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Inverse-matrix formula for a surgery linking form

Statement

Assume the Axiom of Choice. Let $J = J_{1} \cup \dots \cup J_{n}$ be an oriented link of $n \geq 1$ smooth components in $S^{3}$ . Choose pairwise disjoint closed tubular neighborhoods $N (J_{i})$ , let $X = S^{3} ∖ int (⋃_{i} N (J_{i}))$ , and let $μ_{i}$ be the positive meridian of $J_{i}$ . Let $λ_{i}$ be the longitude parallel to $J_{i}$ with zero self-linking. Alexander duality identifies $H_{1} (X; Z)$ with $Z^{n}$ on the meridian basis $[μ_{1}], \dots, [μ_{n}]$ ; for $i \neq j$ , define ${lk}_{S^{3}} (J_{i}, J_{j})$ to be the $j$ -th coordinate of $[λ_{i}]$ , the usual oriented linking number. Use the boundary-torus intersection convention $μ_{i} \cdot λ_{i} = + 1$ . Give $J_{i}$ an integer framing $f_{i}$ and fill along $δ_{i} = λ_{i} + f_{i} μ_{i}$ . Let $A$ be the symmetric integer matrix with $A_{i i} = f_{i}$ and $A_{i j} = {lk}_{S^{3}} (J_{i}, J_{j})$ for $i \neq j$ . Assume $\det A \neq 0$ , and let $Y$ be the resulting oriented surgery manifold, oriented as the boundary of the 4-dimensional surgery handlebody with outward normal first. Then $H_{1} (Y; Z) ≅ Z^{n} / A Z^{n}$ . For torsion classes $[C], [D] \in H_{1} (Y; Z)$ represented by oriented 1-cycles, choose $m > 0$ and an integral 2-chain $F$ with $\partial F = m C$ and define $λ_{Y} ([C], [D]) = (F \cdot D) / m (m o d Z)$ , where the intersection is signed by the orientation of $Y$ . If $x, y \in Z^{n}$ are meridian-coordinate representatives of $[C], [D]$ , then

λ_{Y} ([x], [y]) = - x^{T} A^{- 1} y (m o d Z) .

Statement references

Finite disjoint unions of smooth oriented circles in $S^{3}$

The link exterior and its connected complement

Alexander-duality meridian coordinates

Positive meridian normalization

Homology exactness for the surgery filling

Przytycki–Yasuhara, Theorem 1.1 and Corollary 1.2

Facts & Assumptions

Given: Full AC, the framed link, its surgery slopes, the symmetric linking matrix $A$ , and $\det A \neq 0$ .

[A1]

Full AC is the choice-function axiom required by Alexander duality. The Axiom of Choice

[F1]

A finite oriented link in $S^{3}$ is compact and locally contractible; Alexander duality identifies the first homology of its complement with $Z^{n}$ , with the positive meridians as coordinate generators. Oriented links in the three-sphere and ambient isotopy, Alexander duality for compact locally contractible subsets of a sphere

[F2]

The open complement of a nonempty smooth link is connected. The complement of an oriented link is a connected smooth three-manifold

[F3]

The positive meridians use the total-linking normalization: each positive component meridian maps to $1$ under the sum of the Alexander-duality component coordinates. The one-variable Alexander module of an oriented link

[F4]

Mayer–Vietoris is exact for the collar-thickened cover by the exterior and the attached solid tori. Mayer–Vietoris sequence in singular homology

Proof

Given: The data and boundary-torus intersection convention in the Statement. Write $X = S^{3} ∖ int N (J)$ for the compact exterior and use $[μ_{1}], \dots, [μ_{n}]$ as its meridian-coordinate basis.

1.1A1F1F2givenconstruct

In each chosen tubular neighborhood, normalize the normal-disk radius to one and set $H_{t} (s, v) = (s, ((1 - t) + t / ∣ v ∣) v)$ for $0 < ∣ v ∣ \leq 1$ and $0 \leq t \leq 1$ . Its radius is $(1 - t) ∣ v ∣ + t$ , so this homotopy stays off $J$ , fixes the disk boundary, and glues to the identity on $X$ . It is a strong deformation retraction from $S^{3} ∖ J$ onto $X$ ; in particular $X$ is connected by [F2]. By [F1], Alexander duality identifies $H_{1} (S^{3} ∖ J; Z)$ with ${\tilde{H}}^{1} (J; Z) = ⨁_{i} H^{1} (J_{i}; Z) ≅ Z^{n}$ . The locally constructed retraction makes the inclusion of $X$ a homotopy equivalence and hence a homology isomorphism. Thus $H_{1} (X; Z) ≅ Z^{n}$ .

2.1F3step 1.1givenconstruct

The boundary of a small positive normal disk to $J_{i}$ maps to the $i$ -th Alexander-duality coordinate generator: it meets $J_{i}$ once positively and misses every other component. Hence $[μ_{1}], \dots, [μ_{n}]$ are the coordinate basis, with the positive orientation normalized as in [F3]. The $i$ -th coordinate of $[λ_{i}]$ is zero by its zero self-linking convention, and for $j \neq i$ its $j$ -th coordinate is the defined linking number ${lk}_{S^{3}} (J_{i}, J_{j})$ . Therefore $[δ_{i}] = [λ_{i} + f_{i} μ_{i}] = \sum_{j} A_{i j} [μ_{j}]$ .

