Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 14 (3-manifolds)

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.

Fix the algebraic intersection number of curves on an oriented surface by declaring $u\cdot v=+1$ when the ordered tangent pair $(u,v)$ agrees with the surface orientation. Orient an annular neighbourhood of $\gamma$ with coordinates $(\theta ,r)$ and orientation $d\theta \wedge dr$. A right-handed Dehn twist is represented there by $(\theta ,r)\mapsto (\theta +2\pi f(r),r)$, where $f$ increases from zero to one and is constant near the two boundary circles; it is the identity outside this annulus. A transverse arc gains one oriented copy of $\gamma$ per signed crossing. Consequently its homology action of a Dehn twist isReversing the orientation of $\gamma$ changes both factors' signs and leaves this expression unchanged. A separating curve has $[\gamma ]=0$, so its Dehn twist acts trivially on first homology.

The first assertion is true. Choose a basis $(a,b)$ for the first homology group of the torus with $a\cdot b=1$. The positive Dehn twists about these curves have matricesThe mapping class group of the oriented closed torus is $S{L}_2(\mathbb{Z})$, generated by $A,B$ and their inverses; the Euclidean algorithm on a primitive column gives this generation. The displayed relation rewrites those inverses as positive words:Replacing every inverse in a generating word proves the assertion. The closed torus hypothesis matters: a boundary twist is retained when a boundary circle must be fixed pointwise.

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.

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.

Interpret the paired $A,B,C$ disks as the three one-handles of a genus-three handlebody. Attach two-handles on the two displayed $\beta$ curves. Orient the three disk-crossing generators as $a,b,c$. Reading the signed crossings, starting at the upper-left portion of each attaching curve and changing the start point when needed, givesThe first says $abc=bca$ and the second $abc=cab$. Thus the fundamental group presentation isHere $[u,v]=uv{u}^{-1}{v}^{-1}$. Cyclically changing the starting point, reversing an attaching curve, or changing generator orientations gives equivalent presentations. With $h=abc$, these relations make $h$ central; eliminating $c=b^{-1}{a}^{-1}h$ givesFor the topological identification, thicken the diagram's two nested bands and identify the three paired disk mouths. The complement of those bands is the product region of a pair of pants with a circle; its two compressing curves are exactly $[a,bc]$ and $[ab,c]$. Equivalently, the standard cell decomposition of this product has three one-handles and the two commuting two-handle attachments shown. Thus this is a generalized Heegaard diagram of $P\times S^1$.

The Hopf fibration of $S^3$ has three disjoint regular fibers whose removal leaves $P\times S^1$. Its three fibers are the components of the torus link $T_{3,3}$, as is also apparent from the full three-strand twist in the later link diagram. This identifies $Y$ with that link exterior, using the diagram and product structure rather than just its fundamental group. In particular it is a link exterior in the three-sphere.

Both relators have zero exponent sums, so abelianization gives $H_1(Y)={\mathbb{Z}}^3$, with meridian variables $x,y,z$ corresponding to $a,b,c$. The universal abelian cover has deck transformation group ${\mathbb{Z}}^3$ and coefficient group ringThe generalized Heegaard diagram gives a two-dimensional spine with one vertex, three edges and two faces. Its lifted cellular chain complex iswhere chosen lifts of the cells giveThe two columns are the abelianized Fox derivatives of $[a,bc]$ and $[ab,c]$. The Fox calculus identity gives $d_1{d}_2=0$; this can also be checked by multiplying the displayed matrices. There is no three-cell in this spine. One may use the lifted spine because its deformation retraction from $Y$ lifts to the universal abelian cover.

The maximal minors of the Alexander matrix, in row-pair order $(1,2),(1,3),(2,3)$, areTheir greatest common divisor in $R$ is $xyz-1$, since $x-1,y-1,z-1$ have no common nonunit divisor. Accordingly the multivariable Alexander polynomial isThe allowed units are $\pm x^i{y}^j{z}^k$. The single-variable specialization convention can introduce extra $(t-1)$ factors; the answer here is the genuinely multivariable Alexander polynomial.

The dual Thurston polytope is the polar of the Thurston norm unit ball. More intrinsically, in the real dual of $H_2(Y,\partial Y)$ it isThe pairing can be regarded as evaluation of $H^2(Y,\partial Y;\mathbb{R})$ on relative homology. Its definition remains valid when the Thurston norm has a kernel: the polytope then lies in the annihilator of that kernel.

