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Let $\mathfrak{g}$ be a solvable Lie algebra. By Lie's theorem, it is easy to see that any finite dimensional irreducible representation is 1 dimensional. Is it possible to remove the condition that the representation be finite dimensional?

unknownymous
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  • https://math.berkeley.edu/~reb/courses/261/11.pdf claims that the answer is no, but I don't know an example. – Qiaochu Yuan Nov 03 '15 at 05:25
  • Consider the Weyl algebra $W$. It doesn't even have any nontrivial finite-dimensional representations $V$, since then we'd have $\operatorname{tr}_V 1 = \operatorname{tr}_V[\partial, X] = 0$. Take a suitable subalgebra of $W$. – anomaly Nov 03 '15 at 16:56

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The link given by Qiaochu Yuan contains a counterexample if you remove finite dimensional. Consider the 3-dimensional Lie algebra (over a field $k$) generated by $x$, $d/dx$ and $1$. This is a nilpotent Lie algebra (hence solvable), with $1$ central and $$[d/dx,x]=1.$$ This Lie algebra acts on the polynomial ring $k[x]$ without eigenvalues. In fact, the enveloping algebra of this Lie algebra is the Heisenberg algebra, $H$, and $k[x]$ is a faithful irreducible $H$-module.

David Hill
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