Crossing signs, writhe, and linking number

Let ${\gamma }_1,{\gamma }_2:S^1\to {\mathbb{R}}^2$ be the projected component maps. They are transverse at every mixed crossing by [F1]. The map $F:S^1\times S^1\to {\mathbb{R}}^2\times {\mathbb{R}}^2$, $F(s,t)=({\gamma }_1(s),{\gamma }_2(t))$, is transverse to the diagonal $\mathrm{\Delta }=\{(x,x):x\in {\mathbb{R}}^2\}$, and its inverse image of $\mathrm{\Delta }$ has one point for each mixed crossing because there are no triple points. Therefore $I_2(F,\mathrm{\Delta })$ is the number of mixed crossings modulo $2$. Homotope ${\gamma }_1$ linearly to a point $p$ with $\parallel p\parallel$ greater than ${\max }_t\parallel {\gamma }_2(t)\parallel$; the endpoint map misses $\mathrm{\Delta }$. By [F3], the intersection number is $0$, so the number of mixed crossings is even.

A type-I move contributes no mixed crossing because its new crossing is a self-crossing. Under type II, either both crossings are self-crossings or they are both mixed and have opposite signs by [F5], so $S(D)$ is unchanged. Under type III the three crossing signs and over/under data are preserved, so the mixed subset and its signed sum are unchanged. Planar isotopy creates no crossings and preserves the oriented crossing charts. Hence each move preserves $S(D)\mathrm{/}2$.

Each crossing sign is $+1$ or $-1$, both congruent to $1$ modulo $2$. Thus $S(D)$ has the same parity as the number of mixed crossings shown even in step 1.1; consequently $S(D)\mathrm{/}2$ is an integer. Swapping component labels leaves each mixed crossing unchanged, while reversing one strand orientation negates its tangent vector and hence the determinant in [F2]; reversing both vectors preserves the determinant. This proves symmetry and the orientation rules.

If two regular diagrams represent the same oriented link, [F4] gives a finite sequence of the moves and planar isotopies checked in step 1.2. By step 2.1 their half-sums are integers; step 1.2 preserves these values at every move, so the endpoint linking numbers agree. Countable Choice is used through [F3] for homotopy-invariant mod $2$ intersection and through [F4] for the finite-move reduction; the determinant calculations and the move-by-move checks are choice-free.

The component-count check terminates because the input declares a finite component set. It returns invalid-input when that cardinality is not two. Otherwise the finite crossing list has some length $N\in {\mathbb{N}}_0$ by [F1], and the algorithm performs exactly one update for each of those $N$ crossings with no recursive call or unbounded search. Hence the scan halts after exactly $N$ iterations.

In the accepted case, the update in [F2] adds precisely each summand of $S(D)$ and no self-crossing. Thus the returned value is $S(D)\mathrm{/}2$; [F3] shows this is an integer and is the linking number specified in the lemma. All arithmetic is a finite integer sum.

