Therefore d?( 0 ?1 ; 1) L and hence
d?( ?1; 0?1) L; d?(f (g); f (gsi)) L:
This proves that f is L-Lipschitz. Construct a map f : ? ! G in the similar fashion:
f( ) := g?1; (k) 2 g(K );
the same arguments as above show that f is L0-Lipschitz for some L0 < 1.
Suppose that f (g) = ?1; f( ?1) = h. Then
(k) 2 h?1 (K ) () h(k) 2 ?1(K );
(since the actions of G and ? commute). Thus d(f f; id) Const, d(f f; id) Const for some nite constant.
Denition 61. Groups G1; G2 are said to have a common geometric model if
there exists a proper geodesic metric space X such that Gi; G2 both act isometrically,
properly discontinuously, cocompactly on X .
In view of Lemma 16, if groups have a common geometric model then they are
quasi-isometric. The following theorem shows that the converse is false:
Theorem 62. (Mosher, Sageev, Whyte, [43]) Let G1 := Zp Zp; G2 := Zq Zq, where
p; q are distinct primes. Then the groups G1 ; G2 do not have a common geometric
This theorem in particular implies that in Theorem 60 one cannot assume that
both group actions are isometric.
Spaces (or nitely generated groups) X1; X2 are bilipschitz equivalent if there exists
a bilipschitz bijection f : X1 ! X2.
Theorem 63. (Whyte, [59]) Suppose that G1; G2 are non-amenable nitely generated
groups which are quasi-isometric. Then G1 ; G2 are bilipschitz equivalent.
On the other hand, there are examples (Burago, Kleiner, McMullen, [7, 40]) of separated nets in R 2 which are not bi-Lipschitz homeomorphic. I am unaware of examples
of amenable grooups which are quasi-isometric but are not bilipschitz equivalent.
3 Ultralimits of Metric Spaces
Let (Xi) be a sequence of metric spaces. One can describe the limiting behavior of the
sequence (Xi) by studying limits of sequences of nite subsets Yi Xi. Ultralters
are an ecient technical device for simultaneously taking limits of all such sequences
of subspaces and putting them together to form one object, namely an ultralimit of
3.1 Ultralters
Let I be an innite set, S is a collection of subsets of I . A lter based on S is a
nonempty family ! of members of S with the properties:
; 62 !.
If A 2 ! and A B , then B 2 !.
If A1 ; : : : ; An 2 !, then A1 \ \ An 2 !.
If S consists of all subsets of I we will say that ! is a lter on I . Subsets A I which
belong to a lter ! are called !-large. We say that a property (P) holds for !-all i,
if (P) is satised for all i in some !-large set. An ultralter is a maximal lter. The
maximality condition can be rephrased as: For every decomposition I = A1 [ [ An
of I into nitely many disjoint subsets, the ultralter contains exactly one of these
For example, for every i 2 I , we have the principal ultralter i dened as i :=
fA I j i 2 Ag. An ultralter is principal if and only if it contains a nite subset.
The interesting ultralters are of course the non-principal ones. They cannot be described explicitly but exist by Zorn's lemma: Every lter is contained in an ultralter.
Let Z be the Zariski lter which consists of complements to nite subsets in I . An
ultralter is a nonprincipal ultralter, if and only if it contains Z .
Here is an alternative interpretation of ultralters. An ultralter is a nitely additive measure dened on all subsets of I so that each subset has measure 0 or 1. An
ultralter is nonprincipal i the measure contains no atoms: The measure of each
point is zero.
Given a function f : I ! Y (where Y is a topological space) dene the !-limit
f (i)
to be a point y 2 Y such that for every neighborhood U of y the preimage f ?1U
belongs to !.
Lemma 64. Suppose that Y is compact and Hausdor. Then for each function
f : I ! Y the ultralimit exists and is unique.
Proof. To prove existence of a limit, assume that there is no point y 2 Y satisfying
the denition of the ultralimit. Then each point z 2 Y possesses a neighborhood
Uz such that f ?1Uz 62 !. By compactness, we can cover Y with nitely many of
these neighborhoods. It follows that I 62 !. This contradicts the denition of a lter.
Uniqueness of the point y follows, because Y is Hausdor.
Note that if y is an accumulation point of ff (i)gi2I then there is a non-principal
ultralter ! with !-lim f = y, namely an ultralter containing the pullback of the
neighborhood basis of y.
3.2 Ultralimits of metric spaces
Let (Xi)i2I be a family of metric spaces parameterized by an innite set I . For an
ultralter ! on I we dene the ultralimit
X! = !-lim
as follows. Let i Xi be the product of the spaces Xi, i.e. it is the space of sequences
(xi )i2I with xi 2 Xi. The distance between two points (xi); (yi) 2 i Xi is given by
d! (xi ); (yi) := !-lim i 7! dXi (xi ; yi)
where we take the ultralimit of the function i 7! dXi (xi ; yi) with values in the compact
set [0; 1]. The function d! is a pseudo-distance on iXi with values in [0; 1]. Set
(X! ; d! ) := (iXi; d! )= where we identify points with zero d! -distance.
Exercise 65. Let Xi = Y for all i, where Y is a compact metric space. Then X! = Y
for all ultralters !.
If the spaces Xi do not have uniformly bounded diameter, then the ultralimit X!
decomposes into (generically uncountably many) components consisting of points of
mutually nite distance. We can pick out one of these components if the spaces Xi
have base-points x0i . The sequence (x0i )i denes a base-point x0! in X! and we set
X!0 := x! 2 X! j d! (x! ; x0! ) < 1 :
Dene the based ultralimit as
0 ) := (X 0 ; x0 ):
! !
Example 66. For every locally compact space Y with a base-point y0, we have:
!-lim(Y; y0) = (Y; y0):
Lemma 67. Let (Xi)i2N be a sequence of geodesic i-hyperbolic spaces with i tending
to 0. Then for every non-principal ultralter ! each component of the ultralimit X!
is a metric tree.
Proof. We rst verify that between any pair of points x! ; y! 2 X! there is a unique
geodesic segment. Let ! denote the ultralimit of the geodesic segments i := xi yi Xi; it connects the points x! ; y! . Suppose that is another geodesic segment connecting x! to y! . Pick a point p! 2 . Then
1 [d(x ; p ) + d(y ; p ) ? d(x ; y )] = 0:
(xi; yi)pi = !-lim
i i
i i
i i
i 2
Since, by Lemma 24,
d(pi; i) (xi ; yi)pi + 2i ;
d(p! ! ) = 0:
Now, suppose that (x! y! z! ) is a geodesic triangle in X! . By uniqueness of geodesics
in X! , this triangle appears as ultralimit of the i -thin triangles (xi yizi). It follows
that (x! y! z! ) is zero-thin, i.e. each component of X! is zero-hyperbolic.
Exercise 68. If T is a metric tree, ?1 < a < b < 1 and f : [a; b] ! T is a
continuous embedding then the image of f is a geodesic segment in T . (Hint: use
PL approximation of f to show that the image of f contains the geodesic segment
connecting f (a) to f (b).)
Lemma 69. (Morse Lemma) Let X be a {hyperbolic geodesic space, k; c be positive
constants, then there is a function = (k; c) such that for any (k; c)-quasi-isometric
embedding f : [a; b] ! X the Hausdor distance between the image of f and the
geodesic segment [f (a)f (b)] X is at most .
Proof. Suppose that the assertion of lemma is false. Then there exists a sequence
of (k; c)-quasi-isometric embeddings fn : [?n; n] ! Xn to CAT (?1)-spaces Xn such
lim d (f ([?n; n]); [f (?n); f (n)]) = 1
n!1 Haus
where dHaus is the Hausdor distance in Xn.
Let dn := dHaus(f ([?n; n]); [f (?n); f (n)]). Pick points tn 2 [?n; n] such that
jd(tn; [f (?n); f (n)]) ? dnj 1. Consider the sequence of pointed metric spaces
( d1n Xn; fn(tn)), ( d1n [?n; n]; tn ). It is clear that !-lim n=dn > 1=k > 0 (but this
ultralimit could be innite). Let (X! ; x! ) = !-lim( d1n Xn; fn(tn)) and (Y; y) :=
!-lim( d1n [?n; n]; tn ). The metric space Y is either a nondegenerate segment in R
or a closed geodesic ray in R or the whole real line. Note that the Hausdor distance
between the image of fn in d1n Xn and [fn(?n); fn (n)] d1n Xn is at most 1 + 1=dn.
Each map
fn : d1 [?n; n] ! d1 Xn
is a (k; c=n)-quasi-isometric embedding. Therefore the ultralimit
f! = !-lim fn : (Y; y) ! (X! ; x! )
is a (k; 0)-quasi-isometric embedding, i.e. it is a k-bilipschitz map:
jt ? t0j=k d(f! (t); f! (t0)) kjt ? t0j:
In particular this map is a continuous embedding. On the other hand, the sequence
of geodesic segments [fn(?n); fn(n)] d1n Xn also !-converges to a nondegenerate
geodesic X! , this geodesic is either a nite geodesic segment or a geodesic ray
or a complete geodesic. In any case the Hausdor distance between the image L of
f! and is exactly 1, it equals the distance between x! and which is realized as
d(x! ; z) = 1, z 2 . I will consider the case when is a complete geodesic, the other
two cases are similar and are left to the reader. Then Y = R and by Exercise 68
the image L of the map f! is a complete geodesic in X! which is within Hausdor
distance 1 from the complete geodesic . This contradicts the fact that X! is a metric
Historical Remark. Morse [42] proved a special case of this lemma in the case
of H 2 where the quasi-geodesics in question where geodesics in another Riemannian
metric on H 2 , which admits a cocompact group of isometries. Busemann, [9], proved a
version of this lemma in the case of H n , where metrics in question were not necessarily
Riemannian. A version in terms of quasi-geodesics is due to Mostow [44], in the
context of negatively curved symmetric spaces, although his proof is general.
Corollary 70. Suppose that X; X 0 are quasi-isometric geodesic metric spaces and X
is Gromov-hyperbolic. Then X 0 is also Gromov-hyperbolic.
Proof. Let f : X 0 ! X be a (L; A)-quasi-isometry. Pick a geodesic triangle ABC X 0. Its image is a quasi-geodesic triangle whose sides are (L; A)-quasi-geodesic.
Therefore each of the quasi-geodesic sides of f (ABC ) is within distance c =
c(L; A) from a geodesic connecting the end-points of this side. See Figure 10. The
geodesic triangle f (A)f (B )f (C ) is -thin, it follows that the quasi-geodesic triangle
f (ABC ) is (2c + )-thin. Thus the triangle ABC is L(2c + ) + A-thin.
Here is another example of application of asymptotic cones to study quasi-isometries.
Quasi-geodesic triangle
Figure 10: Image of a geodesic triangle.
Lemma 71. Suppose that X = R n or R + , f : X ! X is an (L; A)-quasi-isometric
embedding. Then NC (f (X )) = X , where C = C (L; A).
Proof. I will give a proof in the case of R n as the other case is analogous. Suppose that
the assertion is false, i.e. there is a sequence of (L; A)-quasi-isometries fj : R n ! R n ,
sequence of real numbers rj diverging to innity and points yj 2 R n n Image(f ) such
that d(yj ; Image(f )) = rj . Let xj 2 R n be a point such that d(f (xj ); yj ) rj + 1.
Using xj ; yj as basi-points on the domain and target to fj rescale the metrics on the
domain and the target by 1=rj and take the corresponding ultralimits. In the limit
we get a bi-Lipschitz embedding
f! : R n ! R n ;
whose image misses the point y! 2 R n . However each bilipshitz embedding is necessarily proper, therefore by the invariance of domain theorem the image of f! is both
closed and open. Contradiction.
Remark 72. Alternatively, one can prove the above lemma as follows: Approximate
f by a continuous mapping g. Then, since g is proper, it has to be onto.
3.3 The asymptotic cone of a metric space
Let X be a metric space and ! be a non-principal ultralter on N . The asymptotic
cone Cone! (X ) of X is dened as the based ultralimit of rescaled copies of X :
1 X; x0 ):
Cone! (X ) := X!0 ;
where (X!0 ; x0! ) = !-lim
The limit is independent of the chosen base-point x0 2 X . The discussion in the
previous section implies:
Proposition 73. 1. Cone! (X Y ) = Cone! (X ) Cone! (Y ).
2. Cone! R n = Rn .
3. The asymptotic cone of a geodesic space is a geodesic space.
4. The asymptotic cone of a CAT(0)-space is CAT(0).
5. The asymptotic cone of a space with a negative upper curvature bound is a metric
Remark 74. Suppose that X admits a cocompact discrete action by a group G of
isometries. The problem of dependence of the topological type of Cone! X on the
ultralter ! was open until recently counterexamples were constructed in [52], [15].
However in the both examples the group G is not nitely presentable. Moreover, if a
nitely-repsentable group has an asymptotic cone which is a tree, then the group is
hyperbolic and hence each asymptotic cone is a tree, see [34].
