Alternating diagrams and crossing-number minimality

Let $N(K)$ be the displayed set of crossing counts. By [F1], move an oriented representative off $\mathrm{\infty }$ into ${\mathbb{R}}^3$; [F3] and [A1] give it a regular diagram. Forgetting the diagram's strand arrows gives a regular diagram of the same unoriented type, so $N(K)\mathrm{\ne }\mathrm{\varnothing }$.

The set $N(K)$ is a nonempty subset of $\mathbb{N}$, so [F4] supplies its unique least element. Ambient-isotopic representatives have the same crossing-count set by definition of the underlying unoriented link type. This makes $c(K)$ well-defined.

A sufficiently thin annular neighborhood of the supplied flyping circle contains its distinguished crossing and no other crossing: the crossing set is finite by [F1], and the circle has just the prescribed three intersection points. Transversality lets us narrow this neighborhood so its diagram consists only of the local strands at those points. The specified flyping circle, the chosen fixed disk and the annular crossing determine the local flype replacement: reflect the other disk and reverse each of its crossings, then reconnect its four ends through the modified annular crossing. This also transports any closed tangle components and the component orientations. The move is reversible by the same turnover construction.

Each crossing ball is chosen inside a disjoint crossing neighborhood; its upper and lower hemispheres encode exactly the overpass and underpass of [F1]. The sphere intersections in the standard-position condition are finite transverse surfaces and saddle disks by definition, so the stated crossing-ball convention is well-defined for a supplied diagram and Conway sphere.

The projection graph, its complementary regions, and the cyclic order of crossing visits along each component are finite data by [F1]. Thus connectedness, alternation, and the stated repeated-region condition are properties determined by the given diagram.

Forgetting all strand orientations preserves the projection graph and the over-under data, so each condition is unchanged by reversing component orientations. The definition therefore applies consistently to the underlying unoriented diagram.

Forgetting the component orientations does not change the point set $L$ or the complementary regions of a sphere disjoint from it. Thus the split predicate depends only on the underlying unoriented link type, as required by [F1].

Ambient isotopy carries any witnessing sphere and the two components of its complement to a sphere and two complementary components for the image link. The existence or nonexistence of a witness is therefore an ambient-isotopy invariant.

At each crossing the four sectors alternate in color, so there is one unordered pair of opposite black sectors and one unordered pair of opposite white sectors. The crossing therefore determines one edge for each graph, with endpoints the region vertices named in the definition; if both sectors belong to one region, the edge is a loop.

The crossings form a finite set by [F1], so each construction adds finitely many edges to the finite set of region vertices. The definition allows repeated endpoint pairs and equal endpoints, so it also covers parallel edges and loops without extra choices.

Fix a point $x_0$ in a complementary region $R_0$. First decompose the edges of $P$ into finitely many edge-disjoint simple cycles. In a nonempty remaining even-degree graph, follow unused edges until a vertex first repeats; the segment between its two occurrences is a simple cycle, including a loop edge as a one-edge cycle. Removing that cycle subtracts two incident half-edges at each of its vertices and leaves every degree even. Repeat until no edge remains. This terminates because each removal deletes an edge; a vertex-free circle is already one cycle. Now choose two piecewise-smooth paths $\alpha ,\beta$ from $x_0$ to the same point in a region $R$, avoiding the finitely many vertices and meeting open edges transversely in finitely many points. Their concatenation $\gamma =\alpha \ast {\beta }^{-1}$ is a closed test path; it need not be embedded or be an edge-path of $P$. Its intersections are counted with traversal multiplicity, so $I(\gamma ,P)=I(\alpha ,P)+I(\beta ,P)$ modulo two. For each simple graph cycle, choose small vertex disks disjoint from the compact image of $\gamma$ and round that cycle inside them. The rounded cycle remains embedded and has exactly the same test-path intersections as its original edges. By [F2] it separates $S^2$ into two sides. Each transverse encounter switches the side occupied by $\gamma$, and a complete traversal returns to its initial side, even when the test path self-intersects or retraces an arc. Thus $\gamma$ meets each rounded cycle evenly. The graph cycles partition all open edges and $\gamma$ misses their vertices, so summing their intersection counts proves $I(\gamma ,P)=0$ modulo two. Therefore $\alpha$ and $\beta$ have the same crossing parity. Coloring $R$ by this parity gives a well-defined function, independent also of the chosen endpoint inside $R$, since two such endpoints are joined within $R$ without crossing $P$. All decompositions, rounding choices and perturbations are finite; no choice axiom is used.

