Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 112 / 3 / b

Solution

Orient the diagram and assign its regions an Alexander numbering with

R_{0}

numbered zero. The abelianization

ϕ : π_{1} (E_{K}) \to ⟨ t ⟩

sends

a_{i}

t^{m_{i}}

, where

m_{i}

is the number of

R_{i}

. Since

R_{1}

is adjacent to

R_{0}

m_{1} = \pm 1

A_{1} = (ϕ (\frac{\partial w _{j}}{\partial a _{i}}))_{1 \leq j \leq n - 1, 2 \leq i \leq n}

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

ϕ (a_{i}) - 1

, and the standard presentation of the Alexander module therefore gives

det A_{1} ≐ \frac{t ^{m_{1}} - 1}{t - 1} Δ_{K} (t) ≐ Δ_{K} (t),

because

m_{1} = \pm 1

. Thus

Δ_{K} (t)

det A_{1}

up to a unit

\pm t^{r}

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