Hyperbolic Dehn filling and exceptional slopes

Let $p,q\in \mathbb{Z}$ satisfy $\gcd (p,q)=1$. Bézout's identity gives $r,s\in \mathbb{Z}$ with $ps-qr=1$. The matrix with columns $(p,q)$ and $(r,s)$ lies in $\mathrm{S}\mathrm{L}(2,\mathbb{Z})$, so it induces a homeomorphism of ${\mathbb{R}}^2\mathrm{/}{\mathbb{Z}}^2$. The image of the first coordinate circle is embedded, essential, and represents $p[\mu ]+q[\lambda ]$. Reversing its parameter negates both coefficients, so the unoriented curve is unchanged.

If $\gamma$ is an essential simple closed curve on $T$, it is nonseparating; cutting along it gives an annulus. An arc joining the two annulus boundary circles closes to a simple curve $\delta$ meeting $\gamma$ once, so the algebraic intersection is $[\gamma ]\cdot [\delta ]=\pm 1$. Writing $[\gamma ]=p[\mu ]+q[\lambda ]$ and $[\delta ]=a[\mu ]+b[\lambda ]$ gives $pb-qa=\pm 1$ because the ordered basis has intersection form $\pm 1$. Any common divisor of $p,q$ would divide $pb-qa$, so $\gcd (p,q)=1$.

Suppose ${\gamma }_0,{\gamma }_1$ are essential simple closed curves whose primitive classes agree up to sign. Reverse one orientation if necessary so their classes agree, and isotope them to be transverse. Let $v\in {\mathbb{Z}}^2$ be their common primitive class in the coordinates of [A1]. If they intersect, choose lifts ${\tilde{\gamma }}_0,{\tilde{\gamma }}_1\subset {\mathbb{R}}^2$ through one lift $x$ of an intersection point. Each chosen lift is preserved by the same nonzero deck translation $T_v$, because its projected curve represents $v$. Hence both $x$ and $T_v(x)$ lie on this pair of lifts and are distinct. By [F1] the curves therefore form an embedded bigon in $T$ whose interior misses them. Isotoping one curve across this bigon removes exactly two intersections and creates none. The transverse intersection set is finite, so repetition terminates and makes the curves disjoint. Cutting $T$ along ${\gamma }_0$ gives an annulus, and the disjoint essential curve ${\gamma }_1$ is parallel to its core, so the two curves cobound a subannulus. Isotopy across that annulus carries ${\gamma }_0$ to ${\gamma }_1$. Thus the primitive class modulo simultaneous sign determines exactly one unoriented isotopy class.

Let $c((e^{i\theta },w),s)=((1-s\mathrm{/}8)e^{i\theta },w)$ for $(e^{i\theta },w)\in \mathrm{\partial }{V}^{\mathrm{\prime }}$ and $s\in [0,2]$; this is a collar because its radial coordinate ranges from $1$ to $3\mathrm{/}4$. Set $\rho (s)=1-s$ for $0\le s\le 1$ and $\rho (s)=0$ for $1\le s\le 2$. The map $J(c(x,s))=c(H_{\rho (s)}(x),s)$ on the collar, extended by the identity outside it, is a homeomorphism: its inverse replaces $H_{\rho (s)}$ with $H_{\rho (s)}^{-1}$. It restricts to $H_1$ at the boundary and is isotopic to the identity through the same formula with $H_{u\rho (s)}$ for $u\in [0,1]$. Thus $h_0=J{\mathrm{\mid }}_{\mathrm{\partial }{V}^{\mathrm{\prime }}}\circ h$ sends $\mu$ exactly to ${\mu }^{\mathrm{\prime }}$, and extending $h_0$ suffices because postcomposition with $J^{-1}$ recovers an extension of $h$ without changing its orientation sign.