Crossing one edge changes the parity by one, so adjacent regions receive different colors. The dual adjacency graph of regions is connected because a generic path between any two regions crosses finitely many edges; hence fixing the color of $R_0$ forces every other color. The two choices for the color of $R_0$ give exactly two colorings.

Choose one connected component $P_0$ of $P$. Around each of its finitely many vertices take a small closed disk, and around each edge take a narrow closed strip joining the incident vertex disks; for a crossingless circle take an annular neighborhood. Round the finitely many corners. Because the graph is finite and its distinct components are disjoint compact sets, the disks and strips can be chosen so that their union $N$ is a compact connected subsurface of $S^2$ with smooth finite boundary, contains $P_0$ in its interior, and misses every other component of $P$.

Choose points $p_0\in P_0$ and $p_1$ in another component of $P$. Choose a piecewise-smooth path from $p_0$ to $p_1$ transverse to the finite smooth boundary $\mathrm{\partial }N$. Its endpoints lie respectively in the interior of $N$ and outside $N$, so it crosses $\mathrm{\partial }N$ an odd number of times. The boundary has finitely many disjoint simple closed components; at least one, call it $\gamma$, is met an odd number of times. The curve $\gamma$ lies in $S^2\setminus P$, since $N$ contains $P_0$ in its interior and misses every other component. By [F2], every transverse crossing of $\gamma$ switches sides, so $p_0$ and $p_1$ lie on opposite sides of $\gamma$.

The construction uses finitely many neighborhoods and applies also when $P_0$ is a crossingless circle; therefore it proves the asserted separation for every nonempty disconnected regular projection. Its chosen separating curve is a smooth boundary component of $N$.

Use source coordinates $(u,v)$ with their standard positive orientation and positive projection coordinates $(t,x)$ on the braid box. For $h>0$, the map $(t,x)=(v,-2hu)$ has determinant $2h>0$. It sends the source over direction $(1,1)$ to $(1,-2h)$ and the source under direction $(-1,1)$ to $(1,2h)$, whose determinant is $4h>0$. The source left/right pairs have constant $u$, hence constant braid position $x$ and give the through pairing $I$; the source top/bottom pairs have constant $v$, hence constant braid time $t$ and give the cup-cap $e_i$. The displayed page coordinates $(x,t)$ reverse the positive projection orientation; their visual quarter-turn is not the orientation comparison. The two terms in [F3] are therefore $AI$ and $A^{-1}{e}_i$.

The inverse half twist reverses which diagonal is over. Reflection of the local crossing interchanges the two pairings while retaining the bracket rule that assigns coefficient $A$ to the chart's $A$-smoothing and $A^{-1}$ to its $B$-smoothing. Consequently the through pairing now has coefficient $A^{-1}$ and the cap-cup pairing has coefficient $A$.

Associate to each crossing the edge of a chosen Tait graph. Along any projection edge joining crossings, the region of the chosen color on one side is the same at both ends, so the two corresponding Tait edges share that region vertex. Since the projection graph is connected, a path between any two crossings gives a sequence of Tait edges in which consecutive edges share a vertex. Every region vertex is incident to an edge because every face boundary of a connected projection with crossings contains a crossing. Therefore the Tait graph is connected. The same argument applies to the other color.

The only excluded connected projection is a crossingless circle, which is not covered by this statement; it will be treated directly in the bracket-span proof. This closes the claimed case.

By [F1], the projection graph has finitely many components and the link has compact image. Choose a point of $S^2\setminus P$ as infinity and work in the resulting finite projection plane. Apply the regular-neighborhood construction in the proof of [F3] in this chart; its smooth separating curve $\gamma$ misses the projection and has diagram points on both sides. Choose the bounded closed side as the disk $\mathrm{\Delta }$, using [F4].

Regard $S^2$ as the projection sphere in $S^3$. The product $\mathrm{\Delta }\times [-M,M]$ is a smoothly embedded 3-ball after rounding its corners, where $M$ exceeds the absolute height of every point of the compact link. Its rounded boundary consists of the side annulus over $\gamma$ and the two caps at heights $\pm M$; it is a sphere. Since $\gamma$ misses the projection and the caps lie above and below the link, this sphere misses the link. The two sides meet the link because $\mathrm{\Delta }$ and its complementary disk each contain projection points, so [F2] says that the link is split.

