LemmaStatement: Literature-sourcedProof: AI-generatedPipeline-generatedverified 2026-10-22 (gpt-6.1-sol)

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Local meridian conjugacy at a labeled link crossing

Statement

Let $D$ be a finite oriented diagram of a nonempty labeled closed link or string link, with a chosen smooth realization and pairwise-disjoint product tubes. Choose a crossing whose two projected branches meet transversely and for which no other branch meets a sufficiently small neighborhood, and let $o$ be the over-strand. Orient the projection plane by the rule that $(v_{1}, v_{2})$ is positive exactly when $(n, v_{1}, v_{2})$ is positive in the ambient three-space, where $n$ is the viewing normal. Let $n_{o}$ be the positive in-plane normal characterized by $(v_{o}, n_{o})$ being positive. The source side $s$ is the side from which $n_{o}$ points, and the target side $t$ is the side toward which it points.

Choose an orientation-preserving local crossing chart, whose existence follows from the transversality of the projected branches, in which the over-branch is $(x, 0, 0)$ oriented by $+ \partial_{x}$ , the under-branch is $(0, y, - h)$ for some $h > 0$ , the viewing normal is $+ \partial_{z}$ , and $(\partial_{x}, \partial_{y})$ is the positive projection orientation. Use or radially shrink the chosen product tube near the under-branch to a radius $r$ with $0 < r < h$ . Choose $d, H > 0$ small enough that the chart contains the points below. The source and target points on the upper boundary of the under-tube are $(0, - d, - h + r)$ and $(0, d, - h + r)$ ; join them by $β (y) = (0, y, - h + r)$ for $- d \leq y \leq d$ . Choose a common basepoint in the chosen exterior above the chart and a path $δ$ in the chosen exterior from it to $(0, 0, H)$ . The two sides of the triangle in $x = 0$ from $(0, 0, H)$ to the endpoints of $β$ , together with $δ$ , give the compatible common-upper-collar access paths to the source and target meridian circles; denote them by $a_{s}, a_{t}$ . Let $a_{o}$ be $δ$ followed by an access arc in the punctured triangle to a small positive meridian circle around the over-strand. The oriented outer boundary of the punctured triangle represents this same based meridian. Base the positive over-meridian by the oriented boundary of this triangle after deleting a small disk about its intersection with the over-strand. Write the resulting based positive meridians as $m_{o}, m_{s}, m_{t}$ . The chart, tube, and compatible paths are part of the data. All displayed based loops are in the link or string-link complement group, identified with the group of the chosen compact exterior by radial retraction in the product tubes.

For these compatible accesses, $m_{t} = m_{o}^{- 1} m_{s} m_{o} .$ Equivalently, the crossing relator is $m_{o}^{- 1} m_{s} m_{o} m_{t}^{- 1} = 1$ . Reversing the direction of the under-strand does not change this side-based equation.

For arbitrary access paths $a_{e}^{'}$ to the same chosen meridian circles, let $a_{e}$ be the compatible access, $c_{e} = a_{e}^{'} a_{e}^{- 1}$ , and $m_{e}^{'} = c_{e} m_{e} c_{e}^{- 1}$ , for $e = o, s, t$ . The exact relation for the independently based meridians is $m_{t}^{'} = c_{t} c_{o}^{- 1} (m_{o}^{'})^{- 1} c_{o} c_{s}^{- 1} m_{s}^{'} c_{s} c_{o}^{- 1} m_{o}^{'} c_{o} c_{t}^{- 1} .$ A common change of base path conjugates all displayed meridians simultaneously.

Statement references

Friedl, §5.3, Claim 1a

Charbonneau, §§2–3

Facts & Assumptions

Given: A finite regular oriented diagram and a smooth realization; the chosen pairwise-disjoint product tubes; an oriented crossing; its positive normal-disk meridians; and the common basepoint and compatible triangle accesses from the Statement.

[F1]

At a regular crossing, the two projected branches are smooth and transverse; their oriented tangents determine local planar coordinates, and the chosen product tubes provide normal-disk meridians. Regular oriented link diagrams, Based string links and labeled link peripheral systems

[F2]

The positive meridian is the oriented boundary of a normal disk whose orientation is fixed by the link tangent followed by the disk orientation agreeing with the ambient orientation. Based string links and labeled link peripheral systems

[F3]

The ambient orientation and viewing normal determine the positive orientation of the crossing chart and hence the orientation of the normal $(y, z)$ plane to the over-strand. Oriented links in the three-sphere and ambient isotopy, Regular oriented link diagrams

[F4]

A $C^{1}$ map between open subsets of $R^{2}$ whose derivative is invertible at a point restricts near that point to a $C^{1}$ diffeomorphism. The Euclidean inverse function theorem

Proof

1.1F1F3F4givenconstruct

Let the two projected branches be parametrized near the crossing by $γ_{o} (u)$ and $γ_{s} (v)$ , with linearly independent derivatives at their common image $p$ . The map $(u, v) \mapsto γ_{o} (u) + γ_{s} (v) - p$ has invertible derivative there, so the inverse function theorem gives local planar coordinates $x = u$ , $y = v$ . Choose the sign of $y$ so $(\partial_{x}, \partial_{y})$ agrees with the oriented projection plane, and translate the crossing to $(0, 0)$ . Write the two height functions as $z = h_{o} (x)$ and $z = h_{s} (y)$ , with $h_{o} (0) = 0$ and $h_{s} (0) = - h < 0$ . The coordinate change $z^{'} = z - h_{o} (x) - h_{s} (y) + h_{s} (0)$ has Jacobian determinant $1$ ; it carries the over-branch to $(x, 0, 0)$ and the under-branch to $(0, y, - h)$ , while preserving the ambient orientation. Shrink the chart so it excludes every other branch, as permitted by the isolated-crossing hypothesis in the Statement. The normal-disk product chart for the under-branch supplies the chosen small radius $r < h$ . Since $(\partial_{x}, \partial_{y})$ is positively oriented, $n_{o} = + \partial_{y}$ , so $y < 0$ is the source and $y > 0$ is the target.

