The tau invariant and four-ball genus

Put $t=\tau (C_K)$ and let $[x]$ be its nonzero total homology class in degree $n$. There is a cycle $x\in F_t{C}_{K,n}$ representing $[x]$. Since $H(F_{t-1}{C}_K)\to H(C_K)$ has zero image, every cycle in $F_{t-1}{C}_{K,n}$ is a boundary. Thus $x\mathrm{\notin }{B}_n+F_{t-1}{C}_{K,n}$, where $B_n$ is the subspace of boundaries. Extend a linear functional $\lambda$ that vanishes on $B_n+F_{t-1}{C}_{K,n}$ and satisfies $\lambda (x)=1$. It is a cocycle. For a finite-dimensional chain complex, $\ker ({\mathrm{\partial }}^{\vee })=B^{\perp }$ and $\mathrm{im}({\mathrm{\partial }}^{\vee })=Z^{\perp }$, so evaluation identifies $H(C_K^{\vee })=B^{\perp }\mathrm{/}{Z}^{\perp }$ with $H(C_K{)}^{\vee }$; hence $\lambda$ represents the nonzero dual class. It belongs to $F_{-t}{C}_K^{\vee }$. If $s<-t$, every element of $F_s{C}_K^{\vee }$ vanishes on $F_t{C}_K$, so its induced class pairs trivially with $[x]$ and maps to zero in total dual homology. Therefore $\tau (C_K^{\vee })=-t$.

Specify the corresponding reverse-mirror diagram as follows. Before applying $r$, exchange the handlebodies and attaching systems, keeping the oriented surface and the ordered marks: ${\mathcal{H}}^{\vee }=(\mathrm{\Sigma },\mathit{\beta },\mathit{\alpha },w,z)$. This diagram lies in $-S^3$. Its alpha bridge is the old beta bridge traversed from $w$ to $z$, and its beta bridge is the old alpha bridge traversed from $z$ to $w$, so its knot is $-K$. Applying $r:-S^3\to S^3$ therefore gives $K^{†}$; the compatible reflected Seifert surface has orientation $F^{†}=-r(F)$. This is the corresponding reflected diagram in the Statement, with the two handlebody labels exchanged, rather than surface reversal alone. Periodic domains and their $w$ coefficients are unchanged by attaching-system exchange, so admissibility is preserved. Work on ${\mathcal{H}}^{\vee }$ and push all data forward by $r$ at the end.

For an original strip $u(s,t)$ from $\mathbf{x}$ to $\mathbf{y}$, put $u^{\vee }(s,t)=u(-s,1-t)$ and $J_t^{\vee }=J_{1-t}$. The new boundary conditions are beta at $t=0$ and alpha at $t=1$, its ends are $\mathbf{y},\mathbf{x}$, and ${\mathrm{\partial }}_s{u}^{\vee }+J_t^{\vee }(u^{\vee }){\mathrm{\partial }}_t{u}^{\vee }=-({\mathrm{\partial }}_{s}u+J_t(u){\mathrm{\partial }}_{t}u)(-s,1-t)=0$. Precomposition is invertible and conjugates the linearized real operators, preserving their kernels, cokernels and indices. It reverses translation and hence gives a bijection on translation orbits, preserving regularity and mod-two counts. The domain map reverses both real coordinates, so it preserves orientation and every marked divisor degree: $n_w({\varphi }^{\vee })=n_w(\varphi )$ and $n_z({\varphi }^{\vee })=n_z(\varphi )$. In particular, it bijects the actual $n_w=0$ strips defining the total hat differentials, including those with $n_z>0$. Thus ${\mathrm{\partial }}^{\vee }{\mathbf{y}}^{\vee }$ contains ${\mathbf{x}}^{\vee }$ exactly when $\mathrm{\partial }\mathbf{x}$ contains $\mathbf{y}$, with the same coefficient. This proves the literal transpose identity for the transported regular data.

