Hopf, Hecke, Quantum-Group, and Braid Interfaces: Examples and Counterexamples

The flip $P$ has eigenvalue $1$ on the three-dimensional symmetric subspace, with basis $e_1\otimes e_1$, $e_1\otimes e_2+e_2\otimes e_1$, $e_2\otimes e_2$, and eigenvalue $-1$ on the one-dimensional antisymmetric subspace spanned by $e_1\otimes e_2-e_2\otimes e_1$. Therefore $R(u)=uI+iP$ has eigenvalues $u+i$ and $u-i$ on these two subspaces, while $\check{R}(u)=uP+iI$ has eigenvalues $u+i$ and $i-u$.

At $u=i$, $R(u)$ vanishes on the antisymmetric line and has rank three; at $u=-i$, it vanishes on the symmetric subspace and has rank one. Thus both $R(u)$ and $\check{R}(u)$ are invertible exactly away from $u=i$ and $u=-i$. At $u=0$, $R(0)=iP$ and $\check{R}(0)=iI$ are invertible.

Set $x=i$, $y=0$, $z=0$. Then the checked identity in [F2] has the factors ${\check{R}}_1(i)$, ${\check{R}}_2(i)$, and ${\check{R}}_1(0)$, and both ${\check{R}}_1(i)$ and ${\check{R}}_2(i)$ are singular by step 2.1. Thus these maps cannot be images of braid-group generators in an automorphism group: the required invertibility in [F3] fails.

At $u=0$ the checked operator is $iI$. Its local operators on every tensor power are scalar $iI$ and satisfy the Artin braid and distant-commutativity relations; they are invertible because $i$ is nonzero. Thus the assignment ${\sigma }_j\mapsto iI$ does define a braid-group representation, consistent with [F4].

For any unit $q$, expanding [F1] and putting $S=qT$ gives $S^2=(q^2-1)S+q^2$, equivalently $(S-q^2)(S+1)=0$; thus the multiplicative source parameter is $\xi =q^2$, while all calculations below keep the local $T$ generator. Now set $q=i$. Then [F1] gives $(T-i)(T+i^{-1})=(T-i{)}^2=0$, and [F2] identifies this as the complete specialized presentation. With $ϵ=T-i$, translation $T=ϵ+i$ gives mutually inverse homomorphisms between $H_2(i)$ and $\mathbb{C}[ϵ]\mathrm{/}({ϵ}^2)$. Every class has a unique form $a+bϵ$: terms of degree at least two vanish, and a polynomial of degree at most one lying in $({ϵ}^2)$ has both coefficients zero. Hence the algebra has basis $1,ϵ$ and $ϵ$ is nonzero.

For $q=1$ and $q=-1$, [F1] gives the same relation $T^2-1=0$, using [F2] for the specialization. The elements $e_{+}=(1+T)\mathrm{/}2$ and $e_{-}=(1-T)\mathrm{/}2$ are orthogonal idempotents with sum $1$. The classes $1,T$ are linearly independent because no nonzero polynomial of degree at most one is divisible by $T^2-1$, so $e_{+}$ and $e_{-}$ are nonzero and form a basis. Thus $H_2(q)=\mathbb{C}{e}_{+}\oplus \mathbb{C}{e}_{-}$, and each one-dimensional left ideal is simple. The regular module is semisimple.

The left ideal $ϵH_2(i)$ is one-dimensional, nonzero and proper by step 1.1, so the left regular module is not simple. If it were a direct sum of simple left ideals, its dimension two would force it to be a sum of two one-dimensional simple modules. On any one-dimensional module, left multiplication by $ϵ$ is a scalar $\lambda$ satisfying ${\lambda }^2=0$, hence $\lambda =0$. It would then annihilate both summands and therefore all of $H_2(i)$, contradicting $ϵ\cdot 1=ϵ\mathrm{\ne }0$. Thus the regular module is not semisimple.

Steps 2.1 and 1.2 establish the two different specialization outcomes under the stated normalized relation. They make no claim about higher ranks, centralizers, positive characteristic, or every root of unity. No choice principle is used: every basis, idempotent, and module complement calculation is explicit and finite.