Torus knots and Seifert-fibered exteriors

Suppose $p=\epsilon \in \{1,-1\}$. Write the original oriented curve as $K(s)=(r{e}^{2\pi i\epsilon s},\sin {\theta }_0{e}^{2\pi iqs})$, where $r=\cos {\theta }_0\in (0,1)$. Put ${\theta }_t=(1-t){\theta }_0$ and define $H_t(s)=(\cos {\theta }_t{e}^{2\pi i\epsilon s},\sin {\theta }_t{e}^{2\pi iqs})$. Each $H_t$ lies in $S^3$; its first coordinate is an injective circle parameterization because $\mathrm{\mid }\epsilon \mathrm{\mid }=1$ and $\cos {\theta }_t>0$, and the family is smooth including $t=1$. It begins with the given orientation and ends at the core circle with orientation sign $\epsilon$.

If $q=\epsilon \in \{1,-1\}$, the path $(z_1,z_2)\mapsto (\cos t{z}_1+\sin t{z}_2,-\sin t{z}_1+\cos t{z}_2)$, $0\le t\le \pi \mathrm{/}2$, lies in $SO(4)$ and isotopes the identity to $(z_1,z_2)\mapsto (z_2,-z_1)$. The image of the curve is $(\sin {\theta }_0{e}^{2\pi i\epsilon s},-r{e}^{2\pi ips})$. Let ${\eta }_0\in (0,\pi \mathrm{/}2)$ satisfy $\cos {\eta }_0=\sin {\theta }_0$ and $\sin {\eta }_0=r$, put ${\eta }_t=(1-t){\eta }_0$, and use $H_t(s)=(\cos {\eta }_t{e}^{2\pi i\epsilon s},-\sin {\eta }_t{e}^{2\pi ips})$. Its first coordinate has winding $\epsilon =\pm 1$, so the family is a smooth isotopy through embeddings to the core with orientation sign $\epsilon$. The initial coordinate rotation also preserves the knot orientation because it is an ambient isotopy.

By [F3], each smooth isotopy of the compact embedded circle constructed above extends to an ambient isotopy of $S^3$. Its terminal core is the standard unknot, with orientation sign $\epsilon$. If $\epsilon =-1$, write $z_j=x_j+i{y}_j$ and define $R_{\tau }(x_1,y_1,x_2,y_2)=(x_1,y_1\cos \tau +y_2\sin \tau ,x_2,-y_1\sin \tau +y_2\cos \tau )$. This is a rotation in the $(y_1,y_2)$-plane, so $R_{\tau }$ is an orientation-preserving isometry of $S^3$ from the identity to $R_{\pi }(z_1,z_2)=({\bar{z}}_1,{\bar{z}}_2)$. On the core, $R_{\pi }(e^{2\pi is},0)=(e^{-2\pi is},0)$, so this ambient isotopy reverses its orientation. In either sign case the original oriented curve is ambient isotopic to the standard oriented unknot.

Applying $(z_1,z_2)\mapsto (z_1,\overline{z_2})$ to the parameterization changes its second angular coordinate from $qs$ to $-qs$ and leaves the first coordinate and parameter direction unchanged. Its image is therefore exactly the oriented curve $T(p,-q)$.

Replacing the parameter $s$ by $-s$ traverses the same geometric curve in the opposite direction and changes the angular winding pair $(p,q)$ to $(-p,-q)$. Hence the reversed orientation is the orientation specified for $T(-p,-q)$.

Use the solid-torus side with core the first-coordinate circle, so the second coordinate is the meridional disk coordinate. At longitudinal angle $u\in \mathbb{R}\mathrm{/}\mathbb{Z}$ in the first coordinate, the knot meets this disk at the $p$ points with second-coordinate angles $2\pi q(u+k)\mathrm{/}p$, $k=0,\dots ,p-1$. Place these points at the vertices of a regular $p$-gon. A counterclockwise rotation by $2\pi \mathrm{/}p$, tapered to fix the disk boundary, carries each point to the next and crosses consecutive points once in the order ${\sigma }_1,\dots ,{\sigma }_{p-1}$; each ${\sigma }_i$ is the positive half twist of adjacent points. Thus one such increment is the braid block $\delta ={\sigma }_1\cdots {\sigma }_{p-1}$, and the rotation through $2\pi q\mathrm{/}p$ gives ${\delta }^q$. The lifted curve has winding vector $(p,q)$ in the fixed Heegaard-torus coordinates, exactly the parameterization in [F1]. The braid block permutes the points by a $p$-cycle, so its $q$th power has one cycle because $\gcd (p,q)=1$.

At each positive crossing replace the crossing by the orientation-compatible smoothing and then attach one half-twisted rectangle in the original crossing ball; this is the oriented braid Seifert-surface construction. It gives $p$ disks joined by $q(p-1)$ positively half-twisted bands, one for each crossing. The first block has one band between each adjacent pair of disks, so its incidence graph is a path; attaching bands along a tree joins the disks to a disk. Each of the remaining $q-1$ blocks has $p-1$ bands.

