Suppose a group \Gamma is acting properly and cocompactly on a metric space X, by isometries.

(Understand: proper, cocompact, isometric action)

Claim

There are only finitely many conjugacy classes of the isotropy subgroups in \Gamma

Sketch

Since the action is cocompact, there is a compact set K whose translates cover X.

For each x \in K , choose small (enough) balls B_x .

How small do we want the balls?

Small enough, such that all but finitely many of the translates of the ball is disjoint from the ball.

In the picture, the green blob around x in K is an example of B_x . It is small enough such that \gamma \cdot B_x is disjoint from B_x . The ball B_x is small enough such that most almost \gamma this should happen (except finitely many of them).

How do we know that there is such a ball? This is by definition of proper action. Since the action is proper, by definition, we will get such balls for each x in X.

Intuitively a proper action is proper, that is, most of them (the isometries) move balls considerably.

Examples to keep in mind: Integer translation on the real line is proper. Rational translation on the real line is not proper.

\cup B_x is an open cover of K. Hence it has a finite subcover (as K is compact).

Choose finitely many balls B_{x_1}, \cdots , B_{x_k} , such that they cover K.

These finitely many balls have two properties:

  1. They cover K
  2. \gamma \cdot B_{x_i} \cap B_{x_i} \neq \phi for finitely many \gamma \in \Gamma

Suppose \displaystyle { \Sigma = \{ \gamma | \gamma \cdot B_{x_i} \cap B_{x_i} \neq \phi \} }

Thus each member of \Sigma has this property : hit all of the balls in the previously found cover by it. At least one of them will not move a lot (that is the translate will have a non-empty intersection with the ball)

Also notice that \Sigma is finite (as there are finitely many balls and for each ball there are finitely many \gamma .

Suppose x \in X be any point. There is a \gamma \in \Gamma such that x \in \gamma \cdot K .

Then \gamma^{-1} \cdot x \in K . This is the element of K that goes to x under \gamma

Now consider the isotropy subgroup \Gamma_x , that is group of all isometries that fixes x.

Notice that \gamma^{-1} \Gamma_x \gamma is the isotropy subgroup of \gamma^{-1} \cdot x . Why? Suppose \gamma_1 \in \Gamma_x . We will show that \gamma^{-1} \gamma_1 \gamma ) fixes ( \gamma^-1 \cdot x .

After all \gamma^{-1} \gamma_1 \gamma ( \gamma^{-1} \cdot x) = \gamma^{-1} \cdot x

We can similarly show the converse.

Hence we have shown that if x is pulled back into K by some \gamma^{-1} conjugating the isotropy subgroup of x by that \gamma gives the isotropy subgroup of the pulled back element.

But this is the subgroup that does not move the ball in the cover containing \gamma^{-1} \cdot x in a disjoint manner. Hence it must be inside \Sigma , and hence finite as \Sigma was finite.