Support entropy of a profile equals log 3 minus its entropy product

The Shannon entropy of a profile's grade histogram is $\log 3 - \log E(a)$.

Let $\beta$ be a strictly positive integral ten-class profile, $D = \sum_r c_r \beta_r$, and $a = \beta/D$ the normalised profile. The fourth power $CW_6^{\otimes 4}$ has exactly forty-five supported ordered grade triples $\sigma$ — those with $\sigma_0+\sigma_1+\sigma_2 = 8$ — and the profile assigns to each the multiplicity $\mu_\beta(1,\sigma)$, giving a probability vector on the forty-five cells after dividing by the address length $3D$.

Then

where $H$ is entropy in bits and $E$ is the entropy product of Davie--Stothers.

This is the dictionary between the two ways the same quantity appears in the argument. The extraction counts addresses, so it meets the entropy of the forty-five-cell histogram; the rate formula of Equation (5.3) is written multiplicatively through $E$. The identity says they differ only by the constant $\log 3$, which cancels in every comparison of two profiles on the same marginal fibre. In particular, for two such profiles $a$ and $b$,

which is exactly the exponential rate of the star-degree ratio appearing in the general-profile outer capacity.

Formalization note. The combinatorial input is that the ten cyclic classes partition the forty-five supported triples, with the orbit of class $r$ having exactly $3c_r$ elements; both facts are finite checks. The constant $\log 3$ comes from the normalisation $3D$ rather than $D$, and the cancellation $\sum_r c_r a_r = 1$ is what makes it a constant rather than a profile-dependent term.