To get an idea of the size of the asymptotic cone, note that in the most interesting cases it is homogeneous. We call a metric space X quasi-homogeneous if
diam(X=Isom(X )) is nite.
Proposition 75. Let X be a quasi-homogeneous metric space. Then for every nonprincipal ultralter ! the cone Cone! (X ) is a homogeneous metric space.
Proof. The group of sequences of isometries Isom(X )N acts transitively on the ultralimit
1 X)
which contains Cone! (X ) as a component.
Lemma 76. Let X be a quasi-homogeneous {hyperbolic space with uncountable number of ideal boundary points. Then for every nonprincipal ultralter ! the asymptotic
cone Cone! (X ) is a tree with uncountable branching.
Proof. Let x0 2 X be a base-point and y; z 2 @1 X . Denote by the geodesic in X
with the ideal endpoints z; y. Then Cone! ([x0 ; y)) and Cone! ([x0 ; z)) are geodesic rays
in Cone! (X ) emanating from x0! . Their union is equal to the geodesic Cone! . This
produces uncountably many rays in Cone! (X ) so that any two of them have precisely
the base-point in common. The homogeneity of Cone! (X ) implies the assertion.
3.4 Extension of quasi-isometries of hyperbolic spaces to the
ideal boundary
Lemma 77. Suppose that X is a proper -hyperbolic geodesic space. Let Q X be a
(L; A)-quasigeodesic ray or a complete (L; A)-quasigeodesic. Then there is Q which
is either a geodesic ray (or a complete geodesic) in X so that the Hausdor distance
between Q and Q is C (L; A; ).
Proof. I will consider only the case of quasigeodesic rays : [0; 1) ! Q X as
the other case is similar. Consider the sequence of geodesic segments i = (0)(i).
By Morse lemma, each i is contained within Nc(Q), where c = c(L; A; ). By local
compactness, the geodesic segments i subconverge to a complete geodesic ray Q =
(R + ) which is contained in Nc(Q).
It remains to show that Q is contained in ND (Q ), where D = D(L; A; ). Consider
the nearest-point projection p : Q ! Q. This projection is clearly a quasi-isometric
embedding with the constants depending only on L; A; . Lemma 71 shows that the
image of p is -dense in Q with = (L; A; ). Hence each point of Q is within distance
D = + c from a point of Q.
Observe that this lemma implies that for any divergent sequence tj 2 R + , the
sequence of points (tj ) on a quasi-geodesic ray in X , converges to a point 2 @1X ,
= (1). Indeed, if ; 0 are geodesic rays Hausdor-close to Q then ; 0 are
Hausdor-close to each other as well, therefore (1) = 0 (1).
We will refer to the point as (1). Note that if 0 is another quasi-geodesic ray
which is Hausdor-close to then (1) = 0 (1).
Theorem 78. Suppose that X and X 0 are Gromov-hyperbolic proper geodesic metric
spaces. Let f : X ! X 0 be a quasi-isometry. Then f admits a homeomorphic extension f1 : @1X ! @1X 0. This extension is such that the map f [ f1 is continuous
at each point 2 @1 X .
Proof. First, we construct the extension f1 . Let 2 @1 X , = (1) where is a
geodesic ray in X . The image of this ray 0 := f : R + ! X 0 is a quasi-geodesic
ray, hence we set f1() := 0(1). Observe that f1() does not depend on the choice
of a geodesic ray asymptotic to . Let f be quasi-inverse of f . It is clear from the
construction that (f)1 is inverse to f1. It remains therefore to verify continuity.
Suppose that xn 2 X is a sequence which converges to in the cone topology,
d(xn; ) c. Then d(f (xn); 0) Lc + A and d(f (xn); (0)) C (Lc + A), where
(0 ) is a geodesic ray in X 0 asymptotic to 0(). Thus f (xn ) converges to f1() in
the cone topology.
Finally, let n 2 @1X be a sequence which converges to . Let n be a sequence of
geodesic rays asymptotic to n with n (0) = (0) = x0 . Then, for each T 2 R + there
exists n0 such that for all n n0 and t 2 [0; T ] we have
d((t); n(t)) 2;
where is the hyperbolicity constant of X . Hence
d(f (n(t)); (t)) 2L + A:
Set 0n := f n . Then
(0n)([0; L?1 T ? A]) NC ((0 )([0; LT + A]));
for all n n0. Thus the geodesic rays (0n ) converge to a ray within nite distance
from (0 ). It follows that the sequence f1(n) converges to f1().
Lemma 79. Let X and X 0 be proper geodesic -hyperbolic spaces. In addition we
assume that X is quasi-homogeneous and that @1 X consists of at least four points.
Suppose that f; g : X ! X 0 are (L; A)-quasi-isometries such that f1 = g1. Then
d(f; g) D, where D depends only on L; A; and the geometry of X .
Proof. Let 1 ; 2 be complete geodesics in X which are asymptotic to the points 1 ; 1 ,
2; 2 respectively, where all the points 1; 1, 2; 2 are distinct. There is a point y 2 X
which is within distance r from both geodesics 1; 2. Let G be a group acting
isometrically on X so that the GB = X for an R-ball B in X . Pick a point x 2 X :
Our goal is to estimate d(gf (x); g(x)). By applying an element of G to x we can
assume that d(x; y) R, in particular, d(x; 1) R + r; d(x; 2) R + r. Thus the
distance from f (x) to the quasi-geodesics f (1); f (2) is at most L(R + r) + A. We
now apply the quasi-inverse g the to quasi-isometry g: gf (i) is an (L2; LA + A)quasi-geodesic in X ; since f1 = g1, these quasi-geodesics are asymptotic to the
points i; i, i = 1; 2. Since the Hausdor distance from gf (i) to i is at most C +2
(where C = C (L2 ; LA + A; ) is the constant from Lemma 77) we conclude that
d(gf (x); i) C 0 := C + 2:
See Figure 11.
Since the geodesics 1; 2 are asymptotic to distinct points in @1X , it follows that
the diameter of the set fz 2 X : d(z; i) max(C 0; r + R); i = 1; 2g is at most C 00,
g f(x)
g f(γ )
g f(γ 1)
Figure 11:
where C 00 depends only on the geometry of X and the xed pair of geodesics 1; 2.
Hence d(gf (x); x) C 00. By applying g to this formula we get:
d(g(x); ggf (x)) L(C 00 + A) + A;
d(f (x); ggf (x)) A:
d(f (x); g(x)) 2A + L(C 00 + A):
Remark 80. The line X = R is 0-hyperbolic, its ideal boundary consists of 2 points.
Take a translation f : X ! X , f (x) = x + a. Then f1 is the identity map of
f?1; 1g but there is no bound on the distance from f to the identity.
4 Tits alternative
Theorem 81. \Tits alternative" (Tits, [53]) Let L be a Lie group with nitely many
components and ? L be a nitely generated subgroup. Then either ? is virtually
solvable or ? contains a free nonabelian subgroup.
I will give a detailed proof of this theorem in the case L = SL(2; R ) and will outline
the proof in the general case. Our proof in the SL(2; R ) case does not require ? to
be nitely generated.
The projectivization PSL(2; R) of SL(2; R) is the orientation-preserving subgroup
of the isometry of group of the hyperbolic plane H 2 . If we use the upper half-plane
model of H 2 then PSL(2; R) acts on H 2 via linear-fractional transformations:
az + b :
P ( c d ) : z 7! cz
It is clear that Tits alternative for PSL(2; R) implies Tits alternative for SL(2; R ),
since they dier by nite center.
Classication of isometries of H 2 :
Let A 2 SL(2; R). Then the xed points for the action of P (A) on C correspond
to the eigenvectors of the matrix A. Thus we get:
Case 1. jtr(A)j > 2 () P (A) has 2 distinct xed points on R = R [ 1.
Then = P (A) is called hyperbolic. It acts as a translation along a geodesic in H 2
connecting the xed points of .
Case 2. jtr(A)j < 2 () P (A) has 2 distinct xed points on C n R , one in the
upper and one in the lower half-plane. Then is called elliptic, in the unit disk model,
if we send the xed point to the origin, acts as a rotation around the origin.
Case 3. jtr(A)j = 2 and 6= Id. Then has a unique xed point in C , this xed
point belongs to R . Then is called parabolic. Conjugate in PSL(2; R) so that
the xed point of is innity. Then (z) = z + c, c 2 R , i.e. acts as a Euclidean
This is a complete classication of orientation-preserving isometries of H 2 . If is
an orientation-reversing isometry of H 2 then either:
(a) is a reection in a geodesic L H 2 , or
(b) is a glide-reection, i.e. it is the composition of a reection in a geodesic
L H 2 with a hyperbolic translation along L.
Dynamics: Suppose that is hyperbolic or parabolic. Then the sequence n, n 2 N ,
converges uniformly on compacts in C n Fix( ) to the constant map z 7! , where is one of the xed points of . If is hyperbolic then is the attractive xed point
of .
Lemma 82. (Ping-Pong lemma) Suppose that g; h 2 PSL(2; R) are hyperbolic or
parabolic with disjoint xed point sets. Then there exists n 2 N such that the group
hgn; hni is free of rank 2.
Proof. Is will consider the case when g; h are hyperbolic since the other cases are
similar. Let A? be a neighborhood of the repulsive xed point of g, bounded by a
geodesic in H 2 and disjoint from the axis of h. Similarly, dene B?, a neighborhood of
the repulsive xed point of h, bounded by a geodesic in H 2 and disjoint from the axis
of h and from A?. By taking suciently large n we can assume that the complements
to gn(A?) and hn(B?) in H 2 are domains A+, B+, as in the Figure 12, so that all
four domains A?; A+; B?; B+ are pairwise disjoint. Let denote the domain in H 2
which is the complement to A? [ A+ [ B? [ B+. Set g := gn; h := hn. I claim that
the group G := hg; hi is free of rank 2. To prove this consider a reduced nonempty
word w in the generators g; h. I claim that w() \ = ;. This would imply that w
is a nontrivial element of G which in turn would imply that G is free of rank 2.
Moreover, suppose that the last letter in w is g (or g?1, or h, or h?1 resp.), i.e.
w = w0g. I claim that w() A+ (resp. A?; B+; B?). Let's prove this by induction
on the length of w. I consider the case when w = w0g, where w0 is a reduced word
whose last letter is not g?1. Hence by induction, w0() is in one of the regions
A+; B+; B?, but not in A?. Then it is clear from the action of the isometry g that
g(A+ [ B+ [ B?) A+. Thus w() A+.
Figure 12:
This proves Tits alternative in the case when G PSL(2; R) contains two hyperbolic/parabolic elements which do not share a xed point.
Denition 83. A subgroup of PSL(2; R) is elementary if it either xes a point in C
or preserves a 2-point subset of C .
Corollary 84. Suppose that ? PSL(2; R) is a nonelementary subgroup which
contains a hyperbolic or parabolic element. Then ? contains F 2 .
Proof. Case 1. Suppose rst that ? contains a parabolic element whose xed point
is ; since ? does not x , there exists 2 ? such that = ( ) 6= ; then := ?1
is a parabolic isometry with the xed point 6= . Then Ping-Pong lemma implies
that h n; ni is isomorphic to F 2 for large n.
Case 2. Now, suppose that 2 ? is a hyperbolic isometry with the xed points
; . There exists 2 ? such that ( ) 6= and () 6= and (f; g) 6= f; g. If
(f; g) \ f; g = ;;
then we are done by the Ping-Pong lemma, analogously to the parabolic case above.
Suppose that () = . Dene := ?1: it is a hyperbolic isometry which xes and does not x . It is easy to see that the commutator [; ] is a parabolic isometry
which xes (just assume that = 1, = 0 and then compute the commutator).
Therefore ? contains a parabolic isometry and we are done by Case 1.
The most dicult case is when ? contains only elliptic elements.
Lemma 85. If ? contains only elliptic elements, then ? xes a point in H 2 .
Proof. Suppose that there are elliptic elements ; in ? with distinct xed points
a; b 2 H 2 . By assumption, their product = is also an elliptic element; its xed
point c is necessarily distinct from a and b. Consider the geodesic triangle in H 2 with
the vertices a; b; c; let J = R1; R2 ; R3 denote reections in the sides [ab]; [bc] and [ca]
respectively. Then
= R1 R3; = R2 R1; = R2 R3 :
See Figure 13.
Then [ ?1; ?1] = JJ = (J )2 . Note that J is an orientation-reversing isometry. If J is a reection then [ ?1; ?1] = Id, which would imply that a = b. Thus
J is a glide-reection; it follows that (J )2 is a hyperbolic isometry (a translation
along the axis of of J ). Hence ? contains a hyperbolic element. Contradiction.
Lemma 86. If ? is an elementary subgroup of PSL(2; R), then ? is virtually solvable.
Proof. If ? preserves a 2-point set then its index 2 subgroup xes a point. Therefore
it suces to consider the case when ? xes a point in C .
