Hyperbolic knot complements and volume

Choose a smooth parametrization $h:T\to {\mathbb{R}}^2\mathrm{/}{\mathbb{Z}}^2$, as supplied by saying that $T$ is a smooth torus. It preserves essentiality, and its induced integral homology isomorphism preserves primitivity and equality up to sign. Apply [F1] to $h(c)$ to obtain primitivity of $[c]$. This is one supplied finite-dimensional parametrization, not a set-indexed choice.

If $[c]=\pm [d]$, apply [F1] to their images and conjugate its ambient isotopy by $h$. This gives the required ambient isotopy in $T$. If the original curves are disjoint and distinct, the two complementary annulus parameters in [F1] pull back by $h$; either supplies an embedded annulus with exactly those original curves as boundary. No annulus is inferred for intersecting original curves, and no choice axiom is introduced.

A nonidentity element of $\mathrm{\Gamma }$ is parabolic or loxodromic. Indeed, represent it by a determinant-one complex matrix and examine its eigenvalues. A repeated eigenvalue gives either the identity in the projective group or a nontrivial unipotent matrix, which is conjugate to a translation. Distinct eigenvalues give a diagonal matrix with entries $\lambda ,{\lambda }^{-1}$. If $\mathrm{\mid }\lambda \mathrm{\mid }\mathrm{\ne }1$ it is loxodromic. If $\mathrm{\mid }\lambda \mathrm{\mid }=1$, its powers lie in a compact circle subgroup; either it has finite order, contrary to torsion-freeness, or distinct powers have quotients tending to the identity, contrary to discreteness. This also proves directly that no nontrivial elliptic element occurs.

Suppose that an abelian subgroup contains a loxodromic element. Every matrix commuting with it preserves its two distinct eigenlines individually: interchanging them would conjugate its eigenvalue ratio to its inverse, which is different because its modulus is not one. Its centralizer is therefore the diagonal group, identified with ${\mathbb{C}}^{\ast }$. Every discrete torsion-free subgroup $D$ of this group is cyclic. For each $B>0$, the annulus $e^{-B}\le \mathrm{\mid }z\mathrm{\mid }\le e^B$ is compact, so contains only finitely many elements of $D$ by [L1]. The kernel of $z\mapsto \log \mathrm{\mid }z\mathrm{\mid }$ is a finite subgroup of the unit circle and hence trivial. If $D$ is nontrivial, this map therefore has a smallest positive value $b$: choose one positive value $B$, and minimize among the finitely many in $(0,B]$. Division with remainder by $b$ shows that every value is an integer multiple of $b$, proving the cyclic assertion. Thus $A$ cannot contain a loxodromic element.

Choose a nonidentity element of $A$. By steps 1.1–2.1 it is parabolic; conjugate its unique fixed point to infinity. Every element commuting with it fixes that point and, writing the parabolic as $z\mapsto z+\lambda$ with $\lambda \mathrm{\ne }0$, commutation forces the affine multiplier to be one. Thus $A$ consists of translations. In fact every element of $P$ is a translation. Write such an element as $z\mapsto az+b$. If $\mathrm{\mid }a\mathrm{\mid }\mathrm{\ne }1$, its conjugation action multiplies translation vectors by $a$, so conjugating $z\mapsto z+\lambda$ by positive or negative powers gives nontrivial translations tending to the identity. If $\mathrm{\mid }a\mathrm{\mid }=1$ and $a\mathrm{\ne }1$, it fixes both infinity and the finite point $b\mathrm{/}(1-a)$ and is elliptic, excluded by step 1.1. Hence $a=1$. The common fixed point is unique, since already the chosen nonidentity parabolic has just one.

