Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 16 / 1

The three handle attachments give $H_2(W;\mathbb{Z})\cong {\mathbb{Z}}^3$: there are no one-handles or three-handles. Use the capped core disks of handles as the homology basis, ordered by the framing labels $-2,-1,+1$. The surgery linking matrix gives the intersection form. Orient the three components so that the outer components have linking number $3$ and the middle component has linking number $1$ with each outer component. There are six crossings between the outer components, and two crossings between the middle component and each outer component; counting their signs gives these linking numbers. Reversing a component changes the corresponding row and column signs and does not change the invariants.

Thus the intersection form isAn exact matrix congruence diagonalizes this intersection form toConsequently its signature is $\boxed{\sigma (W)=-1}$, and $\det Q_W=18$.

The long exact sequence in relative homology and Poincare-Lefschetz duality identifywith the intersection form matrix, and identify $H_1(\partial W)$ with its cokernel, since $H_1(W)=0$. The Smith normal form has first invariant factor $1$, because the entries have greatest common divisor $1$, and second invariant factor $1$, because the upper-left two-by-two matrix minor has determinant $1$. The last invariant factor is $18$. Hence

Use generators $a,b,c$ for the three one-handles in the generalized Heegaard diagram. Read a downward crossing of the cut from a left-hand hole to its right-hand partner as the corresponding generator. Traverse the attaching curve around the left-hand $A$ hole, starting at its outgoing downward strand, and then traverse the attaching curve around the left-hand $B$ hole in the same way. Their cut-crossing words areThese are the two relators imposed by the two handle attachments. The remaining boundary is a torus; no third relator should be added. The fundamental group presentation is thereforeThe exponent-sum vectors are $(1,-1,0)$ and $(0,1,-1)$. Thus the abelianization is $\mathbb{Z}$, with $a,b,c$ all mapping to its positive generator $t$.

Apply the Fox derivatives and then this abelianization. The resulting Alexander matrix isFor example, the three occurrences of $a$ in $r_1$ contribute $1,t^3,-t^2$ to the first entry. Every row sums to zero, as the Fox calculus identity predicts.

The cellular boundary in degree one for the infinite cyclic cover is $(t-1,t-1,t-1)$. Its kernel is freely generated by $e_a-e_c,e_b-e_c$. Expressing the two relation boundaries in this basis gives the two-by-two matrix obtained by deleting column $c$ of $F(t)$. Its determinant is the order of the Alexander module of a space. Equivalently, all three maximal matrix minors of $F(t)$ agree up to sign. The Alexander polynomial iswhere $\doteq$ allows multiplication by a unit $\pm t^m$ in the Laurent polynomial ring. Its value at $1$ is $1$.