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LemmaStatement: AI-adaptedProof: AI-generatedPipeline-generatedprecheck pendingjudge pass (gpt-6-luna)
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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
  • AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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Nonvanishing of the fifth special-unitary homotopy group

Statement

For every integer N≥2, π5(SU(N)) is nonzero. The claim is only nonvanishing; no full computation of these groups is asserted.

Facts & Assumptions

Given: The column fibration and exact sequence in ZF + AC.

[F1]

π5(SU(2))≠0 and π4(SU(2))≅Z/2. Low special-unitary homotopy inputs

[F2]

SU(N−1)→SU(N)→S2N−1 is a Hurewicz fibration with the usual homotopy LES. Special-unitary column fibration, Long exact sequence of homotopy groups of a fibration

[F3]

If N≥4, then π5(S2N−1)=π6(S2N−1)=0 because both indices are below sphere dimension 2N−1≥7. Lower-dimensional sphere maps are based nullhomotopic

[F4]

Degree identifies π5(S5) with Z. Based sphere maps are classified by degree

Proof

1.1F1F2F4algebra

For N=2, the assertion is [F1]. For N=3, the LES of [F2] contains π5(SU(3))⟶π5(S5)→∂π4(SU(2)). By [F1], the target of ∂ has two elements. The degree theorem [F4] gives π5(S5)≅Z, and every homomorphism from Z to a group of order two has kernel containing 2Z, which is nonzero. Exactness identifies that kernel with the image of π5(SU(3)), so π5(SU(3))≠0. This argument makes no surjectivity assumption on ∂.

2.1F1F2F3step 1.1algebra∎

For N≥4, [F3] and the LES [F2] make the inclusion-induced map π5(SU(N−1))→π5(SU(N)) an isomorphism: its preceding and following sphere groups are π6(S2N−1) and π5(S2N−1), both zero. Induction from the nonzero N=3 case proves nonvanishing for every N≥4, completing all cases.

Depends on

Dependency tree · 2 levels

Sources