Linking can be computed here using finite rational chains: a rational $1$-cycle $a$ in $S^3$ bounds a rational $2$-chain $C$ by [F3], and $\mathrm{lk}(a,b)=C\cdot b$ for a disjoint rational cycle $b$. Independence of $C$ follows because the difference of two choices bounds a $3$-chain by [F3], whose intersection with the closed cycle $b$ has total boundary zero. In particular linking is zero if $a$ bounds a chain disjoint from $b$, and is unchanged under a homology of $b$ in the complement of $a$. All chains involved are finite. Curves and surfaces used below can be put in finite general position: subdivide their compact parameter domains finely in finitely many coordinate neighborhoods, approximate by polygonal patches within a tubular neighborhood, perturb the finitely many vertices to avoid the finitely many nontransverse affine incidences, and smooth the patches in smaller neighborhoods. Approximation sufficiently close to the original embedded patches preserves embedding and disjointness from the compact boundary $L$. These operations use finitely many choices, and no simultaneous choices indexed by an infinite family.

We record elementary finite-surface facts needed below. Cutting the supplied presentation along a finite family of curves gives finite triangulations of its pieces. A connected triangulated surface with boundary has a graph spine: in the dual graph include a vertex for the boundary, choose a finite spanning tree rooted there, and remove triangles successively across their edge facing the already removed parent. The remaining $1$-complex is a deformation retract, and reversing these elementary collapses shows that the surface is a regular neighborhood of that graph. Consequently its rational first homology vanishes exactly when that graph is a tree, in which case its regular neighborhood is a disk, by successively removing leaf edges. Also, a two-sided simple closed curve $\gamma$ in the interior of $F$ with $[\gamma ]=0$ bounds an orientable subsurface $F_0$ having $\gamma$ as its only boundary. To see this, triangulate with $\gamma$ as an edge cycle and take a rational $2$-chain bounding it using [F4]. Across edges outside $\gamma$ its coefficients agree with orientation signs. They vanish on the component touching $\mathrm{\partial }F$ and jump by one across $\gamma$, so the nonzero component supplies precisely such an oriented $F_0$. The same coefficient argument shows that $H_1(F_0;\mathbb{Q})\to H_1(F;\mathbb{Q})$ is injective: any bounding chain for a cycle in $F_0$ has zero coefficients on the connected complementary surface, which touches $\mathrm{\partial }F$.

Suppose $D$ is a compressing disk for $F$: its interior misses $F$ and its boundary $\gamma$ does not bound a disk in $F$. Its boundary is two-sided in $F$, since a collar of the disk at $\gamma$ gives a side of the normal interval bundle there. The normal transfer of a sufficiently close parallel copy of $\gamma$ misses $D$; in a collar of $\gamma$, move the parallel copy slightly in the direction tangent to $F$ and transverse to $\gamma$, and take its normal lifts still closer to that copy. Thus $\mathrm{lk}(\gamma ,\tau \gamma )=0$ by step 1.1. Definiteness forces $[\gamma ]=0$. Step 1.2 supplies the subsurface $F_0$ bounded by $\gamma$. It is not a disk, and hence $V=H_1(F_0;\mathbb{Q})$ is nonzero and injects into $H_1(F;\mathbb{Q})$. Cap $F_0$ with $D$ and smooth the corner, giving a closed connected orientable embedded surface $\mathrm{\Sigma }$. Capping the unique boundary circle does not change $H_1$, so $H_1(\mathrm{\Sigma };\mathbb{Q})=V$.

