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The restriction image on a cobordism boundary is Lagrangian

Statement

Assume AC. Let $W$ be a compact oriented smooth manifold of dimension $4 k + 1$ , $k \geq 0$ , with boundary $M = \partial W$ carrying the induced orientation and inclusion $i : M ↪ W$ . Let $K = im (i^{*} : H^{2 k} (W; R) \to H^{2 k} (M; R))$ and let $Q_{M}$ be the middle-dimensional intersection form of The middle-dimensional intersection form of a closed oriented 4k-manifold. Then $K = K^{⊥_{Q_{M}}}$ . In particular $K$ is a totally isotropic subspace of dimension one half of $\dim H^{2 k} (M; R)$ , hence is Lagrangian.

Facts & Assumptions

Given: AC; a compact oriented smooth $(4 k + 1)$ -manifold $W$ with boundary $M = \partial W$ , inclusion $i$ , the induced boundary orientation, and the image $K = im i^{*}$ in $V : = H^{2 k} (M; R)$ .

[F1]

$Q_{M} (x, y) = ⟨ x ⌣ y, [M] ⟩$ for $x, y \in H^{2 k} (M; R)$ , and $K^{⊥} = {x \in V : Q_{M} (k, x) = 0 for all k \in K}$ (The middle-dimensional intersection form of a closed oriented 4k-manifold).

[F2]

The long exact cohomology sequence of the pair is $\dots \to H^{2 k} (W, M; R) \overset{j}{\to} H^{2 k} (W; R) \overset{i^{*}}{\to} H^{2 k} (M; R) \overset{δ}{\to} H^{2 k + 1} (W, M; R) \to \dots$ , exact at every term; the connector sends a cocycle class $[x]$ to $[δ \tilde{x}]$ for any cochain extension $\tilde{x}$ of a representative (Long exact sequence of a pair in singular cohomology).

[F3]

The cup product on cochains satisfies the Leibniz identity $δ (φ ⌣ ψ) = δ φ ⌣ ψ + (- 1)^{∣ φ ∣} φ ⌣ δ ψ$ and restricts naturally to subspaces (Cup product Leibniz identity, Singular cup product on cochains).

[F4]

The singular coboundary is defined on chains by $(δ φ) (σ) = φ (\partial σ)$ , so $(δ φ) (z) = φ (\partial z)$ for every finite chain $z$ ; the Kronecker pairings on $W$ and on $M$ evaluate a cocycle class on a cycle class by evaluating representatives and are well defined and independent of the chosen representatives (Singular cohomology with coefficients, Kronecker evaluation pairing, The kronecker pairing is independent of cocycle and cycle representatives).

[F5]

The relative fundamental class satisfies $\partial [W, M] = [M] \in H_{4 k} (M; R)$ (Relative fundamental class and boundary orientation).

[F6]

Relative cap and evaluation: for a relative $p$ -cocycle $α \in Z^{p} (W, M; R)$ , a relative $n$ -cycle $z$ with $\partial z \in C_{n - 1} (M)$ , and an absolute $(n - p)$ -cochain $ξ$ , the front-evaluation/back-face formulas of the cohomology-first convention give the cochain identity $(α ⌣ ξ) (z) = ξ (α \cap z)$ (Relative cap products with quotient domains displayed, Singular cup product on cochains). By the boundary identity $\partial (α \cap z) = (- 1)^{p} (α \cap \partial z - δ α \cap z)$ (Cap product boundary identity) the chain $α \cap z$ is a cycle: the first term vanishes because $α$ vanishes on chains in $M$ , the second because $δ α = 0$ as a relative cocycle. Its class is the relative cap product $α \cap [W, M]$ , which is the Poincaré–Lefschetz map $T_{p} (α)$ of Poincaré–Lefschetz duality.

[F7]

Poincare-Lefschetz duality gives isomorphisms $T_{p} : H^{p} (W, M; R) \to H_{n - p} (W; R)$ , $a \mapsto a \cap [W, M]$ , for every $p$ , in particular an isomorphism $T_{2 k + 1}$ out of $H^{2 k + 1} (W, M; R)$ ; their representative independence and naturality are as stated there (Poincaré–Lefschetz duality).

[F8]

$Q_{M}$ is symmetric, nondegenerate and finite-dimensional, with both adjoints $x \mapsto Q_{M} (x, -)$ , $y \mapsto Q_{M} (-, y)$ isomorphisms onto the full dual; the Kronecker pairing over the field $R$ satisfies $H^{2 k} (W; R) ≅ {Hom}_{R} (H_{2 k} (W; R), R)$ , so a class $z \in H_{2 k} (W; R)$ with $⟨ a, z ⟩ = 0$ for all $a$ is zero (The middle-dimensional intersection form is symmetric and nondegenerate, Cohomology over a field is dual to homology over that field, The kronecker pairing is independent of cocycle and cycle representatives).

[F9]

For a finite-dimensional vector space and a subspace $U$ , the annihilator $U^{\circ} = {f : f ∣_{U} = 0}$ has $\dim U^{\circ} = \dim V - \dim U$ ; a projection onto a finite-dimensional subspace exists without choice, and rank-nullity computes the dimension of a kernel (Assuming choice, $^{\circ} (U^{\circ}) = U$ ; in finite dimension, $\dim U^{\circ} = \dim V - \dim U$ , The annihilator $U^{\circ} \leq V^{*}$ of $U \leq V$ and the preannihilator $^{\circ} S \leq V$ of $S \leq V^{*}$ , Finite-dimensional subspaces admit projections without Choice, Rank-nullity: $\dim_{F} V = nullity T + rank T$ ).

