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I am trying to figure out representations of loop groups in order to understand conformal blocks. I am currently trying to figure out are the weight spaces of a representation with a given lowest weight and get my hands dirty with one. To build the representation I use the following construction. Let $\hat{\mathfrak{g}}=\mathbb{C}\langle d\rangle \oplus \mathbb{C} \langle k\rangle \oplus \mathfrak{g}\[ z, z^{-1} \]$ be the affine Lie algebra associated to the extended Affine group $\mathbb{T}\ltimes \tilde{L}G$. It has Cartan sub-algebra $\hat{\mathfrak{h}}=\mathbb{C}\langle d\rangle \oplus \mathbb{C} \langle k\rangle \oplus \mathfrak{h}$. Let $\delta$ and $\Lambda$ be dual to $d$ and $k$ respectively.

In the notes I am using it builds a representation of weight $\lambda+l\Lambda$ in the following way. Let $V_0$ be an irreducible representation of $S^1 \times G$ of weight $\lambda+l\Lambda$. Then consider: $$ \mathfrak{U}(\hat{\mathfrak{g}})\otimes_{\mathfrak{U}(\hat{\mathfrak{b}}_- + \mathfrak{g})\ }V_0$$ It then goes on to say that this has a maximal proper sub-module and that the quotient is the correct representation. I've been staring at this for a while and I cannot figure out what it the sub-module is. Can anyone give me a set of generators or point me to somewhere I may find them. All the literature I have relates to this is on normal Verma modules and google has't turned up anything.

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  • $\begingroup$ The submodule in question might depend on the actual affine algebra that you are dealing with and in the actual highest weight. I am not sure that you can be very explicit in the general case. Already for finite simple Lie algebras, where the same construction applies, I am not aware of any explicit formula in general, but others will surely correct me if I'm wrong. Having said that, it is clear that the Verma module will have a maximal proper submodule and that the quotient, by definition, will be irreducible and of highest weight. $\endgroup$ Commented Feb 27, 2013 at 7:43
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    $\begingroup$ (cont'd) So what exactly do you need the explicit submodule for? If you want to gain intuition about this construction, perhaps you should look at how it works with a simple Lie algebra of small dimension, such as $\mathfrak{su}(2)$. $\endgroup$ Commented Feb 27, 2013 at 7:44
  • $\begingroup$ You may find some information in Kac-Raina's Bombay lectures. $\endgroup$
    – S. Carnahan
    Commented Feb 27, 2013 at 13:34

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