The modular Hopf-link matrix squares to D² times duality

Since $s$ is invertible, its rows are linearly independent. If $h_i=h_j$, then $s_{ix}=d_i{h}_i(x)$ and $s_{jx}=d_j{h}_j(x)$ make the two rows proportional; therefore $h_i\mathrm{\ne }{h}_j$ for distinct labels. For $i\mathrm{\ne }{j}^{\ast }$, use symmetry and duality from [F1] to write ${\sum }_x{s}_{ix}{s}_{xj}={\sum }_x{s}_{xi}{s}_{x^{\ast }{j}^{\ast }}=d_i{d}_{j^{\ast }}{\sum }_x{h}_i(x)h_{j^{\ast }}(x^{\ast })\mathrm{.}$ The final sum is zero by [F2], since $h_i\mathrm{\ne }{h}_{j^{\ast }}$. Hence $(s^2{)}_{ij}=0$ unless $j=i^{\ast }$.

For the remaining entries, apply the character identity in [F1] with labels $i,i^{\ast }$: ${\sum }_x{s}_{xi}{s}_{x,i^{\ast }}={\sum }_w{N}_{i,i^{\ast }}^w{\sum }_x{d}_x{s}_{xw}\mathrm{.}$ Since $s_{xw}=d_w{h}_w(x)$, the inner sum is $d_w{\sum }_x{h}_w(x)h_0(x^{\ast })$, using $h_0(x^{\ast })=d_{x^{\ast }}=d_x$. By [F2] this vanishes for $w\mathrm{\ne }0$; for $w=0$ it equals ${\sum }_x{d}_x^2=D^2$. Thus the whole sum is $N_{i,i^{\ast }}^0{D}^2=D^2$ by [F3].

Steps 1.1 and 2.1 give $s^2=D^{2}C$. Since $C^2=I$ and $s$ is invertible, $D^2=0$ would force $s^2=0$, a contradiction. Thus $D^2\mathrm{\ne }0$; choosing a square root gives $S=s\mathrm{/}D$, and division of the identity by $D^2$ gives $S^2=C$.