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Normal-surface unknot recognition examples

1 · Prerequisites

2 · Logical flowchart

3 · Definitions, theorems and proofs

None yet.

4 · Examples, counterexamples and false statements

CounterexampleConstruction: AI-adaptedVerification: AI-generatedjudge pass (gpt-6-luna)verified 2026-10-22 (gpt-6.1-sol)Open item page →

A boundary-parallel disk fails the meridian test

Statement refuted

Every connected admissible normal surface in a finite triangulated knot exterior with nonempty boundary and Euler characteristic one passes the odd-meridian-intersection test for an essential normal disk.

Statement references

Normal-surface coordinates and the Haken cone — normal vertex-link triangles and normal-coordinate conventions.

Finite normal-coordinate tests characterize essential disk candidates — the essential-boundary test in the normal-disk filter.

Hass–Lagarias–Pippenger, Lemma 4.2 and §5 — disk recognition and the meridian-intersection condition.

Erickson, “Normal Surfaces and Knots,” §§14.2–14.4 — boundary-parallel disks and normal-surface filtering.

Facts & Assumptions

Given: A finite simplicial triangulation of a compact knot exterior with torus boundary and a marked meridian curve in its boundary triangulation.

Let $E$ be any finite simplicially triangulated knot exterior with boundary torus $T$ and marked meridian $μ$ . For a boundary vertex $v$ , its normal link is a properly embedded boundary-parallel disk: its boundary is the link of $v$ in $T$ , a contractible simple closed curve. The link is connected, has nonempty boundary and Euler characteristic one, but its boundary has even mod-two intersection with $μ$ , so the normal-surface filter rejects this candidate. Rejecting this one boundary-parallel disk does not determine whether the knot is nontrivial.

[F1]

Normal-surface coordinates and the Haken cone: a normal surface meets each tetrahedron in a union of triangle and quadrilateral disks, and its face arcs match. In a simplicial triangulation of a PL $3$ -manifold, the link of a boundary vertex is a triangulated disk.

[F2]

Finite normal-coordinate tests characterize essential disk candidates: an admissible vector is an essential disk candidate exactly when it is connected, has nonempty boundary, Euler characteristic one, and odd mod-two meridian intersection.

Counterexample

1.1givenF1construct

The link $L k_{E} (v)$ of a boundary vertex in a triangulated $3$ -manifold is a closed disk: its local link is a $2$ -ball. In each tetrahedron containing $v$ the link meets that tetrahedron in the small triangle parallel to the opposite face; it misses tetrahedra not containing $v$ . Across each shared face these triangles have matching normal arcs, so [F1] makes the link a normal surface. Its boundary is $L k_{T} (v)$ , a simple closed curve in the small boundary-star disk of $v$ , hence contractible in $T$ . The local half-ball star shows the disk is boundary-parallel.

2.1step 1.1algebraconstruct

The link disk is connected, has nonempty boundary, and has Euler characteristic one. Its boundary is null-homotopic, so its projection to the product-circle factor has degree zero. After a transverse PL perturbation, the parity of intersections with a fibre meridian equals the parity of preimages of a regular point under that projection; a lift with zero net degree crosses the level an even number of times. Hence the intersection parity is zero.

3.1F2step 2.1algebra∎

The disk therefore passes the first three tests in [F2] but fails the odd-intersection test, so the filter rejects this candidate. This conclusion concerns this boundary-parallel disk only; it does not rule out any other normal essential disk in $E$ and therefore does not classify the knot.

ExampleConstruction: AI-adaptedVerification: AI-generatedverified 2026-10-22 (gpt-6.1-sol)Open item page →

Worked case: a normal disk in the zero-crossing unknot exterior

Example

For the crossingless oriented unknot code, triangulate its solid-torus exterior $D^{2} \times S^{1}$ by triangulating $D^{2}$ as one triangle with vertices $a_{0}, a_{1}, a_{2}$ and $S^{1}$ as the three-edge cycle $b_{0} b_{1} b_{2} b_{0}$ . In the prism over $b_{0} b_{1}$ , write $a_{i}^{0} = (a_{i}, b_{0})$ and $a_{i}^{1} = (a_{i}, b_{1})$ and use the tetrahedra $T_{1} = [a_{0}^{0}, a_{1}^{0}, a_{2}^{0}, a_{2}^{1}], T_{2} = [a_{0}^{0}, a_{1}^{0}, a_{1}^{1}, a_{2}^{1}], T_{3} = [a_{0}^{0}, a_{0}^{1}, a_{1}^{1}, a_{2}^{1}] .$ The properly embedded normal disk $D^{2} \times {s_{*}}$ at the midpoint $s_{*}$ of $b_{0} b_{1}$ meets $T_{1}$ in the triangle cutting off $a_{2}^{1}$ , $T_{2}$ in the quadrilateral separating ${a_{0}^{0}, a_{1}^{0}}$ from ${a_{1}^{1}, a_{2}^{1}}$ , and $T_{3}$ in the triangle cutting off $a_{0}^{0}$ ; every other normal coordinate is zero. It is connected, has nonempty boundary and Euler characteristic one, and its boundary meets the marked original meridian $μ = {a_{0}} \times S^{1}$ once mod two.

