Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-6.1-sol)
How statement and proof provenance work

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.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Pontryagin calibration of the basic quaternionic clutchings

Statement

Assume the Axiom of Choice as inherited from the Chern and Pontryagin suppliers. Let VL be the oriented rank-four bundle over S4 clutched by a↦(v↦av) and VR the one clutched by a↦(v↦va), with the base, fibre and Euler-degree conventions of Quaternionic clutching bundles ξh,j over S4. Then e(VL)=e(VR)=u∈H4(S4;Z),p1(VL)=2u,p1(VR)=−2u.

Facts & Assumptions

Given: The basic bundles VL,VR of Quaternionic clutching bundles ξh,j over S4 with the generator u∈H4(S4;Z).

[A1]

The Axiom of Choice is assumed as inherited from the Chern/Pontryagin suppliers (The Axiom of Choice).

[L1]

For the basic left and right quaternionic clutchings the Euler number is +1, so e(VL)=e(VR)=u (Euler number of a clutched bundle as the clutching degree, Quaternionic clutching bundles ξh,j over S4).

[L2]

The top Chern class of a complex rank-n bundle equals the Euler class of its underlying real bundle in the complex orientation (Top Chern class equals Euler class of the underlying real bundle).

[L3]

A complex bundle V has a canonical complex orientation of VR, natural under complex-linear isomorphisms; orientation reversal negates the Euler class (The complex orientation of the underlying real bundle).

[L4]

For a complex bundle V, the conjugate satisfies ci(V‾)=(−1)ici(V), and the complexification of a real bundle is canonically isomorphic to its conjugate (Complexification is conjugation invariant, Pontryagin classes by complexification).

[L5]

Total Chern classes multiply under Whitney sums and vanish in degrees above the rank (Naturality, normalization, and Whitney sum for Chern classes).

Proof

technique · direct
1.1L1given

Write V=VL or V=VR with its complex structure: for VL right multiplication by i commutes with left multiplication by every quaternion and makes VL a complex rank-two bundle; for VR left multiplication by i plays the same role for right multiplication.

2.1step 1.1L4

The complexification VC=V⊗RC carries the complexified complex structure JC; its ±i eigenspace projections are complex subbundles E+≅V and E−≅V‾ (locally over a trivializing chart they are the ±i eigenspaces of the constant matrix J), so VC≅V⊕V‾ as complex bundles.

3.1step 2.1L4L5

By [L5] and step 2.1, c2(VC)=c2(V)+c2(V‾)=c2(V)+c2(V)=2c2(V) because c2 is even in the conjugate sign of [L4], and c1 terms do not contribute in degree four on S4 as H2(S4)=0.

4.1step 3.1L4A1

By the Pontryagin convention p1(VR)=−c2(VC)=−2c2(V) (the sign is the one fixed in [L4]).

5.1step 4.1L1L2L3

The complex orientation of VL is opposite to the fibre orientation: the commuting complex structure is right multiplication by i, whose complex basis (1,j) gives the real ordered basis (1,i,j,−k), the negative of the fixed fibre basis (1,i,j,k); hence by [L2] and [L3] c2(VL)=e(VL,R) in the complex orientation =−e(VL,R) in the fibre orientation =−u.

6.1step 5.1L2L3

The complex orientation of VR agrees with the fibre orientation: the commuting complex structure is left multiplication by i, whose complex basis (1,j) gives the real ordered basis (1,i,j,k); hence c2(VR)=+u.

7.1step 5.1step 6.1∎

Substituting steps 5.1 and 6.1 into step 4.1 gives p1(VL)=−2(−u)=2u and p1(VR)=−2u, while step 1.1's Euler computation gives e(VL)=e(VR)=u, as asserted.

Depends on

Used by

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