Add the remaining blocks in order, and within each block add the bands in the order ${\sigma }_1,\dots ,{\sigma }_{p-1}$. For each added band, a small ball containing it and the corresponding band from the preceding block can be chosen disjoint from the other crossing balls after sliding their strand rectangles along the Seifert disks. The two bands are positive half twists on the same adjacent pair; their union with the two intervening disk strips is an annulus whose boundary components have linking number $+1$, hence a positive Hopf band. The old band meets the already-built surface in a square, the new annulus lies on the opposite side of that square, and the orientations agree because both bands come from positive crossings. Thus each added band is exactly a plumbing along a square as defined in the Statement. Repeating this for all $(q-1)(p-1)$ later bands gives the stated grid of positive Hopf plumbings.

A plumbing replaces the old square and the Hopf annulus by their union, so inclusion-exclusion for Euler characteristic gives $\chi (F_{p,q})=\chi (F_{p,q}^{\mathrm{o}\mathrm{l}\mathrm{d}})+0-1=\chi (F_{p,q}^{\mathrm{o}\mathrm{l}\mathrm{d}})-1$. Starting from the disk in step 2.1, after $(p-1)(q-1)$ plumbings the surface is connected and has $\chi (F_{p,q})=1-(p-1)(q-1)$.

The boundary of the constructed surface is the closure of the braid in step 1.1. Its permutation has one cycle, so the boundary has one component; the construction is connected and oriented at every band attachment. It is therefore a Seifert surface for $T(p,q)$ in the sense of [F2].

Since $(p-1)(q-1)$ is even, the one-boundary-component identity $\chi (F)=1-2g(F)$ and step 4.1 give $g(F_{p,q})=(p-1)(q-1)\mathrm{/}2$.

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.

Abelianizing the presentation in [F1] gives the relation $p[x]=q[y]$. The map $\varphi (x)=q$, $\varphi (y)=p$ is onto because $\gcd (p,q)=1$; its kernel in ${\mathbb{Z}}^2$ is generated by $(p,-q)$, exactly the abelianized relation vector, so the abelianization is $\mathbb{Z}$. The meridian in [F1] maps to $p\alpha +q\beta =1$, so the epimorphism $\varphi$ in the Statement is precisely the positive total-linking homomorphism that classifies the cover in [F4].

For every integer $n$, $(z^n-1)\mathrm{/}(z-1)$ is a finite Laurent sum; for $n<0$ it equals $-(z^{-1}+\cdots +z^n)$. In the lifted attaching word, a positive edge contributes the current deck translate and a negative edge contributes the negative of the preceding translate. The $x^p$ block therefore contributes $(x^p-1)\mathrm{/}(x-1)$ to the $x$-edge coordinate. The following $y^{-q}$ block starts after the $x^p$ translate and contributes $x^p(y^{-q}-1)\mathrm{/}(y-1)$ to the $y$-edge coordinate. These are the complete signed edge contributions of the two-cell in ${\tilde{P}}_{\varphi }$.

The cyclic cover in the Statement has cellular chain groups $C_2=\mathrm{\Lambda }$, $C_1={\mathrm{\Lambda }}^2$, $C_0=\mathrm{\Lambda }$. Its first boundary is $d_1=(t^q-1,t^p-1)$ because lifted $x$- and $y$-edges end $q$ and $p$ deck steps from their starts. Evaluating the two edge contributions in step 1.2 at $x=t^q$, $y=t^p$ gives the second boundary column; the $y$-coordinate is $t^{pq}(t^{-pq}-1)\mathrm{/}(t^p-1)=-(t^{pq}-1)\mathrm{/}(t^p-1)$: $d_2(1)={(\frac{t^{pq}-1}{t^q-1},-\frac{t^{pq}-1}{t^p-1})}^T\mathrm{.}$ This is the cellular boundary of the lifted two-cell.

In $\mathrm{\Lambda }$, $\gcd (t^p-1,t^q-1)\doteq t-1$: for positive exponents the Euclidean algorithm gives this because $\gcd (\mathrm{\mid }p\mathrm{\mid },\mathrm{\mid }q\mathrm{\mid })=1$, and negative exponents differ from the positive ones by units. Hence $a=(t^q-1)\mathrm{/}(t-1)$ and $b=(t^p-1)\mathrm{/}(t-1)$ are coprime. The kernel of $d_1$ is the free rank-one submodule generated by $(b,-a{)}^T$: if $au+bv=0$, unique factorization implies $a\mid v$, and then $(u,v)$ is a multiple of $(b,-a)$. The column in step 2.1 is ${\delta }_{p,q}(b,-a{)}^T$. To show ${\delta }_{p,q}\in \mathrm{\Lambda }$, replace signed exponents by $m=\mathrm{\mid }p\mathrm{\mid }$, $n=\mathrm{\mid }q\mathrm{\mid }$, which changes each factor only by a unit. Set $A=(t^m-1)\mathrm{/}(t-1)$ and $B=(t^n-1)\mathrm{/}(t-1)$. They are coprime; since $m,n$ divide $mn$, both divide $(t^{mn}-1)\mathrm{/}(t-1)$. Unique factorization gives $AB\mid (t^{mn}-1)\mathrm{/}(t-1)$, hence ${\delta }_{m,n}=((t^{mn}-1)\mathrm{/}(t-1))\mathrm{/}(AB)\in \mathrm{\Lambda }$, and the signed quotient is a unit multiple. If $\mathrm{\mid }p\mathrm{\mid }=1$ or $\mathrm{\mid }q\mathrm{\mid }=1$, this quotient is a unit, so the same cyclic presentation gives $A_K=0$.

