Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 112 (Knots)

Let $K_1$ be the connected sum of $2022$ trefoils. The trefoil has Seifert genus one, and additivity of Seifert genus gives

Take the twist knot whose standard diagram has $2020$ half-twists in its twist region and two crossings in its clasp. Every nontrivial twist knot has Seifert genus one. This diagram is a reduced alternating knot diagram, so the Tait crossing-number theorem says that its $2020+2=2022$ crossings realize the crossing number of a knot. Thus this knot has $g_s(K_2)=1$ and $c(K_2)=2022$.

Consider the symmetric Laurent polynomialIt satisfies $f(1)=1$, so the Alexander polynomial realization theorem gives a knot $K_3$ with ${\Delta }_{K_3}(t)\doteq f(t)$. The polynomial is not a unit of $\mathbb{Z}[t^{\pm 1}]$, whereas

Take the untwisted Whitehead double of a trefoil knot. The Whitehead pattern has winding number of a satellite pattern zero and becomes the unknot when its companion is the unknot. The Satellite formula for the Alexander polynomial therefore givesThe Whitehead pattern is geometrically essential in its solid torus, so the satellite knot with nontrivial companion $T_{2,3}$ is nontrivial. Hence $K_4$ is not isotopic to the unknot despite having trivial Alexander polynomial of a knot.

Apply the Kauffman bracket skein relation to every crossing in the two-string tangle $D_1$. Each complete smoothing is a collection of closed circles together with one of the two crossingless pairings of the four boundary points. Rotation through $180^{\circ }$ about the indicated vertical axis preserves both crossingless pairings and the number of closed circles. Gluing the unchanged tangle $D_2$ to corresponding smoothings therefore gives equal terms, including equal loop factors, so $⟨D⟩=⟨D^{\prime }⟩$.

Choose an orientation of $L$, and transport the induced orientations of the four ends of $D_1$ through the rotation to orient $L^{\prime }$. Crossing signs inside the rotated tangle and outside it then have the same total, so the writhe of a link diagram satisfies $w(D)=w(D^{\prime })$. Applying the writhe normalization of the Jones polynomial gives

The component knot types can change under this Conway mutation. To construct an example, place a knotted arc carrying a trefoil knot summand in one strand of $D_1$ and a knotted arc carrying a figure-eight knot summand in one strand of $D_2$, leaving the other two strand portions trivial. Choose the outside pairing so that before rotation the two knotted portions lie on different components, while after rotation they lie on the same component. Then the component multisets arerespectively. A link isotopy preserves the unordered multiset of component knot types, so these links are not isotopic, although part (a) shows that suitable orientations give them the same Jones polynomial.

Give each bounded region $R_i$ a generator $a_i$ and give the unbounded region $R_0$ the identity generator. At every crossing, read the four incident regions cyclically as $a,b,c,d$ and imposeUsing the opposite cyclic convention inverts all such relators and gives the same group. One crossing relation is redundant, leaving $n$ generators and $n-1$ relators; this is the Dehn presentation of a knot group.

Orient the diagram and assign its regions an Alexander numbering with $R_0$ numbered zero. The abelianization $\varphi :{\pi }_1(E_K)\to ⟨t⟩$ sends $a_i$ to $t^{m_i}$, where $m_i$ is the number of $R_i$. Since $R_1$ is adjacent to $R_0$, $m_1=\pm 1$.

Form the square matrixfrom the Fox derivatives with respect to $a_i$ for $i>1$. This is the Alexander matrix with the $a_1$ column deleted. The Fox identity implies that its maximal minors differ by the factors $\varphi (a_i)-1$, and the standard presentation of the Alexander module therefore givesbecause $m_1=\pm 1$. Thus ${\Delta }_K(t)$ is $\det A_1$ up to a unit $\pm t^r$.

Expand the determinant of the matrix $A_1$ from part (b) by the Leibniz formula for determinants. A matrix entry is a signed sum of monomials arising from the possible corners at its crossing. Choosing one summand in every row chooses one corner at every crossing, while choosing distinct columns puts exactly one chosen corner in every region $R_i$ with $i>1$ and none in the two deleted regions $R_0,R_1$. The surviving determinant terms are therefore in bijection with the Kauffman states $s\in \mathcal{S}(D)$.

