Theorem: Signed slack representation

Signed, anchored, marginally compatible, continuous families are in bijection with continuous functionals $c_T$, one for every nonempty labeled set $T$, vanishing on diagonal states, through

Proof. Put $Q_S(\rho )=F_S(\rho )-\mathrm{diag}\rho$. Normalization gives ${\hat{Q}}_S(\rho )(\varnothing )=0$. Compatibility yields, for $T\subseteq S$,

Define

Walsh inversion gives the displayed representation and uniqueness. Continuity is inherited from $Q_T$, and anchoring makes $c_T$ vanish on diagonal states.

Conversely, choose such functionals and use the formula. Every nonempty character sums to zero, giving normalization. For a diagonal $\rho$, all ${\rho }_T$ are diagonal, so all corrections vanish.

To marginalize from $S$ to $U$, characters with an index outside $U$ cancel. For $T\subseteq U$, summation contributes the factor $2^{|S|-|U|}$, converting $2^{|T|-|S|}$ to $2^{|T|-|U|}$. This is exactly the formula for $F_U({\rho }_U)$. Continuity follows from continuous partial trace and the finitely many terms at each fixed size.