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Number the seven crossings from top to bottom. Orient the strand entering the top-left arch downward. With the usual positive braid crossing convention, the crossing signs areTo make the diagram calculation reproducible, its oriented Gauss code isSubscripts record overpassing and underpassing. In this convention a positive trefoil knot has knot signature $-2$.

Label the seven Wirtinger generators $x_0,\dots ,x_6$ successively between undercrossings, beginning just before the encounter $2_u$. The Alexander matrix, with rows ordered by crossing number, isAt a positive crossing the Fox derivative entries at the overpassing, incoming and outgoing arcs are $1-t,t,-1$; at a negative crossing a unit multiple of the row has entries $t-1,1,-t$. Deleting the last row and column gives determinant $t(t^4-5t^3+7t^2-5t+1)$. Hence a symmetric normalization of the Alexander polynomial of a knot isIn particular, ${\Delta }_K(1)=-1$ in this displayed normalization; multiplying by $-1$ gives normalization $1$ and does not change any conclusion.

The Seifert algorithm produces four Seifert circles and seven bands. Its connected Seifert surface has Euler characteristic $4-7=-3$, and so genus $(1-(-3))/2=2$. Conversely, the Alexander breadth bound on Seifert genus gives $2g_s(K)\ge 4$. ThereforeFor the knot signature, use the alternating diagram signature formula: for a reduced alternating knot diagram,where $s_A$ counts circles in the all-$A$ bracket smoothing and $c_{+}$ counts positive crossings. Here the all-$A$ smoothing has four circles, the all-$B$ smoothing has five, and $c_{+}=5$. ThusThe opposite global knot signature convention gives $+2$ instead.

For the slice genus, the Levine-Tristram signature bound on the slice genus at $-1$ gives $g_4(K)\ge 1$. There is also an explicit unknotting crossing: switch crossing $3$. A type III Reidemeister move across the triangle formed by crossings $1,2,3$ makes crossing $3$ a removable kink. Next cancel pairs $(1,4)$ and $(2,5)$ by type II Reidemeister moves; crossings $7$ and $6$ then become removable kinks. This leaves the unknot. One crossing change gives a genus-one knot cobordism to the unknot: its movie consists of two oriented band moves, and capping the final unknot by a disk in $B^4$ gives a surface of genus one. Consequently

Choose an oriented Seifert surface $F$ of genus $g$ and an integral basis $a_1,\dots ,a_{2g}$ of $H_1(F;\mathbb{Z})$. Push $a_i$ in the positive normal direction to $F$. The Seifert matrix isIt represents the Seifert form. Its skew-symmetrization $V-V^T$ is the integral intersection form of $F$. The relation to the Alexander polynomial of a knot isThis follows from the presentation of the Alexander module of a knot obtained by cutting the knot exterior along $F$ and stacking the resulting copies; a proof is not needed here.

The printed genus inequality is false. The useful consequence of the determinant relation is the Alexander breadth bound on Seifert genus:Here the breadth of a Laurent polynomial is its largest exponent minus its smallest exponent, so it is unchanged by multiplying by $\pm t^m$. Indeed, for a minimal-genus Seifert surface, the matrix has size $2g_s(K)$ and each entry of $tV-V^T$ has degree at most one. Its nonzero determinant is an ordinary polynomial of degree at most $2g_s(K)$; its breadth is no larger than that degree. Equivalently, the highest exponent of a symmetrically normalized Alexander polynomial of a knot is at most $g_s(K)$. An unnormalized degree is not invariant under Laurent units.

For a counterexample to the printed inequality and the requested example, take the untwisted Whitehead double of a trefoil knot. The picture specifies the two-strand tangle in the zero Seifert framing; the two exterior caps form a Whitehead double clasp. The inset identifies the trefoil knot used as companion.

Its usual genus-one Seifert surface is a zero-framed annulus following the trefoil knot, joined by the clasp band. Choose the annulus core and a curve traversing the clasp band as the basis of its first homology. Zero annulus twisting gives the first self-linking number $0$. The clasp contributes a Hopf band with self-linking number $-1$ in the positive-clasp convention used here. Plumbing the bands contributes one linking in one push-off direction and none in the other. Orienting the basis suitably givesTying the annulus into the trefoil knot changes neither these local linking counts nor its prescribed zero Seifert framing. ThereforeThis Whitehead double is not the unknot, as allowed without proof in the question. Its Seifert genus is consequently at least one and at most one, hence exactly one. Thus $g_s(K)=1$ while every normalized constant Alexander polynomial of a knot has degree zero: a direct counterexample to the printed inequality.

Orient $S^3$ as the boundary of $B^4$. Push the interior of an oriented Seifert surface $F$ slightly into $B^4$, and let $X_F$ be the double branched covering of $B^4$ over this pushed-in surface. Its boundary is the two-fold branched cover of a knot. Define the knot signature byThe signature of a possibly degenerate intersection form means the number of positive eigenvalues minus the number of negative eigenvalues; its radical contributes zero.

