Unified Framework

\textbf{Balanced (not braided) functor.} The target category $\mathcal{C}=\mathrm{Rep}(\mathrm{SU}(3)\times \mathrm{SU}(2)\times \mathrm{U}(1))$ is symmetric monoidal; we therefore use a \emph{balanced monoidal} (a.k.a. ribbon into symmetric) functor that maps the source braiding to the symmetry (flip) and the twist to the identity. \textbf{Normalization axiom.} We fix the $\mathrm{U}(1)$ normalization by sending the tensor unit to the trivial representation, equivalent to setting $y_0=0$ in the hypercharge functional.

\textbf{Assumptions.} Let $\mathcal{H}$ be a separable Hilbert space with ONB $\{\lvert N,w,T;b\rangle\}$, with $N,w,T\in\mathbb{Z}$ and finite degeneracy $b$. Define (multiplication) operators on the finite-support core $\mathcal{D}_0$: \[ \hat N\lvert N,w,T;b\rangle=N\lvert N,w,T;b\rangle,\quad \hat w\lvert N,w,T;b\rangle=w\lvert N,w,T;b\rangle,\quad \hat T\lvert N,w,T;b\rangle=T\lvert N,w,T;b\rangle, \] which are self-adjoint and strongly commute (simultaneously diagonal). Let $f(N,w,T)=\Lambda_c e^{\lambda_c N}+\alpha_c w+\kappa_c T^2$ (real Borel on $\mathbb{Z}^3$). \textbf{Theorem.} Let $E$ be the joint PVM of $(\hat N,\hat w,\hat T)$. Define \[ M_{\text{topo}} := f(\hat N,\hat w,\hat T) := \int_{\mathbb{Z}^3} f(N,w,T)\, dE(N,w,T) \] with domain \[ \mathcal{D}_f=\Big\{\psi\in\mathcal{H}:\ \int_{\mathbb{Z}^3} \lvert f\rvert^2\, d\mu_\psi<\infty\Big\},\qquad \mu_\psi(A)=\langle\psi,E(A)\psi\rangle. \] Then $M_{\text{topo}}$ is self-adjoint on $\mathcal{D}_f$. \textbf{Proof.} Strong commutativity implies the existence of a joint PVM (joint spectral theorem). Since $f$ is real Borel, $f(\hat N,\hat w,\hat T)$ is self-adjoint with the stated domain by functional calculus. On a countable set $\mathbb{Z}^3$, every function is Borel; $\mathcal{D}_0$ is a core for each multiplication operator. \quad \square

\textbf{Assumptions.} Let $\mathcal{R}$ be the free strict ribbon category on one framed object $X$, with duals, twist $\theta$, and braiding $c$, modulo ribbon relations (Joyal--Street). Let $\mathcal{C}=\mathrm{Rep}(\mathrm{SU}(3)\times \mathrm{SU}(2)\times \mathrm{U}(1))$ with its symmetric monoidal (balanced) structure. Let $g:\mathrm{Ob}(\mathcal{R})\to\mathbb{Z}^3$ record $(N,w,T)$ additively under $\otimes$. \textbf{Construction.} On objects, set \[ F(X^{\otimes n})\ :=\ \bigoplus_{(N,w,T)=g(X^{\otimes n})}\ \rho_{c(N,w,T)}\ \boxtimes\ \sigma_{T_3(N,w,T)}\ \boxtimes\ \chi_{Y(N,w,T)}, \] with color class $c\equiv N+2w+T\!\!\pmod 3$, an isospin ladder producing doublets/singlets, and hypercharge $Y$ as in (3). On morphisms: evaluation/coevaluation $\mapsto$ rigid duality in $\mathcal{C}$; $\theta\mapsto \mathrm{id}$; $c\mapsto$ the symmetry (flip). \textbf{Theorem (by universal property).} The above assignment extends uniquely to a balanced monoidal functor $F:\mathcal{R}\to\mathcal{C}$, since the images satisfy the ribbon relations in the symmetric target. \quad \square \textit{Remark.} $F$ need not be faithful/full on braiding: mapping $c$ to the symmetry forgets non-trivial braiding phases; this is acceptable as only object labels (reps/charges) are used physically.

\textbf{Assumptions.} Doublet invariance: $y_T=y_w$. Color-blind: $y_N=0$. One-family labels as fixed; SM targets $Y(\ell_L)=-1$, $Y(e_R)=-2$, $Y(q_L)=+\tfrac13$, $Y(u_R)=+\tfrac43$, $Y(d_R)=-\tfrac23$. Normalization: tensor unit $\mapsto$ trivial $\mathrm{U}(1)$ rep $\Rightarrow y_0=0$. \textbf{System.} Unknowns $(y_w,y_T,y_L,y_R,y_0)$ with $y_T=y_w$, $y_N=0$: \[ \begin{array}{rcl} 2y_w+y_L+y_0&=&-1 \quad (\ell_L)\\ 2y_w+y_R+y_0&=&-2 \quad (e_R)\\ 4y_w+y_L+y_0&=&+\tfrac13 \quad (q_L)\\ 7y_w+y_R+y_0&=&+\tfrac43 \quad (u_R)\\ 4y_w+y_R+y_0&=&-\tfrac23 \quad (d_R) \end{array} \] \textbf{Solution and Lemma (uniqueness).} Subtracting $(\ell_L)$ from $(q_L)$ gives $2y_w=\tfrac{4}{3}\Rightarrow y_w=y_T=\tfrac{2}{3}$. Then $y_L=-\tfrac{7}{3}$, $y_R=-\tfrac{10}{3}$, and $y_0=0$ by normalization. The $4\times 4$ system in $(y_w,y_L,y_R,y_0)$ has rank $4$, hence the solution is unique. \[ \boxed{\,y_N=0,\ y_w=y_T=\tfrac{2}{3},\ y_L=-\tfrac{7}{3},\ y_R=-\tfrac{10}{3},\ y_0=0\,}. \] \quad \square

\textbf{Convention.} Sum over left-handed Weyl fields; a right-handed field with hypercharge $Y$ contributes as its LH conjugate with hypercharge $-Y$ and conjugate non-abelian representation. Dynkin indices: $T(\mathbf{3})=\tfrac12$ for $\mathrm{SU}(3)$ fundamentals; $T(\mathbf{2})=\tfrac12$ for $\mathrm{SU}(2)$ doublets. \textbf{SU(3)$^2$U(1).} Per color: $q_L$ gives $2\cdot Y(q_L)=\tfrac{2}{3}$, $u_R^c$ gives $- \tfrac{4}{3}$, $d_R^c$ gives $+\tfrac{2}{3}$; sum $0$. With $3$ colors: still $0$. \textbf{SU(2)$^2$U(1).} $3$ quark doublets with $Y=\tfrac13$ plus one lepton doublet with $Y=-1$: $3\cdot \tfrac13 + (-1) = 0$. \textbf{U(1)$^3$.} Leptons: $(-1)^3+(-1)^3+(+2)^3=6$. Quarks per color: $2\cdot(\tfrac13)^3 + (-\tfrac{4}{3})^3 + (+\tfrac{2}{3})^3 = -2$. With $3$ colors: $-6$. Total $6+(-6)=0$. \textbf{Grav$^2$U(1).} Leptons: $(-1)+(-1)+(+2)=0$. Quarks per color: $(\tfrac13)+(\tfrac13)+(-\tfrac{4}{3})+(+\tfrac{2}{3})=0$; with $3$ colors: $0$. Thus all anomaly coefficients vanish per family. \quad \square