[F10]

Cup product is graded commutative, so $u ⌣ v = v ⌣ u$ whenever the degrees are even (Singular cohomology is graded commutative).

Proof

technique · direct; identify the orthogonal complement with the kernel of the cohomology connector by a cochain evaluation computation, then count dimensions

1.1givenF1F2

Exactness at $H^{2 k} (M; R)$ in [F2] reads $im i^{*} = \ker δ$ , so $K = \ker δ$ ; in particular $K \leq V : = H^{2 k} (M; R)$ is a linear subspace and $x \in K$ holds exactly when $δ x = 0$ .

1.2givenF1F2F3F4F5F6F10

Pairing identity: for $x \in H^{2 k} (M; R)$ and $a \in H^{2 k} (W; R)$ , $Q_{M} (i^{*} a, x) = ⟨ a, T_{2 k + 1} (δ x) ⟩$ . Indeed choose cocycle representatives $X \in Z^{2 k} (M; R)$ of $x$ and $A \in Z^{2 k} (W; R)$ of $a$ , an extension $\tilde{X} \in C^{2 k} (W; R)$ of $X$ , and a relative cycle $z$ with $[z] = [W, M]$ and $[\partial z] = [M]$ , possible by [F5]. Since $X$ is a cocycle, $δ \tilde{X}$ vanishes on chains in $M$ , and $δ x$ is the class of the relative cocycle $δ \tilde{X}$ by [F2]; hence by [F6] the chain $δ \tilde{X} \cap z$ is a cycle representing $T_{2 k + 1} (δ x) = δ x \cap [W, M]$ . Then $⟨ a, T_{2 k + 1} (δ x) ⟩ = A (δ \tilde{X} \cap z) = (δ \tilde{X} ⌣ A) (z) = (δ (\tilde{X} ⌣ A)) (z) = (\tilde{X} ⌣ A) (\partial z) = (X ⌣ i^{*} A) (\partial z)$ , where the equalities use, in order, the absolute Kronecker pairing on $W$ [F4], the cochain identity of [F6], the Leibniz rule [F3] with $δ A = 0$ , the definition of the coboundary [F4], and the fact that $\tilde{X}$ restricts to $X$ on $M$ while $A$ restricts to $i^{*} A$ ; since $\partial z$ represents $[M]$ and $X ⌣ i^{*} A$ is a cocycle representing $x ⌣ i^{*} a$ on $M$ , the last value is $⟨ x ⌣ i^{*} a, [M] ⟩$ by [F4]. Graded commutativity [F10] (degrees $2 k, 2 k$ ) and [F1] give $⟨ x ⌣ i^{*} a, [M] ⟩ = ⟨ i^{*} a ⌣ x, [M] ⟩ = Q_{M} (i^{*} a, x)$ , as asserted.

1.3givenF7F8

Tools for the two inclusions: by [F8], $V$ is finite-dimensional and the adjoint map $Φ : V \to V^{*}$ , $Φ (x) = Q_{M} (x, -)$ , is an isomorphism; by [F8], a homology class $z \in H_{2 k} (W; R)$ that pairs to zero with every cohomology class is zero; and by [F7] the map $T_{2 k + 1}$ is injective.

2.1step 1.1step 1.2step 1.3F8

Orthogonal complement equals the kernel: for $x \in V$ , $x \in K^{⊥}$ holds exactly when $Q_{M} (i^{*} a, x) = 0$ for all $a \in H^{2 k} (W; R)$ ; by step 1.2 this is equivalent to $⟨ a, T_{2 k + 1} (δ x) ⟩ = 0$ for all such $a$ , hence by step 1.3 to $T_{2 k + 1} (δ x) = 0$ , hence to $δ x = 0$ by the injectivity in step 1.3, and hence to $x \in K$ by step 1.1. Therefore $K^{⊥} = K$ .

3.1step 2.1F1F9

Dimension and isotropy: by step 2.1, $Φ (K^{⊥}) = K^{\circ}$ is the annihilator of $K$ in $V^{*}$ , so $\dim K = \dim K^{⊥} = \dim K^{\circ} = \dim V - \dim K$ by the annihilator dimension formula [F9]; hence $2 \dim K = \dim V = \dim H^{2 k} (M; R)$ . Since $K = K^{⊥}$ , $Q_{M} (x, y) = 0$ for all $x, y \in K$ , so $K$ is totally isotropic; a totally isotropic subspace of half the dimension of a nondegenerate finite-dimensional form is Lagrangian.

4.1step 2.1step 3.1given∎

If $M = \emptyset$ (in particular if $W$ is empty), $V = H^{2 k} (M; R) = 0$ , so both $K$ and $K^{⊥}$ are zero, so the identity and the dimension count hold trivially; for $k = 0$ the formula counts the oriented boundary points and gives $K = K^{⊥}$ of dimension $\frac{1}{2} \dim H^{0} (M; R)$ . Thus steps 2.1 and 3.1 prove $K = K^{⊥_{Q_{M}}}$ and the Lagrangian property in all cases, AC being used only through the inherited duality and field-dual suppliers.

Depends on

Used by

The signature of an oriented boundary vanishes, so the signature is an oriented cobordism invariant Theorem

Dependency tree · two levels

69 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

Daniel S. Freed, Bordism: Old and New (lecture notes, UT Austin, Fall 2012) (standard reference, not scraped)
John Milnor and James Stasheff, Characteristic Classes (re-typeset scan; original pagination) (standard reference, not scraped)