A discrete additive subgroup $\mathrm{\Lambda }$ of $\mathbb{C}\cong {\mathbb{R}}^2$ has rank at most two. Here are the needed elementary details. Choose a shortest nonzero vector $v$, whose existence follows from finiteness in bounded disks. If every vector is collinear with $v$, division with remainder along that line proves $\mathrm{\Lambda }=\mathbb{Z}v$. Otherwise subtract integer multiples of $v$ to put the parallel component of every vector in $[0,\mathrm{\mid }v\mathrm{\mid })$. Among vectors with positive perpendicular component, a least such component exists: below any fixed positive candidate and with that normalized parallel component, there are finitely many vectors in a compact rectangle. Choose a vector $w$ realizing it. Division with remainder in the perpendicular direction, followed by division along $v$, shows that every vector is in $\mathbb{Z}v+\mathbb{Z}w$. Since $A$ has rank two, the collinear case is impossible. Therefore the translation vectors of $P$ form a rank-two lattice $\mathrm{\Lambda }$. An element centralizing $A$ must centralize its chosen parabolic and hence belong to $P$ by step 3.1; since $P$ is abelian, it is the unique maximal abelian subgroup containing $A$.

Rescale the finite boundary coordinate so that $T:z\mapsto z+1$ belongs to $\mathrm{\Gamma }$. If $g=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]\in \mathrm{\Gamma }$ has $c\mathrm{\ne }0$, then $\mathrm{\mid }c\mathrm{\mid }\ge 1$. To prove this, suppose $0<\mathrm{\mid }c\mathrm{\mid }<1$ and choose a determinant-one lift $g_0$. Define $g_{n+1}=g_{n}T{g}_n^{-1}$ using the lift $T=\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]$. Matrix multiplication gives $g_{n+1}=\left(\begin{array}{cc}1-a_n{c}_n& a_n^2\\ -c_n^2& 1+a_n{c}_n\end{array}\right),c_{n+1}=-c_n^2\mathrm{.}$ Put $r=\mathrm{\mid }{c}_0\mathrm{\mid }<1$. Then $\mathrm{\mid }{c}_n\mathrm{\mid }=r^{2^n}$ and $\mathrm{\mid }{a}_{n+1}\mathrm{\mid }\le 1+r\mathrm{\mid }{a}_n\mathrm{\mid }$, so $\mathrm{\mid }{a}_n\mathrm{\mid }\le \max \{\mathrm{\mid }{a}_0\mathrm{\mid },(1-r{)}^{-1}\}$. Consequently $a_{n+1}\to 1$, $a_n^2\to 1$ and $g_n\to T$. Their nonzero lower-left entries have distinct absolute values, so these are distinct projective matrices. This contradicts discreteness and proves the bound.

The extension of a Möbius matrix to upper half-space has height coordinate $y(g(z,y))=\frac{y}{\mathrm{\mid }cz+d{\mathrm{\mid }}^2+\mathrm{\mid }c{\mathrm{\mid }}^2{y}^2}\mathrm{.}$ This follows by composing translations, rotations and dilations with inversion; these generate the determinant-one matrices by Gaussian elimination. For $c\mathrm{\ne }0$ and $y\ge R$, step 4.2 therefore gives $y(g(z,y))\le 1\mathrm{/}R$. Choose $R>1$. Then every translate of the closed horoball by an element outside $P$ is disjoint from that horoball, whereas translations in $P$ preserve it. Thus $\{y\ge R\}$ is precisely invariant under $P$, and its quotient injects into $M$. The strict inequality $1\mathrm{/}R<R$ gives an embedded boundary collar as well.

This quotient map is a proper embedding. A closed horoball quotient has the coordinates $(\mathbb{C}\mathrm{/}\mathrm{\Lambda })\times [R,\mathrm{\infty })$. If a sequence escapes every compact subset of this product, its heights tend to infinity after passing to a subsequence, because the horizontal factor is compact. For a fixed nonzero $\lambda \in \mathrm{\Lambda }$, the distance between $(z,y)$ and $(z+\lambda ,y)$ is $2\mathrm{arsinh}(\mathrm{\mid }\lambda \mathrm{\mid }\mathrm{/}(2y))$, as follows by the hyperbolic distance formula in a vertical half-plane. Hence the injectivity radii of their images in $M$ tend to zero. On every compact subset of $M$ the injectivity radius is bounded below by a positive constant: each point has an evenly covered small metric ball, and finitely many smaller such balls cover a compact set. Thus this escaping sequence cannot have its image in a compact set. Sequential compactness in metric spaces proves properness. An injective proper local embedding is an embedding, giving the assertion.

