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ExampleConstruction: Literature-sourcedVerification: AI-generatedPipeline-generatedverified 2026-10-22 (gpt-6.1-sol)
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The Borromean string link has nonzero triple Milnor invariant

Example

Assume AC. Let τB be the based 3-string link represented by the pure braid word β=(σ1−1σ2)3=σ1−1σ2σ1−1σ2σ1−1σ2, using the library's braid-group product, in which the right factor is stacked below the left factor. Thus the six factors from bottom to top are σ2,σ1−1,σ2,σ1−1,σ2,σ1−1. Here σi is the positive geometric half twist of adjacent strands in the crossing convention for this library, and labels 1,2,3 are assigned from left to right at the bottom. Close each top endpoint to the corresponding bottom endpoint outside the braid cylinder by crossing-free closure arcs; the matching closure is a Borromean link. Use the bottom meridians and the diagram-induced blackboard product-tube parallels for this based braid. The third string-link longitude and the matching based longitude of its closed link have truncated expansion E(λ3)=1+X1X2−X2X1(modm3), so μτB(1,2,3)=1. Every pairwise linking number of the closed braid is zero, and its based closed-link residue candidate is μˉτ^Bb(1,2,3)=1.

\begin{tikzpicture}[scale=0.85,>=stealth]
  % Six adjacent half-twists, read from bottom to top in the stated product convention.
  % sigma_2
  \draw[gray] (0,0) -- (0,1);
  \draw[thick,dashed,->] (1,0) .. controls (1.25,0.25) and (1.75,0.75) .. (2,1);
  \draw[thick,->] (2,0) .. controls (1.75,0.25) and (1.25,0.75) .. (1,1);
  \node at (0.55,0.5) {$\sigma_2$};
  % sigma_1^{-1}
  \draw[thick,dashed,->] (1,1) .. controls (0.75,1.25) and (0.25,1.75) .. (0,2);
  \draw[thick,->] (0,1) .. controls (0.25,1.25) and (0.75,1.75) .. (1,2);
  \draw[gray] (2,1) -- (2,2);
  \node at (1.45,1.5) {$\sigma_1^{-1}$};
  % sigma_2
  \draw[gray] (0,2) -- (0,3);
  \draw[thick,dashed,->] (1,2) .. controls (1.25,2.25) and (1.75,2.75) .. (2,3);
  \draw[thick,->] (2,2) .. controls (1.75,2.25) and (1.25,2.75) .. (1,3);
  \node at (0.55,2.5) {$\sigma_2$};
  % sigma_1^{-1}
  \draw[thick,dashed,->] (1,3) .. controls (0.75,3.25) and (0.25,3.75) .. (0,4);
  \draw[thick,->] (0,3) .. controls (0.25,3.25) and (0.75,3.75) .. (1,4);
  \draw[gray] (2,3) -- (2,4);
  \node at (1.45,3.5) {$\sigma_1^{-1}$};
  % sigma_2
  \draw[gray] (0,4) -- (0,5);
  \draw[thick,dashed,->] (1,4) .. controls (1.25,4.25) and (1.75,4.75) .. (2,5);
  \draw[thick,->] (2,4) .. controls (1.75,4.25) and (1.25,4.75) .. (1,5);
  \node at (0.55,4.5) {$\sigma_2$};
  % sigma_1^{-1}
  \draw[thick,dashed,->] (1,5) .. controls (0.75,5.25) and (0.25,5.75) .. (0,6);
  \draw[thick,->] (0,5) .. controls (0.25,5.25) and (0.75,5.75) .. (1,6);
  \draw[gray] (2,5) -- (2,6);
  \node at (1.45,5.5) {$\sigma_1^{-1}$};
  \node at (0,6.25) {$1$};\node at (1,6.25) {$2$};\node at (2,6.25) {$3$};
  \node at (0,-0.3) {$1$};\node at (1,-0.3) {$2$};\node at (2,-0.3) {$3$};
  \node at (1,-0.75) {bottom-to-top braid; close matching labels outside};
\end{tikzpicture}

The dashed branch is the under-strand and each arrow points from bottom to top. The labels at the top and bottom agree because the braid is pure. The diagram records the word only; the proofs below use the explicit crossing sequence and group words.

Facts & Assumptions

Given: The based string link, the stated braid diagram, and its product-tube basing.

[A1]

Full AC, written AC, supplies the classifying-space and nilpotent-quotient branch. The Axiom of Choice

[F1]

The positive Artin generator is the positive geometric half twist in the fixed crossing convention, and its inverse reverses the over/under choice. Oriented crossing signs, diagram writhe, and blackboard framing

[F2]

The based string-link longitude is the ordered product of the over-meridians encountered at its undercrossings, with exponent determined by source-to-target direction. Diagram word for a preferred peripheral longitude

[F3]

The finite lower-central quotient identifies the bottom and top peripheral systems, and the Magnus coefficient defines μτB(1,2,3). String-link end inclusions identify finite nilpotent peripheral quotients, Integer Milnor coefficients and closed-link residues

[F4]

The diagram linking-number scan returns one half the signed mixed-crossing sum. Compute oriented linking number from a diagram, The diagram linking-number scan terminates and returns the stated integer

[F5]

External input: The standard closure of the braid word (σ1−1σ2)3 is the Borromean rings. Wolfram MathWorld lists this exact word as a Borromean braid; this input identifies the closure, while the pairwise-linking and Milnor-coefficient calculations are proved below. Wolfram MathWorld, Braid Word

Verification

Given: The braid word and the input conventions in the Example.

1.1F1F4givenconstruct

The rightmost factor is stacked lowest, so the factors from bottom to top are σ2,σ1−1,σ2,σ1−1,σ2,σ1−1. Starting from the bottom positions, the component-position sequence after successive crossings is (1,2,3), (1,3,2), (3,1,2), (3,2,1), (2,3,1), (2,1,3), (1,2,3). The six crossing-label pairs and signs in that order are respectively (2,3,+1), (1,3,−1), (1,2,+1), (2,3,−1), (1,3,+1), and (1,2,−1), as determined by the signed generators in the braid word and [F1]. Thus the closure returns each label to its own endpoint, and each component pair crosses twice with opposite signs. By [F4] each pairwise linking number is half the signed mixed-crossing sum, hence is zero.

2.1F2givenstep 1.1algebra

At crossing 1, component 2 passes under component 3, so its meridian becomes B=x3−1x2x3. At crossing 2, component 3 passes under component 1 and contributes the longitude factor x1−1. At crossing 3, component 1 passes under component 2, so its meridian becomes C=B−1x1B. At crossing 5, component 3 passes under component 1 and contributes the factor C. The other crossings do not put component 3 under. Hence λ3=x1−1C=x1−1B−1x1B.

3.1A1F3step 2.1algebraconstruct

In R2, E(B)≡1+X2(modm2), so E(C)=E(B)−1(1+X1)E(B)=1+X1+X1X2−X2X1(modm3). Since E(x1−1)=1−X1+X12, multiplication gives E(λ3)=E(x1−1)E(C)=1+X1X2−X2X1(modm3). Under [A1], the X1X2 coefficient is +1, so [F3] gives μτB(1,2,3)=1.

4.1F2F3step 1.1step 3.1∎

The computed longitude has zero degree-one term, as required by the chosen untwisted based peripheral system. Closing the endpoints along crossing-free arcs adds no undercrossing factor, and component 3 has self-writhe zero, so the same word is its based closed-link preferred longitude by [F2]. Every proper length-two subsequence in Δ123b is a pairwise linking number, so step 1.1 makes Δ123b=0. Thus the same coefficient gives the closed-link residue candidate 1.

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