The surface $\mathrm{\Sigma }$ has a two-sided collar. Apply [F2] to that collar and $S^3\setminus \mathrm{\Sigma }$. Their overlap has two components, each retracting onto $\mathrm{\Sigma }$. By [F3], the map from the overlap's $H_0={\mathbb{Q}}^2$ into $\mathbb{Q}\oplus H_0(S^3\setminus \mathrm{\Sigma };\mathbb{Q})$ is injective and has cokernel $\mathbb{Q}$. Hence the complement has exactly two components, and the two collar sides belong to different ones. Let $A$ and $B$ be their closures. Thickening them to open sets, [F2] and [F3] give an isomorphism $V\to H_1(A;\mathbb{Q})\oplus H_1(B;\mathbb{Q})$. At least one projection has nonzero kernel: if both were injective, each target would have dimension at least $\dim V>0$, contradicting that their dimensions sum to $\dim V$. Thus a nonzero $a\in V$ bounds a rational $2$-chain on one side, say in $A$. Represent $a$ by an oriented rational multicurve in $\mathrm{int}{F}_0$. Both of its normal pushoffs have the same linking with it: the two normal directions are joined by a half-rotation in the normal plane of each oriented curve, giving a homology of the pushoffs in its complement. The pushoff into $\mathrm{int}B$ misses the bounding chain in $A$, so that linking is zero by step 1.1. The normal transfer therefore has linking zero as well. Step 1.2 makes $a$ nonzero in $H_1(F;\mathbb{Q})$, contradicting definiteness of $q$. Thus $F$ has no compressing disk.

By step 3.1, $F$ is incompressible. Suppose now that a splitting sphere $S$ exists. Make it transverse to $F$ by the finite perturbation of step 1.1, keeping it disjoint from $L$. Their intersection is a finite union of circles. At least one is essential in $F$: otherwise each bounds a disk in $F$, and removing the interiors of the outermost such disks leaves a connected surface containing every component of $\mathrm{\partial }F=L$, disjoint from $S$. Its boundary would lie entirely on one side of $S$, contrary to [F1]. Choose a circle $\gamma$ innermost on $S$ among the circles essential in $F$, and let $D$ be the disk it bounds on $S$ with no other essential circle in its interior.

Every circle of $\mathrm{int}D\cap F$ bounds a disk in $F$. If there are any, choose one $\beta$ whose disk $E\subset F$ is innermost among them. Its interior misses $D$: any intersection circle in that interior would itself be inessential in $F$ and would contradict the choice. The essential curve $\gamma$ cannot lie inside $E$. Replace the disk on $D$ bounded by $\beta$ by a small parallel copy of $E$, choosing the side that joins the remainder of $D$ and smoothing its boundary. Since $\mathrm{int}E$ misses $D$, this is again an embedded disk with boundary $\gamma$. It stays disjoint from $L$ and has strictly fewer intersection circles with $F$; the replacement disk is pushed off $F$ and all circles inside the discarded disk have disappeared. Repeating this finite reduction gives an embedded disk with boundary $\gamma$ and interior disjoint from $F$. This is a compressing disk, contrary to step 3.1. The splitting sphere cannot exist, so $L$ is nonsplit.

Let $\mu$ be the number of components represented by $D$. By [F8], $(-A^3{)}^{-\mathrm{wr}(D)}⟨D⟩{\mid }_{A=1}=(-2{)}^{\mu -1}\mathrm{\ne }0.$ The writhe-normalizing factor is a Laurent monomial and hence a unit, so $⟨D⟩$ is nonzero and its span is defined. Now let $s_A$ and $s_B$ be the numbers of circles in the all-$A$ and all-$B$ states. If a state has $j$ $B$-smoothings, then its weight factor is $A^{c-2j}$. A single smoothing change is a band surgery and changes the circle count by exactly $1$ or $-1$, so a state with $k$ circles has $k\le s_A+j$. Since the highest degree of ${\delta }^{k-1}$ is $2(k-1)$, every state term has degree at most $c-2j+2(k-1)\le c+2(s_A-1)$. Reversing $A$ and $B$ gives minimum degree at least $-c-2(s_B-1)$. Hence ${\mathrm{span}}_A⟨D⟩\le 2c+2(s_A+s_B-2)$.

