Three-manifold surgery presentations and Kirby moves

By [A1] and [F1], choose a finite triangulation $\mathcal{T}$ of $M$. Its $1$-skeleton is connected: if it were disconnected, each simplex would have all its vertices in a single graph component because every pair of vertices of a simplex is joined by an edge; the realization would therefore be a disjoint union of nonempty subcomplexes, contrary to connectedness of $M$.

In the barycentric subdivision, thicken the finite embedded $1$-skeleton by taking disjoint small $3$-balls at its vertices and disjoint narrow tubes around the edge interiors, each tube meeting its endpoint balls in end disks. The vertex balls joined by the tubes along a spanning tree form a $3$-ball: adding a leaf ball and its joining tube joins two balls along a tube and again gives a ball. Each remaining edge tube attaches along two disks to this connected boundary and is one $1$-handle. There are $\mathrm{\mid }E\mathrm{\mid }-\mathrm{\mid }V\mathrm{\mid }+1$ such edges; the cycles formed by each non-tree edge with its spanning-tree path form a basis of the graph's first homology, so this number is $b_1$. The resulting connected orientable manifold is a handlebody of genus $b_1$ by the definition.

In the barycentric dual cell structure, put one vertex in each tetrahedron and join two such vertices through the barycentre of each common triangular face. This dual graph is connected: choose a piecewise-linear path between interior points of any two tetrahedra and perturb its finitely many segments inside simplex stars so it avoids the primal $1$-skeleton and meets triangular faces transversely. Each crossing then passes between tetrahedra sharing that face, producing a finite path in the dual graph.

Take the vertex balls and edge tubes in step 2.1 as the primal barycentric blocks, so their complement is made of the dual blocks. In each tetrahedron the complementary block is a $3$-ball about its barycentre; across each common triangular face, the two balls join by a tube about the dual edge through that face. Thus the dual balls and tubes form the closure of $M\setminus \mathrm{int}{H}_{-}$. The dual graph is connected by step 2.2; joining its vertex balls along a spanning tree gives a $3$-ball, and each other dual edge contributes one $1$-handle. Thus the complementary closure is a handlebody $H_{+}$. Since $M$ is closed, its boundary is the same surface $\mathrm{\Sigma }=\mathrm{\partial }{H}_{-}$ as the boundary of $H_{+}$.

Both boundaries are $\mathrm{\Sigma }$, so the handlebodies have equal genus by the definition of genus as the genus of their boundary. Their induced boundary orientations are opposite because they are the two sides of the oriented manifold $M$; the gluing map is therefore orientation-reversing. By [F2], $M=H_{-}{\cup }_{\mathrm{\Sigma }}{H}_{+}$ is a Heegaard splitting.

Push $c$ a small distance into $H_{-}$ to obtain an embedded knot $K$ and choose a closed tubular neighbourhood $N(K)$ disjoint from $\mathrm{\Sigma }$. The collar between $c$ and the boundary of $N(K)$ is an embedded annulus $A$; its parallel boundary curve on $\mathrm{\partial }N(K)$ is ${\lambda }_A$. Choose the product coordinates on $N(K)$ so that the meridian $\mu$ bounds a normal disk.

Cut $H_{-}\setminus \mathrm{int}N(K)$ along $A$ and reglue the two annular faces after one full turn. In annular coordinates the turn adds $2\pi$ to the angular coordinate between the two ends and is stationary near the seams. Thus on $\mathrm{\Sigma }$ the induced map is ${\tau }_c$ (or its inverse if the turn is reversed). On $\mathrm{\partial }N(K)$, a meridian crossing the annulus once returns with one extra longitude traverse, so the induced homology shear is $[\mu ]\mapsto [\mu ]+[{\lambda }_A]$ or $[\mu ]\mapsto [\mu ]-[{\lambda }_A]$, respectively; the longitude is fixed. The operation fixes the rest of $H_{-}$ and is smooth after rounding the product corners.

Glue the identity on $H_{+}$ to the cut-and-reglued map on the complement of $N(K)$. Along $\mathrm{\Sigma }$ the maps agree because the new gluing is $\phi \circ {\tau }_c^{\pm 1}$; hence they give an orientation-preserving diffeomorphism between the two knot exteriors. The meridian of the replacement solid torus is identified with $[\mu ]\pm [{\lambda }_A]$, exactly the framed surgery slope from step 2.1. By [F3], filling this boundary is the stated integral surgery, so its result is the Heegaard-glued manifold with the twisted gluing map.

The annular parallel and its product framing are part of the supplied data, and the only auxiliary collar/tubular choices are within the ${\mathrm{A}\mathrm{C}}_{\omega }$ interface [A1]. The construction therefore uses an actual framed knot and introduces no rational coefficient.

