Section 5 — Combinatorial core of the r > d truncation branch #
Helper lemmas for the r > d branch of lem:projective-low-rank-sum
(references/ldt-paper/orthonormalization.tex:559-658).
Given a finite family of overlap values f : α → ℝ (with α the pair index set
{(a, i)} carrying r = |α| entries) and a chosen Large ⊂ α of size
d < r such that every Small = Lᶜ entry has overlap ≤ every Large entry,
the paper derives
|Large| * ∑_{Small} f ≤ |Small| * ∑_{Large} f(pairwise double-counting),r * ∑_{Small} f ≤ |Small| * ∑_{α} f(partition bound).
We then combine this with the global R ≤ (1 + 2√ζ) I estimate to obtain
∑_{Small} f ≤ 4√ζ when ζ ≤ 1/4, which is the combinatorial heart of the
paper's 4√ζ truncation error.
These lemmas do not depend on the matrix/spectral scaffolding in
QXPLayer/Core.lean or QXPLayer/RankReduction/LowRank.lean; they are pure
combinatorics over a chosen partition. The per-projector orthonormal range
decomposition is available as MIPStarRE.Quantum.IsProj.rangeONB; the concrete
Large/Small choice and truncated projectors are assembled in
QXPLayer/RankReduction/LowRank.lean.
References #
references/ldt-paper/orthonormalization.texlines 559–658 (proof oflem:projective-low-rank-sum,r > dbranch).
Choose d elements with the largest values of f, breaking ties
arbitrarily. The resulting Large set has the paper's ordering property:
every element outside Large has value at most every element of Large.
Pairwise ordered double-counting.
If every element of S has value at most every element of L, the sum over S
scaled by |L| is dominated by the sum over L scaled by |S|. Matches the
paper's inequality ∑_{Small} o ≤ (|Small|/r) · ∑ o when L ∪ S exhausts the
index set.
Partition-total form of the pairwise bound.
If S = Lᶜ, the paper's r · ∑_{Small} o ≤ |Small| · ∑ o follows from
card_mul_sum_small_le after splitting ∑ x, f x = ∑_L f + ∑_{Lᶜ} f.
Small-overlap sum bound.
Assuming the paper's inputs:
L.card = dwithd < |α|,|α| ≤ (1 + 2√ζ) · d,- every
Small = Lᶜentry has value at most everyLarge = Lentry, ∑ x, f x ≤ 1 + 2√ζ,0 ≤ ζ ≤ 1/4,
this yields ∑_{Small} f ≤ 4√ζ, matching eq:small-overlaps in the paper. The
Lean statement is slightly stronger than the paper's: it does not need f ≥ 0,
since the ordering hypothesis and partition-total estimate already suffice.
See docs/paper-gaps/truncation-combinatorics-f-nonneg.tex.