Knot exteriors and peripheral systems — Examples and Counterexamples

The map $\mathrm{\Phi }(t,w)=(e^{2\pi it}\sqrt{1-\mathrm{\mid }w{\mathrm{\mid }}^2},w)$ extends to the open normal disk bundle $S^1\times \{\mathrm{\mid }w\mathrm{\mid }<1\}$ and is a diffeomorphism onto $\{z_1\mathrm{\ne }0\}\subset S^3$, with inverse $(z_1,z_2)\mapsto (z_1\mathrm{/}\mathrm{\mid }{z}_1\mathrm{\mid },z_2)$. It fixes the zero section and its derivative in $w$ identifies the $z_2$ plane with the normal quotient in [F3], so [F4] makes it a tubular map. Restricting to the closed disk $D_{1\mathrm{/}\sqrt{2}}^2$ with the norm $\mathrm{\mid }w\mathrm{\mid }$ gives the specified $N(U)\cong S^1\times D_{1\mathrm{/}\sqrt{2}}^2$.

The map $(w,e^{i\theta })\mapsto (w,e^{i\theta }\sqrt{1-\mathrm{\mid }w{\mathrm{\mid }}^2})$ from $D_{1\mathrm{/}\sqrt{2}}^2\times S^1$ to $E_U$ is a smooth diffeomorphism, with inverse $(z_1,z_2)\mapsto (z_1,z_2\mathrm{/}\mathrm{\mid }{z}_2\mathrm{\mid })$ since $\mathrm{\mid }{z}_2\mathrm{\mid }\ge 1\mathrm{/}\sqrt{2}$. Hence $E_U$ is a solid torus.

At $(1,0)$, the ordered outward radial vector, positive tangent to $U$, and the positive coordinate vectors in the $z_2$ plane give the standard orientation of ${\mathbb{C}}^2$. Therefore the positive meridian is the loop varying the phase of $z_2$. The loop $t\mapsto (e^{2\pi it}\mathrm{/}\sqrt{2},e^{i{\theta }_0}\mathrm{/}\sqrt{2})$ is the product section of $\mathrm{\partial }N(U)$, has degree one over $U$, and bounds the disk $D_{1\mathrm{/}\sqrt{2}}^2\times \{e^{i{\theta }_0}\}$ in the displayed exterior coordinates.

To meet [F6]'s larger-domain hypothesis, set $h(v)=v\mathrm{/}\sqrt{1+\mathrm{\mid }v{\mathrm{\mid }}^2}$ and $\mathrm{\Psi }(t,v)=\mathrm{\Phi }(t,h(v))$. The map $h:\mathbb{C}\to \{\mathrm{\mid }w\mathrm{\mid }<1\}$ is a smooth diffeomorphism with inverse $w\mapsto w\mathrm{/}\sqrt{1-\mathrm{\mid }w{\mathrm{\mid }}^2}$, fixes zero with derivative the identity, and preserves radial orientations. Hence $\mathrm{\Psi }$ is a tubular map on all of $S^1\times \mathbb{C}$, containing the closed radius-$2$ disk bundle. Its closed radius-$1$ tube is precisely $N(U)$, since $h(D_1^2)=D_{1\mathrm{/}\sqrt{2}}^2$; it has the same boundary projection, positive meridian and product section as $\mathrm{\Phi }$. Applying [F6] with $\mathrm{\Psi }$ and $ϵ=1$ therefore gives that $[\mu ]$ generates $H_1(E_U;\mathbb{Z})$ and the degree-one kernel class is the preferred longitude for the specified exterior and markings. The section in step 2.1 is in the kernel because it bounds the stated disk, so it is $[\lambda ]$ by uniqueness. Since [F5] gives $([\mu ],[\sigma ])$ as a basis and this section is $\sigma$, the pair $([\mu ],[\lambda ])$ is a basis.

By [F1] and [F2], the exterior inclusion and the meridian and longitude classes have the stated meanings. In the basis from [F3], the degree-one property in [F4] writes $[\lambda ]=[\sigma ]+q[\mu ]$ for a unique integer $q$. The curve $\gamma (t)=(e^{2\pi it},e^{2\pi i(q+1)t})$ is embedded because its first coordinate is injective on $S^1$, and it represents $[\sigma ]+(q+1)[\mu ]=[\lambda ]+[\mu ]$. Its projection to the oriented core has degree $+1$.

In the basis $([\mu ],[\lambda ])$ from [F4], the class $[\lambda ]+[\mu ]$ has coordinates $(1,1)$. If it were $md$ for an integer $\mathrm{\mid }m\mathrm{\mid }>1$, both coordinates would be divisible by $m$, which is impossible; hence it is primitive.

The inclusion sends $[\lambda ]+[\mu ]$ to $[\mu ]$ because $[\lambda ]\in \ker i_{\ast }$. The latter is a generator of $H_1(E_K;\mathbb{Z})\cong \mathbb{Z}$, so this image is nonzero. The primitive degree-one class constructed above therefore contradicts the refuted assertion.