Small augmented holonomy deformations have smooth geometric fillings

We first give an isometry-equivariant way of blending maps into ${\mathbb{H}}^3$. Use the Lorentz form $⟨X,X⟩=X_1^2+X_2^2+X_3^2-X_4^2$ and embed upper half-space by $\mathrm{\Phi }(Z,Y)=(\mathrm{Re}Z\mathrm{/}Y,\mathrm{Im}Z\mathrm{/}Y,(\mathrm{\mid }Z{\mathrm{\mid }}^2+Y^2-1)\mathrm{/}(2Y),(\mathrm{\mid }Z{\mathrm{\mid }}^2+Y^2+1)\mathrm{/}(2Y))$. Direct substitution gives $⟨\mathrm{\Phi },\mathrm{\Phi }⟩=-1$, positive last coordinate, and differentiation gives the upper half-space metric $(\mathrm{\mid }dZ{\mathrm{\mid }}^2+d{Y}^2)\mathrm{/}{Y}^2$. Its coordinate matrix in $(\mathrm{Re}Z,\mathrm{Im}Z,Y)$ is $Y^{-2}{I}_3$, so its volume coefficient is the positive square root of $Y^{-6}$, namely $Y^{-3}$. The action of $G={\mathrm{P}\mathrm{S}\mathrm{L}}_2(\mathbb{C})$ is linear on these Lorentz coordinates. Indeed the equivalent Hermitian-matrix coordinate is $Y^{-1}\left(\begin{array}{cc}\mathrm{\mid }Z{\mathrm{\mid }}^2+Y^2& Z\\ \bar{Z}& 1\end{array}\right)$; it is positive definite with determinant one, and $X\mapsto AX{A}^{\ast }$, for $A\in S{L}_2(\mathbb{C})$, is linear on the four real Hermitian coordinates, preserves determinant and positivity, and is unchanged by replacing $A$ by $-A$. For positive weights summing to one, a weighted sum $S$ of future unit timelike vectors is future timelike: writing each vector as $(x,t)$ gives $t>\mathrm{\mid }x\mathrm{\mid }$, and the triangle inequality gives this strict inequality for the sum. Thus $B(S)=S\mathrm{/}\sqrt{-⟨S,S⟩}$ is well defined and smooth. It commutes with every $G$ action. If every input equals $X$, its value is $X$. This is a normalized Lorentz sum, rather than an undefined linear average inside a manifold.

Fix an end and omit its index temporarily. Choose the frames in [F1] so that at zero its original developing map is $(x,y,t)\mapsto (a(0)(x+\tau (0)y),e^t)$ on ${\mathbb{R}}^2\times [0,\mathrm{\infty })$. The deck generators add one to $x$ and $y$. Such a normalization is obtained by a fixed translation and dilation of a lift of the original horoball; it only fixes the origin and the level called $t=0$. Write $\zeta =x+\tau (u)y$, $L=u\zeta$, $w=(1-e^u)\mathrm{/}a(u)$ and $B=\sqrt{(1+\mathrm{\mid }w{\mathrm{\mid }}^2)\mathrm{/}(1+e^{2t}\mathrm{\mid }w{\mathrm{\mid }}^2)}$, choosing the positive square root. Put $E(z)=(e^z-1)\mathrm{/}z$ with $E(0)=1$. Define an end map ${\mathrm{\Psi }}_u=(Z_u,Y_u)$ by $Y_u=e^t\mathrm{\mid }{e}^L\mathrm{\mid }B$ and $Z_u=a(u)\zeta E(L)\mathrm{/}E(u)+e^L(e^{2t}-1)\bar{w}\mathrm{/}((1+e^{2t}\mathrm{\mid }w{\mathrm{\mid }}^2)(1+B))$. All these formulas are smooth in $(x,y,t,\mathrm{Re}u,\mathrm{Im}u)$, and in the other deformation coordinates: $a,\tau$ are holomorphic, $a$ and $E(u)$ are nonzero after shrinking, and every real denominator is positive. At $u=0$ they give precisely $(a(u)(x+\tau (u)y),e^t)$, including when other deformation coordinates are nonzero. At the complete parameter they give the original developing map. Consequently their convergence, with every derivative, is uniform on each fixed compact coordinate collar.