Cut $V$ and $V^{\mathrm{\prime }}$ along the product meridian disks $D=D^2\times \{1\}$ and $D^{\mathrm{\prime }}=D^2\times \{1\}$. Each cut-open solid torus is $D^2\times [0,1]$. After centering the interval at $0$, this convex cylinder is homeomorphic to the unit ball by radial rescaling: in direction $u=(u_1,u_2,u_3)\in S^2$ its boundary radius is $R(u)=\min \{(u_1^2+u_2^2{)}^{-1\mathrm{/}2},(2\mathrm{\mid }{u}_3\mathrm{\mid }{)}^{-1}\}$, with a zero denominator interpreted as $+\mathrm{\infty }$, and $(r,u)\mapsto rR(u)u$ gives the homeomorphism. The boundary of each cut-open torus consists of the annulus obtained by cutting the boundary torus along the meridian and two disk faces. The restriction of $h_0$ maps the annular portions to one another. Its common restriction to $\mu$ extends over each disk face by [F1], using the same disk map on both copies. The radial disk maps send the center of each disk face to the center of its target face. These maps assemble to a homeomorphism $\psi :\mathrm{\partial }B\to \mathrm{\partial }{B}^{\mathrm{\prime }}$ between the sphere boundaries of the cut-open balls, carrying the pair of disk-face centers to that pair in the target.

Identify each cut-open ball with the unit ball in ${\mathbb{R}}^3$. Apply [F1] with $d=3$ to extend $\psi$ radially over the ball. The cylinder-to-ball radial rescaling of step 2.1 carries the central product arc $\{0\}\times [0,1]$ to the diameter joining the two disk-face center points. Since $\psi$ carries those endpoints to the target pair, its radial extension carries this entire diameter to the corresponding target diameter. Thus the ball extension preserves the product core arc. The extension has orientation sign equal to the degree sign of $\psi$. On the open annulus away from the disk faces, $\psi$ is exactly $h_0$; cutting does not change the local boundary orientation there, so this degree sign is the orientation sign of $h_0$.

The two copies of each meridian disk receive the same disk map from [F1], so the ball extension agrees on their identified faces. It therefore descends after regluing to a homeomorphism $V\to V^{\mathrm{\prime }}$ extending $h_0$ and carrying the product core circle to the target core circle. Composing with $J^{-1}$ extends the original $h$; this collar-supported map fixes the core and keeps the same orientation sign because $J$ is isotopic to the identity.

If $k=0$, the collection of germs and every indexed assertion are empty. Both vectors are the unique vector of ${\mathbb{C}}^0$, with norm zero; there is no parameter with $u\mathrm{\ne }0$, so the final limiting assertion is vacuous. This proves the Statement in that case. Henceforth suppose $k\ge 1$. Composition of the two affine maps shows that commutativity is exactly $a_i(e^{v_i}-1)=b_i(e^{u_i}-1)$. Let $E(t)=(e^t-1)\mathrm{/}t$ for $t\mathrm{\ne }0$ and $E(0)=1$; its convergent series is ${\sum }_{n\ge 0}{t}^n\mathrm{/}(n+1)!$, so it is holomorphic and nonzero near zero. Since $a_i(o)\mathrm{\ne }0$, the holomorphic function ${\tau }_i=(b_i\mathrm{/}{a}_i)E(u_i)\mathrm{/}E(v_i)$ is defined near $o$. The commutativity equation becomes exactly $v_i={\tau }_i{u}_i$, and ${\tau }_i(o)=b_i(o)\mathrm{/}{a}_i(o)$. The finitely many nonreal values permit one common neighborhood on which all $\mathrm{Im}{\tau }_i$ are nonzero. Thus all ${\tau }_i$ are nonzero, proving the equivalence of their zero logarithms and the asserted quotient limit. The argument is an identity of holomorphic functions on $Z$, so singular points cause no difficulty.