Put $X=S^3\setminus K$ and ${\mathrm{\Sigma }}^{\circ }=\mathrm{\Sigma }\setminus K$. Since $\mathrm{\Sigma }$ is compact in $S^3$, it is closed; hence ${\mathrm{\Sigma }}^{\circ }$ is a closed embedded oriented surface in $X$. Since $J$ is transverse to $\mathrm{\Sigma }$, its inverse image of $\mathrm{\Sigma }$ is a compact $0$-manifold and hence finite; choose a basepoint $x_0\in J\setminus \mathrm{\Sigma }$. Then $x_0\in X\setminus \mathrm{\Sigma }$. Every based loop class has a smooth representative: on a lift $t\in [0,1]$ with basepoint $0=1$, fix $0<a<1\mathrm{/}2$ and reparametrize by $r(t)=0$ on $[0,a]$, $r(t)=(t-a)\mathrm{/}(1-2a)$ on $[a,1-a]$, and $r(t)=1$ on $[1-a,1]$. The interpolation $(1-u)t+ur(t)$ is a based homotopy, and the reparametrized loop is constant on a closed arc about $x_0$. Apply [F5] relative to that arc, then [F4] to make the smooth representative transverse to ${\mathrm{\Sigma }}^{\circ }$ while keeping it fixed near the arc, which maps into $X\setminus {\mathrm{\Sigma }}^{\circ }$. For such a loop $C$, let $I_{\mathrm{\Sigma }}(C)$ be its oriented intersection number with ${\mathrm{\Sigma }}^{\circ }$, using the local sign that is $+1$ when $C$ crosses from the negative to the positive side under the ambient-induced coorientation. This is the local sign of [F3]: the positive normal followed by a positive basis of $\mathrm{\Sigma }$ gives the ambient orientation. For any based homotopy between transverse loops, reparametrize its interval variable by ${\chi }_a$ from [F7]; it then agrees with the smooth endpoint loops on endpoint collars. Extend it constantly at both ends to a continuous map $\widehat{H}:\mathbb{R}\times S^1\to X$. Put $A=((-\mathrm{\infty },a\mathrm{/}2]\cup [1-a\mathrm{/}2,\mathrm{\infty }))\times S^1$, a closed subset of the boundaryless smooth source $\mathbb{R}\times S^1$. The extension is smooth on an open neighbourhood of $A$, since there it is independent of the first variable and equals a smooth endpoint loop. Apply [F5] on this boundaryless source relative to $A$, then [F4] there relative to $A$ to make the resulting smooth map transverse to the closed submanifold ${\mathrm{\Sigma }}^{\circ }$. The second application is valid because the map is unchanged near the tails and its endpoint loops are transverse. Both operations leave a neighbourhood of $A$ fixed. Restrict the final map to $[0,1]\times S^1$; it and its boundary restrictions are transverse, so the boundary-preimage assertion in [F4] applies to this compact cylinder. In particular its intersection trace is a compact oriented one-manifold whose boundary consists exactly of the endpoint intersections. Equivalently [F3] gives equality of their signed counts, so $I_{\mathrm{\Sigma }}$ is homotopy invariant. Concatenation can be smoothed in a ball about $x_0$ disjoint from $\mathrm{\Sigma }$, so intersection counts add. Thus $I_{\mathrm{\Sigma }}$ is a group homomorphism ${\pi }_1(X,x_0)\to \mathbb{Z}$ and factors through $H_1(X;\mathbb{Z})$ by [F2].

Let $\lambda ={\mathrm{lk}}_D(J,K)$ be the algorithmic linking number of the given regular diagram. At each mixed crossing where $K$ is over $J$, let $ϵ$ be its sign and switch it so that $J$ passes over $K$. The crossing sign changes from $ϵ$ to $-ϵ$, so [F6] changes $\lambda$ by $-ϵ$. In local coordinates $(x,y,z)$ with ambient orientation $(x,y,z)$, straighten the overpassing $K$-arc to the $x$-axis in a small ball. For $ϵ=+1$, the underpassing $J$-arc runs in the positive $y$ direction; switching it above the $K$-arc makes the local loop new-arc minus old-arc run clockwise in the oriented normal $(y,z)$ disk. Its class in the punctured ball is therefore $-[\eta ]$, where $\eta$ is the positive meridian. For $ϵ=-1$ one tangent is reversed, so the local loop is the positive meridian. Thus in both cases $[J_{\mathrm{n}\mathrm{e}\mathrm{w}}]-[J_{\mathrm{o}\mathrm{l}\mathrm{d}}]=-ϵ[\eta ]$. The same local picture applies to any mixed crossing after an orientation-preserving change of coordinates.

Take a sufficiently small normal disk to $K$ at a boundary point and orient its normal plane so that a positive tangent to $K$ followed by the disk orientation gives the ambient orientation; its oriented boundary is the positive meridian $\eta$. In orientation-preserving coordinates $(u,v,w)$, the surface is locally $\{w=0,v\ge 0\}$ with orientation $({\mathrm{\partial }}_u,{\mathrm{\partial }}_v)$ and $K$ tangent to $+{\mathrm{\partial }}_u$. The disk is oriented $({\mathrm{\partial }}_v,{\mathrm{\partial }}_w)$, so its positively oriented boundary at $+{\mathrm{\partial }}_v$ travels in the $+{\mathrm{\partial }}_w$ direction, from the negative to the positive side of $\mathrm{\Sigma }$. It meets $\mathrm{\Sigma }$ once, hence $I_{\mathrm{\Sigma }}(\eta )=1$. By [F1], if $[J]=m[\eta ]$ in $H_1(X;\mathbb{Z})$, then $I_{\mathrm{\Sigma }}(J)=m$.