The volume density is $y^{-3}d{x}_{1}d{x}_{2}dy$. Integrating over one Euclidean fundamental parallelogram and $y\ge R$ gives $\mathrm{area}(\mathbb{C}\mathrm{/}\mathrm{\Lambda }){\int }_R^{\mathrm{\infty }}{y}^{-3}dy=\mathrm{area}(\mathbb{C}\mathrm{/}\mathrm{\Lambda })\mathrm{/}(2R^2)$. This also identifies the compact boundary as a torus and the open interior as its product with a half-line.

The quotient $M$ is complete: a geodesic lifts locally to ${\mathbb{H}}^3$ and its lifted geodesic is defined for all real time, so it projects to a global extension. Hence [F1] makes $M$ proper. Let $C$ be the embedded closed horocusp and $S=\mathrm{\partial }C$. Its interior is a connected component of $M\setminus S$: it is open there, and it is closed there by properness in step 6.1. The other side is connected. Indeed, the outward collar of the connected torus $S$ belongs to a single component; any other component of the complement of $C$ would have no frontier in $M$ and would be both open and closed, contradicting connectedness of $M$. If that other side had noncompact closure, both sides of the compact separator $S$ would be nonrelatively compact components. This contradicts the one-end hypothesis, since in an interior of a compact collared manifold the definition of one end says that exactly one component of the complement of any sufficiently large compact set is nonrelatively compact. More generally this follows from the usual exhaustion definition of one end, applied after adding a compact connected set containing $S$. Thus the other side has compact closure. This proves the last assertion; it does not assert that arbitrary many-ended quotients have compact complement of a single horocusp.

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].

Define the locally finite Borel measure $\omega (E)=\mathrm{area}(h(E))$. This is a measure because a homeomorphism maps disjoint Borel sets to disjoint Borel sets and respects unions. It is Radon because it is the pushforward of planar area by the homeomorphism $h^{-1}$; compact and open approximation transport through that homeomorphism. The open image $h(B(x,r))$ contains $B(h(x),l(x,r))$: otherwise a segment from $h(x)$ to a point of that smaller disk outside the image would first meet the boundary $h(\mathrm{\partial }B(x,r))$, contradicting the definition of $l$. The maximum distance from $h(x)$ in $h(\overline{B(x,r)})$ occurs on its boundary, because distance from a fixed point has no maximum in a nonempty open set. Hence $\mathrm{\mid }h(y)-h(x)\mathrm{\mid }\le L(x,r)\le Hl(x,r)$ for $y\in \overline{B(x,r)}$, and $\pi l(x,r{)}^2\le \omega (B(x,r))$.

For $x\in V$ and $0<\mathrm{\mid }t\mathrm{\mid }<\mathrm{dist}(V,\mathrm{\partial }U)\mathrm{/}2$, use step 1.1 with radius $2\mathrm{\mid }t\mathrm{\mid }$ to obtain $\mathrm{\mid }h(x+t{e}_i)-h(x){\mathrm{\mid }}^2\le (H^2\mathrm{/}\pi )\omega (B(x,2\mathrm{\mid }t\mathrm{\mid }))$. These balls lie in $U$. Tonelli, applied to their incidence indicator, gives ${\int }_V\omega (B(x,2\mathrm{\mid }t\mathrm{\mid }))dx={\int }_U\mathrm{\mid }V\cap B(y,2\mathrm{\mid }t\mathrm{\mid })\mathrm{\mid }d\omega (y)\le 4\pi t^2\omega (U)$. Combining the two inequalities proves the explicit difference-quotient estimate. All measures in this calculation are locally finite Radon measures and the incidence set is open, so no absolute continuity or null-set preservation of $h$ has been presumed.

