Colored signed twist surgery, circumcision, and paired-handle cancellation

Choose actual finite simple inclusions $u_{k,a}:k\to X$ and projections $v^{k,a}:X\to k$ from a direct sum, with $v^{k,a}{u}_{l,b}={\delta }_{kl}{\delta }_{ab}{1}_k$ and ${\sum }_{k,a}{u}_{k,a}{v}^{k,a}=1_X$. Inserting these two identities on either side of any coupon $f:X\to Y$ gives its complete expansion $f={\sum }_{k,a,b}{u}_{k,b}^Y(v_Y^{k,b}f{u}_{k,a}^X)v_X^{k,a}$. Off-diagonal simple terms vanish by Schur's lemma; every remaining middle term is a scalar. Thus this expansion is independent of basis and respects composition/tensor products. Since ${\mathrm{tr}}_k(v^{k,a}{u}_{k,b})=d_k{\delta }_{ab}$, trace-dual projections are $v^{k,a}\mathrm{/}{d}_k$, and the trace-dual resolution is $1_X={\sum }_{k,a}{d}_k{u}_{k,a}(v^{k,a}\mathrm{/}{d}_k)$. Reversing an edge applies the dual functor and reverses the composite, preserving the identity.

Trace multiplicativity and Frobenius reciprocity give $d_i{d}_j={\sum }_k{N}_{ij}^k{d}_k$ and ${\sum }_i{d}_i{N}_{ij}^k=d_k{d}_j$. Summing the balancing expression $s_{ij}=({\theta }_i{\theta }_j{)}^{-1}{\sum }_k{N}_{ij}^k{\theta }_k{d}_k$ therefore gives ${\sum }_i{\theta }_i{d}_i{s}_{ij}=p_{+}{\theta }_j^{-1}{d}_j$. Multiply by $s_{j^{\ast },k}$ and sum over $j$. The identity ${\sum }_j{s}_{ij}{s}_{j^{\ast },k}=D^2{\delta }_{ik}$ gives $p_{+}{\sum }_j{\theta }_j^{-1}{d}_j{s}_{j^{\ast },k}=D^2{\theta }_k{d}_k$. At $k=0$, reindexing $j$ by $j^{\ast }$ yields $p_{+}{p}_{-}=D^2$. Consequently $p_{\pm }\mathrm{\ne }0$ and ${\kappa }^{-1}=p_{-}\mathrm{/}D$.

For the unit label, put $x_a=u_{0,a}$, $y_a=v^{0,a}$. Then $y_a{x}_b={\delta }_{ab}$ and $P_X={\sum }_a{x}_a{y}_a$ is the unit-isotypic projector. It is the identity on $\mathrm{Hom}(\mathbf{1},X)$ and on the opposite dual pairing, because all nonunit components of a map from or to the simple unit vanish. The bending map $f:k\to X\mapsto (f\otimes 1)b_k$ has inverse $z\mapsto (1_X\otimes e_k)(z\otimes 1_k)$ by the two snakes. Closing a bent pair $f,g$ gives ${\mathrm{tr}}_k(gf)$, so the same $d_k$ factors of step 1.1 apply after bending.

The displayed partial trace satisfies ${\mathrm{ptr}}_Z((f\otimes 1)a(g\otimes 1))=f{\mathrm{ptr}}_Z(a)g$ by interchange. Naturality of both braidings consequently makes each $K_{+}(X)$ natural in $X$. On simple $j$, step 1.2 gives $K_{+}(j)=p_{+}{\theta }_j^{-1}{1}_j$. Inserting the full direct-sum resolution gives $K_{+}(X)={\sum }_{j,a}{u}_{j,a}{K}_{+}(j)v^{j,a}=p_{+}{\theta }_X^{-1}$ for every object. For the negative version, balancing gives $M_{j,i}^{-1}=({\theta }_j\otimes {\theta }_i){\theta }_{j\otimes i}^{-1}$, hence ${\mathrm{tr}}_j{K}_{-}(j)={\sum }_{i,k}{d}_i{\theta }_i^{-1}{\theta }_j{\theta }_i{N}_{ji}^k{\theta }_k^{-1}{d}_k={\theta }_j{d}_j{p}_{-}$. Divide by $d_j$ and use the same naturality/resolution to obtain $K_{-}(X)=p_{-}{\theta }_X$.

