denote the bimodule categories with the right and respectively left actions twisted by $F$. More precisely, for $V,W\in{\mathcal{C}}$ and $M\in{^F{\mathcal{M}}}$ we have $$V\cdot_{new} M\cdot_{new} W=F(V)\cdot_{old} M\cdot_{old} W,$$ with ${\mathcal{M}}^F$ defined analogously. Note that if $F$ is an equivalence then ${^F{\mathcal{M}}}\simeq{\mathcal{M}}^{F^{-1}}$.

Let $\#: {{_H\mathcal{M}}}\rightarrow {{_H\mathcal{M}}}$ be the functor which takes a left $H$-module $M$ to $M^\#\in {{_H\mathcal{M}}}$ where $M^\#$ is the same as $M$ as a vector space but the left action is modified by $S^{2}$, i.e., is now given by $h\cdot m= S^{2}(h)m$. If, as we always assume, $S$ is invertible, then $\#$ is an autoequivalence of ${{_H\mathcal{M}}}$. Thus for $i\in\mathbb{Z}$ we can consider ${{^{\#^i}_H\mathcal{M}}}$, i.e., $$V\cdot M\cdot W=V^{\#^i}\ot M\ot W.$$

We can now repeat verbatim the same arguments as in the usual, YD modules Vs center, case. Roughly speaking, let $M$ be in ${\mathcal{Z}}_{{{_H\mathcal{M}}}}{{^{\#^i}_H\mathcal{M}}}$, then $M$ is already a left $H$ module and for every $V\in{{_H\mathcal{M}}}$ we have an isomorphism $\Phi:V^{\#^i}\ot M\xrightarrow\sim M\ot V$. Take $V=H$ and define the right comodule structure on $M$ via $$\rho(m)=\Phi(1\ot m).$$ Conversely, suppose that $M$ is in ${_H\mathcal{YD}}^H_{-i}$. Then for every $V\in{{_H\mathcal{M}}}$ define $\Phi$ by $$\label{mcent}\Phi(v\ot m)=m{~\hspace{-3pt}_{\left<0\right>}}\ot m{~\hspace{-3pt}_{\left<1\right>}}v.$$ Note that the requirement that $\Phi$ be an $H$ module map is exactly the equation 