Apply [F1] with the standard oriented $N=S^3$, target $M$, and empty external boundary. The hypotheses hold: both manifolds are connected, nonempty, compact and oriented, and their boundaries have the unique empty identification. The supplier gives a finite framed link $L\subset S^3$ with disjoint actual attaching tubes and a diffeomorphism $f:Q\to M$ from its simultaneous framed surgery. Give $Q$ the orientation extending the standard orientation on the link exterior; the orientation sign of $f$ remains to be checked. Its normal disk framings are the actual attaching framings of its four-dimensional two-handle trace; no arbitrary rational filling has been introduced.

Orient each of the finitely many components. In a component's oriented framed tube $S^1\times D^2$, the framed parallel is $t\mapsto (t,v)$ for a fixed nonzero normal vector $v$. Its projection to the core has degree $+1$. In the positive-meridian/preferred-longitude basis from [F2], the meridian has degree zero and the preferred longitude has degree one, so this parallel has class $n_i[{\mu }_i]+[{\lambda }_i]$ for an integer $n_i$. This also follows by writing the normal vector in the preferred trivialization: its winding is the integer $n_i$. The surgery meridian of a framed two-handle is precisely this parallel, by the explicit exchanged disk-factor construction in [F1]. Thus every component has exactly the required integer slope, with the full original normal framing retained.

The disjoint attaching tubes in step 1.1 permit these fillings simultaneously; step 2.1 identifies each with the integral convention in [F2]. If $f:Q\to M$ preserves orientation, this is the required presentation. Otherwise reflect the standard $S^3$ by an orientation-reversing diffeomorphism $r$, transport each component orientation, and replace each integer $n_i$ by $-n_i$. The positive meridian and preferred longitude of the reflected component satisfy $[{\mu }_i^{\mathrm{\prime }}]=-r_{\ast }[{\mu }_i]$ and $[{\lambda }_i^{\mathrm{\prime }}]=r_{\ast }[{\lambda }_i]$: reflection reverses the oriented normal disk, while preserving the degree-one and null-homology properties that characterize the longitude. Thus $r_{\ast }(n_i[{\mu }_i]+[{\lambda }_i])=-n_i[{\mu }_i^{\mathrm{\prime }}]+[{\lambda }_i^{\mathrm{\prime }}]$. To see the oriented effect on the filled manifold, write an original attaching map as $g_i:\mathrm{\partial }(D^2\times S^1)\to \mathrm{\partial }E$. On the filling solid torus use the orientation-reversing diffeomorphism $c(z,w)=(z,\bar{w})$, which fixes its oriented meridian. Choose reflected attaching maps $g_i^{\mathrm{\prime }}=r\circ g_i\circ c^{-1}$; these have the required slopes and glue oriented solid tori to the reflected exterior. The maps $r$ on the exterior and $c$ on each filling agree across these seams and their product collars, giving an orientation-reversing diffeomorphism $R:Q\to Q^{\mathrm{\prime }}$ to the reflected surgery manifold. Consequently $f\circ R^{-1}:Q^{\mathrm{\prime }}\to M$ preserves orientation. This covers both orientation signs without strengthening [F1].

If the link is empty, the same reflection argument applies directly to $S^3$. By [F2] its surgery result is $S^3$, and the empty diagram is an allowed presentation of the standard oriented $S^3$. Together with step 3.1 this proves the claim.

For an ambient isotopy, carry each component neighbourhood, meridian, and preferred longitude along the isotopy. The endpoint diffeomorphism of $S^3$ therefore maps every filling meridian to the same framed slope on the corresponding endpoint torus. Extend it over each attached solid torus by the product identification used in the filling. The resulting map is an orientation-preserving diffeomorphism of the filled manifolds.

