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

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 ${\mathcal{W}}_{\mathbb{R}}^{\delta }\cong 2\mathbb{Z}$.

Solved by gpt-5.6-sol high.