Here is why the definition is independent of $F$. Given two choices $F_0,F_1$, glue their branched covers along the common boundary to obtainThis is the double branched covering of $S^4$ over the closed oriented surface obtained by gluing the pushed-in $F_0$ to $-F_1$. Every closed oriented surface in $S^4$ admits a Seifert hypersurface: its complement has an integral meridional class, represent it by a map to the circle, and take a regular value. Near the surface choose the angular map in its trivial normal disk bundle. The closure of a suitable regular fiber is a compact oriented three-manifold bounding the surface. Its trivial normal framing can be chosen to match this construction.

Push the interior of this Seifert hypersurface into $B^5$. The double branched covering of $B^5$ over the resulting properly embedded three-manifold has boundary $X$. The existence of the cover follows from the meridian homomorphism to $\mathbb{Z}/2$; smoothness near the branch set is the local map $(z,u)\mapsto (z^2,u)$. By the hypothesis about boundaries of five-manifolds, $\mathrm{sign}X=0$. Novikov additivity for gluing along an entire closed three-dimensional boundary givesThus the knot signature is independent of the Seifert surface. An isotopy of the knot carries the construction to an equivalent one, so it is also a knot invariant.

To calculate its intersection form, express $F$ as a disk with $2g$ bands. The double branched covering of $B^4$ over the pushed-in disk is again $B^4$. Each band lifts to a two-handle. The capped lifted cores give a basis of $H_2(X_F;\mathbb{Z})$ corresponding to the band-core basis of $H_1(F;\mathbb{Z})$. The framing of the $i$th lifted attaching circle is $2\mathrm{lk}(a_i^{+},a_i)=2V_{ii}$. For distinct bands, the two sheets contribute the two push-off linkings $V_{ij}$ and $V_{ji}$, so the mutual linking number is $V_{ij}+V_{ji}$. HenceA different integral basis changes this matrix by matrix congruence and leaves its signature unchanged.

In Figure 7 the upper knot is a connected sum of knots $T\#T$, while the lower one is $T\#\overline{T}$, where $T$ is the positive trefoil knot. Both knot groups have the same presentationTo see this, the Seifert-van Kampen theorem expresses the knot group of a connected sum of knots as the amalgamated free product of the two summand knot groups, amalgamating their meridians of a knot. A trefoil knot has presentation $⟨a,b\mid aba=bab⟩$ with $a$ a meridian. Reflection gives the same abstract group for the mirror. If it reverses the chosen meridian, send both generators to their inverses; the inverse braid relation is again the same relation. Thus the meridian amalgamation yields the displayed group presentation in both cases.

A band basis for the positive trefoil knot giveswhose symmetrization is negative definite. Therefore $\sigma (T)=-2$ and $\sigma (\overline{T})=2$. Boundary-connected-summing Seifert surfaces makes the Seifert matrix block diagonal, so the knot signature is additive under connected sum of knots. ConsequentlyThe upper knot and lower knot are therefore not isotopic, even though their knot groups are isomorphic. With the opposite global knot signature convention the first value is $+4$, and the distinction is unchanged.

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.

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 fiber monodromy is periodic when isotopy is allowed to move the boundary, in the sense of a periodic surface homeomorphism. Here is a justification from the Seifert fibered space structure, rather than a special formula about torus knots. Its base is a disk with cone-point orders $p,q$. The fiber Seifert surface is an incompressible surface because its inclusion in the mapping torus is injective on fundamental groups. It is also a boundary-incompressible surface: lift a proposed boundary-compressing disk to the infinite cyclic cover $\Sigma \times \mathbb{R}$ and project to $\Sigma$. This would homotope an essential arc of $\Sigma$ into $\partial \Sigma$, contradicting its essentiality. The classification of incompressible surfaces in Seifert fibered spaces therefore applies. A vertical surface in a Seifert fibered space has zero intersection with a regular Seifert fiber, whereas the fibration class evaluates on $f=pqm$ as $pq$. Thus the Seifert surface is isotopic to a horizontal surface in a Seifert fibered space, with $pq$ intersections per regular orbit.

Following the oriented Seifert fibers from one intersection to the next gives a first-return homeomorphism of the horizontal topological surface. It is a representative of the fiber monodromy, and its $pq$-th power is the identity. It has order $pq$: a regular orbit meets the connected horizontal surface in a Seifert fibered space in $pq$ cyclically ordered points. This realizes the periodic case of the Nielsen–Thurston classification theorem. It is neither an Anosov homeomorphism nor a pseudo-Anosov homeomorphism.

There is a boundary convention here. A representative fixed pointwise on the boundary is not periodic relative to the boundary: its $pq$-th power is a boundary Dehn twist (with sign determined by the return-map convention). Capping the boundary, or allowing it to rotate during isotopy, gives the finite-order representative intended by “periodic”. This distinguishes periodic surface type from literal finite order in the mapping class group relative to the boundary.

