Ribbon knots and the slice-ribbon problem

Identify $B^4$ with the unit ball in ${\mathbb{R}}^4$ and use the explicit boundary collar $c:S^3\times [0,\epsilon )\to B^4$, $c(y,s)=(1-s)y$, with $0<\epsilon <1$. On the unit disk set $r(x)=1-\parallel x{\parallel }^2$, so $r=0$ on $\mathrm{\partial }{D}^2$, $r>0$ in the interior, and $dr$ is nonzero in the inward normal direction at the boundary. By [F1], the finitely many preimage arcs form disjoint compact subsets of $D^2$. Choose pairwise disjoint neighborhoods of the interior arcs ${\beta }_i$, all disjoint from every proper arc ${\alpha }_j$ and from $\mathrm{\partial }{D}^2$, and choose smooth bump functions ${\varphi }_i$ supported in those neighborhoods with ${\varphi }_i=1$ on ${\beta }_i$ and $0\le {\varphi }_i\le 1$. This is a finite choice of smooth functions on a disk.

Let $M=\max \{0,{\max }_{i,s}(r({\alpha }_i(s))-r({\beta }_i(s)))\}$ when there is at least one double arc, and let $M=0$ otherwise. Choose $\delta >0$ so that $\delta (2M+3)<\epsilon$, set $a=2\delta (M+1)$, and define $h=\delta r+a{\sum }_i{\varphi }_i$. The supports are disjoint, so $0<h<\epsilon$ on $\mathrm{Int}{D}^2$ and $h=0$ exactly on $\mathrm{\partial }{D}^2$. For every double arc and every parameter $s$, $h({\beta }_i(s))-h({\alpha }_i(s))=a+\delta (r({\beta }_i(s))-r({\alpha }_i(s)))\ge \delta (M+2)>0$. Near $\mathrm{\partial }{D}^2$ all bumps vanish, so $h=\delta r$ there and has positive inward derivative.

Define $\tilde{f}(x)=c(f(x),h(x))$. If $\tilde{f}(x)=\tilde{f}(y)$, injectivity of the collar coordinates gives $f(x)=f(y)$ and $h(x)=h(y)$. By [F1], either $x=y$ or $x,y$ are the two preimages of one of the double arcs; the strict height inequality in step 2.1 excludes the latter. Thus $\tilde{f}$ is injective.

The map $f$ is an immersion, so its derivative is injective; adjoining the height coordinate preserves injectivity of the derivative, and the collar map $c$ is a local diffeomorphism. Hence $\tilde{f}$ is an immersion. A continuous injection from compact $D^2$ to the Hausdorff space $B^4$ is a homeomorphism onto its image, so this injective immersion is a smooth embedding. Since $h^{-1}(0)=\mathrm{\partial }{D}^2$, its preimage of $S^3=\mathrm{\partial }{B}^4$ is exactly $\mathrm{\partial }{D}^2$; the nonzero inward derivative established in step 2.1 gives the proper boundary collar condition in [F2]. The disk orientation transports through the embedding and its boundary remains the original oriented knot $K$.

Choose $p\in S^3\setminus K$ and a smooth path $p_t$ from $p$ to $\mathrm{\infty }$, constant near both endpoints of $I$. This is an isotopy of embeddings of the compact $0$-manifold consisting of one point. By [F2] it extends to an ambient isotopy $H_t$ of $S^3$. Set $K^{\mathrm{\prime }}:=H_1(K)$, so $K^{\mathrm{\prime }}\subset S^3\setminus \{\mathrm{\infty }\}={\mathbb{R}}^3$, and take a regular diagram of $K^{\mathrm{\prime }}$ by [F1]. Delete a short open subarc of an edge incident to the unbounded complementary region. Its remaining projected arc can be placed in a closed planar disk with its two endpoints on the disk boundary, all its other points in the disk interior, and the deleted closing arc outside that disk: draw the boundary through the deleted edge along the adjacent unbounded region and round it away from the diagram. Parameterize the remaining immersed planar arc by $\gamma :[0,1]\to {\mathbb{R}}^2$, straight near its endpoints. It has nonzero derivative, finitely many transverse crossing pairs $(u_i,v_i)\in (0,1{)}^2$, no triples, and no endpoint multiple points. The closing arc has no crossing. At each crossing mark which of $u_i,v_i$ is over according to the diagram of $K^{\mathrm{\prime }}$. Since $H_t$ is orientation-preserving, the conjugate isotopy $r{H}_t{r}^{-1}$ shows $K^{\mathrm{\prime }†}$ is oriented ambient-isotopic to $K^{†}$. By [F3], the oriented connected-sum type $K^{\mathrm{\prime }}\mathrm{\#}{K}^{\mathrm{\prime }†}$ is therefore $K\mathrm{\#}{K}^{†}$ under the fixed gluing convention.

