# Isometric group of hyperbolic 3-dim manifolds

In the book : Foundation of Hyperbolic Manifolds

There is a theorem that any finite subgroup of $Isom(\mathbb{E}^n)$ fixes a point.

And I hope to solve the following question :

Any finite subgroup of $Isom_{+}(\mathbb{H}^3)$ fixes a point in $\mathbb{H}^3$

($Isom_{+}(\mathbb{H}^3)$ is Lie group of orientation-preserving isometric group $Isom(\mathbb{H}^3)$ and $Isom_{+}(\mathbb{H}^3) \simeq PSL_2(\mathbb{C})$ )

Some books states that

Every finite subgroup $\Gamma$ of Isometric group $Isom(M)$ fixes a point, precisely fixes the centroid of Orbit $\Gamma x$ where $x \in M$

But I don't get it. Element in $Isom_{+}(H^3)$ is a Mobius transfomer which is not linear in general.

Could you give me some hints ?

• Oh,it is a notation. Element $\begin{pmatrix}a & b\\c & d\end{pmatrix}$ represents the isometry $f(z) = \frac {az+b}{cz+d}$ @DietrichBurde Commented Jul 19, 2015 at 11:06

In any CAT(0) space $X$, a bounded set $S \subset X$ has a well-defined center, ie., there exists a unique point $c$ such that $S \subset \bar{B}(c,r)$ where $r= \inf \{ s> 0 \mid \exists x \in X, \ X \subset \bar{B}(x,s) \}$. See Proposition II.2.7 in Bridson and Haefliger's book, Metric spaces of nonpositive curvature.

Now, if $G$ acts on $X$ by isometries and has a bounded orbit $S$ (eg. if $G$ is finite), then $G$ cleary fixes the center $c$ of $S$: because the center is defined metrically, for all $g \in G$, $g \cdot c$ is the center of $g \cdot S =S$, hence $g \cdot c =c$ by the uniqueness of the center.

Therefore, we deduce:

Proposition: Let $G$ be a group acting by isometries on a CAT(0) space $X$. If $G$ has a bounded orbit, then $G$ has a global fix point.

In your case, $\mathbb{H}^3$ is obviously a CAT(-1) space, and a fortiori a CAT(0) space.

Each finite group of isometries of hyperbolic space $\mathbb{H}^n$ has a global fixed point in $\mathbb{H}^n$, because it is conjugate to a finite group of orthogonal transformations of the ball $B^n$ (which are exactly the isometries of $\mathbb{H}^n$ which fix the origin in $B^n$).

In hyperbolic $3$-space, each non-identical Moebius transformations fixes $1$ or $2$ points (it is called parabolic, if it fixes exactly one point in $\widehat{\mathbb{C}}$). However, you want that all elements of a finite subgroup fix a common point.

• It is conjugate to the $O(B^n)$ ? Amazing, it is sufficient to get what I want. Commented Jul 19, 2015 at 13:43