Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 24 / 4

Choose an essential simple closed curve $c$ in the oriented topological surface and an annular neighborhood $N(c)\cong S^1\times [0,1]$. A positive Dehn twist about $c$ is represented on this annulus bywhere $\rho$ is zero near $0$, one near $1$, and increases between them. Since a full turn is the identity on the boundary circle, this map is the identity near both boundary components of the annulus and extends by the identity outside it. It preserves orientation and turns one end relative to the other by one full turn. An inverse twist uses $-\rho$.

More intrinsically, $h$ is a Dehn twist if its mapping class is that of this annulus-supported homeomorphism, for some curve $c$ and the chosen handedness. Choices of annular coordinates and interpolation give isotopic maps. The genus hypothesis permits essential curves, but the definition also makes sense for boundary-parallel curves on surfaces with boundary.

The Lickorish-Dehn theorem says that every orientation-preserving homeomorphism of a compact connected orientable surface, fixing its boundary pointwise, is isotopic relative to the boundary to a finite product of Dehn twists and their inverses about simple closed curves. Boundary-parallel curves are permitted. For a closed surface of genus $g\ge 1$, a finite collection of curves suffices to generate its mapping class group. If boundary components are allowed to be permuted, twists alone do not express those permutations; the boundary condition is part of the precise statement.

The fundamental theorem of Dehn surgery asserts that every closed connected orientable three-manifold is integer surgery on a finite framed link in $S^3$. To prove it, first choose a Heegaard splitting of $M$ of genus $g$. Existence of such splittings is a standard background result: a triangulation, with spanning trees collapsed in its primal and dual one-skeletons, gives complementary regular neighborhoods of finite graphs, hence two handlebodies. Choose a standard genus-$g$ splitting of $S^3$, formed by stabilizing its genus-zero splitting. Identify the two pairs of handlebodies. If $f_0$ is the sphere's gluing map and $f$ is the gluing map for $M$, then $f_0^{-1}f$ is an orientation-preserving self-homeomorphism of the common surface. The Lickorish-Dehn theorem writes it, up to isotopy, as a product ${\tau }_{c_1}^{{ϵ}_1}\cdots {\tau }_{c_m}^{{ϵ}_m}$, ${ϵ}_i=\pm 1$.

Here is the local surgery mechanism. Place $c$ on an intermediate level of a product collar of the surface. Its tubular neighborhood has longitude $\lambda$ given by that surface and meridian $\mu$ around the normal disk. Remove the neighborhood and fill with meridian slope $\mu \pm \lambda$. Cut the remaining annular neighborhood of $c$ along a transverse arc. Going around the new meridian now makes one extra positive or negative turn in the longitude direction. On gluing the cut annulus back, the two collar levels are therefore identified by $(\theta ,s)\mapsto (\theta \pm \rho (s),s)$, precisely the Dehn twist from part (a). Outside this annulus the identification is unchanged. This proves unit surface-framed surgery realizes a Dehn twist; its surgery coefficient is $+1$ or $-1$ relative to the surface framing, with sign fixed by the chosen normal convention.

Put the curves $c_i$ on distinct collar levels and order those levels to realize the indicated composition. They are disjoint as curves in the three-manifold even if their surface projections intersect. Performing the prescribed surgeries replaces $f_0$ by $f$, up to isotopy. An isotopy of a gluing map extends across a collar, so the filled manifold is homeomorphic to $M$.

Finally each surface framing differs from the preferred longitude framing in $S^3$ by an integer number of twists. Thus its coefficient $\pm 1$ becomes an integer coefficient in the usual knot longitude convention. These finitely many disjoint curves form the required framed link $L$:The ingredients used were Heegaard-splitting existence, the stated twist-generation theorem, and the explicit local unit-surgery construction.

Each component is an unknot, so its closed exterior $S^3\setminus \mathrm{int}{U}_i$ is a solid torus. Choose the usual boundary basis $({\mu }_i,{\lambda }_i)$, where ${\mu }_i$ is a meridian of the component and ${\lambda }_i$ its Seifert longitude. In the exterior's fundamental group, ${\mu }_i$ generates $\mathbb{Z}$ and ${\lambda }_i$ is trivial, since it bounds a meridian disk of that exterior.

Every essential embedded circle on a torus represents a primitive homology class $a{\mu }_i+b{\lambda }_i$ with $\gcd (a,b)=1$. If it bounds a disk in the exterior, it is null-homotopic there, so $a=0$. Primitivity then forces $b=\pm 1$. Essential simple curves of the same primitive unoriented slope on a torus are isotopic: lift them to the plane and straighten their parallel primitive lines, or cut the torus along one curve to obtain an annulus. ConsequentlyThis is the preferred longitude of an unknot. The disk is only required to lie in the complement of its own solid torus; it may meet the other link component, so no additional disjoint-disk assertion is being used.

For the intended standard Hopf link, identify $S^3$ with the unit sphere in ${\mathbb{C}}^2$ and its two components with the coordinate circles $z_1=0$ and $z_2=0$. Outside small neighborhoods of those circles, both coordinates are nonzero, and their arguments together with $|z_1|$ giveOn this product the longitude of either component is the meridian of the other. Therefore the two attaching meridians become the two coordinate curves of $T^2$, intersecting exactly once. Adding the product collar does not change a gluing of solid tori. The meridian-intersection-one argument in Question 2(c) identifies the result with the standard genus-one splitting of the sphere. Thus zero surgery on the Hopf link gives

There is a genuine link-type qualification in the wording: merely specifying two linked but individually unknotted components does not determine the Hopf link. For example the two components of the torus link $T(2,4)$ are unknotted but have linking number $2$. Filling their preferred longitudes is zero surgery with linking matrixso the resulting first homology is its cokernel, $\mathbb{Z}/2\oplus \mathbb{Z}/2$, rather than the zero group of $S^3$. The cokernel formula follows because zero longitudes impose twice the other meridian as the homology relations. Thus the sphere answer applies to the standard once-linked Hopf interpretation; the literal unspecified link description alone is insufficient to force a single manifold.