The auxiliary nu-minus threshold is bounded by smooth slice genus

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.