Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 141 (3-manifolds)

Label the unbounded region $r_0$ and the three lobe regions $a,b,c$ in cyclic order, and label the central region $d$. The Dehn presentation of a knot group Heegaard diagram is obtained by thickening the projection graph: take one alpha compressing curve for each of $a,b,c,d$, one beta curve for each crossing, and place the basepoint in $r_0$. Orient and order the crossing curves so that their cyclic region words areThis intersection data, together with $r_0=1$, specifies the requested pointed Heegaard diagram up to the usual isotopies and handle slides.

The generators are the alpha-compressing disks, equivalently the four bounded regions. Each beta curve supplies the relator read from its signed intersections with the alpha curves. Since the basepoint region has generator $r_0=1$, the presentation isThe relations say $d=ba=cb=ac$. Eliminating $c,d$ leaves $bab=aba$, the standard two-generator presentation of the trefoil knot group, which checks the region conventions.

The Alexander numbering of the three lobe regions is one and that of the central region is two, relative to the unbounded region numbered zero. Equivalently, abelianizing the three relators givesThus $A=B=C$ and $D=2A$. The class $A$ is the positively oriented meridian of a knot, so

Forthe Fox calculus product rule and $\partial d^{-1}/\partial d=-d^{-1}$ give, in generator order $(a,b,c,d)$,Multiplication of every row by the unit $t$ gives the equivalent Alexander matrixFor example, deleting the $a$-column gives determinant $-(t^2-t+1)$, the Alexander polynomial of a knot of the trefoil up to a unit.

Star the unbounded region $r_0$ and the adjacent lobe region $a$. The remaining columns of the row-scaled matrix are $(b,c,d)$:Choose at crossings $1,2,3$ the corners in regions $b,c,d$, respectively. Every unstarred region then contains one chosen corner, so this is a Kauffman state of a knot diagram. Circle the entriesTheir signed determinant product is $-1$, one summand inThe other two states give $t$ and $-t^2$, and the total differs from $t^2-t+1$ only by the unit $-1$.

Orient the alpha curves and choose generators $x,y$ dual to ${\alpha }_1,{\alpha }_2$, respectively. Following ${\beta }_1$ once around the edge identifications and through the attached annulus gives, up to cyclic permutation and simultaneous inversion conventions,The dotted blue arcs pass underneath the attached annulus, so their apparent crossings with the red arc on the annulus are not intersections. The fundamental group presentation is thereforeThe exponent sums of $x,y$ in $r$ are $2,0$. Henceand abelianization sends $x$ to the order-two generator and $y$ to the infinite cyclic generator.

The curve $\mu$ crosses the alpha disk ${\alpha }_2$ once and is disjoint from ${\alpha }_1$, so in the presentation it represents $y$ up to inversion and conjugacy. The Dehn filling adds the relation $y=1$, after which the beta relator becomes $x^2$. Thus the filled genus-one Heegaard diagram has two alpha–beta intersections andCompressing the displayed genus-two diagram along $\mu$ visibly cancels the annular handle and leaves that genus-one Heegaard diagram. It is therefore the lens space

Two explicit fibrations of $L(2,1)={\mathbb{RP}}^3$ are supplied by Seifert fibrations of real projective 3-space:

Thus these are Seifert fibered descriptions of the same filled manifold with different numbers of exceptional fibers.

The free part of the abelianization is generated by $[y]$, so the fibration class sends $y\mapsto 1$ and $x\mapsto 0$. Applying Fox calculus to the relator and then substituting $x=1$, $y=t$ givesThis has breadth two. Since $K$ is assumed fibered, the breadth is twice the genus, so its fiber is a genus-one surface with one boundary component:

The monodromy can also be read directly from the relator. Put $x_i=y^{i}x{y}^{-i}$. Reidemeister rewriting givesThus the kernel is freely generated by $(x_{-1},x_0)$, and conjugation by $y$ sendsOn first homology, in the ordered basis $([x_{-1}],[x_0])$, a representative isIts characteristic polynomial is $t^2-4t+1$, agreeing up to the unit $-t$ with $4-t-t^{-1}$.

