Cross-Cut Rank Is Not Monotone Under Stabilizer Extension¶
Claim/Theorem¶
The intrinsic cross-cut stabilizer rank
is not monotone under enlarging the stabilizer space. That is, there exist stabilizer spaces
on the same qubit set and a cut \(L\) such that
Concrete example:
- take two qubits with cut \(L=\{1\}\) and \(R=\{2\}\),
- let
so \(\mathcal S\) is generated by one cross-cut stabilizer, - let
Then \(\mathcal S\subsetneq \mathcal T\), but
Indeed, using [[cross-cut-stabilizer-rank-rank-formula.md]]:
- for \(\mathcal S\), one has \(\operatorname{rank}(H_L)=1\), \(\operatorname{rank}(H_R)=1\), and \(\operatorname{rank}(H)=1\), hence
- for \(\mathcal T\), with generator matrix
one has \(\operatorname{rank}(H_L)=1\), \(\operatorname{rank}(H_R)=1\), and \(\operatorname{rank}(H)=2\), hence
So adding an \(L\)-local stabilizer can destroy the earlier cross-cut rank contribution.
For Conjecture 3 this is a real proof-theoretic warning: one cannot lower-bound the global intrinsic cut rank merely by exhibiting a favorable subfamily of disjoint local blocks inside the full stabilizer space. Any globalization of [[small-side-local-cut-gives-full-local-cross-rank.md]] must control the entire row space, not just a convenient subspace.
Dependencies¶
- [[cross-cut-stabilizer-rank-rank-formula.md]]
Conflicts/Gaps¶
- This is a negative structural fact, not a lower bound.
- It rules out a tempting monotonicity shortcut, but does not say what replacement argument should work.
- The explicit-family frontier is therefore tighter: globalizing local cut-rank contributions requires a theorem about the full stabilizer space or parity-check matroid, not just a selected subset of rows.
Sources¶
10.48550/arXiv.2109.1459910.48550/arXiv.0805.2199