Torus knots and Seifert-fibered exteriors — Examples and Counterexamples

For $(p,q)=(2,3)$ the group relation is $x^2=y^3$, and this common element $h$ is central. The equation $2\alpha +3\beta =1$ has solution $(\alpha ,\beta )=(2,-1)$, so [F1] gives $\mu =y^2{x}^{-1}$ and $\lambda =h{\mu }^{-6}$.

Substituting $(2,3)$ into [F2] gives $\mathrm{\Delta }(t)\doteq \frac{(t^6-1)(t-1)}{(t^2-1)(t^3-1)}=\frac{t^3+1}{t+1}=t^2-t+1,$ and $g=(2-1)(3-1)\mathrm{/}2=1$.

For $(p,q)=(1,5)$, the relation $x=y^5$ eliminates $x$ and leaves the infinite cyclic group generated by $y$. The meridian equation has solution $(\alpha ,\beta )=(1,0)$, so [F1] gives $\mu =y$ and $\lambda =h{\mu }^{-5}=1$ in the group.

Full AC [A1] supplies the countable-choice hypothesis of [F3] by [L1], so [F3] identifies $T(1,5)$ with the unknot. In [F2] the quotient cancels to $1$ and the genus expression is $0$, giving the stated Alexander polynomial and genus.

The greatest common divisor of $\mathrm{\mid }2\mathrm{\mid }$ and $\mathrm{\mid }2\mathrm{\mid }$ is $d=2>1$, so the pair $(2,2)$ is outside the coprime hypothesis in [F2].

Full AC [A1] supplies the countable-choice hypothesis of [F1] by [L1]. The associated torus link then consists of $d=2$ disjoint parallel copies of the primitive slope $(2\mathrm{/}2,2\mathrm{/}2)=(1,1)$. The two copies are separate embedded circles, so the result has two components and is not a knot.

The knot genus formula in [F2] has no application to this pair: its coprimality hypothesis fails and its knot-genus input is not a single-component knot. Thus this example refutes extending that knot formula to all nonzero integer pairs.