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Hello everyone! Please excuse me if this question is too elementary...

Let $M$ and $E$ be modules living in category $\mathcal{O}$, $E$ is finite dimensional, hence $M\otimes E$ also lives in $\mathcal{O}$.

I'm wondering if the weight spaces of $M\otimes E$ look like this:

$( M\otimes E )_{\lambda}=\bigoplus_{\mu+\nu=\lambda} M_{\mu}\otimes E_{\nu}$.

The inclusion "from right to left" is obvious, but the other one?

I would think that this has to be well known if it is true - but I could not find it here, on math.stackexchange.com, somewhere else in the web or in the books I have access to.

Thank you very much in advance - any pointers would be helpful to me.

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up vote 3 down vote accepted

In general, if $V = \bigoplus_{\mu \in \Lambda} V_\mu$ and $W = \bigoplus_{\nu\in \Lambda} W_\nu$ are $\Lambda$-graded vector spaces (for some abelian group $\Lambda$), then

$V \otimes W = (\bigoplus_\mu V_\mu) \otimes (\bigoplus_\nu W_\nu) = \bigoplus_{\mu,\nu} V_\mu \otimes W_\nu = \bigoplus_\lambda \left( \bigoplus_{\mu + \nu = \lambda} V_\mu \otimes W_\nu\right)$

shows that $V \otimes W$ is again a $\Lambda$-graded vector space, and that its $\lambda$-component is $\bigoplus_{\mu + \nu = \lambda} V_\mu \otimes W_\nu$.

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Ah - right, thanks. That was much easier than expected :) –  Sh4pe Mar 27 '12 at 13:45
    
@Konstantin: As you point out, the grading is a formal matter when you know that the direct sums involved are finite. While the original BGG category has strong finiteness conditions (assuring for instance existence of enough projectives), analogous categories for Kac-Moody algebras or quantum versions raise more challenging questions about the notion of tensoring. See for instance the 1993-1994 series of papers in J. Amer. Math. Soc. by Kazhdan & Lusztig on Tensor structures arising from affine Lie algebras. –  Jim Humphreys Mar 27 '12 at 15:48
    
@Humphreys: I don't understand how the direct sums in my case are finite and what you mean by finiteness. Do you mean that the summands are finite dimensional? Or that there are only finitely many summands? (The latter is not true in my case if I'm not mistaken...). Thank you! –  Sh4pe Mar 28 '12 at 9:39
    
@SH4pe: My wording was awkward. All I wanted to emphasize is that the grading on tensor products described by Konstantin is purely formal for graded vector spaces, but to apply it you have to know which tensor products of modules in the BGG category are actually in that category. In that category the most natural examples come from tensoring by finite dimensional modules. –  Jim Humphreys Mar 28 '12 at 12:04
    
@Humphreys: Thank you! :) –  Sh4pe Mar 29 '12 at 8:54
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