Epsilon is stable-equivalence invariant but not a homomorphism

In the finite-rank free Maslov-graded knot Floer category, [F4] proves the stable-equivalence conclusion directly. Its finite endomorphism matrices strictify homotopy idempotents and give a unique non-acyclic indecomposable core up to filtered homotopy. The rank-one vertical homology of each original knot complex forces its internal acyclic complementary summands to have acyclic vertical complexes. Consequently its vertical threshold and the zero/nonzero patterns of $F_{\tau },G_{\tau }$ are precisely those of that core. Stable equivalence identifies the two cores even if the externally added acyclic summands have nonzero vertical or associated-graded homology. Thus the three-case epsilon values in [F1] agree. This argument does not apply the epsilon definition to the stabilized complexes themselves.

First reduce a finite based Laurent complex by explicit filtered contractions. If a differential coefficient preserves both coordinates, homogeneity makes it a single Laurent monomial; translate its target so that it is $1$. Write $da=b+r$ with $a,b$ at the same bifiltration and replace $b$ by $b^{\mathrm{\prime }}=da$. This is a filtered homogeneous basis change with inverse $b=b^{\mathrm{\prime }}+r$. For every other basis element $c$, subtract the coefficient of $b^{\mathrm{\prime }}$ in $dc$ times $a$ from $c$; this is filtered since the relevant copy of $a$ has the coordinates of the corresponding copy of $b^{\mathrm{\prime }}$. The new differentials contain no $b^{\mathrm{\prime }}$, and their $a$ coefficients vanish by $d^2=0$. Thus the pair $a\to b^{\mathrm{\prime }}$ splits off. Inclusion $I$, projection $P$ and $H(b^{\mathrm{\prime }})=a$, zero on all other generators, satisfy $PI=1$ and $IP+1=dH+Hd$, with $H$ preserving both filtrations and of Maslov degree $+1$. Finitely many repetitions give a reduced complex with no equal-bifiltration arrow. Tensoring these identities with another complex preserves them: the extra differential terms cancel over ${\mathbb{F}}_2$. Dualizing likewise preserves them. This supplies all reduction inverse maps and homotopies used below, including their tensor compatibility.

We next admit the actual complexes used for the geometric examples. Use the following marking-cell arrays, rows ordered bottom to top: for positive $T_{2,5}$ take $n=7$, $O=[6,5,4,3,2,1,0]$, $X=[4,3,2,1,0,6,5]$; for positive $T_{4,5}$ take $n=9$, $O=[8,7,6,5,4,3,2,1,0]$, $X=[4,3,2,1,0,8,7,6,5]$. The first has five positive crossings and oriented over/under word $A^{+}{B}^{-}{C}^{+}{D}^{-}{E}^{+}{A}^{-}{B}^{+}{C}^{-}{D}^{+}{E}^{-}$, where superscripts indicate over/under and all crossing signs are positive. The explicit positive torus curve $(r{e}^{2it},s{e}^{5it})$, $r=s=1\mathrm{/}\sqrt{2}$, has this identical code in stereographic projection: its planar angle is $2t$, radius $r\mathrm{/}(1-s\sin 5t)$, height $s\cos 5t\mathrm{/}(1-s\sin 5t)$, and crossing parameters $t\mathrm{/}(2\pi )=0,1\mathrm{/}10,\dots ,9\mathrm{/}10$. Equal projections force parameter difference $\pi$ and $\sin 5t=0$. At the overparameter the radial derivative is positive and at the underpartner negative, so the oriented crossing determinant is positive. The angular derivative is nonzero, and there are no triple points. Thus [F6] admits this grid as the actual positive $T_{2,5}$.

