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Extension detects trivial kernel bundles

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QAlgorithms.Papers.Independent.Intakethird100.P93.proposition_3_7

by Gohan · Nov 1, 2026 · Mathlib 0df444a (Lean v4.33.1)

prove2quantum

Let t:G→Ht:G\to Ht:G→H be a topological crossed module, with continuous action α:H→Aut⁡(G)\alpha:H\to\operatorname{Aut}(G)α:H→Aut(G). Thus t(α(h)g)=ht(g)h−1t(\alpha(h)g)=ht(g)h^{-1}t(α(h)g)=ht(g)h−1 and α(t(g))k=gkg−1\alpha(t(g))k=gkg^{-1}α(t(g))k=gkg−1. Put G1=ker⁡tG_1=\ker tG1​=kert and give both G/G1G/G_1G/G1​ and H/t(G)H/t(G)H/t(G) their quotient topologies. Assume their quotient projections admit continuous local sections. All actions, structure maps and isomorphisms below are continuous, all principal bundles are locally trivial, and a bispace has a nonempty principal homogeneous underlying right space.

For any principal G1G_1G1​-bundle Q→MQ\to MQ→M, form the actual quotient extension ι(Q)=Q×G1G\iota(Q)=Q\times_{G_1}Gι(Q)=Q×G1​​G with structure map ψ([q,g])=t(g)\psi([q,g])=t(g)ψ([q,g])=t(g). Then

Q is trivial as a G1-bundle  ⟺  ι(Q) is trivial as an (H,G)-bibundle.Q\text{ is trivial as a }G_1\text{-bundle} \iff\iota(Q)\text{ is trivial as an }(H,G)\text{-bibundle}.Q is trivial as a G1​-bundle⟺ι(Q) is trivial as an (H,G)-bibundle.

On the right the trivialization must preserve the structure map as well as the base and right action. The conclusion therefore detects the original kernel-bundle triviality, rather than the weaker triviality of an underlying GGG-bundle.

Preamble
import Definitions.Def_QAlgorithms_Papers_Independent_Intakethird100_P93
Formal statement
namespace QAlgorithms.Papers.Independent.Intakethird100.P93

universe u

variable {G H : Type u} [Group G] [TopologicalSpace G] [IsTopologicalGroup G]
  [Group H] [TopologicalSpace H] [IsTopologicalGroup H]

variable (C : CrossedModule G H)

/-- Proposition 3.7, PDF p.11: ordinary G1-bundle triviality is equivalent to
bibundle triviality of the actual quotient extension, including its structure map. -/
theorem proposition_3_7 {M Q : Type u} [TopologicalSpace M] [TopologicalSpace Q]
    (R : PrincipalBundle C.t.ker M Q) :
    IsTrivialPrincipal R ↔ IsTrivialIota C R := by sorry

end QAlgorithms.Papers.Independent.Intakethird100.P93
Source
On the existence of bibundles, arXiv:1102.4388v2, PDF p. 11, Proposition 3.7
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What the Lean code literally says, in plain math · gpt-6.1-sol

Let G,HG,HG,H be topological groups, with a continuous homomorphism t:G→Ht:G\to Ht:G→H and a homomorphism α:H→Aut⁡(G)\alpha:H\to\operatorname{Aut}(G)α:H→Aut(G) whose evaluation is continuous, satisfying t(αh(g))=h t(g) h−1t(\alpha_h(g))=h\,t(g)\,h^{-1}t(αh​(g))=ht(g)h−1 and αt(g)(k)=gkg−1\alpha_{t(g)}(k)=gkg^{-1}αt(g)​(k)=gkg−1 for all h∈Hh\in Hh∈H and g,k∈Gg,k\in Gg,k∈G. Assume that both quotient maps G→G/ker⁡tG\to G/\ker tG→G/kert and H→H/t(G)H\to H/t(G)H→H/t(G) admit continuous sections on open neighborhoods of every target point, meaning that the quotient of each section is the inclusion of its neighborhood. Put K=ker⁡tK=\ker tK=kert. For arbitrary topological spaces M,QM,QM,Q, let RRR consist of a continuous right KKK-action on QQQ, satisfying the identity and multiplication laws, and a continuous invariant projection π:Q→M\pi:Q\to Mπ:Q→M, with, around every m∈Mm\in Mm∈M, an open neighborhood UUU and a homeomorphism π−1(U)≅U×K\pi^{-1}(U)\cong U\times Kπ−1(U)≅U×K preserving the projection and carrying the action to right multiplication on KKK. Let E=(Q×G)/∼E=(Q\times G)/{\sim}E=(Q×G)/∼, where (q,g)∼(q⋅k,k−1g)(q,g)\sim(q\cdot k,k^{-1}g)(q,g)∼(q⋅k,k−1g) for k∈Kk\in Kk∈K, and write [q,g][q,g][q,g] for a class. Then the following two existence statements are equivalent: there is a homeomorphism a:Q→M×Ka:Q\to M\times Ka:Q→M×K satisfying pr⁡Ma(q)=π(q)\operatorname{pr}_M a(q)=\pi(q)prM​a(q)=π(q) and a(q⋅k)=(pr⁡Ma(q),pr⁡Ka(q) k)a(q\cdot k)=(\operatorname{pr}_M a(q),\operatorname{pr}_K a(q)\,k)a(q⋅k)=(prM​a(q),prK​a(q)k) for every q,kq,kq,k; and there is a homeomorphism e:E→M×Ge:E\to M\times Ge:E→M×G satisfying, for all q∈Qq\in Qq∈Q and g,b∈Gg,b\in Gg,b∈G,

pr⁡Me([q,g])=π(q),e([q,gb])=(pr⁡Me([q,g]),pr⁡Ge([q,g]) b),t(pr⁡Ge([q,g]))=t(g).\operatorname{pr}_M e([q,g])=\pi(q),\qquad e([q,gb])=\bigl(\operatorname{pr}_M e([q,g]),\operatorname{pr}_G e([q,g])\,b\bigr),\qquad t\bigl(\operatorname{pr}_G e([q,g])\bigr)=t(g).prM​e([q,g])=π(q),e([q,gb])=(prM​e([q,g]),prG​e([q,g])b),t(prG​e([q,g]))=t(g).

The equivalence includes the empty-base case, when QQQ and EEE are necessarily empty.

Human review
  • Endorsed by Shuze Chen · Nov 1, 2026

    Confirmed by the moderator at approval.

  • Endorsed by Gohan · Nov 1, 2026

    Confirmed by the mission captain (proposal self-audit).

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