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I read the article by Dijkhuizen and Noumi (http://arxiv.org/pdf/q-alg/9605017v1.pdf). Here they describe in section $4$ what the algebra of functions on the total space of a family of quantum (2n−1)-spheres is. If one now has a look on aticles which consider representations of odd quantum groups, one almost everytime find another presentation of this algebra, namely one without the elements $c$ and $d$. If one has a representation of quantum $SU(n)$ on a Hilbert space $\mathcal{H}$ is it possible to "lift" this to a presentation of the algebra of functions on the total space of a family of quantum (2n−1)-spheres? I can find representations of the odd quantums sphere without $c$ and $d$ but are there articles about representations with $c$ and $d$?

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This two-parameter family of "quantum spheres" does not contain any quantum sphere. By this I mean that while $c$ is central $d$ is not: thus $c$ can be specialized to a real number but $d$ cannot. So, in this sense, there is no relation between this 1-parameter family of quantum spheres and the usual odd-dimensional quantum sphere.

Obviously, since the 1-parameter family ${\cal A}(\tilde S)(c,d)$ sits inside the tensor product of the standard quantum group ${\cal F}_q(SU(n))$ with $\mathbb C[\theta^{-1}]$ and the usual quantum sphere sits there as well (in the layer corresponding to $\theta^0$) the two are related in this sense.

Irreps of ${\cal F}_q(SU(n))\otimes \mathbb C[\theta^{-1}]$ are easily classified (since irreps of ${\cal F}_q(SU(n))$ are) and one can argue whether from here one gets all irreps of the 1-parameter family of quantum spheres. I do not see a reason, however, why they should be related to irreps of the usual quantum sphere.

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  • $\begingroup$ Thanks a lot Nicola. And what is if we deform the quantum SU(n) also by a parameter (for example if we assume that K acts by a multiple of the identity)? $\endgroup$ Commented Nov 6, 2015 at 18:24
  • $\begingroup$ I am not sure I get your idea, here. Quantum SU(n) tensored by $\mathbb C[\theta^{-1}]$ is already a "deformation", in the sense of being a $S^1$ family of $SU(n)$'s... $\endgroup$ Commented Nov 6, 2015 at 18:27

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