In the pointed Morse construction for ${\mathcal{H}}^{\vee }$, replace the compatible Morse function $f$ by $-f$ and negate the generator vector field, carrying its local trajectory modifications along with it. The ambient orientation and vector-field direction both reverse, so the oriented perpendicular plane, and hence its first Chern class, retain their physical orientations. Carry the prescribed zero-framed cap as well: its zero relative Euler number remains zero, and multiplying its boundary frame by a constant $-1$ changes no winding. The capped Seifert surface has the opposite orientation, so the absolute formula in [F1] gives $A_{F^{†}}({\mathbf{x}}^{\vee })=-A_F(\mathbf{x})$, with no additive shift. Pushforward by $r$ carries this pairing and the strip counts unchanged. Sending ${\mathbf{x}}^{\vee }$ to the dual basis functional is therefore a chain isomorphism by step 2.1. Its level is $-A_F(\mathbf{x})$, so its sublevel $s$ consists exactly of functionals annihilating the generators at levels at most $-s-1$, which is the displayed dual filtration. This proves the filtered chain isomorphism; step 1.1 gives its one-dimensional total homology and the tau sign formula. The literal isomorphism uses the corresponding transported diagram, whereas [F4] only states the homotopy-type comparison for general normalized models.

Write $t=\tau (C)$ and $u=\tau (D)$. For each degree $n$, let $B_n=\mathrm{im}({\mathrm{\partial }}_{n+1})$ and $Z_n=\ker ({\mathrm{\partial }}_n)$. Finite-dimensional linear algebra gives complements $H_n,L_n$ with $Z_n=B_n\oplus H_n$ and $C_n=Z_n\oplus L_n$; the differential maps $L_n$ isomorphically onto $B_{n-1}$. Thus each complex is the direct sum of homology representatives with contractible two-term complexes. If $h$ contracts one such summand $A$, then $h\otimes 1$ contracts $A\otimes E$: the two cross terms cancel by the graded tensor signs, while $(\mathrm{\partial }h+h\mathrm{\partial })\otimes 1=1$. Hence $H(C\otimes D)\cong H(C)\otimes H(D)\cong {\mathbb{F}}_2$. By [F1], choose cycles $x\in F_{t}C$ and $y\in F_{u}D$ representing the nonzero homology classes. Their tensor is a nonzero class in $H(C\otimes D)$ and lies in $F_{t+u}(C\otimes D)$, so $\tau (C\otimes D)\le t+u$.

Let $n$ be the homological degree of $[x]$. The image of $H(F_{t-1}C)$ in $H(C)$ is zero by minimality of $t$, so every cycle in $F_{t-1}{C}_n$ is a boundary in $C$. The cycle $x$ cannot lie in $B_n+F_{t-1}{C}_n$; otherwise $x=b+v$ would make $v=x-b$ a cycle in $F_{t-1}C$ representing the same nonzero class. Extend the linear functional that is zero on $B_n+F_{t-1}{C}_n$ and one on $x$ to a functional $\lambda :C_n\to {\mathbb{F}}_2$. It vanishes on boundaries, so is a cocycle. Construct $\mu$ on the homological degree of $y$ in $D$ in the same way. The tensor cocycle $\lambda \otimes \mu$ evaluates to one on $x\otimes y$ and vanishes on every $F_{t+u-1}(C\otimes D)$: each summand $F_{p}C\otimes F_{q}D$ with $p+q\le t+u-1$ has either $p<t$ or $q<u$. Therefore no class in that lower filtration maps to the nonzero total homology class, and $\tau (C\otimes D)\ge t+u$. Together with step 1.1 this proves equality.

By [F2], the total complex has one generator and zero differential, so its only nonzero homology class is represented by that generator.

By [F1] the generator has Alexander level $0$. For every $s<0$ its filtered subcomplex is zero, while at $s=0$ the inclusion is the identity on the one-dimensional homology. Thus the first level with nonzero image is $0$, which is the asserted value.

