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Relative to the meridian of a knot $\mu$ and Seifert longitude $\lambda$, an integer surgery coefficient $n$ means that the filling meridian of a solid torus is identified with $n\mu +\lambda$.

A meridian is an oriented simple closed curve on the boundary of a knot exterior that bounds a disk in the removed tubular neighborhood and has linking number one with the knot.

Viewed inside the removed tubular neighborhood, this is a meridian of a solid torus.

A Heegaard splitting of a closed oriented three-manifold is a decomposition $Y=H_0{\cup }_{\Sigma }{H}_1$ into two handlebodies, with their common boundary the Heegaard surface $\Sigma$. With boundary present, the corresponding pieces are compression bodies, whose negative boundaries account for $\partial Y$.

For the given triangulation, take the barycentric subdivision. A regular neighbourhood $H_0$ of the original one-skeleton is a handlebody: thicken vertices to balls and edges to one-handles, then contract a spanning tree. The complementary region is a regular neighbourhood $H_1$ of the dual one-skeleton, whose vertices are tetrahedron centres and whose edges cross triangular faces. It too is a handlebody. Both graphs are connected, and their common boundary supplies the Heegaard splitting. If the original triangulation has $v$ vertices and $e$ edges, its Heegaard surface has genus $e-v+1$; the dual count gives the same number by $v-e+f-t=0$.

Choose an oriented meridian of a knot $\mu$ and the Seifert longitude $\lambda$ supplied by a Seifert surface for the null-homologous knot. For relatively prime integers $p,q$, rational Dehn surgery removes the interior of a tubular neighborhood and attaches a solid torus with its meridian of a solid torus on the unoriented slope $p\mu +q\lambda$. The choices $(p,q)$ and $(-p,-q)$ describe the same slope; $q=0$ is the original meridional filling. The boundary gluing reverses boundary orientation so that the oriented three-manifold extends across the filling.

For integral coefficients $n_i$, attach two-handles to $B^4$ along the components of the framed link, with their indicated Seifert framing shifts. The boundary operation removes $S^1\times D^2$ and inserts $D^2\times S^1$, with meridian $n_i{\mu }_i+{\lambda }_i$. Thus the compact oriented surgery trace satisfiesFor a finite rational coefficient, use a negative continued fractionReplace that component by an integrally framed chain of successive meridians with coefficients $a_0,\dots ,a_k$. Repeated slam-dunk moves give back $p/q$. Perform this replacement for every rationally framed component, leaving meridional fillings out. The resulting integral framed link has the same filled boundary, so its surgery trace proves that every such rational filling bounds a compact oriented four-manifold.

The drawn component $L_1$ is an unknot. Its exterior is a solid torus $V$, in which $L_2$ is a pattern of a satellite knot. A meridian of a solid torus of $V$ is ${\ell }_1$, whereas its longitudinal direction is $m_1$. Thus the prescribed gluing sends these directions to $m$ and $\ell$, respectively. This identifies $V$ with a tubular neighborhood of $K$ with its zero Seifert framing.

Removing the pattern before making this identification givesEquivalently, meridionally filling the remaining boundary first removes $L_2$ from the construction and leaves $E(K)\cup V=S^3$; its filling core is the required satellite knot $C(K)$.

In the original link diagram, the two parallel passages through a meridional disk of $V$ run in the same direction; closing them uses a single interchange. The winding number of a satellite pattern is therefore two. Untwisting the surrounding disk shows that $P(U)$ is the unknot (the pattern is the $(2,1)$ cable knot, up to the harmless sign of its single interchange). In particular ${\Delta }_{P(U)}=1$. The Satellite formula for the Alexander polynomial now gives the concise answerHere $\doteq$ allows multiplication by $\pm t^j$. Reversing the winding orientation replaces $t^2$ by $t^{-2}$, giving the same Alexander polynomial of a knot up to such a unit.

Write $V=M_{S^2}(p^{\ast }/q,\ast )$. Filling one end of $S^1$ times an annulus gives a solid torus, so $V$ is a solid torus. The prescribed algebraic intersection number of curves on an oriented surface convention gives ${\tilde{f}}_2\cdot {\tilde{h}}_2=1$. HenceIn $H_1(V;\mathbb{Z})$, the first Dehn filling imposes $p^{\ast }f-q{h}_1=0$, while $h_1+h_2=0$. ThereforeThe class ${\lambda }_2$ is primitive because $\gcd (p^{\ast },q)=1$, a consequence of $p{p}^{\ast }-q{q}^{\ast }=1$. It is therefore the disk-bounding meridian of a solid torus of $V$. Since ${\mu }_2$ meets it once, Dehn filling $V$ along ${\mu }_2$ glues two solid torus pieces with disk boundaries meeting once. This is the standard genus-one Heegaard splitting of $S^3$, soTo identify the regular fiber geometrically, put it on their common boundary torus. Invert the displayed basis change:Here ${\mu }_2$ bounds a disk in the newly attached solid torus and ${\lambda }_2$ bounds a disk in $V$. Thus the curve has winding numbers $p$ and $q$ in the two complementary solid torus pieces. Both coefficients are positive and $\gcd (p,q)=1$, so it is the positive $(p,q)$ torus knot. This identifies the torus knot directly from the genus-one splitting, without invoking a theorem specifically about torus knots.

