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In a compact connected orientable three-manifold with no spherical boundary components, every embedded knot is homotopic to a knot with an excellent three-manifold as exterior. In particular this can preserve the generator class in $S^1\times S^2$, or the winding number of a satellite pattern one class in a solid torus. Homotopy here is weaker than isotopy.

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.

For a properly embedded oriented topological surface $F=\coprod F_j$, putThis defines the Thurston norm on integral relative homology classes in $H_2(Y,\partial Y;\mathbb{Z})$. Homogeneity extends it to rational classes, and continuity extends it to real classes. It is a seminorm: nonzero classes carried by spheres, disks, annuli or tori can have zero value.

Not every integral class has a connected embedded representative. In $Y=S^1\times S^2$, the class $2[\{\ast \}\times S^2]$ is a counterexample. A connected embedded oriented surface either separates, in which case it is null-homologous, or has connected complement. In the latter case join its two local sides through its complement to obtain a loop intersecting it exactly once. Its homology class is consequently a primitive lattice element. The proposed double has all intersection numbers even, so cannot have such a representative. Two disjoint parallel spheres do represent it.

For the knot exterior in an integral homology sphere, $H_2(Y,\partial Y;\mathbb{Z})=\mathbb{Z}$ is generated by a Seifert surface. Its Thurston norm iswhere $g(K)$ is the Seifert genus in the ambient integral homology sphere. To justify using a one-boundary-component Seifert surface, simplify a minimizing representative's boundary to parallel essential longitudes. The algebraic sum of these longitudes is one. Join oppositely oriented pairs by boundary annuli and push inward; this preserves Euler characteristic and does not increase ${\chi }_{-}$. Discard closed components, which represent zero because the ambient integral homology sphere has $H_2=0$. The component retaining the single boundary is a Seifert surface. This proves the formula, including the disk case.

When $\Vert y{\Vert }_T>0$, $g(K)\ge 1$. In zero Dehn surgery the longitude bounds a meridional disk in the filling solid torus. Cap a minimizing Seifert surface with that disk. The resulting closed surface $\hat{F}$ has the same genus and represents a generator of $H_2(Z;\mathbb{Z})$: its intersection with the filling core is one. ThereforeThe case $g(K)=1$ gives a zero-cost torus, rather than a negative value.

Here is a rigorous family of strict examples with $Z=S^1\times S^2$. Use the excellent knot representative theorem to choose a knot $J$ representing the generator of ${\pi }_1(Z)$ with an excellent three-manifold as exterior $E_0$. Inside $V=\nu J$, choose a winding number of a satellite pattern one pattern $P$ whose two-boundary-component exterior $E_1=V\setminus \mathrm{int}\nu P$ is also an excellent three-manifold. Both choices are available because neither ambient manifold has a spherical boundary component. Put $\Gamma =P(J)$ and $E=Z\setminus \mathrm{int}\nu \Gamma =E_0{\cup }_{\partial V}{E}_1$.

The interface is an incompressible surface. On a Thurston norm minimizing surface in $E$ with boundary the meridian ${\mu }_{\Gamma }$, arrange all interface intersections to be essential. The winding number of a satellite pattern one condition forces the oriented boundary on that interface to have net class ${\mu }_J$. The piece in $E_0$ has odd meridional boundary sum and costs at least one: a zero-cost representative would require an essential disk or annulus, forbidden by excellence. The piece in $E_1$ has opposite net meridians on its two boundary tori. It costs at least two: zero-cost components are boundary-parallel annuli or tori and carry no such boundary class, and its total meridional boundary count is even, so its nonzero negative Euler characteristic has even magnitude. Cutting along the interface adds the costs, since the essential pieces have no disk or sphere components. ThusHere the notation denotes the relative class whose boundary is ${\mu }_{\Gamma }$, not the meridian curve itself. This is the usual Thurston norm gluing along an incompressible torus argument.

Because $\Gamma$ represents the generator of $H_1(Z)$, $H_1(E)=\mathbb{Z}$, its meridian is null-homologous in $E$, and a longitudinal curve ${\lambda }_{\Gamma }$ generates $H_1(E)$. Fill $E$ along ${\lambda }_{\Gamma }$ to obtain $X$. The Mayer–Vietoris sequence gives $H_1(X)=0$, so $X$ is an integral homology sphere. Let $K$ be the filling core. Its preferred longitude is ${\mu }_{\Gamma }$; zero Dehn surgery on $K$ consequently recovers $Z$. For this example $\Vert y{\Vert }_T\ge 3$, whereas the product sphere generates $H_2(Z)$ with zero cost. HenceThe construction needs a nontrivial winding-one pattern; merely tying a local knot into the product core would not provide this lower bound.

The Mayer--Vietoris sequence for the splice identifies the companion meridian with the pattern longitude and kills exactly the relation already killed when passing from $X_{U,P}$ to $X_P$. It therefore gives a canonical isomorphismApply multiplicativity of Turaev torsion to the torus union defining $X_{CP}$, and use part (b) to replace the torsion of $X_{U,P}$. The factors $1-[{\lambda }_U]$ cancel against the peripheral factor from the companion exterior. Usingon the three knot exteriors leavesUnder a one-variable identification, ${\iota }_{\ast }^C$ sends the companion variable to $t^w$, where $w=[U]\in H_1(X_P)\cong \mathbb{Z}$ is the winding number of a satellite pattern. Thus this is also the usual Satellite formula for the Alexander polynomial.

Take the untwisted Whitehead double of a trefoil knot. The Whitehead pattern has winding number of a satellite pattern zero and becomes the unknot when its companion is the unknot. The Satellite formula for the Alexander polynomial therefore givesThe Whitehead pattern is geometrically essential in its solid torus, so the satellite knot with nontrivial companion $T_{2,3}$ is nontrivial. Hence $K_4$ is not isotopic to the unknot despite having trivial Alexander polynomial of a knot.

Let $V=S^1\times D^2$ contain the pattern of a satellite knot $P$. Its class in $H_1(V;\mathbb{Z})\cong \mathbb{Z}$ is the winding number of a satellite pattern. Winding number zero therefore makes $P$ null-homologous in $V$, so $P$ bounds an oriented embedded surface $F_P\subset V$. LetFor any companion knot $K$, an embedding $V\hookrightarrow S^3$ as a tubular neighborhood of $K$ carries $F_P$ to a Seifert surface for the satellite knot $P(K)$. Henceand the bound depends only on the pattern.