5CAT(0) spaces and groups

IV Topics in Geometric Group Theory Gromov’s link condition is a criterion that makes it very easy to write down
interesting examples of non-positively-curved metric spaces.
Definition
(Euclidean cell complex)
.
A locally finite cell complex
X
is Euclidean
if every cell is isometric to a convex polyhedron in Euclidean space and the
attaching maps are isometries from the lower-dimensional cell to a face of the
new cell.
Such a complex
X
has a natural length metric which is proper and geodesic
by Hopf–Rinow. What we’d like to do is to come up with a condition that
ensures X is CAT(0).
Example.
The usual diagram for a torus gives an example of a Euclidean
complex.
Example. We can construct a sphere this way, just by taking a cube!
We know that
T
2
is non-positively curved (since it is flat), but the cube is
not, because by Cartan–Hadamard, it would be CAT(0), hence contractible, but
it is not.
Definition
.
Let
X
be a Euclidean complex, and let
v
be a vertex of
X
,
and let 0 < ε shortest 1-cell. Then the link of v is
Lk(v) = S
v
(ε) = {x X : d(x, v) = ε}.
Note that
Lk
(
V
) is a cell complex: the intersection of
Lk
(
v
) with a cell of
X
of dimension is a cell of dimension n 1.
Example. In the torus, there is only one vertex. The link looks like
1
.
Example.
If we take the corner of a cube, then
Lk
(
v
) is homeomorphic to
S
1
,
but it is a weird one, because it is made up of three right angles, not two.
How can we distinguish between these two
S
1
’s? Angle puts a metric on
Lk
(
v
). We can do this for general metric spaces, but we only need it for Euclidean
complexes, in which case there is not much to do.
Restricted to each cell, the link is just a part of a sphere, and it has a natural
spherical metric, which is a length metric. These then glue together to a length
metric on Lk(v). Note that this is not the same as the induced metric.
Example.
In the torus, the total length of the link is 2
π
, while that of the cube
is
3π
2
.
The important theorem, which we are not going to prove, is this:
Theorem
.
A Euclidean complex
X
is non-positively-
curved iff for every vertex v of X, Lk(v) is CAT(1).
Note that by definition, we only have to check the CAT(1) inequality on
triangles of perimeter < 2π.
Exercise. Check these in the case of the torus and the cube.
Thus, given a group, we can try to construct a space whose
π
1
is it, and then
put a Euclidean structure on it, then check Gromov’s link criterion.
In general, Gromov’s link condition might not be too useful, because we still
don’t know how to check if something is CAT(1)! However, it is very simple in
dimension 2.
Corollary.
If
X
is a 2-dimensional Euclidean complex, then for all vertices
v
,
Lk
(
v
) is a metric graph, and
X
is CAT(0) iff
Lk
(
v
) has no loop of length
<
2
π
for all v.
Example (Wise’s example). Consider the group
W = ha, b, s, t | [a, b] = 1, s
1
as = (ab)
2
, t
1
bt = (ab)
2
i.
Letting
c
=
ab
, it is easy to see that this is the
π
1
of the following Euclidean
complex:
b
a
c
s
t
This is metrized in the obvious way, where all edges have length 2 except the
black ones of length 1. To understand the links, we set
α = sin
1
1
4
, β = cos
1
1
4
.
Then the triangles each has angles 2α, β, β.
We show that
W
is non-positively curved, and then use this to show that
there is a homomorphism
W W
that is surjective but not injective. We say
W
is non-Hopfian. In particular, this will show that
W
is not linear, i.e. it is
not a matrix group.
To show that X is non-positively curved, we check the link condition:
a
a
c c
b
b
t
t
s
s
To check the link condition, we have to check that there are no loops of length
<
2
π
in
Lk
(
v
). Note that by trigonometric identities, we know
α
+
β
=
π
2
. So
we can just observe that everything is fine.
To see that W is non-Hopfian, we define
f : W W
a 7→ a
2
b 7→ b
2
s 7→ s
2
t 7→ t
We check that this is a well-defined homomorphism, and is surjective, since we
observe
a = sa
2
b
2
s
1
, b = ta
2
b
2
t
1
,
and so a, b, s, t im t. The non-trivial claim is that ker f 6= 0. Let
g = [scs
1
, tct
1
].
Note that
f(g) = [f(scs
1
), f(tct
1
)] = [sc
2
s
1
, tc
2
t
1
] = [a, b] = 1.
So the crucial claim is that
g 6
= 1. This is where geometry comes to the
rescue. For convenience, write
p
=
scs
1
, q
=
tct
1
. Consider the following local
geodesics in the two squares:
c
s
t
The left- and right-hand local geodesics represent p and q respectively. Then
g = p · q · ¯p · ¯q.
We claim that this represents
g
by a local geodesic. The only place that can
go wrong is the points where they are joined. By the proof of the Gromov link
condition, to check this, we check that the three “turns” at the vertex are all of
angle π.
a
a
c c
b
b
t
t
s
s
q
+
q
p
p
+
Here
p
+
is the point where
p
ends and
p
is the point where
p
starts, and
similarly for
q
. Moreover, the distance from
p
±
, q
±
to the top/bottom left/right
vertices is β. So we can just check and see that everything works.
Recall that every homotopy class of paths in
X
contains a unique locally
geodesic representative. Since the constant path is locally geodesic, we know
g 6= 1.
Definition
(Residually finite group)
.
A group
G
is residually finite if for every
g G \ {
1
}
, there is a homomorphism
ϕ
:
G finite group
such that
ϕ
(
g
)
6
= 0.
Theorem
(Mal’cev)
.
Every finitely generated linear subgroup (i.e. a subgroup
of GL
n
(C)) is resudially finite.
Proof sketch.
If the group is in fact a subgroup of
GL
n
(
Z
), then we just reduce
mod
p
for
p
0. To make it work over
GL
n
(
C
), we need a suitable version of
the Nullstellensatz.
Theorem
(Mal’cev)
.
Every finitely generated residually finite group is Hopfian.
Proof. Finding a proof is a fun exercise!
We know that Wise’s example is not Hopfian, hence not residually finite,
hence not a linear group.
Contrast this with the amazing theorem of Sela that all hyperbolic groups
are Hopfian. However, a major open question is whether all hyperbolic groups
are residually finite. On the other hand, it is known that not all hyperbolic
groups are not linear.
How are we supposed to think about residually finite groups?
Lemma
(Scott’s criterion)
.
Let
X
be a cell complex, and
G
=
π
1
X
. Then
G
is
residually finite if and only if the following holds:
Let
p
:
˜
X X
be the universal cover. For all compact subcomplexes
K
˜
X
,
there is a finite-sheeted cover
X
0
X
such that the natural covering map
p
0
:
˜
X X
0
is injective on K.
we have a map
f
:
K X
that may be complicated, and in particular is not
injective, then we might hope that there is some “finite resolution”
X
0
X
such
that
f
lifts to
X
0
, and the lift is injective. Of course, this is not always possible,
and a necessary condition for this to work is that the lift to the universal cover
is injective. If this is not true, then obviously no such resolution can exist. And
residual finiteness says if there is no obvious reason to fail, then it in fact does
not fail.