Special product-charge uniqueness and actual block-permutation invariance

We first prove the exact relative uniqueness assertion. Let $(A,Z_0)$ be one of the constructed lower-dimensional ordered conditions on $Y$, let $C$ be elliptic, and let $B=A_C$ be its actual noetherian AP heart on $Y\times C$. Suppose a condition $\tau$ on this product has charge $b+c+i(d-a)$ and, on EVERY fibre, its inverse slicing is exactly the same lower condition $(A,Z_0)$. This last hypothesis will be constructed inductively using F2, not assumed from equality of charges.

Here is the all-fibre AP detection needed for the comparison. Point pushforward $i_{\ast }$ is t-exact from $A$ to $B$, directly from the large-twist AP criterion $R{p}_{\ast }{i}_{\ast }F(k)=F$ up to a one-dimensional scalar line. Locally at a point of $C$, the Koszul triangle for its parameter $u$ gives, for every bounded $E$, the EXACT sequence in the AP heart $0\to \mathrm{coker}(u:H_B^{q}E\to H_B^{q}E)\to i_{\ast }{H}_A^q(L{i}^{\ast }E)\to \ker (u:H_B^{q+1}E\to H_B^{q+1}E)\to 0.$ Thus there is no unproved general spectral-sequence degeneration premise. Every nonzero $Q\in B$ has a fibre where that cokernel is nonzero. Otherwise multiplication by its parameter is surjective in the local noetherian heart and hence injective: stabilize its increasing kernels and lift along a sufficiently large surjective power. It would be invertible, so coherent derived Nakayama would make $Q$ vanish near that fibre. Applying this at a point in its nonempty ordinary support is a contradiction. If all $L{i}^{\ast }E$ have $A$-cohomology in degrees $\le r$, the highest nonzero $B$-cohomology outside that range has a nonzero such cokernel and violates the bound. If all have degrees $\ge r$, the same cokernel for the lowest bad cohomology injects into a forbidden fibre degree. Consequently ALL-fibre upper and lower bounds detect the corresponding AP bounds.

Every $E\in B$ has $\tau$-HN phases in $(-1,1]$. If its highest factor $F$ had phase $>1$, compatibility on each fibre gives $\mathrm{Hom}(F,i_{\ast }G[k])=0$ for all $G\in A,k\le 0$. By left adjunction $\mathrm{Hom}(L{i}^{\ast }F,G[k])=0$, so $L{i}^{\ast }F\in D_A^{\le -1}$ for every fibre. step 2.1 makes $F\in D_B^{\le -1}$ and hence $\mathrm{Hom}(F,E)=0$. Its first HN map is nonzero: the strict lower phases in the remaining triangle make $\mathrm{End}(F)\to \mathrm{Hom}(F,E)$ injective. Contradiction.

If its lowest factor $Q$ had phase $\le -1$, then $\mathrm{Hom}(i_{\ast }G[k],Q)=0$ for all $G\in A,k\ge -1$, since those source phases are strictly above $-1$. The actual divisor adjunction is $i^{!}Q=L{i}^{\ast }Q[-1]$ tensored with the one-dimensional normal fibre. It gives $\mathrm{Hom}(G,L{i}^{\ast }Q[j])=0$ for every $j\le 0$, hence $L{i}^{\ast }Q\in D_A^{\ge 1}$. step 2.1 makes $Q\in D_B^{\ge 1}$ and $\mathrm{Hom}(E,Q)=0$, contradicting its nonzero last HN quotient map. This proves the stated phase interval, including exclusion of its lower endpoint.

These phase bounds give exactly one HRS tilt. Indeed $B\subset D_{\tau }^{[0,1]}$ implies $D_B^{\le -1}\subset D_{\tau }^{\le 0}\subset D_B^{\le 0}$ by bounded extension generation and orthogonality. Thus the $\tau$ heart lies in $D_B^{[-1,0]}$. For $E\in B$, its two $\tau$ truncations $T\to E\to F$ belong to $B$: the possible $B$-cohomology of $T$ is $[-1,0]$ and that of $F$ is $[0,1]$; the exact cohomology sequence kills the endpoint degrees. They give an exact torsion decomposition in $B$, with $T\in T_{\tau }=B\cap A_{\tau }$ and $F\in F_{\tau }=B\cap A_{\tau }[-1]$. Conversely these decompositions generate the $\tau$ heart as $⟨T_{\tau },F_{\tau }[1]⟩$. Put $I=d-a$. For nonzero $T_{\tau }$ one has $I\ge 0$, and at $I=0$ its real charge $b+c$ is strictly negative. For nonzero $F_{\tau }$ one has $I\le 0$, and at $I=0$ its real charge is strictly positive. These are the actual boundary signs of a stability heart.

No nonzero $F\in F_{\tau }$ has $c=0$. The AP inequalities then give $a\le 0,d\ge 0$; combined with $I\le 0$ they force $a=d=0$. But its nonzero boundary real charge would be $b<0$, whereas $F_{\tau }$ requires it positive. Thus every nonzero $F_{\tau }$ has $c>0$ and finite weak slope $\nu \le 0$. If a weak-semistable object has positive finite slope, its $F_{\tau }$ quotient must vanish by the weak semistable quotient inequality. For weak slope infinity, its total $c=0$ makes every quotient have $c=0$, again ruling out a nonzero $F_{\tau }$. Hence all positive weak-semistables lie in $T_{\tau }$.