The matrix in part (d) has trace $4$ and determinant $1$. Since $|\mathrm{tr}{\varphi }_{\ast }|>2$, the once-punctured-torus case of the Nielsen–Thurston classification theorem says that $\varphi$ is pseudo-Anosov. The Hyperbolization of a pseudo-Anosov mapping torus therefore gives

Begin with the product cobordism $(X_0\bigsqcup X_1)\times [0,1]$. In its outgoing boundary choose one embedded $n$-ball in each component and attach an $(n+1)$-dimensional one-handle $D^1\times D^n$ along $S^0\times D^n$. On the outgoing boundary this deletes those two balls and joins their boundary spheres by $D^1\times S^{n-1}$, which is precisely the connected-sum construction. With the product orientation,This is the one-handle cobordism from a disjoint union to a connected sum.

Take a collar ${\partial }_{i}X\times [0,\epsilon ]$. In the quotient, the two annuli $A_1,A_2$ are folded together by $f$, while their common boundary curves ${\beta }_1,{\beta }_2$ become the boundary of a meridional disk. Because $f({\alpha }_1)={\alpha }_2$, the two halves of $\alpha$ join across this disk and bound it. The quotient of the collar is therefore a solid torus whose meridian is attached along $\alpha$; outside the collar nothing changes. Hence the annulus-quotient model of Dehn filling gives

Choose Seifert longitudes ${\lambda }_i$ disjoint from the small arcs removed in forming the connected sum of knots. Join the two resulting longitude arcs through the same connecting tube used for the knots. Their oriented band sum is a zero-linking parallel of $K_1\#K_2$, and therefore

Let $X_{\#}$ be the connected-sum exterior and fill it along its Seifert longitudeThe decomposing annulus in $X_{\#}$ has two parallel boundary circles, each meeting ${\lambda }_{\#}$ once. Applying part (b) cuts the filling back into $X_1$ and $X_2$ and identifies the peripheral curves byup to the signs required by the orientation-reversing torus gluing.

In $Y_1^{\prime }=X_1({\lambda }_1)$, the core $K_1^{\prime }$ has meridian ${\mu }_1^{\prime }={\lambda }_1$. By the orientation stipulated in the question its longitude is ${\lambda }_1^{\prime }={\mu }_1$. Thus the displayed gluing sends the meridian and longitude of $K_1^{\prime }$ to the longitude and meridian of $K_2$, respectively, which is exactly the splice of knots. Therefore the required integer surgery is zero surgery on the connected sum:

Write $X_{U,P}=S^3\setminus \mathrm{int}(\nu U\bigsqcup \nu K_P)$. Thenwhere the splice gluing exchanges meridian and longitude. Filling the remaining boundary along ${\mu }_P$ restores $\nu K_P$, so the pattern side becomes $X_U=S^3\setminus \mathrm{int}\nu U$, a solid torus. Under the splice gluing its meridian is attached to the meridian of $K_C$, because the meridian and longitude are exchanged twice in the two descriptions. This is the meridional filling of $X_C$, which restores the original three-sphere. Hence

The manifold $X_P$ is obtained from $X_{U,P}$ by filling the $U$-boundary along ${\mu }_U$. The filling core represents ${\iota }_{\ast }^{\partial X_U}[{\lambda }_U]$ in $H_1(X_P)$. The Turaev-torsion Dehn-filling formula therefore givesSince a knot exterior has first Betti number one, its normalized Turaev torsion isSubstitution proves

The Mayer--Vietoris sequence for the splice identifies the companion meridian with the pattern longitude and kills exactly the relation already killed when passing from $X_{U,P}$ to $X_P$. It therefore gives a canonical isomorphismApply multiplicativity of Turaev torsion to the torus union defining $X_{CP}$, and use part (b) to replace the torsion of $X_{U,P}$. The factors $1-[{\lambda }_U]$ cancel against the peripheral factor from the companion exterior. Usingon the three knot exteriors leavesUnder a one-variable identification, ${\iota }_{\ast }^C$ sends the companion variable to $t^w$, where $w=[U]\in H_1(X_P)\cong \mathbb{Z}$ is the winding number of a satellite pattern. Thus this is also the usual Satellite formula for the Alexander polynomial.

The slope-$4/3$ pattern has winding number $4$, so the Satellite formula for the Alexander polynomial givesUsing the stated torus knot formula in each factor,Up to a Laurent unit this may be simplified to $(1-t)(1-t^{24})/((1-t^8)(1-t^3))$.