Case 1. 2= @1H 2 = R . Then ? xes a point in the hyperbolic plane H 2 (either or its complex conjugate). By using the unit disk model we can assume that ? xes
the origin in the unit disk. Then ? SO(2); since the latter is abelian it follows that
? is abelian as well.
Figure 13:
Case 2. 2 @1H 2 = R . We can assume that = 1; then ? is contained in the
group S of ane transformations z 7! az + b. The group S contains abelian subgroup
A which consists of translations z 7! z + b. The group A = [S; S ] is the commutator
subgroup of S . Therefore S is solvable. It follows that ? is solvable as well.
Outline of the proof of Tits' alternative in the general case. By taking a homomorphism L ! ad(L) y Lie(L), where Lie(L) is the Lie algebra of L, it suces to
prove Tits alternative for subgroups ? GL(n; R ). Let G denote Zariski closure of
? in GL(n; R ), i.e. the smallest algebraic subgroup (i.e. subgroup given by algebraic
equations) of GL(n; R ) which contains ?. If the identity component of G happens
to be solvable then we are done. Otherwise the identity component of G has nontrivial semisimple part; by dividing G by its solvable radical we can assume that G
is semisimple, i.e. its Lie algebra is a direct sum of simple Lie algebras. It suces
of course to treat the case when G is simple (by considering projections of ? to the
simple components of G). There are two cases which can occur:
(A) G is noncompact.
(B) G is compact.
(A) First, let's consider the noncompact case. There is a Riemannian manifold
X , called symmetric space, associated with G on which G acts isometrically and
transitively: X = G=K , where K is a maximal compact subgroup of G. The most
important feature of X is that X has nonpositive sectional curvature and moreover,
the sectional curvature is negative in certain directions. Thus one can use X as a
replacement of the hyperbolic plane as we have done it in the case of SL(2; R ). There
is a classication of isometries of X similar to the classication of isometries of H 2 :
There are hyperbolic, parabolic and elliptic isometries. The elliptic ones x points in
X , hyperbolic isometries act as translations along certain geodesics in X . The fact
that G is the Zariski closure of ? then implies that ? contains hyperbolic isometries.
Then one can run a version of Ping-Pong lemma as we did in the case of H 2 to show
that ? contains F 2 .
(B) The noncompact case is much more complicated. Let 1; :::; m denote generators of ? and consider the eld F in R generated by the matrix entries of the
generators. If the eld F happens to be a transcendental extension of Q one can
show that there are homomorphisms j : ? ! G which converge (on each generator)
to the identity embedding so that j (?) have the property: The elds Fj associated
with j (?) as above are algebraic extensions of Q . The reason for that is that we
can assume that G is dened over Q (i.e. is given by equations with rational coefcients), thus the variety Hom(?; G) is dened over Q as well; therefore algebraic
points are dense in this variety. Because ? was Zariski dense in G, there exists j such
that j (?) is Zariski dense as well and we are reduced to the case where the eld F
is contained in Q . Let G(F ) denote the group of F -points in G (i.e. points whose
coordinates belong to F ). Consider the action of the Galois group Gal(Q =Q ) on the
eld F . Every such 2 Gal(Q =Q ) will induce (a discontinuous!) automorphism of the complexication G(C ) of the group G, and therefore it will send the groups
? G(F ) to (?) G((F )) G(C ). The homomorphism : ? ! ?0 := (?) is
1 ? 1 and therefore, if for some the group G((F )) happens to be a non-relatively
compact subgroup of G(C ) we are back to the noncompact case (A).
However it could happen that for each the group G((F )) is relatively compact
and thus we seemingly have gained nothing. There is a remarkable construction which
saves the proof.
Adeles. (See [39, Chapter 6].) The ring of adeles was introduced by A. Weil in
1936. For the eld F consider various norms j j : F ! R + . A norm is called
nonarchimedean if instead of the usual triangle inequality one has:
ja + bj max(jaj; jbj):
For each norm we dene F to be the completion of F with respect to this norm.
For each nonarchimedean norm the ring of integers O := fx : jxj 1g is an open
subset of F : If jxj = 1, jyj < 1=2, then for z = x + y we have: jzj max(1; jyj ) =
1. Therefore, if z belongs to a ball of radius 1=2 centered at x, then z 2 O .
Example 87. (A). Archimedean norms. Let 2 Gal(Q =Q ), then the embedding
: F ! (F ) C denes a norm on F by restriction of the norm (the usual
absolute value) from C to (F ). Then the completion F is either isomorphic to R
or to C . Such norms (and completions) are archimedean and each archimedean norm
of F appears in this way.
(B). Nonarchimedean norms. Let F = Q , pick a prime number p 2 N . For
each number x = q=pn 2 Q (where both numerator and denominator of q are not
divisible by p) let p(x) := pn. One can check that is a nonarchimedean norm and
the completion of Q with respect to this norm is the eld of p-adic numbers.
Let Nor(F ) denote the set of all norms on F which restrict to either standard or
one of the p-adic norms on Q F . Note that for each x 2 Q , x 2 Op (i.e. p-adic
norm of x is 1) for all but nitely many p's, since x has only nitely many primes
in its denominator. The same is true for elements of F : For all but nitely many
2 Nor(F ), (x) 1.
Product formula: For each x 2 Q n f0g
(x) = 1:
Indeed, if x = p is prime then jpj = p for the archimedean norm, (p) = 1 if 6= p is
a nonarchimedean norm and p(p) = 1=p. Thus the product formula holds for prime
numbers x. Since norms are multiplicative functions from Q to R + , the product
formula holds for arbitrary x 6= 0. A similar product formula is true for an arbitrary
algebraic number eld F :
( (x))N = 1;
2Nor(F )
where N = [F : Q ], see [39, Chapter 6].
Denition 88. The ring of adeles is the restricted product
F ) :=
2Nor(F )
F ;
i.e. the subset of the direct product which consists of points whose projection to F
belongs to O for all but nitely many 's.
We topologize A (F ) via the product topology. For instance, if F = Q then A (Q )
is the restricted product
Q p:
p is prime
Now a miracle happens:
Theorem 89. (See [39, Chapter 6, Theorem 1].) The image of the diagonal embedding F ,! A (F ) is a discrete subset in A (F ).
Proof. It suces to verify that 0 is an isolated point. Take the archimedean norms
1 ; :::; m (there are only nitely many of them) and consider the open subset
2Nor(F )nf1 ;:::;mg
fx 2 Fi : i(x) < 1=2g O
of A (F ). Then for each (x ) 2 U ,
2Nor(F )
(x ) < 1=2 < 1:
Hence, by the product formula, the intersection of U with the image of F in A (F )
consists only of f0g.
Thus the embedding F ,! A (F ) induces a discrete embedding
? G(F ) ,! G(A (F )):
For each norm 2 Nor(F ) we have the projection p : ? ! G(F ). If the image p (?)
is relatively compact for each then ? is a discrete compact subset of G(A (F )), which
implies that ? is nite, a contradiction! Thus there exists a norm 2 Nor(F ) such
that the image of ? in G(F ) is not relatively compact. If happens to be archimedean
we are done as before. The more interesting case occurs if is nonarchimedean.
Then one can dene a metric space X on which the group G(F ) acts isometrically,
faithfully and cocompactly (although the quotient is not a point but a Euclidean
simplex). The space X is called a Euclidean building, it is a nonarchimedean analogue
of the symmetric space. It has nonpositive curvature in the sense that the geodesic
triangles in X are \thinner" than geodesic triangles in the Euclidean plane. The
space X is covered by isometrically embedded copies of the Euclidean space E r ,
called apartments, so that each pair of points in X belongs to an apartment. The
number r is called rank of the space X .
Example 90. If r = 1 then X is a simplicial metric tree where each edge has unit
We note that the homomorphism ? ! G(F ) ! Aut(X ) is an embedding. The
isometries of X admit a classication similar to the isometries of H 2 : Each isometry
is either hyperbolic (i.e. a translation along a geodesic contained in one of the apartments) or elliptic, i.e. xes a point in X . The group ? is Zariski dense in G(F ),
therefore it contains hyperbolic isometries. This allows one to run an analogue of
Ping-Pong type arguments in X and show that ? contains F 2 .
5 Growth of groups and Gromov's theorem
Let X be a metric space of bounded geometry and x 2 X is a base-point. We dene
the growth function
X;x(R) := jB (x; R)j;
the cardinality of R-ball centered at x. We introduce the following asymptotic inequality between functions : R + ! R + :
if there exist constants C1; C2 such that (R) C1(C2R) for suciently large R.
We say that two functions are equivalent, , if
and :
Lemma 91. (Equivalence class of growth is QI invariant.) Suppose that f : (X; x) !
(Y; y) is a quasi-isometry. Then X;x Y;y .
Proof. Let f be a coarse inverse to f , assume that f; f are L-Lipschitz. Then both
f , f have multiplicity m (since X and Y have bounded geometry). Then
f (B (x; R)) B (y; LR):
It follows that jB (x; R)j mjB (y; LR)j and jB (y; R)j mjB (x; LR)j.
Corollary 92. X;x X;x for all x; x0 2 X .
Henceforth we will suppress the choice of the base-point in the notation for the
growth function.
Denition 93. X has polynomial growth if X (R) Rd for some d. X has exponential growth if eR X (R). X has subexponential growth if for each c > 0,
X (R) ecR for all suciently large R.
Example 94. Show that for each (bounded geometry) space X , X (R) eR .
For a group G with nite generating set S we sometimes will use the notation S (R)
for G(R), where S is used to metrize the group G. Since G acts transitively on itself,
this denition does not depend on the choice of a base-point.
Example 95. Suppose that G = F r is a free nonabelian group. Show that G has
exponential growth.
Suppose that H is a subgroup of G. It is then clear that
H G:
Note that if : G ! F r is an epimorphism, then its admits a left inverse : F r ! G.
Hence G contains F r and if r 2 it follows that G has exponential growth.
The main objective of this chapter is to prove
Theorem 96. (Gromov, [27]) If G is a nitely generated group of polynomial growth
then G is virtually nilpotent.
We will also verify that all virtually nilpotent groups have polynomial growth.
Corollary 97. Suppose that G is a nitely generated group which is quasi-isometric
to a nilpotent group. Then G is virtually nilpotent.
Proof. Follows directly from Gromov's theorem since polynomial growth is a QI invariant.
Remark 98. An alternative proof of the above corollary (which does not use Gromov's
theorem) was recently given by Y. Shalom [50].
5.1 Nilpotent and solvable groups
Given a group G and a subgroup S G dene [G; S ] as the subgroup generated by
the commutators [g; s]; g 2 G; s 2 S . Dene the lower central series of G:
G = G0 [G; G] = G1 [G; G1] = G2 [G; G2] = G3:::
and the upper central series of G:
G = G0 [G; G] = G1 [G1 ; G1] = G2 [G2; G2] = G3 :::
The group G is called nilpotent, resp. solvable, if the lower, resp. upper, central series
of G terminates at the trivial group. The group G is called s-step nilpotent if its
lower central series is
G0 G1 :::Gs?1 1;
where Gs?1 6= 1.
Given a nilpotent group G we note that all the subgroups Gi are nitely generated,
their generators are the iterated commutators of the generators of G. We also have
nitely generated abelian groups Ai := Gi=Gi+1. After passing to a nite index
subgroup in G we can assume that each Ai is torsion-free. Let be an automorphism
of G, then it preserves the lower central series and induces automorphisms of the free
abelian groups Ai. Each such automorphism i is given by a matrix with integer
coecients. After taking suciently high power of we can assume that none of
these matrices have a root of unity (dierent from 1) as an eigenvalue. If each i has
only 1 as an eigenvalue then after \rening" the lower central series we can assume
that each i is trivial. Then the extension G~ of G by is again a nilpotent group.
Theorem 99. Suppose that 1 ! G ! G~ ! Z ! 1 is an extension of G by and
at least one eigenvalue of one of the i 's is dierent from 1. Then G~ has exponential
Proof. We begin with
Lemma 100. Let A be a nitely generated free abelian group and 2 Aut(A). Then:
If has an eigenvalue such that jj 2 then there exists a 2 A such that
0a + 1 (a) + ::: + m m(a) + ::: 2 A
(where i 2 f0; 1g and i = 0 for all but nitely many i's) are distinct for dierent
choices of the sequences (i ).
Proof. The transpose matrix T also has as its eigenvalue. Hence there exists a
nonzero linear function : A ! C such that = . Pick any a 2 A n Ker( ).
