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These questions might be elementary for I just started to learn affine Kac-Moody algebra.

It is well known that if we consider finite dimensional Lie algebra, we have the folloing projection:

$R(\lambda)\otimes R(\nu)\rightarrow R(\lambda+\nu)$

where $\lambda$ and $\nu$ are dominant highest weights, $R(\lambda)$ and $R(\nu)$ are fnite dimensional irreducible representations.

First question:

I wonder whether this projection(or some analogue)still survives in affine Kac-Moody case? To be precisely, if we have $R(\lambda)$ and $R(\nu)$ as irreducible integral highest weight representations. Do we have the projection in the same form? (in classical case, we can prove this by using Weyl Character formula or equivalent multiplicity formula, does the affine case follows from the Weyl-Kac formula. Of course I can try to write the argument, but if somebody can point out the reference, I will be happy.

Second question:

It is related to the question I asked several days ago on extreme vector here:tensor product of extreme vector, Now I think I know the proof based on the comment of Peter Tingley and Jim.

Argument based on Peter

$e_{w\lambda}\otimes e_{w\nu}$ has the weight $w(\lambda+\nu)$ which belongs to the same weight space as $e_{w(\lambda+\nu)}$. So if one need to show $e_{w\lambda}\otimes e_{w\nu}=ke_{w(\lambda+\nu)}$ for some nonzero invertible elements $k$, it is sufficient to show the multiplicity of weight space is 1($dimV[w(\lambda+\nu)]$=1). But this just follows from multiplicity of weight space of highest weight is equal to 1 and the fact that multiplicity of weight space is invariant under action of Weyl group.

My second question is:

Can this argument extend to affine Kac-Moody case? Jim made a comment here, he pointed out the extreme vector can be extend to the integral representations of affine Kac-Moody algebra. So if we pick up two extreme vector of intergal highest weight, then take the tesnor product, can we do the similar argument as above(for finite dimensional Lie algebra)to obtain the same result?

Comments: Jim pointed out the reference by S.Kumar on solutions of PRV conjectures. But I think the situation there is slightly different.(maybe much more difficult), because the situation there, one need to consider multiplicity of $V(\lambda+w\nu)$, where just one weight moved by $w$, but in the situation here, it is much simpler.

All comments are welcome. Thanks!

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    $\begingroup$ Perhaps you know this, but to show that $R(\lambda)\otimes R(\mu)$ maps to $R(\lambda + \mu)$, you don't need anything as sophisticated as a character formula. Here is a more robust argument, which might extend to other contexts: Simply note that if $e_{\lambda}$ and $e_{\mu}$ are the highest weight vectors, then $e_{\lambda} \otimes e_{\mu}$ is a highest weight vector in the tensor product, which must then generate a copy of $R(\lambda + \mu)$ inside the tensor product. The existence of a projection now follows from semi-simplicity of the representation category. $\endgroup$
    – Emerton
    Commented Apr 22, 2010 at 16:00
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    $\begingroup$ 1) More care is needed about the meaning of tensor product of two irreducible representations in the Kac-Moody case, even when the modules involved are "integrable" and exhibit some of the classical character behavior. Tensoring infinite dimensional modules is quite a subtle question. (I'm not at all a specialist, however.) 2) As Emerton points out, you don't need Weyl's character formula in the classical case, but of course you do need well-behaved tensor products. $\endgroup$
    – Jim Humphreys
    Commented Apr 22, 2010 at 17:35
  • $\begingroup$ @Jim, thank you for your caution on tensor products of infinite dimensional space. $\endgroup$
    – Shizhuo Zhang
    Commented Apr 22, 2010 at 17:47
  • $\begingroup$ @Jim- What's wrong with the tensor product in the usual case? It's still an integrable module with finite dimensional weight spaces (of course, it can have infinitely many isotypic components, which is a bit weird). $\endgroup$
    – Ben Webster ♦
    Commented Apr 23, 2010 at 2:17
  • $\begingroup$ @Ben: My first comment was indeed misguided; I was thinking of a more complicated problem. The highest weights behave well enough, though the analysis of infinitely many "composition factors"is problematic. $\endgroup$
    – Jim Humphreys
    Commented Apr 23, 2010 at 12:16

1 Answer 1

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First question: Peter's (and Emerton's) argument works not only for affine algebras, but for arbitrary Kac-Moody algebras. Decompose your integrable representations into weights, and take the tensor product as the sum of tensor products of weight spaces. It is straightforward to check that the result is still an integrable representation. The tensor products live in Category O, where irreducible quotients are unique for each highest weight. See Kac, Infinite dimensional Lie algebras, sections 9.1-9.3.

Second question: Yes. See previous paragraph.

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