Zero-meridian pairs are generated by Kirby moves in the three-sphere

Framing bookkeeping from bands. For disjoint oriented components $X,Y$, write $\mathrm{lk}(X,Y)$ for the coefficient of $[Y]$ in $H_1(E_X;\mathbb{Z})=\mathbb{Z}[{\mu }_X]$, using the positive meridian convention in [F1] and the homology calculation in [F3]. A parallel representing integer framing $a$ has class $[{\lambda }_X]+a[{\mu }_X]$, so its linking with $X$ is $a$. This linking number is symmetric: by [F4], choose transverse oriented 2-chains $A,B$ in $S^3$ with boundaries $X,Y$. Intersection with $A$ defines a homomorphism on $H_1(E_X)$ because its boundary lies outside $E_X$, and it takes the value $+1$ on the positive meridian. By [F3], its value on $Y$ is therefore the meridian coefficient defining $\mathrm{lk}(X,Y)$; likewise intersection with $B$ computes $\mathrm{lk}(Y,X)$. The boundary formula for the intersection 1-chain $A\cap B$ equates these two counts, since its total signed boundary count is zero. For oriented components $A,B$ with framings $a,b$ and linking number $l$, a slide of $A$ over the signed parallel $sB$, $s=\pm 1$, has framing $a+b+2sl$. Indeed the band sum and its framing parallel are homologous, in the complement of the other curve, to the sums of the corresponding old parallels; additivity gives the two cross-linkings $sl,sl$. Its linking with a third component $C$ is $\mathrm{lk}(A,C)+s\mathrm{lk}(B,C)$, because the band sum represents $[A]+s[B]$ in the complement of $C$. These are consequences of the actual parallel-and-band construction, not permissions to replace a geometric link by a matrix. We use bands with no added full twist; an existing twist of any component is carried with its strip.

The zero-meridian crossing operation. A slide over $U$ can switch a crossing of a selected strand with the strand of $K$ encircled by $U$. Here is the local construction. Take the disk of $U$ transverse to the latter strand, and take a thin band from the selected strand to a disk-parallel copy of $U$. Put the interior of the band off the disk and all the link. In a cross-section perpendicular to the encircled strand, the band sum replaces a short detour on one side of the origin by that detour followed by one signed circuit about the origin. Round its two band corners. The result is the detour on the other side: the region between the circuit with its return segment and this detour is a disk in the punctured cross-section, so pushing across its product with the short strand interval is an ambient isotopy relative to the ends. Thus precisely the chosen crossing changes. Use the opposite band side for the opposite crossing. The disk-parallel framing of $U$ has no twist, so the strip is the transported framing of this operation, including any change of self-linking prescribed by step 1.1. The operation is valid when the selected strand is another part of $K$: use distinct disjoint short arcs as the two band ends. The encircled arc is unchanged, so $U$ remains its small zero-framed meridian. To use a different arc, move this meridian along a thin tubular neighbourhood of $K$; a meridional disk slides along the product tube, so this is an isotopy of $U$ in the complement of $K\cup L$, with its disk framing. The bands and their corners have smooth rounded product neighbourhoods; the disk pushes extend by a cutoff in those neighbourhoods to smooth orientation-preserving ambient isotopies.

Signed-unit blowdown as allowed moves. Suppose an unknotted component $J$ has framing $\epsilon =\pm 1$ and a spanning disk $D$ transverse to the other strands. At a piercing of sign $d$, slide that strand over $-\epsilon d$ times the oriented framed parallel of $J$. Route its band beside $D$. The parallel makes exactly $\epsilon$ full turns around the disk-normal direction, so this banded detour removes that piercing and gives one signed twist of the remaining disk-piercing strands. This is a geometric disk operation: cut the strands at their small piercing intervals, carry their cut ends along disjoint narrow radial strips of $D$ to its boundary, and use the framed parallel there to join the ends on the same side of $D$. Nested radial strips are used one at a time; corners are rounded in the disk-normal product. After all piercings have been removed, $J$ bounds a disk missing the remaining link, so a small neighbourhood of that disk is a ball separating $J$. Delete it by [F2]. The remaining strands have acquired $-\epsilon$ full twist about $D$; reversing this construction gives the corresponding blowup. The framing/linking updates are $a_i^{\mathrm{\prime }}=a_i-\epsilon l_i^2$ and $l_{ij}^{\mathrm{\prime }}=l_{ij}-\epsilon l_i{l}_j$, where $l_i=\mathrm{lk}(A_i,J)$: apply step 1.1 successively to the signed slides, or count the crossings of the parallel strips in the displayed full twist. In the applications below each remaining component crosses $D$ only once. A full twist of one such strand is a curl of an unknotted arc; of two such strands it is their two-strand full twist. Thus this operation has an explicit geometric meaning, as well as the displayed integer update.