( i i(a)) = ( i i(a)) (a):
Suppose that
i i(a) =
ii(a) = 0;
where jij 1 for each i. Let N be the maximal value of i for which i 6= 0. Then
N ?1
= jj ? 1 (jjN ? 1)=2:
We now can prove theorem 99. Suppose there is i such that i has an eigenvalue
which is not a root of unity. After taking appropriate iteration of and possibly
replacing with ?1 we can assume that such that jj 2. Let x 2 Gi be an element
which projects to a 2 Ai under the homomorphism Gi ! Gi=Gi+1. Let z 2 G~ denote
the generator corresponding to the automorphism . Dene elements
x0 (zx1 z?1 ):::(zm xm z?m ) 2 Gi; i 2 f0; 1g:
After canceling out z's we get:
x0 zx1 zx2 z:::zxm z?m
The norm of each of these elements in G~ is at most 3(m + 1). These elements
are distinct for dierent choices of (i)'s, since their projections to Ai are distinct
according to the above lemma. Thus we get 2m distinct elements of G~ whose word
norm is at most 3(m + 1). This implies that G~ has exponential growth.
Proposition 101. Suppose that G is a group of subexponential growth, which ts
into a short exact sequence
1 ! K ! G ! Z ! 1:
Then K is nitely generated. Moreover, if G (R) Rd then K (R) Rd?1 .
Proof. Let 2 G be an element which projects to the generator of Z, let g1 ; :::; gk be
elements such that fg1; :::; gk ; g generate G. Dene
S := fm;i := m gi ?m; m 2 Z; i = 1; :::; kg:
Then the (innite) set S generates K . Given i consider products of the form:
m ; 2 f0; 1g:
We have 2m+1 words like this, each of length 2m. Hence subexponential growth of
G implies that two of these words are equal:
m = 1 ::: m ;
m 6= m . It follows that
m;i = w(0;i; :::; m?1;i) 2 h0;i; :::; m?1;ii:
m+1;i = m;i ?1 = w(0;i; :::; m?1;i) ?1 = w0(1;i; :::; m;i);
the latter by inserting pairs ?1 ; between each pair of letters in the word w and
using the fact that
j+1;i = j;i ?1; j = 0; :::; m ? 1:
However w0 2 h0;i; :::; m?1;ii, since
m;i 2 h0;i; :::; m?1;ii:
Thus m+1;i 2 h0;i; :::; m?1;ii too. We continue by induction: It follows that n;i 2
h0;i; :::; m?1;ii for each n. Therefore K = h0;i; :::; m?1;ii. This proves the rst
Now let's prove the second estimate. Let Y denote a nite generating set of K and
X := Y [ f g, where is as above. Let
H := fh1 ; :::; hY ([n=2]) g
be distinct elements of K of the norm [n=2] (with respect to the generators Y ).
Then we get pairwise distinct elements
hi j ; ?[n=2] j [n=2]; hj 2 H;
the length of each of these elements is at most n. Hence
nY ([n=2]) X (n) Cnd:
It follows that
Y ([n=2]) Cnd?1 :
5.2 Growth of nilpotent groups
Consider an s-step nilpotent group G with the lower central series
G0 G1 :::Gs?1 1;
and the abelian quotients Ai = G=Gi+1 . Let di denote the rank of Ai (or, rather, rank
of its free part). Dene
d(G) :=
G(R) R G) .
(i + 1)di:
Theorem 102. (Bass, [2])
Example 103. Prove Bass' theorem for abelian groups.
Our goal is to prove only that G has polynomial growth without getting a sharp
For the proof I introduce the notion of distortion for subgroups which is another
useful concept of the geometric group theory. Let H be a nitely generated subgroup
of a nitely generated group G, let dH ; dG denote the respective word metrics on H
and G, let BG(e; r) denote r-ball centered at the origin in the group G.
Denition 104. Dene the distortion function (R) = (H : G; R) as
(R) := maxfdH (e; h) : h 2 BG (e; R)g:
The subgroup H is called undistorted (in G) if (R) R.
Example 105. Show that H is undistorted i the embedding : H ! G is a quasiisometric embedding.
In general, distortion functions for subgroups can be as bad as one can imagine, for
instance, nonrecursive.
Example 106. Let G := ha; b : aba?1 = bpi, p 2. Then the subgroup H = hbi is
exponentially distorted in G.
Proof. To establish the lower exponential bound note that:
gn := anba?n = bpn ;
hence dG(1; gn) = 2n + 1, dH (1; gn) = pn, hence
(R) p[(R?1)=2] :
It will leave the upper exponential bound as a exercise (compare the proof of Theorem
Recall that each subgroup of a nitely generated nilpotent group is nitely generated itself.
Theorem 107. Let G be a nitely generated nilpotent group, then every subgroup
H G has polynomial distortion.
I will prove only that the commutator subgroup G1 := [G; G] has at most polynomial distortion. As the equivalence class of the distortion function is a QI invariant, it
suces to consider the case where all abelian quotients Ai = Gi=Gi+1 are free abelian.
Consider rst the case where G is 2-step nilpotent.
Let T0 := fx1 ; :::; xng be the generators of G and let T1 := f[xi 1 ; xj 1 ]; i; j g denote
generators of H = G1. Consider a word
w = xi11 :::ximm
of length m, so that w 2 H . This means that the total exponent of each generator
in the word w is zero. Let's reorder the letters in w so that the new word starts with
the powers of x1 , then come the powers of x2, etc, by moving rst all appearances of
x1 to the left, then of x2 to the left, etc. This involves at most m2 \crossings" of the
letters in w. Each \crossing" (with i > j ) involves introducing a commutator of the
corresponding generators:
xi xj ! xj xi[x?i 1 ; x?j 1 ]
which increases the length of the word (in terms of the generators T0 [ T1) by one
unit. Note that since G is 2-step nilpotent, the elements of T1 are in the center of
G, therefore crossing xk 's with the elements of T1 does not change the length of the
word. After all moves to the left are made, in the new word the generators from T0
cancel out and it leaves us with a new word w0 on the generators from T1 , so that
dH (1; w0) m2 . Hence (m) m2 is at most polynomial.
Now, let's consider the case where G is 3-step nilpotent, we have generators T0 of
G0, T1 of G1 and the generators
T2 := f[xk 1 ; [xj 1; xi 1 ]]g
of G2 . Note that G2 now is in the center of G. Take a word w 2 G and argue the same
way as above: First move the powers of x1 to the left, then powers of x2 , etc. This
introduces m2 commutators, i.e. generators from T1 . However these generators no
longer commute with the generators from T0. Therefore, to move xk to the left we
have to cross it with the generators from T1 , it involves introducing an extra iterated
commutator from T2 :
[xi ; xj ]xk ! xk [xi ; xj ][([xi ; xj ])?1 ; x?k 1] = xk [xi; xj ] [[x?j 1 ; x?i 1]; x?k 1 ]:
The total number of double commutators introduced this way is m3 . The elements
of T2 are in the center, therefore crossing xk 's with elements of T2 does not change the
length of the word. After we are done moving the elements of T0 to the left they all
cancel out and we get a word in the generators T1 [ T2 of the length m2 + m3 m3 .
Thus the subgroup G1 has distortion R3 in G.
Continue inductively on the number of steps of the nilpotent group...
Theorem 108. Each nilpotent group has at most polynomial growth.
Proof. The proof is by induction on the number of steps in the nilpotent group. The
assertion is clear is G is 1-step nilpotent (i.e. abelian). Suppose that each s ? 1-step
nilpotent group has at most polynomial growth. Consider s-step nilpotent group G:
G = G0 G1 :::Gs?1 1:
By the induction hypothesis, G1 has growth Rd and, according to Theorem 107,
the distortion of G1 in G is at most RD . Let r denote the rank of the abelinization
of G.
Consider an element 2 BG (e; R), then can be written down as a product w0w1
where w0 is a word on T0 of the form:
xk11 :::xknn ;
and w1 is a word on T1 . Then kw0k R, and kw1k kw0k + k k 2R. The number
of the words w0 of length R is Rr . Since G1 has distortion RD in G, the length
of the word w1 on the generators T1 is (2R)D . Since G1 Rd we conclude that
G(R) Rr (2R)dD RdD+r :
Corollary 109. A solvable group G has polynomial growth i G is virtually nilpotent.
Proof. It remains to show that if G is a solvable and has polynomial growth then G
is virtually nilpotent group. By considering the upper central series of G we get the
short exact sequence
1 ! K ! G ! Z:
Suppose that G has polynomial growth Rd , then K is nitely generated, solvable
and has growth Rd?1. By induction, we can assume that K is virtually nilpotent.
Then Theorem 99 implies that G is also virtually nilpotent.
Corollary 110. Suppose that G is a nitely generated linear group. Then G either
has polynomial or exponential growth.
Proof. By Tits alternative either G contains a nonabelian free subgroup (and hence
G has exponential growth) or G is virtually solvable. For virtually solvable groups
the assertion follows from Corollary 109.
R. Grigorchuk [26] constructed nitely generated groups of intermediate growth, i.e.
their growth is superpolynomial but subexponential. Existence of nitely-presented
groups of this type is unknown.
5.3 Elements of the nonstandard analysis
Our discussion here follows [25], [57].
Let I be a countable set. Recall that an ultralter on I is a nitely additive measure
with values in the set f0; 1g dened on the power set 2I . We will assume that ! is
nonprincipal. Given a set S dene its ultraproduct
S := S I =!
as the collection of equivalence classes of maps f : I ! S where f and g are equivalent
i they agree for !-all i 2 I . Note that if G is a group (ring, eld...) then G has
a natural group (ring, eld...) structure. If S is totally ordered then S is totally
ordered as well: [f ] [g] (for f; g 2 S I ) i f (i) g(i) for !-all i 2 I .
For subsets P S we have the canonical embedding P ,! P^ S given by sending
x 2 P to the constant function f (i) = x.
Thus we dene ordered semigroup N (nonstandard natural numbers) and ordered
eld R (nonstandard real numbers). An element R 2 R is called innitely large if
given any r 2 R R , one has R r. Note that given any R 2 R there exists
n 2 N such that n > R.
Denition 111. A subset W S1 ::: Sn is called internal if \membership in W
can be determined by coordinate-wise computation", i.e. if for each i 2 I there is a
subset Wi S1 ::: Sn such that for f1 2 S1I ; :::; fn 2 SnI
([f1]; ::::; [fn]) 2 W () (f1(i); :::; fn(i)) 2 Wi for ! ? all i 2 I:
The sets Wi are called coordinates of W .
Using this denition we can also dene internal functions S1 ! S2 as functions
whose graphs are internal subsets of S1 S2 . Clearly the image of an internal function
is an internal subset of S2.
Lemma 112. Suppose that A S is innite subset. Then A^ S is not internal.
Proof. Suppose that Ai ; i 2 I; are coordinates of A^. Let a1 ; a2 ; :::: be an innite
sequence of distinct elements of A. Dene the following function f 2 S I :
Case 1. f (n) = aj , where j = maxfj 0 : aj 2 Ang if the maximum exists,
Case 2. f (n) = an+j , where j = minfj 0 : an+j 2 Ang if the maximum above does
not exist.
Note that for each n 2 I , f (n) 2 An, therefore [f ] 2 A^. Since A^ consists of (almost)
constant functions, there exists m 2 N such that f (n) = am for !-all n 2 I .
It follows that the Case 2 of the denition of f cannot occur for !-all n 2 I . Thus
for almost all n 2 I the function f is dened as in Case 1. It follows that for almost
all n 2 I , am+1 2= An. Thus am+1 2= A, which is a contradiction.
Corollary 113. N is not an internal subset of N .
Suppose that (X; d) is a metric space. Then X has a natural structure of R -metric
space where the \distance function" d takes values in R + :
d([f ]; [g]) := [i 7! d(f (i); g(i))]:
We will regard d as a generalized metric, so we will talk about metric balls, etc. Note
that the \metric balls" in X are internal subsets.
A bit of logic. Let be a statement about elements and subsets of S . The nonstandard interpretation of is a statement obtained from by replacing:
1. Each entry of the form \x 2 S " with \x 2 S ".
2. Each entry of the form \A S " with \A an internal subset of S ".
Theorem 114. (Los) A statement about S is true i its nonstandard interpretation
about S is true.
As a corollary we get:
Corollary 115. 1. (Completeness axiom) Each nonempty bounded from about internal subset A R has supremum. (Note that R R does not have supremum.)
2. (Nonstandard induction principle.) Suppose that S N is an internal subset
such that 1 2 S and for each n 2 S , one has n + 1 2 S . Then S = N . (Note that
this fails for S = N N .)
Example 116. 1. Give a direct proof of the completeness axiom for R .
2. Use the completeness axiom to derive the nonstandard induction principle.
Suppose we are given an 2 R , where n 2 N . Using the nonstandard induction
principle on can dene the nonstandard products:
a1:::an ; n 2 N ;
as an internal function f : N ! R given by f (1) = a1 , f (n + 1) = f (n)an+1.
5.4 Regular growth theorem
A metric space X is called doubling if there exists a number N such that each R-ball
in X is covered by N balls of radius R=2.
Exercise 117. Show that doubling implies polynomial growth for spaces of bounded
Although there are spaces of polynomial growth which are not doubling, the Regular
Growth Theorem below shows that groups of polynomial growth exhibit doubling-like
Our discussion here follows [57].