Continuity makes $h$ square integrable on every relatively compact $U$. The uniform estimate in step 2.1 and [F1], with $p=2$ and complex-valued functions, give $h\in W^{1,2}(V)$. Exhausting the plane proves local Sobolev regularity. Apply [F2] on a countable collection of rectangles. On almost every coordinate line, its ACL representative agrees with the continuous representative almost everywhere by Fubini, hence everywhere on each compact line interval because both are continuous there. Thus the actual map $h$ is ACL and has finite coordinate partial derivatives almost everywhere, agreeing with the weak derivatives.

By the general-homeomorphism interface [F3], $h$ is totally differentiable almost everywhere. At such a point $x$, write $h(x+v)=h(x)+Av+o(\mathrm{\mid }v\mathrm{\mid })$; the remainder is uniform over $\mathrm{\mid }v\mathrm{\mid }=r$ after division by $r$. If $A$ has rank one, there are unit vectors $u,v$ with $Au=0$ and $\mathrm{\mid }Av\mathrm{\mid }>0$. Then $l(x,r)=o(r)$ whereas $L(x,r)\ge r\mathrm{\mid }Av\mathrm{\mid }+o(r)$, contradicting the finite circular bound. If $A$ is nonsingular with singular values $s_1\ge s_2>0$, the same uniform expansion gives $L(x,r)\mathrm{/}r\to s_1$ and $l(x,r)\mathrm{/}r\to s_2$, so $s_1\mathrm{/}{s}_2\le H$. Its determinant is positive by [F3], since $h$ preserves orientation. The remaining possibility is $A=0$.

At almost every point the classical partials of the total derivative equal the weak derivatives by step 3.1. At the nonsingular points, [F5] and step 4.1 give $\mathrm{\mid }{h}_{\bar{z}}\mathrm{\mid }\mathrm{/}\mathrm{\mid }{h}_z\mathrm{\mid }\le (H-1)\mathrm{/}(H+1)$; at zero derivatives the asserted inequality also holds. Together with $W_{\mathrm{l}\mathrm{o}\mathrm{c}}^{1,2}$ and ACL regularity, this is exactly analytic $H$-quasiconformality. The proof uses only the locally proved total-differentiability interface [F3]; it does not invoke an infinitesimal metric-to-quasiconformal criterion.

For $\mathrm{\mid }w\mathrm{\mid }<1$, integrating the uniformly convergent geometric series on compact subdisks gives $-\log (1-w)={\sum }_{n\ge 1}{w}^n\mathrm{/}n$. Taking real parts at $w=r{e}^{2iu}$, where $0<r<1$, yields $-\log \mathrm{\mid }1-r{e}^{2iu}\mathrm{\mid }={\sum }_{n\ge 1}{r}^n\cos (2nu)\mathrm{/}n$. This series is uniformly absolutely convergent in $u$ for fixed $r$, so integrating from $0$ to $\theta$ gives $-{\int }_0^{\theta }\log \mathrm{\mid }1-r{e}^{2iu}\mathrm{\mid }du=\frac{1}{2}{\sum }_{n\ge 1}{r}^n\sin (2n\theta )\mathrm{/}{n}^2$.

For $r\ge 1\mathrm{/}2$, $\mathrm{\mid }1-r{e}^{2iu}{\mathrm{\mid }}^2=(1-r{)}^2+4r{\sin }^{2}u$, so $\sqrt{2}\mathrm{\mid }\sin u\mathrm{\mid }\le \mathrm{\mid }1-r{e}^{2iu}\mathrm{\mid }\le 2$. Therefore $\mathrm{\mid }\log \mathrm{\mid }1-r{e}^{2iu}\mathrm{\mid }\mathrm{\mid }$ is bounded by a constant plus $\mathrm{\mid }\log \mathrm{\mid }\sin u\mathrm{\mid }\mathrm{\mid }$, which is integrable on bounded intervals. Dominated convergence sends the integral in step 1.1 to $\mathrm{\Lambda }(\theta )$ as $r\uparrow 1$; the series converges uniformly in $\theta$ because $\sum n^{-2}<\mathrm{\infty }$. This proves the Fourier expansion on $[0,\pi ]$ and continuity there.

