Upsilon additivity, mirror symmetry, and four-ball genus bound

By [F1], each fixed Maslov-zero cycle has filtration equal to the maximum of finitely many affine functions of $t$, and there are finitely many such cycles. Thus ${\mathrm{{\rm Y}}}_K$ is piecewise linear with finitely many breakpoints. The supplied model comparison [F8] makes this calculation a knot invariant. At $t=0$, $F_0$ is the algebraic filtration; the normalization in [F1] says the tower generator first appears at level zero, so ${\nu }_0=0$ and ${\mathrm{{\rm Y}}}_K(0)=0$. Exchanging the two bifiltration coordinates sends $F_t$ to $F_{2-t}$; [F2] therefore gives ${\mathrm{{\rm Y}}}_K(t)={\mathrm{{\rm Y}}}_K(2-t)$.

Under the supplied geometric tensor comparison [F5], the connected-sum complex is the convolution-filtered Laurent tensor product. For a fixed $t$, apply the proved filtered splitting [F4] separately to the two factors. Their rank-one summands tensor to a zero-differential tower whose normalized generator has filtration ${\nu }_t(K)+{\nu }_t(J)$. Every other tensor summand contains an acyclic Laurent factor and is acyclic. To justify that last assertion without a filtered contraction assumption, normalize the factor's basis to Maslov degrees zero and one: its differential is given by finite binary matrices from degree one to zero and from degree zero to $U$ times degree one. Acyclicity makes the resulting finite parity complex acyclic, so finite vector-space complements give a contracting homotopy; the Laurent degree shifts insert only powers of $U$ or $U^{-1}$ to make it a chain homotopy. Tensoring this contraction with any other complex is a contraction, since the two cross terms cancel over ${\mathbb{F}}_2$. Thus only the displayed tower carries total homology. The splitting is filtered, so no representative of its nonzero class can have a smaller filtration than that tower generator. This proves ${\nu }_t(K\mathrm{\#}J)={\nu }_t(K)+{\nu }_t(J)$ and Upsilon additivity.

First derive the actual minus normalization from [F6]. Since $B$ is finite free over ${\mathbb{F}}_2[U]$ with homogeneous generators, its chain groups vanish above a fixed Maslov degree. The exact sequence $0\to UB\to B\to B\mathrm{/}UB\to 0$ and $H(B\mathrm{/}UB)={\mathbb{F}}_2$ in degree zero imply that multiplication by $U$ is an isomorphism between adjacent even-degree homology groups below zero and between adjacent odd-degree groups; boundedness above gives $H_{>0}(B)=0$, $H_0(B)={\mathbb{F}}_2$, and all odd groups zero. Thus $H(B)={\mathbb{F}}_2[U]$ generated in degree zero. Every normalized minus cycle projects to the hat generator. Conversely, lift a Maslov-zero hat generator cycle to $c\in B$. Write $\mathrm{\partial }c=Ud$; then $\mathrm{\partial }d=0$ and $M(d)=1$. Since $H_1(B)=0$, write $d=\mathrm{\partial }e$. The polynomial cycle $c+Ue$ lifts the SAME hat cycle and has the same leading $i=0$ terms. Therefore the least Alexander level among the $i=0$ parts of normalized minus cycles is exactly $\tau (K)$. Every normalized minus representative has a nonzero $i=0$ part, since a representative wholly in $UB$ would make the top homology generator $U$-divisible. The finite Maslov-zero Laurent cycle set admits a common positive $t$ interval in which any term with $i\ge 1$ excludes a cycle from minimizing, and the maximum on a cycle with $i\le 0$ comes only from its $i=0$ terms. Hence ${\nu }_t(K)=t\tau (K)\mathrm{/}2$ on that interval and ${\mathrm{{\rm Y}}}_K^{\mathrm{\prime }}(0+)=-\tau (K)$. No genus estimate is used.

By the actual normalized geometric duality [F2], the reverse-mirror complex is the filtered dual. Dualize the fixed-$t$ filtered splitting in [F4]. A filtration-preserving change of basis with filtration-preserving inverse has the same property after dualizing and negating the basis filtrations. The dual of the minimizing tower has normalized generator at filtration $-{\nu }_t(K)$. The dual complementary factor is acyclic, because the unfiltered finite Laurent contraction in step 1.2 dualizes to a contraction. Therefore the dual minimax equals $-{\nu }_t(K)$, and ${\mathrm{{\rm Y}}}_{-K}(t)=-{\mathrm{{\rm Y}}}_K(t)$. This uses no unproved identification of the homology of a dual torsion module.

Let $s={\nu }^{-}(K)$ as defined in [F9], and put $g=g_4^{\mathrm{s}\mathrm{m}}(K)$. Since $\delta (\mathcal{A}(K,s))=0$, there is a non-torsion homology class of Maslov grading zero represented by a cycle $z$ in $\mathcal{A}(K,s)$. For every finite basis generator $x_r$, all sufficiently large powers $U^n{x}_r$ satisfy $A(x_r)-n\le s$; hence localizing the inclusion $\mathcal{A}(K,s)\hookrightarrow C^{-}(K)$ gives the full ${\mathrm{C}\mathrm{F}\mathrm{K}}^{\mathrm{\infty }}(K)$ complex and induces an isomorphism on localized homology. The non-torsion class $[z]$ therefore maps to a nonzero degree-zero element of $H_{\ast }(C^{-}(K)){\otimes }_{{\mathbb{F}}_2[U]}{\mathbb{F}}_2[U,U^{-1}]={\mathbb{F}}_2[U,U^{-1}]$, whose only nonzero degree-zero element is the normalized generator. Thus $z$ represents that generator in $C^{-}(K)$ and in $C(K)$. Each term of $z$ has $i\le 0$ and $j\le s$, so $F_t(z)=(1-t\mathrm{/}2)i+(t\mathrm{/}2)j\le ts\mathrm{/}2$ for $0\le t\le 1$. Therefore ${\nu }_t(K)\le ts\mathrm{/}2\le tg\mathrm{/}2$ by [F9], and ${\mathrm{{\rm Y}}}_K(t)\ge -tg$. A linear reflection extending the fixed reflection of $S^3$ to $B^4$ sends any smooth genus-$g$ surface bounded by $K$ to a genus-$g$ surface bounded by its mirror; reversing the surface orientation changes that boundary to the reverse mirror. Applying this diffeomorphism in both directions gives $g_4^{\mathrm{s}\mathrm{m}}(-K)=g$. Apply the same threshold argument to $-K$ and use step 2.1 to obtain ${\mathrm{{\rm Y}}}_K(t)\le tg$. Combining both inequalities proves the bound, using the externally supplied genus upper estimate in [F9].