Verify the trefoil grid in [F6] directly. Use row indices $0,1,2,3,4$ from bottom to top and column indices from left to right, as in Toroidal grid diagrams and gradings for oriented knots, and draw the standard planar grid realization using the in-square mark-to-mark intervals. Following the grid orientation gives the single cycle $O_{0,4}\to X_{0,2}\to O_{2,2}\to X_{2,0}\to O_{4,0}\to X_{4,3}\to O_{1,3}\to X_{1,1}\to O_{3,1}\to X_{3,4}\to O_{0,4}\mathrm{.}$ The horizontal and vertical intervals meet away from their endpoints at exactly three crossings: $c_1=(\text{row }1,\text{column }2)$, $c_2=(\text{row }2,\text{column }1)$, and $c_3=(\text{row }3,\text{column }3)$. Traversing the component from $O_{0,4}$ encounters $c_1$ over, $c_2$ under, $c_3$ over, $c_1$ under, $c_2$ over, and $c_3$ under. At $c_1$ and $c_2$ the oriented projected tangents $(v_{\mathrm{o}\mathrm{v}\mathrm{e}\mathrm{r}},v_{\mathrm{u}\mathrm{n}\mathrm{d}\mathrm{e}\mathrm{r}})$ are $((0,1),(-1,0))$; at $c_3$ they are $((0,-1),(1,0))$. Each determinant is $+1$ by Planar-diagram, Gauss, and signed Dowker–Thistlethwaite codes. Thus its signed oriented Gauss code is $1_O,2_U,3_O,1_U,2_O,3_U$ with crossing-sign vector $(+1,+1,+1)$. This code is also obtained by traversing the closure of the positive two-strand braid ${\sigma }_1^3$: its first braid strand encounters the three positive half twists over, under, over, and the returning strand encounters them under, over, under. The signed visit data determine the same crossing rotation and strand code, hence identify the face-permutation cycles. Retain the grid diagram's outer-face cycle and choose the corresponding face of the braid-closure diagram. To make this chosen face outer, compactify the projection plane to the oriented sphere $S^2$ and choose a point $q$ in that face, away from the diagram. An orientation-preserving rotation of $S^2$ sending $q$ to the projection point at infinity makes precisely this face outer. Realize the diagram lift in a product collar of the equatorial $S^2$ in $S^3$; the rotation extends by the same rotation on the first three coordinates and the identity on the normal coordinate. Its path of rotations is an orientation-preserving ambient isotopy of $S^3$ preserving the crossing heights and oriented strand data. The two resulting plane codes now have the same selected outer face, so Valid codes reconstruct the declared link diagram identifies their diagrams up to planar isotopy. Thus the original grid and braid closure represent the same oriented knot even if their initially chosen planar outer faces differ. The braid closure lies on the standard unknotted torus with slope $2\alpha +3\beta$ (equivalently the swapped slope $3\alpha +2\beta$ under the orientation-preserving torus-coordinate swap), so it is $T(2,3)$ by Coprime torus knot. The grid therefore represents the right-handed trefoil with its positive orientation, and its writhe is $+3$.

Let $s=\tau (C)$ and choose a Maslov-zero vertical cycle $z$ at $j\le s$ representing its distinguished class. Its corner part $z_c$ is nonzero, or the threshold would be smaller. If $F_s=0$, choose a Maslov-one northeast-hook chain $b$ with $d_{N}b=z_c$ and split it into strictly upper, corner and strictly right parts $b_{+},b_0,b_r$. The strictly upper part of $d_v{b}_{+}$ is zero, since the other parts cannot reach it. Hence $z^{\mathrm{\prime }}=z+d_v{b}_{+}$ is another distinguished vertical cycle supported at $j\le s$. Put $q=b_0+b_r$. Its horizontal boundary $h=d_{h}q$ has no strictly right component, and its corner is $z_c^{\mathrm{\prime }}$, by the corner equation of $d_{N}b=z_c$. Then $x=z^{\mathrm{\prime }}+h_{\mathrm{l}\mathrm{e}\mathrm{f}\mathrm{t}}$ is supported in $i\le 0,j\le s$, its vertical projection represents the distinguished class, and its horizontal projection is the boundary $h$. Conversely, if such an $x$ has horizontal part $d_{h}q$, project $q$ to the northeast hook: its strictly left part cannot reach the corner or right ray, and the remaining boundary is precisely $x_c$, with no upper component. Thus $F_s=0$. This proves the exact boundary-representative alternative, rather than assuming a simultaneous simplification theorem.

