Simple lie algebra that holds just four generators?

[PhysSub]

Homework Statement

Show that there is no simple lie algebra with just four generators.

Relevant Equations

bracket product

I’m reading Weinberg’s QFT books, and stacking how to solve problem 15.4.
Weinberg says there is no simple lie algebra with just four generators, but I have no idea how to approach this problem. If the number of generators are only one or two, it can easy to say there is not such a simple lie algebra because we can’t take the relevant structure constant. I tried to apply the same discussion to four or five generators, but I think it does not work.
Does anyone know more about this problem? We would appreciate any references or tips.

 

[fresh_42]

The solution to this question strongly depends on which results you want to use. E.g. the smallest simple Lie algebra is three-dimensional. Then every Lie algebra $\mathfrak{g}$ can be written as the semidirect product of a semisimple subalgebra $\mathfrak{h}\leq \mathfrak{g}$ and its radical $\mathfrak{R}\trianglelefteq \mathfrak{g}$ which is a solvable ideal, i.e.
$$
\mathfrak{g}= \mathfrak{h} \ltimes \mathfrak{R}
$$
A four-dimensional Lie algebra is therefore solvable or has a one-dimensional radical. In both cases, there is a one-dimensional ideal. That is what we know because the result is true. We could use this for an indirect proof: Given a four-dimensional Lie algebra $\mathfrak{g}$. If we now show that it contains a one-dimensional ideal (and we already know that there is one), then it cannot be simple.

Another idea is: If $\mathfrak{g}$ is a four-dimensional simple Lie algebra, then it contains a copy of $\mathfrak{sl}(2)$ as subalgebra and we can write $\mathfrak{g}=\mathfrak{sl}(2) + \mathbb{F}\cdot Z$ as a sum of vectorspaces. Both are already subalgebras. If $[\mathfrak{sl}(2),Z]=0$ then $Z\in \mathfrak{Z(g)}$ is a central element and we are done. So all we have to do is to show that $Z$ can be chosen such that $\mathbb{F}Z$ is a one-dimensional $\mathfrak{g}-$module, an ideal. (In this case it will be the radical and $\mathfrak{g}$ could not be simple.) I guess that $[\mathfrak{g},\mathfrak{g}]=\mathfrak{g}$ is helpful here.

In short: We have to show that every four-dimensional Lie algebra is either solvable or has a one-dimensional center.

 

[fresh_42]

I think this is a plan:

Let $\mathfrak{g}$ be simple and four-dimensional. Then $\mathfrak{g}=\mathfrak{sl}(2)\oplus \mathbb{F}Z$ as a sum of subalgebras. Thus
$$
\mathfrak{sl}(2)\oplus \mathbb{F}Z=\mathfrak{g}=[\mathfrak{g},\mathfrak{g}]=[\mathfrak{sl}(2)\oplus \mathbb{F}Z,\mathfrak{sl}(2)\oplus \mathbb{F}Z=[\mathfrak{sl}(2),\mathfrak{sl}(2)]+[\mathfrak{sl}(2),\mathbb{F}Z]=\mathfrak{sl}(2)+[\mathfrak{sl}(2),\mathbb{F}Z]
$$
and $Z$ can be chosen such that $Z\in [\mathfrak{sl}(2),\mathbb{F}Z]$. Therefore $\mathbb{F}Z$ is a one-dimensional $\mathfrak{sl}(2)$-module, i.e. a one-dimensional ideal of $\mathfrak{g}$, which is impossible if $\mathfrak{g}$ is simple.