For any fibered knot, its Alexander module is $H_1(\Sigma ;\mathbb{Z})$ with the deck transformation given by homological monodromy, so its Alexander polynomial of a knot is $\det (tI-{\varphi }_{\ast })$ up to a unit. The general Turaev torsion relation $\Delta (t)\doteq (1-t)\tau (X)$ and part (b) therefore giveThis rational expression is a polynomial: in its factorization into cyclotomic polynomials, $\gcd (p,q)=1$ cancels all denominator factors. It is monic, has constant coefficient one, and has polynomial degree $pq+1-p-q=(p-1)(q-1)$. The characteristic polynomial has polynomial degree $2g$, is monic, and has constant coefficient $\det {\varphi }_{\ast }=1$, because homological monodromy preserves the intersection form. These facts remove the unit ambiguity and giveThe sign change between $\det ({\varphi }_{\ast }-tI)$ and $\det (tI-{\varphi }_{\ast })$ is trivial because $\mathrm{rank}{H}_1(\Sigma )=2g$ is even. As a separate geometric check, the orbifold Euler characteristic of the base is $1/p+1/q-1$, and its $pq$-sheeted horizontal surface in a Seifert fibered space has Euler characteristic $\chi (\Sigma )=p+q-pq=1-2g$.

Choose a fiber monodromy representative fixing the boundary of the fiber Seifert surface pointwise. The boundary of each fiber is the Seifert longitude, precisely the slope used by zero Dehn filling. Attach the filling solid torus as $D^2\times S^1$, matching $\partial D^2\times \{\theta \}$ with the boundary of the fiber at $\theta$. The fibers cap off to closed topological surfaces $\hat{\Sigma }$ of the same genus $g$; extend the fiber monodromy across the disk by the identity. ThusThis is a foliation by fibers of a bundle over $S^1$, and every leaf is compact with genus $(p-1)(q-1)/2$.

Choose a point in the interior of the capping disk, which $\hat{\varphi }$ fixes. Its suspension is an embedded closed curve $\{x\}\times S^1$ in the mapping torus. It is transverse to the foliation and meets every fiber once. Hence every leaf meets a closed transversal to a foliation, which is the defining criterion for a taut foliation. Therefore the fiber foliation is taut.

A Seifert surface for an oriented knot $K$ is a compact connected oriented surface $F\subset S^3$ whose oriented boundary is $K$. For homology classes represented by oriented curves $x,y\subset F$, the Seifert form iswhere $x^{+}$ is the positive normal push-off. Choosing a basis of $H_1(F;\mathbb{Z})$ gives a Seifert matrix $A$.

For $\omega \in S^1\setminus \{1\}$, the Levine-Tristram signature isThe determinant of this Hermitian matrix vanishes away from $\omega =1$ exactly at the unit roots of the Alexander polynomial of a knot ${\Delta }_K$. Consequently the signature is locally constant on their complement.

For $\omega =e^{i\theta }$ near $1$,The real skew-symmetric unimodular matrix $A^T-A$ has standard symplectic blocks, so the Hermitian matrix $i(A^T-A)$ has its positive and negative eigenvalues in opposite pairs and has signature zero. Thus ${\sigma }_{\omega }(K)=0$ near $1$. If ${\Delta }_K$ has no unit roots, then $S^1\setminus \{1\}$ contains no singular point of the signature form and is connected, so local constancy gives ${\sigma }_{\omega }(K)=0$ everywhere. With the usual convention ${\sigma }_1(K)=0$, the signature vanishes identically.

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.

Boundary-connected-summing minimal Seifert surfaces for $K$ and $K^{\prime }$ givesFor the reverse inequality, let $F$ be a minimal-genus Seifert surface for the connected sum of knots and let $S$ be its standard splitting sphere. A minimal-genus Seifert surface is incompressible in the knot exterior: a compression either lowers its genus or separates off a closed component that can be discarded. Put $F$ and $S$ in transverse position and minimize the number of intersection circles. An innermost circle on $S$ either gives a compression of $F$ or bounds a disk on $F$ across which it can be removed. Both alternatives contradict minimality, so $F\cap S$ consists only of the single arc joining the two points of $S\cap \partial F$.

Cutting $F$ along this arc gives Seifert surfaces $F_1,F_2$ for $K,K^{\prime }$. Their Euler characteristics satisfywhich, since all three surfaces have one boundary component, is equivalent toThis proves additivity.

The torus knot $T_{2,3}$ bounds a once-punctured torus, and its degree-two Alexander polynomial of a knot forces every Seifert surface to have genus at least one. Thus $g_s(T_{2,3})=1$. If it were a composite knot, both nontrivial summands would have positive Seifert genus, and additivity would give genus at least two. Hence $T_{2,3}$ is a prime knot.

The Seifert framing pushes a knot in the direction tangent to a Seifert surface and normal to its boundary. The knot and its push-off then have linking number zero.

The Seifert genus is the smallest genus of a Seifert surface for the knot.

It minimizes the genus over all choices of Seifert surface.