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Worked case: the unknot exterior

Example

Assume countable choice. Give the unit sphere $S^{3} = {(z_{1}, z_{2}) \in C^{2} : ∣ z_{1} ∣^{2} + ∣ z_{2} ∣^{2} = 1}$ its standard boundary orientation, and orient the standard unknot $U (t) = (e^{2 π i t}, 0)$ by increasing $t \in R / Z$ . For $r > 0$ , write $D_{r}^{2} = {w \in C : ∣ w ∣ \leq r}$ . Choose the closed tubular neighbourhood $N (U) = {(z_{1}, z_{2}) \in S^{3} : ∣ z_{2} ∣ \leq 1 / \sqrt{2}}$ , using the norm $∣ w ∣$ on its normal disk fibres. Its exterior $E_{U} = {(z_{1}, z_{2}) \in S^{3} : ∣ z_{2} ∣ \geq 1 / \sqrt{2}}$ is a solid torus diffeomorphic to $D_{1 / \sqrt{2}}^{2} \times S^{1}$ . On its boundary, the positive meridian is the phase loop in $z_{2}$ , while the preferred longitude has representative $t \mapsto (e^{2 π i t} / \sqrt{2}, e^{i θ_{0}} / \sqrt{2})$ for fixed $θ_{0}$ ; consequently $H_{1} (E_{U}; Z) ≅ Z$ is generated by $[μ]$ and $[λ] = 0$ .

Example references

Cattabriga and Francaviglia, Example 4.1.2 and §4.4

Hatcher, Theorem 3.44 and Corollary 3.45

Facts & Assumptions

Given: Countable choice, the standard oriented sphere and unknot above, and the chosen closed tube $N (U)$ with its product fibre norm $∣ w ∣$ .

[A1]

Countable choice, written ${A C}_{ω}$ , is The Axiom of Countable Choice ( ${A C}_{ω}$ ).

[F1]

Knot exterior defines the exterior as $S^{3} ∖ int N (U)$ and retains its boundary torus.

[F2]

Meridian and preferred longitude orients the meridian by the normal disk and characterizes the preferred longitude as a degree-one boundary class that maps to zero in exterior homology.

[F3]

Normal and conormal bundles of an embedded submanifold defines the normal fibre at a point of $U$ as the quotient of the ambient tangent space by the knot tangent; here it is represented by the $z_{2}$ plane.

[F4]

Tubular neighbourhoods of embedded submanifolds defines a tubular neighbourhood by a smooth embedding on an open normal-bundle neighbourhood that fixes the zero section.

[F5]

The knot exterior has a peripheral torus basis gives the product-section and fibre-meridian basis of the boundary torus homology.

[F6]

Homology and preferred longitude of a knot exterior states that the positive meridian generates $H_{1} (E_{U}; Z)$ and that there is a unique degree-one kernel class, represented by a simple preferred longitude.

Proof

technique · direct

1.1A1F3F4givenconstruct

The map $Φ (t, w) = (e^{2 π i t} \sqrt{1 - ∣ w ∣^{2}}, w)$ extends to the open normal disk bundle $S^{1} \times {∣ w ∣ < 1}$ and is a diffeomorphism onto ${z_{1} \neq 0} \subset S^{3}$ , with inverse $(z_{1}, z_{2}) \mapsto (z_{1} / ∣ z_{1} ∣, z_{2})$ . 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 $D_{1 / \sqrt{2}}^{2}$ with the norm $∣ w ∣$ gives the specified $N (U) ≅ S^{1} \times D_{1 / \sqrt{2}}^{2}$ .

1.2F1constructalgebra

The map $(w, e^{i θ}) \mapsto (w, e^{i θ} \sqrt{1 - ∣ w ∣^{2}})$ from $D_{1 / \sqrt{2}}^{2} \times S^{1}$ to $E_{U}$ is a smooth diffeomorphism, with inverse $(z_{1}, z_{2}) \mapsto (z_{1}, z_{2} / ∣ z_{2} ∣)$ since $∣ z_{2} ∣ \geq 1 / \sqrt{2}$ . Hence $E_{U}$ is a solid torus.

2.1F1F2step 1.2construct

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 $C^{2}$ . Therefore the positive meridian is the loop varying the phase of $z_{2}$ . The loop $t \mapsto (e^{2 π i t} / \sqrt{2}, e^{i θ_{0}} / \sqrt{2})$ is the product section of $\partial N (U)$ , has degree one over $U$ , and bounds the disk $D_{1 / \sqrt{2}}^{2} \times {e^{i θ_{0}}}$ in the displayed exterior coordinates.

3.1A1F5F6step 1.1step 2.1constructalgebra∎

To meet [F6]'s larger-domain hypothesis, set $h (v) = v / \sqrt{1 + ∣ v ∣^{2}}$ and $Ψ (t, v) = Φ (t, h (v))$ . The map $h : C \to {∣ w ∣ < 1}$ is a smooth diffeomorphism with inverse $w \mapsto w / \sqrt{1 - ∣ w ∣^{2}}$ , fixes zero with derivative the identity, and preserves radial orientations. Hence $Ψ$ is a tubular map on all of $S^{1} \times C$ , containing the closed radius- $2$ disk bundle. Its closed radius- $1$ tube is precisely $N (U)$ , since $h (D_{1}^{2}) = D_{1 / \sqrt{2}}^{2}$ ; it has the same boundary projection, positive meridian and product section as $Φ$ . Applying [F6] with $Ψ$ and $ϵ = 1$ therefore gives that $[μ]$ generates $H_{1} (E_{U}; Z)$ 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 $[λ]$ by uniqueness. Since [F5] gives $([μ], [σ])$ as a basis and this section is $σ$ , the pair $([μ], [λ])$ is a basis.

Peripheral square for the unknot exterior

Use product coordinates $(t, θ) \in R^{2} / Z^{2}$ on $T_{U}$ , where $t$ is the oriented core parameter and $θ$ is the positive meridian angle. Opposite edges are identified with their arrows in the same direction. The horizontal loop is the section $λ$ , which bounds the disk $D_{1 / \sqrt{2}}^{2} \times {e^{i θ_{0}}}$ in $E_{U}$ ; the vertical loop is $μ$ , which is the $S^{1}$ factor of $D_{1 / \sqrt{2}}^{2} \times S^{1}$ . The drawing records these product coordinates and boundary classes; the proof above verifies them from the explicit parametrizations.

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Sources

Alessia Cattabriga and Stefano Francaviglia, Knot Theory and Applications: Lecture Notes (standard reference, not scraped)
Allen Hatcher, Algebraic Topology (standard reference, not scraped)