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A properly embedded topological surface is boundary-incompressible when no disk in the ambient three-manifold joins an essential arc on the topological surface to an arc on the ambient boundary, with the disk's interior disjoint from both.

A properly embedded topological surface is boundary-parallel if it cobounds a product region with a topological surface in the ambient boundary, with the remaining boundary of that product contained in the ambient boundary. An annulus parallel into one boundary component has zero net oriented boundary class on that component.

Let a knot lie in a fiber of a mapping torus of an oriented topological surface. Surgery with coefficient $-1$ relative to its surface framing replaces the monodromy by its composition with a right-handed Dehn twist. Coefficient $+1$ gives the inverse twist. The composition side depends on the choice of seam and time orientation; the existence of the new fiber bundle does not.

A genus-$g$ handlebody is a three-ball with $g$ one-handles attached. Its boundary is a closed orientable topological surface of genus $g$.

A horizontal surface is a properly embedded topological surface transverse to every Seifert fiber. Its projection to the base orbifold is an orbifold covering. The covering degree is its intersection number with a regular Seifert fiber.

An incompressible surface is a properly embedded two-sided topological surface with no compressing disk, subject to the usual exclusion of inessential spherical components. A compressing disk has boundary an essential closed curve on the topological surface and interior disjoint from it.

Represent one-dimensional homology classes on a closed oriented topological surface by transverse oriented curves and sum their signed crossing points. This gives a bilinear form that is skew-symmetric. A symplectic handle basis has matrix $\left(\begin{array}{cc}0& I_g\\ -I_g& 0\end{array}\right)$, which is unimodular. In particular, a class detected with intersection $1$ or $-1$ by another integral class is a primitive homology class: the detecting homomorphism splits its rank-one subgroup.

An orientable three-manifold is irreducible if every embedded two-sphere bounds a three-ball. This makes innermost-disk simplifications of embedded topological surfaces possible without changing the ambient three-manifold.

For a compact topological surface with cone-point orders ${\alpha }_1,\dots ,{\alpha }_k$, the orbifold Euler characteristic is ${\chi }_{\mathrm{orb}}=\chi (\text{underlying surface})-{\sum }_i(1-1/{\alpha }_i)$. An orbifold covering multiplies this quantity by its degree.

The second assertion is true. Separate the points $p_i$ in the circle direction by a small isotopy if necessary. Cutting $S^1\times S$ at fibers just before and after $p_i\times {\gamma }_i$ exposes two copies of $S$. The surface framing is precisely the stated push-off $p_i^{\prime }\times {\gamma }_i$: the two normal directions give homotopic nonzero normal fields along the curve.

For a single fiber-curve surgery, a cut and regluing of these copies by a Dehn twist changes the curve identified with the transverse meridian by one copy of the surface longitude. In the usual convention that negative surface-framed surgery produces a positive Dehn twist, the filling slope $\mu +\lambda$ implements ${\tau }_{{\gamma }_i}^{-1}$. One can see the coefficient locally by tracing a transverse arc across the twist annulus: it acquires one reverse turn along ${\gamma }_i$, and reversing this cut-and-glue identifies the compressible slope as $\mu +\lambda$. Reversing the circle parameter reverses the monodromy convention, without changing the existence of the fiber bundle.

Every filled block is therefore a product of $S$ with an interval, with a modified endpoint identification. Their cyclic assembly is the mapping torus of the product of these inverse Dehn twists, in the order of the $p_i$ along the oriented circle. It has the original closed topological surface $S$ as fiber, hence fibers over the circle. Intersections between the ${\gamma }_i$ do not obstruct this argument: their twists occur in different fibers, and their ordered product need not commute.

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 topological surface is a Hausdorff, second-countable space in which every point has a neighbourhood homeomorphic to an open subset of ${\mathbb{R}}^2$.

For $S_1$, interior points and points in the interiors of paired edges plainly have disc neighbourhoods. The corner identifications form two classes; in each class two quarter-discs are glued to make a half-disc, and the adjacent identified edge neighbourhoods complete a disc. Equivalently, this polygon is a cell decomposition of the real projective plane, hence a quotient of $S^2$ by the antipodal group. Thus $S_1$ is a topological surface.

In $S_2$, three distinct sides labelled $a$ are identified. A point in the interior of their common image has a neighbourhood made from three half-discs meeting along their diameters. Removing the common diameter leaves three local sides rather than two, so this neighbourhood is not a disc or half-disc. Therefore