Choose product coordinates $T_i^2\times [0,\mathrm{\infty })$ on the ends, and include their collars through $t=2$ in a compact set $K$. The given complete developing map identifies the universal cover of $M$ with ${\mathbb{H}}^3$, with deck action ${\rho }_0$. Cover $K$ by finitely many simply connected evenly covered coordinate sets $U_j$ and choose one lift ${\tilde{U}}_j$ for each. On an open neighborhood $V$ of $K$ covered by these sets, use [F2] to choose a subordinate smooth partition ${\varphi }_j$. Pull each function to each component ${\rho }_0(\gamma ){\tilde{U}}_j$ and extend it by zero; denote this lifted function by ${\varphi }_{\gamma j}$. Its support stays inside that component. These functions are locally finite and sum to one on the inverse image of $V$. Put $D_{\gamma j}(x,u)={\rho }_u(\gamma ){\rho }_0(\gamma {)}^{-1}x$ and let $D_u^{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}(x)$ be the Lorentz blend of the points $D_{\gamma j}(x,u)$ with weights ${\varphi }_{\gamma j}(x)$. The sums are locally finite and smooth. The deck identities for the weights and $D_{\delta \gamma ,j}({\rho }_0(\delta )x,u)={\rho }_u(\delta )D_{\gamma j}(x,u)$ show exact ${\rho }_u$ equivariance. At zero every input is $x$, so $D_0^{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}(x)=x$.

When $u\mathrm{\ne }0$, put $c=1\mathrm{/}w=a\mathrm{/}(1-e^u)$ and $C=\mathrm{\mid }c\mathrm{\mid }$. The identity $(1-B)\mathrm{/}w=(e^{2t}-1)\bar{w}\mathrm{/}((1+e^{2t}\mathrm{\mid }w{\mathrm{\mid }}^2)(1+B))$ rewrites the preceding formula as $Z_u=c(1-B{e}^L)$, $Y_u=e^t\mathrm{\mid }{e}^L\mathrm{\mid }B$. On adding one to $x$, these become $Z\mapsto e^{u}Z+a$, $Y\mapsto e^{\mathrm{Re}u}Y$. Commutativity gives $a(e^v-1)=b(e^u-1)$, so $c(1-e^v)=b$; adding one to $y$ gives the other peripheral isometry. When $u=0$, the translation formula in step 1.2 gives the same equivariance because $v=0$ and $b=a\tau$. Compose ${\mathrm{\Psi }}_u$ with the inverse of the frame from [F1]; this gives a map $D_u^{\mathrm{e}\mathrm{n}\mathrm{d}}$ on a chosen lifted end, equivariant under its peripheral subgroup. Extend to all its deck translates by $D_u^{\mathrm{e}\mathrm{n}\mathrm{d}}({\rho }_0(\gamma )x)={\rho }_u(\gamma )D_u^{\mathrm{e}\mathrm{n}\mathrm{d}}(x)$. This is well defined because two descriptions differ by a peripheral deck transformation.

On the compact quotient $K$, the last construction converges to the identity in every derivative as $u\to 0$, in fixed local lift charts. To see uniformity without an assertion about infinitely many deck transformations, cover $K$ by finitely many smaller relatively compact chart pieces and lift each piece once. Only finitely many of the locally finite functions ${\varphi }_{\gamma j}$ meet those compact lifts. Only these finitely many matrices ${\rho }_u(\gamma ){\rho }_0(\gamma {)}^{-1}$ enter the derivative estimates there, and each converges smoothly to the identity. Invertibility of a matrix is open; compactness and [F2] therefore give one neighborhood of zero on which $D_u^{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}$ is a local diffeomorphism at all points over $K$. Its pullback metric descends by equivariance, is hyperbolic, and converges smoothly to the original metric on $K$.

