Epsilon and Upsilon concordance invariants

The quotient and inclusion constructions defining $F_s,G_s$ are chain maps between the stated subquotient complexes, since the differential cannot increase either coordinate. The vertical complex is finite-dimensional: translate each of the finitely many homogeneous Laurent basis generators to algebraic level zero. Its remaining basis has finitely many integer Alexander levels. Since its homology is one-dimensional by the Given data, some finite cycle represents its nonzero class. Below the least basis level the filtered subcomplex is zero, and above the greatest level it is the whole vertical complex. Thus the minimum defining $\tau (C)$ exists.

Let $s=\tau (C)$ and let $\xi$ be the nonzero vertical homology class, with a cycle representative supported at $j\le s$. If $F_s=0$, [F1] produces a southwest-hook cycle whose image under $G_s$ is $\xi$, so $G_s\mathrm{\ne }0$. Hence the two maps cannot both vanish. Their remaining zero/nonzero possibilities are exactly the three displayed cases, proving the algebraic trichotomy and the value set $\{-1,0,1\}$. This argument assumes no horizontal-representative equivalence, and supplies no model independence.

For each of the finitely many basis generators $x_r$, at most one Laurent translate has Maslov grading zero, since $M(x_r)-2k=0$ determines $k$ uniquely when it has an integer solution. Thus the Maslov-zero chain group is finite-dimensional over ${\mathbb{F}}_2$. A finite chain represents ${\xi }_K$, so ${\mathcal{Z}}_0(C)$ is nonempty; being a subset of a finite vector space, it is finite. The minimum defining ${\nu }_t$ therefore exists for every $t\in [0,2]$. The algebraic argument here does not establish knot-model independence; the current grid comparison theorem supplies the normalized comparison for its stated grid and analytic models.

If $f:C\to C^{\mathrm{\prime }}$ is a bifiltered chain homotopy equivalence, then it sends every Maslov-zero cycle representing ${\xi }_K$ to one representing the normalized generator of $C^{\mathrm{\prime }}$. Each term of $f(z)$ has both filtration coordinates no larger than the corresponding input term, so $F_t(f(z))\le F_t(z)$ for $0\le t\le 2$. Hence ${\nu }_t(C^{\mathrm{\prime }})\le {\nu }_t(C)$. Applying the same argument to a filtered inverse gives the reverse inequality, so ${\nu }_t(C^{\mathrm{\prime }})={\nu }_t(C)$.

For a fixed cycle support, $F_t(z)$ is the maximum of finitely many affine functions of $t$. Since ${\mathcal{Z}}_0(C)$ is finite, ${\nu }_t$ is the minimum of finitely many such finite maxima. A finite minimum or maximum of affine functions is continuous and piecewise linear with finitely many breakpoints. Therefore ${\mathrm{{\rm Y}}}_K(t)=-2{\nu }_t(C)$ is continuous and piecewise linear on $[0,2]$.

For a differential term $U^{k}y$ in $\mathrm{\partial }x$, the $i$-filtration gives $k\ge 0$ and the $j$-filtration gives $A(y)-A(x)\le k$. The Maslov grading gives $M(y)-M(x)+1=2k$, so its exponent in ${\mathrm{\partial }}_{t}x$ is ${\mathrm{gr}}_t(y)-{\mathrm{gr}}_t(x)+1=2k-t(A(y)-A(x))\ge (2-t)k\ge 0$. For a two-step path $x\to y\to z$, the two exponents sum to ${\mathrm{gr}}_t(z)-{\mathrm{gr}}_t(x)+2$, independent of $y$. Thus the coefficient of each $z$ in ${\mathrm{\partial }}_t^{2}x$ is the coefficient of $z$ in ${\mathrm{\partial }}^{2}x$ multiplied by the same power of $v$; since ${\mathrm{\partial }}^2=0$, ${\mathrm{\partial }}_t^2=0$.

The given annulus makes $K$ smoothly slice in exactly the sense of [F3]. Use [F2] to identify the input with its normalized finite homogeneous model. Apply [F3] at these fixed data, with the finite/countable specialization [F4], to obtain $C(K)\simeq \mathrm{\Lambda }\oplus D$ with all maps and homotopies preserving both filtrations and Maslov grading. Its displayed unknot factor has the same origins as the Statement; no Alexander translation or positive $U$-power is inserted.