Choose a smooth function $q:[0,1]\to (0,\mathrm{\infty })$ so that $h(s)=\sqrt{s(1-s)q(s)}$ has the prescribed strict over-minus-under sign at every crossing pair. This is a finite interpolation: assign positive heights $2$ and $1$ at the over and under parameters, respectively, and prescribe $q(s)=h(s{)}^2\mathrm{/}(s(1-s))$ at these finitely many distinct parameters; interpolate the positive $q$ values by disjoint nonnegative smooth bumps against a positive constant. Distinct crossing parameters are disjoint because there are no triples. Make $q$ constant near the endpoints. Put $\mathrm{\Omega }=\{(s,t):0\le s\le 1,t^2\le s(1-s)q(s)\}$. The map $(s,v)\mapsto (s,v\sqrt{q(s)})$ is a smooth diffeomorphism from the round disk $v^2\le s(1-s)$ onto $\mathrm{\Omega }$, including its two endpoint tips. Thus $\mathrm{\Omega }$ is a smooth disk.

Define $f:\mathrm{\Omega }\to {\mathbb{R}}^3\subset S^3$ by $f(s,t)=(\gamma (s),t)$. Since ${\gamma }^{\mathrm{\prime }}(s)\mathrm{\ne }0$, its two derivative vectors $({\gamma }^{\mathrm{\prime }}(s),0)$ and $(0,0,1)$ are independent, so $f$ is an immersion, including in the smooth disk charts at the tips. The only possible multiple points occur at the crossing pairs of $\gamma$. For a pair $(u,v)$, assume $h(u)<h(v)$. The image double arc is $\{(\gamma (u),t):\mathrm{\mid }t\mathrm{\mid }\le h(u)\}$, with preimages $\{u\}\times [-h(u),h(u)]$ and $\{v\}\times [-h(u),h(u)]$. The first is properly embedded in $\mathrm{\Omega }$, whereas the second lies entirely in its interior because $h(u)<h(v)$. The image sheets are transverse: their planar tangent directions are independent, and both contain the vertical tangent. Double arcs from different crossings are disjoint because their planar crossing points are distinct. There are no triple points. No two boundary points have the same image, since boundary heights at a crossing are $\pm h(u)$ and $\pm h(v)$, all distinct. Thus $f$ satisfies exactly [F4].

The plane $t=0$, completed by the point at infinity, is an equatorial sphere of $S^3$. It cuts the boundary of $f(\mathrm{\Omega })$ at exactly the two tips. The upper boundary arc is $(\gamma (s),h(s))$ and the lower is its reflection $(\gamma (s),-h(s))$ with opposite boundary orientation. Both are smooth proper ball-arc tangles: near a tip the equation $t^2=s(1-s)q(s)$ solves smoothly for $s$ as a function of $t^2$, so the arc is transverse to the equatorial plane. Close the upper arc by the deleted trivial arc in the equatorial sphere and round its two corners in crossing-free endpoint neighborhoods. This closure represents $K^{\mathrm{\prime }}$: it has exactly the projected crossings of the diagram in step 1.1 and their original height ordering. Explicitly, after choosing smooth endpoint roundings, interpolate its heights to those of the regular diagram realization of $K^{\mathrm{\prime }}$ while keeping the planar immersion fixed. At every crossing, each endpoint height difference is positive in its marked ordering, so their convex interpolation stays positive; the planar immersion ensures immersion of every slice. This gives a smooth isotopy of embedded circles, and [F2] makes it ambient. The endpoint rounding takes place in disjoint balls without other strands and changes no knot type. Consequently [F3] identifies the oriented boundary of $f(\mathrm{\Omega })$ with $K^{\mathrm{\prime }}\mathrm{\#}{K}^{\mathrm{\prime }†}$, which represents $K\mathrm{\#}{K}^{†}$ by step 1.1. Together with step 3.1, this is the required ribbon immersion.

Start with a no-maximum disk movie. On its stipulated boundary collar, $\rho (c_D(y,s))=(1-s{)}^2$, so there are no radial critical points for $0<s<\epsilon$. The critical set is therefore a compact discrete subset of the disk interior, hence finite by [F2]. On each regular radial interval, radial projection to the unit three-sphere gives an isotopy of the level link, extended by [F3]. A local minimum has the usual disk model and creates a split round unknot in a small ball; a local saddle has the rectangle model and changes the link by its framed band move. These assertions follow in coordinates from the Morse equations $x^2+y^2$ and $x^2-y^2$: the first level is a small circle born in a ball, and in the second the two attaching sides of the rectangle are replaced by the other two sides. Critical points sharing a value have disjoint small ambient neighborhoods, so the corresponding local events may be processed successively. Thus there is a finite sequence of births, band attachments, and boundary-link isotopies, whose abstract disk and band handles give exactly the original disk. We will realize this same sequence by immersed surfaces in $S^3$.

For the reverse implication, start with a ribbon immersion $f:D^2\to S^3$. Remove narrow disjoint strip neighborhoods of all its proper preimage arcs. Choose them disjoint from every interior preimage arc; this is possible because all preimage arcs are disjoint compact sets. Cutting along a proper arc of a disk separates it into two disks, so cutting along all these strips leaves finitely many disk components. Their images are embedded and pairwise disjoint, since every original multiple point had one preimage in a removed strip. The removed strips themselves are embedded bands with pairwise disjoint images, and are disjoint from the disk boundaries except along their attaching edges: their remaining intersections with the disk components are exactly the original ribbon arcs, away from these boundaries. Rounding the cut corners gives a finite presentation by disjoint embedded disks and disjoint embedded bands. The boundaries of the disks form an unlink. Indeed each disk has a disjoint small product neighborhood, and shrinking concentric circles in its disk coordinates gives an isotopy to a small round unknot in that neighborhood; [F3] extends these disjoint isotopies. The band surgeries on this unlink recover the original boundary $K$.