The end map is a local diffeomorphism on its entire domain. For $u=0$ this follows directly from the translation formula and $\mathrm{Im}\tau >0$. For $u\mathrm{\ne }0$ introduce Fermi coordinates about the vertical geodesic with endpoints $c,\mathrm{\infty }$: $Z=c+e^s\tanh r{e}^{i\theta }$, $Y=e^s\mathrm{/}\cosh r$. Direct differentiation using [F3] gives $d{r}^2+{\sinh }^{2}rd{\theta }^2+{\cosh }^{2}rd{s}^2$. The map in step 2.2 is precisely $r(t)=\mathrm{arsinh}(C{e}^{-t})$, $s=\log \sqrt{C^2+1}+\mathrm{Re}L$, $\theta =\arg (-c)+\mathrm{Im}L$. The choice of the constant argument is immaterial modulo $2\pi$. The real linear map $(x,y)\mapsto (\mathrm{Re}L,\mathrm{Im}L)$ has determinant $\mathrm{Im}(\bar{u}v)=\mathrm{\mid }u{\mathrm{\mid }}^2\mathrm{Im}\tau >0$, and $r^{\mathrm{\prime }}(t)\mathrm{\ne }0$. Thus the differential is invertible everywhere for finite $t$. For $t\ge T$ its image is the whole punctured tube $0<r\le R(T)$, where $R(T)=\mathrm{arsinh}(C{e}^{-T})$, and $(s,\theta )$ before angular reduction range over all ${\mathbb{R}}^2$. No discreteness of the full peripheral image is required for this local assertion.

We now join the core and the end maps. Take a fixed smooth function $\beta (t)$ equal to zero near $t=0$ and one for $t\ge 1$. On the end collar $0\le t\le 1$, Lorentz blend $D_u^{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}$ and $D_u^{\mathrm{e}\mathrm{n}\mathrm{d}}$ with weights $1-\beta$ and $\beta$; use the core map below this collar and the end map above it. Both maps have the same ${\rho }_u$ equivariance, so the blend does too. The lifted end components are disjoint; their deck translates give the construction on all of the universal cover. At the complete parameter both inputs are the identity, and steps 3.1 and 1.2 show smooth convergence on the compact transition collars. Therefore, after one common shrinking for the finitely many ends, the blended differential is invertible throughout these collars. Above them step 3.2 applies; below them step 3.1 applies. We have constructed a global smooth equivariant local diffeomorphism $D_u:\tilde{M}\to {\mathbb{H}}^3$. Its pullback of the hyperbolic metric is invariant under the original deck group and hence defines $g_u$ on $M$, with holonomy ${\rho }_u$. Local inverses from [F2] give its hyperbolic charts. The construction is smooth in parameters on compact subsets and gives $g_0$ exactly.

If $u_i=0$, the metric above the collar is the translation-cusp metric $d{t}^2+e^{-2t}\mathrm{\mid }{a}_i(u){\mathrm{\mid }}^2\mathrm{\mid }dx+{\tau }_i(u)dy{\mathrm{\mid }}^2$. It has volume $\mathrm{\mid }{a}_i(u){\mathrm{\mid }}^2\mathrm{Im}{\tau }_i(u)e^{-2T}\mathrm{/}2$ for $t\ge T\ge 1$, by integration on $[0,1{]}^2\times [T,\mathrm{\infty })$. Its end is complete: the length of any path dominates the total change in $t$, so reaching arbitrarily large $t$ requires unbounded length. This proves all assertions about unfilled ends, independently of what happens at the other cusps.