Set $A=0$ and $B=D$. The zero complex is finite-rank free and totally acyclic, as is $D$ by [F3]. The equivalence in step1.1 is therefore the exact stable equivalence requested. No associated-graded acyclicity or filtration-preserving contraction of $D$ is claimed. This closes the original Statement by the explicit external slice-splitting input; the earlier locally attempted movie admission is not represented as a proof.

Put $\nu ={\nu }_t(C)$ and choose a Maslov-zero representing cycle $\alpha$ with $F_t(\alpha )=\nu$, using [F1]. Let $C_0,C_1$ be the finite-dimensional chain groups in Maslov degrees zero and one. Set $B_0=\mathrm{\partial }{C}_1$ and $L=\{z\in C_0:F_t(z)<\nu \}$, including zero. The maximum-filtration rule makes $L$ a subspace. One has $\alpha \mathrm{\notin }{B}_0+L$: otherwise $\alpha =b+\mathrm{\ell }$ would make $\mathrm{\ell }$ a cycle representing the same nonzero class at a smaller filtration. Finite-dimensional linear algebra therefore gives a functional $\lambda :C_0\to {\mathbb{F}}_2$ with $\lambda (\alpha )=1$ and $\lambda (B_0+L)=0$. Translate each homogeneous Laurent basis vector by the unique power of $U$ putting its Maslov degree in $\{0,1\}$. Define a $\mathrm{\Lambda }$-linear map $P:C\to \mathrm{\Lambda }\alpha$ by $P(x)=\lambda (x)\alpha$ on these degree-zero basis vectors and $P(y)=0$ on the degree-one ones. Boundaries from degree one are annihilated by $\lambda$, and the differential of a degree-zero vector lies in $U{C}_1$, on which $P$ is zero. Thus $P$ is a chain map and $P(\alpha )=\alpha$. If $P(x)\mathrm{\ne }0$, then $F_t(x)\ge \nu$ because $\lambda (L)=0$; the $U$-translation rule gives the same inequality for every translate. Hence $P$ preserves filtration.

Write $\alpha$ in the degree-zero basis and choose a term of largest filtration, which has filtration $\nu$. Replacing that basis vector by $\alpha$ is invertible and preserves filtration in both directions, since all other terms in $\alpha$ have filtration at most $\nu$. For every remaining degree-zero basis vector $x$, replace it by $x+\lambda (x)\alpha$. When the extra term is present, $F_t(x)\ge \nu$, so this and its inverse preserve filtration. The degree-one basis is unchanged. The resulting basis consists of $\alpha$ and a basis of $\ker P$. Since $P$ is a chain map, this is a filtered graded direct-sum decomposition $C=T\oplus A$, with $T=\mathrm{\Lambda }\alpha$ and $A=\ker P$. Its total homology on $T$ already maps isomorphically to $H(C)=\mathrm{\Lambda }$, so $H(A)=0$. This proves the first assertion by an explicit filtered projection, without assuming that an arbitrary unfiltered acyclic complex is filtered contractible.

Return to the original basis $x_r$ at algebraic level zero, and put $m_r=M(x_r)$ and $a_r=A(x_r)$. Its differential has coefficients $c_{rs}{U}^{(m_s-m_r+1)\mathrm{/}2}$ with $c_{rs}\in {\mathbb{F}}_2$. Extend scalars to $\mathcal{K}$ by $U=v^2$, and write $e_r=v^{m_r}{x}_r$. The differential is then $\mathrm{\partial }{e}_r=v{\sum }_s{c}_{rs}{e}_s$. Let $W$ be the finite ${\mathbb{F}}_2$-space on these formal $e_r$, with differential $d_0{e}_r={\sum }_s{c}_{rs}{e}_s$. It squares to zero, since the two-step coefficient in the original differential has the same $U$ exponent for every intermediate index. Separate $W$ by the parity of $m_r$. The even space identifies with $C_0$ by $e_r\mapsto U^{m_r\mathrm{/}2}{x}_r$, and the odd space with $C_1$ by $e_r\mapsto U^{(m_r-1)\mathrm{/}2}{x}_r$. The two matrices of $d_0$ are respectively those of $\mathrm{\partial }:C_1\to C_0$ and $U^{-1}\mathrm{\partial }:C_0\to C_1$. Consequently $H(W{)}_{\mathrm{e}\mathrm{v}\mathrm{e}\mathrm{n}}={\mathbb{F}}_2$ and $H(W{)}_{\mathrm{o}\mathrm{d}\mathrm{d}}=0$, by the given total graded homology. Finite vector-space splittings extend to $\mathcal{K}$; since $v$ is invertible, $\mathrm{\partial }=v{d}_0$ has one-dimensional homology over $\mathcal{K}$, detected by the same even homology functional.