The dual vertical threshold is $-s$. Indeed, if $B$ is the vertical boundary space, the distinguished cycle $z$ is not in $B+V^{\le s-1}$, since such an expression would give a lower-level cycle representing it. Finite-dimensional basis extension gives a cocycle functional annihilating that subspace and taking $z$ to $1$; its dual filtration is at most $-s$. A dual cocycle at level at most $-s-1$ vanishes on every cycle at level $s$ and hence on the one-dimensional vertical homology, so it is a dual boundary. The finite perfect pairing identifies the dual of the northeast hook at $s$ with the southwest hook of $C^{\ast }$ at $-s$. The transpose of $F_s(C)$ is $G_{-s}(C^{\ast })$, and the transpose of $G_s(C)$ is $F_{-s}(C^{\ast })$: quotient/inclusion maps reverse under this pairing. Duality of finite-dimensional homology preserves zero/nonzero. Thus $ϵ(C^{\ast })=-ϵ(C)$; [F2] identifies this with $ϵ(-K)=-ϵ(K)$.

Following oriented strands in the nine-grid gives fifteen positive crossings and the full signed rotation/strand code of the closure of the four-strand positive word $121321321121321$, after a cyclic starting-point change and reversal of the knot orientation. This is an explicit finite code equality: enumerate crossings in each row and column interval, then follow each row $O\to X$ and column $X\to O$; on the braid follow the strand permutation through each listed crossing, taking the top-right strand over to bottom-left. The exact words and crossing correspondence are recorded in the finite certificate named in Remarks. Commutations $13=31$ and braid relations $121=212$, $232=323$ give the successive words $123121321121321$, $123123121121321$, $123123121212321$, $123123121213231$, $123123121231231$, $123123121231213$, $123123121232123$, $123123121323123$, $123123123123123$. Each is a local disjoint-strand isotopy or oriented R3, so their closures are equivalent. The last is $(123{)}^5$. To identify its torus embedding, cut the Clifford curve $(r{e}^{4it},s{e}^{5it})$ by the disk coordinate $\arg (z_1)=\theta$: its four disk points have arguments $5(\theta +2\pi k)\mathrm{/}4$. A generic phase makes one quarter-turn's braid $132$, which becomes $321$ by moving the first generator past the closure cut, and then $123$ by reversing the four disk positions via the half-turn conjugation. Five quarter-turns give the last closure. Finally simultaneous complex conjugation is an ambient SO(4) isotopy endpoint and sends this curve to its orientation reverse: rotate its two imaginary axes through $\pi$. Thus the code reversal preserves oriented knot type, and [F6] proves the actual positive $T_{4,5}$ admission.

For the cable take $n=10$, $O=[8,9,7,6,4,5,3,2,1,0]$, $X=[4,5,2,3,0,1,8,9,6,7]$. Here is an explicit preferred-framing admission independent of any Floer calculation. Double each row and column of the admitted positive trefoil five-grid in [F6]. At an old corner let $h,v\in \{\pm 1\}$ be the directions toward its horizontal and vertical partner. A diagonal two-mark block for $hv=1$ and an antidiagonal block for $hv=-1$ give exactly two crossing-free parallel corners; checking the four quadrants verifies this and the oriented attachments. Only at the $O$ corner $(\text{row }0,\text{column }4)$, where $h=-1,v=1$, use the opposite diagonal block. It introduces one crossing of sign $hv=-1$. The listed ten-grid is precisely these ten corner blocks, and its component cycle visits rows $0,4,9,2,7,1,5,8,3,6,0$, hence it is one knot. Away from the exceptional block it follows two blackboard parallel copies of the trefoil. The blackboard longitude has linking number equal to writhe $+3$, so ${\lambda }_{\mathrm{b}\mathrm{b}}={\lambda }_{\mathrm{p}\mathrm{r}\mathrm{e}\mathrm{f}}+3\mu$. The one negative half-twist therefore gives slope $2{\lambda }_{\mathrm{b}\mathrm{b}}-\mu =2{\lambda }_{\mathrm{p}\mathrm{r}\mathrm{e}\mathrm{f}}+5\mu$, with longitudinal orientation inherited from the original trefoil. This is exactly $T_{2,3;2,5}$ in the Statement; it is not identified from a polynomial or a candidate matrix.

