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The Kronecker pairing is multiplicative under cross products

Statement

Let $X, Y$ be spaces, $R$ a commutative unital ring, $p, q \geq 0$ , $α \in H^{p} (X; R)$ , $β \in H^{q} (Y; R)$ , $c \in H_{p} (X; R)$ and $d \in H_{q} (Y; R)$ . Then $⟨ α \times β, c \times d ⟩ = ⟨ α, c ⟩ ⟨ β, d ⟩ \in R .$ The same identity holds for $c \in H_{p} (X; Z)$ and $d \in H_{q} (Y; Z)$ , using coefficient extension for the homology inputs. If $c$ and $d$ instead have degrees $m, n$ with $m + n = p + q$ , the left side is zero unless $(m, n) = (p, q)$ . Pairings in unequal total degrees are not asserted. Both the additive shuffle convention and the external cup-product convention of the cohomology cross product give this identity. No AC is required.

Facts & Assumptions

Given: Spaces $X, Y$ , a commutative unital ring $R$ , cocycles $φ, ψ$ of degrees $p, q$ , and cycles $z, w$ over $R$ of degrees $m, n$ with $m + n = p + q$ . In the multiplicative identity take $m = p, n = q$ . Integral input cycles are extended along $Z \to R$ .

[F1]

Kronecker evaluation pairing and The kronecker pairing is independent of cocycle and cycle representatives define $⟨ [φ], [c] ⟩ = φ (c)$ by evaluation and prove it independent of both representatives and biadditive over $R$ .

[F2]

Additive singular cohomology cross product represents $α \times β$ by the composite $J (φ, ψ) T$ , where $J (φ, ψ) (x \otimes y) = φ (x) ψ (y)$ for $∣ x ∣ = p, ∣ y ∣ = q$ and vanishes on the other bidegrees of total degree $p + q$ , and $T : C (X \times Y; R) \to C (X; R) \otimes_{R} C (Y; R)$ is a natural chain homotopy inverse of the shuffle $S$ . By The additive singular cohomology cross product is well-defined the functional satisfies $δ J (φ, ψ) = J (δ φ, ψ) + (- 1)^{p} J (φ, δ ψ)$ , so it is a cocycle when $φ, ψ$ are cocycles, the composite with the chain map $T$ defines a class, and the product is $R$ -bilinear and natural.

[F3]

The singular chain cross product on generators gives the shuffle expansion of the chain cross product $S (c \otimes d) = c \times d$ , and The singular chain cross product satisfies the boundary formula gives $\partial (a \times b) = \partial a \times b + (- 1)^{i} a \times \partial b$ for $a \in C_{i}$ , so a cross product of cycles is a cycle and $S$ is a chain map. Singular chain cross products are natural makes it natural. These integral chain identities extend $R$ -bilinearly by scalar extension.

[F4]

The homology cross product for tensor complexes and The Kunneth cross-product map is well defined and natural make $[x] \times [y] = [x \otimes y]$ well defined and natural on homology. For cycles $z \in C_{m} (X; R)$ and $w \in C_{n} (Y; R)$ , the singular class $[z] \times [w]$ is represented by $S (z \otimes w) \in C_{m + n} (X \times Y; R)$ .

[F5]

Singular product chain equivalence by simplex models supplies, for the shuffle $S$ , its natural inverse $T$ and natural homotopies $S T ≃ 1$ and $T S ≃ 1$ ; in particular there is a natural chain homotopy $K$ with $d K + K d = T S - 1$ on $C (X; R) \otimes_{R} C (Y; R)$ .

[F6]

Cohomological Kunneth cross product is a ring isomorphism uses the external product $a \times b = {pr}_{X}^{*} a ⌣ {pr}_{Y}^{*} b$ . Singular cup product on cochains gives its front/back cochain formula, and Alexander--Whitney and shuffle are natural chain-homotopy inverses supplies the Alexander-Whitney map $A$ and a natural homotopy $A - T = d H + H d$ with the shuffle inverse of [F2], without AC.

Proof

1.1givenF2F3F4F5

By [F2] the class $α \times β$ is represented by the cocycle $J (φ, ψ) T$ , and by [F4] the class $c \times d$ is represented by the cycle $S (z \otimes w)$ , which is a cycle because $S (z \otimes w)$ has boundary $\partial z \times w + (- 1)^{m} z \times \partial w = 0$ by [F3]. The shuffle equivalence [F5] supplies the inverse $T$ and a homotopy $K$ with $d K + K d = T S - 1$ on the tensor complex, so the pairing can be computed on these representatives.

2.1F2F5step 1.1

Evaluation gives $(J T) (S (z \otimes w)) = J (T S (z \otimes w))$ . Since $z \otimes w$ is a cycle, $T S (z \otimes w) - z \otimes w = d K (z \otimes w)$ . The cocycle $J = J (φ, ψ)$ annihilates this boundary by [F2], so the value is $J (z \otimes w)$ . This equals $φ (z) ψ (w)$ when $(m, n) = (p, q)$ , and is zero for every other bidegree with $m + n = p + q$ , by the defining bidegree support of $J$ .

3.1F1F2F6step 2.1∎

For $(m, n) = (p, q)$ , [F1] identifies these values with the two Kronecker pairings, proving the identity. Representative independence follows from [F1, F2]. The front/back formula in [F6] identifies $J A$ with ${pr}_{X}^{*} φ ⌣ {pr}_{Y}^{*} ψ$ : only the $(p, q)$ cut survives. Since $J d = 0$ , its difference from $J T$ is $J (A - T) = J H d = δ (J H)$ , so these cochains represent the same class. This comparison uses no additive Kunneth bijectivity. Extension of integral cycles to $R$ commutes with shuffle and evaluation, proving the integral-input version. This includes degree zero, empty spaces and the zero ring.

Depends on

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Sources

Allen Hatcher, Algebraic Topology (standard reference, not scraped)
Allen Hatcher, Algebraic Topology (standard reference, not scraped)