The initial elliptic uniqueness does not require a new point-base product theorem. For a condition on $C$ with charge $-\mathrm{deg}+i\mathrm{rk}$ and every point stable of phase one, use the preceding phase-bound proof with $Y$ a point, $A=\mathrm{V}\mathrm{e}\mathrm{c}\mathrm{t}$ with its point in phase one, and $B=\mathrm{Coh}C$. Its all-fibre detection is ordinary coherent Nakayama, so it differs from $B$ by one tilt as above. The coefficients are $a=-\mathrm{rk},c=d=0,b=-\mathrm{deg}$. For a nonzero object in $F_{\tau }$, $I=\mathrm{rk}\le 0$ forces rank zero, so it is a positive-length torsion sheaf and $b<0$, contradicting its required positive real boundary. Therefore $F_{\tau }=0$ and the heart is $\mathrm{Coh}C$, with precisely ordinary slope semistability. The condition is unique. For the empty product, $X=\mathrm{Spec}\mathbb{C}$ and $H=0$, so the displayed charge is $Z(F)=-\chi (F)$, equal to minus dimension on finite vector spaces. Its phase-one point is $\mathbb{C}$; the shift axiom forces $\mathbb{C}[k]$ to phase $1+k$. Every bounded complex splits into a finite direct sum of these shifts, and their inclusions and projections, together with slicing Hom vanishing, force its semistable summands to have those phases. Hence the entire slicing is uniquely $\mathcal{P}(1+k)={\mathrm{V}\mathrm{e}\mathrm{c}\mathrm{t}}_{\mathbb{C}}^{\mathrm{f}\mathrm{d}}[k]$ and zero otherwise. It is locally finite with support constant one on the rank-one Euler-class lattice. This is exactly the zero-step ordered condition; all factor permutations, geometric automorphisms and flat-line twists act identically on the point.

A negative weak-semistable object has no nonzero $T_{\tau }$ subobject, since it would have nonnegative or infinite weak slope and would destabilize. It therefore lies in $F_{\tau }$. At the INCLUDED zero slope, a $T_{\tau }$ subobject cannot have $c=0$ (that would have infinite weak slope). It has $c>0,I=0$ and is itself weak-semistable of slope zero, since it is an equal-slope subobject of a weak-semistable. The ACTUAL F1 inequality gives $bc-ad\ge 0$, and $d=a$ gives $b\ge a^2\mathrm{/}c\ge 0$. Thus $b+c>0$, contrary to the strictly negative $I=0$ boundary for $T_{\tau }$. So every nonpositive weak-semistable lies in $F_{\tau }$. Finite weak HN gives $T_{\nu }\subseteq T_{\tau },F_{\nu }\subseteq F_{\tau }$. Their torsion decompositions force equality: a nonzero $F_{\nu }$ quotient of a $T_{\tau }$ object would be a prohibited map $T_{\tau }\to F_{\tau }$, and conversely for the other inclusion. Hence the heart is EXACTLY the F1 Liu heart. Equal heart and equal charge give equal semistable inequalities and HN slicings. This proves relative uniqueness over the already tilted base.

Prove the Statement's special-charge uniqueness by induction on the number of elliptic factors, including the empty product computed in step 6.2. Given such a condition on $Y\times C$, its torsion-flat-line invariance includes every pullback from the last elliptic curve. F2 therefore constructs its compatible condition on EVERY fibre $Y\times \{c\}$. The restriction retains stable skyscrapers of phase one, and retains the actual torsion-flat-line invariance from the remaining factors because those tensor functors commute with $i_{\ast }$. Its charge is the lower product charge: for a degree-one class $h$ on $C$, $\mathrm{ch}({\mathcal{O}}_c)=h$ by $0\to {\mathcal{O}}_C(-c)\to {\mathcal{O}}_C\to {\mathcal{O}}_c\to 0$, so direct external-product calculation gives $Z_{Y\times C}(i_{\ast }E)=Z_Y(E)$, since $h^2=0$ and ${\int }_{C}h=1$. Its support lattice is the actual F2 coimage, with no full-$K_{\mathrm{n}\mathrm{u}\mathrm{m}}$ identification needed. Induction identifies EVERY fibre condition with the same constructed lower ordered condition. The relative uniqueness construction then identifies the original condition with the actual next Liu condition. Existence is already F1.

Reorder the elliptic factors, or apply a geometric permutation of identical factors preserving the degree-one numerical classes. The transported ordered condition has the SAME product-exponential charge, ACTUAL torsion-flat-line invariance, and stable point sheaves of phase one. step 8.1 therefore makes it equal to the original condition as a heart and entire slicing. This is a consequence of the constructed special-charge uniqueness, not an inference from a symmetric formula alone. Factor automorphisms already preserve it by F1. Thus each generator of the finite wreath groups in the named families preserves the condition, and so does the whole group.

Finally replace the F1 support lattice $v$ by the coimage of $v_G=(v\circ g^{\ast }{)}_{g\in G}$ in the finite direct sum of its translates, modulo torsion. It is genuinely equivariant by component permutation. Each translate of a semistable remains semistable with the same charge magnitude, so $\mathrm{\parallel }{v}_G(E)\mathrm{\parallel }\le \sqrt{\mathrm{\mid }G\mathrm{\mid }}C\mathrm{\mid }Z(E)\mathrm{\mid }$. Its identity component recovers the charge. This gives the actual finite equivariant support interface for induced wreath-product stability and completes every claimed conclusion.