= Solution
Replace the open <knot complement> by the compact <knot exterior> $E_K$; they have the same <homotopy type>. The <abelianization>
$$
\pi_1(E_K)\longrightarrow H_1(E_K;\mathbb Z)\cong\mathbb Z
$$
sends an oriented <meridian of a knot> to $1$. The connected <covering space> corresponding to its kernel is the <infinite cyclic cover of a knot exterior> $\widetilde E_K$. Choose a generator $t$ of its <deck transformation group>. Then
$$
\mathcal A_K=H_1(\widetilde E_K;\mathbb Z)
$$
is the <Alexander module of a knot> over $\Lambda=\mathbb Z[t,t^{-1}]$. It is a finitely presented torsion <module>, and its order, equivalently the greatest common divisor of the appropriate maximal <matrix minors> of an <Alexander matrix>, is the <Alexander polynomial of a knot>, up to a <unit> $\pm t^m$.
For a <fibered knot>, the <infinite cyclic cover of a knot exterior> is $\Sigma\times\mathbb R$. Write the <mapping torus> with the gluing convention in the question. The <deck transformation>
$$
D(x,s)=(\phi(x),s-1)
$$
has exactly that quotient: $(x,1)$ and $(\phi(x),0)$ are identified. Choose $t=D$. Its action on $H_1(\Sigma;\mathbb Z)$ is $\phi_*$. If one instead chooses the generator projecting in the positive circle direction, its action is $\phi_*^{-1}$; this replaces $t$ by $t^{-1}$ and produces an associate for a knot.
Choose an integral basis for $H_1(\Sigma)\cong\mathbb Z^{2g}$ and let $\Phi$ be the <matrix> of $\phi_*$. The <Alexander module of a knot> has the finite free module presentation
$$
\Lambda^{2g}\xrightarrow{\,tI-\Phi\,}\Lambda^{2g}\longrightarrow\mathcal A_K\longrightarrow0.
$$
There are no further relations: using $t v=\Phi v$ and the invertibility of $\Phi$, every Laurent combination reduces uniquely to an integral combination of the chosen basis elements. Taking the order of this square presentation gives the \b[monodromy formula]
$$
\boxed{\Delta_K(t)\doteq\det(tI-\phi_*).}
$$
For a disk fiber both sides are $1$, using the empty <determinant> convention.
Back to article page
