d = 4 counter-example: an even surgical twist destroys an oval
A concrete, fully reproducible demonstration that the §4.4 invocation of Haas Thm 10.6.0.5
in "121 patchworked curves" is not justified for surgical twists: applying the §4.3 surgical
twist for an even Harnack split (0,3)–(2,0) to η on the triangulation T below
yields a sign distribution σ' whose patchworked curve C(T, σ') has only 3 ovals, vs. C(T, η)'s 4.
The M-curve property is destroyed.
1. The setup: triangulation T of 4·Δ₂
Unimodular triangulation T with 16 triangles. The edge (0,3)–(2,0) (red) is a primitive interior edge shared between two triangles.
T is a unimodular triangulation of 4·Δ₂ with all 15 lattice points as vertices and 16 triangles of unit area each. The edge (0,3)–(2,0) is included as an interior edge — this is the precondition for the surgical twist to be defined on T.
Verified independently: 5 distinct unimodular T's at d=4 contain this edge, all give the same (p,n) result reported here.
2. The split S = (0,3)–(2,0) and its zone Z
The simple Harnack split S highlighted in red; zone Z = surrounded region (yellow) = triangle (0,0)–(0,3)–(2,0).
S is a simple Harnack split: a single primitive lattice edge with endpoints of distinct parities.
Parities: (0,3) has parity (0,1); (2,0) has parity (0,0). So the parity pair is {α, β} = {(0,0), (0,1)}.
Since one parity is (0,0), this is an even Harnack split, giving ε = α₁β₂ + α₂β₁ = 0.
The "zone surrounded by S" is the corner triangle (0,0)–(0,3)–(2,0), i.e. the smaller of the two regions S cuts 4·Δ₂ into.
3. The surgical twist (§4.3 formula)
σ'(x) = σ(x) if x ∉ Z,
σ'(x) = ε ⊕ σ(sji(x)) if x ∈ Z
with ε = α₁β₂+α₂β₁, i = β₁+β₂, j = α₁+α₂
For our split: α = (0,0), β = (0,1), so (ε, i, j) = (0, 1, 0). The canonical extension of σ to A^◇ gives σ(sji(x)) = σ(x) ⊕ (j·x₁ ⊕ i·x₂) = σ(x) ⊕ x₂. So inside Z, σ' differs from σ at points with odd x₂.
η at each lattice point. Filled blue = σ=+ (parity (0,0)), red = σ=−.
σ' = twist(η, S). The two flipped points are (0,1) and (1,1) (black outline) — both in zone Z, both with odd x₂.
4. The patchworked curves via Viro construction
For each triangle of T, the curve crosses an edge iff the two vertex signs differ. The midpoint of each such edge is a crossing; in each triangle the curve consists of arcs connecting these crossings. We render this in Q1, then reflect to the four quadrants of (R*)² to display the full real curve (the (p,n) classification is invariant under further compactification to RP²).
C(T, η) — 4 ovals, scheme ⟨4⟩
Standard d=4 Harnack M-curve. 4 disjoint ovals, all "outside" each other. (p, n) = (4, 0).
C(T, σ') — 3 ovals, scheme ⟨3⟩
Same triangulation T, signs after the even surgical twist. Only 3 ovals. (p, n) = (3, 0) — this is no longer an M-curve (d=4 M-curve has 4 ovals).
Observation. The surgical twist by an even Harnack split changed the patchworked curve from an M-curve (4 ovals) to a non-M-curve (3 ovals). In particular, (p,n) is not preserved. This contradicts the §4.4 paper's implicit reading of Haas Thm 10.6.0.5 as "even surgical twists are (p,n)-preserving no-ops".
5. What this does and doesn't say
Haas Thm 10.6.0.5 itself is correct and not contradicted. Haas's theorem is about removing even edges from a zone decomposition Δ and recomputing the zone-wise Harnack distribution K(Δ'). That is a different operation from the §4.3 surgical twist on a sign distribution σ.
The 121 paper's §4.4 invocation of Haas 10.6.0.5 is unjustified for surgical twists. A bridge lemma "applying the surgical twist for an even Harnack split S to σ = K(Δ) yields σ' = K(Δ \ S)" is needed and not provided. Our test shows that no such bridge can hold in this form: σ' here does not match K(Δ') for any Δ', because the result isn't even an M-curve.
For the σ-DAG of M-curve sign distributions, even splits must be included as live edges. Restricting to odd splits only (as the §4.4 reduction suggests) genuinely misses reachable σ's.
6. Reproducibility
The σ' is computed bit-for-bit by the surgical-twist formula. The triangulation T is one of 5 unimodular triangulations at d=4 containing the edge (0,3)–(2,0); all 5 give the same ⟨4⟩ → ⟨3⟩ shift. The Viro classification is independent infrastructure (the standard isotopy classifier used throughout the project). See thread 5a2865 in the research repo for the full discussion and reference notes (.references/1997:Haas:RealAlgCurvesCombinatorialConstructions/NOTES.md).