A full-turn peripheral kernel fills as a smooth geodesic core

In cylindrical coordinates $(r,\theta ,z)$ about a geodesic in ${\mathbb{H}}^3$, the metric is $d{s}^2=d{r}^2+{\sinh }^2(r)d{\theta }^2+{\cosh }^2(r)d{z}^2$. The full-turn condition makes $\theta$ have period $2\pi$, while the axial translation identifies $z$ with $z+L$. Thus the developed tube quotient is $(0,\epsilon )\times (\mathbb{R}\mathrm{/}2\pi \mathbb{Z})\times (\mathbb{R}\mathrm{/}L\mathbb{Z})$, and its meridian is exactly the $\theta$ circle representing $\mu$.

Set $x=r\cos \theta$ and $y=r\sin \theta$. Then $d{r}^2+{\sinh }^2(r)d{\theta }^2=d{x}^2+d{y}^2+\frac{({\sinh }^{2}r-r^2)}{r^4}(xdy-ydx{)}^2$. The coefficient has the convergent even-power expansion $1\mathrm{/}3+O(r^2)$ at $r=0$, and ${\cosh }^2(r)$ is smooth in $x^2+y^2$. Hence the metric extends smoothly across $x=y=0$; the added set is the circle $\mathbb{R}\mathrm{/}L\mathbb{Z}$ and is the core of a solid torus. The extended metric has curvature $-1$ on the dense punctured tube by [F2], so its smooth curvature tensor has curvature $-1$ on the core by continuity.

First use the model attaching map from step 1.1; by step 2.1 it gives a smooth hyperbolic extension over that filled solid torus. For any specified ordinary filling map with the same meridian slope, [F4] supplies an oriented homeomorphism from that filling to the model filling, equal to the identity on the complement. Transport the smooth structure and extended metric across this homeomorphism. The transported structure agrees with the original smooth structure on the complement, so this gives the asserted extension on the specified topological filling. For finitely many ends the tube neighborhoods are disjoint, and these identifications and extensions can be made independently.