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A representation $V$ of a Lie algebra $\mathfrak{g}$ is "locally finite-dimensional" if $\dim U(\mathfrak{g}) v < \infty$ for every $v \in V$. I want to show that this condition holds if and only if $V$ is a direct sum of finite-dimensional modules $V_i$. The $\Leftarrow$ direction is easy: any $v \in \bigoplus V_i$ is some $\oplus v_i$ for finitely many non-zero $v_i$, and so applying $U(\mathfrak{g})$ we get finite $\times$ finite $=$ finite. However, I'm a bit stuck in the other direction.

So far, here's what I have. Take a basis $\{v_1,v_2,\ldots \}$ of $V$, and let $U(\mathfrak{g})$ identify what it may. I'm hoping that the result gives a partition of this basis into finite-dimensional clumps. By contradiction, suppose there was an infinite sequence $v_{i(m)}$ such that $$v_{i(m+1)} = x_m v_{i(m)}$$ for some $x_m \in U(\mathfrak{g})$. I want to conclude that this violates $\dim U(\mathfrak{g}) v_{i(1)}< \infty$. I'm not quite sure how to do this, since it seems I would need $$\cdots x_m \cdots x_2 x_1 \in U(\mathfrak{g})$$ which I don't think should be true. Any advice would be greatly appreciated!

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1 Answer 1

Whoops - this shouldn't be direct sum" but simplysum'', in which case the problem is obvious. Sorry folks!

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