A quasiconformal conjugacy of finite-covolume Kleinian groups is Möbius

The quotient $X$ is a second-countable smooth manifold: left translation by a discrete subgroup is free, and each point of $G$ has a small open neighborhood disjoint from its nontrivial left translates. The projection restricts to a diffeomorphism there, giving the quotient charts. The subgroup is countable, since a discrete subgroup of the second-countable manifold $G$ is discrete and closed and its distinct points can be assigned distinct elements of a countable neighborhood basis. A Borel set in $X$ is $\nu$-null exactly when its full inverse image is Haar null, by the local quotient definition of $\nu$, countable quotient charts and countability of $\mathrm{\Gamma }$. Both measures are sigma-finite: $G$ and $X$ have countable relatively compact chart covers, and their Radon measures are finite on compact sets.

Let $\mu$ be the Beltrami coefficient of $h$ in the standard finite source coordinate, defined arbitrarily on its null exceptional set, with $\mathrm{\mid }\mu \mathrm{\mid }\le k<1$. Finite source and target charts suffice outside at most finitely many points. Since holomorphic coordinate changes obey [F3], the conjugacy identity gives $\mu (\gamma z)\overline{{\gamma }^{\mathrm{\prime }}(z)}\mathrm{/}{\gamma }^{\mathrm{\prime }}(z)=\mu (z)$ almost everywhere, for every $\gamma \in \mathrm{\Gamma }$. Consider the evaluation map $p:G\to \widehat{\mathbb{C}}$, $p(g)=g(0)$. It is a smooth submersion: infinitesimal translations span the tangent plane at $0$, and left translation transports that statement to every $g$. Its local coordinates are a product of a sphere chart and a four-dimensional fiber chart; Haar measure has a smooth positive density in these coordinates. Thus inverse images under $p$ of area-null sets are Haar null by Fubini; conversely a positive-area set has inverse image of positive Haar measure in some such product chart.

Define $R_{g}f(x)=f(xg)$ on $L^2(X,\nu )$. Right invariance of $\nu$ makes every $R_g$ unitary, and $R_g{R}_k=R_{gk}$. This representation is strongly continuous. Indeed, if $f\in C_c(X)$ and $g$ stays in a compact identity neighborhood, the supports of $R_{g}f-f$ lie in one compact subset of $X$. Joint continuity of the action and compactness give uniform convergence $R_{g}f\to f$ there as $g\to e$, so convergence holds in $L^2$. For a general $v\in L^2$, approximate by $f\in C_c(X)$ using [F1] and use $\mathrm{\parallel }{R}_{g}v-v{\mathrm{\parallel }}_2\le 2\mathrm{\parallel }v-f{\mathrm{\parallel }}_2+\mathrm{\parallel }{R}_{g}f-f{\mathrm{\parallel }}_2$.

Outside the Haar-null set where $g(0)=\mathrm{\infty }$, define $F(g)=\mu (g(0))\overline{g^{\mathrm{\prime }}(0)}\mathrm{/}{g}^{\mathrm{\prime }}(0)$, and put it equal to zero on that exceptional set. This is bounded and measurable. Step 1.2 and the holomorphic chain rule give $F(\gamma g)=F(g)$ for almost every $g$, for each $\gamma$. Because $\mathrm{\Gamma }$ is countable, remove the union of the exceptional sets and all their $\mathrm{\Gamma }$ translates; this is an invariant Haar-null set. Redefining $F$ as zero on it gives an exactly left-$\mathrm{\Gamma }$-invariant representative. It descends to a measurable bounded function on $X$: on each quotient chart use a local section of the covering projection, and exact invariance makes the resulting definitions agree. Finite measure gives $F\in L^2(X,\nu )$.

