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The Poincaré sphere is a nontrivial integral homology sphere

Statement

Assume the Axiom of Choice. Let $I^{*}$ be the binary icosahedral group, the group of $120$ unit quaternions acting on the unit sphere $S^{3}$ by left multiplication, and put $P = S^{3} / I^{*}$ . The quotient is a closed connected oriented $3$ -manifold with universal cover $S^{3}$ and deck group $I^{*}$ . Then $H_{*} (P; Z) ≅ H_{*} (S^{3}; Z)$ and $π_{1} (P) ≅ I^{*}$ is nontrivial.

Statement references

Algebraic Topology, Winter Semester 2021/22, §4.5.12

Arun Debray, Group Actions on $S^{3}$ , §§1–2

Facts & Assumptions

Given: AC, the binary icosahedral group presentation $I^{*} = ⟨ s, t ∣ (s t)^{2} = s^{3} = t^{5} ⟩$ of order $120$ , its free orientation-preserving action on $S^{3}$ , and the quotient $P = S^{3} / I^{*}$ .

[A1]

The Axiom of Choice: every family of nonempty sets has a choice function. The local Poincaré-duality supplier below assumes this axiom.

[F1]

$S^{n}$ is simply connected for every $n \geq 2$ : $S^{3}$ is simply connected.

[F2]

Universal covering spaces: a universal cover is a covering map with simply connected total space.

[F3]

For a path-connected locally path-connected semilocally simply connected base, the deck group of a universal cover is isomorphic to the fundamental group: for a universal cover, its deck group is isomorphic to the fundamental group of its path-connected, locally path-connected, semilocally simply connected base.

[F4]

The first Hurewicz map is abelianization: for a path-connected space, $H_{1} (-; Z)$ is the abelianization of $π_{1}$ .

[F5]

Poincaré duality for oriented topological manifolds: under AC, compact oriented boundaryless $P$ has cap-duality isomorphisms $H_{i} (P; Z) ≅ H^{3 - i} (P; Z)$ in every degree.

[F6]

The universal coefficient theorem for cohomology over a PID: under AC, the free integral singular chain complex of $P$ gives $0 \to {Ext}^{1} (H_{0} (P), Z) \to H^{1} (P; Z) \to Hom (H_{1} (P), Z) \to 0$ .

[F7]

Debray, Group Actions on $S^{3}$ , §§1–2, pp.1–2: $I^{*}$ is the order- $120$ preimage of the icosahedral rotation group in $S U (2)$ , has presentation $⟨ s, t ∣ (s t)^{2} = s^{3} = t^{5} ⟩$ , and acts freely by left multiplication; the finite-group quotient is a smooth covering $S^{3} \to S^{3} / I^{*}$ .

Proof

1.1F1F2F3F7given

By [F7], the order- $120$ subgroup of unit quaternions acts freely by left multiplication, so $S^{3} \to P$ is a covering with deck group $I^{*}$ . Each left multiplication is isotopic to the identity because $S^{3}$ is path-connected; hence the action preserves orientation and the quotient is oriented. Since $S^{3}$ is simply connected by [F1], this cover is universal by [F2]. Then [F3] gives $π_{1} (P) ≅ I^{*}$ , which is nontrivial because its order is $120$ .

1.2givenF4F7algebra

In the abelianization, write the images of $s, t$ additively. The presentation in [F7] gives the relation $(s t)^{2} = s^{3} = t^{5}$ , hence $2 s + 2 t = 3 s = 5 t$ . The first equality yields $s = 2 t$ ; substituting into $3 s = 5 t$ gives $6 t = 5 t$ , hence $t = 0$ and $s = 0$ . Thus $(I^{*})^{a b} = 0$ , and [F4] gives $H_{1} (P; Z) = 0$ .

2.1givenA1F6step 1.2algebra

The quotient is connected, compact, and oriented, so $H_{0} (P; Z) = H^{0} (P; Z) = Z$ . By [F6], $H^{1} (P; Z)$ is an extension of $Hom (0, Z) = 0$ by ${Ext}^{1} (Z, Z) = 0$ , so $H^{1} (P; Z) = 0$ . The AC assumption is the one in [A1], needed for [F6].

3.1A1F5givenstep 1.1step 1.2step 2.1∎

By [F5] and step 2.1, $H_{2} (P; Z) = 0$ and $H_{3} (P; Z) ≅ H^{0} (P; Z) = Z$ . For $i > 3$ , [F5] identifies $H_{i} (P; Z)$ with negative-degree cohomology, which is zero. Together with steps 1.2 and 2.1, this is exactly the integral homology of $S^{3}$ . Step 1.1 shows that $P$ is not simply connected, so it is not $S^{3}$ . The use of [F5] requires [A1].

Depends on

Used by

A null-homologous unknot in a homology sphere can have reducible exterior Counterexample

Dependency tree · two levels

39 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Algebraic Topology, Winter Semester 2021/22, §4.5.12 (standard reference, not scraped)
Arun Debray, Group Actions on S^3 (standard reference, not scraped)