Splitting the pair from the old link. Choose a regular finite projection of the compact smooth link, keeping a small meridional arc marked. Such a projection can be obtained by replacing finitely many short arcs inside disjoint tubular charts by polygonal arcs and rounding, followed by a small generic change of direction; the replacements are isotopies in their product tubes, and finitely many nonadjacent projected line segments have only finitely many transverse intersections after that change. At every mixed crossing of $K$ with $L$, use step 2.1 on the strand of $L$ to put $K$ above that strand. Its framing does not change: that component has linking zero with the small meridian $U$, and $U$ has framing zero, so step 1.1 gives the same integer. The local replacement changes no crossing between two strands of $L$. Consequently the old framed link, considered by itself, is still its original framed isotopy class. Once $K$ is above $L$ at every mixed crossing, lift $K$ and its very small meridian in the projection direction above a slab containing $L$, and move them into a ball disjoint from $L$. This lift has no obstruction at a mixed crossing because their height order was chosen precisely this way. From now on all operations are in that ball.

Unknotting the companion. Fix a base point away from the meridional arc. Switch self-crossings of $K$ by step 2.1 until its based diagram is descending: on traversing from the base point, the first encounter with each crossing is the overpass. A descending diagram is an unknot by the following explicit disk construction. Cut at the base point, and straighten the arc in traversal order, lifting each initial segment above all later segments; each lift is possible because the initial segment is the overpass at every first encounter. Join the endpoints by the short base-point arc outside the projection. This gives a simple planar circle after a final height isotopy, hence an embedded spanning disk. There are finitely many crossings, so these are finitely many moves and disk pushes. Keep $U$ tiny at an unchanged arc during each crossing operation and carry it by the subsequent isotopies. The result is the standard Hopf pair $K_n,U_0$ in the disjoint ball, for some integer $n$. Changes of $n$ have been carried by the strips; we do not assert that crossing changes preserve $n$.

Reducing the Hopf framing. The positive meridian convention in [F1] gives $\mathrm{lk}(K,U)=+1$; the orientations are transported through the moves, so no component is reoriented. Sliding $K$ over either signed disk-parallel of $U$ changes $n$ by $+2$ or $-2$ by step 1.1. Choose the band in the standard Hopf solid torus $S^3\setminus \mathrm{int}N(U)=D^2\times S^1$, where $K$ is a core. A disk-framed parallel of $U$ is a meridian of this exterior solid torus. The band running radially from the core to that meridian makes the new curve, after pushing slightly inside, $(r{w}^{\pm 1},w)$, $w\in S^1$, for small fixed $r>0$: the added meridian is the single turn of its disk coordinate. The curves $((1-t)r{w}^{\pm 1},w)$, $0\le t\le 1$, are embedded for every $t$, since the second coordinate is $w$. They give an isotopy to the core, supported away from the exterior boundary and hence away from $U$. Extend its velocity in a small tube by a cutoff to obtain an ambient isotopy; the full normal strip is transported. Thus the pair is again a Hopf pair, rather than merely having the same linking matrix. Repeating gives $K_0,U_0$ if $n$ is even and $K_1,U_0$ if $n$ is odd. Negative $n$ is covered by choosing the increasing sign.

The odd Hopf pair. In $K_1,U_0$, the disk of $K$ has just one piercing, by $U$. Apply step 2.2 with $J=K$ and $\epsilon =1$. The remaining component is the unknot $U$ with framing $-1$: the single strand receives one curl, which is removed by an ambient isotopy while retaining its changed normal framing. It is split from $L$, so delete it. This removes the odd pair by allowed moves alone.

The even Hopf pair. In $K_0,U_0$, introduce a split $+1$ unknot $V$ in the same otherwise empty ball. Slide $U$ over $V$ along the untwisted band between their spanning disks, orienting the band so the new linking of $U$ and $V$ is $+1$. The band sum of these two unknots is still an unknot: its spanning disk is the band sum of their disjoint disks. Its framing is $1$. The three components are now the standard chain $K_0,U_1,V_1$, with adjacent linkings $+1$ and with $K,V$ unlinked. The disk of the middle component meets each end component once in separate straight product arcs. Blow down $U_1$ by step 2.2. Its two-strand twist makes the ends a standard Hopf pair of linking $-1$, with framings $K_{-1},V_0$. Each end remains unknotted, since its own arc has only received a curl; their mixed two crossings are the Hopf clasp. Blow down $K_{-1}$. The remaining $V$ is the unknot with framing $0-(-1)(-1{)}^2=1$, again with only a single-arc curl. Delete this split $+1$ unknot. Thus the even pair also disappears. The disk, band and full-twist descriptions identify every intermediate knot and not just its algebraic linking data.

Steps 3.1 and 4.1 reduce the original configuration to the original framed $L$ and an isolated Hopf pair. Steps 5.1 and 6.1 and 6.2 remove that pair. All bands, disk pushes and lists are finite, and every deletion was a split signed-unit unknot deletion or its explicitly proved slide expansion. Orientations of components and their normal strips are transported at each move. Reversing the finite list proves the insertion assertion.