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First Steenrod square relation for w2

Statement

Assume full AC. Let E→B be a numerable finite-rank real vector bundle, meaning it has a linear trivializing cover with a locally finite subordinate partition of unity. Let B be paracompact and Hausdorff, compactly generated (a subset is closed exactly when every continuous map from a compact Hausdorff space has closed inverse image of it), and weak Hausdorff (every continuous map from a compact Hausdorff space has closed image); also require B to be homotopy equivalent to a CW complex. With Stiefel–Whitney classes in H∗(B;F2) and Steenrod squares in singular cohomology, one has Sq1w2(E)=w1(E)⌣w2(E)+w3(E). The conventions wj(E)=0 for j above the rank include ranks zero and one.

Facts & Assumptions

Given: A numerable finite-rank real vector bundle on the stated admissible base; assume the full AC required by the cited splitting/class construction inputs.

[F1]

Real splitting principle with mod-two injective pullback: the real flag pullback q∗ on mod-two cohomology is injective and q∗E=⨁jLj.

[F2]

Real flag bundle and Stiefel–Whitney roots: set tj=w1(Lj)∈H1(Fl⁡(E);F2) for the split line summands.

[F3]

Whitney sum formula for Stiefel–Whitney classes: w(⨁jLj)=∏jw(Lj)=∏j(1+tj), so w1=∑jtj, w2=∑i<jtitj, and w3=∑i<j<ktitjtk after pullback.

[F4]

Naturality of Stiefel–Whitney classes: the classes wi commute with bundle pullback.

[F5]

Steenrod squares are well-defined and natural: Sq1 is a natural cohomology operation.

[F6]

Steenrod normalization, instability, suspension, and top square: for a degree-one class t, Sq1(t)=t2.

[F7]

Cartan formula for Steenrod squares: Sq1(xy)=Sq1(x)y+xSq1(y).

[F8]

Cup-i coboundary identity: the cup-one coboundary identity implies that, for mod-two cocycles, the two orders of the cup product represent the same cohomology class; hence the mod-two cohomology product is commutative.

[F9]

Cup product is natural, unital and associative: every continuous pullback is a ring map, so q∗(xy)=q∗x q∗y.

[A1]

Full AC is assumed as defined in The Axiom of Choice, exactly as required by [F1] and the Stiefel–Whitney construction.

Proof

1.1F1F2F3F4F5F6F7A1

On the flag bundle, write q∗E=L1⊕⋯⊕Ln and ti=w1(Li) by [F1]–[F2]. The Whitney formula [F3] gives q∗w2(E)=∑i<jtitj using naturality [F4]. Apply [F7] to each summand and use [F6]: Sq1q∗w2(E)=∑i<j(ti2tj+titj2). By [F5], this equals q∗Sq1w2(E).

2.1F1F3F4F8F9A1step 1.1algebra

By [F9], q∗(w1(E)w2(E))=q∗w1(E) q∗w2(E)=(∑ktk)(∑i<jtitj), using [F3]. Reorder factors using mod-two cup-product commutativity [F8]. Terms for which k=i or k=j give exactly ∑i<j(ti2tj+titj2), the expression in step 1.1. For each triple of distinct indices i<j<k, the monomial titjtk occurs three times, once for each choice of the singled-out factor tℓ in w1; three equals one in F2. Therefore q∗(w1w2)=Sq1q∗w2+∑i<j<ktitjtk=Sq1q∗w2+q∗w3. The empty pair sum for ranks zero and one and the empty triple sum for ranks below three obey the same expansion.

3.1F1A1step 2.1algebra∎

Rearranging the mod-two identity in step 2.1 gives q∗(Sq1w2+w1w2+w3)=0. Since q∗ is injective by [F1] and subtraction equals addition over F2, the class inside the pullback is zero. This is the asserted identity for all ranks in the stated scope.

Depends on

Dependency tree · 2 levels

Sources