Theorem 118. (Regular growth theorem) Suppose that G is a nitely generated group
such that G (R) Rd . Then there exists an innitely large 2 R such that for all
i 2 N n f1g the following assertion P (; i) holds:
If x1 ; :::; xt 2 B (e; =2) G and the balls B (xj ; =i) are pairwise disjoint (j =
1; :::; t) then t id+1 .
Here e is the identity in G .
Proof. Start with an arbitrary innitely large R 2 R (for instance, represented by
the sequence n, n 2 N = I ). I claim that the number can be found within the
interval [log R; R] (here logarithm is taken with the base 2). Suppose to the contrary,
that for each 2 [log R; R] there exists i 2 N n f1g such that P (; i) fails. Then we
dene the function
: [log R; R] ! N ; () is the smallest i for which P (; i) fails:
Since the nonstandard distance function is an internal function, the function is
internal as well, therefore its image has to be nite (see Lemma 112). Therefore there
exists K 2 N such that
() 2 [2; K ]; 8 2 [log R; R]:
We now dene (using the nonstandard induction) the following elements of G:
1. x1 (1); :::; xt1 (1) 2 B (1; R=2) such that t1 = id1+1 , (for i1 = (R)) and the balls
B (xj (1); R=i1) are pairwise disjoint.
2. Each nonstandard ball B (xj (1); R=i1) is isometric to B (e; R=i1). Therefore
failure of P (R=i1; i2 ) (i2 = (R=i1 )) implies that in each ball B (xj (1); R=(2i1)) we can
nd points
x1 (2); :::; xt2 (2); t2 = id2+1;
so that the balls B (xj (2); R=(i1i2 )) are pairwise disjoint.
We continue via the nonstandard induction. Given u 2 N such that the points
x1 (u); :::; xtu (u) are constructed we construct the next generation of points x1 (u +
1); :::; xtu+1 (u + 1) within each ball B (xj (u); R=(2i1:::iu )) so that the balls B (xj (u +
1); R=(i1:::iu+1 )) are pairwise disjoint and tu+1 = idu+1 .
This induction process continues as long as R=(i1 :::iu+1) log R. Recall that ij 2,
R=(i1 :::iu) 2?uR:
Therefore, if u > log R ? log log R then
R=(i1:::iu ) < log R:
Thus there exists u 2 N such that
R=(i1 :::iu) log R; but R=(i1:::iu+1 ) < log R:
Let's count the \number" (nonstandard of course!) of points xi(k) we have constructed between the step 1 of induction and the u-th step of induction:
We get id1+1id2+1 :::idu+1 points; since
R=(K log R) R=(iu+1 log R) < i1:::iu ;
we get:
(R=(K log R))d+1 (i1i2 :::idu+1 )d+1 :
What does this inequality really mean? Recall that R and u are represented by
sequences of real and natural numbers Rn ; un respectively. The above inequality thus
implies that for !-all n 2 N , one has:
Rn d+1 jB (e; R )j:
K log Rn
Since jB (e; R)j CRd, we get:
Rn Const(log(Rn))d+1 ;
for !-all n 2 N . If Rn = 2n , we obtain
2n=(d+1) Constn;
where !-lim n = 1. Contradiction.
Lemma 119. Suppose that = (n) is a sequence satisfying the assertion of the
regular growth theorem. Then the asymptotic cone X! constructed from the Cayley
graph of G by rescaling via n , is (a) locally compact, (b) has covering dimension
d + 1.
Proof. (a) Since X! is complete and G! acts transitively on X! it suces to show
that the metric ball B (e! ; 1) is totally bounded. Let > 0. Then there exists i 2 N ,
i 2, such that 1=(2i) < . In the ball B (e; ) G consider the maximal t (t id+1 )
for which there are points x1 ; :::; xt 2 B (e; =2) G so that the balls B (xj ; =i) are
pairwise disjoint. Then the points x1 ; :::; xt form =(2i)-net in B (e; ). By passing to
X! we conclude that the corresponding points x1! ; :::; xt! 2 B (e! ; 1) form an -net.
Since t is nite we conclude that B (e! ; 1) is totally bounded and therefore compact.
(b) To prove the dimension bound I rst recall the concept of Hausdor dimension
for compact metric spaces. Let K be a compact metric space and > 0. The
-Hausdor measure (K ) is dened as
lim inf
where the inmum is taken over all nite coverings of K by balls B (xi; ri), ri r
(i = 1; :::; N ). Then the Hausdor dimension of K is dened as:
dimHaus (K ) := inf f : (K ) = 0g:
Theorem 120. (Hurewicz-Wallman, [31]) dim(K ) dimHaus (K ), where dim stands
for the covering dimension.
We will verify that the Hausdor dimension of B (e! ; 1) is at most d + 1. Pick
> d + 1; for each i consider the covering of B (e! ; 1) by the balls
B (xj! ; 2=i); j = 1; :::; t id+1:
Therefore we get:
(2=i) 2 id+1=i = 2id+1? :
j =1
limi!1 2id+1?
Since > d + 1,
= 0. Hence (B (e! ; 1)) = 0. It follows that
dim(B (e! ; 1)) d + 1.
Theorem 121. (Montgomery-Zippin, [41]) Suppose that X is a topological space
which is connected, locally connected, nite-dimensional and locally compact. Suppose
that L is a group which acts on X by homeomorphisms and the action is transitive.
Then L is a Lie group with nitely many connected components.
5.5 Proof of Gromov's theorem
The proof is by induction on the degree of polynomial growth. If G(R) R0 = 1
then G is nite and there is nothing to prove. Suppose that each group of growth at
most Rd?1 is virtually nilpotent. Let G be a a (nitely generated group) of growth
Rd. Find a sequence n satisfying the conclusion of the regular growth theorem
and construct the asymptotic cone X! of the Cayley graph of G via rescaling by the
sequence n. Then X! is connected, locally connected, nite-dimensional and locally
compact. The group G! acts on X! isometrically and transitively. Then Isom(X! )
is a Lie group L with nitely many components. We get the diagonal homomorphism
` : G ! G! L.
We have the following cases:
(a) The image of ` is not virtually solvable. Then by Tits' alternative, `(G) contains
a free nonabelian subgroup; it follows that G contains a free nonabelian subgroup as
well which contradicts the assumption that G) has polynomial growth.
(b) The image of ` is virtually solvable and innite. Then, after passing to a nite
index subgroup in G, we get a homomorphism from G onto Z. Then, according to
Proposition 101, K = Ker() is a nitely generated group of growth Rd?1. Thus,
by the induction hypothesis, K is a virtually nilpotent group. Since G has polynomial
growth, Theorem 99 implies that the group G is virtually nilpotent as well.
(c) `(G) is nite.
To see that the latter case can occur consider an abelian group G. Then the
homomorphism ` is actually trivial. How to describe the kernel of `? For each g 2 G
dene the displacement function (g; r) := maxfd(gx; x) : x 2 B (e; r)g. Then
Ker(`) = fg 2 G : !-lim (g; n)=n = 0g:
Fix a generating set g1; :::; gm of a subgroup G0 in G. Dene
D(G0; r) := j=1
(g ; r):
;:::;m j
(This is an abuse of notation, this function of course depends not only on G but also
on the choice of the generating set.)
Given a point in the Cayley graph, p 2 ?G, we dene another function
D(G0; p; r) := maxfd(gj x; x); x 2 B (p; r); j = 1; :::; m; g:
Clearly D(G0; e; r) = D(G0; r) and for p 2 G ?G,
D(G0; p; r) = D(p?1G0p; r);
where we take the generators p?1gj p; j = 1; :::; m for the group p?1G0p. The function
D(G0; p; r) is 2-Lipschitz as a function of p.
Lemma 122. Suppose that D(G0; r) is bounded as a function of r. Then G is virtually
Proof. Suppose that d(gj x; x) C for all x 2 G. Then
d(x?1gj x; 1) C;
and therefore the conjugacy class of gj in G has cardinality G(C ) = N . Hence
the centralizer ZG(gj ) of gj in G has nite index in G: Indeed, if x0 ; :::; xN 2 G then
there are 0 i 6= k N such that
x?i 1 gj xi = x?k 1 gj xk ) [xk x?i 1 ; gj ] = 1 ) xk x?i 1 2 ZG(gj ):
Thus the intersection
A :=
j =1
ZG(gj )
has nite index in G; it follows that A \ G0 is an abelian subgroup of nite index in
G0 .
We now assume that `(G) is nite and consider the subgroup of nite index G0 :=
Ker(`) G. Let g1; :::; gm be generators of G0. By the previous lemma it suces
to consider the case when the function D(G0; r) is unbounded but has \sublinear
growth", i.e.
!-lim (gj ; n)=n = 0; j = 1; :::; m:
Lemma 123. Let > 0. Then there exists xn 2 G such that
!-lim Dn (x?n 1 Gxn; n) = :
Proof. For !-all n 2 N we have D(G0 ; n) n=2. On the other hand, for the given
n, there exists qn 2 G such that
D(G0; qn; n) > 2n:
Therefore, since G is connected, there exists yn 2 ?G such that
D(G0; yn; n) = n :
Pick a point xn 2 G nearest to yn, then
jD(G0; xn; n) ? nj 2:
It follows that jD(x?n 1Gxn ; n) ? nj 2 and therefore
!-lim Dn (x?n 1Gxn ; n) = :
Now, given > 0 we dene a homomorphism ` : G0 ! G! L by sending each
g 2 G0 to the sequence
[g] := [x?n 1 gxn] 2 G! :
Note that since D(x?n 1 Gxn; n) = O(), the elements `(gj ) belong to G! , hence
` : G ! G ! .
We topologize the group L via compact-open topology with respect to its action on
X! , thus -neighborhood of the identity in L contains all isometries h 2 L such that
(h; 1) ;
where is the displacement function of h on the unit ball B (e! ; 1). By our choice
of xn , there exists a generator h = gj of G0 such that (`(h); 1) = . If there is an
N 2 N such that the order j`(h)j of `(h) is at most N for all , then L contains
arbitrarily small nite cyclic subgroups h`(h)i, which is impossible since L is a Lie
group. Therefore
j` (h)j = 1
!0 If for some > 0, `(G0) is innite we are done as above. Hence we assume that `(G0)
is nite for all > 0. We then use
Theorem 124. (Jordan) Let L be a Lie group with nitely many connected components. Then there exists a number q = q(L) such that each nite subgroup F in L
contains an abelian subgroup of index q.
We prove this theorem in section 5.6.
For each consider the preimage G0 in G0 of the abelian subgroup in `(G0 ) which
is given by Jordan's theorem. The index of G0 in G0 is at most q. Let G00 be the
intersection of all the subgroups G0; > 0. Then G00 has nite index in G and G00
admits homomorphisms onto nite abelian groups of arbitrarily large order. Since
all such homomorphisms have to factor through the abelinization (G00)ab , the group
(G00)ab has to be innite. Since (G00 )ab is nitely generated it follows that it has
nontrivial free part, hence G00 again admits an epimorphism to Z. Thus we are again
done by induction.
5.6 Proof of Jordan's theorem
In this section I outline a proof of Jordan's theorem, for the details see [47, Theorem
8.29]. I will be assuming that L = GLn(R ), this does not restrict us too much
since each connected Lie group embeds, modulo the center, in the group of linear
automorphisms of its Lie algebra.
Given a subset L dene inductively subsets (i) as (i+1) = [
; (i) ], (0) := .
Lemma 125. There is a neighborhood of 1 2 L such that
lim (i) = f1g:
Proof. Let A; B 2 L be near the identity; then A = exp(); B = exp( ) for some ; in the Lie algebra of L. Therefore
[A; B ] = [1 + + 1 2 + :::; 1 + + 1 2 + :::] =
(1 + + 21 2 + :::)(1 + + 12 2 + :::)(1 ? + 21 2 ? :::)(1 ? + 21 2 ? :::)
By opening the brackets we see that the linear term in the commutator [A; B ] is zero
and each term in the resulting innite series involves both nonzero powers of and
of . Therefore
k1 ? [A; B ]k C k1 ? Ak k1 ? B k:
Therefore, by induction, if Bi+1 := [A; Bi], B1 = B , then
k1 ? [A; Bi]k C ik1 ? Aki k1 ? B k:
By taking be such that k1 ? Ak < C for all A 2 , we conclude that
lim k1 ? Bik = 0:
Lemma 126. (Zassenhaus lemma) Let ? L be a discrete subgroup. Then the set
? \ generates a nilpotent subgroup.
Proof. There exists a neighborhood V of 1 in L such that V \ ? = f1g; it follows
from the above lemma that all the iterated commutators of the elements of ? \ converge to 1. It thus follows that the iterated m-fold commutators of the elements
in ? \ are trivial for all suciently large m. Therefore the set ? \ generates a
nilpotent subgroup in ?.
