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Gts odd factor even total r3 parity

Example

Explicit CY4 degree-minus-one self-dual loop changes factor parity while an admitted total remains even. Compute the parity of every subsum and expose the unauthorized odd 5.10 application, without claiming a full counterexample.

Example references

Bojko, Wall-crossing for Calabi–Yau fourfolds: framework, tools, and applications, Assumptions 5.1 and 5.10; Definition 5.3; §6.2; §6.3; Assumption 6.9; Remark 6.10; Example 6.11; Theorem 6.12; Lemma 6.13; Equation (6.14); §7.2.

Bojko, Kuhn, Liu, and Thimm, Wall-crossing for equivariant DT4 invariants, §2.2.8, §3.2, Assumption 4.2.4, and Theorem 1.2.1; used only under their stated hypotheses.

Setup

The ambient category is Rep⁡C(b→a), with dimension lattice Γ=Zea⊕Zeb. Use the CY4 completion with one self-dual degree-minus-one loop at a and the usual degree-minus-two dual to the degree-zero arrow. Modulo 2, the middle loop contributes da2≡da, while vertex and ordinary-arrow/dual terms are even. The diagnostic admitted monoid is Γeven={(da,db)∈N2:da is even}; this is a parity model, not an assertion that every source 5.1 decomposition hypothesis holds.

Use the source-style filtration 0⊂Zea⊂Γ with Z1(ea)=i and Z2(eb)=−1. It gives phase 1/2 when db=0 and phase 1 when db>0, with weak see-saw equal to the maximum. For any representation f:B→A, (B,im⁡f) is a phase-one surjective-arrow subrepresentation and (0,A/im⁡f) has phase one-half, so this is an HN filtration after omitting zero terms. The canonical diagnostic framing is Fr(B,A)=A⊕B, exact on the ambient representation category; no 5.3 uniformity or CY4 virtual class is asserted. Let the torus T=Gm act trivially on the degree-zero quiver and loop.

Let T0=(Cb→1Ca) and S=(Cb→0). Then T0 is stable of odd self-Euler parity and S is stable of even parity. The admitted total is E=T0⊕6⊕S⊕4, of dimension (da,db)=(6,10). The actual ambient subrepresentations of E are pairs (B′⊆C10,A′⊆C6) with π(B′)⊆A′, where π is projection. In the diagnostic inherited exact structure, only sequences whose three class vectors have even a-coordinate are admitted. The split summands T0 and odd direct subsums are ambient objects, not admitted 5.10 classes. The physical torus is trivial and the calculation uses no orientation-dependent virtual class.

For every direct subsum T0⊕u⊕S⊕v with 0≤u≤6 and 0≤v≤4, its self-Euler parity is u(mod2). Thus odd subsums are exactly those with odd u, while the total u=6 is even. The example tests an odd-subsum proof shortcut; it does not claim the diagnostic datum satisfies all original 5.1(a) or 5.10 hypotheses.

Facts & Assumptions

Given: the exact data and all unresolved suppliers retain their stated status.

[F1]

Gts weak phase framing moduli datum: exact planned interface: “Specify original weak stability, exact phase category, actual ordinary-vs-admissible maps, original concentration opens and exact Fr_k, common-line framing, even emergent class domain, torus linearizations and orientations. No sum closure, cofinal-index availability, or abelian phase category is implicit.” The link records scope, not proof acceptance.

Context only — no mathematical supplier use

lem-gts-parity-safe-admitted-total-closed-orbit-reductivity name the held construction interfaces discussed by this diagnostic. Their conclusions are not assumed or used to establish the calculations below; this example does not prove those interfaces.

Verification

1.1givenF1algebra

The CY4 Euler contributions are 2da2+2db2 from vertices and −2dadb from the ordinary arrow and its dual; each is even. The single self-dual middle loop contributes da2. Hence χ(d,d)≡da(mod2). Therefore T0 is odd, S is even, and E is even because da(E)=6.

1.2givenF1algebra

The representation T0 has only the proper sink subobject (Ca,0), of phase 1/2, while its quotient has phase 1; so it is stable. The representation S is simple. Their direct sum has surjective arrow C10→C6. For any proper subrepresentation with nonzero source space, its source quotient is nonzero: if the source spaces agreed, surjectivity would force the target spaces to agree too. Thus both it and the quotient have phase one. A source-zero subobject has phase 1/2 and source-positive quotient. This verifies semistability directly.

2.1givenF1step 1.1algebra

In the stabilizer of this surjection, choose splittings C10=C6⊕C4. The subgroup acting by (I60NI4), for arbitrary N:C6→C4, preserves the arrow. It is nonzero and unipotent because N2=0. The full stabilizer consists of ga=A∈GL6 and gb=(A0ND) with D∈GL4; hence the displayed additive subgroup Hom⁡(C6,C4) is normal, by block-matrix conjugation. It is therefore a nontrivial connected normal unipotent subgroup, so the stabilizer is nonreductive. Thus the total even parity does not alone imply a reductive stabilizer.

3.1givenF1step 1.2step 2.1algebra∎

This is a diagnostic against inferring odd-factor reductivity from total parity; the full source hypotheses are not all checked, so it is not a counterexample to the original theorem.

Depends on

Used by

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