Let $g$ be a simple Lie algebra and $U_q(g)$ the corresponding quantum group. What are the relation between integrable representations and highest weight representations of $U_q(g)$? Are all highest weight representations of $U_q(g)$ integrable? Thank you very much.

Edit: here an integrable module is a weight module such that the actions of $e_i$ and $f_i$ are locally nilpotent (i.e. for any vector $v$ in the module, there exists a positive integer $k$, possibly dependent on $v$, such that ${\displaystyle e_{i}^{k}.v=f_{i}^{k}.v=0}$ for all $i$).

A weight module is a module with a basis of weight vectors. A weight vector is a nonzero vector $v$ such that $k_{\lambda} . v = d_{\lambda} v$ for all $\lambda$, where $d_{\lambda}$ are complex numbers for all weights $\lambda$ such that $$ {\displaystyle d_{0}=1}, \\ {\displaystyle d_{\lambda }d_{\mu }=d_{\lambda +\mu }}, $$ for all weights $\lambda, \mu$.

  • $\begingroup$ Could you include the definition of integrable you are working with? This seems to differ somewhat between authors (and I can't recall if they are all equivalent). $\endgroup$ – Tobias Kildetoft Jun 10 '16 at 5:46
  • $\begingroup$ @Tobias Kildetoft, thank you for your comments. I edited the post. $\endgroup$ – LJR Jun 10 '16 at 7:29
  • $\begingroup$ Ok, then I don't think all highest weight modules are integrable. For example, induce a $1$-dimensional module from the "Borel" (I forgot what this is called for quantum groups) to the full quantum group. This is a highest weight module by the usual arguments, but we can "go down" indefinitely by applying the $f$'s, so these do not act locally nilpotently. In fact, I think that a highest weight integrable module is finite dimensional. But it has been a long time since I looked at anything like this, so I might be missing some details. $\endgroup$ – Tobias Kildetoft Jun 10 '16 at 7:37
  • $\begingroup$ @Tobias Kildetoft, thank you very much. $\endgroup$ – LJR Jun 10 '16 at 7:47
  • $\begingroup$ But do take what I write with a grain of salt. I am mainly familiar with the corresponding concepts for Lie algebras, where this definition of being integrable would correspond (at least if we also assume the module to be finitely generated) to considering the parabolic subcategory of category $\mathcal{O}$ with the parabolic subalgebra being the entire Lie algebra. And in this case we do indeed get the category if finite dimensional modules. $\endgroup$ – Tobias Kildetoft Jun 10 '16 at 8:06

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