If $c>0$, the connected projection has finite disk faces. To see this, thicken a spanning tree to a disk, then add the remaining edge strips one at a time. The core of each added edge lies in one complementary disk and is a properly embedded arc there. Its strip cuts that disk along the arc; after rounding corners, the two new boundary curves are smooth Jordan curves, and [F7] makes their closed sides disks. At each stage the new edge lies in one old region and splits it into exactly the two regions adjacent to the new edge, so these disk complements correspond one-to-one to the projection regions. Thus each complementary region has a disk core disjoint from the graph. Now suppose $D$ is reduced and alternating. Color its regions by [F3]. At one crossing choose the color name so the $A$-smoothing follows the two black sectors. Moving along a projection arc to the next crossing, alternation switches which branch is over; the chart definition of the $A$-smoothing therefore continues to follow the same color. Connectedness propagates this choice to every crossing. Thus the all-$A$ state circles trace the black disk cores and the all-$B$ circles trace the white disk cores. Embed the black Tait graph by placing vertices in black cores and drawing each crossing edge through its crossing disk; its complementary faces are precisely the white disk cores. Hence it is connected by [F6], has $c$ edges, $s_A$ vertices, and $s_B$ faces by [F4]. Euler's formula [F5] gives $s_A-c+s_B=2$.

Suppose $c>0$. Put the all-$A$ state on the lower level $S^2\times \{0\}$ and the all-$B$ state on the upper level $S^2\times \{1\}$. Outside the crossing disks, join corresponding state arcs by product rectangles; inside each crossing disk, join the two local smoothings by one saddle cobordism. The resulting compact surface $\mathrm{\Sigma }$ has boundary the $s_A+s_B$ state circles and exactly $c$ index-one saddles, hence $\chi (\mathrm{\Sigma })=-c$. Form a bipartite incidence graph with one vertex for each state circle and one for each saddle, joining a saddle to every state circle containing one of its attaching arcs. This graph is connected: each projection edge lies on one all-$A$ and one all-$B$ state circle, both incident to the saddles at its endpoints, and the projection is connected. Every state circle contains a projection edge, so it is incident to a saddle. Thus $\mathrm{\Sigma }$ is connected. Cap each boundary circle by a disk, obtaining a connected closed surface $\widehat{\mathrm{\Sigma }}$ with $\chi (\widehat{\mathrm{\Sigma }})=s_A+s_B-c$. Give it a finite cellulation and take cellular chains $C_i$ over ${\mathbb{F}}_2$, with boundary maps ${\mathrm{\partial }}_i$. If $r_i=\mathrm{rank}{\mathrm{\partial }}_i$ and $h_i=\dim H_i$, then rank-nullity gives $\dim C_i=h_i+r_i+r_{i+1}$; alternating the sum over $i=0,1,2$ cancels the $r_i$ terms and yields $\chi (\widehat{\mathrm{\Sigma }})=h_0-h_1+h_2$. Connectedness gives $h_0=1$. A 2-cycle has equal coefficients on the two faces adjacent to every edge; the dual graph of a connected closed surface is connected, so all face coefficients agree and $h_2=1$. Therefore $\chi (\widehat{\mathrm{\Sigma }})=2-h_1\le 2$, and $s_A+s_B-c\le 2$. For $c=0$, a connected projection is one circle and $s_A=s_B=1$, giving the same inequality. Combining this bound with step 1.1 proves ${\mathrm{span}}_A⟨D⟩\le 4c$.

At any crossing, reducedness in [F2] says the two black sectors belong to distinct black regions and the two white sectors belong to distinct white regions. The all-$A$ smoothing circles trace the black regions, so changing one crossing from all-$A$ joins two distinct state circles by a band and merges them; the all-$B$ change similarly merges two distinct white state circles. If a state changes $j\ge 1$ crossings from all-$A$, the first change lowers the circle count to $s_A-1$, and each later change can increase it by at most one; hence its circle count is at most $s_A+j-2$. Its term's degree is therefore at most $c-2j+2(s_A+j-3)=c+2(s_A-3)$, four below the all-$A$ term's top degree $c+2(s_A-1)$. The all-$A$ term has top coefficient $(-1{)}^{s_A-1}\mathrm{\ne }0$, so it cannot cancel. The same argument at all-$B$ proves a nonzero bottom coefficient. When $c=0$, the single state has bracket $1$, and the same conclusions hold vacuously.

Using $s_A+s_B=c+2$ from step 1.2, the highest degree minus the lowest degree is $(c+2(s_A-1))-(-c-2(s_B-1))=2c+2(s_A+s_B-2)=4c$. The extreme coefficients are nonzero by step 2.2, so this is the actual bracket span. The $c=0$ case has span $0=4c$.