The nite subgroup F L is clearly discrete, therefore the subgroup hF \ i is
nilpotent. Then log(F \ ) generates a nilpotent subalgebra in the Lie algebra of
L. Since F is nite, it is also compact, hence, up to conjugation, it is contained in
the maximal compact subgroup K = O(n) GL(n; R ) = L. The only nilpotent Lie
subalgebras of K are abelian subalgebras, therefore the subgroup F 0 generated by
F \ is abelian. It remains to estimate the index. Let U be a neighborhood
of 1 in K such that U U ?1 (i.e. products of pairs of elements xy?1, x; y 2 U ,
belong to ). Let q denote V ol(K )=V ol(U ), where V ol is induced by the biinvariant
Riemannian metric on K .
Lemma 127. jF : F 0j q.
Proof. Let x1 ; :::; xq+1 2 F . Then
V ol(xi U ) = (q + 1)V ol(U ) > V ol(K ):
Hence there are i 6= j such that xi U \ xj U 6= ;. Thus x?j 1 xi 2 UU ?1 . Hence
x?j 1 xi 2 F 0.
This also proves Jordan's theorem.
6 Quasiconformal mappings
Denition 128. Suppose that D; D0 are domains in R n , n 2, and let f : D ! D0
be a homeomorphism. The mapping f is called quasiconformal if the function
fd(f (z); f (x)) : d(x; z) = rg
Hf (x) = lim sup sup
r!0 inf fd(f (z ); f (x)) : d(x; z ) = rg
is bounded from above in X . A quasiconformal mapping is called K -quasiconformal
if the function Hf is bounded from above by K a.e. in X .
The notion of quasiconformality does not work well in the case when the domain
and range are 1-dimensional. It is replaced by
Denition 129. Let C S1 be a closed subset. A homeomorphism f : C ! f (C ) S1 is called quasimoebius if there exists a constant K so that for any quadruple of
mutually distinct points x; y; z; w 2 S1 their cross-ratio satises the inequality
K ?1 (jf (x) (:jxf (:yy) :: fz (:zw) j:)f (w)j) K
where (t) = j log(t)j + 1.
Note that if f is K -quasimoebius then for any pair of Moebius transformations ; the composition f is again K -quasimoebius.
Recall that a mapping f : S n ! S n is Moebius if it is a composition of inversions.
Equivalently, f is Moebius i it is the extension of an isometry H n+1 ! H n+1 . Yet
another equivalent denition: Moebius mappings are the homeomorphisms of S n
which preserve the cross-ratio.
Here is another (analytical) description of quasiconformal mappings. A homeomorphism f : D ! D0 is called quasiconformal if it has distributional partial derivatives
in Lnloc(D) and the ratio
Rf (x) := kf 0(x)k=jJf (x)j1=n
is uniformly bounded from above a.e. in D. Here kf 0(x)k is the operator norm of
the derivative f 0(x) of f at x. The essential supremum of Rf (x) in D is denoted
by KO (f ) and is called the outer dilatation of f . Let us compare Hf (x) and Rf (x).
Clearly it is enough to consider positive-denite diagonal matrices f 0(x). Let be
the maximal eigenvalue of f 0(x) and be the minimal eigenvalue. Then kf 0(x)k = ,
Hf (x) = = and
Rf (x) Hf (x) Rf (x)n:
Two denitions of quasiconformality (using Hf and Rf ) coincide (see for instance
[48], [58], [56]) and we have:
KO (f ) K (f ) KO (f )n:
In particular, quasiconformal mappings are dierentiable a.e. and their derivative is
a.e. invertible.
Note that quasiconformality of mappings and the coecients of quasiconformality K (f ), KO (f ) do not change if instead of the Euclidean metric we consider a
conformally-Euclidean metric in D. This allows us to dene quasiconformal mappings on domains in S n, via the stereographic projection.
1. If f : D ! D0 is a conformal homeomorphism the f is quasiconformal. Indeed,
conformality of f means that f 0(x) is a similarity matrix for each x, hence Rf (x) = 1
for each x. In particular, Moebius transformations are quasiconformal.
2. Suppose that the homeomorphism f extends to a dieomorphism D ! D0 and
the closure D is compact. Then f is quasiconformal.
3. Compositions and inverses of quasiconformal mappings are quasiconformal.
Moreover, KO (f g) KO (f )KO (g), KO (f ) = KO (f ?1).
Theorem 131. (Liouville's theorem for quasiconformal mappings, see [48], [44].)
Suppose that f : S n ! S n, n 2, is a quasiconformal mapping which is conformal
a.e., i.e. for a.e. x 2 S n, Rf (x) = 1. Then f is Moebius.
Conformality of f at x means that the derivative f 0(x) exists and is a similarity
matrix (i.e. is the product of a scalar and an orthogonal matrix).
Historical remark. Quasiconformal mappings for n = 2 were introduced in 1920s by Groetch as a generalization of conformal mappings. Quasiconformal mappings
in higher dimensions were introduced by Lavrentiev in 1930-s for the purposes of
application to hydrodynamics. The discovery of relation between quasi-isometries
of hyperbolic spaces and quasiconformal mappings was made by Efremovich and
Tihomirova [16] and Mostow [44] in 1960-s.
Theorem 132. Suppose that f : H n ! H n is a (k; c)-quasi-isometry. Then the
homeomorphic extension h = f1 of f to @1 H n constructed in Theorem 78 is a quasiconformal homeomorphism (if n 3) and quasimoebius (if n = 2).
Proof. I will verify quasiconformality of h for n 3 and will leave the case n = 2 to
the reader. According to the denition, it is enough to verify quasiconformality at
each particular point x with uniform estimates on the function Hh(x). Thus, after
composing h with Moebius transformations, we can take x = 0 = h(x), h(1) = 1,
where we consider the upper half-space model of H n .
Take a Euclidean sphere Sr (0) in R n?1 with the center at the origin. This sphere
is the ideal boundary of a hyperplane Pr H n which is orthogonal to the vertical
geodesic L H n , connecting 0 and 1. Let xr = L \ Pr . Let L : H n ! L be the
nearest point projection. The hyperplane Pr can be characterized by the following
equivalent properties:
Pr = fw 2 H n : L (w) = xr g
Pr = fw 2 H n : d(w; xr ) = d(w; L)g:
Since quasi-isometric images of geodesics in H n are uniformly close to geodesics, we
conclude that
diam[L (f (Pr ))] Const
where Const depends only on the quasi-isometry constants of f . The projection L
extends naturally to @1H n . We conclude:
diam[L (h(Sr (0)))] Const:
Thus h(Sr (0)) is contained in a spherical shell
fz 2 R n?1 : 1 jzj 2 g
where log[1=2 ] Const. This implies that the function Hh(0) is bounded from
above by K := exp(Const). We conclude that the mapping h is K -quasiconformal.
7 Quasi-isometries of nonuniform lattices in H n .
Recall that a lattice in a Lie group G (with nitely many components) is a discrete
subgroup ? such that the quotient ? n G has nite volume. Here the left-invariant
volume form on G is dened by taking a Riemannian metric on G which is leftinvariant under G and right-invariant under K , the maximal compact subgroup of G.
Thus, if X := G=K , then this quotient manifold has a Riemannian metric which is
(left) invariant under G. Hence, ? is a lattice i ? acts on X properly discontinuously
so that vol(? n X ) is nite. Note that the action of ? on X is a priori not free. A
lattice is called uniform if ? n X is compact and ? is called nonuniform otherwise.
Note that each lattice is nitely-generated (this is not at all obvious), in the case
of the hyperbolic spaces nite generation follows from the thick-thin decomposition
above. Thus, if ? is a lattice, then it contains a torsion-free subgroup of nite index
(Selberg lemma). In particular, if ? is a nonuniform lattice in H 2 then ? is virtually
free of rank 2.
Example 133. Consider the subgroups ?1 := SL(2; Z) SL(2; R), ?2 := SL(2; Z[i])
SL(2; C ). Then ?1 ; ?2 are nonuniform lattices. Here Z[i] is the ring of Gaussian
integers, i.e. elements of Z iZ. The discreteness of ?1 ; ?2 is clear, but niteness of
volume requires a proof.
Let's show that ?i, i = 1; 2, are not uniform. I will give the proof in the case of ?1,
the case of ?2 is similar.
Note that the symmetric space SL(2; R)=SO(2) is the hyperbolic plane. I will use
the upper half-plane model of H 2 . The group ?1 contains the upper triangular matrix
A := 10 11 :
This matrix acts on H 2 by the parabolic translation : z 7! z + 1 (of innite order).
Consider the points z := (0; y) 2 H 2 with y ! 1. Then the length of the geodesic
segment z (z) tends to zero as y diverges to innity. Hence the quotient S := ?1 n H 2
has injectivity radius unbounded from below (from zero), hence S is not compact.
More generally, lattices in a Lie group can be constructed as follows: let h : G !
GL(N; R ) be a homomorphism with nite kernel. Let ? := h?1(GL(N; Z)). Then ?
is an arithmetic lattice in G.
Recall that a horoball in H n (in the unit ball model) is a domain bounded by a
round Euclidean ball B H n , whose boundary is tangent to the boundary of H n
in a single point (called the center or footpoint of the horoball). The boundary of a
horoball in H n is called a horosphere. In the upper half-space model, the horospheres
with the footpoint 1 are horizontal hyperplanes
f(x1; :::; xn?1; t) : (x1 ; :::; xn?1) 2 R n?1 g;
where t is a positive constant.
Theorem 134. (Thick-thin decomposition) Suppose that ? is a nonuniform lattice in
Isom(H n ). Then there exists an (innite) collection C of pairwise disjoint horoballs
C := fBj ; j 2 J g, which is invariant under ?, so that (H n n [j Bj )=? is compact.
The quotient (H n n [j Bj )=? is called the thick part of M = H n =? and its (noncompact) complement in M is called thin part of M .
Hn / Γ
T xR +
Figure 14: Truncated hyperbolic space and thick-thin decomposition.
The complement := H n n [j Bj is called a truncated hyperbolic space. Note that
the stabilizer ?j of each horosphere @Bj acts on this horosphere cocompactly with the
quotient Tj := @Bj =?j . The quotient Bj =?j is naturally homeomorphic to Tj R + ,
this product decomposition is inherited from the foliation of Bj by the horospheres
with the common footpoint j and the geodesic rays asymptotic to j . In the case ?
is torsion-free, orientation preserving and n = 3, the quotients Tj are 2-tori.
Denition 135. Let ? G be a subgroup. The commensurator of ? in G, denoted
Comm(?) consists of all g 2 G such that the groups g?g?1 and ? are commensurable,
i.e. their intersection has nite index in the both groups.
Here is an example of the commensurator: let ? := SL(2; Z[i]) SL(2; C ). Then
the commensurator of ? is the group SL(2; Q (i)). In particular, the group Comm(?)
is nondiscrete in this case. There is a theorem of Margulis, which states that a lattice
in G is arithmetic if and only if its commensurator is discrete. We note that each
element g 2 Comm(?) determines a quasi-isometry f : ? ! ?. Indeed, the Hausdor
distance between ? and g?g?1 is nite. Hence the quasi-isometry f is given by
composing g : ? ! g?g?1 with the nearest-point projection to ?.
The main goal of the remainder of the course is to prove the following
Theorem 136. (R. Schwartz [49].) Let ? Isom(H n ) is a nonuniform lattice,
n 3. Then:
(a) For each quasi-isometry f : ? ! ? there exists 2 Comm(?) which is within
nite distance from f . The distance between these maps depends only on ? and on
the quasi-isometry constants of f .
(b) Suppose that ?; ?0 are non-uniform lattices which are quasi-isometric to each
other. Then there exists an isometry g 2 Isom(H n ) such that the groups ?0 and g?g ?1
are commensurable.
(c) Suppose that ?0 is a nitely-generated group which is quasi-isometric to a
nonuniform lattice ? above. Then the groups ?; ?0 are weakly commensurable, i.e.
there exists a nite normal subgroup F ?0 such that the groups ?; ?0=F contain
isomorphic subgroups of nite index.
The above theorem fails in the case of the hyperbolic plane (except for the last
7.1 Coarse topology of truncated hyperbolic spaces
On each truncated hyperbolic space we put the path-metric which is induced by
the restriction of the Riemannian metric of H n to . This metric is invariant under
? and since the quotient =? is compact, is quasi-isometric to the group ?. Note
that the restriction of this metric to each peripheral horosphere is a at metric.
The following lemma is the key for distinguishing the case of the hyperbolic plane
from the higher-dimensional hyperbolic spaces (of dimension 3):
Lemma 137. Let is a truncated hyperbolic space of dimension 3. Then each
peripheral horosphere does not coarsely separate .
Proof. Let R < 1 and let B be the horoball bounded by . Then the union of
NR () [ B is a horoball B 0 in H n (where the metric neighborhood is taken in H n ).