After finitely many switches, every mixed crossing has $J$ over $K$. The loop $J$ can then be translated in the viewing-normal direction until it lies in a half-space above the compact image of $K$: at each projected coincidence it starts above $K$ and remains above, and away from those finitely many coincidences the projections are disjoint. This isotopy stays in $X$, so the final class is zero because the half-space is contractible. Write ${\gamma }_J,{\gamma }_K$ for the final projected loops. Each remaining crossing sign is $\mathrm{sgn}\det ({\dot{\gamma }}_J,{\dot{\gamma }}_K)$. Orient $S_J^1\times S_K^1$ by the $J$ factor first, orient ${\mathbb{R}}^2\times {\mathbb{R}}^2$ by its coordinate blocks, and orient the diagonal $\mathrm{\Delta }=\{(x,x):x\in {\mathbb{R}}^2\}$ by $x\mapsto (x,x)$. By the local-sign definition [F3], the intersection sign of $F=({\gamma }_J,{\gamma }_K)$ with $\mathrm{\Delta }$ is $-\mathrm{sgn}\det ({\dot{\gamma }}_J,{\dot{\gamma }}_K)$: the determinant of the columns $({\dot{\gamma }}_J,0),(0,{\dot{\gamma }}_K),(e_1,e_1),(e_2,e_2)$ is $-\det ({\dot{\gamma }}_J,{\dot{\gamma }}_K)$ for a positive basis $(e_1,e_2)$ of the plane. Fix $a\in (0,1\mathrm{/}2)$ and replace the linear homotopy of ${\gamma }_J$ to a point $p$ outside the second projected loop by ${\gamma }_t=(1-{\chi }_a(t)){\gamma }_J+{\chi }_a(t)p$, using [F7]. The map $((t,s,r)\mapsto ({\gamma }_t(s),{\gamma }_K(r)))$ is constant in $t$ on endpoint collars, transverse to $\mathrm{\Delta }$ there at the initial end, and misses $\mathrm{\Delta }$ near the terminal end. Extend it constantly in $t$ to the boundaryless source $\mathbb{R}\times S_J^1\times S_K^1$. It is smooth across the former endpoints because of those constant collars. The set $A=((-\mathrm{\infty },a\mathrm{/}2]\cup [1-a\mathrm{/}2,\mathrm{\infty }))\times S_J^1\times S_K^1$ is closed, and the extended map is already transverse to the closed diagonal on a neighbourhood of $A$: at the initial tail transversality follows from the regular diagram, while on the terminal tail the image misses the diagonal. Apply [F4] on this boundaryless source fixing a neighbourhood of $A$. Restrict the resulting transverse map to $[0,1]\times S_J^1\times S_K^1$; its unchanged endpoint restrictions are transverse, and [F4] supplies a compact oriented one-dimensional preimage whose boundary is precisely the endpoint preimages. Thus [F3] applies to the restricted family. The endpoint map at $t=1$ misses $\mathrm{\Delta }$, so its intersection number is zero, the same is true at $t=0$, and hence the final mixed signed sum is zero. Thus both the final linking number and the final meridian coefficient are zero. Since step 1.2 shows their changes agree at every switch, the original $\lambda$ equals the original coefficient $m$.

Step 2.2 proves ${\mathrm{lk}}_D(J,K)=m$, and step 2.1 proves $I_{\mathrm{\Sigma }}(J)=m$. Since $\mathrm{lk}(J,K)$ was defined to be $m$, all three integers agree. The Axiom of Choice is used through Alexander duality in [F1] and relative transversality in [F4]; no surface-existence result is used.

Choose a tubular chart $\mathrm{\Phi }:\mathrm{\Omega }\to U$ as in [F1]. The graph map $x\mapsto (x,\epsilon s(x))$ is an embedding because its projection to the base $K$ is the identity; it is disjoint from the zero section because $s(x)\mathrm{\ne }0$. Since $\mathrm{\Phi }$ is a diffeomorphism on $\mathrm{\Omega }$, its image $K_{\epsilon s}$ is embedded and disjoint from $K$ whenever the graph is contained in $\mathrm{\Omega }$. The zero section is compact, so an open neighbourhood of it contains all sufficiently small scalar multiples of $s$ over $K$; hence such $\epsilon$ exist. The orientation of $K_{\epsilon s}$ is transported from the base orientation by its projection to $K$.

