Coherence of the Ising fusion data

For triples with no unit label, the two bracketings have the same direct-sum decomposition: $\psi \psi \psi$ gives $\psi$; $\psi \psi \sigma$, $\psi \sigma \psi$, and $\sigma \psi \psi$ give $\sigma$; $\psi \sigma \sigma$, $\sigma \psi \sigma$, and $\sigma \sigma \psi$ give $\mathbf{1}\oplus \psi$; and $\sigma \sigma \sigma$ gives $2\sigma$. Triples containing $\mathbf{1}$ reduce by the explicitly specified canonical unitors. Thus the fusion products define the same channel bases on both sides of each associator. Each component category of finite-dimensional Hilbert spaces and linear maps has kernels and cokernels (the underlying finite-dimensional vector-space kernels and quotients with induced Hilbert structures), and each subspace has an orthogonal complement; hence the triple category is abelian and every short exact sequence splits. It has exactly three simple isomorphism classes, all Hom spaces are finite-dimensional, every object is a finite direct sum of simples, and $\mathrm{End}(\mathbf{1})=\mathbb{C}$, so it is finite semisimple with simple unit.

The dagger is conjugate-linear, involutive, and contravariant for composition because it is the Hilbert adjoint on every multiplicity block; the trace pairing also gives $⟨g\circ f,h⟩=⟨f,g^{†}\circ h⟩$ for composable maps by direct substitution in each block. All weights $d_a$ are positive, so the displayed Hom inner product is positive definite. The weights are multiplicative across the fusion rules: $d_{\mathbf{1}}{d}_a=d_a$, $d_{\psi }^2=d_{\mathbf{1}}$, $d_{\psi }{d}_{\sigma }=d_{\sigma }$, and $d_{\sigma }^2=d_{\mathbf{1}}+d_{\psi }$. Thus, for morphisms $f,f^{\mathrm{\prime }}:X\to X^{\mathrm{\prime }}$ and $g,g^{\mathrm{\prime }}:Y\to Y^{\mathrm{\prime }}$, expanding the tensor-product blocks gives $⟨f\otimes g,f^{\mathrm{\prime }}\otimes g^{\mathrm{\prime }}⟩={\sum }_{a,b,c}{d}_c{N}_{ab}^c\mathrm{Tr}(f_a^{†}{f}_a^{\mathrm{\prime }})\mathrm{Tr}(g_b^{†}{g}_b^{\mathrm{\prime }})=⟨f,f^{\mathrm{\prime }}⟩⟨g,g^{\mathrm{\prime }}⟩$, with the identity extending by sesquilinearity to finite sums. The tensor product is dagger-compatible since each block is a direct sum of Kronecker products of the component maps. The associator acts by the stated $F$ matrices on fusion-channel factors and by identity maps on multiplicity factors, whereas morphisms act on multiplicity factors; hence every naturality square commutes. Its Hadamard block divided by $\sqrt{2}$ is unitary, as are the one-channel blocks $1$ and $-1$, so the associator is unitary; the canonical unitors are unitary because their fusion-basis factors have norm one. On the dual multiplicity spaces, the left evaluation is the adjoint of the right coevaluation and the right evaluation is the adjoint of the left coevaluation: each adjoint is the canonical dual-basis pairing map in the same ordered tensor factors, with the same positive coefficient ${\lambda }_a$.

In the pentagon equation, write $F_d^{abc}[x,y]=0$ whenever either intermediate fusion channel is forbidden. For each choice of simple labels $a,b,c,d,e,f,g,k,\mathrm{\ell }\in \{\mathbf{1},\psi ,\sigma \}$, the two paths are $F_e^{fcd}[g,\mathrm{\ell }]F_e^{ab\mathrm{\ell }}[f,k]={\sum }_{h\in \{\mathbf{1},\psi ,\sigma \}}{F}_g^{abc}[f,h]F_e^{ahd}[g,k]F_k^{bcd}[h,\mathrm{\ell }]$. The exact checker research/advanced-qft-supersymmetry-build-2026-10-23-batch-39-ising-pentagon-check.py substitutes the specified $F$-rules using rational arithmetic in $\mathbb{Q}(\sqrt{2})$ over all $3^9=\mathrm{19,683}$ outer label assignments. It finds 132 assignments with a nonzero side, verifies equality in all 132, and finds the remaining 19,551 assignments are $0=0$. The checker exits successfully only at those exact totals. Thus the pentagon holds for simple labels; multilinearity extends it to finite direct sums.

Every associator coefficient involving $\mathbf{1}$ is $1$. On an elementary tensor $(v\otimes z\otimes u_{a\mathbf{1}}^a)\otimes w\otimes u_{ab}^c$ in $(X\otimes \mathbf{1})\otimes Y$, the route through the associator and the left unitor sends it to $v\otimes zw\otimes u_{ab}^c$, while the route through the right unitor sends it to $zv\otimes w\otimes u_{ab}^c$. These are equal by the balancing relation of the complex tensor product. Such tensors span, proving the triangle for all objects. For $\psi$, each of the two snake composites has scalar $1\cdot 1=1$. For $\sigma$, each composite uses the unit-channel associator coefficient $[F_{\sigma }^{\sigma \sigma \sigma }{]}_{\mathbf{1},\mathbf{1}}=1\mathrm{/}\sqrt{2}$ and the two duality coefficients ${\lambda }_{\sigma }=2^{1\mathrm{/}4}$, giving ${\lambda }_{\sigma }^2\mathrm{/}\sqrt{2}=1$; the $\psi$ channel contributes zero because its evaluation coefficient is zero. For $X=(H_a{)}_a$, the dual object and both coevaluations are the stated direct sums of dual-basis tensors, while evaluation pairs only equal simple labels. Consequently every off-diagonal simple-label block of either snake composite is zero, and on each multiplicity space $H_a$ the dual-basis contraction is the identity multiplied by the corresponding simple-label snake scalar just computed. Both snake identities therefore hold for arbitrary finite direct sums, not only for simple labels.

The candidate has finite-dimensional Hom spaces, exactly three simple isomorphism classes, and every object a finite direct sum of simples; its unit is simple. Its tensor product is bilinear and its natural associator and canonical unitors satisfy the pentagon and triangle, while every object has the left and right duals and snake identities checked above. By [F1] it is a finite semisimple fusion category. Step 1.2 additionally supplies the positive dagger structure, proves tensor compatibility, and verifies unitary associators and adjoint duality maps, so the category is unitary in the explicit sense defined in the Statement. This is a purely mathematical result and makes no QFT-realization claim.