The previous Proposition (Tait correspondence on partial tire dual)
stated equality between non-equivalent 4-vertex-colorings of T and
non-equivalent 3-edge-colorings of D(T). This is wrong as
empirically verified on the octahedron (n=m=3, O=C_3, spoke-only):
- Octahedron: 96 4-vertex-colorings -> 4 classes mod S_4.
- Partial tire dual C_6 ∘ K_1: 66 3-edge-colorings -> 11 classes
mod S_3.
Replaces that proposition with a variant on the COMPLETE tire dual
D*(T) that incorporates non-annular constraints:
Definition 1.13 (Complete tire dual): Quotient D(T)'s leaves into
non-annular-face vertices. Outer leaves merge into a single
outer-face vertex v_out of degree n; for each bounded face F of
O interior to B_in, the corresponding inner leaves merge into
v_F of degree |F|. Equivalently, D*(T) is the planar dual of T.
Proposition 1.14 (Tait correspondence on complete tire dual): the
number of non-equivalent 4-vertex-colorings of T (mod S_4) equals
the number of non-equivalent Tait colorings of D*(T) (mod S_3).
A Tait coloring is an edge labelling by the three nonzero elements
of Z_2 x Z_2 with XOR-to-0 at every vertex of D*(T).
Remark 1.16 (octahedron verification): For octahedron tire,
D*(T) is the cube Q_3. Octahedron has 4 vertex-coloring classes;
Q_3 has 24 proper 3-edge-colorings -> 4 Tait-coloring classes.
Empirically verified via Sage:
- chromatic_polynomial(octahedron)(4) = 96
- chromatic_polynomial(L(Q_3))(3) = 24
The partial tire dual definition (Def 1.7) and its corona-graph
structure proposition (Prop 1.8) are unchanged.
Co-Authored-By: Claude Opus 4.7 <noreply@anthropic.com>
didericis
93ae55bd42