Put $a_t=\left[\begin{array}{cc}{e}^{t\mathrm{/}2}& 0\\ 0& e^{-t\mathrm{/}2}\end{array}\right]$, $n_{+}(z)=\left[\begin{array}{cc}1& z\\ 0& 1\end{array}\right]$ and $n_{-}(z)=\left[\begin{array}{cc}1& 0\\ z& 1\end{array}\right]$. If $v\in L^2(X)$ satisfies $R_{a_t}v=v$ for every $t$, then $\mathrm{\parallel }{R}_{n}v-v{\mathrm{\parallel }}_2=\mathrm{\parallel }{R}_{a_{t}n{a}_{-t}}v-v{\mathrm{\parallel }}_2$ by unitarity and this invariance. For $n=n_{+}(z)$ the conjugate is $n_{+}(e^{t}z)$ and tends to the identity as $t\to -\mathrm{\infty }$; for $n=n_{-}(z)$ it is $n_{-}(e^{-t}z)$ and tends to the identity as $t\to +\mathrm{\infty }$. Strong continuity from step 2.1 forces $R_{n}v=v$ in both cases.

These two unipotent subgroups generate $G$. To check this without any structural rigidity theorem, define $w(\lambda )=n_{+}(\lambda )n_{-}(-{\lambda }^{-1})n_{+}(\lambda )=\left[\begin{array}{cc}0& \lambda \\ -{\lambda }^{-1}& 0\end{array}\right]$ for $\lambda \mathrm{\ne }0$. Then $w(\lambda )w(-1)=\mathrm{diag}(\lambda ,{\lambda }^{-1})$. Every determinant-one matrix with upper left entry $a\mathrm{\ne }0$ factors as $n_{-}(c\mathrm{/}a)\mathrm{diag}(a,a^{-1})n_{+}(b\mathrm{/}a)$, where $b,c$ are its other off-diagonal entries. If $a=0$, its lower left entry is nonzero, so left multiplication by $n_{+}(1)$ makes that upper left entry nonzero. The factorization descends to $G$. Thus every vector in step 3.1 is invariant under every $R_g$.

A vector invariant under all $R_g$ is constant almost everywhere. Take a Borel representative $v$ and its lift $\tilde{v}$ to $G$. For each fixed $k\in G$, right invariance in $L^2(X)$ and the null-set equivalence in step 1.1 give $\tilde{v}(gk)=\tilde{v}(g)$ for Haar-almost every $g$. Tonelli on compact exhaustions of $G\times G$ gives this equality for almost every pair $(g,k)$. The smooth change of variables $(g,k)\mapsto (g,y=gk)$ preserves the Haar product measure class, since for fixed $g$ left translation preserves Haar measure. Hence $\tilde{v}(y)=\tilde{v}(g)$ for almost every $(g,y)$. Fubini supplies one $g$ whose section has full measure, proving $\tilde{v}$, and therefore $v$, almost everywhere constant. The preceding argument has proved the complete contraction assertion needed below.

In the finite source coordinate, $a_t(z)=e^{t}z$, so $a_t$ fixes $0$ and has positive real derivative there. Consequently $F(g{a}_t)=F(g)$ for Haar-almost every $g$, for every $t$; the redefinition in step 2.2 changes this only on a null set and its right translate. Thus $R_{a_t}F=F$ in $L^2(X)$, and steps 3.1–5.1 imply $F=c$ almost everywhere for a complex constant $c$. Now let $m_{\theta }(z)=e^{i\theta }z$. The same computation gives $F(g{m}_{\theta })=e^{-2i\theta }F(g)$ almost everywhere. Right translation preserves the measure class, so the constant satisfies $c=e^{-2i\theta }c$ for every $\theta$. Taking $\theta =\pi \mathrm{/}2$ gives $c=-c$, hence $c=0$.

Since the derivative ratio in step 2.2 has modulus one, $F=0$ Haar-almost everywhere implies $\mu \circ p=0$ Haar-almost everywhere. The positive-area converse in step 1.2 forces $\mu =0$ area-almost everywhere. By [F4], $h$ is holomorphic in every finite source and target chart. The finitely many omitted chart points cause no issue: use alternate source and target charts there and the conformal coefficient transformation law [F3]; their exceptional sets remain null. Thus $h$ is a biholomorphic sphere homeomorphism and is Möbius by [F4].