Fix a filled cusp, so $u\mathrm{\ne }0$ by its full-turn equation. Let $m=(p,q)$ be its primitive filling vector and choose integers $p^{\mathrm{\prime }},q^{\mathrm{\prime }}$ with $p{q}^{\mathrm{\prime }}-q{p}^{\mathrm{\prime }}=1$, using the integer Euclidean algorithm. Put $d=(p^{\mathrm{\prime }},q^{\mathrm{\prime }})$ and $W=p^{\mathrm{\prime }}u+q^{\mathrm{\prime }}v=A+i{B}_0$. Under the Fermi linear map in step 3.2, $m$ maps to $(0,2\pi )$ and $d$ to $(A,B_0)$. The determinant identity gives $-2\pi A=\mathrm{\mid }u{\mathrm{\mid }}^2\mathrm{Im}\tau >0$. In particular $A\mathrm{\ne }0$ and $\mathrm{\mid }A\mathrm{\mid }=\mathrm{\mid }u{\mathrm{\mid }}^2\mathrm{Im}\tau \mathrm{/}(2\pi )$. The vectors $m,d$ are an integral basis, so the image of the deck lattice is exactly the lattice generated by $(0,2\pi )$ and $(A,B_0)$. It follows that the end $t\ge T\ge 1$ with its actual metric is isometric to the punctured radius-$R(T)$ hyperbolic tube modulo the screw motion $(r,\theta ,s)\mapsto (r,\theta +B_0,s+A)$. This is a bijective local isometry: the real linear map is invertible, $t\mapsto r$ is one-to-one onto $(0,R(T)]$, and its only identifications after reducing $\theta$ modulo $2\pi$ and then by the screw are precisely the original integral deck lattice. This checks the global quotient, beyond the local developing map.

Add the axis $r=0$ to this tube before taking its screw quotient. Its smooth structure and metric are ordinary hyperbolic ones. Explicitly, with $X=r\cos \theta$, $Y^{\mathrm{\prime }}=r\sin \theta$, the upper-half-space formula becomes $Z=c+e^s(\tanh r\mathrm{/}r)(X+i{Y}^{\mathrm{\prime }})$ and $Y=e^s\mathrm{/}\cosh r$. The even smooth functions $\tanh r\mathrm{/}r$ and $1\mathrm{/}\cosh r$ extend at $r=0$ with value one. The map has invertible differential there and gives standard hyperbolic coordinates across the axis. The screw extends smoothly, freely and properly: it translates $s$ by the nonzero number $A$, so no nonidentity power fixes a point and only finitely many powers move one compact set to meet another. Its closed radius-$R(T)$ quotient is compact, as a closed disk times an $s$ interval of length $\mathrm{\mid }A\mathrm{\mid }$ with its ends identified by a rotation. More explicitly $(r,\theta ,s)\mapsto (r{e}^{i(\theta -B_{0}s\mathrm{/}A)},e^{2\pi is\mathrm{/}A})$ identifies the quotient smoothly with a product solid torus, including at $r=0$. The angular vector $m$, which changes $\theta$ by exactly $2\pi$ while fixing $s$, is its meridian. Thus the boundary identification furnished in step 5.2 attaches an ordinary solid torus with meridian $p\mu +q\lambda$, exactly as in [F4]. Its metric agrees on a whole punctured collar with $g_u$, so the attachment is smooth. The core is a geodesic of length $\mathrm{\mid }A\mathrm{\mid }={\mathrm{\ell }}_i$; no nonidentity screw power fixes a point on that core because the same screw has nonzero axial translation.

