Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 112 / 2

After replacing a class by a nonsingular representative, let $A$ represent a Seifert form over a field $F$ of characteristic different from two. SetA direct calculation gives $T^{T}QT=Q$, so $(F^{2g},Q,T)$ is an isometric structure. A metabolizer for $A$ corresponds to a $T$-invariant metabolizer for $Q$, and stabilization gives the canonical homomorphismConversely, for an isometric structure with $I+T$ invertible, defineThen $A+A^T=Q$ and $A^{-1}{A}^T=T$. These constructions respect orthogonal sums and metabolic structures and are inverse on Witt classes, proving ${\mathcal{A}\mathcal{C}}_F\cong {\mathcal{W}}_F$.

For an irreducible symmetric Laurent polynomial $\delta$, the primary component of an isometric structure isfor large $N$. The primary decomposition is orthogonal, so restriction of $Q$ and $T$ to $V_{\delta }$ defines the projection

Now take $F=\mathbb{R}$ and let $\delta$ have roots $\omega ,\overline{\omega }$ on the unit circle, with $\omega =e^{i\theta }$ in the upper half-plane. The isomorphism ${\mathcal{W}}_{\mathbb{R}}^{\delta }\cong 2\mathbb{Z}$ sends a class to the even signature jumpFor the class of a knot, this is precisely the jump of its Levine-Tristram signature at the root $\omega$; reversing the choice of side changes the overall sign convention.

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Over $\mathbb{R}$, the relevant part of the Alexander polynomial of a knot of $T_{2,5}$ has the two irreducible symmetric factorsTheir upper-half-plane roots are $\alpha =e^{\pi i/5}$ and ${\alpha }^3=e^{3\pi i/5}$. The supplied determinant shows that the Levine-Tristram signature can jump only at these roots and their conjugates.

For the supplied Seifert matrix, direct inertia calculations on successive arcs of the upper semicircle giveChanging the orientation convention reverses all signs but changes no conclusion. Thus the jumps at both $\alpha$ and ${\alpha }^3$ are $-2$. It follows from part a thatand in both nonzero cases the image is a generator of ${\mathcal{W}}_{\mathbb{R}}^{\delta }\cong 2\mathbb{Z}$.

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In fact the conclusion holds for every amphichiral knot; the hypothesis on the Arf invariant of a knot is unnecessary. Let $Y={\Sigma }_2(K)$ be the two-fold branched cover of a knot. Amphichirality gives an orientation-reversing self-homeomorphism of $Y$, so its linking form of a branched cover satisfiesFix an odd prime $p$ and pass to the $p$-primary subgroup. The standard filtration by powers of $p$ decomposes its linking form into nonsingular symmetric forms over ${\mathbb{F}}_p$. On a graded piece of dimension $d$, an anti-isometry has a matrix $P$ satisfyingTaking determinants givesIf $p\equiv 3(\mathrm{mod}4)$, then $-1$ is not a square in ${\mathbb{F}}_p$, so every such $d$ is even. The sum of these graded dimensions is the exponentIt is therefore even, as required.

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For the supplied Seifert matrix $A$,and the symmetric form $Q=A+A^T$ has signature $-2$. ThusSince the Levine-Tristram signature is an additive homomorphism on the algebraic concordance group, $K$ has infinite algebraic-concordance order.

Over ${\mathbb{Q}}_3$, reduction of the Alexander polynomial givesThe factors are coprime, nonsymmetric, and exchanged by reciprocity. Hensel lifting therefore decomposes the local isometric structure into a reciprocal pair, which is metabolic. Its class in ${\mathcal{W}}_{{\mathbb{Q}}_3}$ is zero and in particular does not have order four.

For $p=11$, diagonalization givesThe second residue at $11$ is the one-dimensional formBecause $11\equiv 3(\mathrm{mod}4)$, $W({\mathbb{F}}_{11})\cong \mathbb{Z}/4$ and this one-dimensional form is a generator. The P-adic algebraic-concordance obstruction therefore has exact order four, so the image of $K$ in ${\mathcal{W}}_{{\mathbb{Q}}_{11}}$ has order four.

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The Satellite formula for the Levine-Tristram signature applied to the $(2,5)$ cable knot givesAt $\omega =-1$, the second term is ${\sigma }_1(K)=0$, whereas part b givesThe Levine-Tristram signature bound on the slice genus now yieldsHence $g_4(K_{2,5})\ge 2$, so the cable cannot bound a punctured torus in $B^4$.

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