The Hidden Bit in the Steane Code

The Steane code is a $[[7,1,3]]$ quantum error-correcting code: 7 physical qubits protecting 1 logical qubit, correcting any single-qubit error. Its stabilizer structure is defined by the Fano plane — the same 7-point, 7-line projective geometry from the golden zeta writeup.

I computed the entanglement entropy of every possible 3-qubit subsystem. There are $\binom{7}{3} = 35$ of them. The result is cleaner than it has any right to be: the 35 triples partition into exactly three classes, and the partition is controlled by the Fano plane.

The partition

The logical $|0_L\rangle$ state of the Steane code is a uniform superposition over the 8 even-weight codewords of the classical Hamming(7,4) code:

$|0_L\rangle = \frac{1}{\sqrt{8}} \sum_{x \in C_{\text{even}}} |x\rangle$

For each of the 35 ways to pick 3 qubits from 7, I traced out the other 4 and computed the von Neumann entropy $S$ and the tripartite mutual information $I_3$:

$I_3(A:B:C) = S(A) + S(B) + S(C) - S(AB) - S(BC) - S(AC) + S(ABC)$

Every triple falls into one of three bins. No exceptions.

triples 35 Play

7

Fano

6

Anti-Fano

22

Generic

35

Total

triple	class	S (bits)	$I_3$	$|\Phi|$
{0, 1, 5}	Fano	3	0	0
{0, 2, 4}	Fano	3	0	0
{0, 3, 6}	Fano	3	0	0
{1, 2, 3}	Fano	3	0	0
{1, 4, 6}	Fano	3	0	0
{2, 5, 6}	Fano	3	0	0
{3, 4, 5}	Fano	3	0	0
{0, 1, 4}	Anti	2	-1	2
{0, 2, 5}	Anti	2	-1	2
{1, 2, 6}	Anti	2	-1	2
{1, 3, 5}	Anti	2	-1	2
{2, 3, 4}	Anti	2	-1	2
{4, 5, 6}	Anti	2	-1	2
{0, 1, 2}	—	3	0	2
{0, 1, 3}	—	3	0	2
{0, 1, 6}	—	3	0	2
{0, 2, 3}	—	3	0	2
{0, 2, 6}	—	3	0	2
{0, 3, 4}	—	3	0	2
{0, 3, 5}	—	3	0	2
{0, 4, 5}	—	3	0	2
{0, 4, 6}	—	3	0	2
{0, 5, 6}	—	3	0	2
{1, 2, 4}	—	3	0	2
{1, 2, 5}	—	3	0	2
{1, 3, 4}	—	3	0	2
{1, 3, 6}	—	3	0	2
{1, 4, 5}	—	3	0	2
{1, 5, 6}	—	3	0	2
{2, 3, 5}	—	3	0	2
{2, 3, 6}	—	3	0	2
{2, 4, 5}	—	3	0	2
{2, 4, 6}	—	3	0	2
{3, 4, 6}	—	3	0	2
{3, 5, 6}	—	3	0	2

7 + 7 + 21 = 35. The bar heights show entropy: Anti-Fano triples are shorter (S=2 bits) than the rest (S=3 bits).

| Type | Count | Entropy $S$ | $I_3$ | Associator $|\Phi|$ | |---|---|---|---|---| | Fano lines | 7 | 3 bits | 0 | 0 | | Anti-Fano triples | 7 | 2 bits | $-1$ | 2 | | Generic triples | 21 | 3 bits | 0 | 2 |

Seven triples have maximal entropy and zero tripartite correlation — these are the 7 Fano lines. Seven triples have sub-maximal entropy and $I_3 = -1$ — genuine 3-party entanglement that can’t be decomposed into pairs. The remaining 21 are generic.

The 7 + 7 + 21 = 35 partition is exact. And it’s the Fano plane that decides which is which.

Fano lines vs Anti-Fano triples

The Fano plane has 7 lines. But there’s a dual structure hiding in the complement. Each Anti-Fano triple’s 4-qubit complement contains exactly one Fano line:

Anti-Fano triple	Complement	Fano line in complement
$\{0, 1, 4\}$	$\{2, 3, 5, 6\}$	$\{2, 3, 5\}$
$\{0, 2, 5\}$	$\{1, 3, 4, 6\}$	$\{3, 4, 6\}$
$\{1, 2, 6\}$	$\{0, 3, 4, 5\}$	$\{0, 4, 5\}$
$\{1, 3, 5\}$	$\{0, 2, 4, 6\}$	$\{0, 2, 6\}$
$\{2, 3, 4\}$	$\{0, 1, 5, 6\}$	$\{1, 5, 6\}$
$\{4, 5, 6\}$	$\{0, 1, 2, 3\}$	$\{0, 1, 3\}$

This is a canonical bijection: 7 Fano lines $\leftrightarrow$ 7 Anti-Fano triples, paired through complementation. The Fano plane is dual to itself, but the duality swaps the roles: lines become “anti-lines.”

Select 3 points to classify a triple

0/3 selected

show all

7 Fano 7 Anti-Fano All 35

● 7 Fano lines: S=3, I₃=0, Φ=0

● 7 Anti-Fano: S=2, I₃=−1, |Φ|=2

● 21 Generic: S=3, I₃=0, |Φ|=2

Click 3 points to select a triple. The Fano plane has exactly 35 triples: 7 lines, 7 anti-lines, 21 generic.

The octonionic connection

The Fano plane isn’t just a combinatorial object — it’s the multiplication table of the octonions. The 7 imaginary units $e_1, \ldots, e_7$ multiply according to the Fano lines: if $\{i, j, k\}$ is a line with the right cyclic orientation, then $e_i \cdot e_j = e_k$.