3.1F4step 1.1step 2.1givenalgebra

In Mayer–Vietoris for the filling, the map $H_{1} (\partial V_{i}) \to H_{1} (V_{i})$ is onto and its kernel is generated by the meridian-disk boundary $δ_{i}$ . The map $H_{0} (⨆_{i} \partial V_{i}) \to H_{0} (X) \oplus H_{0} (⨆_{i} V_{i})$ is injective: in coordinates it sends $(a_{1}, \dots, a_{n})$ to $(\sum_{i} a_{i}, - a_{1}, \dots, - a_{n})$ . Exactness therefore gives $H_{1} (Y; Z) = Z^{n} / A Z^{n}$ .

4.1step 3.1givenalgebra

Since $A$ is invertible over $Q$ , for every $x \in Z^{n}$ some positive integer $m$ clears the denominators of $A^{- 1} x$ , so $m x \in A Z^{n}$ and each class is torsion. If $F, F^{'}$ bound $m C, m^{'} C$ , pass to a common multiple $M$ ; the difference $(M / m) F - (M / m^{'}) F^{'}$ is a 2-cycle $Z$ . If $r [D] = 0$ , write $r D = \partial G$ ; the chain boundary identity gives $r (Z \cdot D) = Z \cdot \partial G = \partial Z \cdot G = 0$ , hence $Z \cdot D = 0$ . Thus the quotient is independent of the chosen bounding chain and multiple modulo integers. Replacing either cycle representative by a homologous one adds an integer by the same boundary identity.

4.2step 2.1step 3.1givenconstructalgebra

Let $B$ be the slope-coordinate matrix $[δ] = B [μ]$ in $H_{1} (X; Q)$ ; step 2.1 gives $B = A$ . In the reverse filling, the slope-coordinate matrix is $B^{- 1}$ . Let ${\tilde{J}}_{i}$ denote the core of the attached solid torus, oriented so the boundary longitude $ℓ_{i}$ mapping to it satisfies $δ_{i} \cdot ℓ_{i} = 1$ . Since $δ_{i} \cdot μ_{i} = - 1$ , the image of $μ_{i}$ in that solid torus is $- [{\tilde{J}}_{i}]$ . Define $H_{i j} = {lk}_{Y} ({\tilde{J}}_{i}, {\tilde{J}}_{j})$ for $i \neq j$ , and let $H_{i i}$ be the reverse-surgery framing determined by the slope $μ_{i}$ . For $i \neq j$ , clear denominators in $[μ_{i}] = \sum_{k} (B^{- 1})_{i k} [δ_{k}]$ and cap each $δ_{k}$ by its meridian disk. The chain in $X$ realizing this homology relation misses every core; with $δ_{k} \cdot ℓ_{k} = + 1$ , the capped disk in $V_{k}$ meets the oriented core ${\tilde{J}}_{j}$ once positively if $k = j$ and not at all otherwise. Hence $(B^{- 1})_{i j} = {lk}_{Y} (μ_{i}, {\tilde{J}}_{j}) = - H_{i j}$ . The reverse diagonal framing is the reverse slope coefficient divided by $δ_{i} \cdot μ_{i} = - 1$ , so $H_{i i} = - (B^{- 1})_{i i}$ . Therefore $H = - B^{- 1} = - A^{- 1}$ .

5.1step 3.1step 4.1step 4.2algebra∎

Each old meridian $μ_{i}$ represents $- {\tilde{J}}_{i}$ in its attached solid torus. Hence for distinct indices their linking equals the core-linking entry $H_{i j}$ ; for a self-pairing, a parallel meridian maps to the parallel core specified by the reverse framing, also giving $H_{i i}$ . Thus $λ_{Y} ([μ_{i}], [μ_{j}]) = - (A^{- 1})_{i j}$ . The chain formula is additive in both cycles, by taking common multiples and summing bounding chains, so its matrix on the meridian basis gives $λ_{Y} ([x], [y]) = - x^{T} A^{- 1} y (m o d Z)$ . If $x$ is replaced by $x + A z$ , the value changes by $- z^{T} y$ ; if $y$ is replaced by $y + A z$ , it changes by $- x^{T} z$ , both integers because $A$ is symmetric and integral.

Remarks

The reverse filling restores $S^{3}$ by surgery on the core link. The diagram records the two surgery descriptions; the slope-basis calculation in steps 4.2–5.1 supplies the inverse-matrix sign.

diagram failed to render:
\begin{tikzcd}
S^3 \arrow[r, "\text{forward surgery}"] & Y \arrow[r, "\text{reverse surgery}"] & S^3
\end{tikzcd}

Depends on

Used by

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Sources

Jozef H. Przytycki and Akira Yasuhara, Linking Numbers in Rational Homology 3-Spheres, Cyclic Branched Covers and Infinite Cyclic Covers (standard reference, not scraped)