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From Erdmann and Wildon's Introduction to Lie algebras.

If $L$ is a non-abelian 3-dimensional Lie algebra over a field $F$, then we know only that the derived algebra $L'$ is non-zero. It might have dimension 1 or 2 or even 3.

I don't understand how $L'$ could have less than 3 dimensions. I did the following computations:

Consider a Lie algebra over the vector space $L$ over $F$ such that $B=\{x,y,z\}$ spans $L$, i.e. $\operatorname{dim}L=3$. Note $[x,y],[x,z],[y,z] \neq 0$ because $L$ is non-abelian, i.e. $[a,b]=0 \Leftrightarrow b=ca$, for some $c \in F$; that is, $a$ and $b$ are pairwise dependent of each other. However, if $[x,y]=\alpha[x,z]$, then $[x,y]-\alpha[x,z]=0 \Leftrightarrow [x,y]-[x,\alpha z]=0 \Leftrightarrow [x, y-\alpha z]=0$. Since $L$ is non-abelian, $x=y-\alpha z$. This means $B$ is not linearly independent which is a contradiction. Hence, $\forall \alpha \in F$, $[x,y] \neq \alpha[x,z]$.

For this reason, I don't understand how $L'$ could be of dimension 1.

But looking at $$[x,y]=c_1[x,z]+c_2[y,z]=[c_1x+c_2y,z]$$ makes me think that there is no reason why $L'$ cannot have 2 dimensions.

Can anyone please elaborate this passage on the book. Thanks!

TheLast Cipher
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    I think you are assuming that "non-Abelian" means more than it does. – Angina Seng May 08 '19 at 06:43
  • Now, I hate myself. – TheLast Cipher May 08 '19 at 06:50
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    Non-Abelian means that there are $a$ and $b$ with $[a,b]\ne0$. It doesn't mean that there are no Abelian subalgebras of dimension $2$. – Angina Seng May 08 '19 at 06:55
  • Oh yes, I just figured, as well. Thank you very much my Lord. – TheLast Cipher May 08 '19 at 06:57
  • Actually, calling "strongly non-abelian" a Lie algebra with no abelian 2-dimensional Lie subalgebra, one can prove that a strongly non-abelian finite-dimensional Lie algebra (over a field of characteristic zero) has dimension $\le 3$ (these are those that are either of dimension $\le 1$, non-abelian of dimension 2, or simple of dimension 3). This is way too restrictive. – YCor May 08 '19 at 10:11
  • @YCor: you mean there exists a derived algebra of dimension $\leq 3$ for a finite-dimensional "strongly non-abelian" Lie algebra? Or do you actually mean any finite-dimensional non-abelian Lie algebra has dimension $\leq 3$..? – TheLast Cipher May 08 '19 at 12:48
  • No, I don't mean this. [My previous comment was erased, presumably by a moderator. I'll not further comment.] – YCor May 10 '19 at 08:11

1 Answers1

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For one dimension, one can have generators $x$, $y$, $z$ with $[x,y]=y$, $[x,z]=[y,z]=0$. For two dimensions: $[x,y]=y$, $[x,z]=z$, $[y,z]=0$.

Angina Seng
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