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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.

mavzolej
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2 Answers2

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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.

Hanno
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  • Thanks for response. First, I want to make sure that you are using the word 'regular' in the meaning 'usual', pointing to the one I suggested above; and it has nothing to do with the regular representations of groups :)

    I do understand now why this representation is not irreducible by a slightly more primitive argument - simply because all the matrices have zeros in the bottom row and none of them can change the third component of the vector in the linear space where they act. (deep analogies with the generators of translations...)

    Now, coming back to your logic.

     – mavzolej Oct 03 '16 at 21:49
  • Any element of the derived algebra is traceless (just by the property of the trace), and $z$ is, of course, the one.

    However, $z$ has to be proportional to the unit element (since it commutes with everyone) - due to the Shur lemma.

    Contradiction!

    So, I think I got (with your help!) the answer for the main question.

    The only remaining thing is: why is the provided representation not faithful? Basically, what stops us from saying that the suggested mapping is injective?

     – mavzolej Oct 03 '16 at 21:50
  • Double check please: 1) The matrix representation above of the Heisenberg algebra is faithful but not irreducible 2) In the adjoint representation of any Lie algebra, all the elements from the centre are represented by zero matrices. 3) Any algebra with the non-zero centre cannot have faithful finite-dimensional irreducible representations. – mavzolej Oct 03 '16 at 22:07
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    @mavzolej: (1) and (2) are correct, but (3) is not: For example, ${\mathfrak g}{\mathfrak l}(n)$ has a faithful finite-dimensional irreducible representation, namely the matrix representation. The crucial point here is that $Z(L)\cap [L,L]\neq {0}$. – Hanno Oct 03 '16 at 22:15
  • To be honest, I am still shocked by the fact that none of the $[L,L]$ elements of $\mathfrak{gl}(n)$ are in its centre... So, are you saying that: the algebra can have faithful finite-dimensional irredicible representations iff $Z(L)\cap[L,L] = {0}$? – mavzolej Oct 04 '16 at 06:08
  • @mavzolej: I have extended the answer to keep this comments sections from overflowing :) I hope it makes things clearer. – Hanno Oct 04 '16 at 06:46
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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.

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Dietrich Burde
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