Torus-knot Alexander polynomial and genus formulas

Apply [F2] to identify the exterior group presentation and its positive-meridian homomorphism. The Fox-module calculation [F3] applied to that presentation gives ${\mathrm{\Delta }}_K\doteq {\delta }_{p,q}$ and proves ${\delta }_{p,q}\in \mathrm{\Lambda }$.

Suppose $a,b\ge 2$. For the four possible sign patterns, $T(p,q)$ is obtained from $T(a,b)$ by no operation, the mirror, orientation reversal, or both, according to [F5]. The genus invariance in [F5] therefore gives $g(K)=g(T(a,b))$.

For every positive integer $n$, $t^{-n}-1=-t^{-n}(t^n-1)$, so each signed factor $t^p-1$, $t^q-1$, and $t^{pq}-1$ differs from its absolute-exponent factor by a unit of $\mathrm{\Lambda }$. Therefore ${\delta }_{p,q}$ is a unit multiple of ${\delta }_{a,b}$ as a rational function. By [F3] this places ${\delta }_{a,b}$ in $\mathrm{\Lambda }$. Its defining rational function is finite at $t=0$, because numerator and denominator both have constant term $1$; a Laurent polynomial finite at $0$ has no negative powers. Thus ${\delta }_{a,b}\in \mathbb{Z}[t]$, and ${\mathrm{\Delta }}_K\doteq {\delta }_{a,b}$.

By [F6], $T(a,b)$ bounds a connected oriented surface of genus $g_0=(a-1)(b-1)\mathrm{/}2$, so the minimum in [F7] satisfies $g(K)=g(T(a,b))\le g_0$.

If $a=1$ or $b=1$, the quotient defining ${\delta }_{a,b}$ cancels to $1$. By [F4], $K$ is ambient isotopic to the standard unknot, which bounds a disk; an ambient isotopy carries spanning surfaces to spanning surfaces of the same genus in both directions. Since genus is nonnegative by [F7], the minimum is $g(K)=0=(a-1)(b-1)\mathrm{/}2$.

Since ${\delta }_{a,b}\in \mathbb{Z}[t]$, its degree is $ab+1-a-b=(a-1)(b-1)$: the numerator and denominator have leading coefficient $1$, and the quotient is polynomial by step 2.1. Its constant coefficient is $1$, since the numerator and denominator both have constant term $1$. Hence its exponent width is $(a-1)(b-1)=2g_0$, and multiplication by a unit does not change width.

By the Seifert-determinant theorem and the genus-bound corollary [F8], $\mathrm{width}({\mathrm{\Delta }}_K)\le 2g(K)$. Step 2.1 gives $\mathrm{width}({\mathrm{\Delta }}_K)=2g_0$, so $2g_0\le 2g(K)$. Together with step 2.2 this yields $g(K)=g_0$. The formulas hold in the nonunit case, and step 3.1 handles the unit case.