These constructions are simultaneous and give the required topology. Choose the same level $T\ge 1$ in the finitely many disjoint end products. Retain the compact truncated core, and attach the solid tori of step 6.1 on the filled boundary components, using the explicit boundary diffeomorphisms of step 5.2. Retain the translation ends on the unfilled components. This produces the ordinary partial filling with the indicated meridians, with its orientation extended from the core. The boundary maps reverse the induced boundary orientations because they identify the two sides of the same product collar; the constructed metric agrees there and fixes the oriented structure. Equivalently the removed open punctured solid torus in each completed tube has exactly the original $T^2\times [T,\mathrm{\infty })$ marking. Thus no separate identification of slopes, covering degree or core topology is being inferred merely from trivial peripheral holonomy. For precision concerning arbitrary supplied attaching maps, write $f_i:\mathrm{\partial }{V}_i\to T_i$ for our explicit attachment and $h_i:\mathrm{\partial }{W}_i\to T_i$ for a supplied orientation-reversing boundary homeomorphism with the same meridian slope. By the ambient-isotopy convention in [F4], choose an ambient isotopy $J_{i,t}$ of $T_i$, starting at the identity, carrying $h_i(\mathrm{\partial }{D}^2\times \{1\})$ onto $f_i(\mathrm{\partial }{D}^2\times \{1\})$. For $e_i=f_i^{-1}{h}_i$, the conjugate isotopy $f_i^{-1}{J}_{i,t}{f}_i$ therefore satisfies exactly the meridian hypothesis of [F4]. Extend $e_i$ to a core-preserving homeomorphism ${\bar{e}}_i:W_i\to V_i$. It is orientation preserving because both $f_i,h_i$ reverse boundary orientation. The identity on the truncated manifold and these extensions glue to an orientation-preserving, core-preserving homeomorphism $P$ from the supplied filling to our constructed filling: on the attaching boundary, $f_i{\bar{e}}_i=h_i$. On the supplied topological filling declare charts to be $(P^{-1}U,\phi \circ P)$ for every chart $(U,\phi )$ of the constructed smooth filling. These charts have the same smooth transition functions as the constructed atlas; in this transported smooth structure $P$ is a diffeomorphism, and the metric defined by those same chart coefficients is $P^{\ast }{g}_u$. It makes $P$ an isometry. Because $P$ is the identity on the truncated original manifold, its original smooth structure and metric there are retained. Completeness, orientation, volume and core lengths are preserved by this isometry. This is transport of a smooth structure, not a claim that a given arbitrary boundary homeomorphism is smooth for some independently specified atlas. Smooth convergence on compact subsets of the marked original $M$ follows from step 4.1, since the explicit realized metrics agree with $g_u$ there.

The extended metric is complete and has finite volume. The union of the truncated core and the finitely many closed filled tubes is compact. The other ends are the explicit complete translation cusps of step 5.1. For metric completeness, define $f=0$ on that compact part and $f=t-T$ on each retained cusp. It is continuous and $1$-Lipschitz for the path metric: on cusp segments $\mathrm{\mid }df\mathrm{\mid }=1$, on the compact part it is zero, and splitting paths at the common boundary gives the global inequality. A Cauchy sequence has bounded $f$ and therefore stays in a compact union of that part and finitely many truncated cusps. A smooth Riemannian metric induces the manifold topology locally, since on a small relatively compact coordinate ball its positive-definite coefficients bound Euclidean lengths above and below; hence that compact union is metric compact. The sequence has a convergent subsequence, and being Cauchy converges. This proves completeness directly. The compact part has finite volume, each retained cusp has the finite volume in step 5.1, and each filled tube has finite volume: its Fermi density is $\sinh r\cosh rdrd\theta ds$, so its volume is $\pi \mathrm{\mid }A\mathrm{\mid }{\sinh }^{2}R(T)$. Curvature is $-1$ everywhere because the extension charts of step 6.1 are actual hyperbolic charts.

Finally the same formulas give convergence of volume, including for varying simultaneous slope tuples. On a filled end ${\sinh }^{2}R(T)=\mathrm{\mid }c{\mathrm{\mid }}^2{e}^{-2T}$ and $\mathrm{\mid }A\mathrm{\mid }=\mathrm{\mid }u{\mathrm{\mid }}^2\mathrm{Im}\tau \mathrm{/}(2\pi )$, so its completed tail volume is exactly $\mathrm{\mid }a(u){\mathrm{\mid }}^2\mathrm{Im}\tau (u)e^{-2T}\mathrm{/}(2\mathrm{\mid }E(u){\mathrm{\mid }}^2)$, since $\mathrm{\mid }1-e^u{\mathrm{\mid }}^2=\mathrm{\mid }u{\mathrm{\mid }}^2\mathrm{\mid }E(u){\mathrm{\mid }}^2$. On an unfilled end this formula has the same value by step 5.1 and $E(0)=1$. Thus, for any fixed $T\ge 1$, each tail volume tends to its original complete-cusp volume as the full parameter tends to zero. On the compact part $t\le T$, step 4.1 gives smooth metric convergence, hence convergence of volume densities and their integrals; finitely many coordinate charts and a partition of unity reduce this to uniform convergence of smooth positive determinants on compact sets. Adding the finitely many tail volumes proves convergence of total volume. The displayed core-length formula in step 5.2 tends to zero as well, because $\tau$ stays bounded. All assertions of the Statement follow.