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I tried one problem from "Introduction to Lie Algebras” by Karin Erdmann and Mark Wildon and there is one question that I am not sure how to do.

"Let $L$ be a Heisenberg lie algebra with basis $f,g,z$ such that $[f,g]=z$, and z lies in the center of $L$. Show that there is no finite dimensional faithful irreducible representation of L."

My attempt so far is as follows:

Assume that $V$ is a finite dimensional faithful irreducible representation of $L$. So there is an injective lie algebra homomorphism $\phi:L\rightarrow gl(V)$. It is easy to check that the set $z\cdot V=\{\phi(z)(v)|v\in V\}$ is an $L$ submodule of $V$. Since $V$ is irreducible representation of $L$, it follows that either $z\cdot V=\{0\}$ or $z\cdot V=V$. However, $z\cdot V\neq\{0\}$ because $\phi$ is an injective map. Therefore, $z\cdot V=V$.

I am not sure how to continue from here. Are there any hints how to proceed from here?

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    $\begingroup$ I think the book you refer to is the one by Karin Erdmann and Mark Wildon? I’m sure you didn’t intend any disrespect, but it seemed odd to omit one author and refer to the other just by her first name, so I’ve edited your question. $\endgroup$
    – Jeremy Rickard
    Commented Mar 3, 2018 at 19:47

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Since the Heisenberg Lie algebra is nilpotent, it is also solvable. By Lie's theorem, every finite-dimensional irreducible representation of a solvable Lie algebra is $1$-dimensional. On the other hand, the minimal dimension of a faithful representation of the Heisenberg Lie algebra is $3$. Hence there is no finite-dimensional faithful irreducible representation of the Heisenberg Lie algebra.

References:

Are all irreducible representations of solvable Lie algebras 1-dimensional?

Faithful representation of the Heisenberg group

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  • $\begingroup$ I think Lie's theorem is only for algebraically closedfield with 0 characteristic. But I can modify the argument by employing Engel's theorem instead. $\endgroup$
    – KnobbyWan
    Commented Mar 4, 2018 at 0:32
  • $\begingroup$ You can first apply Lie's theorem over an algebraically closure, and then apply the "Lefschetz's principle" sometimes. On the other hand, Erdmann and Wilson anyway assume $K=\mathbb{C}$, right? $\endgroup$
    – Dietrich Burde
    Commented Mar 4, 2018 at 19:54
  • $\begingroup$ after thinking about it, i don't think we can use Engel's theorem $\endgroup$
    – KnobbyWan
    Commented Mar 23, 2018 at 4:09
  • $\begingroup$ @DietrichBurde could you please elaborate on why the result holds for other fields? $\endgroup$
    – MathPanda
    Commented May 22, 2022 at 13:25

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