Torus-knot exterior group and peripheral system

The round-metric tube is an admissible invariant tube. Full AC supplies the countable-choice hypothesis for the smooth normal-bundle structure by [L1]. Regard $S^3$ as the unit sphere in ${\mathbb{R}}^4$ with its round metric, and identify each quotient normal plane $T_x{S}^3\mathrm{/}{T}_{x}K$ with its orthogonal complement to $T_{x}K$ in $T_x{S}^3$. For $v\in {\nu }_x(K)$ let $\widehat{v}$ be that representative and define $\mathrm{\Phi }(x,v)=\cos (\mathrm{\parallel }v\mathrm{\parallel })x+\frac{\sin (\mathrm{\parallel }v\mathrm{\parallel })}{\mathrm{\parallel }v\mathrm{\parallel }}\widehat{v},$ with $\mathrm{\Phi }(x,0)=x$. This is the round normal exponential map. Its differential along the zero section sends a tangent-normal pair $(u,w)$ to $u+w$, an isomorphism. By [F5] it is a local diffeomorphism at every zero vector. Compactness of $K$ gives a uniform disk neighbourhood on which its differential stays invertible. After shrinking that radius, $\mathrm{\Phi }$ is injective there: otherwise there would be distinct $(x_n,v_n),(y_n,w_n)$ with $\mathrm{\parallel }{v}_n\mathrm{\parallel },\mathrm{\parallel }{w}_n\mathrm{\parallel }\to 0$ and equal images; compactness gives subsequences converging to $(x,0),(y,0)$, equality gives $x=y$, and a local inverse near $(x,0)$ contradicts distinctness. Thus, for some $R>0$, $\mathrm{\Phi }$ is a diffeomorphism from the open normal disk bundle of radius $R$ onto an open neighbourhood of $K$. Choose $0<ϵ<R\mathrm{/}2$ also smaller than $\pi$ and than the distance from $K$ to either coordinate circle. The closed metric $ϵ$-neighbourhood of $K$ equals $\mathrm{\Phi }(\{v:\mathrm{\parallel }v\mathrm{\parallel }\le ϵ\})$: a nearest point of $K$ exists for every point in that neighbourhood, and minimizing round distance is equivalent to maximizing the Euclidean dot product with that point; differentiating along $K$ shows the corresponding short great-circle segment is normal to $K$. Conversely, each normal vector of length at most $ϵ$ gives a geodesic of that length, and injectivity on the radius-$R$ disk bundle rules out a closer nearest point. Hence this $N(K)$ is the closed tubular neighbourhood required by [F2], is disjoint from the two coordinate circles, and is invariant under ${\rho }_t$, since these maps are orthogonal and preserve $K$. Finally, the local product charts of $\mathrm{\Phi }$ and compactness of $K$ allow $ϵ$ to be decreased so that $N(K)\cap T_r$ is an annular neighbourhood of $K$ in the Heegaard torus; its two halves are boundary collars on the two solid-torus sides.

The primitive slope $p\alpha +q\beta$ is essential on the Heegaard torus. Removing a small annular neighborhood of $K$ from that torus leaves a closed annulus $A$. On either side, removing the corresponding half of the tube from a solid torus only removes a boundary collar, so the two pieces retract to solid tori $W_x,W_y$ and their overlap retracts to $A$. Thicken the pieces by half-collars to obtain path-connected open sets covering $E_K$ whose intersection retracts to $A$. The core of $A$ winds $p$ times around the core generator $x$ on one side and $q$ times around $y$ on the other. Full AC [A1] supplies the countable-choice hypothesis of [F2] by [L1].

The generic orbits of ${\rho }_t$ have slope $(p,q)$ and trivial stabilizer: if both coordinates are nonzero, a stabilizing parameter $t\in \mathbb{R}\mathrm{/}\mathbb{Z}$ satisfies $pt,qt\in \mathbb{Z}$, hence $t=0$ because $\gcd (p,q)=1$. The stabilizer orders on the coordinate circles $z_1=0$ and $z_2=0$ are $\mathrm{\mid }q\mathrm{\mid }$ and $\mathrm{\mid }p\mathrm{\mid }$, respectively. The orbit space is a sphere: on each intermediate torus $\mathrm{\mid }{z}_1\mathrm{\mid }=r$ with $0<r<1$, quotienting by the primitive circle orbit leaves a circle, and at $r=0,1$ that quotient circle collapses to the corresponding coordinate-orbit point. The invariant tube from step 1.1 lies in the free-orbit region, and its quotient is a disk: every orbit in its normal disk bundle meets the fiber over one chosen point of $K$ exactly once. Removing its interior therefore removes a disk from the orbit sphere, leaving a disk. The quotient map is a Seifert fibration of $E_K$ with the two stated core multiplicities. Multiplicity one is regular; multiplicity greater than one is exceptional.