When $u_i\mathrm{\ne }0$, real linear independence of $u_i$ and $v_i={\tau }_i{u}_i$ follows from $\mathrm{Im}{\tau }_i\mathrm{\ne }0$. Hence the real linear map $(p,q)\mapsto p{u}_i+q{v}_i$ is invertible and gives the unique real pair. Its equation is $(p_i+q_i{\tau }_i)u_i=2\pi i$. Since $\mathrm{Im}{\tau }_i^0\mathrm{\ne }0$, the denominator $p_i+q_i{\tau }_i^0$ cannot vanish: if it did, its imaginary part would give $q_i=0$, then its real part $p_i=0$, contrary to the full-turn equation. Therefore ${\tilde{u}}_i\mathrm{/}{u}_i=(p_i+q_i{\tau }_i)\mathrm{/}(p_i+q_i{\tau }_i^0)$ and $({\tilde{u}}_i-u_i)\mathrm{/}{u}_i=q_i({\tau }_i-{\tau }_i^0)\mathrm{/}(p_i+q_i{\tau }_i^0)$. The elementary bound $\mathrm{\mid }{p}_i+q_i{\tau }_i^0\mathrm{\mid }\ge \mathrm{\mid }{q}_i\mathrm{\mid }\mathrm{\mid }\mathrm{Im}{\tau }_i^0\mathrm{\mid }$ proves the estimate, including $q_i=0$ where its left side is zero.

On a sufficiently small fixed neighborhood all ${\tau }_i$ are bounded, so step 2.1 gives $\mathrm{\mid }{\tilde{u}}_i\mathrm{\mid }\le C\mathrm{\mid }{u}_i\mathrm{\mid }$. Thus extending ${\tilde{u}}_i$ by zero across $u_i=0$ is continuous there; elsewhere its formula is continuous because it comes from an invertible real linear system. Put $\eta (z)={\max }_i\mathrm{\mid }{\tau }_i(z)-{\tau }_i^0\mathrm{\mid }\mathrm{/}\mathrm{\mid }\mathrm{Im}{\tau }_i^0\mathrm{\mid }$. Finiteness of $k$ and continuity give $\eta (z)\to 0$ as $z\to o$. Applying the scalar estimates coordinatewise, also where $u_i=0$, gives $\mathrm{\parallel }\tilde{u}-u\mathrm{\parallel }\le \eta (z)\mathrm{\parallel }u\mathrm{\parallel }$. This proves the final assertion. No claim that $u$ is a coordinate system or has nonvanishing derivative has entered the argument.

Put $G={\mathrm{P}\mathrm{S}\mathrm{L}}_2(\mathbb{C})$, $K=\mathrm{P}\mathrm{S}\mathrm{U}(2)$ and $V={\mathfrak{s}\mathfrak{l}}_2(\mathbb{C})$. Choose $X_1=\frac{1}{2}\left(\begin{array}{cc}i& 0\\ 0& -i\end{array}\right)$, $X_2=\frac{1}{2}\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)$ and $X_3=\frac{1}{2}\left(\begin{array}{cc}0& i\\ i& 0\end{array}\right)$ in $\mathfrak{k}$, and put $Y_j=i{X}_j$. These $Y_j$ are an orthonormal tangent basis for curvature $-1$ hyperbolic space; for example $\exp (t{Y}_1)$ moves the basepoint distance $\mathrm{\mid }t\mathrm{\mid }$, and the $K$ action rotates this basis. The matrix commutators give $[X_i,X_j]={ϵ}_{ijk}{X}_k$ and $[Y_i,Y_j]=i{ϵ}_{ijk}{Y}_k$. On $V$ let $R_j=\mathrm{ad}{Y}_j$. Each $Y_j$ is Hermitian, so $R_j$ is self-adjoint for $⟨A,B⟩=\mathrm{tr}(A{B}^{\ast })$, by cyclicity of trace. The $K$ action is unitary for this product.

