Smooth concordance implies stable filtered equivalence of knot Floer complexes

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.