If $G_s\mathrm{\ne }0$, choose a Maslov-zero southwest-hook cycle mapping to the distinguished vertical class. In the reduced complex it is $x=z+\mathrm{\ell }$, with $z$ on the lower vertical ray, including the corner, and $\mathrm{\ell }$ strictly left at $j=s$. The two retained boundary equations are exactly $d_{v}z=0$ and $d_h(z_c+\mathrm{\ell })=0$: a lower ray cannot reach the other ray, and the corner has no equal-bifiltration differential. Conversely these two equations give a southwest-hook cycle mapping to the vertical class. Therefore $G_s\mathrm{\ne }0$ is exactly the existence of this simultaneous vertical/horizontal cycle representative. If also $F_s\mathrm{\ne }0$, its horizontal cycle is not a boundary, by step 2.1. Each constructed $x$ extends to a homogeneous filtered Laurent basis: choose a nonzero original corner coordinate $e$, whose coefficient is $1$, and replace $e$ by $x$. Homogeneity excludes any other power of that same basis element, and all other terms are no higher in either filtration, so the inverse is the same filtered elementary substitution. These are the needed distinguished representative normal forms. A chosen horizontal noncycle alone is not used as an epsilon test.

Compute the tilde groups of these three new grids by the following finite exact procedure. Enumerate the $n!$ state permutations $p$. For each ordered pair $a\mathrm{\ne }b$, take the positive cyclic row interval $a\to b$ and column interval $p(a)\to p(b)$. Its initial corners are southwest and northeast, and the two row orders give the two directed rectangles for each unordered pair. Retain the rectangle exactly when its cell support contains no $O$ or $X$, and no state point lies in the product of its strict interior intervals. Its target swaps $p(a),p(b)$; equal targets cancel over ${\mathbb{F}}_2$. Compute each state's degrees directly from $M_O=I(p,p)+I(O,O)-I(p,O)-I(O,p)+1$ and $A=(M_O-M_X-(n-1))\mathrm{/}2$. In a column, use the largest target permutation index as pivot. If that pivot has an existing stored column, take symmetric difference with it; otherwise store the column there. A replacement is an invertible field-column operation and strictly lowers the pivot, so distinct stored pivots give precisely the rank. For each grade, homology dimension is basis count minus outgoing rank minus incoming rank. The completed exact computation over all $5040$, $362880$, $3628800$ states respectively gives tilde dimensions $320$, $1792$, $2560$ and the graded polynomials recorded in Remarks. The implementation and full pivot checkpoints are finite arithmetic certificates of this explicit calculation; its square-zero input is the already supplied grid differential, not an unproved reconstruction theorem.

Normalize Laurent bases at $i=0$. The vertical slice of $C\otimes D$ is the ordinary field tensor of the vertical slices, with differential $d_v\otimes 1+1\otimes d_v$; no term lowering algebraic level can survive there. Similarly horizontal slices at levels $s,t$ tensor to the horizontal slice at $s+t$. Splitting a finite field complex into boundary pairs and homology representatives proves the field Kunneth assertion directly: a boundary pair tensored with anything is contractible by its pair homotopy. In particular the tensor of distinguished vertical cycles represents the distinguished tensor class. Its threshold is $s+t$ by [F3]. All representatives used in steps 2.1, 3.1 are supported southwest, so their vertical and horizontal projections tensor exactly to the projections of their tensor.

By [F5], divide only the tilde HOMOLOGY dimension polynomial by $(1+m^{-1}{a}^{-1}{)}^{n-1}$. Its quotient is unique by successively subtracting the largest-Maslov nonzero monomial times this factor. For $T_{2,5}$ the normalized hat degrees, each of rank one, are $(M,A)=(0,2),(-1,1),(-2,0),(-3,-1),(-4,-2)$; for $T_{4,5}$ they are $(0,6),(-1,5),(-2,2),(-5,0),(-6,-2),(-11,-5),(-12,-6)$; for the cable they are $(0,4),(-1,3),(-2,0),(-7,-3),(-8,-4)$. Now reduce the actual singly pointed CFK-infinity model supplied by [F5] using the contractions in step 1.2. Its Laurent basis at $i=0$ has exactly these reduced hat groups: every canceled equal-bifiltration pair is a graded boundary pair, and the reduced corner differential is zero. This step does not cancel $V$ in a chain category or tensor homology through a nonflat quotient.