Represent $E$-valued forms by horizontal $K$-equivariant $V$-valued forms on $Q=\mathrm{\Gamma }\mathrm{\backslash }G$, expressing their coefficients in the left Maurer–Cartan frame. Indeed a frame $g$ identifies the fiber over $gK$ with $V$ by $\mathrm{Ad}g$, and replacing $g$ by $gk$ changes these coefficients by $\mathrm{Ad}{k}^{-1}$ and rotates the tangent indices. In these coefficients write ${\epsilon }_j$ for exterior multiplication by the dual of $Y_j$ and ${\iota }_j$ for contraction by $Y_j$. The ordinary covariant differential of the $K$ connection is $D={\sum }_j{\epsilon }_j{Y}_j$, because the horizontal part of $[Y_i,Y_j]$ is zero. Differentiating the fiber change $\mathrm{Ad}g$ gives $d_E=D+T$, where $T={\sum }_j{\epsilon }_j{R}_j$. These are actual coefficient formulas on all degrees, not definitions of a different connection.

The invariant volume on $Q$ is the product of hyperbolic volume and normalized compact-fiber volume. Integration in a local bundle chart and then a subordinate smooth partition of unity [F2] shows that integration of a $K$-invariant coefficient norm on $Q$ equals its integration on $M$. The left invariant fields $Y_j$ have zero divergence for this volume: $G$ is unimodular, since $\mathrm{tr}(\mathrm{ad}Z)=0$ for the displayed Lie algebra basis, hence for all $Z$, and $\det \mathrm{Ad}(\exp Z)=1$. Right translation therefore preserves its Haar volume. Integration by parts on compact supports now gives $D^{\ast }=-{\sum }_j{\iota }_j{Y}_j$ and $T^{\ast }={\sum }_j{\iota }_j{R}_j$. The relations ${\epsilon }_i{\iota }_j+{\iota }_j{\epsilon }_i={\delta }_{ij}$ show that $D{T}^{\ast }+T^{\ast }D+T{D}^{\ast }+D^{\ast }T=0$: the two anticommutator sums are respectively ${\sum }_j{R}_j{Y}_j$ and its negative, since coefficient differentiation commutes with the constant matrices $R_j$. Thus $d_E{\delta }_E+{\delta }_E{d}_E=D{D}^{\ast }+D^{\ast }D+H$, where $H=T{T}^{\ast }+T^{\ast }T$. All statements are $K$-equivariant and descend from $Q$ to $M$.

On one-form coefficients $a=(a_1,a_2,a_3)\in V^3$, the exterior-contraction relations give $(Ha{)}_i={\sum }_j{R}_j^2{a}_i+{\sum }_j[R_i,R_j]a_j$. This finite-dimensional self-adjoint operator is positive definite. If $Ha=0$, positivity of $T{T}^{\ast }+T^{\ast }T$ gives $Ta=0$, namely $R_i{a}_j=R_j{a}_i$ for all $i,j$. Since $[R_i,R_j]=i{ϵ}_{ijk}{R}_k$, the commutator sum in the displayed formula is a multiple of $R_k{a}_j-R_j{a}_k$ and vanishes. Consequently $0=⟨Ha,a⟩={\sum }_{i,j}\mathrm{\parallel }{R}_j{a}_i{\mathrm{\parallel }}^2$. Each $a_i$ commutes with all $Y_j$, which span ${\mathfrak{s}\mathfrak{l}}_2(\mathbb{C})$ over $\mathbb{C}$. A matrix commuting with all traceless matrices is scalar, and a traceless scalar is zero. Thus $a=0$. Compactness of the unit sphere in $V^3$ supplies $c>0$ with $H\ge cI$.