The defining integrand is even, so $\mathrm{\Lambda }$ is odd. Its derivative away from the integral's singularities is $-\log \mathrm{\mid }2\sin \theta \mathrm{\mid }$, a $\pi$-periodic function; the Fourier formula gives $\mathrm{\Lambda }(0)=\mathrm{\Lambda }(\pi )=0$, so the integral of this derivative over each period is zero. Hence $\mathrm{\Lambda }(\theta +\pi )=\mathrm{\Lambda }(\theta )$ for every real $\theta$. Substitution $\theta =\pi \mathrm{/}2$ in the Fourier series gives $\mathrm{\Lambda }(\pi \mathrm{/}2)=0$.

The identity $\mathrm{\mid }2\sin (2u)\mathrm{\mid }=\mathrm{\mid }2\sin u\mathrm{\mid }\mathrm{\mid }2\sin (u+\pi \mathrm{/}2)\mathrm{\mid }$ and substitution $v=2u$ in the defining integral give $\mathrm{\Lambda }(2\theta )=2\mathrm{\Lambda }(\theta )+2\mathrm{\Lambda }(\theta +\pi \mathrm{/}2)-2\mathrm{\Lambda }(\pi \mathrm{/}2)$. Step 3.1 gives the stated duplication identity.

In the Fourier series at $\theta =\pi \mathrm{/}3$, $\sin (2n\pi \mathrm{/}3)$ cycles through $\sqrt{3}\mathrm{/}2,-\sqrt{3}\mathrm{/}2,0$, so grouping successive triples gives the stated positive series. For $a=3k+1$, $a^{-2}-(a+1{)}^{-2}=(2a+1)\mathrm{/}(a^2(a+1{)}^2)\le 2\mathrm{/}{a}^3\le 2\mathrm{/}(27k^3)$ for $k\ge 1$. For $N\ge 1$, the tail after $k=0,\dots ,N$ is at most $\frac{\sqrt{3}}{4}\frac{2}{27}{\sum }_{k>N}{k}^{-3}\le \frac{\sqrt{3}}{4}\frac{2}{27}{\int }_N^{\mathrm{\infty }}{x}^{-3}dx=\frac{\sqrt{3}}{108N^2}\le \frac{\sqrt{3}}{72N^2}$.

Since $\mathrm{Im}z>0$, the imaginary parts of $z$, $z^{\mathrm{\prime }}=1\mathrm{/}(1-z)$ and $z^{\mathrm{\prime }\mathrm{\prime }}=1-1\mathrm{/}z$ are all positive: the second is $\mathrm{Im}z\mathrm{/}\mathrm{\mid }1-z{\mathrm{\mid }}^2$, and the third is $\mathrm{Im}z\mathrm{/}\mathrm{\mid }z{\mathrm{\mid }}^2$. Thus their principal arguments lie in $(0,\pi )$.

Multiplication gives $z{z}^{\mathrm{\prime }}{z}^{\mathrm{\prime }\mathrm{\prime }}=z\cdot (1-z{)}^{-1}\cdot (z-1)z^{-1}=-1$. The sum of the three arguments is therefore an odd multiple of $\pi$; because it lies strictly between $0$ and $3\pi$, it must equal $\pi$. The arguments are the dihedral angles by the edge cross-ratio convention in [F1].

For $\omega =e^{i\pi \mathrm{/}3}$, $1-\omega ={\omega }^{-1}$ and $\omega -1={\omega }^2$. Hence $1\mathrm{/}(1-\omega )=\omega$ and $(\omega -1)\mathrm{/}\omega =\omega$; each parameter has argument $\pi \mathrm{/}3$. Each of the three parameter arguments occurs on a pair of opposite edges, proving the regular-angle assertion.