Suppose first that there is at least one crossing. Build a small regular neighborhood of the projection graph from disjoint crossing disks and narrow edge strips. To see that its complementary faces are disks, choose a finite spanning tree of the graph; its neighborhood is a disk, by adding each tree strip between two distinct disk components. Add the remaining edge strips one at a time. Each edge core is a properly embedded arc in one complementary disk; its strip cuts that disk along the arc. After rounding corners the two new boundary curves are smooth Jordan curves, and [F4] makes both closed sides disks. At each step the added edge lies in one old face and splits it into the two new faces, so these complementary disks correspond one-to-one to the regions of the projection. Their closures are disjoint from the graph and serve as disk cores. Put a disk in each core of the chosen color. At every crossing join the two appropriate disk cores by a narrow band with one half-twist; choose its direction so that its boundary strands have the diagram's prescribed overpass and underpass. The neighborhoods are disjoint, so the resulting surface is embedded. Its boundary is the usual height realization of $D$. Collapsing each disk to a vertex and each band to its core retracts $F$ onto the Tait graph $G$. This graph is connected by [F2], so $F$ is connected and has a supplied finite disk-and-band presentation.

Measure the signed half-twist of a band relative to the planar framing of its two disk ends; call this incidence sign ${ϵ}_e$. This is the unoriented crossing incidence, not the oriented link crossing sign. At two crossings joined by a projection edge, alternation changes which branch is over, while the chosen-color sector changes sides along that edge. The two changes cancel in the half-twist chart, giving the same incidence sign at the two crossings. Since the projection graph is connected, every band has one common sign $ϵ$. Reversing the chosen orientation of a graph edge reverses its cycle coefficient, without changing this incidence sign.

We compute the normal pairing in the disk-and-band presentation. After clearing rational denominators, represent a cycle by finitely many oriented parallel strands through each band, joined by arcs in the vertex disks; any extraneous arc pairs can be canceled inside those disks. The signed number of strands through band $e$ is its graph coefficient $a_e$. Normal transfer gives two local pushoffs of each strand, exchanged at a one-sided passage. In an untwisted planar disk the contributions of the two normal sides to the mixed-crossing count cancel. In a band with a signed half-turn, each ordered pair of an original strand and a transferred strand contributes ${ϵ}_e$ to the total pairing: one normal side passes from one side of the original strand to the other during the half-turn, and the other normal side makes the complementary passage; the two mixed crossings have the same incidence sign, and the linking count is half their signed sum. Reversing either strand orientation reverses the corresponding contribution. Consequently the contribution of that band is ${ϵ}_e({\sum }_i{s}_i)({\sum }_j{s}_j)={ϵ}_e{a}_e^2$, where $s_i=\pm 1$ are the strand orientations. This is also a direct computation with the half-twisted strip $(u,v)\mapsto (u,v\cos (\pi u),v\sin (\pi u))$, reversing the half-turn for the other sign. The strips' end collars are included in this count; crossings of joining arcs in the planar disks cancel between the two normal sides. Linking here can be computed by the signed mixed-crossing count because a finite oriented smoothing of the first curve gives spanning disks and crossing bands, whose intersections with the second curve give that count. This finite construction uses the supplied drawing and does not use a link-diagram invariance theorem. Summing over all bands and restoring rational coefficients gives $\mathrm{lk}(c,\tau c)={\sum }_e{ϵ}_e{a}_e^2$.

By [F3] and the retraction in step 1.1, a homology class has a unique graph cycle vector. Thus $q(a)=ϵ{\sum }_e{a}_e^2$ is a well-defined quadratic form on $H_1(F;\mathbb{Q})$, and step 3.1 proves the required normal-transfer formula for every representative. If $a\mathrm{\ne }0$, some edge coefficient is nonzero, so ${\sum }_e{a}_e^2>0$. Therefore $q$ is positive definite if $ϵ=1$ and negative definite if $ϵ=-1$.

If the connected diagram has no crossings, it is a single circle. Either complementary disk is an embedded connected spanning disk with $H_1=0$. Every cycle in that disk bounds there and has zero linking with its normal transfer, so the zero-dimensional form has the asserted formula and definiteness. This also covers the zero-dimensional cycle space of a Tait tree in the preceding case.

