define $``\rightharpoonup"$ as follows: For any $\psi\in H^0$, $y, z\in Y$, $$\psi\rightharpoonup \Theta_{y, z}=\Theta_{[\psi_{(2)}\triangleright y]\,, \,[S^0 (\psi_{(1)}^* )\triangleright z]} .$$

\[lem:identity\]With the above notations, for any $\psi\in H^0$ and $T\in {}_C {\mathbf B}(Y)$, $$[\psi\rightharpoonup T](x)=\psi_{(2)}\triangleright T(S^0 (\psi_{(1)})\triangleright x)$$ for any $x\in Y$.

Since ${}_C {\mathbf B}(Y)$ is the linear span of the rank-one left $C$-module maps on $Y$, it suffices to show that $$[\psi\rightharpoonup \Theta_{y, z}](x)=\psi_{(2)}\triangleright \Theta_{y, z}(S^0 (\psi_{(1)})\triangleright x)$$ for any $\psi\in H^0$ and $x, y, z\in Y$. For any $\psi\in H^0$ and $x, y, z\in Y$, $$\begin{aligned}
[\psi\rightharpoonup\Theta_{y, z}](x) & =\Theta_{[\psi_{(2)}\triangleright y] , [S^0 (\psi_{(1)}^* )\triangleright z]}(x)
={}_C {\langle}x \, , \, S^0 (\psi_{(1)}^* )\triangleright z {\rangle}\cdot [\psi_{(2)}\triangleright y] \\
& ={}_C {\langle}\psi_{(2)}S^0 (\psi_{(1)})\triangleright x \, , \, S^0 (\psi_{(3)}^* )\triangleright z {\rangle}\cdot [\psi_{(4)}\triangleright y] \\
& =[\psi_{(2)}\cdot_{\widehat{\rho}}{}_C {\langle}S^0 (\psi_{(1)})\triangleright x \, , \, z {\rangle}]\cdot [\psi_{(3)}\triangleright y] \\
& =\psi_{(2)}\triangleright ({}_C {\langle}S^0 (\psi_{(1)})\triangleright x \, ,\, z {\rangle}\cdot y) \\
& =\psi_{(2)}\triangleright \Theta_{y, z}(S^0 (\psi_{(1)})\triangleright x) .\end{aligned}$$ Hence we obtain the conclusion.

\[lem:action\]With the above notations, the bilinear map $`` \rightharpoonup "$ from $H^0 \times D$ to $D$ is an action of $H^0$ on $D$.

Let $T, R\in {}_C {\mathbf B}(Y)$ and $\psi, \chi \in H^0$. For 