Let $C_t$ be the exact $t$-modified polynomial-basis complex in the definition, and put $g_r=m_r-t{a}_r$. The $\mathcal{K}$-linear map $\mathrm{\Phi }(x_r)=v^{t{a}_r}{x}_r$ from its scalar extension to the scalar extension of $C$ is a grading-preserving chain isomorphism. Indeed, a differential term on the source has coefficient $v^{m_s-m_r+1-t(a_s-a_r)}$, and multiplying its target by $v^{t{a}_s}$ gives exactly $v^{t{a}_r}{v}^{m_s-m_r+1}$, the differential of the image. The weighted filtration on the latter scalar extension is defined by $F_t(v^{\beta }{x}_r)=t{a}_r\mathrm{/}2-\beta \mathrm{/}2$, extended to a series by its least nonzero exponent and to a finite basis expansion by the maximum. Thus the image of $C_t$ is exactly the lattice $F_t\le 0$: its $r$th coefficient has least exponent at least $t{a}_r$. This is the explicit coordinate change, and it uses $U=v^2$.

A cycle in $C_t$ is non-torsion in homology exactly when its class remains nonzero over $\mathcal{K}$. One direction follows by inverting a nonzero annihilating coefficient. Conversely, if the cycle becomes a boundary over $\mathcal{K}$, the bounding chain has finitely many coefficients; a single monomial $v^c$ with $c\ge 0$ shifts all their least exponents into $[0,\mathrm{\infty })$ by [F2]. Multiplying the bounding equation by $v^c$ gives an annihilator of the original class over $\mathcal{R}$. Now a homogeneous cycle of degree $g$ in $C_t$ has the form $z={\sum }_r{ϵ}_r{v}^{g_r-g}{x}_r$, with ${ϵ}_r\in {\mathbb{F}}_2$, all occurring exponents nonnegative. Its image is $v^{-g}{\sum }_r{ϵ}_r{e}_r$, so the vector $\zeta ={\sum }_r{ϵ}_r{e}_r$ is a $d_0$-cycle. If $z$ is non-torsion, step 3.1 shows that the even part of $\zeta$ represents the nonzero even homology class. Hence $z_0={\sum }_{m_r\text{ even}}{ϵ}_r{U}^{m_r\mathrm{/}2}{x}_r$ is an original Maslov-zero cycle representing the normalized generator. For every term in $z_0$, the nonnegative-exponent condition gives $g\le m_r-t{a}_r$, and therefore $F_t(z_0)\le -g\mathrm{/}2$. It follows that every homogeneous non-torsion degree satisfies $g\le -2\nu$.

Conversely write a minimizing original Maslov-zero cycle as $\alpha ={\sum }_{m_r\text{ even}}{ϵ}_r{U}^{m_r\mathrm{/}2}{x}_r$ and put $g=-2\nu$. Each exponent $m_r-t{a}_r-g$ in its support is nonnegative by the definition of $F_t(\alpha )=\nu$. Thus $z_t={\sum }_r{ϵ}_r{v}^{m_r-t{a}_r-g}{x}_r$ is a homogeneous cycle in $C_t$, since its image under step 4.1 is $v^{-g}\alpha$ and $\mathrm{\Phi }$ is injective. Its scalar-extended homology class is nonzero by step 3.1, so it is non-torsion by step 5.1. Its degree is $g=-2\nu$, attaining the upper bound. Finally the filtered projection and direct sum in steps 1.1–2.1 extend over $\mathcal{K}$ and restrict to the lattice $F_t\le 0$. On $T$ this lattice is $\mathcal{R}{v}^{2\nu }\alpha$, with zero differential and generator degree $-2\nu$; the complementary lattice has only torsion homology, since its scalar extension is acyclic. By step 4.1 this is precisely the non-torsion tower in the $t$-modified complex. This proves the complete tower/minimax identification, including irrational $t$ and both endpoints.