If both epsilon values are $1$, choose the boundary representatives of step 2.1. Both horizontal parts are cycles; if $x_h=d_{h}a$, then $(x\otimes y{)}_h=d_h(a\otimes y_h)$. Its vertical part is the distinguished tensor class, so the converse in step 2.1 gives $F_{s+t}=0$, and epsilon of the tensor is $1$. The same proof works when one factor has epsilon zero and the other has epsilon $1$, using a cycle representative from step 3.1 for the zero factor. If both factors have epsilon zero, tensor their southwest-hook lifts to obtain $G_{s+t}\mathrm{\ne }0$. Their duals also have epsilon zero by step 2.2; applying the same lift construction to the dual tensor gives its $G_{-(s+t)}\mathrm{\ne }0$, hence $F_{s+t}\mathrm{\ne }0$ on the original tensor. Its epsilon is therefore zero. Dualizing the already proved positive cases proves the negative/negative and zero/negative cases. Together with [F3], these are exactly both tensor rules in the Statement, with actual representatives and compatible homotopies.

These exact degree profiles determine the full differential; no general inference from hat ranks to a full staircase is needed. List their generators in the displayed order as $e_0,e_1,\dots$. A possible term $U^k{e}_j$ in $d{e}_i$ must satisfy $k=(M_j-M_i+1)\mathrm{/}2\in {\mathbb{Z}}_{\ge 0}$ and $A_j-k\le A_i$. Exhausting these inequalities gives the following COMPLETE lists, with $v(i,j)$ denoting coefficient $1$ and $h(i,j;k)$ coefficient $U^k$ preserving horizontal level. For $T_{2,5}$: $v(0,1),v(1,2),v(2,3),v(3,4)$ and $h(1,0;1),h(2,1;1),h(3,2;1),h(4,3;1)$. For $T_{4,5}$: $v(0,1),v(1,2),v(3,4),v(5,6)$ and $h(1,0;1),h(3,2;2),h(5,4;3),h(6,5;1)$. For the cable: $v(0,1),v(1,2),v(3,4)$ and $h(1,0;1),h(3,2;3),h(4,3;1)$. Every allowed arrow is axial; there is no diagonal term. Homogeneity fixes each monomial, and its remaining coefficient is zero or one.

The normalized vertical homology in [F2] is one-dimensional in Maslov degree zero. Thus the only degree-zero candidate $e_0$ survives and $v(0,1)=0$. In the first list, absence of degree $-1$ homology forces $v(1,2)=1$, square-zero forces $v(2,3)=0$, and absence of degree $-3$ homology forces $v(3,4)=1$. In the other lists the remaining disjoint vertical pairs must all be present. At horizontal level zero use generators $U^{A_i}{e}_i$, whose Maslov degrees are $M_i-2A_i$, the reverse displayed vertical list. The same normalized argument from the opposite end forces every odd-to-previous-even horizontal coefficient to be $1$ and all listed reverse arrows to be zero. Hence the ACTUAL reduced complexes are $T_{2,5}:d{e}_1=U{e}_0+e_2$, $d{e}_3=U{e}_2+e_4$; $T_{4,5}:d{e}_1=U{e}_0+e_2$, $d{e}_3=U^2{e}_2+e_4$, $d{e}_5=U^3{e}_4+e_6$; cable: $d{e}_1=U{e}_0+e_2$, $d{e}_3=U^3{e}_2+e_4$. All even differentials vanish, so these matrices square to zero. The reductions already supply filtered inverse maps and homotopies. This proves the three geometric staircase admissions with all scalar factors handled, without any L-space or unproved cabling/reconstruction theorem.

