5CAT(0) spaces and groups

IV Topics in Geometric Group Theory

5.4 Alexandrov’s lemma
Alexandrov’s lemma is a lemma that enables us to glue CAT(0) space together
to obtain new examples.
Lemma
(Alexandrov’s lemma)
.
Suppose the triangles
1
= ∆(
x, y, z
1
) and
2
= ∆(
x, y, z
2
) in a metric space satisfy the CAT(0) condition, and
y
[
z
1
, z
2
].
z
1
z
2
x
y
Then ∆ = ∆(x, z
1
, z
2
) also satisfies the CAT(0) condition.
This is the basic result we need if we want to prove “gluing theorems” for
CAT(0) spaces.
Proof.
Consider
¯
1
and
¯
2
, which together form a Euclidean quadrilateral
¯
Q
with with vertices
¯x, ¯z
1
, ¯z
2
, ¯y
. We claim that then the interior angle at
¯y
is
180
. Suppose not, and it looked like this:
¯x
¯y
¯z
1
¯z
2
If not, there exists ¯p
i
[¯y, ¯z
i
] such that [¯p
1
, ¯p
2
] [¯x, ¯y] = {¯q} and ¯q 6= ¯y. Now
d(p
1
, p
2
) = d(p
2
, y) + d(y, p
2
)
= d(¯p
1
, ¯y) + d(¯y, ¯p
1
)
= d(¯p, ¯y) + d(¯y, ¯p
2
)
> d(¯p
1
, ¯q) + d(¯q, ¯p
2
)
d(p
1
, q) + d(q, p
2
)
d(p
1
, p
2
),
Thus, we know the right picture looks like this:
¯x
¯y
¯z
1
¯z
2
To obtain
¯
, we have to “push”
¯y
out so that the edge
¯z
1
¯z
2
is straight, while
keeping the lengths fixed. There is a natural map
π
:
¯
¯
Q
, and the lemma
follows by checking that for any a, b
¯
∆, we have
d(π(a), π(b)) d(a, b).
This is an easy case analysis (or is obvious).
A sample application is the following:
Proposition.
If
X
1
, X
2
are both locally compact, complete CAT(0) spaces and
Y
is isometric to closed, subspaces of both
X
1
and
X
2
. Then
X
1
Y
X
2
, equipped
with the induced length metric, is CAT(0).