Signed local cancellation models for the Khovanov Reidemeister moves

Meaning of every local map. Use rectangles with top boundary points $1,\dots ,n$ and bottom boundary points $1^{\mathrm{\prime }},\dots ,n^{\mathrm{\prime }}$, both listed left to right. Let $I$ join $a$ to $a^{\mathrm{\prime }}$. Let $E_i$ join top $i$ to $i+1$, bottom $i^{\mathrm{\prime }}$ to $(i+1{)}^{\mathrm{\prime }}$, and all other $a$ to $a^{\mathrm{\prime }}$. In a product $UV$, $U$ is above $V$ and equal middle labels are joined. Thus $E_i{E}_i$ is $E_i$ with one disjoint middle circle, and $E_1{E}_2{E}_1$ is boundary-fixing planar isotopic to $E_1$; likewise with $1,2$ interchanged. For the latter assertion, trace top $1,2$ to one cap, bottom $1^{\mathrm{\prime }},2^{\mathrm{\prime }}$ to one cup, and top $3$ to bottom $3^{\mathrm{\prime }}$ along the remaining arc; there is no closed component. Contracting the intervening bends gives the stated isotopy. Close any of these matchings by a fixed exterior resolution. Order the resulting circle factors by the least arc label on each circle, using a fixed total order on all local and exterior arcs; factor permutations have no Koszul signs. A dot on an arc means multiplication by $X$ on the circle containing that arc. A saddle replacing an occurrence of $I$ by $E_i$, or conversely, means precisely $m$ on the two ordered circle factors if they merge, and $\mathrm{\Delta }$ if they split, followed by the indicated factor permutation. All unmentioned factors have the identity map. This defines the maps below on every tensor basis $\{1,X{\}}^{\otimes k}$, even when several local arcs belong to the same global circle. In the based case use $M$ for the distinguished factor and [F2] for a saddle meeting it.

Circle coordinates and neck cutting. For a disjoint middle circle put its factor first and write $w=1\otimes w_1+X\otimes w_X$. If a saddle splits that circle from an arc carrying dot operator $x$, its map in these coordinates is $u\mapsto (xu,u)$; if it merges the circle into an arc carrying dot operator $y$, its map is $(w_1,w_X)\mapsto w_1+y{w}_X$. These follow from $\mathrm{\Delta }(a)=1\otimes Xa+X\otimes a$ and $m(a\otimes b)=ab$. Two successive opposite saddles between the same two local arcs equal the sum of a dot on each arc. If the arcs are on distinct circles, this is the identity $\mathrm{\Delta }m(a\otimes b)=Xa\otimes b+a\otimes Xb$: on $1\otimes 1,1\otimes X,X\otimes 1,X\otimes X$, both sides are $X\otimes 1+1\otimes X,X\otimes X,X\otimes X,0$. If they are on one circle it is $m\mathrm{\Delta }(a)=2Xa$. The same formulas with the distinguished factor reduced modulo $X$ prove the assertion in that case too. These are the exact tensor interpretations of the local neck-cutting identities, including when two dot positions are on the same circle.

An explicit cancellation rule. Suppose a differential between consecutive degrees has block matrix $\left(\begin{array}{cc}\varphi & \beta \\ \gamma & \delta \end{array}\right):A_0\oplus R\to B_0\oplus S$, with $\varphi$ an isomorphism. Set $L(a,r)=(a+{\varphi }^{-1}\beta r,r)$ and $T(b,s)=(b,s-\gamma {\varphi }^{-1}b)$. These invertible integral changes of coordinates give $Td{L}^{-1}=\mathrm{diag}(\varphi ,\delta -\gamma {\varphi }^{-1}\beta )$. Conjugate the adjacent differentials too. The equation $d^2=0$ and invertibility of $\varphi$ imply that the incoming differential has zero $A_0$ coordinate and the outgoing differential vanishes on $B_0$. Thus there is a direct summand $A_0\stackrel{\varphi }{\to }{B}_0$, contracted by $h{\mathrm{\mid }}_{B_0}={\varphi }^{-1}$, $h=0$ elsewhere. Inclusion and projection of the remaining coordinates, conjugated by $L,T$, are chain maps $i,p$ with $pi=1$ and $1-ip=dh+hd$. This supplies actual chain maps and a homotopy for every cancellation used below, rather than merely an acyclicity assertion. If the blocks are homogeneous, these maps have quantum degree zero and $h$ has cohomological degree $-1$.