Every term in ${\mathrm{\partial }}_t^{2}x$ corresponding to a two-step path $x\to y\to z$ has exponent $({\mathrm{gr}}_t(y)-{\mathrm{gr}}_t(x)+1)+({\mathrm{gr}}_t(z)-{\mathrm{gr}}_t(y)+1)={\mathrm{gr}}_t(z)-{\mathrm{gr}}_t(x)+2$, which is independent of the intermediate generator $y$. Its coefficient is therefore the coefficient of $z$ in ${\mathrm{\partial }}^{2}x$, multiplied by that common power of $v$. Since ${\mathrm{\partial }}^2=0$, ${\mathrm{\partial }}_t^2=0$. For every differential term the exponent is nonnegative: writing its $U$-power as $k$, the Maslov relation gives $M(y)-M(x)+1=2k$, and the Alexander filtration gives $A(y)-A(x)\le k$, hence $2k-t(A(y)-A(x))\ge 0$ for $0\le t\le 2$. Thus $D_G^t$ is a chain complex over $\mathcal{R}$.

The quotient identifies finitely many marking variables in a complex with finitely many grid states, so $D_G$ has a finite polynomial basis. Its t-modification is a finite homogeneous complex over the coefficient ring proved in The real-exponent coefficient series form a valuation domain. The explicit finite matrix reduction of Finite homogeneous complexes over the real-series ring have finite cyclic splittings gives a finite sum of free homology generators and cyclic torsion summands, and proves that, if a free summand is present, its largest homogeneous non-torsion grading exists. Thus the asserted maximum is justified without a Noetherian assumption on $\mathcal{R}$. The current grid comparison [F4] supplies a nonzero free tower for this actual diagonal quotient.

The quotient convention and the symmetric extension for $t>1$ are part of the definition, not consequences of the comparison theorem. No grid-independence or agreement with the knot Floer Upsilon invariant is asserted here; those are theorems on this page.

A finite homogeneous generating set bounds the Maslov degrees of $D$, and hence of $H_{\ast }(D)$, above. Therefore the integer degrees of homogeneous non-torsion classes, if nonempty, have a maximum, so $\delta (D)$ is well-defined on the stated domain. Write $M_r=M(x_r)$ and $A_r=A(x_r)$. The terms of $\mathcal{A}(K,s)$ on the $r$th $U$-tower are $U^n{x}_r$ with $n\ge \max (0,A_r-s)$, so their Maslov gradings are at most $M_r-2\max (0,A_r-s)$. By [F1], this bound tends to $-\mathrm{\infty }$ as $s$ tends to $-\mathrm{\infty }$; since the basis is finite, choose $s_0$ so that it is strictly negative for every $r$ when $s\le s_0$. Then $H_{\ast }(\mathcal{A}(K,s))$ has no class of degree zero and $\delta (\mathcal{A}(K,s))\mathrm{\ne }0$. For all $s\ge {\max }_r{A}_r$, $\mathcal{A}(K,s)=C^{-}(K)$ and its normalized tower gives $\delta =0$. Thus the set defining ${\nu }^{-}$ is nonempty and bounded below, and its least element exists. This argument does not prove that the least element is nonnegative.

Put $R={\mathbb{F}}_2[U]$, $B=C^{-}(K)$ and $A_s=\mathcal{A}(K,s)$. The latter has finite free basis $U^{\max (0,A_r-s)}{x}_r$. Its actual inclusion $v_s:A_s\to B$ becomes the identity of $C(K)$ after localization. Therefore the kernel of its homology map is $U$-torsion and its image is a nonzero homogeneous ideal $U^{V_s}R$, with $V_s\ge 0$. A homogeneous polynomial-torsion class is $U$-torsion: the different powers of $U$ have distinct Maslov degrees, so a polynomial annihilator implies a monomial annihilator for that class. Consequently homogeneous non-torsion classes are exactly the homogeneous classes having nonzero image in the localized tower. Their maximal grading is the top grading of the image ideal, and hence $\delta (A_s)=-2V_s$.

Reuse the supplied Maslov-zero Laurent-linear analytic coordinate-swap equivalence $\mathrm{\Phi }:C(K)\to C(K)$, preserving the normalized total generator and swapping the two filtrations, with its swap-filtered inverse and homotopies. For $s\in \mathbb{Z}$, $h_s=\mathrm{\Phi }{U}^s:A_s\to B$ is an actual polynomial-target map: a point with $i\le 0,j\le s$ goes to $(j-s,i-s)$, whose first coordinate is at most zero. Its Maslov degree is $-2s$. It localizes to an isomorphism, so its homology image is $U^{H_s}R$, $H_s\ge 0$. The same non-torsion-domain calculation gives $-2H_s=\delta (A_s)-2s$, and thus $H_s=V_s+s$. This argument permits negative $s$ in the Laurent intermediate complex and proves that the final target lies in the polynomial subcomplex; it does not treat $U^s$ alone as an endomorphism of $B$ for negative $s$.

