For the quantum group ${\cal O}_q(SL(n))$, $q\in \mathbb{R}$, I have read, without a proof, that for $p>1$, there exists a $q\in (0,1)$ such that $$ {\cal O}_p(SL(n)) \simeq {\cal O}_q(SL(n)). $$ What is this isomorphism, i.e. how does it act specifically on the standard generators $x_{ij}$? Moreover, for the compact real form ${\cal O}_q(SU(n))$, does this isomorphism still hold?

1$\begingroup$ Is this just $q=p^{1}$? $\endgroup$ – Theo JohnsonFreyd Apr 16 '16 at 14:19

$\begingroup$ Probably, but what are the isomorphism? $\endgroup$ – Alesandro Levi Apr 16 '16 at 14:32
The isomorphism between $\mathcal{U}_q(\mathfrak{g})$ and $\mathcal{U}_{q^{1}}(\mathfrak{g})$ of the form presented in Lemma 2.4.2 of the NeshveyevTuset book gives you the isomorphism of Hopf *algebras $\mathcal{O}_q({\rm SU}(n))$ and $\mathcal{O}_{q^{1}}({\rm SU}(n))$. The elements $x_{ij}$ are matrix coefficients of the defining representation, so the presentation on this side can be worked out too.