Why no faithful finite-dimensional irreducible representation of the Heisenberg algebra

Question

Apologies for asking a stupid question. Working on the exercise 7.8 from Introduction to Lie Algebras. It says:

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.

First of all, what is wrong with the traditional upper-diagonal representation? Namely,

$$ f = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{pmatrix} \quad,\quad g = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0\end{pmatrix} \quad,\quad z = \begin{pmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{pmatrix} \quad. $$

I would also be great to know why these three requirements for the representation (faithful, finite-dimensional, irreducible) are incompatible for the Heisenberg algebra.

Thanks.

Answer 1

You've already described below why the matrix representation is not irreducible (sorry, I mistakenly talked about the regular representation in the first place).

To prove that there is no finite-dimensional faithful irreducible representation, consider the action of $z$: On the one hand, it must be of trace $0$ (why?) on the other hand, it must be a scalar (why?). Can you fill in the details?

Remarks from the comments: The crucial point here is that ${\mathfrak g}$ has the property ${\mathfrak z}({\mathfrak g})\cap [{\mathfrak g},{\mathfrak g}]\neq \{0\}$, but this property does not distinguish Lie algebras possessing a finite-dimensional faithful irreducible representation:

• No reductive Lie algebra ${\mathfrak g}$ like ${\mathfrak g}{\mathfrak l}(n)$ has this property (even though it may have nontrivial center), since ${\mathfrak g} = {\mathfrak z}({\mathfrak g})\oplus [{\mathfrak g},{\mathfrak g}]$ for reductive ${\mathfrak g}$. Nonetheless, it may or may not possess a faithful finite-dimensional irreducible representation: Namely, if ${\mathfrak z}({\mathfrak g})$ is $1$-dimensional, the adjoint representation $[{\mathfrak g},{\mathfrak g}]\to{\mathfrak g}{\mathfrak l}([{\mathfrak g},{\mathfrak g}])$ extends to a faithful finite-dimensional irreducible representation of ${\mathfrak g}$ by letting ${\mathfrak z}({\mathfrak g})$ acts nontrivially by some scalar, while on the other hand, no irreducible finite-dimensional representation can be faithful on ${\mathfrak z}({\mathfrak g})$ by Schur's Lemma if $\dim{\mathfrak z}({\mathfrak g})>1$.

• A solvable Lie algebra may or may not have this property, but (over an algebraically closed field of characteristic $0$) it never has a finite-dimensional faithful irreducible representation by Lie's Theorem.

Btw, it's not a stupid question.

Answer 2

By Lie's theorem, finite-dimensional irreducible representations of solvable Lie algebras over $K$ are $1$-dimensional. Here we assume that $K$ is algebraically closed of characteristic zero. Since the Heisenberg Lie algebra is nilpotent, it is solvable. But a $1$-dimensional representation for the Heisenberg Lie algebra cannot be faithful, because the minimal dimension for a faithful representation of the Heisenberg Lie algebra is $3$ - see this MSE-question.