2.1F2F3step 1.1givenconstruct

The triangle boundary is the loop from its apex to the source endpoint, along $β$ to the target endpoint, and back to the apex. Its interior lies above the under-tube boundary and meets the over-branch exactly at $(0, 0, 0)$ . With the boundary orientation induced by $(\partial_{y}, \partial_{z})$ , its transverse intersection sign is positive because $(\partial_{x}, \partial_{y}, \partial_{z})$ is a positive ambient basis. Removing a small disk about that intersection leaves an annulus in the crossing complement from the triangle boundary to a positive normal meridian of the over-strand. Conjugating by the common path $δ$ therefore identifies $a_{s} β a_{t}^{- 1}$ with the stated $m_{o}$ .

3.1F2step 1.1step 2.1givenconstructalgebra

The product tube around the under-branch carries its positive meridian circles along the branch by an annulus. The top arc $β$ is the path on that tube boundary joining the chosen source and target circle basepoints. If $μ_{s}, μ_{t}$ denote the meridian loops based at those points, the annulus gives $μ_{s} = β μ_{t} β^{- 1}$ . With the access paths $a_{s}, a_{t}$ of the Statement, it follows that $m_{s} = (a_{s} β a_{t}^{- 1}) m_{t} (a_{s} β a_{t}^{- 1})^{- 1} .$ The punctured-triangle argument in step 2.1 identifies $a_{s} β a_{t}^{- 1}$ with $m_{o}$ . Solving gives the asserted conjugation relation.

4.1step 3.1algebra

Multiplying the conjugation equality by $m_{t}^{- 1}$ gives the displayed crossing relator. Reversing the under-strand orientation reverses both under-meridian loops, so inverting the equality leaves the same side-based equation.

4.2givenstep 3.1algebra

For arbitrary access paths, $a_{e}^{'} = c_{e} a_{e}$ gives $m_{e}^{'} = c_{e} m_{e} c_{e}^{- 1}$ . Substitute $m_{o} = c_{o}^{- 1} m_{o}^{'} c_{o}$ and $m_{s} = c_{s}^{- 1} m_{s}^{'} c_{s}$ into $m_{t} = m_{o}^{- 1} m_{s} m_{o}$ , then conjugate by $c_{t}$ . This gives exactly the correction formula in the Statement; simultaneous base-path change conjugates all meridians by the same loop.

5.1givenF2construct∎

On a chosen product tube with normal coordinate $v \neq 0$ , the homotopy $v ⟼ ((1 - s) ∣ v ∣ + s) \frac{v}{∣ v ∣}, 0 \leq s \leq 1,$ pushes the punctured tube radially to its boundary and fixes that boundary. Applying it on each disjoint tube and fixing the exterior identifies the based fundamental group of the open complement with that of the compact exterior. Thus the relation proved in the open complement is the same relation in the chosen exterior group.

Remarks

The source/target labels determine which conjugation formula is used. Incoming/outgoing labels depend on the direction of the under-strand and should be derived from this equation, not substituted for it.

\begin{tikzpicture}[>=stealth,line width=0.9pt]
  \draw[->,line width=1.2pt] (-1,-1)--(1,1);
  \draw[->] (1,-1)--(0.15,-0.15);
  \draw[->] (-0.15,0.15)--(-1,1);
  \node at (0.8,-0.7) {$s$};
  \node at (-0.8,0.8) {$t$};
  \node at (-0.8,-0.8) {$o$};
\end{tikzpicture}

The $x = 0$ cross-section of the local chart shows the triangle access used in the proof. The symbol $⊙$ marks the over-strand in its positive $x$ direction.

\begin{tikzpicture}[scale=1.0,>=stealth]
  \coordinate (apex) at (0,1.8);
  \coordinate (src) at (-1.35,-0.75);
  \coordinate (tgt) at (1.35,-0.75);
  \draw[->] (-1.9,-1.65) -- (1.9,-1.65) node[right] {$y$};
  \draw[->] (0,-1.8) -- (0,2.1) node[above] {$z$};
  \draw[gray,thick] (-1.7,-1.0) -- (1.7,-1.0) node[right] {$z=-h$};
  \draw[gray,dashed] (-1.7,-0.75) -- (1.7,-0.75) node[right] {$z=-h+r$};
  \draw[->,thick] (apex) -- (src) node[midway,left] {$a_s$};
  \draw[->,thick] (src) -- (tgt) node[midway,below] {$\beta$};
  \draw[->,thick] (tgt) -- (apex) node[midway,right] {$a_t^{-1}$};
  \fill (0,0) circle (2pt) node[right] {$\odot\,o$};
  \fill (apex) circle (1.5pt) node[above] {apex};
  \fill (src) circle (1.5pt) node[below left] {$s$};
  \fill (tgt) circle (1.5pt) node[below right] {$t$};
\end{tikzpicture}

Depends on

Used by

Dependency tree · two levels

24 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

Stefan Friedl, Knot Theory, §5.3, proof of the Wirtinger Presentation Proposition (standard reference, not scraped)
Cole Charbonneau, Computing the Wirtinger Presentation of a Knot Group from a Knot Diagram (standard reference, not scraped)