Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

share|cite|improve this question

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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.