Apply the external connected-sum theorem [F4] to the chosen doubly pointed models for $(K,F_K)$ and $(J,F_J)$. Its $CF{K}^{\mathrm{\infty }}$ conclusion gives the required bifiltered chain homotopy equivalence, and its $CF{K}^0$ conclusion gives the finite hat equivalence for the boundary-connected-sum surface. The theorem's maps and inverse homotopies are over the stated coefficient ring; base change from $\mathbb{Z}$ to ${\mathbb{F}}_2$ preserves each chain-map and homotopy identity. The source's generator filtration is the sum of the input Alexander levels, so the induced increasing filtration is exactly ${\sum }_{p+q\le s}{F}_p{C}_K\otimes F_q{C}_J$.

A filtered chain homotopy equivalence between finite complexes with one-dimensional total homology preserves the threshold. Indeed, each filtered map commutes with the inclusion into total homology, and the induced total map is an isomorphism; a nonzero class from one filtered subcomplex therefore maps to a nonzero class from the corresponding subcomplex of the other. Applying this in both directions shows that the least such filtration level agrees.

The two total homology groups are ${\mathbb{F}}_2$, so their tensor has one-dimensional total homology. Apply the finite tensor-threshold formula [F3] to the equivalence in step 1.1, obtaining additivity for the specified diagramwise thresholds. Hom's external theorem [F5] identifies the unqualified $\tau (K)$ with the common standard knot invariant and states its additivity under connected sum. This gives the final invariant-level formula without treating any unverified local analytic factorization as a premise.

Choose a smooth embedded arc in $S$ joining its two boundary components. After orientation-preserving isotopies of the endpoint knots inside their product collars, which preserve their knot types, choose a standard reference arc with both endpoints at the same point of $S^3$; its tubular complement is $(S^3\setminus \mathrm{int}{D}^3)\times [0,1]$, a four-ball after rounding corners. Since $S^3\times [0,1]$ is simply connected, the chosen arc is homotopic rel endpoints to this reference. A generic one-parameter homotopy is an isotopy through embedded arcs: the double-point condition has parameter-domain dimension $1+1+1=3<4$, and the nonimmersion condition has dimension $1+1=2<4$. Extend its velocity field from the compact arc family to a time-dependent ambient vector field using a tubular neighborhood and a bump function; integrating this field gives an ambient isotopy carrying the arc neighborhood to the reference neighborhood. Delete that tubular neighborhood from the ambient product and delete the corresponding open rectangular strip from $S$. The remaining surface is connected and has one boundary component, the oriented sum $K_1\mathrm{\#}(-K_0)$ under [F2]. Since $\chi (S)=-2g$, cutting along this arc raises Euler characteristic by one to $1-2g$; the resulting surface has one boundary component, so its genus is $g$.

Apply Hom's imported genus bound [F3] to the surface from step 1.1: $\mathrm{\mid }\tau (K_1\mathrm{\#}(-K_0))\mathrm{\mid }\le g_4^{\mathrm{s}\mathrm{m}}(K_1\mathrm{\#}(-K_0))\le g$. Hom's homomorphism statement [F3] gives $\tau (K_1\mathrm{\#}(-K_0))=\tau (K_1)-\tau (K_0)$, proving the asserted inequality. For a concordance annulus $g=0$, so the endpoint values agree. If one endpoint is the standard unknot, [F4] gives $\tau (U)=0$, and the inequality is the four-ball genus estimate for the other endpoint.

The threshold in [F1] is Hom's standard $\tau$ definition: both use the normalized Alexander-filtered knot Floer complex and the first level whose inclusion maps nontrivially to total homology. The target is one-dimensional, so the nonzero-image and surjectivity formulations agree. Thus Hom's $\tau (K)$ is the common normalized threshold asserted here.

Hom's homomorphism theorem [F2] gives diagram independence, invariance under smooth concordance, and $\tau (K\mathrm{\#}J)=\tau (K)+\tau (J)$.

Hom's genus theorem [F3], with the identical genus convention [F4], gives $\mathrm{\mid }\tau (K)\mathrm{\mid }\le g_4^{\mathrm{s}\mathrm{m}}(K)$. This proves the remaining conclusion.