Dense k-Connected Set Forces Balanced Cut Rank¶
Claim/Theorem¶
Let \(M\) be a binary matroid, or equivalently the column matroid of a full-row-rank binary matrix, on a ground set \(E\) of size \(n\). Fix a balance parameter \(0<\beta\le 1/2\). Suppose \(Z\subseteq E\) is a \(k\)-connected set of size
in the sense of [[large-k-connected-set-gives-balanced-cut-rank.md]], and assume
Then every \(\beta\)-balanced cut \(L\subseteq E\), meaning
satisfies
Equivalently, if \(M\) is the column matroid of a stabilizer matrix for a stabilizer space \(\mathcal S\), then every \(\beta\)-balanced qubit cut satisfies
Combining this with [[stabilizer-cut-rank-functional.md]] yields
for any hardware family whose bottleneck separator cuts are \(\beta\)-balanced.
Proof sketch:
- Since \(|E-L|\le (1-\beta)n\),
- By symmetry,
- Applying the \(k\)-connected-set inequality from [[large-k-connected-set-gives-balanced-cut-rank.md]] gives
So once the highly connected set is dense enough, the old alignment gap disappears: every hardware-balanced cut automatically splits that set on both sides by at least k-1 elements.
Dependencies¶
- [[large-k-connected-set-gives-balanced-cut-rank.md]]
- [[stabilizer-cut-rank-functional.md]]
Conflicts/Gaps¶
- This is a sufficient condition, not yet a theorem about Quantum Tanner parity-check matroids.
- It requires a dense \(k\)-connected set, stronger than merely having some large \(k\)-connected set somewhere in the matroid.
- The live intrinsic gap can now be stated more sharply: prove that the relevant Quantum Tanner parity-check matroid contains a \(k\)-connected set \(Z\) with both \(k=\Omega(n)\) and density \(|Z|/n\) above the hardware balance threshold.
Sources¶
10.37236/1246710.48550/arXiv.2109.1459910.48550/arXiv.0805.2199