\begin{tikzpicture}[scale=1]
\coordinate (A0) at (0,0); \coordinate (A1) at (2,0); \coordinate (A2) at (1,1.5);
\coordinate (B0) at (0.6,0.7); \coordinate (B1) at (2.6,0.7); \coordinate (B2) at (1.6,2.2);
\draw (A0)--(A1)--(A2)--cycle; \draw (B0)--(B1)--(B2)--cycle;
\draw (A0)--(B0) (A1)--(B1) (A2)--(B2);
\draw[dashed] (A0)--(B1) (A0)--(B2) (A1)--(B2);
\coordinate (M0) at (0.3,0.35); \coordinate (M1) at (2.3,0.35); \coordinate (M2) at (1.3,1.85);
\draw[red,very thick] (M0)--(M1)--(M2)--cycle;
\node[below left] at (A0) {$a_0^0$}; \node[below right] at (A1) {$a_1^0$}; \node[left] at (A2) {$a_2^0$};
\node[left] at (B0) {$a_0^1$}; \node[right] at (B1) {$a_1^1$}; \node[above] at (B2) {$a_2^1$};
\node[red] at (1.35,0.8) {$D^2\times\{s_*\}$};
\end{tikzpicture}

The dashed edges indicate the staircase prism subdivision specified by $T_{1}, T_{2}, T_{3}$ ; the red section is the geometric disk whose intersections with those tetrahedra give the stated triangle–quadrilateral–triangle vector. The formula for the tetrahedra, not the schematic projection, fixes the exact triangulation.

Example references

Normal-surface coordinates and the Haken cone — triangle and quadrilateral normal disk types.

Linear-size good exterior triangulation from a planar knot code — fixed crossingless solid-torus exterior model and marked meridian path.

Hass–Lagarias–Pippenger, §4 and §5 — unknot exterior and normal disk coordinates.

Facts & Assumptions

Given: The product triangulation $D^{2} \times S^{1}$ of the exterior and its marked meridian path.

[F1]

Normal-surface coordinates and the Haken cone: a triangle or quadrilateral normal disk meets tetrahedral faces in matching normal arcs; admissibility allows at most one quadrilateral type per tetrahedron.

[F2]

Linear-size good exterior triangulation from a planar knot code: at zero crossings the exterior is a fixed triangulated solid torus with the original meridian represented by a three-edge fibre cycle.

Verification

1.1givenF2construct

For the prism over $b_{0} b_{1}$ , write $a_{i}^{0} = (a_{i}, b_{0})$ and $a_{i}^{1} = (a_{i}, b_{1})$ . Use the staircase tetrahedra $T_{1} = [a_{0}^{0}, a_{1}^{0}, a_{2}^{0}, a_{2}^{1}], T_{2} = [a_{0}^{0}, a_{1}^{0}, a_{1}^{1}, a_{2}^{1}], T_{3} = [a_{0}^{0}, a_{0}^{1}, a_{1}^{1}, a_{2}^{1}] .$ Their common faces are triangulated by the same ordered diagonals, so the three tetrahedra form the triangular-prism block; repeating this order on the three-cycle of $S^{1}$ gives the product triangulation in [F2].

2.1F1step 1.1construct

The level disk at the midpoint of the product edge separates the bottom vertices $a_{i}^{0}$ from the top vertices $a_{i}^{1}$ . In $T_{1}$ it cuts off the single top vertex $a_{2}^{1}$ , giving one triangle; in $T_{2}$ it separates the pair ${a_{0}^{0}, a_{1}^{0}}$ from ${a_{1}^{1}, a_{2}^{1}}$ , giving one quadrilateral; in $T_{3}$ it cuts off the single bottom vertex $a_{0}^{0}$ , giving one triangle. It misses all tetrahedra in the other two prism blocks. Its intersections on shared faces agree because they are the same level section of the product, so the matching equations hold. Only one quadrilateral coordinate is nonzero, so [F1] gives an admissible normal disk vector.

3.1givenF2step 2.1construct

The geometric surface is exactly $D^{2} \times {s_{*}}$ , hence is a connected disk. Its boundary is the nonempty circle $\partial D^{2} \times {s_{*}}$ . The disk itself has Euler characteristic $1$ . The marked meridian ${a_{0}} \times S^{1}$ passes through exactly one point of this boundary at $(a_{0}, s_{*})$ , in the interior of the chosen fiber edge; the intersection is transverse and has parity $1$ .

4.1F1F2step 2.1step 3.1algebra∎

Thus the explicit vector satisfies the matching, quadrilateral, connectedness, boundary, Euler-characteristic and mod-two meridian checks claimed in the statement.

Normal-surface unknot recognition examples

2 · Logical flowchart

3 · Definitions, theorems and proofs

4 · Examples, counterexamples and false statements

A boundary-parallel disk fails the meridian test

Statement refuted

Statement references

Facts & Assumptions

Counterexample

Worked case: a normal disk in the zero-crossing unknot exterior

Example

Example references

Facts & Assumptions

Verification

Sources