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I'm reading over the classification of 3-dimensional complex Lie algebras, and have come to the classification of a particular Lie algebra spanned by $\{x,y,z\}$ satisfying the relations $$[x,y] = y, \; [x,z] = y+z, \; [y,z]=0$$

Seeking to find a concrete example of such a Lie algebra (in particular, a subalgebra of $\mathfrak{gl}(n,\mathbb{C})$), I adapted another example to obtain the subalgebra spanned by \begin{align*} x=\begin{pmatrix} 0 & -1 & 0 \\ 0 & 0 & -1 \\ 0 & 0 & -1 \end{pmatrix}, \; \; \; y=\begin{pmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}, \; \; \; z = \begin{pmatrix} 0 & 0 & 1 \\ 0 & 0 & -1 \\ 0 & 0 & 0 \end{pmatrix} \end{align*} which satisfies the above relations. However, I was wondering if a "nicer" example existed, and/or how to derive such an example.

Edit: I managed to find the following subalgebra of $\mathfrak{gl}(n,\mathbb{C})$, derived by considering the matrices of $\text{ad}(x)$, $\text{ad}(y)$ and $\text{ad}(z)$:

\begin{align*} x=\begin{pmatrix} 0 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{pmatrix}, \; \; \; y=\begin{pmatrix} 0 & 0 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}, \; \; \; z = \begin{pmatrix} 0 & 0 & 0 \\ -1 & 0 & 0 \\ -1 & 0 & 0 \end{pmatrix} \end{align*}

lokodiz
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    not a representation in $\mathfrak{gl}(n)$, but: you can represent your Lie algebra $\mathfrak g$ by linear operators acting on (say smooth) functions on $\mathbb R$, via $x\cdot f=df/dt$, $y\cdot f=e^tf$, $z\cdot f=te^tf$ ($t$ is the variable of $f$) (to make the space smaller, take for $f$'s just polynomials in $t$ and $e^t$). Another representation is by vector fields on $\mathbb R^2$: $y\mapsto\partial/\partial s$, $z\mapsto\partial/\partial t$, $x\mapsto(s+t)\partial/\partial s+t\partial/\partial t$ ($s,t$ the coordinates on $\mathbb R^2$) – user8268 Sep 25 '13 at 11:04
  • What does "nicer" mean? – RghtHndSd Sep 25 '13 at 12:58
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    My first thought would be to put the bracket $[a,b]=ab-ba$ on the associative ring $\Bbb C[x,y,z]/(xy-yx-y,xz-zx-y-z,yz-zy,x^2,z^2)$ and then find matrices representing it. That gave me exactly the second representation you gave :) That seems like a general strategy with algebras and relations that would work in most cases. – rschwieb Sep 25 '13 at 13:14
  • I don't feel confident enough to say so because I'm not very familiar with Lie theory, but maybe one can conclude all Lie algebras satisfying those relations look like that quotient ring? – rschwieb Sep 25 '13 at 13:17
  • @ rghthndsd: yeah, I thought that might get brought up when I asked the question. Roughly speaking, I meant an example where the matrices looked simpler, or one where it was clear where the matrices came from (as per my edit).

    @ rschwieb: I see how that works, but where do you get that quotient ring from?

     – lokodiz Sep 25 '13 at 14:17

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