Alphabeta Math
ExampleConstruction: AI-generatedVerification: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-6-sol)audited 2026-09-27
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.

Two adjacent noncomposable Gaussian pivots in either finite order

Example

Let X∙ be a cochain complex with objects X0=A,X1=B⊕C,X2=D1⊕D2⊕E,X3=F⊕G,X4=H,Xj=0 (j∉{0,1,2,3,4}), differentials d0=(p;q), d1=(ψβxγyδ) (rows D1,D2,E, columns B,C), d2=(zμλwνφ) (rows F,G, columns D1,D2,E) and d3=(r s), with ψ:B→D1 and φ:E→G isomorphisms and with d1d0=0, d2d1=0 and d3d2=0; this is the Lemma A.2 shape of Clark–Morrison–Walker. The two pivots ψ and φ are adjacent but not composable, so the cancellation order is a genuine choice, and the example computes the reduction in each order.

Facts & Assumptions

Given: The cochain complex X∙ displayed above, with the invertible entries ψ:B→D1 of d1 and φ:E→G of d2, and the two cancellation orders: cancel ψ first and then φ in the reduced complex, or cancel φ first and then ψ.

[L1]

At a pivot φ of a block (abcφ) in a decomposition Xn=A′⊕U, Xn+1=B′⊕V, the candidate reduction keeps Xj for j∉{n,n+1} and the components of the neighbouring differentials along the retained summands A′ and B′, and replaces the differential by the Schur complement a−bφ−1c (An invertible cochain differential block and its candidate reduction, Triangular basis changes diagonalize an invertible differential block).

[L2]

Each single cancellation in its current complex gives cochain maps p,ı and a degree-(−1) homotopy h with pı=1, 1−ıp=dh+hd, ph=0, hı=0, h2=0 (Explicit strong deformation retract from Gaussian cancellation).

[L3]

A finite sequence of cancellations with invertible current pivots composes: the composite data is p=p2p1, ı=ı1ı2, h=h1+ı1h2p1 and satisfies the same five identities, and reductions obtained from different valid choices are homotopy equivalent (Finite iteration of current invertible-block cancellations).

Verification

1.1

Complex check. The composite d2d1 has (i,j)-entry the sum over D1,D2,E of the products of entries of d2 and d1, and each of the three displayed matrix identities d1d0=0, d2d1=0, d3d2=0 is exactly the hypothesis that consecutive differentials of X∙ vanish; the remaining composites are zero because X−1=X5=0.

L1algebra
1.2

Cancelling ψ first. Write X1=C⊕B and X2=(D2⊕E)⊕D1, so that the pivot block of d1 is ψ:B→D1, with a=(γ;δ):C→D2⊕E, b=(x;y):B→D2⊕E and c=β:C→D1. The Schur complement of the pivot is a−bψ−1c=(γ−xψ−1β; δ−yψ−1β):C→D2⊕E; the incoming arrow of the reduction is the C-component q of d0, and the outgoing arrow is the restriction of d2 to the rows D2,E, namely (μλνφ), in which the pivot φ:E→G still appears unchanged.

L1L2algebra
1.3

Cancelling φ first. In the decomposition X2=(D1⊕D2)⊕(E) and X3=(F)⊕(G), the pivot φ:E→G has complement blocks z,μ (row F) and w,ν, so the Schur complement is the arrow D1⊕D2→F with entries z−λφ−1w and μ−λφ−1ν; the incoming arrow is the restriction of d1 to the rows D1,D2, namely (ψβxγ), and the outgoing arrow is r:F→H. Cancelling ψ second gives the middle differential γ−xψ−1β on C→D2, the incoming arrow q and the outgoing arrow r.

L1L2algebra
2.1

Cancelling φ second. In the complex of step 1.2 the pivot φ:E→G sits in the block of (μλνφ) with retained summands D2 of the degree-2 object and F of the degree-3 object, so the new middle differential is the Schur complement μ−λφ−1ν:D2→F, while the incoming arrow loses its E-component and becomes γ−xψ−1β:C→D2; the outgoing arrow is r:F→H.

L1step 1.2algebra
3.1

Order comparison. By steps 1.2 and 2.1, cancelling ψ then φ leaves the cochain complex A→qC→γ−xψ−1βD2→μ−λφ−1νF→rH; by step 1.3, cancelling φ then ψ leaves the same objects and the same four arrows. In each order the composite of the single-cancellation data is by [L3] the strong deformation retract data p=p2p1, ı=ı1ı2, h=h1+ı1h2p1 of X∙ onto that reduction, satisfying pı=1, 1−ıp=dh+hd, ph=0, hı=0 and h2=0; [L3] also gives that the two reductions are homotopy equivalent.

L3step 2.1step 1.3algebra
4.1

A concrete instance over Q. Take every entry of d1 equal to 1, so ψ=β=x=γ=y=δ=1; take d2=(11−21−21), d0=(1;−1) and d3=(0 0). Then d1d0=(1−1;1−1;1−1)=0, d2d1=(1+1−21+1−21−2+11−2+1)=0 and d3d2=(0 0), so X∙ is a complex. The two reduced middle arrows are γ−xψ−1β=1−1=0 and μ−λφ−1ν=1−(−2)(1)−1(−2)=1−4=−3, and the end arrows are q=−1 and r=0; both orders give the reduced complex Q→−1Q→0Q→−3Q→0Q, whose composites 0⋅(−1)=0 and (−3)⋅0=0 vanish. ∎

L1L3step 3.1algebra

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

7 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