level_diff
plain-language theorem explainer
On the polarized diamond, any 4-neighbour edge whose endpoints have unequal charge has charge difference exactly +1 or -1. Cosmology cost ledgers cite this to pin every bichromatic interface to a single phi-rung gap. The proof unfolds L1 adjacency and the sign-of-x charge, then case-splits and finishes by integer arithmetic.
Claim. Fix $t \in \mathbb{N}$ and vertices $a,b$ of the L1 diamond ball of radius $t$. If $a$ and $b$ are 4-neighbours (L1 distance exactly one) and their polarized charges differ, then $\mathrm{charge}(a)-\mathrm{charge}(b) \in \{+1,-1\}$.
background
The module weights the Phase-54 edge-count ledger by recognition cost $J(x)=(x+x^{-1})/2-1$, evaluated at the phi-rung gap between cell charges. Carried monochromatic edges cost zero; interface edges should cost exactly $J(\varphi)$.
Vertices are lattice points in the diamond ball. Adjacency is the 4-neighbour relation: L1 distance exactly one. The polarized charge is forced by the $x$-coordinate: $+1$ on $x>0$, $-1$ on $x<0$, and $0$ on the spine $x=0$.
A unit L1 step changes $x$ by at most one, so it cannot jump from the fine half to the coarse half without landing on the spine. That geometric fact is what this lemma records algebraically.
proof idea
Unfold the adjacency hypothesis to the L1-distance equation. Simplify both the unequal-charge hypothesis and the goal with the definition of polarized charge (nested if on the sign of $x$). Case-split on those conditionals; each surviving branch is a small integer constraint on the two $x$-coordinates. omega closes every branch, yielding difference $+1$ or $-1$.
why it matters
This is the geometric pin that turns the interface ledger into a pure multiple of $J(\varphi)$. Downstream, edgeCost_interface applies it (via absolute-value-one of the gap) to prove every bichromatic edge costs exactly $J(\varphi)$. Graded-rung work uses it to show the polarized field satisfies the unit-step property, which feeds the Phase-56 headline: carried bulk costs zero and total cost equals interface count times $J(\varphi)$. That is the compute-watch law in cost units: cost scales with the codimension-1 interface, not bulk volume. It sits under the unique $J$ from the recognition composition law (T5) and the forced $\varphi$ fixed point (T6).
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