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I am working on the same problem as the original asker in here. The question is as follows.

Let $L$ be the Heisenberg algebra with basis $f, g, z$ such that $[f,g]=z$ and $z$ is central. Show that $L$ does not have a faithful finite-dimensional irreducible representation.

The proof, as far as I know, uses Schur's lemma (more precisely a lemma following from it). These are given in the same book as the following.

Schur's lemma:

Let $L$ be a complex Lie algebra and let $S$ be a finite-dimensional irreducible $L$-module. A map $\theta:S\to S$ is an $L$-module homomorphism if and only if $\theta$ is a scalar multiple of the identity transformation; that is, $\theta = \lambda 1_S$ for some $\lambda\in\mathbb{C}$.

Lemma 7.14:

Let $L$ be a complex Lie algebra and let $V$ be an irreducible $L$-module. If $z\in Z(L)$, then $z$ acts by scalar multiplication on $V$; that is, there is some $\lambda\in\mathbb{C}$ such that $z\cdot v = \lambda v$ for all $v\in V$.

My question is: what happens if the base field of the Heisenberg algebra is not $\mathbb{C}$? Can we, say for example, still claim the same result for $\mathbb{R}$, or fields of non-zero characteristic?

I apologise if it's a dumb question. But thanks in advance!

Plue
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1 Answers1

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Remark: Another MSE-question with the same question is here:

Representation of Heisenberg Algebra

Schur's Lemma need not be true for the real numbers. Consider the irreducible representation of the $1$-dimensional real Lie algebra $L=\mathbb{R}$ in the $2$-dimensional real vector space $V=\mathbb{C}$, where $\lambda\in L$ acts as multiplication with $\lambda i$. Then ${\rm End}_L(V)=\mathbb{C}\cdot id\neq \mathbb{R}\cdot id$.

Dietrich Burde
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