Let $\widetilde{U_q(sl_n)}$ denote a deformation of the algebra $U_q(sl_n)$. In particular, $\widetilde{U_q(sl_n)}$ is defined by the same generators and relations and $*$-operations as $U_q(sl_n)$ exepted the relation between $F_i$ and $E_i$. We say: $$[E_i,F_i]=(q-q^{-1})^{-1}(K_i^2-\sigma_iK_i^{-2})$$ where $\sigma_i=-1$ if $i=1$ and $\sigma_i=1$ otherwise. What I wanted do do is to find a $*$-representation of this algebra. In particular, I want to find a Hilbert space $H$ and a $*$-homomorphism $\psi:\widetilde{U_q(sl_n)}\rightarrow B(H)$. I thought that $H=\ell^2(\mathbb{N}^n)$ will be a good candidate for $H$, but I dont't know. Maybe one can start with a highest weight vector, say we want to find $x\in H$ such that $K_ix=q^{<x,\alpha_i>}x$ and $E_i=0$ for each $i$. But therefore we have to construct the operators which belongs to $K_i$ and $E_i$ and which satisfied also the relations of the algebra. Can maybe someone help me with this? Thank you very much.
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$\begingroup$ I think one obtain a pre-Hilbert space by $M(\lambda)=\widetilde{U_q(sl_n)}\otimes_{U_q(b^+)}\mathbb{C}_{\lambda}$ (see quantum bounded symmetric domains). By completion one obtains (I think) \ell^2(\matbb{N})^{\otimes (n-1)}. But how to obtain the operators? $\endgroup$– Kendra HaynesCommented Oct 5, 2015 at 14:13
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