For a compactly supported smooth one-form, integration by parts in step 3.1 and step 4.1 give $\mathrm{\parallel }{d}_E\alpha {\mathrm{\parallel }}_2^2+\mathrm{\parallel }{\delta }_E\alpha {\mathrm{\parallel }}_2^2=\mathrm{\parallel }D\alpha {\mathrm{\parallel }}_2^2+\mathrm{\parallel }{D}^{\ast }\alpha {\mathrm{\parallel }}_2^2+(H\alpha ,\alpha )\ge c\mathrm{\parallel }\alpha {\mathrm{\parallel }}_2^2$. We next extend this inequality to the joint weak graph domain. On each cusp choose a smooth cutoff equal to one for $t\le R$, zero for $t\ge 2R$, with derivative bounded by $C\mathrm{/}R$, and set it equal to one on the compact core. These cutoffs ${\chi }_R$ have compact support and $\mathrm{\mid }d{\chi }_R\mathrm{\mid }\le C\mathrm{/}R$. For a weak graph-domain form $v$, the product formulas $d_E({\chi }_{R}v)={\chi }_R{d}_{E}v+d{\chi }_R\wedge v$ and ${\delta }_E({\chi }_{R}v)={\chi }_R{\delta }_{E}v-{\iota }_{\mathrm{\nabla }{\chi }_R}v$ show convergence to $v$ in the joint graph norm. The same conclusion holds separately for each differential and codifferential and for forms of every degree. In the compact case take the cutoff identically one.

Here are the local smoothing details needed to finish that graph-density argument. In a finite collection of trivializing charts around the support of ${\chi }_{R}v$, every differential or codifferential is a first-order matrix operator $P={\sum }_j{A}_j(x){\mathrm{\partial }}_j+B(x)$ with smooth coefficients. Multiplication by a subordinate smooth partition [F2] reduces to forms supported inside individual charts. For convolution with a smooth unit-mass kernel ${\rho }_{ϵ}$, the difference $P(v\ast {\rho }_{ϵ})-(Pv)\ast {\rho }_{ϵ}$ is a sum of integral operators with kernels $(A_j(x)-A_j(y)){\mathrm{\partial }}_j{\rho }_{ϵ}(x-y)+({\mathrm{\partial }}_j{A}_j)(y){\rho }_{ϵ}(x-y)$ and a zero-order coefficient commutator. The $L^1$ kernel bounds are uniform on the compact chart, by the mean-value bound for $A_j$ and scaling of ${\rho }_{ϵ}$; thus these commutators are uniformly bounded on $L^2$. They tend to zero on smooth functions by differentiation of convolutions, hence on all $L^2$ functions by the smooth-density fact in [F2] and the uniform bound. Therefore local mollification converges in each asserted graph norm, also for the direct sum operator $(d_E,{\delta }_E)$. The partition contributes only fixed zero-order commutators, so a finite sum completes the approximation by compactly supported smooth forms. Together with step 5.1 this proves the coercive inequality on the maximal joint graph domain. It also proves that maximal differentials are the adjoints of maximal formal codifferentials, since their compact-support integration-by-parts identities extend by this graph density.

Let ${\mathcal{H}}_p=L^2({\mathrm{\Lambda }}^p{T}^{\ast }M\otimes E)$ and let $d_p$ be the maximal weak differential. It is closed: convergence of a form and its differential in $L^2$ preserves the weak test-function identity. Flatness gives $d_1{d}_0=0$ distributionally. Put $Z=\ker d_1$, a closed subspace of ${\mathcal{H}}_1$, and $\delta =d_0^{\ast }$. Every $w\in Z^{\perp }$ belongs to $\mathrm{Dom}\delta$ with $\delta w=0$, because it is orthogonal to $\mathrm{ran}{d}_0\subseteq Z$. Hence the orthogonal projection $P_Z$ maps $\mathrm{Dom}\delta$ into itself and $\delta P_{Z}v=\delta v$. On $Z\cap \mathrm{Dom}\delta$, steps 5.1 and 6.1 give $\mathrm{\parallel }v{\mathrm{\parallel }}_2\le c^{-1\mathrm{/}2}\mathrm{\parallel }\delta v{\mathrm{\parallel }}_2$. In particular $A=\delta {\mathrm{\mid }}_{Z\cap \mathrm{Dom}\delta }$ is injective with closed range: a Cauchy sequence of its images is, by this estimate, the image under the closed operator $\delta$ of a convergent sequence in $Z$.

