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Ribbon knots and the slice-ribbon problem — Examples and Counterexamples
1 · Prerequisites
- Abelian Categories
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2 · Logical flowchart
3 · Definitions, theorems and proofs
None yet.
4 · Examples, counterexamples and false statements
The square knot is a nontrivial ribbon knot
Example
Assume ACω. Let be the standard right-handed trefoil, represented by the closure of the standard two-strand braid with three positive half-twists, and let be its reverse mirror. Choose the connected-sum balls on arcs colored in the Fox 3-colorings described below. The square knot is nontrivial, is ribbon, and bounds a smooth properly embedded disk in .
Example references
Facts & Assumptions
Given: ACω, the standard right-handed trefoil , its reverse mirror , and the connected-sum datum specified in the Example.
Under ACω, every smooth oriented knot has bounding a ribbon immersion in , and separately bounding a smooth proper disk whose squared radial function is Morse with only index- and index- critical points. (A reverse-mirror sum admits an explicit ribbon immersion, A knot summed with its reverse mirror bounds a ribbon disk)
Resolving a ribbon immersion gives a smooth proper disk with the same boundary. (Local resolution of ribbon singularities)
Under ACω, a ribbon immersion yields a no-maximum disk movie with the stipulated radial product boundary collar. (Ribbon immersions and disk movies without maxima are equivalent)
The reverse mirror is reflection of the knot followed by reversal of its knot orientation, and the connected sum uses the fixed orientation-matching gluing convention. (Reverse mirror of an oriented knot, Oriented connected sum and prime knot)
A Fox 3-coloring assigns an element of to each diagram arc and satisfies at every crossing. (Quandles and knot-diagram colorings, The dihedral operation is a quandle)
For , the number of Fox -colorings is unchanged by planar isotopy and every oriented Reidemeister move. (Dihedral Fox coloring counts are preserved by Reidemeister moves)
Under ACω, two regular diagrams represent the same oriented knot exactly when connected by a finite sequence of planar isotopies and oriented Reidemeister moves. (Reidemeister's theorem for oriented diagrams)
ACω is the sequence-choice axiom used by [F1] and [F5]. (The Axiom of Countable Choice ())
Verification
Given: ACω, the standard right-handed trefoil, , and the stated connected-sum datum.
Apply [F1] with . Its explicit ribbon immersion has oriented boundary , so the square knot is ribbon by the A-page definition. Apply [F6] to that immersion to obtain a smooth properly embedded disk in with the same boundary, proving smooth sliceness as well. Applying [F7] to the same immersion also gives a no-maximum disk movie with its required radial product boundary collar.
In the two-strand braid chart with parameter increasing from bottom to top, let be the color transfer for the positive generator whose overstrand runs from lower left to upper right. By [F3], ; the inverse crossing has transfer . Direct calculation gives and in . Hence is also the identity. Starting each trefoil tangle with the ordered pair therefore satisfies the braid closure conditions for both the trefoil and its mirror. The first arc in each closure has color ; choose those arcs for the connected-sum datum. The colorings glue because the shared arc colors agree, and the color- arcs remain, giving a nonconstant Fox 3-coloring of the square knot.
A crossing-free diagram of the unknot has one arc and exactly three Fox 3-colorings, one for each residue. The square-knot diagram has those three constant colorings and the nonconstant coloring from step 1.2, so its coloring count is at least four. If it were isotopic to the unknot, [F5] would give a finite Reidemeister sequence between their diagrams, and [F4] would preserve the coloring count, a contradiction. Thus the square knot is nontrivial.
Diagram
The schematic depicts one positive generator and its inverse in the two-strand braid. The braid parameter increases upward; the continuous strand is over, and the understrand has a gap. The top and bottom labels are ordered by left and right position. The calculation in step 1.2, rather than the drawing, verifies the three-crossing closure colorings.
Remarks
The Fox-coloring calculation proves nontriviality. The direct ribbon immersion in [F1] proves ribbonness, and its resolution in [F6] proves smooth sliceness. The general conversion between arbitrary no-maximum disk movies and ribbon immersions is supplied separately on the A page; this example also has the explicit immersion from [F1].
Under full AC, the right-handed trefoil is not ribbon
Statement refuted
Assume the full Axiom of Choice. The standard right-handed trefoil is a ribbon knot.
Statement references
Facts & Assumptions
Given: Full AC and the standard right-handed trefoil .
Full AC is the choice-function principle. (The Axiom of Choice)
Under full AC, the standard right-handed trefoil does not bound a smooth properly embedded disk in . (The trefoil does not bound a smooth proper disk in the four-ball)
Under ACω, every ribbon knot bounds a smooth properly embedded disk in . (Every ribbon knot is smoothly slice)
Full AC implies ACω. (AC implies DC implies countable choice)
Counterexample
The standard right-handed trefoil has no smooth proper disk in , so it cannot admit a ribbon disk whose resolution is such a disk.
The given full AC is the choice-function principle in [A1]. Applying [F1] under that hypothesis, no smooth properly embedded disk in has boundary .
If were ribbon, [F3] would supply the ACω hypothesis of [F2], and [F2] would give a smooth properly embedded disk in with boundary .
Step 1.2 contradicts step 1.1. Therefore the trefoil is not ribbon, refuting the statement.
Diagram
The box denotes the three-crossing positive two-strand braid whose closure is the standard right-handed trefoil. The closure arcs join equal endpoint positions outside the box. This encoded diagram fixes the standard trefoil used in the counterexample; the proof relies on the nonsliceness supplier, not on the drawing.