Modular Gauss sums, the Kirby color, and its slide identity

Pivotal trace additivity and tensor multiplicativity give $d_i{d}_j={\sum }_k{N}_{ij}^k{d}_k$. Frobenius reciprocity and the change of variable in the fusion sum then give ${\sum }_i{d}_i{N}_{ij}^k={\sum }_i{d}_i{N}_{k,j^{\ast }}^i=d_k{d}_{j^{\ast }}=d_k{d}_j\mathrm{.}$ Using the balancing formula from [F1] and summing over $i$ yields ${\sum }_i{\theta }_i{d}_i{s}_{ij}={\theta }_j^{-1}{\sum }_{i,k}{d}_i{N}_{ij}^k{\theta }_k{d}_k={\theta }_j^{-1}{d}_j{p}_{+}\mathrm{.}$

A simple strand $j$ encircled by a zero-framed simple loop $i$ is the scalar $s_{ij}\mathrm{/}{d}_j$ on $j$ by [F5]. Hence the $\mathrm{\Omega }$-meridian acts by $\frac{1}{d_j}{\sum }_i{d}_i{s}_{ij}=\frac{(s^2{)}_{0j}}{d_j}=D^2{\delta }_{j,0},$ where $d_j\mathrm{\ne }0$ by [F3]. Thus $D^{-2}$ times this meridian is the projector onto the unit channel. This proves the projector identity from modular inversion.

Multiply the last identity by $s_{j^{\ast },k}$ and sum over $j$. By [F1]–[F2], ${\sum }_j{s}_{ij}{s}_{j^{\ast },k}={\sum }_x{s}_{i^{\ast },x}{s}_{xk}=(s^2{)}_{i^{\ast },k}=D^2{\delta }_{i,k}\mathrm{.}$ Therefore $p_{+}{\sum }_j{\theta }_j^{-1}{d}_j{s}_{j^{\ast },k}=D^2{\theta }_k{d}_k$. Reindexing $j$ by $j^{\ast }$ and taking $k=0$ gives $p_{+}{p}_{-}=D^2$; since $p_{+}{p}_{-}=D^2\mathrm{\ne }0$, both Gauss sums are nonzero and ${\kappa }^{-1}=p_{-}\mathrm{/}D$.

By [F4], a $\pm 1$-framed $\mathrm{\Omega }$ unknot evaluates to ${\sum }_i{d}_i{\theta }_i^{\pm 1}{d}_i=p_{\pm }$. For a $+1$ blow-up, the normalization changes by $D^{-1}{\kappa }^{-1}{p}_{+}=1$; for a $-1$ blow-up it changes by $D^{-1}\kappa p_{-}=p_{+}{p}_{-}\mathrm{/}{D}^2=1$. The empty surgery diagram gives $D^{-1}$. By [F5], a zero-framed unknot colored by $\mathrm{\Omega }$ evaluates to ${\sum }_i{d}_i^2=D^2$, so its one-component normalization is $D^{-2}{D}^2=1$, the $S^2\times S^1$ value.

Apply [F6] to the full coupon-bearing graph. Its finite coend identity preserves the entire band-slide morphism with every boundary factor and exterior coupon fixed, for either two surgery components or an open colored strand slid over a surgery component. It is independent of the simple-meridian computation of step 1.2. A surgery-component slide changes the linking matrix by the integral basis congruence $B\mapsto P^{\mathsf{t}}BP$, preserving signature even when $B$ is singular, and preserves the number of surgery components. An open graph-strand slide changes neither surgery matrix nor component count. Therefore every normalization factor is unchanged, proving the full normalized slide identities. Together with step 3.1 this proves all the stated local move identities; it imports no relative topological move-completeness theorem.