Fix $a\in (0,1\mathrm{/}2)$ and use the smooth plateau ${\chi }_a$ of [F6]. For two sufficiently small radii ${\epsilon }_0,{\epsilon }_1>0$ in the same chart, define $\epsilon (u)=(1-{\chi }_a(u)){\epsilon }_0+{\chi }_a(u){\epsilon }_1$. The family of embeddings of $P:=S^1\bigsqcup S^1$ whose first component is $K$ and whose second is $K_{\epsilon (u)s}$ is smooth, stays embedded and disjoint by step 1.1, and is constant on endpoint collars. Here $P$ is compact and boundaryless, so [F4] extends this family to an ambient isotopy; [F5] shows that the linking number is unchanged. If $s_v$ is a smooth homotopy through nonzero sections, compactness of $K\times [0,1]$ gives one small scalar $\epsilon$ for which all graphs lie in $\mathrm{\Omega }$. Replacing $v$ by ${\chi }_a(u)$ in the family $K\cup K_{\epsilon s_v}$ again gives a smooth family of embeddings of the same compact boundaryless source, stationary near its endpoints; [F4] and [F5] show that the linking number is independent of the representative section.

Let ${\mathrm{\Phi }}_0:{\mathrm{\Omega }}_0\to S^3$ and ${\mathrm{\Phi }}_1:{\mathrm{\Omega }}_1\to S^3$ be two normalized tubular charts. By [F2], choose a smooth bundle metric on $\nu K$. Since $K$ is compact and ${\mathrm{\Omega }}_0\cap {\mathrm{\Omega }}_1$ is an open neighbourhood of the zero section, finitely many bundle trivializations give a common radius $r>0$ such that the open metric disk bundle ${\mathrm{\Omega }}_r=\{v\in \nu K:\parallel v\parallel <r\}$ is contained in both domains. Restrict both charts to this same ${\mathrm{\Omega }}_r$. They now satisfy the common-domain hypothesis of [F3]. After its permitted shrinking, [F3] gives an ambient isotopy $H_t$ with $H_1\circ {\mathrm{\Phi }}_0={\mathrm{\Phi }}_1$ on a neighbourhood of the compact zero section. Choose $\epsilon >0$ small enough that the graph of $\epsilon s$ lies in this neighbourhood. Then $H_1$ carries the oriented pair formed from $K$ and its ${\mathrm{\Phi }}_0$-push-off to the corresponding pair for ${\mathrm{\Phi }}_1$, so [F5] makes their linking numbers equal. Steps 1.1–2.1 prove independence of chart, sufficiently small radius and representative section. Countable Choice is used through the metric and tubular-neighbourhood suppliers; the finite linking computations are choice-free.

The zero section of $\nu K$ is compact, so after shrinking the parameter interval there is a $\delta >0$ such that $\rho (t,\epsilon ):=\pi \mathrm{\Phi }(K(t),\epsilon [b_D(t)])$ is smooth on $S^1\times (-\delta ,\delta )$, with $\rho (t,0)=\gamma (t)$. The normalized-chart condition says that ${\mathrm{\partial }}_{\epsilon }\rho (t,0)$ represents $R{\widehat{T}}_{\gamma }(t)$ modulo the tangent line spanned by ${\gamma }^{\mathrm{\prime }}(t)$. Therefore ${\gamma }^{\mathrm{\prime }}(t)$ and ${\mathrm{\partial }}_{\epsilon }\rho (t,0)$ are linearly independent at every $t$. By [F4], $(t,\epsilon )\mapsto \rho (t,\epsilon )$ is locally injective near every $(t,0)$. Compactness of $S^1$ and [F7] give an open diagonal neighbourhood $V_{\mathrm{\Delta }}\subseteq S^1\times S^1$ and a uniform ${\epsilon }_0>0$ such that, for $0<\epsilon <{\epsilon }_0$, whenever $(s,t)\in V_{\mathrm{\Delta }}$, the points $\rho (s,0)$ and $\rho (t,\epsilon )$, and the points $\rho (s,\epsilon )$ and $\rho (t,\epsilon )$, lie in one common injectivity chart. Shrink $V_{\mathrm{\Delta }}$, if necessary, so that it contains no off-diagonal crossing pair; this is possible because there are finitely many such pairs. Hence there are no projected coincidences between the original and offset branches with close parameters, and $t\mapsto \rho (t,\epsilon )$ is locally injective. Choose disjoint small open neighbourhoods of the finitely many ordered crossing pairs of $\gamma$ outside $V_{\mathrm{\Delta }}$; by [F1], these are the only off-diagonal equalities $\gamma (s)=\gamma (t)$. The compact complement of $V_{\mathrm{\Delta }}$ and those crossing neighbourhoods contains no such equality, so $\parallel \gamma (s)-\gamma (t)\parallel$ has a positive minimum there. Smoothness gives uniform convergence $\rho (\cdot ,\epsilon )\to \gamma$, excluding original/offset and offset/offset projected coincidences on this compact complement for sufficiently small $\epsilon$. Thus every projected coincidence lies in one of the chosen crossing neighbourhoods.