The sign of a permutation in the determinant together with the corner signs gives $(-1{)}^{ϵ(s)}$, and multiplying the corner monomials gives $t^{\delta (s)}$. Since part (b) identifies this determinant with the Alexander polynomial of a knot up to a unit,

A Wirtinger presentation from a connected knot diagram has one generator per arc and one relator per crossing, with one relator redundant. Its presentation complex is a finite two-dimensional CW complex $X$ with one zero-cell, $n$ one-cells, and $n-1$ two-cells, and the usual diagrammatic construction gives a homotopy equivalence $X\simeq E_K$.

The abelianization of the knot group is $H_1(E_K;\mathbb{Z})\cong \mathbb{Z}$, generated by a meridian of a knot. Every homomorphism to the cyclic group $\mathbb{Z}/2$ factors through this abelianization, and reduction modulo two is its unique surjection. Thus the requested map $\alpha$ is unique.

Its kernel determines a two-sheeted covering space $\hat{X}$. The nontrivial deck transformation acts on cellular chains and homology, giving them module structures over the group ringLift one copy of every cell of $X$; its two deck translates form a free $\hat{R}$-basis. Hence $C_0\cong \hat{R}$, $C_1\cong {\hat{R}}^n$, and $C_2\cong {\hat{R}}^{n-1}$. If $\tilde{X}$ is the infinite cyclic cover, its cellular chains are free over $R=\mathbb{Z}[t^{\pm 1}]$, and imposing $t^2=1$ gives

For an odd prime $p$, the two idempotents $e_{+}=(1+t)/2$ and $e_{-}=(1-t)/2$ split the group algebraThe plus summand is the cellular chain complex of $X$ with ${\mathbb{F}}_p$ coefficients, while the minus summand is $C_{-}=C_{\ast }^{\mathrm{cell}}(\tilde{X}){\otimes }_R{R}_{-}$. Therefore

On the minus summand, the boundary $t-1:C_1\to C_0$ becomes multiplication by $-2$, which is invertible in ${\mathbb{F}}_p$, so $H_0(C_{-})=0$. After the corresponding cancellation, the remaining square boundary matrix is an Alexander matrix specialized at $t=-1$. It is singular over ${\mathbb{F}}_p$ exactly whenor equivalently when $p$ divides the knot determinant $\det K$. Thus $H_{\ast }(C_{-})$ is nonzero exactly in that case.

Finally, $\det K=|{\Delta }_K(-1)|$ is a nonzero odd integer, so the minus complex is acyclic over $\mathbb{Q}$. Since a knot exterior has the rational homology of a circle,

A framing of an embedded sphere $C:S^{k-1}\hookrightarrow N^{n-1}$ is a trivialization of its rank-$(n-k)$ normal bundle. The standard complex line ${\mathbb{CP}}^1\subset {\mathbb{CP}}^2$ has normal bundle of Euler number $+1$, so this embedded $S^2$ has no framing.

If one framing $f_0$ exists, every other orientation-compatible framing is obtained from it by a map $S^{k-1}\to SO(n-k)$. Consequently the set of homotopy classes is a torsor forFor $k=2$ and $n=4$, this is ${\pi }_1(SO(2))\cong \mathbb{Z}$; the integer is the winding number of one framing relative to $f_0$.

Distinct components of the positively oriented torus link $T_{n,n}$ have linking number one. Since every component also has framing $+1$ relative to the Seifert framing, the surgery linking matrix and hence the intersection form of the surgery trace $W({\hat{L}}_1)$ areIts Smith normal form is $\mathrm{diag}(1,0,\dots ,0)$. The surgery exact sequence, equivalently the kernel and cokernel of $Q_1$, gives

In Kirby calculus, sliding the components over one chosen component diagonalizes the framed link to a split $+1$-framed unknot and $n-1$ zero-framed unknots. Hencewhere $♮$ denotes boundary connected sum.

With zero framings, the surgery linking matrix isIts eigenvalues are $n-1$ on the span of $(1,\dots ,1)$ and $-1$ on the complementary subspace, so $\det Q_2=(-1{)}^{n-1}(n-1)$. For $n>1$, its Smith normal form is $\mathrm{diag}(1,\dots ,1,n-1)$, and therefore

Handle slides reduce this surgery diagram to the standard surgery diagram of the lens space $L(n-1,1)$; equivalently, the generator of the cokernel has linking pairing $1/(n-1)$. Thusup to the orientation convention for surgery. When $n=1$, the matrix is $(0)$ and the exceptional answer is $S^1\times S^2$, with $H_1\cong H_2\cong \mathbb{Z}$.