gauge_invariant_seed
plain-language theorem explainer
The gauge-invariant Maxwell seed on the cube is $4\pi$ times the cycle rank of the 1-skeleton. Anyone testing whether $\alpha^{-1}$ can arise from a genuine U(1) coupling normalization on $Q_3$ cites this quantity. It is a one-line definition: the QED prefactor $4\pi$ times the first Betti number $b_1$.
Claim. Define the gauge-invariant Maxwell seed on the cube by $4\pi\cdot b_1$, where $b_1=E-V+1$ is the cycle rank (first Betti number) of the cube 1-skeleton.
background
Module M11 asks whether the $\alpha$ seed $4\pi\cdot 11$ promotes from a ledger identification to a theorem about U(1) coupling normalization on the cube $Q_3$. QED writes $\alpha^{-1}=(4\pi)\cdot(\mathrm{stiffness})/e^2$ with $e^2=1$; a genuine gauge-invariant stiffness counts independent plaquette field strengths, not ledger channels.
The cycle rank of the cube 1-skeleton is $b_1=E-V+1=12-8+1=5$ (equivalently six faces minus one Bianchi relation). Gauge fixing removes $V-1=7$ link redundancies, again leaving five physical link modes. The upstream definition cube_cycle_rank packages exactly this $b_1$.
By contrast the seed's $11=E-1$ drops only the active edge and ignores the seven gauge redundancies, so it is a recognition-channel count, not a photon stiffness.
proof idea
One-line definition: multiply $4\pi$ by the real cast of the cycle rank. No tactics; the value is fixed once cube_cycle_rank is fixed.
why it matters
This is the quantity that makes the M11 verdict sharp. Downstream, gauge_invariant_seed_eq_20pi evaluates it to $20\pi\approx 62.8$, and gauge_invariant_seed_excluded shows $20\pi<63<137.030<\alpha^{-1}$, so the genuine gauge-invariant U(1) seed cannot source the electromagnetic coupling. The verdict certificate U1NormalizationVerdict bundles that exclusion with the mismatch $11\neq 5$ between ledger channels and gauge degrees of freedom.
In the broader RS constants pipeline the $\alpha^{-1}$ band sits near $137$, forced by the ledger reading $4\pi\cdot 11$, not by Maxwell normalization on $Q_3$. The definition therefore closes the make-or-break test negatively: the seed remains a cross-consistent ledger number (also appearing in $\Omega_\Lambda=11/16$ and related counts) but is not a derived U(1) coupling on the cube.
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