The octonions are non-associative. The associator

$\Phi(a, b, c) = (ab)c - a(bc)$

measures how badly associativity fails. For triples of basis elements:

• Fano line $\{i, j, k\}$: $\Phi(e_i, e_j, e_k) = 0$. The triple generates a quaternionic (associative) subalgebra.
• Non-Fano triple: $|\Phi| = 2$. Associativity breaks.

The Steane code’s entropy partition mirrors this exactly. Fano lines = associative = maximal entropy. Anti-Fano triples = non-associative = sub-maximal entropy with irreducible 3-party entanglement.

But the octonion associator alone doesn’t distinguish Anti-Fano from generic — all 28 non-Fano triples have $|\Phi| = 2$. What separates the 7 Anti-Fano triples from the 21 generic ones is the Hamming code structure: Anti-Fano triples are the ones where the XOR parity $x_i \oplus x_j \oplus x_k$ is constant across all 8 codewords of $|0_L\rangle$. This causes the reduced density matrix to have rank 4 instead of 8.

The hidden bit

Here’s the punchline. Three quantities — from algebra, entropy, and information theory — all evaluate to the same number:

$\log_2 |\Phi| = \Delta S = |I_3| = 1 \text{ bit}$

log₂|Φ| = ΔS = |I₃| = 1 bit

Replay

Three different quantities — algebraic, entropic, information-theoretic — all encode the same hidden bit.

• $|\Phi| = 2$ is the associator magnitude, so $\log_2(2) = 1$
• $\Delta S = 3 - 2 = 1$ is the entropy deficit of Anti-Fano triples
• $|I_3| = 1$ is the magnitude of the tripartite mutual information

One bit. The same bit, expressed three different ways. The logical qubit protected by the Steane code is stored in the non-associativity of the octonions.

Zero pairwise, nonzero tripartite

The Anti-Fano triples have a striking property: all pairwise correlations vanish, but the 3-party correlation doesn’t.

For the Anti-Fano triple $\{0, 1, 4\}$:

$S(0) = S(1) = S(4) = 1$ $S(01) = S(14) = S(04) = 2$ $S(014) = 2$

So:

$I(0:1) = 1 + 1 - 2 = 0$ $I(1:4) = 1 + 1 - 2 = 0$ $I(0:4) = 1 + 1 - 2 = 0$

But:

$I_3 = 1 + 1 + 1 - 2 - 2 - 2 + 2 = -1$

No pairwise measurement can access the hidden bit. The information is entirely in the irreducible 3-party correlation. You need all three qubits to see it. This is also why the associator needs three elements — $(ab)c$ vs $a(bc)$ — you can’t detect non-associativity from pairs alone.

The associator encodes the logical qubit

On an Anti-Fano triple, the two bracketings of the octonion product give opposite signs:

$(e_i e_j) e_k = +X, \qquad e_i(e_j e_k) = -X$

That sign flip — left-association vs right-association — is exactly 1 bit. And it distinguishes the logical states: measuring the XOR parity $Z_i Z_j Z_k$ on any Anti-Fano triple gives $+1$ for $|0_L\rangle$ and $-1$ for $|1_L\rangle$.

Fano lines can’t do this. Their associator is zero — no sign to carry information. The logical qubit lives in the non-associative part of the algebra.

Hardware and software

The picture that emerges:

• Fano lines = the hardware. Associative subalgebras. Maximal entropy (thermal, boundary-like). They provide the stabilizer structure — the redundancy that enables error correction. But they carry no logical information.
• Anti-Fano triples = the software. Non-associative. Sub-maximal entropy (coherent, bulk-like). They carry the logical qubit in their irreducible 3-party entanglement.

Error correction works because these roles are separated. Errors on the “hardware” qubits can be detected and corrected without disturbing the “software” correlations. The associative/non-associative boundary is the boundary between the code’s structure and its content.

Why 7?

The number 7 isn’t a design choice. It’s forced by deep mathematics:

• Hurwitz’s theorem: Normed division algebras over $\mathbb{R}$ exist only in dimensions 1, 2, 4, 8. The octonions ($\dim = 8$) are the last one, with 7 imaginary dimensions.
• Adams’ theorem: Parallelizable spheres exist only for $S^0, S^1, S^3, S^7$.
• Hopf fibrations: $S^{2n-1} \to S^n$ with fiber $S^{n-1}$ exist only for fibers $S^0, S^1, S^3, S^7$.
• Cross products satisfying $|a \times b|^2 = |a|^2|b|^2 - (a \cdot b)^2$ exist only in dimensions 0, 1, 3, 7.

The pattern is always $2^n - 1$ for $n = 0, 1, 2, 3$: namely $0, 1, 3, 7$. Seven is the maximum. The Steane code’s 7 qubits, the Fano plane’s 7 points, and the octonions’ 7 imaginary units are all reflections of this single constraint.

Open questions

• Is the hidden bit identity $\log_2 |\Phi| = \Delta S = |I_3| = 1$ a coincidence, or does it generalize? What happens for larger codes built from higher-dimensional projective geometries?
• The Fano/Anti-Fano duality looks like a toy version of holographic bulk/boundary correspondence. The boundary (Fano lines) is thermal; the bulk (Anti-Fano triples) carries protected information. How far does this analogy go?
• An observer restricted to associative operations sees the hidden bit averaged to zero — it looks like thermal noise. Is this a model for how semiclassical observers lose quantum information at horizons?