By [F5], $H_1$ of the cyclic cover of $P_{p,q}$ is $\ker d_1\mathrm{/}\mathrm{im}{d}_2\cong \mathrm{\Lambda }\mathrm{/}({\delta }_{p,q})$. The presentation complex and $E_K$ have the same fundamental group and the same homomorphism $\varphi$; the fundamental groups of their cyclic covers are the corresponding kernels, and the deck action on their abelianizations is conjugation by a preimage of $1\in \mathbb{Z}$. Hurewicz therefore identifies their first-homology $\mathrm{\Lambda }$-modules. Full AC [A1] supplies the countable-choice hypothesis of [F3] by [L1]; the deformation retraction then identifies the cyclic-cover homology of $E_K$ with that of $S^3\setminus K$. Thus $A_K\cong \mathrm{\Lambda }\mathrm{/}({\delta }_{p,q})$.

The one-by-one presentation $[{\delta }_{p,q}]$ of $\mathrm{\Lambda }\mathrm{/}({\delta }_{p,q})$ has $E_0$ generated by its sole $1\times 1$ minor, namely ${\delta }_{p,q}$. Isomorphism invariance of the Alexander module and the definition [F6] give $E_0(A_K)=({\delta }_{p,q})$ and hence ${\mathrm{\Delta }}_K\doteq {\delta }_{p,q}$.

Apply [F2] to identify the exterior group presentation and its positive-meridian homomorphism. The Fox-module calculation [F3] applied to that presentation gives ${\mathrm{\Delta }}_K\doteq {\delta }_{p,q}$ and proves ${\delta }_{p,q}\in \mathrm{\Lambda }$.

Suppose $a,b\ge 2$. For the four possible sign patterns, $T(p,q)$ is obtained from $T(a,b)$ by no operation, the mirror, orientation reversal, or both, according to [F5]. The genus invariance in [F5] therefore gives $g(K)=g(T(a,b))$.

For every positive integer $n$, $t^{-n}-1=-t^{-n}(t^n-1)$, so each signed factor $t^p-1$, $t^q-1$, and $t^{pq}-1$ differs from its absolute-exponent factor by a unit of $\mathrm{\Lambda }$. Therefore ${\delta }_{p,q}$ is a unit multiple of ${\delta }_{a,b}$ as a rational function. By [F3] this places ${\delta }_{a,b}$ in $\mathrm{\Lambda }$. Its defining rational function is finite at $t=0$, because numerator and denominator both have constant term $1$; a Laurent polynomial finite at $0$ has no negative powers. Thus ${\delta }_{a,b}\in \mathbb{Z}[t]$, and ${\mathrm{\Delta }}_K\doteq {\delta }_{a,b}$.

By [F6], $T(a,b)$ bounds a connected oriented surface of genus $g_0=(a-1)(b-1)\mathrm{/}2$, so the minimum in [F7] satisfies $g(K)=g(T(a,b))\le g_0$.

If $a=1$ or $b=1$, the quotient defining ${\delta }_{a,b}$ cancels to $1$. By [F4], $K$ is ambient isotopic to the standard unknot, which bounds a disk; an ambient isotopy carries spanning surfaces to spanning surfaces of the same genus in both directions. Since genus is nonnegative by [F7], the minimum is $g(K)=0=(a-1)(b-1)\mathrm{/}2$.

Since ${\delta }_{a,b}\in \mathbb{Z}[t]$, its degree is $ab+1-a-b=(a-1)(b-1)$: the numerator and denominator have leading coefficient $1$, and the quotient is polynomial by step 2.1. Its constant coefficient is $1$, since the numerator and denominator both have constant term $1$. Hence its exponent width is $(a-1)(b-1)=2g_0$, and multiplication by a unit does not change width.

By the Seifert-determinant theorem and the genus-bound corollary [F8], $\mathrm{width}({\mathrm{\Delta }}_K)\le 2g(K)$. Step 2.1 gives $\mathrm{width}({\mathrm{\Delta }}_K)=2g_0$, so $2g_0\le 2g(K)$. Together with step 2.2 this yields $g(K)=g_0$. The formulas hold in the nonunit case, and step 3.1 handles the unit case.