If $D$ is empty, the represented link $L$ is empty, so no sphere has link points on both sides; $L$ is nonsplit by [F4]. Henceforth assume $D$ is nonempty. Apply [F2] to the supplied connected alternating diagram. The resulting surface $F$ is connected, has the represented link as its entire boundary, and carries the stated definite normal pairing. If the diagram has no crossings, [F1] identifies it as a single circle and [F2] uses a spanning disk, so the same supplied-surface conclusions hold.

The surface from step 1.1 satisfies every hypothesis of [F3], which therefore says that no sphere of the kind in [F4] splits its boundary $L$. Thus $L$ is nonsplit, with no reduction or choice assumption required.

If $D$ is empty, it has zero crossings. It is a regular diagram of $L$, so [F1] gives $0\le c(L)\le c(D)=0$, proving the claim in this case without evaluating a bracket. Henceforth $D$ and the represented link $L$ are nonempty.

Let $E$ be any regular diagram representing the same underlying unoriented link type as $L$. Since $L$ is nonempty, so is $E$. If its projection were disconnected, [F2] would make $L$ split, contradicting [F3] applied to $D$ and [F6]. Hence every competing diagram $E$ is connected. Orient each component of $E$ so an ambient isotopy to $L$ preserves its orientation.

By [F4], ${\mathrm{span}}_A⟨D⟩=4c(D)$ and ${\mathrm{span}}_A⟨E⟩\le 4c(E)$. The normalized bracket multiplies each raw bracket by a monomial, which shifts all exponents equally and preserves span. By [A1] and [F5], the normalized bracket values for the nonempty oriented diagrams $D$ and $E$ agree. Therefore $4c(D)\le 4c(E)$ and $c(D)\le c(E)$.

The diagram $D$ is one of the regular diagrams in the nonempty crossing-count set of [F1], so $c(L)\le c(D)$. Step 3.1 shows that $c(D)$ is no larger than any member of that set, and therefore $c(D)\le c(L)$. Combining the inequalities proves $c(D)=c(L)$.

Identify the two represented oriented links with one smooth link $L$ by the given isotopy. It is nonsplit by [F1]. Full AC supplies the choice hypotheses of [F2] and [F5]. The identification retains the actual oriented component correspondence to be used in all capped factors.

In either diagram choose a smooth simple circle meeting the diagram transversely in two ordinary points whenever both sides contain crossings. By [F7] its two sides are smooth disks, so cap each cut arc by a crossing-free arc in its cap disk, carrying the endpoint pairing and orientation. Both capped shadows are connected: an additional shadow component on either side would already have been a separate component of the original shadow. Their crossing sets partition that of the piece being cut. The capped diagrams are alternating. The one proper projected arc has an even number of crossing encounters: its self-crossings contribute two visits each, and its intersections with each closed projected component are even. For the latter, decompose the closed component's finite mod-two edge set into simple cycles. Each cycle separates by [F7], and the proper arc begins and ends on the disk boundary outside those cycles, so it crosses each cycle evenly. The first and last encounters on the proper arc therefore have opposite over-under status. Closing its ends introduces no new encounter.

If $L$ is empty, both diagrams are empty, and the identity sphere planar isotopy proves the conclusion. Henceforth assume $L$ is nonempty. If $L$ is the unknot, [F2] forces both diagrams to have zero crossings. A connected crossing-free diagram is one embedded oriented circle on the projection sphere, and the two such circles are carried to one another by a sphere planar isotopy with their orientations transported. This proves the conclusion in this case. Henceforth $L$ is nontrivial and both diagrams have positive crossing count.

The capped diagrams are reduced. Indeed, a nugatory separator through an old crossing in a capped diagram can be moved off the crossing-free cap disk: each part in that disk lies in one of the two complementary disks of its cap arc and can be pushed to the boundary collar. It would then give a nugatory separator through the same crossing in the original piece, contrary to reducedness. The cut and caps have smooth product collars. In the represented link they are the usual twice-punctured decomposition sphere and its standard trivial-arc cap balls.

Repeat these cuts in the capped pieces, retaining their cap-ball parameters and attachment collars. A later cut is moved off the interiors of earlier cap balls using their crossing-free trivial-arc product collars; equivalently shorten each artificial cap into its boundary collar before choosing the next cut. Lifting the resulting finite diagrammatic connected-sum decomposition gives pairwise disjoint decomposition spheres in the original link. Termination is finite: if a piece with $a+b$ crossings is replaced by two pieces with $a,b>0$ crossings, the sum of squared crossing counts falls by $2ab$. The final positive-crossing capped diagrams are therefore connected, reduced, alternating and have no two-point splitting circle with crossings on both sides. Crossing-free trivial connectors are suppressed while retaining all other attachment collars.

