Rank-one Frobenius defect sewing with every coefficient equal to one

Every finite-dimensional vector space is a finite direct sum of copies of $\mathbb{C}$, and a subspace has a linear complement by finite basis extension. The unit is simple, its endomorphisms are scalars, and there is exactly one simple isomorphism class. The flip obeys the braiding identities on pure tensors; its square is the identity. Ordinary evaluations and coevaluations satisfy both duality snakes on each basis vector. The canonical pivotal trace is the ordinary trace, hence $d_0=1$. The identity twist obeys the ribbon identities because the double flip and its dual are identities. The unnormalized Hopf-link matrix is $(\mathrm{tr}({\mathrm{id}}_{\mathbb{C}}))=(1)$ and is invertible, so this is a modular ribbon fusion category.

Under the selected tensor identifications, all four algebra maps have scalar $1$. Associativity, unitality, coassociativity, counitality, and the three Frobenius composites are consequently equal maps with scalar $1$ on their common one-dimensional source and target. In particular $m\mathrm{\Delta }=\mathrm{id}$, $ϵ\eta =1=d_A$, and $d_A\mathrm{\ne }0$. Both transposes of the pairing $a\otimes b\mapsto ab$ send $a$ to the functional $b\mapsto ab$, proving symmetry. Scalar bimodule endomorphisms of $A$ form $\mathbb{C}\mathrm{id}$, so $A$ is simple as a bimodule; the same holds for $B$. Scalar actions are associative, unital and commuting, verifying every module and bimodule clause for $M_{\mathrm{\ell }},M_r,X$.

The two balancing maps $X\otimes B\otimes M_{\mathrm{\ell }}\rightrightarrows X\otimes M_{\mathrm{\ell }}$ both send $x\otimes b\otimes m$ to the scalar $xbm$. Their coequalizer is therefore the identity of the one-dimensional target, with the canonical identification $X{\otimes }_B{M}_{\mathrm{\ell }}=\mathbb{C}$. Also $e_B=\mathrm{\Delta }\eta$ sends $1$ to $1\otimes 1$, so the explicit projector $P_{X,M_{\mathrm{\ell }}}$ sends $x\otimes m$ to $x\otimes m$. Choose its inclusion and restriction to be the identity under the selected tensor identifications. The descended $A$-action is scalar multiplication. The boundary projector on $M_r^{\vee }\otimes (X{\otimes }_B{M}_{\mathrm{\ell }})$ is identically $1$ by the same calculation.

Since every simple label is $0=\bar{0}$, both bulk defect-field spaces in [F1] are ${\mathrm{Hom}}_{\mathbb{C}\mathrm{\mid }\mathbb{C}}(\mathbb{C},\mathbb{C})=\mathbb{C}$; the induced left and right actions are scalar multiplication, including the inverse-braiding convention since both flips are identity on the unit. Choose the basis morphism $1\mapsto 1$ in each space. Both boundary-defect spaces, and both defect-free boundary spaces, are likewise ${\mathrm{Hom}}_{\mathbb{C}}(\mathbb{C},\mathbb{C})=\mathbb{C}$, with that same basis. The composition-trace pairing of two basis morphisms is $1$, so these bases are trace dual as well as composition dual.

In the unit-only RT handle decomposition there is one handle color, one simple summand at each cut, and every Hom tensor factor is $\mathrm{Hom}(\mathbb{C},\mathbb{C})=\mathbb{C}$ with basis $1$. Thus every state space specified in the Example is one-dimensional with the asserted tensor identifications. Every unit-colored closed ribbon, crossing, twist, evaluation or coevaluation has value $1$; a scalar coupon contributes precisely its scalar. The Kirby color has one summand of coefficient $d_0=1$. Moreover ${\sum }_i{d}_i^2=1$, the positive and negative Gauss sums are $1$, and the chosen $D=1$ gives $S_{00}=1$ and $\kappa =1$. In the normalized weighted surgery formula of [F2], every power of $D$ and every outgoing dimension factor equals $1$. The full anomaly factor is ${\kappa }^{n+a(R;\lambda ,\alpha )-\sigma (B)}=1$, since $\kappa =1$. Here $a(R;\lambda ,\alpha )\in \mathbb{Z}$ is the cap-reference correction in the exponent; its value need not be $1$ (and it is $0$ for the identity with matching reference). Only its multiplicative contribution ${\kappa }^{a(R;\lambda ,\alpha )}$ is $1$. The graph value is exactly the product of the scalar coupons, even for disconnected bordisms. In particular all allowed Lagrangian choices and Maslov corrections give this same scalar.

Composing two selected bordisms multiplies their scalar coupon products, just as composing their assigned linear maps does; the Maslov contribution is $1$ by step 5.1. The identity bordism has empty product $1$. Disjoint union concatenates coupon lists, agreeing with tensor product on $\mathbb{C}$ under $1\otimes 1\mapsto 1$. Evaluation and coevaluation on these state spaces both have scalar $1$ and satisfy the snakes, so the functor is compatible with duality. These checks verify directly the selected trivial normalized weighted TQFT, including its normalized tube maps, rather than assuming a general anomaly cancellation result.

Every coupon in either connecting sphere or disk two-point graph has scalar $1$. Hence its RT vector is $1$ by step 5.1. Its chosen block basis is also $1$, so the coefficient matrix is $(1)$ with inverse $(1)$. Both induced pairing maps $\mathbb{C}\to {\mathbb{C}}^{\ast }$ send $1$ to the functional taking $1$ to $1$, and are isomorphisms. Similarly the marked-torus state space is $\mathbb{C}$, and the FRS projector graph consists only of the selected scalar-$1$ structural coupons and ribbons. It acts as $1$; its image basis and its dual cap basis are both $\{1\}$. The source's normalization $(d_p{d}_q{)}^2\mathrm{/}(d_A{d}_X)$ is $1$, with nonzero denominator $d_A{d}_X=1$ in this example.

A connecting graph with the selected coupons, before or after cutting and capping, has correlator $1$ by step 5.1. The gluing map has scalar $1$ by step 6.1, regardless of the integer weight chosen to match the composite. In the bulk sum the only labels are $p=q=0$, the only field indices are $\alpha =\beta =1$, and $d_p{d}_q(c^{\mathrm{d}\mathrm{e}\mathrm{f}}{}^{-1}{)}_{\alpha \beta }=1\cdot 1\cdot 1=1$. In the boundary sum the only label is $i=0$ and $d_i(c^{\mathrm{b}\mathrm{d}\mathrm{e}\mathrm{f}}{}^{-1}{)}_{\beta \alpha }=1\cdot 1=1$. Each sewing identity is therefore the scalar equality $1=1$. Replacing the local boundary projector by its split inclusion/restriction changes only an identity map into the composite of two identity maps, so the defect-free boundary reduction also holds directly. This calculation establishes the claimed instance and uses no unresolved general factorization argument.