Fix a weakly closed $\alpha \in {\mathcal{H}}_1$, so $\alpha \in Z$. On the closed range of $A$ define $\mathrm{\ell }(Av)=⟨v,\alpha ⟩$. It is well defined and bounded, with norm at most $c^{-1\mathrm{/}2}\mathrm{\parallel }\alpha {\mathrm{\parallel }}_2$, by step 7.1. Extend it to ${\mathcal{H}}_0$ by zero on the orthogonal complement of that range and use [F1] to obtain $b\in {\mathcal{H}}_0$ with $⟨Av,b⟩=⟨v,\alpha ⟩$ and $\mathrm{\parallel }b{\mathrm{\parallel }}_2\le c^{-1\mathrm{/}2}\mathrm{\parallel }\alpha {\mathrm{\parallel }}_2$. For arbitrary $v\in \mathrm{Dom}\delta$, replace $v$ by $P_{Z}v$; its $\delta$ image is unchanged and its pairing with $\alpha$ is unchanged. Thus $⟨\delta v,b⟩=⟨v,\alpha ⟩$ on the full domain of $\delta$. By the adjoint definition and the graph-density conclusion of step 6.1, $b\in \mathrm{Dom}{d}_0$ and $d_{0}b=\alpha$. This proves the primitive assertion in the weak sense without an external closed-range or harmonic-representative theorem.

If $\alpha$ is smooth, the primitive $b$ is smooth. In a flat parallel frame on a coordinate ball, $d_{E}b=\alpha$ is the ordinary componentwise distributional equation $db=\alpha$. A smooth closed one-form on a smaller star-shaped coordinate ball has a smooth primitive $b_0$, given by integrating along straight radial segments; differentiating the integral and using closedness verifies $d{b}_0=\alpha$. Then $d(b-b_0)=0$ distributionally. Mollification on still smaller balls makes this difference a sequence of constants, so its distributional limit is constant as well. Thus $b=b_0+\text{constant}$ is smooth locally and hence globally. Its $L^2$ norm and the bound from step 8.1 remain unchanged.

Shrink to a closed convex polydisc $B_{ϵ}=\{\mathrm{\parallel }u{\mathrm{\parallel }}_{\mathrm{\infty }}\le ϵ\}$ inside the holomorphy domain. Its compactness and continuity of the derivatives give a common Lipschitz constant $L$ for all ${\tau }_i$ there, by integrating their derivatives along line segments. Put $A={\max }_i\mathrm{\mid }\mathrm{Im}{\tau }_i(0){\mathrm{\mid }}^{-1}$ and shrink $ϵ$ so that $ALϵ\le 1\mathrm{/}4$ and every $\mathrm{Im}{\tau }_i(u)$ stays nonzero on $B_{ϵ}$. For any nonzero real pair $(p,q)$, $p+q{\tau }_i(0)\mathrm{\ne }0$ and $\mathrm{\mid }q\mathrm{/}(p+q{\tau }_i(0))\mathrm{\mid }\le A$. Moreover $\mathrm{\mid }p+q{\tau }_i(0)\mathrm{\mid }\ge c\sqrt{p^2+q^2}$ for one common $c>0$: the real-linear maps $(p,q)\mapsto p+q{\tau }_i(0)$ are invertible, and their moduli have a positive minimum on the unit circle.

