A rank-one Hecke algebra with a perfect symmetrizing form and degenerate regular trace

Since $i^{-1}=-i$, the defining polynomial is $(T-i{)}^2$, and the standard basis is $(1,T)$ by [F2]. Expanding the relation gives $T^2=1+2iT$.

Let $ϵ:=T-i$. The basis $(1,T)$ implies $ϵ\mathrm{\ne }0$, while the defining polynomial gives ${ϵ}^2=0$. For any simple left $H$-module $S$, $ϵS$ is a submodule because $H$ is commutative. If $ϵS\mathrm{\ne }0$, simplicity [F4] gives $ϵS=S$, and applying $ϵ$ again yields $0={ϵ}^{2}S=ϵS=S$, a contradiction. Thus $ϵ$ annihilates every simple module.

In the basis $(1,T)$, the columns of $L_T$ are $(0,1{)}^{\mathsf{T}}$ and $(1,2i{)}^{\mathsf{T}}$, so $[L_T]=\left(\begin{array}{cc}0& 1\\ 1& 2i\end{array}\right)$. Hence the regular-trace Gram matrix is $G=\left(\begin{array}{cc}\mathrm{Tr}(L_1)& \mathrm{Tr}(L_T)\\ \mathrm{Tr}(L_T)& \mathrm{Tr}(L_{T^2})\end{array}\right)=\left(\begin{array}{cc}2& 2i\\ 2i& -2\end{array}\right)$. Its determinant is $-4-(2i{)}^2=0$, so the regular-trace pairing is degenerate and the regular-trace determinant vanishes.

Define $\tau (1)=1$ and $\tau (T)=0$. From $T^2=1+2iT$, for $x=a+bT$ and $y=c+dT$ one obtains $\tau (xy)=ac+bd=\tau (yx)$. Its Gram matrix in $(1,T)$ is $\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)$, so the pairing $\tau (xy)$ is nondegenerate and perfect.

If the regular module $H$ were semisimple, it would be a direct sum of simple submodules by [F5], each annihilated by $ϵ$ by step 1.2; then $ϵ$ would annihilate all of $H$. But $ϵ\cdot 1=ϵ\mathrm{\ne }0$. Therefore $H$ is not semisimple, despite the perfect symmetrizing pairing in step 2.2. This refutes the statement.