Proposition

Lemma 6 ([The cap identity]). For every coloring of the annulus, H_ b -H_ w = 1_ T . Consequently the signed difference between the two inner cap states is transferred to the same signed difference at the outer cycle, multiplied by the …

Lemma 6 ([The cap identity]). For every coloring of the annulus, Hb -Hw = 1T . Consequently the signed difference between the two inner cap states is transferred to the same signed difference at the outer cycle, multiplied by the indicator of four-interface transmission. The identity remains valid under composition across several cycles.

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Lemma 5 ([Exponent without an amplitude]). Under Assumption 1 ([Scalar Cardy…Proposition 4 ([Calibration]). There are constants c,B∈(0,∞) such that m_ q_…Proposition
Walls in a triangulation neither branch nor cross. If fewer than four inner ends are connected to outer ends, the wall pairing visible from the outer cycle is independent of the inner cap: there is no choice of pairing on zero or two surviving ends. Thus the left side of is zero. If all four are transmitted, their cyclic order is preserved, with any possible winding retained. The intervening monochromatic bands carry black connectivity to black connectivity and white connectivity to white connectivity. Hence H_ b =1 and H_ w =0 . This is also the usual Hex duality proof after each run is contracted to a vertex. Composing the planar pairings proves the last assertion; any closed walls created in the composition have weight one.
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