For a parameter tuple with finite pairs outside a sufficiently large radius $R$, define $a_i=q_i\mathrm{/}(p_i+q_i{\tau }_i(0))$ and the $w_i$ in the Statement. At an infinite coordinate put $a_i=w_i=0$. On $B_{ϵ}$ define $T_i(u)=w_i\mathrm{/}[1+a_i({\tau }_i(u)-{\tau }_i(0))]$. Step 1.1 makes every denominator have modulus at least $3\mathrm{/}4$, hence $\mathrm{\parallel }T(u){\mathrm{\parallel }}_{\mathrm{\infty }}\le (4\mathrm{/}3)\mathrm{\parallel }w{\mathrm{\parallel }}_{\mathrm{\infty }}$. Also $\mathrm{\parallel }T(u)-T(u^{\mathrm{\prime }}){\mathrm{\parallel }}_{\mathrm{\infty }}\le (16\mathrm{/}9)AL\mathrm{\parallel }w{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{\parallel }u-u^{\mathrm{\prime }}{\mathrm{\parallel }}_{\mathrm{\infty }}$. Since step 1.1 makes $\mathrm{\parallel }w{\mathrm{\parallel }}_{\mathrm{\infty }}\le 2\pi \mathrm{/}(cR)$, choose $R$ so large that $\mathrm{\parallel }w{\mathrm{\parallel }}_{\mathrm{\infty }}\le ϵ\mathrm{/}2$ and $(16\mathrm{/}9)AL\mathrm{\parallel }w{\mathrm{\parallel }}_{\mathrm{\infty }}\le 1\mathrm{/}2$. The closed polydisc is nonempty and complete, and $T$ maps it into itself, so [F1] supplies its unique fixed point. Its defining equations are exactly the simultaneous full-turn equations at finite coordinates, and $u_i=0$ at infinite ones. At a finite coordinate $w_i\mathrm{\ne }0$, so its solution $u_i$ cannot vanish. Every solution in $B_{ϵ}$ is such a fixed point and is therefore this one.

The fixed-point bound in step 2.1 gives $\mathrm{\parallel }u{\mathrm{\parallel }}_{\mathrm{\infty }}\le (4\mathrm{/}3)\mathrm{\parallel }w{\mathrm{\parallel }}_{\mathrm{\infty }}$. Its coordinate equation also gives $\mathrm{\mid }{u}_i-w_i\mathrm{\mid }\le (4\mathrm{/}3)A\mathrm{\mid }{w}_i\mathrm{\mid }\mathrm{\mid }{\tau }_i(u)-{\tau }_i(0)\mathrm{\mid }\le (16\mathrm{/}9)AL\mathrm{\mid }{w}_i\mathrm{\mid }\mathrm{\parallel }w{\mathrm{\parallel }}_{\mathrm{\infty }}$. This proves the stated uniform expansion, including every $w_i=0$ coordinate exactly.

Define the coefficient map $\mathrm{\Phi }$ on the open polydisc by the equations in the Statement. If $u_i\mathrm{\ne }0$, the real-linear equation $p_i+q_i{\tau }_i(u)=2\pi i\mathrm{/}{u}_i$ has a unique real pair, continuously depending on $u$. If $u_i\to 0$, the norms of these pairs tend to infinity uniformly with the other coordinates in $B_{ϵ}$: boundedness of ${\tau }_i$ gives $2\pi \mathrm{/}\mathrm{\mid }{u}_i\mathrm{\mid }=\mathrm{\mid }{p}_i+q_i{\tau }_i(u)\mathrm{\mid }\le C\sqrt{p_i^2+q_i^2}$. Thus $\mathrm{\Phi }$ is continuous also at its infinite coordinates, with $\mathrm{\Phi }(0)=(\mathrm{\infty },\dots ,\mathrm{\infty })$. Let $V$ be the product neighborhood specified using $R$ in step 2.1, and put $U={\mathrm{\Phi }}^{-1}(V)$ inside the open polydisc. This is an open neighborhood of $0$. Every parameter in $V$ has its unique solution in the interior by step 2.1, so $\mathrm{\Phi }:U\to V$ is bijective.

