Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 141 / 4

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.

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.