Let $Q$ be a fundamental parallelogram of area $A$. Integrating the density from [F1] over $Q\times [1,\mathrm{\infty })$ gives $\mathrm{Vol}(C)={\int }_Q{\int }_1^{\mathrm{\infty }}{t}^{-3}dtdxdy=A[-1\mathrm{/}(2t^2){]}_1^{\mathrm{\infty }}=A\mathrm{/}2$. The improper integral converges because $t^{-3}$ is integrable on $[1,\mathrm{\infty })$.

Along any piecewise smooth path with height coordinate $t$, its hyperbolic length is at least $\int \mathrm{\mid }dt\mathrm{\mid }\mathrm{/}t$, since the metric contains the term $d{t}^2\mathrm{/}{t}^2$. A path escaping the end has $t\to \mathrm{\infty }$, so this lower bound diverges as $\log t$. Thus the end is at infinite distance and is complete toward its ideal end.

Each supplied cusp group $P_i$ is the full stabilizer of its ideal point. By [F1] the full stabilizer is a translation lattice $Q_i$ containing $P_i$. Every translation in $Q_i$ preserves the entire source horoball. If some were outside $P_i$, two points of that horoball distinct modulo $P_i$ would have the same image in $M$, contradicting the supplied cusp embedding. Thus $P_i=Q_i$, and [F1] makes it maximal abelian. Conversely, let $A$ be any maximal abelian rank-two subgroup. Its proper embedded horocusp from [F1] has a connected tail missing the given compact core: properness makes its inverse image of that core compact. That tail lies in one of the finitely many supplied cusp components. Its inclusion on fundamental groups therefore puts $A$, up to conjugacy, inside that component's $P_i$. Maximality gives equality. In the compact case such a proper horocusp is impossible, so no rank-two abelian subgroup occurs. An isomorphism preserves abelianness, rank two and maximality, proving the peripheral assertion in both directions.

We construct a smooth equivariant map on a neighborhood of the lifted compact core, without assuming a homotopy realization theorem. Enlarge the core to include cusp depths $r\le 2$. Cover this compact set by finitely many evenly covered coordinate neighborhoods $U_j$ and choose smooth nonnegative bumps ${\psi }_j$ compactly supported in them whose positive sets cover the enlarged core. Such bumps are obtained by the ordinary explicit Euclidean bump function in smaller chart balls and a finite subcover. Choose one lift ${\tilde{U}}_j$ of each chart. On its translate $\gamma {\tilde{U}}_j$ put ${\varphi }_{\gamma j}(x)={\psi }_j(\pi x)$, and extend it by zero. This is smooth, since the bump support stays inside the chart. The family is locally finite: for a fixed $j$ the lifted chart components are disjoint, and over the compact support a sufficiently small evenly covered base neighborhood meets only the corresponding local component near each point. There are only finitely many $j$. Their sum is positive on a neighborhood of the lifted enlarged core.

For each matched cusp, choose upper half-space coordinates in source and target so that $\theta (P_i)$ is the corresponding target translation lattice, with any necessary conjugacy included in the choice of its lift. A group isomorphism of the two lattices extends uniquely to an invertible real-linear map $B_i:\mathbb{C}\to \mathbb{C}$: prescribe its values on a lattice basis. The map $C_i(z,y)=(B_{i}z,c_{i}y)$, $c_i=R_i^{\mathrm{\prime }}\mathrm{/}{R}_i>0$, is equivariant and preserves depth $r=\log (y\mathrm{/}{R}_i)$. Its differential and inverse have uniformly bounded hyperbolic operator norms, because horizontal tangent vectors are changed by $B_i\mathrm{/}{c}_i$ and vertical logarithmic tangent vectors are unchanged. Propagate this map to every translate of the chosen horoball by $C_i(\gamma x)=\theta (\gamma )C_i(x)$. Equivariance under $P_i$ makes this independent of the representative of the coset. Disjointness of the lifted cusp interiors makes the resulting map well defined there. These maps for ${\theta }^{-1}$ can be chosen to be exact inverses on corresponding cusps.

