The ordered-basis Hecke operator and its normalization

Let $F=\mathbb{Q}(q)$, where $q$ is an indeterminate, let $N\ge 1$, and let $V_N$ have ordered basis $e_1<\cdots <e_N$. Put $c=q-q^{-1}$. Define $T\in {\mathrm{End}}_F(V_N\otimes V_N)$ by

$T(e_a\otimes e_b)=\{\begin{array}{ll}q{e}_a\otimes e_a,& a=b,\\ e_b\otimes e_a,& a<b,\\ e_b\otimes e_a+c{e}_a\otimes e_b,& a>b\mathrm{.}\end{array}$

For an invertible operator $R$ on $V_N\otimes V_N$, the phrase Hecke symmetry of normalized parameter $q$ means that the operators $R_1=R\otimes 1$ and $R_2=1\otimes R$ on $V_N^{\otimes 3}$ satisfy $R_1{R}_2{R}_1=R_2{R}_1{R}_2$, and that $(R-qI)(R+q^{-1}I)=0$. This is a property in the definition, not an assertion that every invertible operator has it. The next item proves that the displayed $T$ has this property.

The alternative generator $S=qT$ has parameter $Q=q^2$ and satisfies $(S-QI)(S+I)=0$ whenever $T$ satisfies its normalized quadratic relation. At $q=1$ the displayed $T$ is the ordinary flip. At $q=-1$ the quadratic relation becomes $T^2=I$, while the displayed operator is not the ordinary flip.