Fix one crossing disk with disjoint parameter intervals around the over branch $\alpha$ and under branch $\beta$, and with no other branch entering. For $i,j\in \{0,1\}$, the map $G_{ij}(s,t,\epsilon ):=(\rho (s,i\epsilon )-\rho (t,j\epsilon ),\epsilon )$ has invertible derivative at the original crossing and $\epsilon =0$: its first two columns are ${\alpha }^{\mathrm{\prime }}(0)$ and $-{\beta }^{\mathrm{\prime }}(0)$, which are linearly independent by transversality, and its last coordinate is $\epsilon$. By [F4], each of the four branch pairs has exactly one transverse projected intersection near this crossing for every sufficiently small $\epsilon$. They are, respectively, the original crossing, the two mixed crossings, and the crossing of the two offset branches. The local injectivity from step 1.1 excludes intersections between an original branch and its own offset; it also implies that the four listed crossing points are distinct, since a coincidence between any two would place an original branch and its own offset at the same projected point. Thus there are no triple points in this crossing disk. The strict height order of $\alpha$ over $\beta$ at the original crossing persists for all four nearby crossings when $\epsilon$ is small. The projected tangent determinants are nonzero and vary continuously, so both mixed crossings have sign $\mathrm{sgn}(c)$ by [F2].

There are finitely many crossing disks, so choose $\epsilon$ below all local bounds from steps 1.1 and 2.1. By [F6], the spatial push-off is an embedded circle disjoint from $K$. The projection of $K\cup K_{\epsilon }$ has only the finitely many transverse double points described above and no triple points, so it is a regular diagram. Each original crossing disk contributes exactly two mixed crossings, both with the original sign, and step 1.1 excludes all other mixed crossings. Countable Choice is used only through [F6] for the tubular push-off; the finite projected-crossing argument uses no choice.

By [F3], the mixed crossing list for $K\cup K_{b_D}$ has two crossings of sign $\mathrm{sgn}(c)$ for each crossing $c$ of $D$ and no others. Therefore its signed mixed sum is $2{\sum }_c\mathrm{sgn}(c)=2\mathrm{wr}(D)$.

By [F2], the linking number of $K$ with its blackboard push-off is half this mixed sum; by [F1] it is the self-linking number. Thus $\mathrm{sl}(K,b_D)=\mathrm{wr}(D)$. For a link diagram $D$ and each component $K_i$, apply the same calculation to its regular component subdiagram with induced blackboard framing $b_{D,i}$; crossings with other components are not mixed crossings of that component and its push-off, so precisely its self-crossings contribute.

Apply [F1] to two regular diagrams of the same oriented link. By [F4], the integer linking number is unchanged by planar isotopy and by every move in the resulting finite sequence. Hence the two diagrams have equal linking number.

A type-I move creates or deletes exactly one self-crossing and no mixed crossing by [F2]. Its contribution to the writhe in [F3] is therefore the sign $+1$ or $-1$ of that crossing. A type-II move either contributes no mixed crossings or creates a mixed pair of opposite signs; in both cases its total writhe contribution is zero. A type-III move preserves each crossing sign, so it preserves writhe.

By [F5], the knot component's blackboard self-linking is equal to the writhe of its diagram before and after each move. Thus a type-I move changes that self-linking by the same $+1$ or $-1$ as writhe, while type II and III preserve it.

Let $D_0$ be a crossing-free diagram of the unknot, so $\mathrm{wr}(D_0)=0$. Add a positive type-I kink to obtain $D_1$. By [F1], the two diagrams represent ambient-isotopic unframed knots. The calculation in step 1.2 gives $\mathrm{wr}(D_1)=1$. Hence writhe is not an invariant of unframed ambient isotopy; step 2.1 gives the same change for the blackboard self-linking.