If $s<0$, $V_s=H_s-s\ge -s>0$. Therefore $\delta (A_s)\mathrm{\ne }0$ for every negative $s$. Together with step1.1 this proves existence of the threshold and ${\nu }^{-}(K)\ge 0$. A direct local alternative for the upper inequality would require an actual surface input. Sufficient would be an $R$-linear Maslov-zero map from the unknot tower to $B$, with image in $A_g$, sending the homology generator to the normalized generator1. Its image cycle would force $V_g=0$ and ${\nu }^{-}(K)\le g$. A map with value a positive $U$-power, a localized nonzero map, or a tau inequality does not suffice.

The finite free basis of $A_s$ in step1.2 is exactly the basis in the external Definition2.12, since its exponent condition is $n\ge \max (0,A_r-s)$, and the normalized top minus grading is zero in both conventions. Thus [F4] applies to the SAME threshold, not to a shifted plus invariant. It gives ${\nu }^{-}(K)\le g_4^{\mathrm{s}\mathrm{m}}(K)$ at the fixed data with [F5]. Combine this with step3.1 to obtain both original inequalities. The upper bound is explicitly external; the local exponent/nonnegativity argument is retained.

In the smooth concordance group, the annulus gives $[K]=[J]$, so $K\mathrm{\#}(-J)$ is smoothly slice. Also $J\mathrm{\#}(-J)$ is slice by the group inverse law.

Apply the exact stable form of [F5] to the slice knots $K\mathrm{\#}(-J)$ and $J\mathrm{\#}(-J)$. Write $C_L=CF{K}^{\mathrm{\infty }}(L)$. It gives acyclic complexes $A_0,B_0,A_1,B_1$ with $C_{K\mathrm{\#}(-J)}\oplus A_0\simeq \mathrm{\Lambda }\oplus B_0$ and $C_{J\mathrm{\#}(-J)}\oplus A_1\simeq \mathrm{\Lambda }\oplus B_1$. It does not supply an unstabilized direct-sum splitting of either slice complex, and none is assumed. These equivalences are supplied by the exact slice stable-splitting input [F5]. The supplied model and disk comparisons [F3]–[F4] do not replace the geometric slice input.

Tensor the first equivalence with $C_J$ and the second with $C_K$. Under the supplied geometric tensor comparison [F2], their three-factor terms are filtered-equivalent to the same complex $X=C_K\otimes C_{-J}\otimes C_J$, by associativity and the graded flip, which is sign-free over ${\mathbb{F}}_2$. Put $P=A_0\otimes C_J$, $Q=B_0\otimes C_J$, $P^{\mathrm{\prime }}=A_1\otimes C_K$ and $Q^{\mathrm{\prime }}=B_1\otimes C_K$. Then $X\oplus P\simeq C_J\oplus Q$ and $X\oplus P^{\mathrm{\prime }}\simeq C_K\oplus Q^{\mathrm{\prime }}$. Each of $P,Q,P^{\mathrm{\prime }},Q^{\mathrm{\prime }}$ has zero total homology. Indeed a finite free graded acyclic Laurent complex, after translating generators to degrees zero and one, has a finite acyclic parity matrix over ${\mathbb{F}}_2$. Splitting its cycles and boundaries in these finite vector spaces gives a contraction; inserting the required $U$ or $U^{-1}$ degree shifts gives a Laurent-linear contraction. Tensoring with any free complex is still a contraction, because the cross terms cancel in characteristic two. No filtration-preserving contraction is claimed or required.

Add $P^{\mathrm{\prime }}$ to the first equivalence of step 3.1 and $P$ to the second. The common middle term is $X\oplus P\oplus P^{\mathrm{\prime }}$, so composing and permuting direct summands gives $C_J\oplus (Q\oplus P^{\mathrm{\prime }})\simeq C_K\oplus (Q^{\mathrm{\prime }}\oplus P)$. All added factors are finite-rank free and totally acyclic by step 3.1, and every equivalence preserves both filtrations and Maslov grading. This proves exactly the original stable Statement, without strengthening the slice supplier or cancelling a common complex by an unjustified cancellation law.

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.