Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 24 / 5

A Kleinian group is a discrete subgroup of the orientation-preserving isometry group of ${\mathbb{H}}^3$, identified with ${\mathrm{PSL}}_2(\mathbb{C})$. For an abstract finitely generated group this means a faithful representation into ${\mathrm{PSL}}_2(\mathbb{C})$ whose image is discrete; the chosen image is the geometric realization. Its boundary action is by Möbius transformations on $S_{\infty }^2\cong \mathbb{C}\cup \{\infty \}$.

Discreteness means the identity is isolated in the group topology, and for a subgroup it also excludes an infinite sequence of distinct elements converging anywhere in the isometry group. Torsion-freeness is not part of this definition: elliptic Möbius transformations of finite order are allowed. Therefore a Kleinian quotient need not be a smooth hyperbolic manifold.

Fix an interior point $o\in {\mathbb{H}}^3$. The limit set of a Kleinian group is the set of boundary accumulation points of its orbit, orThe boundary compactification may be realized as the closed unit ball. A discrete isometry group has only finitely many orbit points in each compact interior region: isometries moving $o$ into a compact set form a compact family after accounting for the compact rotation stabilizer, and infinitely many distinct group elements there would contradict discreteness. The stabilizer of $o$ is finite for the same reason. Thus an infinite group has an infinite orbit escaping every compact interior set and a nonempty boundary accumulation set. A finite group has empty limit set. This also shows the displayed intersection does not introduce ordinary interior accumulation points.

For another point $o^{\prime }$, the corresponding points satisfy $d(\gamma o,\gamma o^{\prime })=d(o,o^{\prime })=:C$. In the Poincaré ball model the distance identity isIf ${\gamma }_{j}o\to \xi \in S_{\infty }^2$, bounded distance impliesSo ${\gamma }_j{o}^{\prime }\to \xi$ as well. Exchanging $o,o^{\prime }$ proves equality of the two accumulation sets. Hence the definition is independent of the interior base point.

It is closed as an intersection of the compact orbit closure with the closed boundary, and it is $\Gamma$-invariant because multiplying an orbit by an element of $\Gamma$ permutes that orbit. Changing ball coordinates or conjugating the geometric realization transports the set by the induced boundary homeomorphism. Different nonconjugate representations of an abstract group are not asserted to have the same subset of one fixed sphere. None of these arguments requires finite generation.

Let $F\subseteq S_{\infty }^2$ be nonempty, closed and $\Gamma$-invariant. We first show it cannot be a singleton when $\Gamma$ is a non-elementary Kleinian group.

Suppose every element fixes one boundary point, conjugated to $\infty$. Each boundary map then has the affine form $z\mapsto az+b$. If all elements have $|a|=1$, their action in upper half-space keeps the height of $o$ fixed. An orbit sequence escaping compact interior sets must have its horizontal coordinate tending to infinity along any convergent boundary subsequence, so its only possible boundary limit is $\infty$. If some element has $|a|\ne 1$, move its second fixed point to $0$, making it $f(z)=az$. For any $g(z)=cz+d$, its conjugates are $f^{n}g{f}^{-n}(z)=cz+a^{n}d$. Choose $n\to +\infty$ or $-\infty$ so $a^n\to 0$. If $d\ne 0$, these are distinct group elements converging to $z\mapsto cz$, contradicting discreteness. Thus $d=0$ for every group element: all fix both $0$ and $\infty$, and their only possible interior-orbit boundary limits are these two points. In both cases the Kleinian limit set has at most two points. This contradicts non-elementarity. Therefore $F$ has at least two distinct points.

Use the Klein ball model, in which hyperbolic geodesics are straight chords, and take the closed Euclidean convex hull $C$ of $F$ in the closed ball. It contains an interior point $o$, since the open chord between two distinct points of $F$ lies in the open ball. The interior part of this hull is hyperbolically convex and invariant under $\Gamma$: isometries preserve geodesic segments, so they preserve the hyperbolic convex hull of the invariant set. Thus $\Gamma o\subseteq C$.

The only points of $C$ on the boundary sphere are the points of $F$. Indeed, if $\xi \in S_{\infty }^2\setminus F$, compactness of $F$ gives $\xi \cdot z\le c<1$ for every $z\in F$, because $\xi \cdot z=1$ on the unit sphere only when $z=\xi$. The same linear inequality holds throughout $C$, excluding $\xi$ from it. Conversely $F\subset C$, so $C\cap S_{\infty }^2=F$.

Every boundary accumulation point of $\Gamma o$ consequently lies in $F$. Base-point independence from part (b) identifies that accumulation set with $\Lambda (\Gamma )$, proving minimality of the Kleinian limit set:This proves the stated containment for every nonempty closed invariant set, not merely for sets containing an arbitrarily chosen orbit on the boundary.

Write $N={\Gamma }^{\prime }$. As a subgroup of a discrete group, it is a Kleinian group, even if it is not finitely generated. Since it is infinite, part (b) gives a nonempty closed Kleinian limit set $\Lambda (N)$. For $\gamma \in \Gamma$, conjugation and base-point independence giveOne can see this directly by applying $\gamma$ to $No$: the image is the orbit of $\gamma o$ under the conjugate subgroup, and the choice of interior point does not affect its limit set. Normality makes the conjugate subgroup equal to $N$, so $\Lambda (N)$ is $\Gamma$-invariant. Applying part (c) yields $\Lambda (\Gamma )\subseteq \Lambda (N)$.

Conversely $No\subseteq \Gamma o$, so every boundary accumulation point of the smaller orbit is one of the larger orbit: $\Lambda (N)\subseteq \Lambda (\Gamma )$. Therefore the limit set of an infinite normal Kleinian subgroup satisfiesInfinitude supplies nonemptiness, normality supplies invariance, and non-elementarity supplies minimality. Properness of the subgroup is not needed, and no finite-generation assumption on it has been inserted.

Take $\Gamma ={\mathrm{PSL}}_2(\mathbb{Z})$ as a subgroup of ${\mathrm{PSL}}_2(\mathbb{C})$. It is discrete because representatives have integer entries, so a bounded set of matrices contains only finitely many such representatives, modulo the two choices of sign. It is finitely generated byThe Euclidean algorithm on the first column of an integer determinant-one matrix reduces it using these two transformations, proving the generation statement.

It is non-elementary. The powers of $T$ move an interior upper-half-space point towards $\infty$, so $\infty \in \Lambda (\Gamma )$. Conjugating $T$ by $S$ gives a parabolic with fixed point $0$, and conjugating that parabolic by $T$ gives one with fixed point $1$. Their interior orbits approach those respective points. Thus $0,1,\infty$ are three distinct points in the Kleinian limit set.

However $S^2=1$ in the projective group, and $S$ is a nonidentity elliptic Möbius transformation fixing the hyperbolic geodesic with boundary endpoints $i,-i$. It is a half-turn around that axis. Near an ordinary point of the axis, quotienting identifies the transverse angle modulo $\pi$, giving cone angle $\pi$ instead of $2\pi$. The elliptic stabilizer obstruction to a hyperbolic manifold quotient therefore givesThe obstruction is the nonfree action, not finite generation or elementary behavior. An underlying topological quotient can still be a manifold; the conical quotient metric is what prevents this example from being a hyperbolic manifold quotient.