RII: all four modules and signs. Take two strands and order the crossing with states $(I,E)$ first, the opposite crossing with states $(E,I)$ second. The four states $00,01,10,11$ are $U=E$, $V=I$, $W=E$ with a middle circle, and $Z=E$, respectively. The local differential is $d^0(u)=(su,xu,u),d^1(v,w_1,w_X)=tv-w_1-y{w}_X,$ where $s:E\to I$ and $t:I\to E$ are the two saddles, $x$ is the dot on the arc from which the middle circle splits, and $y$ is the dot on the arc into which it merges. Here $C^1=V\oplus W_1\oplus W_X$; the minus signs on the $10\to 11$ edge are exactly the cube sign for the second crossing. Step 2.1 gives $ts=x+y$, so these displayed differentials square to zero. Define $i(v)=(v,tv,0),p(v,w_1,w_X)=v-s{w}_X,$ with $i,p$ zero in other local degrees, and define $h^1(v,w_1,w_X)=w_X,h^2(z)=(0,-z,0)\mathrm{.}$ Then $di=0$, $pd=0$, and $pi=1$. In degree zero $hd=1$, in degree two $dh=1$, and in degree one $dh+hd$ sends $(v,w_1,w_X)$ to $(s{w}_X,w_1-tv+(x+y)w_X,w_X)$, which is exactly $1-ip$. The surviving $V$ has local degree one and quantum grading increased by one; in the shift conventions of the question this is $[-1]\{-1\}$. Thus both acyclic pairs and the signed quotient are explicitly identified. The other crossing order follows by swapping the two cube coordinates with state sign $(-1{)}^{v_1{v}_2}$.

RIII: ordered states and the first cancellation. Let each crossing have states $(I,E_i)$, and form the word $121$ from top to bottom, with crossings ordered in that order. Write $P=E_1{E}_2$ and $Q=E_2{E}_1$. In degrees $0,1,2,3$ the ordered states are $I;(E_1^{100},E_2^{010},E_1^{001});(P^{110},(E_1\bigsqcup ◯{)}^{101},Q^{011});E_1^{111}\mathrm{.}$ In the last term use the boundary-fixing isotopy $E_1{E}_2{E}_1\cong E_1$ of step 1.1. Denote a degree-one element by $(a,b,c)$ and the circle coordinates in state $101$ by $(w_1,w_X)$. Introduce four precisely located saddles: $u:E_1\to P$ inserts $E_2$ below $E_1$, $v:E_2\to P$ inserts $E_1$ above $E_2$, $w:E_1\to Q$ inserts $E_2$ above $E_1$, and $z:E_2\to Q$ inserts $E_1$ below $E_2$. Let $x$ and $y$ be the dot operators on the bottom cup and top cap of $E_1$, respectively. The degree-one differential is $d^1(a,b,c)=(-ua+vb,-xa+yc,-a+c,wc-zb),$ with output coordinates $(P,W_1,W_X,Q)$. Indeed the six cube edges are $100\to 110$ with sign $-$, $010\to 110$ with sign $+$, $100\to 101$ with sign $-$, $001\to 101$ with sign $+$, $010\to 011$ with sign $-$, and $001\to 011$ with sign $+$. The two edges to $101$ split the middle circle from the bottom cup or top cap, giving exactly the two displayed circle-coordinate pairs by step 2.1. The $101\to 111$ edge has sign $-$ and merges that middle circle into the remaining through arc of $E_1{E}_2{E}_1$. Its $W_1\to E_1^{111}$ component is therefore $-1$, with its $W_X$ component minus the dot on that through arc. Cancel this $-1$ component by step 2.2. There are no remaining degree-three terms; the differential into $(P,W_X,Q)$ is the corresponding projection of the displayed $d^1$.

