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The second homotopy group of SO(3) vanishes

Statement

$π_{2} (S O (3), I) = 0$ . More precisely, for the two-sheeted covering homomorphism $ρ : S^{3} \to S O (3)$ , $ρ (q) (v) = q v q^{- 1}$ , from the unit quaternions onto the rotations of $Im H$ , the induced homomorphism $ρ_{*} : π_{2} (S^{3}, 1) \to π_{2} (S O (3), I)$ is a bijection and $π_{2} (S^{3}, 1) = 0$ . Consequently also $π_{2} (O (3)) = 0$ and $π_{2} (V_{2} (R^{3})) = 0$ : the map $(u, v) \mapsto (u, v, u \times v)$ is a homeomorphism $V_{2} (R^{3}) \to S O (3)$ , and $O (3)$ is the disjoint union of the two cosets of $S O (3)$ , each homeomorphic to $S O (3)$ .

Facts & Assumptions

Given: The quaternion double cover $ρ : S^{3} \to S O (3)$ , the identity matrix $I \in S O (3)$ , and the standard frames $(e_{1}, e_{2})$ of $V_{2} (R^{3})$ , $(e_{1}, e_{2}, e_{3})$ of $R^{3}$ .

[F1]

$ρ (q) (v) = q v q^{- 1}$ defines a continuous surjective group homomorphism with kernel ${\pm 1}$ , it is a two-sheeted covering map, and for every covering $p : E \to B$ , every $e_{0}$ and every $n \geq 2$ , the induced map $p_{*} : π_{n} (E, e_{0}) \to π_{n} (B, p (e_{0}))$ is an isomorphism. The quaternion double cover generates the third homotopy group of SO(3)

[F3]

$π_{n}$ is computed by based cubes and $π_{0}$ is the pointed set of path components; $π_{2}$ is a group and based homotopy equivalences induce isomorphisms. Cubical classes agree with based sphere-map classes. Cubical and spherical models of higher homotopy agree, Higher homotopy group by based cubes, Higher homotopy groups are functorial and based homotopy invariant

[F4]

For $0 \leq k < r$ , every continuous based map $(S^{k}, a) \to (S^{r}, b)$ is nullhomotopic through maps fixing $a$ . Lower-dimensional sphere maps are based nullhomotopic

[F5]

$V_{2} (R^{3}) = {(u, v) : ⟨ u, u ⟩ = ⟨ v, v ⟩ = 1, ⟨ u, v ⟩ = 0}$ with the subspace topology; $S O (3)$ is the group of real $3 \times 3$ matrices with $R^{T} R = I$ and $\det R = 1$ . Stiefel spaces, Grassmannians, and tautological bundles, The quaternion double cover generates the third homotopy group of SO(3). Write $O (3) = {A : A^{T} A = I}$ with its matrix subspace topology.

[F6]

The cross product is bilinear, alternating, orthogonal to both factors, and satisfies the scalar triple product identity $⟨ x \times y, z ⟩ = \det [x y z]$ . The cross product in $R^{3}$ , The cross product is bilinear, alternating, and orthogonal to both factors

Proof

1.1F5F6

The map $ϕ : V_{2} (R^{3}) \to S O (3)$ , $ϕ (u, v) = (u ∣ v ∣ u \times v)$ , is a homeomorphism. Its image lies in $S O (3)$ : for orthonormal $u, v$ the vector $u \times v$ is orthogonal to $u$ and $v$ by [F6] and is unit, since expanding its coordinates gives $∥ u \times v ∥^{2} = ∥ u ∥^{2} ∥ v ∥^{2} - ⟨ u, v ⟩^{2} = 1$ , so the three columns are orthonormal, and $\det (u ∣ v ∣ u \times v) = ⟨ u \times v, u \times v ⟩ = 1$ by the triple product identity [F6]. It is injective because the first two columns determine the argument. It is surjective: for $R \in S O (3)$ with columns $c_{1}, c_{2}, c_{3}$ , the vector $c_{1} \times c_{2}$ is a unit vector orthogonal to $c_{1}$ and $c_{2}$ by [F6], hence equals $\pm c_{3}$ , and the sign is $+$ because $⟨ c_{1} \times c_{2}, c_{3} ⟩ = \det (c_{1} ∣ c_{2} ∣ c_{3}) = \det R = 1$ ; thus $R = ϕ (c_{1}, c_{2})$ with $(c_{1}, c_{2}) \in V_{2} (R^{3})$ . Both $ϕ$ and the projection $R \mapsto (c_{1}, c_{2})$ to the first two columns are continuous, so $ϕ$ is a homeomorphism.

1.2F1

$ρ_{*} : π_{2} (S^{3}, 1) \to π_{2} (S O (3), I)$ is an isomorphism: $ρ$ is a two-sheeted covering map by [F1], and covering projections induce isomorphisms on $π_{n}$ for $n \geq 2$ by the second clause of [F1] applied with $n = 2$ , $p = ρ$ , $e_{0} = 1$ .

1.3F3F4

$π_{2} (S^{3}, 1) = 0$ : every continuous based map $S^{2} \to S^{3}$ is nullhomotopic through based maps by [F4] with $k = 2 < r = 3$ , so every element of $π_{2} (S^{3}, 1)$ equals the class of the constant map, the distinguished element of the group [F3].

2.1F3step 1.2step 1.3

Hence $π_{2} (S O (3), I) = 0$ : an isomorphism of groups carries the distinguished element to the distinguished element, so the triviality of the source in step 1.3 forces the triviality of the target.

3.1F3F5step 2.1

$O (3)$ is the disjoint union of its two cosets: every $A \in O (3)$ has $\det A = \pm 1$ by $A^{T} A = I$ , so $A \in S O (3)$ or $A \in R_{0} S O (3)$ for the reflection $R_{0} = diag (1, 1, - 1)$ , and the two cosets are disjoint and each is homeomorphic to $S O (3)$ by left translation. Since the square $I^{2}$ is connected, every based cube $I^{2} \to O (3)$ and every boundary-fixed homotopy of such cubes lies in the single component of the basepoint, so evaluating cubical representatives identifies $π_{2} (O (3), J)$ with $π_{2}$ of the component of $J$ , which after left translation is $π_{2} (S O (3), I)$ and hence is $0$ by step 2.1.

4.1F1F3step 1.1step 2.1step 3.1∎

$π_{2} (V_{2} (R^{3}), e) = 0$ at every basepoint $e = (u, v)$ : the homeomorphism of step 1.1 satisfies $ϕ (u, v) = R$ and is a based homotopy equivalence, so it induces an isomorphism $π_{2} (V_{2} (R^{3}), e) ≅ π_{2} (S O (3), R)$ [F3]; the left translation $A \mapsto R^{- 1} A$ is a homeomorphism of $S O (3)$ carrying $R$ to $I$ , hence induces an isomorphism $π_{2} (S O (3), R) ≅ π_{2} (S O (3), I)$ , which is $0$ by step 2.1. Together with steps 2.1 and 3.1 this proves all three claimed vanishings at every basepoint, the argument uses the unconditional topological covering statement [F1] and no Lie-group structure or choice principle.

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Sources

Allen Hatcher, Algebraic Topology, §1.3 (covering isomorphisms on higher homotopy groups) and §4.2–4.3 (standard reference, not scraped)
Ralph L. Cohen, Immersions of Manifolds and Homotopy Theory (Harvard CMSA Math-Science Literature Lecture write-up, June 30 2022), §2.1 eversion paragraph (standard reference, not scraped)