Identify $Y$ with the pair of pants product $P\times S^1$. A regular Seifert fiber has homology class $h={\mu }_1+{\mu }_2+{\mu }_3$. Let $e_i$ be the relative homology class corresponding by Poincare-Lefschetz duality to the homomorphism taking the $i$th meridian to one and the other two to zero. A spanning disk for $L_1$ punctured once by each of $L_2,L_3$ is an embedded pair of pants $F$ representing $e_1$, with $c(F)=1$.

Take two arcs in $P$, one joining boundary one to boundary two, the other joining boundary one to boundary three. Their products with $S^1$ are embedded vertical surfaces in a Seifert fibered space, namely annuli $A_{12},A_{13}$. Orient them so that their relative classes are $e_1-e_2$ and $e_1-e_3$. They cost zero. Thus the three required inequalities, with ${\alpha }_i=\alpha (e_i)$, areFor completeness they give the whole polytope. Oriented cut-and-paste of copies of $F,A_{12},A_{13}$ gives the upper bound $\Vert a{e}_1+b{e}_2+c{e}_3{\Vert }_T\le |a+b+c|$. For the reverse bound, compress a minimizing surface and use the classification of incompressible surfaces in Seifert fibered spaces. Its horizontal components cover $P$ and have negative Euler characteristic equal to their unsigned covering degree; its vertical components have zero cost and zero intersection with a regular Seifert fiber. The total signed horizontal degree is $a+b+c$, so its cost is at least $|a+b+c|$. HenceIt is a line segment, because this Thurston norm has a two-dimensional kernel.

Orient the two components coherently through the twist region and assign meridian variables $x,y$. The link diagram is the torus link $T_{2,2n}$. It is obtained from the three-component torus link of part 2 by $-1/(n-1)$ rational Dehn surgery on the third component: removing its meridional disk adds $n-1$ full twists to the original single full twist. For $n=1$, this just means meridionally deleting the third component.

Write $z$ for the removed component's meridian. Its longitude is homologous to ${\mu }_1+{\mu }_2$, so the filling imposes $z=(xy{)}^{n-1}$. Under this substitution, the polynomial of part 2 becomes $(xy{)}^n-1$. The filling core is homologous, up to sign, to ${\mu }_1+{\mu }_2$. The Turaev-torsion Dehn-filling formula therefore removes the factor $xy-1$, givingFor positive $n$ this is the Laurent polynomial $1+xy+\cdots +(xy{)}^{n-1}$. For $n=1$ it is one, as for a Hopf link; for $n=0$ it is zero, as for the two-component unlink. For negative $n$ the displayed quotient is still a Laurent polynomial and agrees with the mirrored positive-twist answer up to a unit. These checks also fix the twist count: the exponent is $n$, rather than $n+1$.

Give the three Hopf fibration components coherent orientations, so their pairwise linking numbers are one. In the link exterior, their longitudes satisfy $[{\lambda }_i]={\sum }_{j\ne i}{\mu }_j$. Filling along $p_i{\mu }_i+q_i{\lambda }_i$ imposes that relation on first homology. For the three rational coefficients the presentation matrix isIts determinant is $30$, and the greatest common divisor of its two-by-two minors is one (for example, $-5$ and $-3$ occur). Its Smith normal form is $\mathrm{diag}(1,1,30)$, soA nonzero rational coefficient with numerator one does not by itself make a multi-component surgery an integral homology sphere: the nonzero linking numbers must be included.

For the integer filling, use the Seifert fibered space structure $P\times S^1$. Its central regular fiber is $h=abc$, and each preferred longitude is $h{\mu }_i^{-1}$. The three filling relations are consequentlySubstituting $b=h$ into $abc=h$ gives $c=a^{-1}$. The remaining equations say $a^2=a^4$, hence $a^2=1$ and $h=1$. Thus its fundamental group is cyclic of order two.

Geometrically, a filling coefficient $r$ attaches a Seifert fibered space solid torus with multiplicity $|r-1|$, the distance of $r\mu +\lambda$ from the regular fiber $\mu +\lambda$. The multiplicities are $2,1,4$; the middle filling creates no exceptional fiber. The result is a Seifert fibered space over the sphere with at most two exceptional fibers, hence a union of two solid tori, or a lens space. A lens space with fundamental group of order two is $L(2,1)$. ThereforeAs a separate arithmetic check, the integer surgery linking matrix has determinant $-2$, in agreement with its first homology of order two.