The horoball B 0 does not separate H n . Therefore, for each pair of points x; y 2 n B 0 ,
there exists a PL path p connecting them within H n n B 0 . If the path p is entirely
contained in , we are done. Otherwise, it can be subdivided into nitely many
subpaths, each of which is either contained in or connects a pair of points on the
boundary of a complementary horoball Bj H n n . The intersection of NR(
) (B 0)
with j = @Bj is a metric ball in the Euclidean space j (here NR(
) is the metric
neighborhood taken within ). Note that a metric ball does not separate R n?1 ,
provided that n ? 1 2. Thus we can replace pj = p \ Bj with a new path p0j which
connects the end-points of pj within the complement j n NR(
) (B 0). By making these
replacements for each j we get a path connecting x to y within n NR(
) ().
Let now ; 0 be truncated hyperbolic spaces (of the same dimension), f : ! 0
be a quasi-isometry. Let be a peripheral horosphere of , consider its image f ()
in 0 .
Proposition 138. There exists a peripheral horosphere 0 @ 0 which is within
nite Hausdor distance from f ().
Proof. Note that , being isometric to R n?1 , has bounded geometry and is uniformly
contractible. Therefore, according to Theorem 58, f () coarsely separates H n ; however it cannot coarsely separate 0 , since f is a quasi-isometry and does not coarsely
separate . Let R < 1 be such that NR (f ()) separates H n into (two) deep components X1; X2 . Suppose that for each complementary horoball Bj0 of 0 (bounded by
the horosphere 0j ),
N?R (Bj0 ) := Bj0 n NR(0j ) X1 :
Then the entire 0 is contained in NR (f ()). It follows that f () does not coarsely
separate H n , a contradiction. Thus there are complementary horoballs B10 ; B20 for 0
such that N?R (B10 ) X1; N?R (B20 ) X2 . If either 1 or 2 is not contained in
Nr (f ()) for some r then f () coarsely separates 0. Thus we found a horosphere
0 := 01 such that
0 Nr (f ()):
Our goal is to show that f () N (0) for some < 1. The nearest-point projection
0 ! f () denes a quasi-isometric embedding h : 0 ! . However Lemma 71
proves that a quasi-isometric embedding between two Euclidean spaces of the same
dimension is a quasi-isometry. Thus there exists < 1 such that f () N (0 ).
Lemma 139. dHaus(f (); 0)) r, where r is independent of .
Proof. The proof is by inspection of the arguments in the proof of the previous proposition. First of all, the constant R depends only on the quasi-isometry constants of the
mapping f and the uniform geometry/uniform contractibility bounds for R n?1 and
H n . The inradii of the shallow complementary components of NR (f ()) again depend
only on the above data. Therefore there exists a uniform constant r such that 1 of 2
is contained in Nr (f ()). Finally, the upper bound on such that N(Image(h)) = 0
(coming from Lemma 71) again depends only on the quasi-isometry constants of the
projection h : 0 ! .
7.2 Hyperbolic extension
The main result of this section is
Theorem 140. f admits a quasi-isometric extension f~ : H n ! H n .
Proof. We will construct the extension f~ into each complementary horoball B H n n . Without loss of generality we can use the upper half-space model of H n so
that the horoballs B and B 0 are both given by
f(x1; :::; xn?1 ; 1) : (x1 ; :::; xn?1 ) 2 R n?1 g:
We will also assume that f () 0 . For each vertical geodesic ray (t); t 2 R + , in Bj
we dene the geodesic ray 0 (t) to be the vertical geodesic ray in B 0 with the initial
point f ((0)). This gives the extension of f into B :
f~(rho(t)) = 0 (t):
Let's verify that this extension is coarsely Lipschitz. Let x and y 2 B be points
within the (hyperbolic) distance 1. By the triangle inequality it suces to consider
the case when x; y belong to the same horosphere Ht (of the Euclidean height t)
with the footpoint at 1 (if x and y belong to the same vertical ray we clearly get
d(f~(x); f~(y)) = d(x; y)). Note that the distance from x to y along the horosphere H
does not exceed , which is independent of t. Let x; y denote the points in such that
x; y belong to the vertical rays in Bj with the initial points x; y respectively. Then
d(x; y) = tdHt (x; y) t:
Hence, since f is (L; A)-coarse Lipshitz,
d(f (x); f (y)) Lt + A:
It follows that
d(f~(x); f~(y)) L + A=t L + A:
This proves that the extension f~ is coarse Lipschitz in the horoball B . Since the
coarse Lipschitz is a local property, the mapping f~ is coarse Lipschitz on H n . The
same argument applies to the hyperbolic extension f~0 of the coarse inverse f 0 to the
mapping f . It is clear that the mapping f~ f~0 and f~0 f~ have bounded displacement.
Thus f~ is a quasi-isometry.
Since f~ is a quasi-isometry of H n , it admits a quasiconformal extension h : @1H n !
@1H n . Let ; 0 denote the sets of the footpoints of the peripheral horospheres of
; 0 respectively. It is clear that h() = 0.
7.3 Zooming in
Our main goal is to show that the mapping h constructed in the previous section is
Moebius. By the Liouville's theorem for quasiconformal mappings, h is Moebius i
for a.e. point 2 S n?1, the derivative of h at is a similarity. We will be working
with the upper half-space of the hyperbolic space H n .
Proposition 141. Suppose that h is not Moebius. Then there exists a quasi-isometry
F : ! 0 whose extension to the sphere at innity is a linear map which is not a
Proof. Since h is dierentiable a.e. and is not Moebius, there exists a point 2
S n?1 n such that Dh( ) exists, is invertible but is not a similarity. By pre- and
post-composing f with isometries of H n we can assume that = 0 = h( ). Let
L H n denote the vertical geodesic through . Since is not a footpoint of a
complementary horoball to , there exists a sequence of points xj 2 L \ which
converges to . For each t 2 R + dene t : z 7! tz, a hyperbolic translation along L.
Let tj be such that tj (x1 ) = xj . Set
f~j := t?j 1 f~ tj ;
the quasiconformal extensions of these mappings to @1H n are given by
hj (z) = h(ttj z) :
By the denition of dierentiability,
lim h
j !1 j
= A = Dh(0);
where the convergence is uniform on compacts in R n?1 . Let's verify that the sequence
of quasi-isometries f~j subconverges to a quasi-isometry of H n . Indeed, since the quasiisometry constants of all f~j are the same, it suces to show that ff~j (x1)g is a bounded
sequence in H n . Let L1 ; L2 denote a pair of distinct geodesics in H n through x1 , so
that the point 1 does not belong to L1 [ L2 . Then the quasi-geodesics f~j (Li) are
within distance C from geodesics L1j ; L2j in H n . Note that the geodesics L1j ; L2j
subconverge to geodesics in H n with distinct end-ponts (since the mapping A is 11). The point f~j (x1 ) is within distance C from L1j ; L2j . If the sequence f~j (x1 ) is
unbounded, we get that L1j ; L2j subconverge to geodesics with a common end-point
at innity. Contradiction.
We thus pass to a subsequence such that f~j converges to a quasi-isometry f1 :
H ! H n . Note however that f1 in general does not send to 0 . Recall that
=?; 0 =?0 are compact. Therefore there exist sequences j 2 ?; j0 2 ?0 such that
j (xj ), j0 (f~(xj )) belong to a compact subset of H n . Hence the sequences j :=
t?j 1 j?1; j0 := t?j 1 j0?1 is precompact in Isom(H n ) and therefore they subconverge
to isometries 1; 10 2 Isom(H n ). Set
j := t?j 1
= t?j 1 j?1
= j ;
0j := t?j 1
0 = j0 0 ;
then f~j : j ! 0j . On the other hand, the sets j ; 0j subconverge to the sets
; 10 0 and f~1 is a quasi-isometry between 1
and 10 0. Since 1
10 0 are isometric copies of and 0 the assertion follows.
The situation when we have a linear mapping (which is not a similarity) mapping
to 0 seems at the rst glance impossible. Here however is an example:
0 := SL(2; Z[p?2]). Then = Q (i); 0 =
Q ( ?2).
Dene a real linear mapping A : C ! C by sending 1 to 1 and i to ?2. Then A
is not a similarity, however A() = 0).
Thus to get a contradiction we have to exploit the fact that the linear map in
question is quasiconformal extension of an isometry between truncated hyperbolic
7.4 Inverted linear mappings
Let A : R n?1 ! R n?1 be an (invertible) linear mapping and I be the inversion in the
unit sphere about the origin, i.e.
I (x) = jxxj2 :
Denition 143. An inverted linear map is the composition h := I A I , i.e.
jxj2 A(x):
h(x) = jAx
jxj2 is asymptotically constant, i.e. the gradient
Lemma 144. The function (x) = jAx
of converges to zero as jxj ! 1.
Proof. The function is a rational function of degree zero, hence its gradient is a
rational vector-function of degree ?1.
Note however that is not a constant mapping unless A is a similarity. Hence h is
linear i A is a similarity.
Corollary 145. Let R be a xed positive real number, xj 2 R n?1 , jxj j ! 1. Then
the function h(x ? xj ) ? h(xj ) converges (uniformly on the R-ball B (0; R)) to a linear
function, as j ! 1.
We would like to strengthen the assertion that is not constant (unless A is a
similarity). Let G be a discrete group of Euclidean isometries acting cocompactly on
R n?1 . Fix a G-orbit Gx, for some x 2 R n?1 .
Lemma 146. There exists a number R and a sequence of points xj 2 Gx diverging
to innity such that the restrictions to B (xj ; R) \ Gx are not constant for all j .
Proof. Let P be a compact fundamental domain for G, containing x. Let denote
diam(P ). Pick any R 4. Then B (x; R) contains all images of P under G which
are adjacent to P . Suppose that the sequence xj as above does not exist. This means
that there exists r < 1 such that the restriction of to B (xj ; R) is constant for each
xj 2 Gx n B (0; r). It follows that the function is actually constant on Gx n B (0; r).
Note that the set
fy=jyj; y 2 Gx n B (0; r)g
is dense in the unit sphere. Since (y=jyj) = (y) it follows that is a constant
We now return to the discussion of quasi-isometries.
Let A be an invertible linear mapping (which is not a similarity) constructed in the
previous section, by composing A with Euclidean translations we can assume that 0 =
A(0) belongs to both and 0. Then 1 = I (0) belongs to both I () and I (0). To
simplify the notation we replace ; ; 0; 0 with I (); I (
); I (0); I (
0) respectively.
Then the truncated hyperbolic spaces ; 0 have complementary horoballs B1; B10 .
Given x 2 R n?1 dene h (x) := h(?1x). Let ?1, ?01 be the stabilizers of 1 in
?; ?0 respectively. Without loss of generality we can assume that 1 2 ; 0, hence
?1, ?01 act cocompactly (by Euclidean isometries) on R n?1 .
Lemma 147. (Scattering lemma) Suppose that A is not a similarity. Then for each
x 2 R n?1 , h (x) is not contained in nitely many ?01-orbits.
Proof. Let xj = j x 2 ?1x and R < 1 be as in Lemma 146, where G = ?1 . We
have a sequence of maps j0 2 ?01 such that j0 h(xj ) is relatively compact in R n?1 .
Then the mapping hjB (xj ; R) \ ?1x is not linear for each j (Lemma 146). However
the sequence of maps
j0 h j := hj
converges to an ane mapping h1 on B (x; R) (since h is asymptotically linear). We
conclude that the union
[j=1hj (?1x \ B (x; R))
is an innite set.
Theorem 148. Suppose that h is an inverted linear map which is not a similarity.
Then h admits no quasi-isometric extension ! 0 .
Proof. Let x be a footpoint of a complementary horoball B to , B 6= B1 . Then,
by the scattering lemma, h (x) is not contained in a nite union of ?01-orbits. Let
j 2 ?1 be a sequence such that the ?01-orbits of the points x0j := hj (x) are all
distinct. Let Bj0 denote the complementary horoball to 0 whose footpoint is x0j . It
follows that the Euclidean diameters of the complementary horoballs Bj0 converge to
zero. Let Bj be the complementary horoball to whose footpoint is j x. Then
dist(Bj ; B1) = dist(B1; B1) = ? log(diam(B1 )) = D;
dist(Bj0 ; B10 ) = ? log(diam(Bj0 )) ! 1:
If f : ! 0 is an (L; A) quasi-isometry whose quasiconformal extension is h then
dist(Bj0 ; B10 ) L(D + Const) + A:
Therefore we have proven
Theorem 149. Suppose that f : ! 0 is a quasi-isometry of truncated hyperbolic
spaces. Then f admits an (unique) extension to S n?1 which is Moebius.
7.5 Proof of Theorem 136
(a) For each quasi-isometry f : ? ! ? there exists 2 Comm(?) which is within
nite distance from f .