Applying [A1] to the family of values of any sequence $(X_n)$ in [F8] gives a choice function on that range; composing it with $n\mapsto X_n$ supplies the choice function required by [F8]. Thus the ${\mathrm{A}\mathrm{C}}_{\omega }$ hypothesis of [F3]–[F6] holds. For a diagram $D$, form its $4$-dimensional surgery trace from $S^3\times [0,1]$ by attaching one $2$-handle for each framed link component. Orient the product by $dt$ followed by the fixed orientation of $S^3$, and orient each handle so its attaching identification reverses the induced boundary orientations of the handle and cylinder, as required by [F5]; this is possible because each attaching region $S^1\times D^2$ is connected. The trace is connected. On the outgoing face, the unaltered part $E_D=S^3\setminus {\bigcup }_i\mathrm{int}N(K_i)$ has the standard $S^3$ orientation from the upper product face, exactly the orientation that the filling conventions [F1, F7] extend over the attached solid tori. The link exterior is nonempty and connected (paths in $S^3$ can be perturbed off the one-dimensional link; for the empty diagram $E_D=S^3$). Thus the identity-on-$E_D$ identification in [F6] preserves the surgery orientation. The link slide of [F2] is exactly the $k=2$ handle slide of [F3], including its full transported normal framing. Since $1\le 2\le 4-2$, [F4] supplies a diffeomorphism between the two traces relative to the incoming $S^3$ face. Both traces have the product orientation there and the diffeomorphism fixes that face; its tangent derivative is the identity and its normal derivative preserves the interior side, so it preserves the trace orientation. Its restriction preserves the induced orientations on the outgoing faces. Composing with the identifications in [F6] gives the required orientation-preserving diffeomorphism of surgery results.

Suppose a split unknot is added with coefficient $\epsilon \in \{1,-1\}$ in a ball disjoint from the rest of the link. The separating sphere shows that the filled manifold is the connected sum of the old surgery result with the filling of the unknot exterior along $\epsilon [{\mu }_U]+[{\lambda }_U]$. For the standard unknot, take $E_U=\{\mathrm{\mid }{z}_2\mathrm{\mid }\ge 1\mathrm{/}\sqrt{2}\}\subset S^3\subset {\mathbb{C}}^2$, identified with $D^2\times S^1$ by $(z_1,z_2)\mapsto (z_1,z_2\mathrm{/}\mathrm{\mid }{z}_2\mathrm{\mid })$. Its preferred longitude ${\lambda }_U$ is the meridian of this solid torus, and ${\mu }_U$ is its core longitude. The diffeomorphism $g_{\epsilon }(z,w)=(w^{-\epsilon }z,w)$ of $D^2\times S^1$ preserves orientation, fixes $[{\lambda }_U]$, and sends $[{\mu }_U]$ to $[{\mu }_U]-\epsilon [{\lambda }_U]$; hence it sends the slope $\epsilon [{\mu }_U]+[{\lambda }_U]$ to the unoriented meridional slope $[{\mu }_U]$. Choose the $S^3$ gluing with meridian $[{\mu }_U]$ and the $\epsilon$-filling gluing as its pullback by $g_{\epsilon }^{-1}$. The map $g_{\epsilon }$ on the exterior and the identity on the attached solid torus then give an orientation-preserving diffeomorphism between the two fillings. Filling along $[{\mu }_U]$ restores the original unknot neighbourhood, so the $\epsilon$-framed unknot filling is orientation-preservingly diffeomorphic to $S^3$. Hence the connected sum is the original manifold: deleting a ball and regluing the complementary ball of $S^3$ restores the same manifold. Removing a split $\epsilon$-framed unknot is the inverse operation.

Each allowed move preserves the oriented surgery manifold by steps 1.1–1.3. Composing the finitely many orientation-preserving diffeomorphisms along a move sequence proves the assertion for Kirby-move-equivalent diagrams.

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.

If $D_0,D_1$ are related by the stated finite sequence, apply [F2] to each move and compose its orientation-preserving diffeomorphisms. The result is an orientation-preserving diffeomorphism $M_0\to M_1$.

Conversely, let $f:M_0\to M_1$ be an orientation-preserving diffeomorphism. Use the two given diagrams as undecorated presentations based in the same standard oriented $S^3$. Take empty external boundary, empty ribbon graph and arbitrary equal auxiliary weights; the global supplier's undecorated comparison is independent of its later decoration and evaluation clauses. Every component framing is its actual normal two-disk framing, as in [F1]. Thus [F3] applies to this specified $f$ and yields a finite O1/O2/O3 and framed-isotopy sequence from $D_0$ to $D_1$. In particular it compares the given endpoint diagrams; it does not merely produce some new presentations of their output manifolds.

The O1 and O2 operations in this sequence are precisely the split signed-unit-unknot moves and fully framed slides of [F1]. Replace each O3 insertion or deletion by the finite sequence supplied by [F4]. That lemma has no restriction on the companion knot's linking with the rest of the diagram, so it applies at every intermediate stage. Its integer normal framings agree with [F1]. This replacement need not preserve the particular marked map $f$, which the present claim does not require, but it preserves the exact endpoint framed diagrams. A finite list with finitely many finite replacements is finite.

The resulting sequence uses only ambient isotopy, fully framed handle slides and split $\pm 1$ unknot stabilization or destabilization. It joins $D_0$ to $D_1$, proving the converse. Together with step 1.1 this establishes the equivalence with all stated orientation, smoothness and finiteness conventions.