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In the book "Representations of Semisimple Lie Algebras in the BGG Category O" by Humphrey.

Exercise 1.5: When $\mathfrak{g}=\mathfrak{sl}(2,\mathbb{C})$, show that $M(\lambda)\otimes M(\mu)$ cannot lie in $\mathcal{O}$.

If $M,N\in\mathcal{O}$, why not $M\otimes N\in\mathcal{O}$ in general?

Here is my "argument" (should be wrong) to show $M,N\in\mathcal{O}\implies M\otimes N\in\mathcal{O}$:

  1. Since $M,N$ are both finitely generated as $U(\mathfrak{g})$-module, then $M\otimes N$ is finitely generated as $U(\mathfrak{g})$-module.

  2. Clearly, $\bigoplus_{\lambda\in\mathfrak{h}^*}(M\otimes N)_\lambda\subseteq M\otimes N$. Conversely, suppose $v\otimes w\in M\otimes N$, then $v=v_1+\cdots+v_k$, where $v_i\in M_{\lambda_i}$ and $w=w_1+\cdots+w_l$, where $w_i\in N_{\mu_i}$.

Then $v\otimes w=\sum_{i=1}^k\sum_{j=1}^l v_i\otimes w_j\in \bigoplus_{\lambda\in\mathfrak{h}^*}(M\otimes N)_\lambda$ since $h.(v_i\otimes w_j)=(h.v_i)\otimes w_j+v_i\otimes (h.w_j)=(\lambda_i v_i)\otimes w_j+v_i\otimes (\mu_jw_j)=(\lambda_i+\mu_j)v_i\otimes w_j$.

Hence $M\otimes N=\bigoplus_{\lambda\in\mathfrak{h}^*}(M\otimes N)_\lambda$.

  1. For each $v\otimes w\in M\otimes N$, $U(\mathfrak{n}).(v\otimes w)=(U(\mathfrak{n}).v)\otimes w+v\otimes (U(\mathfrak{n}).w)$ is a finite dimensional since $U(\mathfrak{n}).v$ and $U(\mathfrak{n}).w$ are finite dimensional.

Where do I make a mistake? And how to do the exericse 1.5?

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  • $\begingroup$ Certainly the notation used in your question can be found in Humphrey's book. But the reader who doesn't have access to the book would like to know what $M(\lambda)$, $\mathcal{O}$ and $U(\mathfrak{g})$ are. $\endgroup$
    – Paul Frost
    Oct 2, 2018 at 13:41
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    $\begingroup$ Point 1 is wrong. The tensor products of two finitely generated $\mathfrak{g}$ modules may not be finitely generated. $\endgroup$
    – Joppy
    Oct 2, 2018 at 13:53

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