Show that $\pi(Z)$ acts as a scalar over $\mathbb{g}$

Question

Let $(\pi, V)$ be a finite dimensional irreducible representation of $\mathbb{g}$

$V$ is a vector space of homogeneous polynomials in 3 variables of degree d over $\mathbb{R}$

$\mathbb{g}=\begin{bmatrix} 0 & a & b\\ 0 & 0 & c \\ 0 & 0 & 0 \end{bmatrix} $ where $a, b, c \in \mathbb{C}$

For example, $X=\begin{bmatrix} 0 & 1 & 0\\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} $, $Y=\begin{bmatrix} 0 & 0 & 0\\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix} $ and $Z=\begin{bmatrix} 0 & 0 & 1\\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} $

The basis for $\mathbb{g}$ is $\{X, Y, Z \}$

Show that $\pi(Z)$ acts as a scalar.

To do this I think we need to show that $\pi(Z)$ is an intertwining operators between $\pi$ and itself, then the result follows from Shur's lemma

But how do we show it is an intertwining operator? Thanks

Answer 1

The Lie algebra is the complex $3$-dimensional Heisenberg Lie algebra $\mathfrak{h}_1(\mathbb{C})$. Since it is solvable, by Lie's theorem every finite-dimenisonal irreducible representation of $\mathfrak{h}_1(\mathbb{C})$ is $1$-dimensional.