For arbitrary coupons $a:A\to X$ and $b:X\to B$, step 2.2 says $b{\theta }_X^{\pm 1}{K}_{\pm }(X)a=p_{\pm }ba$. Take $X$ to be the tensor product of every oriented strand through the local surgery annulus, using duals for reversed strands. Repeated hexagons expand $M_{X,i}$ into the full encircling strand word; repeated balancing expands ${\theta }_X$ into all individual twists and pairwise crossings in its full bunch twist. This explicitly identifies the complete signed twist-surgery composite even if exterior coupons join different members of the bunch. If its eliminated framing is $ϵ=\pm 1$ and linking vector is $\mathrm{\ell }$, its matrix $B=\left[\begin{array}{cc}ϵ& {\mathrm{\ell }}^t\\ \mathrm{\ell }& A\end{array}\right]$ is congruent to $[ϵ]\oplus (A-ϵ\mathrm{\ell }{\mathrm{\ell }}^t)$ by replacing each remaining vector $z$ with $z-ϵ{\mathrm{\ell }}_{z}u$. Thus $m$ decreases by1 and signature by$ϵ$. Its graph scalar is $p_{ϵ}$, and $D^{-1}{\kappa }^{-ϵ}{p}_{ϵ}=1$, proving the normalized signed-move equality with its exact framing/signature factors.

The same naturality makes $E_X$ natural. On simple $j$ its scalar is $d_j^{-1}{\sum }_i{d}_i{s}_{ji}=D^2{\delta }_{j0}$ by $s^2=D^{2}C$. Inserting step 1.1 proves $E_X=D^2{P}_X$ on every object, not just a simple strand. Thus $D^{-2}y{E}_{X}x=yx$ for any $x:\mathbf{1}\to X$, $y:X\to \mathbf{1}$, with arbitrary coupon networks defining $x,y$.

Rigidity gives $\mathrm{Hom}(\mathbf{1},i^{\ast }\otimes j)\cong \mathrm{Hom}(i,j)$, zero unless $i=j$. When $i=j$ its generator is $b_i^R$, and $e_i{b}_i^R=d_i$ by sphericality. Hence $P_{i^{\ast }\otimes j}={\delta }_{ij}{b}_i^R{e}_i\mathrm{/}{d}_i$. Insert this into $R_{ij}(E_{i^{\ast }\otimes j})$. The right snake $(r_i\otimes 1_i)(1_i\otimes b_i^R)=1_i$ retracts its first pair, and the left snake $(e_i\otimes 1_{i^{\ast }})(1_{i^{\ast }}\otimes b_i)=1_{i^{\ast }}$ retracts the remaining pair. The resulting full morphism is ${\delta }_{ij}(D^2\mathrm{/}{d}_i)1_{i\otimes i^{\ast }}$. Tensoring $g$ such disjoint pieces gives zero unless the entire color tuples agree and gives $D^{2g}\mathrm{/}{\prod }_r{d}_{i_r}$ times the identity otherwise, with every multiplicity-space coupon retained.

For circumcision introduce a framed knot $K$ and a small zero-framed meridian $U$ meeting only $K$. In the finite $\mathrm{\Omega }$ sum, $U$ acts on the simple $K$-color $j$ by $D^2{\delta }_{j0}$. The surviving coefficient is $d_0=1$, and the unit-colored knot, with all its framing and crossings, evaluates as the empty graph. Thus the full arbitrary-coupon exterior evaluation is multiplied by exactly $D^2$. If $K$ has framing $f$ and links the old surgery link by $\mathrm{\ell }$, the new matrix is $B^{\mathrm{\prime }}=\left[\begin{array}{ccc}A& \mathrm{\ell }& 0\\ {\mathrm{\ell }}^t& f& 1\\ 0& 1& 0\end{array}\right]$. Subtracting ${\mathrm{\ell }}_z$ times the new meridian vector from each old vector gives $B^{\mathrm{\prime }}\cong A\oplus \left[\begin{array}{cc}f& 1\\ 1& 0\end{array}\right]$. The last block has determinant$-1$ and signature0. Two new components contribute $D^{-2}$ and cancel $D^2$, with no signature or weight change. This proves the complete stated local identities, while leaving the genuine relative move-completeness theorem separate.