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The signature is additive under disjoint union and negates under orientation reversal

Statement

Assume AC. Let $M, N$ be closed oriented smooth $4 k$ -manifolds. Then $σ (M ⊔ N) = σ (M) + σ (N)$ and $σ (- M) = - σ (M)$ , where $- M$ is $M$ with the reversed orientation. More generally $σ$ is additive over disjoint unions with arbitrary orientation signs.

Facts & Assumptions

Given: AC; closed oriented smooth $4 k$ -manifolds $M, N$ with middle forms $Q_{M}, Q_{N}$ and signatures $σ$ .

[F1]

$σ (M) = p - q$ is the inertia difference of the nondegenerate symmetric form $Q_{M}$ on $H^{2 k} (M; R)$ (The signature of a closed oriented manifold of dimension divisible by four).

[F2]

$Q_{M} (x, y) = ⟨ x ⌣ y, [M] ⟩$ (The middle-dimensional intersection form of a closed oriented 4k-manifold).

[F3]

For $M = M_{1} ⊔ \dots ⊔ M_{r}$ the fundamental class is $\sum_{j} (i_{j})_{*} [M_{j}]$ and the middle form is orthogonal direct sum: $Q_{M} (x, y) = \sum_{j} Q_{M_{j}} (i_{j}^{*} x, i_{j}^{*} y)$ under $H^{2 k} (M; R) ≅ ⨁_{j} H^{2 k} (M_{j}; R)$ (The middle-dimensional intersection form is symmetric and nondegenerate, Fundamental class of a compact oriented manifold, The singular homology of a disjoint union is the direct sum).

[F4]

If a nondegenerate symmetric bilinear form is an orthogonal direct sum of forms $B_{1}, B_{2}$ , then the inertia triples add: $p = p_{1} + p_{2}$ , $q = q_{1} + q_{2}$ , $r = r_{1} + r_{2}$ ; this follows because the union of diagonalizing bases diagonalizes the sum, and by Sylvester's law the inertia is intrinsic (Two real symmetric bilinear forms are congruent if and only if they have the same inertia).

[F5]

Reversing the orientation negates the fundamental class, $[- M] = - [M]$ , while the underlying smooth manifold and its tangent data are unchanged; orientability is componentwise (Fundamental class of a compact oriented manifold, Every manifold is F2-orientable and orientability is componentwise).

Proof

technique · direct; the form splits orthogonally over components and changes sign under orientation reversal

1.1givenF1F3F4

Disjoint union: for $M ⊔ N$ the form is the orthogonal direct sum $Q_{M} \oplus Q_{N}$ under $H^{2 k} (M ⊔ N; R) ≅ H^{2 k} (M; R) \oplus H^{2 k} (N; R)$ by [F3], and both summands are nondegenerate. By [F4] the inertia data add, so $p (M ⊔ N) = p (M) + p (N)$ and $q (M ⊔ N) = q (M) + q (N)$ ; hence $σ (M ⊔ N) = p (M) + p (N) - q (M) - q (N) = σ (M) + σ (N)$ by [F1].

1.2givenF1F2F5

Orientation reversal: by [F5], $[- M] = - [M]$ , so by [F2] $Q_{- M} (x, y) = ⟨ x ⌣ y, [- M] ⟩ = - ⟨ x ⌣ y, [M] ⟩ = - Q_{M} (x, y)$ . Multiplication by $- 1$ is an isomorphism of $H^{2 k} (M; R)$ carrying positive-definite subspaces of $Q_{M}$ to negative-definite subspaces of $- Q_{M}$ and conversely, so $p (- M) = q (M)$ and $q (- M) = p (M)$ ; hence $σ (- M) = q (M) - p (M) = - σ (M)$ by [F1].

2.1step 1.1step 1.2given∎

More generally, for a finite disjoint union with signs $⨆_{j} ε_{j} M_{j}$ , where $ε_{j} M_{j}$ means $M_{j}$ with its given orientation when $ε_{j} = + 1$ and the reversed orientation when $ε_{j} = - 1$ , steps 1.1 and 1.2 applied successively give $σ (⨆_{j} ε_{j} M_{j}) = \sum_{j} ε_{j} σ (M_{j})$ ; the empty union has signature $0$ and the zero-dimensional case $k = 0$ is the signed count of components, consistent with both steps.

Depends on

Used by

Dependency tree · two levels

31 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 and James Stasheff, Characteristic Classes (re-typeset scan; original pagination) (standard reference, not scraped)
Daniel S. Freed, Bordism: Old and New (lecture notes, UT Austin, Fall 2012) (standard reference, not scraped)
Tom Weston, An Introduction to Cobordism Theory (lecture notes, Stanford) (standard reference, not scraped)