Its inverse is continuous on the whole product neighborhood, including partially unfilled tuples. If parameter tuples converge, their fixed-point maps $T$ from step 2.1 converge uniformly on $B_{ϵ}$. For a finite limiting coordinate, $a_i,w_i$ converge in the displayed rational formulas. For an infinite limiting coordinate, $w_i\to 0$ while $\mathrm{\mid }{a}_i\mathrm{\mid }\le A$, so the corresponding $T_i$ tends uniformly to zero by the $3\mathrm{/}4$ denominator bound, regardless of the directions of the real pairs. If $u,u^{\mathrm{\prime }}$ are the two fixed points and $T,T^{\mathrm{\prime }}$ their maps, then $\mathrm{\parallel }u-u^{\mathrm{\prime }}{\mathrm{\parallel }}_{\mathrm{\infty }}\le \frac{1}{2}\mathrm{\parallel }u-u^{\mathrm{\prime }}{\mathrm{\parallel }}_{\mathrm{\infty }}+{\sup }_{B_{ϵ}}\mathrm{\parallel }T-T^{\mathrm{\prime }}{\mathrm{\parallel }}_{\mathrm{\infty }}$, proving continuity. This establishes the homeomorphism. No primitivity was used in this analytic proof; primitivity enters only when a later geometric construction identifies a smooth solid-torus meridian.

In $S^3=\{(z_1,z_2)\in {\mathbb{C}}^2:\mathrm{\mid }{z}_1{\mathrm{\mid }}^2+\mathrm{\mid }{z}_2{\mathrm{\mid }}^2=1\}$, the standard Hopf link is $H=\{z_1{z}_2=0\}$. On its complement, $f(z_1,z_2)=z_1{z}_2\mathrm{/}\mathrm{\mid }{z}_1{z}_2\mathrm{\mid }$ is a submersion to $S^1$. Writing $r=\mathrm{\mid }{z}_1{\mathrm{\mid }}^2$, $\theta =\arg z_1$, and $\varphi =\arg z_2$, a fiber is $0<r<1$, $\varphi =-\theta +\text{constant}$, hence an annulus. For base angle $s\in [0,2\pi ]$, transport by ${\theta }_s=\theta +\chi (r)s$, ${\varphi }_s=\varphi +(1-\chi (r))s$, where $\chi =1$ near $r=0$ and $\chi =0$ near $r=1$, fixes both boundary circles at $s=2\pi$ and makes one full annular twist between them. The sign is determined by the signed linking of the positive normal push-off of the core with the core: changing the handedness of the Hopf band reverses the twist. With $a{\cdot }_{F}b=+1$, the positive twist about $a$ acts by $b\mapsto a+b$, while the negative twist about $b$ acts by $a\mapsto a+b$. Hence the two band monodromies have the stated actions $R_a$ and $L_b$.

The square plumbing is the Murasugi sum in Construction 5.37 of [F3]. Theorem 5.38 of [F3] gives a fibration with fiber $F$ and return map the composition of the two Hopf-band monodromies extended by the identity; in the stipulated order, step 1.1 identifies it as $R_a\circ L_b$. Example 5.39 of [F3] identifies the boundary of the opposite-handed sum as the standard figure-eight knot. Deleting the binding neighborhood as in [F1] therefore gives the stated exterior as the mapping torus.

Each annulus has Euler characteristic $0$, and the plumbing square has Euler characteristic $1$, so $\chi (F)=0+0-1=-1$. The plumbing is connected, oriented, and has one boundary component by the stated gluing; hence $\chi (F)=1-2g(F)$ gives $g(F)=1$. Thus $F$ is a once-punctured torus.

In the ordered basis $([a],[b])$, the columns of $R_a$ are $(1,0)$ and $(1,1)$, while the columns of $L_b$ are $(1,1)$ and $(0,1)$. Their product is $\left(\begin{array}{cc}1& 1\\ 0& 1\end{array}\right)\left(\begin{array}{cc}1& 0\\ 1& 1\end{array}\right)=\left(\begin{array}{cc}2& 1\\ 1& 1\end{array}\right)$. Inclusion of the once-punctured torus into the torus obtained by capping its sole boundary induces an isomorphism on $H_1$, since attaching a disk along the boundary adds no 1-cycle or 2-cycle relation in degree one. Extending the boundary-fixing monodromy by the identity on that disk therefore has the same displayed action.