kappa_sph
plain-language theorem explainer
The dimensionless sphaleron prefactor is fixed by Q₃ topology as three Hamiltonian cycles on K₄ times four edges each, divided by the square of the even sign-flip count in three dimensions, equaling 3/4. Cosmologists writing Γ_sph/T⁴ = κ α_W⁵ cite this as the O(1) combinatorial coefficient. The body is a direct arithmetic quotient of three named combinatorial constants, not a derived inequality.
Claim. Define the sphaleron rate prefactor by $\kappa_{\mathrm{sph}}=(N_{\mathrm{Ham}}\cdot E_{\mathrm{cyc}})/|G_{\mathrm{even}}|^2$, where $N_{\mathrm{Ham}}=3$ is the number of Hamiltonian cycles on $K_4$, $E_{\mathrm{cyc}}=4$ is edges per cycle, and $|G_{\mathrm{even}}|=2^{3-1}=4$ is the order of the even sign-flip subgroup in spatial dimension three. Equivalently $\kappa_{\mathrm{sph}}=12/16=3/4$.
background
In the electroweak plasma above the phase transition, baryon-number violating transitions are mediated by sphalerons. The standard thermal rate per unit volume is written $\Gamma_{\mathrm{sph}}/T^4=\kappa_{\mathrm{sph}},\alpha_W^5$, with $\alpha_W$ the weak coupling and $\kappa_{\mathrm{sph}}$ a dimensionless O(1) prefactor. Lattice estimates place $\kappa_{\mathrm{sph}}$ roughly in $0.1$–$1$.
Recognition Science fixes that prefactor from Q₃ topology. A sphaleron path is a topologically nontrivial trajectory through SU(2) configuration space that changes all three winding numbers at once. On the cube, such paths are Hamiltonian cycles through the even sign-flip subgroup $(\mathbb{Z}/2\mathbb{Z})^2$, which has order $2^{D-1}$. For $D=3$ that order is $4$; the complete graph $K_4$ on those four vertices has three distinct Hamiltonian cycles (up to direction), each with four edges.
The three ingredients are therefore: Hamiltonian-cycle count $3$, edges per cycle $4$, and even sign-flip count $2^{3-1}=4$. Their combination is the definition under study.
proof idea
Definitional, not a proof. The real is the product of the two natural-number constants (three Hamiltonian cycles on $K_4$, four edges per cycle), cast to $\mathbb{R}$, divided by the square of the even sign-flip count at $D=3$. Unfolding those three defs and normalizing yields $12/16=3/4$, which is recorded separately as the equality lemma.
why it matters
This constant is the combinatorial heart of the module: every rate formula and certificate in Cosmology.SphaleronRate reads it. Downstream, the dimensionless rate is $\kappa_{\mathrm{sph}},\alpha_W^5$; positivity and the bound $\kappa_{\mathrm{sph}}<1$ are one-line rewrites of the equality to $3/4$. The provenance structure packages $\kappa_{\mathrm{sph}}=3/4$ together with positivity of $\alpha_W$ and of the rate, marking $\kappa$ as structural (from Q₃ cycles) while $\alpha$ still carries a boundary datum.
In the broader RS chain this sits on the $D=3$ forcing (T8) and the cube gauge construction: the even sign-flip subgroup of order $2^{D-1}$ is exactly the vertex set whose Hamiltonian cycles count the sphaleron channels. The numerical value $3/4=0.75$ lands inside the lattice window $0.1$–$1$, so the combinatorial prediction is phenomenologically viable. It also feeds the baryon-asymmetry derivation path, where the sphaleron rate sets the washout scale above the electroweak transition.
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