Lemma 33 ([Centered block concentration]). Fix a surviving incidence type and one of its same or adjacent layer blocks. Assign independent samples from Lemma 28 ([Finite product sampler]) to its unoriented edges. Let X_ ik be its …

An edge entry is W_eλ(v_e)-𝔼[W_eλ(v_e)] and has norm at most 2M_η . Double the block off the diagonal to make it selfadjoint. Its absolute row and column coefficient sums are at most C_*M_ηD . We first make precise why a fixed finite-test exponent is available before choosing s. Require H_0≥s , so D≤2^ .60005H_i , A_i=2^ H_i+.2s ≤2^ 1.2H_i . After raising H_0 to absorb fixed constants, the row bound is at most 2^ 2H_i , and the number of base and matrix indices is at most 2^ 3H_i . The coordinates of every possible voltage and inverse have the fixed exponent of Lemma 28 ([Finite product sampler]). Apply Lemma 30 ([Heisenberg boxes and regular norm]) with a boundary exponent larger than four. Its box error tends to zero exponentially even after the row loss, and its total finite compression dimension is at most 2^ K_1H_i for a fixed K_1 . Enlarge H_0 further so the box error is at most ηq_ τ,ik /2 . These enlargements do not change K_1 . Now fix C>2(K_1+6) , increasing it strictly if needed, and put h=(CH_i)/(_2d). Every no-singleton support in the 2h -moment has at most h base edges. It lies in the two fixed layers of this block. Their base girth is at least .9H_i/_2q after increasing H_0 , since adjacent H-values differ by only b. Because _2q/_2d<1.25 , the ratio of support length to this girth is at most 2C , again increasing H_0 for the ceiling. Lemma 31 ([Length versus cycle rank]) gives a rank bound depending on C alone. This is a use of base girth, not the as-yet unproved voltage girth. Let X denote the finite selfadjoint compression. Expand 𝔼Tr(X^ 2h ) along base vertices and individual edges. Any summand with a singleton unoriented edge has expectation zero: after conditioning on the other edge draws it is a fixed left factor, one centered compressed entry or its adjoint, and a fixed right factor. Compression does not change this independence. For every other summand its absolute trace on the deck and matrix indices is at most their dimension times (2M_η)^ 2h . Lemma 32 ([Repeated-edge walk count]), with Δ≤C_*D , proves 𝔼Tr(X^ 2h ) ≤2^ K_1H_i (C_1D)^h, where C_1 is fixed after C and the coefficient data, but before s. Since X is selfadjoint, its largest absolute eigenvalue to the power 2h is at most this nonnegative trace. Markov's inequality and give X_ ik >ηq_ τ,ik ≤2^ K_1H_i ((C_2)/(η^2d))^h. Choose s large enough that C_2/η^2≤d^ 1/2 . The last expression is at most 2^ (K_1-C/2)H_i ≤2^ -6H_i , by the choice of C. The potentially large dependence of C_2 on C is not circular: s is chosen after both constants. This proves the claim.