Understanding proof of Collins' Weingarten calculus formula

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I am trying to understand Collins' paper on Weingarten calculus. You can find it here: click!

On page 7, between the formulas $(2.6)$ and $(2.7)$, he claims that the action $\pi^{\otimes q} \otimes \Psi$ is on $M_d(\mathbb{C})^{\otimes q}$, but, as far as I understand, since both $\pi^{\otimes q}$ and $\Psi$ are on $M_d(\mathbb{C})^{\otimes q}$, the tensor product action should be on $M_d(\mathbb{C})^{\otimes q} \otimes M_d(\mathbb{C})^{\otimes q}$, right? However, the minimal invariant projections $I_\lambda$ are multiplicated by a $\tilde{B}$, which is an element of $M_d(\mathbb{C})^{\otimes q}$, confirming that the afore-mentioned action should be on $M_d(\mathbb{C})^{\otimes q}$. Am I missing something?

Moreover, he claims that we know, from representation theory, that an algebra is abelian and that the minimal invariant projections are indexed by partitions of $q$. Could you provide me with a reference on that? I would be very interested by a reference carrying out concrete examples for small $q$.

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I mean, sure, technically putting a $\otimes$ symbol between $\pi^{\otimes q}$ and $\Psi$ ought to imply an action on the tensor product $M_d(\mathbb{C})^{\otimes q}\otimes M_d(\mathbb{Q})^{\otimes q}$, but that's not what they mean. Since the action of $\Psi$ normalizes the action of $\mathbb{U}_d^q$ (adopting their notation of $\mathbb{U}_d$), so there is an action of the semidirect product $\mathbb{U}_d^q\rtimes S_q$, which is known as the wreath product $\mathbb{U}_d\wr S_d$, on $M_d(\mathbb{C})$. Maybe you want to say $\pi^{\otimes q}\cdot\Psi$ instead of $\pi^{\otimes q}\otimes\Psi$.

The keyword to start searching is "Schur-Weyl duality."