Given a monoidal category $M$ one can consider its Drinfeld centre $Z(M)$. Objects of the Drinfeld centre are pairs $(m, \alpha)$ where $m$ is an object and $\alpha$ is an isomorphism $\alpha: - \otimes m \to m \otimes -$ satisfying some "obvious" conditions.


A simple and important example of a monoidal category is the category of $G$-equivariant sheaves of $k$-vector spaces on $Y \times Y$ where $Y$ is a finite $G$-set. The monoidal structure is given by 


$V\otimes W = p\_{13\*} (p\_{12}^\*V \otimes p\_{23}^\* W)$


where $p\_{ij} : Y \times Y \times Y \to Y \times Y$ denotes the projection map (in a hopefully obvious notation).


For example, if $Y$ is a point then one recovers the (tensor) category of representations of $G$. If $Y = G$ then one recovers a monoidal category equivalent vectors spaces graded by $G$. If $G$ is the trivial group then one obtains the tensor category of ``matrices of vector spaces'' over $Y$.


Now there is a result, which I have heard (by Ostrik) called ``Muerger's Morita invariance of Drinfeld centre''. It should have the consequence that, with $G$ and $Y$ as above:



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> The Drinfeld centre $Z(Sh\_G(Y \times Y))$ does not depend on $Y$ up to equivalence.
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(I guess the baby example of $G$ the trivial group explains the term ``Morita invariance''.)


My question is:



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> Where can I read about these results in the literature?
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Morita invariance of Drinfeld centre
To expand on Noah's answer, Müger shows that if two fusion categories are Morita equivalent, in the sense that there is an invertible bimodule between them, then their Drinfeld centers are equivalent. In fact, the reverse implication is also true: this is Theorem 3.1 in [a different Etingof-Nikshych-Ostrik paper](http://arxiv.org/abs/0809.3031). So two fusion categories having equivalent centers is the same as Morita equivalence.


In your case, if $Y$ and $Y'$ are two $G$-sets, then $Sh\_G(Y \times Y')$ is a bimodule from $Sh\_G(Y \times Y)$ to $Sh\_G(Y' \times Y')$ with the obvious left and right actions. This bimodule is invertible, with inverse $Sh\_G(Y' \times Y)$. In other words, $$Sh\_G(Y \times Y') \boxtimes\_{Sh\_G(Y' \times Y')} Sh\_G(Y' \times Y) \cong Sh\_G(Y \times Y)$$ as $Sh\_G(Y \times Y)$-bimodules, and similarly with $Y$ and $Y'$ switched.