The inductive invariant is a compact oriented ribbon-immersed surface with the current oriented level link as embedded boundary. Begin with the empty surface. An isotopy event acts on the whole immersion by [F3]. For a birth, its ball is disjoint from the old boundary link. The old immersed surface has empty interior in $S^3$: in each immersion chart its image is a smooth two-dimensional sheet, and finitely many such charts cover its compact domain. Its image is also closed. Hence there is a small ball inside the birth ball disjoint from the entire old image. Add an embedded disk there. Its boundary is a split unknot isotopic within the birth ball to the required newborn component, and apply [F3] to match the new boundary link. This introduces no multiple point.

For a band event, first shrink its attaching intervals in their boundary collars and slide them along the boundary, if needed, to avoid the finitely many endpoints of old ribbon arcs. This changes the banded boundary by an isotopy supported in its attaching collars. The immersion is embedded on these collars. Keep the prescribed normal framing of the band core. Perturb the core relative to these collars, as in [F4], to avoid the old double arcs and meet each regular old sheet transversely. Away from its endpoints it is compact and disjoint from the old boundary, so all resulting intersections with the old surface lie in its regular interior sheets; transversality makes their inverse image a finite discrete set. The perturbation can be arbitrarily small, so the core remains embedded, the band framing persists, and the boundary band move is unchanged up to isotopy. Shrink the framed band about that core, including its attaching intervals. This shrinking is an isotopy of the banded boundary: in the rectangle coordinates contract the width, extend the same contraction along the two attaching link intervals, and round corners in their disjoint collars. At an intersection with an old sheet the transverse core parameter is a local defining coordinate for that sheet, so the implicit function theorem expresses its intersection with a sufficiently narrow band as a single arc running across the full band width. Compactness away from these finitely many intersections allows one width small enough to exclude every other intersection. Each new double arc consequently has one proper preimage in the new band and one interior preimage in the old surface. The new band is embedded, it avoids all old double arcs, and the neighborhoods of its new intersections are disjoint, so no triple point is introduced. Glue and smooth the band at its attaching collars. The old proper preimage arcs remain proper because those collars avoided all their endpoints. The new proper arcs have endpoints on the two free band sides, hence on the new surface boundary. This proves the invariant at the saddle.

Finite induction over the movie now gives a ribbon-immersed oriented surface with boundary $K$. It is the original abstract surface: births added its zero-handles, saddles added its one-handles in the same order and with the same oriented attaching data, and boundary isotopies changed neither handle topology nor orientation. Thus its domain is a disk, not merely a surface of Euler characteristic one. Step 3.1 also shows that its only multiple points are finitely many disjoint transverse ribbon double arcs and its boundary is embedded. This proves the implication from the disk movie to a ribbon immersion.

Realize that presentation in a radial shell $S^3\times [a,1]\subset B^4$, $a>0$. At early distinct radii create a minimum disk for each unlink component, using its embedded disk to give the trace down to its cap. Between these caps and the bands keep product annuli on the unlink. At later radii attach the disjoint bands by the standard saddle traces of [F2]. Each saddle trace meets the existing surface only at its attaching collars: below that radius the surface contains only the product boundary annuli at those levels, and the band interior is disjoint from the current boundary link. The interiors of the zero-handle disks lie at earlier radii, so band intersections with their projections in $S^3$ cause no intersection in the four-ball. Distinct bands have disjoint images and disjoint attaching intervals, so they can also be placed at successive radii. Finish with a product collar up to radius one. Smooth the finitely many seams in their product charts. This is a smooth embedded disk because its abstract disks and bands reconstruct the original domain disk of step 1.2. Its radius has exactly the chosen minimum and saddle models; squared radius is their composition with an increasing smooth function having positive derivative on $[a,1]$, so it has the same nondegenerate critical points and indices. Its boundary is $K$, and the final product collar makes it proper and neat. This proves the converse and the equivalence.

Choose a ribbon immersion $f:D^2\to S^3$ for $K$ as in [F1]. Applying [F2] gives a smooth proper embedding $\tilde{f}:D^2\to B^4$ that agrees with $f$ on $\mathrm{\partial }{D}^2$; hence its boundary is exactly the oriented knot $K$.

The image of $\tilde{f}$ is a connected orientable disk, so its Euler characteristic is $1$ and its genus is $0$. By [F3], this disk is an admissible smooth spanning surface in the set defining $g_4^{\mathrm{s}\mathrm{m}}(K)$; since that set contains $0$ and consists of nonnegative integers, $g_4^{\mathrm{s}\mathrm{m}}(K)=0$. Thus $K$ is smoothly slice. The argument from [F1] to [F2] itself is choice-free; ACω is present only for the library's slice-genus convention [F3].