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

),

which is a contradiction.

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).