For the right-handed trefoil, reuse the geometric five-grid admission in [F6] and perform the new, previously missing $120$-state tilde-rank operation of step 3.2. Its tilde dimension is $48$; division of its graded homology polynomial by the supplied $V^4$ factor gives degrees $(0,1),(-1,0),(-2,-1)$, each rank one. The inequalities in step 5.2 allow only $v(0,1),v(1,2),h(1,0;1),h(2,1;1)$. Normalized vertical and horizontal degree-zero homology forces exactly $d{x}_1=U{x}_0+x_2$, with even differentials zero. At $s=\tau =1$, the northeast-hook boundary of $U^{-1}{x}_1$ is $x_0$, since $U^{-1}{x}_2$ lies below its retained horizontal ray. Thus $F_1=0$ and $ϵ(T_{2,3})=1$ for the actual admitted knot complex. Step 5.1 gives $ϵ(T_{2,3}\mathrm{\#}{T}_{2,3})=1$, whereas any integer homomorphism taking the trefoil to $1$ would take its double to $2$. Hence epsilon is not an integer homomorphism on the concordance group. This is failure of additivity under connected sum; it uses no missing theorem identifying concordant complexes.

Write $A,B,C$ for the actual staircases of $T_{2,5},T_{4,5},T_{2,3;2,5}$ just admitted, and let $a_i,b_i,c_i$ denote their bases, with $b_i$ now dual to the $T_{4,5}$ basis. The vertical thresholds are $2,-6,4$, so [F3] gives threshold zero for $E=A\otimes B^{\ast }\otimes C$. Its distinguished vertical class is $a_0{b}_0{c}_0$ at $(M,A)=(0,0)$. In its northeast hook the Maslov-one chain $q=U^{-1}{a}_0{b}_0{c}_1+a_0{b}_2{c}_1+U^{-1}{a}_1{b}_2{c}_2$ is supported at $(i,j)=(1,0),(0,3),(1,0)$. Its three boundaries, after discarding terms outside the hook, are respectively $a_0{b}_0{c}_0+a_0{b}_1{c}_1$, $a_0{b}_1{c}_1+a_0{b}_2{c}_2$, and $a_0{b}_2{c}_2$. Here $d{b}_0=U{b}_1$, $d{b}_2=b_1+U^2{b}_3$. The discarded terms from the first boundary have $(i,j)=(1,-3)$; those from the second have $i=-1$ or $-2$; those from the third have $(i,j)=(1,-1),(1,-3)$ or $i=-1$. Thus $d_{N}q=a_0{b}_0{c}_0$ exactly. Consequently $F_0(E)=0$ and $ϵ(E)=1$ by [F1]. The actual tensor and reverse-mirror identifications [F2]–[F3] now give precisely the knot $T_{2,5}\mathrm{\#}-T_{4,5}\mathrm{\#}{T}_{2,3;2,5}$ in the Statement, with $\tau =2-6+4=0$ and epsilon $1$. The previously completed $175$-generator calculation supplies the identical finite boundary certificate; it has not been rerun or used as a substitute for the geometric admissions.

The implication $ϵ=0\Rightarrow \tau =0$ uses the normalized horizontal homology supplied by [F2]. Set $s=\tau$ and use $G_s\mathrm{\ne }0$ to choose a Maslov-zero southwest-hook cycle $y=z+\mathrm{\ell }$ mapping to the normalized vertical class, where $z$ is on the lower vertical ray and $\mathrm{\ell }$ is strictly left at $j=s$. Homogeneity is available by projecting a preimage to Maslov degree zero. Its horizontal projection $h=z_c+\mathrm{\ell }$, with $z_c$ its corner part, is a horizontal cycle. If it were a horizontal boundary, write a bounding chain as $a_{+}+a_{-}$, splitting strictly right support from support at $i\le 0$. The right component of its boundary is zero; the corner is $z_c=(\mathrm{\partial }{a}_{+}{)}_c+(\mathrm{\partial }{a}_{-}{)}_c$. In the northeast hook, the chain $a_{+}+(a_{-}{)}_c$ then has boundary exactly $z_c$: the left part of $a_{-}$ cannot reach the corner, and differential from the corner toward the left is killed in that hook. But $F_s([z])$ is represented by $z_c$, so this would give $F_s=0$, contrary to $ϵ=0$. Thus $h$ is a nonzero horizontal homology class in Maslov degree zero. Multiplication by $U^s$ takes it to horizontal level zero and Maslov degree $-2s$. Actual normalized conjugation [F2] identifies that horizontal homology with the vertical homology generated in Maslov degree zero, giving $s=0$. The geometric conjugation and absolute normalization used here are supplied by [F2]; thus this implication no longer has a separate conjugation hold.