Algebraic braid relations do not imply unitarity

The map $R=2{\mathrm{id}}_{X\otimes X}$ is invertible. In the $\mathbb{C}$-linear strictification used by [F1], the transported map is $2\mathrm{id}$ on $X^{\mathrm{\prime }}\otimes X^{\mathrm{\prime }}$. Hence both local operators on $X^{\mathrm{\prime }\otimes 3}$ are $2\mathrm{id}$, and each side of the Yang–Baxter equation is $8\mathrm{id}$. Thus [F1] applies without identifying the tensor power with the original object $\mathbb{C}$.

By [F2] and [F3], ${\sigma }_1$ and ${\sigma }_2$ act by $2I$ and their braid relator holds, since both sides are $8I$. At the root Hecke parameter $v=2$, the valid specialization $\mathbb{Z}[v^{\pm 1}]\to \mathbb{C}$ sends each $T_i$ to $2$ and satisfies the root quadratic equation $(2-2)(2+1)=0$. Thus the scalar action also satisfies the Hecke quadratic relation in the stated normalization.

For any positive-definite inner product and any nonzero $z\in \mathbb{C}$, [F6] and nonnegativity of the norm give $\mathrm{\parallel }2z\mathrm{\parallel }=2\mathrm{\parallel }z\mathrm{\parallel }\mathrm{\ne }\mathrm{\parallel }z\mathrm{\parallel }$. By [F5], the generator $2I$ is not unitary for any such inner product. Therefore these algebraic braid and Hecke relations do not force unitarity.