Proof. The quasi-isometry f extends to a quasi-isometry of the hyperbolic space
f~ : H n ! H n . The latter quasi-isometry extends to a quasiconformal mapping
h : @1H n ! @1H n . This quasiconformal mapping has to be Moebius according to
Theorem 149. Therefore f~ is within nite distance from an isometry of H n (which is
an isometric extension of h to H n ). It remains to verify that belongs to Comm(?).
We note that sends the peripheral horospheres of within (uniformly) bounded
distance of peripheral horospheres of . The same is of course true for all mappings
of the group ?0 := ? ?1. Thus, if 0 2 ?0 xes a point in (a footpoint of a
peripheral horosphere ), then it has to preserve : Otherwise by iterating 0 we
would get a contradiction. The same applies if 0 (1) = 2, where 1; 2 2 are in
the same ?-orbit: 0(1 ) = 2 , where i is the footpoint of the peripheral horosphere
i . Therefore we modify as follows: Pick peripheral horospheres 1 ; :::; m with
disjoint ?-orbits and for each 0 2 ?0 such that 0(i) is not contained in , we replace
the peripheral horosphere parallel to 0(i ) with the horosphere 0(i ). As the result
we get a new truncated hyperbolic space 0 which is invariant under both ? and ?0.
Observe now that the group ?00 generated by ?; ?0 acts on properly discontinuously
and cocompactly. Therefore the projections
=? ! =?00; =?0 ! =?00
are nite-to-one maps. It follows that j?00 : ?j and j?00 : ?0j are both nite. Therefore
the groups ?; ?0 are commensurable and 2 Comm(?).
To prove a uniform bound on the distance d(f; gj) we notice that f and g have
the same extension to the sphere at innity. Therefore, by 79, the distance d(f; g) is
uniformly bounded in terms of the quasi-isometry constants of f .
(b) Suppose that ?; ?0 are non-uniform lattices which are quasi-isometric to each
other. Then there exists an isometry g 2 Isom(H n ) such that the groups ?0 and
g?g?1 are commensurable.
Proof. The proof is analogous to (a): The quasi-isometry f is within nite distance
from an isometry g. Then the elements of the group g?g?1 have the property that
they map the truncated hyperbolic space 0 of ?0 within (uniformly) bounded distance
from . Therefore we can modify 0 to get a truncated hyperbolic space 00 which is
invariant under both ?0 and g?g?1. The rest of the argument is the same as for (a).
(c) Suppose that ?0 is a nitely-generated group which is quasi-isometric to a
nonuniform lattice ? above. Then the groups ?; ?0 are weakly commensurable, i.e.
there exists a nite normal subgroup K ?0 such that the groups ?; ?0=K contain
isomorphic subgroups of nite index.
Proof. Let f : ? ! ?0 be a quasi-isometry and let f 0 : ?0 ! ? be its quasi-inverse.
We dene the set of uniform quasi-isometries
?0f := f 0 ?0 f
of the truncated hyperbolic space of the groups ?. Each quasi-isometry g 2 ?0f
is within a (uniformly) bounded distance from a quasi-isometry of induced by an
element g of Comm(?). We get a map
: 0 7! f 0 0 f 7! (f 0 0 f ) 2 Comm(?):
I claim that this map is a homomorphism with nite kernel. Let's rst check that
this map is a homomorphism:
d(f 0 10 20 f; f 0 10 f f 0 20 f ) < 1;
hence the above quasi-isometries have the same Moebius extension to the sphere at
innity. Suppose that 0 2 Ker( ). Then the quasi-isometry f 0 0 f has a bounded
displacement on . Since the family of quasi-isometries
ff 0 0 f; 0 2 K g
has uniformly bounded quasi-isometry constants, it follows that they have uniformly
bounded displacement. Hence the elements 0 2 K have uniformly bounded displacement as well. Therefore the normal subgroup K is nite. The rest of the argument
is the same as for (a) and (b): The groups ?; ?00 := (?0 ) Comm(?) act on a
truncated hyperbolic space 0 which is within nite distance from . Therefore the
groups ?00; ? are commensurable.
8 A quasi-survey of QI rigidity
Given a group G one denes the abstract commensurator Comm(G) as follows. The
elements of Comm(G) are equivalence classes of isomorphisms between nite index
subgroups of G. Two such isomorphisms : G1 ! G2; : G01 ! G02 are equivalent if their restrictions to further nite index subgroups G001 ! G002 are equal. The
composition and the inverse are dened in the obvious way, making Comm(G) a
Let X be a metric space or a group G. Call X strongly QI rigid if each (L; A)quasi-isometry f : X ! X is within nite distance from an isometry : X ! X or
an element of Comm(G) and moreover d(f; ) C (L; A).
Call a group G QI rigid if any group G0 which is quasi-isometric to G is actually
weakly commensurable to G.
Call a class of groups G QI rigid if each group G which is quasi-isometric to a
member of G is actually weakly commensurable to a member of G .
Theorem 150. (Pansu, [45]) Let X be a quaternionic hyperbolic space H nH (n 2)
or the hyperbolic Cayley plane H 2Ca . Then X is strongly QI rigid.
Theorem 151. (Tukia, [54] for the real-hyperbolic spaces H n ; n 3 and Chow [11]
for complex-hyperbolic spaces H nC ; n 2). Let X be a symmetric space of negative
curvature which is not the hyperbolic plane H 2 . Then the class of uniform lattices in
X is QI rigid.
Theorem 152. (Combination of the work by Gabai [21], Casson and Jungreis [10]
and Tukia [55]) The fundamental groups of closed hyperbolic surfaces are QI rigid.
Theorem 153. (Stallings, [51]) Each nonabelian free group is QI rigid. Thus each
nonuniform lattice in H 2 is QI rigid.
Theorem 154. (Kleiner, Leeb, [37]) Let X be a symmetric space of nonpositive
curvature such that each deRham factor of X is a symmetric space of rank 2.
Then X is strongly QI rigid.
Theorem 155. (Kleiner, Leeb, [37]) Let X be a Euclidean building such that each
deRham factor of X is a Euclidean building of rank 2. Then X is strongly QI rigid.
Theorem 156. (Kleiner, Leeb, [37]) Let X be a symmetric space of nonpositive
curvature without Euclidean deRham factors. Then the class of uniform lattices in X
is QI rigid.
Theorem 157. (Eskin, [17]) Let X be an irreducible symmetric space of nonpositive
curvature of rank 2. Then each nonuniform lattice in X is strongly QI rigid and
QI rigid.
Theorem 158. (Kleiner, Leeb, [38]) Suppose that ? is a nitely-generated groups
which is quasi-isometric to a Lie grooup G with the nilpotent radical N and semisimple
quotient G=N = H . Then ? ts into a short exact sequence
1 ! K ! ? ! Q ! 1;
where K is quasi-isometric to N and Q is weakly commensurable to a uniform lattice
in H .
Problem 159. Prove an analogue of the above theorem for all Lie groups G (without
assuming that the sol-radical of G is nilpotent).
Theorem 160. (Bourdon, Pajot [4]) Let X be a thick hyperbolic building of rank 2
with right-angled fundamental polygon and whose links are complete bipartite graphs.
Then X is strongly QI rigid.
Problem 161. Construct an example of a hyperbolic group with Menger curve
boundary, which is QI rigid.
Problem 162. Let G be a random k-generated group, k 2. Is G QI rigid?
Randomness can be dened for instance as follows. Consider the set B (n) of presentations
hx1 ; :::; xk jR1; :::; Rl i
where the total length of the words R1; :::; Rl is n. Then a class C of k-generated
groups is said to consist of random groups if
jB (n) \ C j = 1:
n!1 jB (n)j
Here is another notion of randomness: x the number l of relators, assume that all
relators have the same length n; this denes a class of presentations S (k; l; n). Then
lim jS (k; l; n) \ C j = 1:
n!1 jS (k; l; n)j
Theorem 163. (Kapovich, Kleiner, [32]) There is a 3-dimensional hyperbolic group
which is strongly QI rigid.
Theorem 164. Each nitely generated abelian group is QI rigid.
Theorem 165. (Farb, Mosher, [18]) Each solvable Baumslag-Solitar group
BS (1; q) = hx; y : xyx?1 = yq i
is QI rigid.
Theorem 166. (Whyte, [60]) All non-solvable Baumslag-Solitar groups
BS (p; q) = hx; y : xypx?1 = yq i;
jpj =6 1; jqj =6 1 are QI to each other.
Theorem 167. (Farb, Mosher, [19]) The class of non-polycyclic abelian-by-cyclic
groups, i.e. groups ? which t into an exact sequence
with A is an abelian group, is QI rigid.
Theorem 168. (Dyubina, [14]) The class of nitely generated solvable groups is not
QI rigid.
Problem 169. Is the class of nitely generated polycyclic groups QI rigid?
Example 170. Let S be a closed hyperbolic surface, M is the unit tangent bundle
over S . Then we have an exact sequence
1 ! Z ! G = 1 (M ) ! Q := 1 (S ) ! 1:
This sequence does not split even after passage to a nite index subgroup in G, hence
G is not weakly commensurable with Q Z. However G is quasi-isometric to Q Z.
More generally, if Q is a hyperbolic group, then all groups G which t into an exat
1 ! Z ! G ! Q ! 1;
are quasi-isometric.
Example 171. There are uniform lattices in H n ; n 3, which are not weakly commensurable.
Indeed, take an arithmetic and a nonarithmetic lattice in H n .
Example 172. The product of free groups G = Fn Fm , (n; m 2) is not QI rigid.
Proof. The group G acts discretely, cocompactly, isometrically on the product of
simplicial trees X := T T 0 . However there are examples [61], [8], of groups G0 acting
discretely, cocompactly, isometrically on X so that G0 contains no proper nite index
subgroups. Then G is quasi-isometric to G0 but these groups are clearly not weakly
Problem 173. Suppose that G is (a) a Mapping Class group, (b) Out(Fn), (c) an
Artin group, (d) a Coxeter group, (e) the fundamemtal group of one of the negatively
curved manifolds constructed in [30], (f) 1 (N ), where N is a nite covering of the
product of a hyperbolic surface by itself S S , ramied over the diagonal (S S ).
Is G QI rigid?
One has to exclude, of course, Artin and Coxeter groups which are commensurable
with the direct products of free groups.
Theorem 174. (Kapovich, Leeb, [36]) The class of fundamental groups G of 3dimensional Haken 3-manifolds, which are not Sol-manifolds1, is QI rigid.
Theorem 175. (Papasoglou, [46]) The class of nitely-presented groups which split
over Z is QI rigid. Moreover, quasi-isometries of 1-ended groups G preserve the JSJ
decomposition of G
Theorem 176. (Kapovich, Kleiner, Leeb, [35]) Quasi-isometries preserve deRham
decomposition of the universal covers of closed nonpositively curved Riemannian manifolds.
Problem 177. Are there nitely generated (amenable) groups which are quasiisometric but not bi-Lipschitz equivalent?
Problem 178. Suppose that G is a nitely-presented group. Does the topology of
the asymptotic cone of G depend on the scaling sequence/ultralter?
Theorem 179. (Gersten, [22] The cohomological dimension (over an arbitrary ring
R) is a QI invariant within the class of nitely-presented groups of type FP (over
I refer to [6] for the denitions of cohomological dimension and the type FP .
Theorem 180. (Shalom, [50]) The cohomological dimension (over Q ) of amenable
groups is a QI invariant.
Problem 181. Is the cohomological dimension of a group (over Q ) a QI invariant?
I.e. excluding
which are polycyclic but not nilpotent.
Recall that a group G has property (T) if each isometric ane action of G on
a Hilbert space has a global xed point, see [13] for more thorough discussion. In
particular, such groups cannot map onto Z.
Theorem 182. The property (T) is not a QI invariant.
Proof. This theorem should be probably attributed to S. Gersten and M. Ramachandran; the example below is a variation on the Raghunathan's example discussed in
Let ? be a hyperbolic group which satises property (T) and such that H 2(?; Z) 6=
0. To construct such a group, start for instance with an innite hyperbolic group F
satisfying Property (T) which has an aspherical presentation complex (see for instance
[1] for the existence of such groups). Then H 1(F; Z) = 0 (since F satises (T)), if
H 2(F; Z) = 0, add enough random relations to F , keeping the resulting groups F 0
hyperbolic, innite, 2-dimensional. Then H 1(F 0; Z) = 0 since F 0 also satises (T).
For large number of relators we get a group ? = F 0 such that (?) > 0 (the number
of relator is larger than the number of generators), hence H 2(?; Z) 6= 0. Now, pick a
nontrivial element ! 2 H 2(?; Z) and consider a central extension
with the extension class !. The cohomology class ! is bounded since ? is hyperbolic;
hence the groups G and G0 := Z ? are quasi-isometric, see [23]. The group G0 does
not satisfy (T), since it surjects to Z. On the other hand, the group G satises (T),
see [13, 2.c, Theorem 12].
However the following question is still open:
Problem 183. A group G is said to be a-T-menable if it admits a proper isometric
ane action on a Hilbert space. Is a-T-menability a QI invariant?
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