The three fundamental groups in this cover are ${\pi }_1(W_x)=⟨x⟩\cong \mathbb{Z}$, ${\pi }_1(W_y)=⟨y⟩\cong \mathbb{Z}$, and ${\pi }_1(A)=⟨a⟩\cong \mathbb{Z}$, with inclusion maps $a\mapsto x^p$ and $a\mapsto y^q$. By [F4], ${\pi }_1(E_K)$ is their pushout. A group receiving maps from the two cyclic factors is determined by images of $x,y$; the maps agree on the intersection exactly when $x^p=y^q$. Hence the pushout has presentation $⟨x,y\mid x^p=y^q⟩$.

A regular orbit parallel to $K$ represents the annulus core $x^p=y^q=h$. This element commutes with $x$ because it is $x^p$, and commutes with $y$ because it is $y^q$; as $x,y$ generate the group from step 3.1, $h$ is central.

Choose integers $c,d$ with $pd-qc=1$ and let $L=c\alpha +d\beta$. With the coordinate orientation determined by $\alpha$ followed by $\beta$, the ordered intersection number $K\cdot L=pd-qc=1$, so $L$ is a simple curve meeting $K$ once. At the intersection, the boundary $m$ of a small normal disk splits into arcs ${\gamma }_x,{\gamma }_y$ on the two solid-torus sides, while $L$ splits into its crossing arc and complementary arc $L_1$. Inserting $L_1{L}_1^{-1}$ into $m={\gamma }_x{\gamma }_y$ gives $m=({\gamma }_x{L}_1)(L_1^{-1}{\gamma }_y)$. The first factor winds $c$ times around the $x$ core and the second winds $-d$ times around the $y$ core, hence $m=x^c{y}^{-d}$. To calibrate signs geometrically, the positively oriented coordinate cores representing $x$ and $y$ bound the hemispherical disks $D_x=\{\mathrm{Im}{z}_2=0,\mathrm{Re}{z}_2\ge 0\}\cap S^3$ and $D_y=\{\mathrm{Im}{z}_1=0,\mathrm{Re}{z}_1\ge 0\}\cap S^3$, oriented by their respective core boundaries. The knot meets $D_x$ transversely in $\mathrm{\mid }q\mathrm{\mid }$ points, each with intersection sign $\mathrm{sgn}(q)$, and meets $D_y$ in $\mathrm{\mid }p\mathrm{\mid }$ points, each with sign $\mathrm{sgn}(p)$. For the first sign, at a point where $z_2$ is positive real, the ambient orientation is the complex $z_1$ orientation followed by increasing $\mathrm{Im}{z}_2$; the component of the knot tangent transverse to $D_x$ has sign $q$. For the second, at positive real $z_1$, the ambient orientation is increasing $\mathrm{Im}{z}_1$ followed by the complex $z_2$ orientation, and the transverse tangent component has sign $p$. Deleting small disks about these intersections gives oriented surfaces in the knot complement with boundary relations $[x]=q[\mu ]$ and $[y]=p[\mu ]$, where $\mu$ is the positive normal-disk meridian. Thus the abelianization $x\mapsto q$, $y\mapsto p$ sends this geometric meridian to $1$: [F3] makes $[\mu ]$ a generator, and the displayed relations and $\gcd (p,q)=1$ fix its sign. The disk-boundary curve $m$ has abelianization $cq-dp=-1$, so its orientation is opposite to that positive meridian. Thus $\mu =m^{-1}=y^d{x}^{-c}$. Setting $\alpha =d$ and $\beta =-c$ gives $p\alpha +q\beta =1$ and the stated word; another Bezout solution changes the exponents by $(qk,-pk)$, so the word is unchanged because $y^{qk}=h^k=x^{pk}$ and $h$ is central.

Abelianizing the presentation in step 3.1 gives the relation $p[x]=q[y]$. The map $x\mapsto q$, $y\mapsto p$ is surjective because $\gcd (p,q)=1$, and its kernel in ${\mathbb{Z}}^2$ is generated by $(p,-q)$, exactly the relation vector; hence it identifies this quotient with $\mathbb{Z}$. Step 4.2 sends the positive meridian to $1$, so by [F3] it is the positive generator of $H_1(E_K;\mathbb{Z})$. Thus $[h]=p[x]=pq[\mu ]$. The regular orbit $h$ is a degree-one parallel of $K$ under the equivariant normal-bundle projection from step 1.1, and $\mu$ has degree zero there, so $\lambda =h{\mu }^{-pq}$ has degree one and zero image in $H_1(E_K)$. Uniqueness in [F3] identifies it with the preferred longitude; [F3] also proves that $[\mu ],[\lambda ]$ form a basis of $H_1(\mathrm{\partial }{E}_K;\mathbb{Z})$, so each is primitive.