The second cancellation and the common residual matrix. In the resulting complex, the $c\to W_X$ component is $+1$. Cancel it by step 2.2: its row is $-a+c$, so the residual coordinate substitution is $c=a$. The retained degree-one coordinates are $(a,b)$, the retained degree-two coordinates are $(P,Q)$, and the differentials become $I\stackrel{\left(\genfrac{}{}{0px}{}{s_1}{s_2}\right)}{\to }{E}_1\oplus E_2\stackrel{\left(\begin{array}{cc}-u& v\\ w& -z\end{array}\right)}{\to }P\oplus Q,$ where $s_i:I\to E_i$ is the saddle inserting $E_i$. The first map is obtained from the original $d^0=(s_1,s_2,s_1)$ by deleting the canceled coordinate; it is compatible with $c=a$. Repeat both cancellations for $212$ with $1$ and $2$ interchanged. Its retained coordinates before reordering are $(E_2,E_1)$ and $(Q,P)$, with matrix $\left(\begin{array}{cc}-z& w\\ v& -u\end{array}\right)$. Swapping both pairs, without a sign, gives exactly the preceding matrix and the preceding $d^0$. The four entries refer to the same saddles on the same labelled boundary matchings: for instance $E_2\to P$ inserts $E_1$ above, and $E_1\to P$ inserts $E_2$ below, on both sides. Hence these are actual equal tensor maps by step 1.1, not identifications of unlabelled spaces. Their kernels, composites and signs need no further $\tau$ identities. Composing the two explicit cancellation maps and their inverses gives the RIII equivalence; the homotopy for a composite of retractions $(i_1,p_1,h_1)$ and $(i_2,p_2,h_2)$ is $h_1+i_1{h}_2{p}_1$.

Quantum degrees. A state of local height $r$ has quantum degree increased by $r$. In RII, $W_1$ has the underlying grading of $E$ increased by two, $W_X$ has the grading of $E$, and $Z$ has the grading of $E$ increased by two. Thus both canceled isomorphisms and the maps of step 3.1 preserve quantum degree. In RIII, $W_1^{101}$ and $E_1^{111}$ both have the grading of $E_1$ increased by three; $W_X^{101}$ and $E_1^{001}$ both have the grading of $E_1$ increased by one. The common residual complex has quantum increases $0,1,2$ in its three degrees. Every change of coordinates in step 2.2 is therefore homogeneous.

Exterior differentials and the marked quotient. Dots commute with every exterior saddle: $m(Xa,b)=Xm(a,b)=m(a,Xb)$ and $\mathrm{\Delta }(Xa)=(X\otimes 1)\mathrm{\Delta }(a)=(1\otimes X)\mathrm{\Delta }(a)$. Two saddles in disjoint local disks commute before cube signs, including when their participating arcs lie on a common global circle, by associativity, coassociativity and the Frobenius identity in [F1]. A middle circle being delooped is contained in the local disk, so no exterior saddle meets it. Boundary-fixing isotopies of the matchings preserve the labelled exterior arcs. All matrix changes, inverses of $\pm 1$ blocks, projections, inclusions and contractions above therefore commute with unsigned exterior saddles and with $X$ at a fixed exterior marked point. The total differential, with local crossings first, is $d_{\mathrm{l}\mathrm{o}\mathrm{c}}+(-1{)}^r{d}_{\mathrm{e}\mathrm{x}\mathrm{t}}$ on local degree $r$. A local homotopy lowers $r$ by one, so its two exterior terms in $dh+hd$ cancel; local degree-zero maps commute with the signed total differential. These identities prove the same retractions for the total cube. The identities descend modulo $\mathrm{im}{X}_p$ because every map and every homotopy commutes with $X_p$; this argument does not assume that every local saddle acts trivially on the distinguished interval. It proves all assertions, integrally and choice-free.