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The boundary middle form is well defined and glues

Statement

Assume the Axiom of Choice as inherited from Poincare-Lefschetz duality. The boundary middle form $Q_{W}$ of Boundary middle form and boundary signature is well defined and symmetric on its image $I$ and is nondegenerate there. If $W, W^{'}$ are compact oriented eight-manifolds with identified oriented boundary $M$ satisfying $H^{3} (M; Z) = H^{4} (M; Z) = 0$ , then $N = W \cup_{M} (- W^{'})$ is closed oriented and $σ (N) = σ (W) - σ (W^{'}) .$

Facts & Assumptions

Given: Compact oriented eight-manifolds $W, W^{'}$ with common oriented boundary $M = \partial W = \partial W^{'}$ satisfying the integral vanishing, and the classes $A = H^{4} (W, M; R)$ , $B = H^{4} (W; R)$ , $j : A \to B$ , $I = im j$ .

[A1]

The Axiom of Choice is assumed (The Axiom of Choice).

[L1]

Poincare-Lefschetz duality identifies $A$ with $H_{4} (W; R)$ , and evaluation identifies $B$ with its full linear dual (Cohomology over a field is dual to homology over that field). Since these spaces are finite-dimensional by [L3], this gives a perfect pairing $A \times B \to R$ by $T (a, x) = ⟨ a ⌣ x, [W, M] ⟩$ , and the relative cap/evaluation identity identifies it with the cap pairing determined by $[W, M]$ (Poincaré–Lefschetz duality, Relative cap and cup evaluation identity).

[L2]

The relative middle-degree cup product is symmetric, because both factors have degree four: $a ⌣ b = b ⌣ a$ (Relative degree-four cup products are symmetric).

[L3]

$A$ and $B$ are finite-dimensional over $R$ (Compact oriented manifolds with boundary have finite-dimensional field cohomology), and integral vanishing of $H^{3}, H^{4}$ of $M$ implies the corresponding real vanishing (Vanishing integral middle cohomology of a closed oriented seven-manifold implies real vanishing).

[L4]

The collared gluing of $W$ and $W^{'}$ along $M$ gives the closed oriented $N = W \cup_{M} (- W^{'})$ , with excision and evaluation comparisons and the restrictions $[W, M]$ and $- [W^{'}, M]$ of $[N]$ (Collared gluing has relative excision and evaluation maps).

[L5]

The closed middle intersection form and closed signature of a closed oriented four- $k$ -manifold are defined by Poincare duality, and Sylvester's law of inertia makes the signature additive on orthogonal direct sums with a sign for negated summands (The middle-dimensional intersection form of a closed oriented 4k-manifold, The signature of a closed oriented manifold of dimension divisible by four, Sylvester's law of inertia: every real symmetric form is congruent to $diag (I_{p}, - I_{q}, 0_{r})$ , and $(p, q, r)$ is unique).

[L6]

Mayer-Vietoris computes $H^{*} (N; R)$ from the collar cover $U, V$ (Mayer vietoris sequence in singular cohomology).

Proof

technique · direct

1.1L1L2L3A1

By [L1] and [L3] the pairing $T$ on the finite-dimensional spaces $A \times B$ is perfect; by [L2] it satisfies $T (a, j b) = T (b, j a)$ for all $a, b \in A$ , since $j$ is restriction of the second factor to $I$ .

2.1step 1.1

Well-definedness of $Q_{W}$ : if $j (a - a^{'}) = 0$ , then for every $b \in A$ , $T (a - a^{'}, j b) = T (b, j (a - a^{'})) = 0$ by step 1.1 and additivity; since $j b$ ranges over $I$ , the functional $T (a - a^{'}, \cdot)$ vanishes on $I$ , so $Q_{W} (x, y) = T (a, y)$ is independent of the lift $a$ of $x$ ; symmetry of $Q_{W}$ follows from the same identity with $x = j a$ , $y = j b$ .

3.1step 2.1L1

Nondegeneracy: suppose $x = j (a) \in I$ satisfies $Q_{W} (x, y) = 0$ for all $y \in I$ ; then $T (a, j b) = 0$ for every $b \in A$ , so $T (b, j a) = T (a, j b) = 0$ for every $b$ , and perfectness of $T$ in the $B$ variable forces $x = j a = 0$ ; hence $I$ has zero radical and $Q_{W}$ is already nondegenerate on $I$ , so the quotient in the definition is the identity.

4.1step 3.1L3

By [L3] the integral vanishing on $M$ implies the real vanishing, so the maps $j_{W} : H^{4} (W, M; R) \to H^{4} (W; R)$ and $j_{W^{'}}$ are isomorphisms; hence $I = B$ on both sides.

5.1step 4.1L4L6

By [L6] and [L4] the restriction $H^{4} (N; R) \to H^{4} (W; R) \oplus H^{4} (W^{'}; R)$ is an isomorphism, and the closed middle form of $N$ restricts to the block form $Q_{W} \oplus (- Q_{W^{'}})$ : same-side products evaluate as the two boundary forms by the evaluation comparison of [L4], while the cross products vanish because the corresponding relative product lies in $H^{8} (N, N) = 0$ .

6.1step 5.1L5

By [L5] and step 5.1 the signature of the closed form on $N$ is the signature of $Q_{W} \oplus (- Q_{W^{'}})$ , which by Sylvester inertia is $σ (W) - σ (W^{'})$ ; hence $σ (N) = σ (W) - σ (W^{'})$ .

7.1step 3.1step 6.1L4∎

Therefore the boundary middle form is well defined, symmetric and nondegenerate on $I$ , and the glued closed manifold has signature the difference of the two boundary signatures, as asserted.

Depends on

Used by

The Milnor lambda invariant is well defined modulo seven Theorem

Cited to discharge well-definedness by Boundary middle form and boundary signature.

Dependency tree · two levels

73 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

John Milnor, On Manifolds Homeomorphic to the 7-Sphere, Annals of Mathematics 64 (1956), 399-405 (standard reference, not scraped)
Allen Hatcher, Algebraic Topology, Cambridge University Press 2002 (complete book) (standard reference, not scraped)