All classes in the nontrivial homological equalities are taken in $H_1(X_{(p,q)};\mathbb{Z})$, where $X_{(p,q)}=M_{S^2}(p^{\ast }/q,-q^{\ast }/p,\ast )$. The two internal gluing torus components still have inclusion maps into this knot exterior. Interpreting the maps instead as maps into the closed $S^3$ would make every displayed class zero.

Before the two Dehn fillings, $H_1(S^1\times \text{pair of pants};\mathbb{Z})$ has generators $f,h_1,h_2,h_3$ and relation $h_1+h_2+h_3=0$. The Dehn fillings add $p^{\ast }f-q{h}_1=0$ and $-q^{\ast }f-p{h}_2=0$. Put $m=-h_3$. Eliminating the relations using $p{p}^{\ast }-q{q}^{\ast }=1$ givesFor completeness, the three relation rows in generators $(f,h_1,h_2)$ are $(p^{\ast },-q,0)$, $(-q^{\ast },0,-p)$, $(0,1,1)$ when computing the quotient by $m$. Their determinant is $p{p}^{\ast }-q{q}^{\ast }=1$, so $m$ really generates the entire first homology group, and no finite torsion or index is hidden in the elimination.

Orient each exceptional-fiber longitude by ${\mu }_i\cdot {\lambda }_i=1$. In the boundary basis $(m_i,{\tilde{f}}_i)=(-{\tilde{h}}_i,{\tilde{f}}_i)$, write ${\mu }_i={\alpha }_i{m}_i+{\beta }_i{\tilde{f}}_i$ and ${\lambda }_i=u_i{m}_i+v_i{\tilde{f}}_i$, where ${\alpha }_i{v}_i-{\beta }_i{u}_i=1$. ThenThe filled meridian of a solid torus maps to zero, and $({\alpha }_1,{\alpha }_2)=(q,p)$, yieldingAdding multiples of ${\mu }_i$ to ${\lambda }_i$ has no effect. The word “any” therefore means any longitude with this compatible orientation; negating a longitude would negate the corresponding equality.

To compute Turaev torsion, use the general product formula $\tau (S^1\times F)\doteq (1-[f]{)}^{-\chi (F)}$ and multiplicativity under gluing along torus components. For a pair of pants, $\chi (F)=-1$, so the unfilled piece contributes $1-[f]$. The two filling solid torus pieces contribute $(1-[{\lambda }_1]{)}^{-1}$ and $(1-[{\lambda }_2]{)}^{-1}$. With $t=[m]$, the equalities above give $[f]=t^{pq}$, $[{\lambda }_1]=t^p$, and $[{\lambda }_2]=t^q$. ThereforeHere $\doteq$ allows the unit $\pm t^n$: a fully refined Turaev torsion requires an Euler structure and a homology orientation, neither of which the statement specifies. The displayed rational function is the representative whose expansion at $t=0$ starts with $1$, equivalently the standard nonnegative-exponent normalization.

For nonzero coprime integers $p,q$, the torus knot $T_{p,q}$ lies on a standard unknotted torus in the three-sphere and winds $p$ times meridionally and $q$ times longitudinally. Here a meridian of a solid torus goes around the small cross-section of the doughnut, while the longitudinal direction goes around its central hole. The integers count signed turns, so changing one sign gives the mirror of a link of the torus knot. In this article $p$ is always the meridional count and $q$ the longitudinal count.

An explicit parametric curve in three-dimensional Euclidean space iswhere $R>r>0$. Its cylindrical radius $\rho =\sqrt{x^2+y^2}$ satisfies $(\rho -R{)}^2+z^2=r^2$, the equation of the supporting torus. Its meridional angle is $pt$ and its longitudinal angle is $qt$: this checks the convention directly rather than relying on the picture. Its squared speed isIf two parameter values give the same point, their difference $\delta$ satisfies $p\delta ,q\delta \in 2\pi \mathbb{Z}$. Bézout's identity for the coprime integers $p,q$ then gives $\delta \in 2\pi \mathbb{Z}$. Thus the loop closes and is an injective function on the parameter circle, with nonzero derivative; it is a smooth embedding.

For example, $R=2$, $r=0.7$, $p=2$, $q=3$ produces a trefoil knot. During one circuit the meridional angle changes by $4\pi$ and the longitudinal angle by $6\pi$. The supporting torus is a guide for construction, not a constraint on later ambient isotopies: the knot may move off it. This family makes arithmetic visible as geometry: the coprime integers determine whether a single traversal visits the whole loop without repetition.

The coprime integers hypothesis matters. If the greatest common divisor is $d=\gcd (|p|,|q|)>1$, the same parametric curve retraces a single component $d$ times; it does not automatically draw $d$ different components. The corresponding torus link is instead the full set of points on the torus whose angles satisfy $q\varphi -p\theta \in 2\pi \mathbb{Z}$, with meridional angle $\varphi$ and longitudinal angle $\theta$. It has $d$ disjoint components, each with winding counts $(p/d,q/d)$. For instance, the torus link $T_{2,2}$ consists of two parallel $(1,1)$ loops and is a Hopf link, whereas substituting $(2,2)$ into ${\gamma }_{p,q}$ simply traces one of those loops twice.