Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 20 / 1

Let $x,y$ be the generators associated with the two pairs of one-handle feet. Following the attaching circle across the identifications marked $a,b$ and $c,d$ gives the word $xy{x}^{-1}{y}^{-1}$, up to reversal and changing generator orientations. Thus the van Kampen theorem givesThe handle decomposition has cellular groups $C_0=\mathbb{Z}$, $C_1={\mathbb{Z}}^2$, $C_2=\mathbb{Z}$, with no higher groups. The boundary of the two-cell is the abelianized attaching word, which is zero. HenceThe zero framing is the product framing of the commutator handlebody of the torus, so the represented manifold is $T^2\times D^2$ and its intersection form is $(0)$. In particular its boundary is $T^2\times S^1=T^3$.

The boundary homology can also be obtained without using the product identification. By Poincare-Lefschetz duality, $H_3(W,\partial W)=H^1(W)={\mathbb{Z}}^2$ and $H_2(W,\partial W)=H^2(W)=\mathbb{Z}$. The map $H_2(W)\to H_2(W,\partial W)$ is its zero intersection form. The long exact sequence of a pair then gives split sequencesThe boundary is connected and oriented, so its zeroth and third homology are both $\mathbb{Z}$. Therefore

There are no one-handles, so attaching the three two-handles to a four-ball cannot create a fundamental group generator: ${\pi }_1(W)=1$. The handle chain complex has $C_2={\mathbb{Z}}^3$, $C_1=0$ and $C_0=\mathbb{Z}$, givingTo calculate the boundary, orient the two long components down their central strands, and orient the small component clockwise. The six mutual crossings of the long components have negative sign, giving linking number $-3$. Each long component has two positive crossings with the small component, giving linking number $1$. The surgery linking matrix, in the order of framings $4,3,2$, is consequentlyChanging a component orientation changes the corresponding row and column signs, and does not change the resulting homology. The long exact sequence of a pair and Poincare-Lefschetz duality identify $H_2(\partial W)=\ker Q$ and $H_1(\partial W)=\mathrm{coker}Q$. The determinant is nonzero, so the kernel vanishes. The gcd of matrix entries is one; the two-by-two minors include $-6$ and $7$, so their gcd is also one. The Smith normal form therefore has diagonal $1,1,7$, andIn particular the sign of the braid crossings matters: replacing $-3$ by $+3$ while leaving both meridian linkings positive would represent a different link and would give the wrong determinant.

Start with a disk in the zero-handle and attach two untwisted bands passing through the two one-handles. The four passages around its boundary occur in the order $x,y,x^{-1},y^{-1}$, giving an embedded once-punctured torus whose boundary is the attaching circle of the two-handle. Push its interior into the existing four-dimensional handlebody, keeping its boundary on the attaching locus, and cap it with the core disk of a handle. The matching product framing makes the union smooth after rounding its corner. This gives an embedded closed torus $T\subset W$.

The torus uses the core of the two-handle exactly once. Equivalently, it intersects that handle's relative co-core disk transversely in one point. Its cellular two-chain is therefore the generator of $C_2=\mathbb{Z}$, which is a cycle and is not a boundary because there are no three-cells. ThusThis is the zero section in the commutator handlebody of the torus $T^2\times D^2$.

Use the boundary-orientation convention in which $-p/q$ surgery on the unknot gives $L(p,q)$. The negative continued fraction isAttach three two-handles to a zero-handle along a chain of unknots, with adjacent linking numbers $+1$ and framings $-3,-2,-2$. Equivalently, take the plumbing of oriented disk bundles over three spheres with Euler numbers $-3,-2,-2$. Successive slam-dunk moves change the boundary surgery coefficient to $-3-1/(-2-1/(-2))=-7/3$, proving that its oriented boundary is $L(7,3)$ in this convention.

The zero sections are a basis of second homology, and the intersection form isIts successive leading principal minors are $-3,5,-7$. Equivalently those of $-Q$ are $3,5,7>0$, so Sylvester's criterion proves that $Q$ is negative definite. This explicitly constructs the required filling.

If lens spaces are instead labeled by positive surgery with its boundary orientation, use the chain with framings $-2,-4$, whose coefficient is $-7/4$; its boundary is orientation-preservingly the positive-surgery $L(7,3)$, since orientation reversal changes $q=4$ to $q=3$. Its negative-definite form is $\left(\begin{array}{cc}-2& 1\\ 1& -4\end{array}\right)$. This is only a convention change in the named boundary, and specifies a valid filling in either labeling system.