Logical quotient and sector distances

Question. Why does the scalar label [[15,1,3]] hide a seven-versus-three asymmetry between logical X and logical Z?

Let L=row(H_X) and G=L+span{1} as on the construction page. The CSS normalizer splits into two classical quotient problems.

1. X normalizers and the logical-X coset

An X-support vector x commutes with every Z stabilizer exactly when x in S_Z^perp. Since S_Z=G^perp, finite-dimensional duality gives

S_Z^perp = G = L + span{1}.

Quotienting by X stabilizers L leaves two cosets: the trivial coset L and the logical coset 1+L. Complete enumeration gives

The constructor declares the all-one representative, but multiplying it by any nonzero X stabilizer replaces it by the complement of a weight-eight word and reduces its weight to seven. Therefore the pure-X distance is

d_X=7.

The logical-basis routine selects XXXXXXX________ in its deterministic symplectic Gram–Schmidt convention; it is in the same coset as the all-one operator.

2. Z normalizers as zero-sum label sets

A Z-support vector z commutes with all four X checks when

H_X z^T = 0.

The four syndrome bits are the coordinate-wise XOR of all labels in the support. Thus z is a Z normalizer exactly when its selected nonzero vectors sum to 0000.

The Z stabilizer space imposes one further condition: z dot 1=0, meaning the support size is even. Consequently:

• even zero-sum supports are Z stabilizers;
• odd zero-sum supports are the nontrivial logical-Z coset.

This parity quotient has dimension one, matching the encoded-qubit count.

3. Exact lower and upper bounds meet at three

No weight one. A one-point support sums to its nonzero label, not to zero.

No weight two. Two labels sum to zero over F_2 only when they are equal, but a binary support cannot contain the same coordinate twice.

A weight-three witness. For distinct nonzero labels a and b, the third label a+b is nonzero and distinct from both. The support

{a,b,a+b}

sums to zero and has odd cardinality, so it commutes with every X check but is not a Z stabilizer. In the declared order, {0001,0010,0011} gives ZZZ____________.

The lower exclusion and upper witness prove d_Z=3. There are exactly thirty-five such three-point supports: each two-dimensional subspace of F_2^4 contains precisely its three nonzero points, and the Gaussian binomial count is

((2^4-1)(2^4-2))/((2^2-1)(2^2-2)) = 35.

The reproduction transcript also enumerates all C(15,3) supports and finds the same thirty-five witnesses without relying on that counting formula.

4. Total distance and error correction

For a CSS code, a minimum logical Pauli may be chosen pure X or pure Z, so

d = min(d_X,d_Z) = min(7,3) = 3.

The exact verifier searches supports in increasing weight, tests both CSS sectors, and returns (3, exact-css-enum). This establishes correction of an arbitrary single-qubit Pauli error. It does not establish comparable logical rates in the two sectors under biased noise: rates also depend on the decoder, syndrome circuit, and physical error distribution.

5. Full Z-normalizer enumerator

The kernel of H_X has dimension eleven and therefore contains 2048 support vectors. Exact enumeration separates it by parity:

The two rows each contain 1024 words. The odd row becomes the logical-error numerator in the idealized distillation model on the protocol ledger. These coefficients are constructor-derived facts; no experimental or circuit-level rate is being inferred from them.