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Adem relations for Steenrod squares

Statement

Assume AC. Let a,b be positive integers with 0<a<2b. For every space X, every q0, and every xHq(X;F2),

SqaSqb(x)=j=0a/2(bj1a2j)Sqa+bjSqj(x).

The binomial coefficients are reduced modulo two. A binomial coefficient is zero when its lower index is negative or exceeds its nonnegative upper index, and a Steenrod square with an index outside its defining nonnegative range is zero.

Facts & Assumptions

Given: AC, positive integers a,b with 0<a<2b, a space X, a nonnegative degree q, and a class xHq(X;F2).

[F1]

In each degree 2s1+b with 2s>a, the finite-cellular cyclic squares satisfy the displayed Adem formula (Adem double-power comparison).

[F2]

On a finite oriented regular cell complex, a cohomology isomorphism from singular to cellular cohomology intertwines every singular square with the finite-cellular cyclic square (Finite-cellular cyclic squares agree with singular cup-i squares).

[F3]

Under AC, a natural singular-cohomology identity that is zero on all finite regular cell complexes is zero on every space (Natural singular-cohomology identities are detected on finite regular complexes).

[F4]

Singular Steenrod squares obey the external Cartan formula (Cartan formula for Steenrod squares).

[F5]

They satisfy Sq0=id, instability, and Sqnz=zz for a degree-n class (Steenrod normalization, instability, suspension, and top square).

[F6]

Every square is additive and natural, with its outside-range values zero (Steenrod squares are well-defined and natural).

[F7]

Under AC, external product is a cohomology isomorphism over a PID when one factor has finite-free homology in every degree (Cohomological Kunneth cross product is a ring isomorphism).

[F8]

Cellular homology is naturally isomorphic to singular homology for every CW complex and coefficient group (Cellular homology computes singular homology).

[F9]

AC supplies a choice function for every family of nonempty sets (The Axiom of Choice).

Proof

Proof technique: transfer the high-degree cellular calculation to singular squares, descend degrees by external product with the circle generator, and then detect the resulting natural identity on finite regular complexes.

1.1

Define the residual operation in input degree r.

givenF6

Rr(u)=SqaSqb(u)+j=0a/2(bj1a2j)Sqa+bjSqj(u).

Addition is subtraction over F2. By [F6], for fixed r this is a natural map Hr(;F2)Hr+a+b(;F2). All its sums are finite, and every operation appearing in it has a nonnegative index.

1.2

Compute the circle class used for descent. [F2, F8] Regard S1 as the boundary of a triangle, with vertices v0,v1,v2 and edges e01,e12,e02. Over F2 its cellular boundary is

eij=vi+vj.

Hence ker(:C1C0) is generated by e01+e12+e02, the image has dimension two, and there are no cells above degree one. Thus cellular homology is F2 in degrees zero and one and zero otherwise. By [F8], the same is true of singular homology, so all the singular homology groups of this S1 are finite free.

The dual cellular coboundary sends a vertex function (λ0,λ1,λ2) to (λ0+λ1,λ1+λ2,λ0+λ2). Its image is the plane of edge functions whose three values sum to zero. Consequently the edge function taking value one on e01 and zero on the other two edges represents the unique nonzero vˉHcell1(S1;F2). By [F2], there is a unique nonzero vHsing1(S1;F2) with ιv=vˉ, and Hd(S1;F2)=0 for d>1.

2.1

Establish the residual identity in unbounded finite degrees. [F1, F2, step 1.1] Fix s with 2s>a, put rs=2s1+b, let K be a finite oriented regular cell complex, and let uHsingrs(K;F2). Write ι for the isomorphism of [F2]. Applying it successively to each square in step 1.1 gives

ιRrs(u)=Rrscyc(ιu).

The right side is zero by [F1]. Since ι is injective, Rrs(u)=0. Thus Rrs=0 on every finite regular complex for every s with 2s>a.

2.2

Show that square compositions preserve the circle factor. [F4, F5, step 1.2] For a finite regular complex K, a class uHd(K;F2), and k0, external Cartan gives

Sqk(u×v)=i+j=kSqi(u)×Sqj(v).

Here Sq0(v)=v. The top-square formula and step 1.2 give Sq1(v)=vv=0 in H2(S1;F2), and instability gives Sqj(v)=0 for j>1. Therefore

Sqk(u×v)=Sqk(u)×v.

Applying this equality twice covers every two-square composition occurring in R. Linearity then yields

Rd+1(u×v)=Rd(u)×v.
3.1

Descend the identity by one degree. [F7, step 1.2, step 2.2] Suppose r1 and Rr=0 on every finite regular complex. The product of two finite regular cell complexes is finite regular, so for every such K and every uHr1(K;F2),

0=Rr(u×v)=Rr1(u)×v.

Apply [F7] over the PID F2. Its finite-free hypothesis holds for the circle by step 1.2, so external product identifies the last class with Rr1(u)v. Tensoring an F2-vector space with the nonzero vector v is injective on the first factor. Hence Rr1(u)=0, proving the one-degree descent.

4.1

Prove the finite-regular identity in every degree. [step 2.1, step 3.1] Given q0, choose the least s with both 2s>a and 2s1+bq. Step 2.1 gives the identity in degree rs=2s1+b. Apply step 3.1 exactly rsq times. This proves Rq=0 on every finite regular complex. The construction works separately for each q and uses no limit or simultaneous choice.

5.1

Extend from finite regular complexes to every space. [F3, F6, step 1.1, step 4.1] For the fixed input degree q, step 1.1 and [F6] make Rq a natural singular-cohomology operation, and step 4.1 makes it zero on every finite regular complex. The detection theorem [F3] therefore gives Rq=0 on every space. Expanding its definition is exactly the formula in the statement.

6.1

Empty-space and zero-class inputs give zero, while both finite-sum endpoint indices remain included. [F1, F2, F3, F5, F6, F7, F9, step 1.1, step 1.2, step 2.1, step 2.2, step 3.1, step 4.1, step 5.1] The empty space and the zero class give zero on both sides by additivity. On a point, all positive-degree input groups vanish, while in degree zero the instability clauses in [F5] make every term zero because a,b>0. The endpoints j=0 and j=a/2 are retained. The strict inequality a<2b makes every upper binomial index bj1 nonnegative; the stated convention handles every oversized lower index. The values Sq0=id used at j=0 and on the circle are explicit, and all above-degree or negative-index squares have the conventions stated in [F5] and [F6]. Ordinary singular cohomology, including degenerate singular simplices, is used in [F2], [F3], and [F6], so no normalized-chain identification is hidden. The theorem is an equality, not a biconditional.

AC from [F9] is used exactly through four suppliers: [F1]'s cyclic and wreath carrier comparisons and the field duality used to identify the cyclic basis on its finite regular QN; [F2]'s carrier fillings and complements in the cellular-to-singular mapping cone; [F3]'s complement used to make cohomology evaluation injective; and [F7]'s additive Kunneth bijectivity. The triangle calculation, each product, the least integer s, and the finite sequence of descents are explicitly prescribed and require no further choice. ∎

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