Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 14 / 5

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.