Every positive-crossing terminal factor is nontrivial by [F2]; otherwise its reduced alternating crossing count would be zero. It is nonsplit by [F1] and prime by [F4]. Thus the two finite systems just constructed, with their actual cap parameters, factor diagrams and oriented component labels, satisfy exactly the decomposition hypotheses of [F5].

Apply the comparison part of [F5] to those two systems in $L$. It provides the corresponding capped prime factors and actual oriented factor isotopies needed by its lift, including all refinements and suppressed trivial connectors. The correspondence transports the fixed global component names and attachment data from step 1.1.

For each corresponding pair of prime factor diagrams needed in that comparison, apply the imported theorem [F3] to the orientation-preserving pair map induced by its actual capped-factor isotopy. When an intermediate topological planar map $\varphi :S^2\to S^2$ carries one decorated diagram to the next, it gives a finite graph isomorphism preserving the cyclic order at every crossing, the over-under data, and the oriented strand directions. Realize that same graph isomorphism smoothly as follows. Take disjoint crossing disks and thin edge strips for the two connected projection graphs. On each crossing disk choose an orientation-preserving smooth map carrying the four branch germs to their matched germs; on each edge strip choose a smooth monotone arc map that agrees with these maps on smaller endpoint collars. The connected graph's regular neighborhood has disk complements: thicken a spanning tree to a disk, then add the other edge strips one at a time. Each new edge lies in one old disk face and cuts it along a proper arc; the two rounded boundary curves are smooth Jordan curves whose disk sides, by [F7], are exactly the two new complementary disks. By [F7] extend the matched boundary maps over these disk faces with product collars. Gluing produces an orientation-preserving smooth sphere diffeomorphism $g$ inducing exactly $\varphi$'s decorated graph map and component bijection. By [F8], $g$ is joined to the identity by a jointly smooth orientation-preserving isotopy. Thus each topological planar map supplied by [F3] is realized by a sphere planar isotopy in the definition, with the prescribed oriented component correspondence.

The lift in [F5] now applies with its exact marked prime input supplied by step 6.1. It realizes the attachment transfers and prime sequences by finitely many allowed flypes and sphere planar isotopies, giving $D_0$ to $D_1$ with the fixed oriented component correspondence. Together with the crossing-free case, this proves the stated theorem for connected reduced alternating diagrams, including composite links.

By [F1], the input is a connected reduced alternating diagram, so the finite state-merging conclusion in [F2] applies. It establishes both A-adequacy and B-adequacy without invoking the separate Jones axiom branch, and gives ${\mathrm{span}}_A⟨D⟩=4c$.

Suppose $D$ represents a knot $K$. Let $E$ be any $c^{\mathrm{\prime }}$-crossing diagram of the same knot, with its component oriented to represent the chosen orientation of $K$. The projection of a one-component diagram is connected because it is the continuous image of a circle; thus [F2] gives ${\mathrm{span}}_A⟨E⟩\le 4c^{\mathrm{\prime }}$. Multiplication by the writhe-normalizing monomial $(-A^3{)}^{-\mathrm{wr}(E)}$ shifts all exponents equally and does not change span. By [A1] and [F3], substituting $t=A^{-4}$ divides the exponent span by $4$, so ${\mathrm{span}}_t{V}_K(t)\le c^{\mathrm{\prime }}$. Applying this to $D$ and using step 1.1 gives ${\mathrm{span}}_t{V}_K(t)=c$.

By [A1] and [F1], the crossing-number definition [F2] applies. Since $D$ is a connected reduced alternating diagram by [F4], [F3] gives ${\mathrm{span}}_t{V}_K(t)=c(D)$.

For every regular diagram $E$ of the same underlying knot type, choose its single-component orientation to match $K$. By [F3], $c(D)\le c(E)$ because the common Jones span is $c(D)$ and is at most $c(E)$. The diagram $D$ itself is one of the competitors in [F2], so $c(K)\le c(D)$; the preceding inequality for every $E$ gives $c(D)\le c(K)$. Thus $c(K)=c(D)$, and step 1.1 gives the asserted Jones span.