Use the hyperboloid model $\{v\in {\mathbb{R}}^{3,1}:⟨v,v⟩=-1,v_0>0\}$ for the target. The usual upper half-space identification is $(z,y)⟼(\frac{\mathrm{\mid }z{\mathrm{\mid }}^2+y^2+1}{2y},\frac{\mathrm{Re}z}{y},\frac{\mathrm{Im}z}{y},\frac{\mathrm{\mid }z{\mathrm{\mid }}^2+y^2-1}{2y})\mathrm{.}$ Its pullback metric is $(\mathrm{\mid }dz{\mathrm{\mid }}^2+d{y}^2)\mathrm{/}{y}^2$ by differentiation. Hyperbolic isometries act by Lorentz-linear maps: the upper half-space translations, rotations, dilations and inversion give Lorentz maps in this formula by substitution, and generate the Möbius group by matrix fraction division. Fix target points $p_j$ and put $S(x)={\sum }_{\gamma ,j}{\varphi }_{\gamma j}(x)\theta (\gamma )p_j$. This sum is locally finite and future timelike. Indeed, for future unit vectors $v,w$, $-⟨v,w⟩\ge 1$ by the distance formula, so for nonnegative weights $a_j$ with positive sum one has $-⟨\sum a_j{v}_j,\sum a_j{v}_j⟩\ge (\sum a_j{)}^2>0$. Thus $F_0(x)=S(x)\mathrm{/}\sqrt{-⟨S(x),S(x)⟩}$ is smooth. Reindexing the sum gives $F_0(\gamma x)=\theta (\gamma )F_0(x)$.

On each cusp take one fixed smooth function $\chi (r)$ equal to zero for $r\le 1\mathrm{/}2$ and one for $r\ge 1$. Where both $F_0$ and $C_i$ are defined, normalize their future-timelike linear combination $(1-\chi )F_0+\chi C_i$ as in step 2.2. Use $F_0$ below that overlap and $C_i$ above it. Constancy of the cutoff near its endpoints gives a smooth global map $F$, and all formulas are equivariant. On the compact quotient of the region outside the deep cusp tails, its derivative norm has a finite maximum by continuity. On the tails it is bounded by the finitely many affine constants in step 2.1. Integrating the derivative bound along hyperbolic segments proves that $F$ is globally Lipschitz. Repeating the construction for ${\theta }^{-1}$, with the inverse affine cusp maps, gives a globally Lipschitz map $G$.

The composites $GF$ and $FG$ are the identity on all sufficiently deep cusp tails. On the remaining compact quotient cores their displacement functions $d(GF(x),x)$ and $d(FG(y),y)$ are invariant under the respective deck groups and continuous, so have finite global bounds $D,D^{\mathrm{\prime }}$. If $L_G,L_F\ge 1$ are Lipschitz constants, the triangle inequality gives $d(x,x^{\mathrm{\prime }})\le 2D+L_{G}d(Fx,F{x}^{\mathrm{\prime }})$, and similarly in the other direction. Thus $F$ is a quasi-isometric embedding, and $d(FGy,y)\le D^{\mathrm{\prime }}$ proves coarse surjectivity. This proves the quasi-isometry assertion with explicit finite constants obtained from the constructed maps. On the quotients the maps are proper, since on each tail they preserve the depth coordinate and permute the finitely many cusp ends, and their remaining domains lie in compact cores.

Unique hyperbolic geodesics give an equivariant homotopy from $GF$ to the identity and from $FG$ to the identity. Their tracks have length at most $D,D^{\mathrm{\prime }}$. On the quotients these homotopies are proper: the inverse image of a compact target set is contained in the closed $D$ or $D^{\mathrm{\prime }}$ neighborhood of that set in the proper complete hyperbolic quotient, crossed with $[0,1]$. Completeness follows by lifting and extending geodesics in ${\mathbb{H}}^3$, and properness is Hopf–Rinow as in [F1]'s proof. Hence the descended maps are proper homotopy equivalences.