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This question is pretty basic but I am but a simple physics postgrad studying Lie Algebras for the first time and want to check my understanding...

I am interested in showing that a rank one lie algebra is abelian.

$$ [x,x] = 0 \quad\forall x \in \mathfrak{g} $$

If I have proved this successfully am I also correct that if the Lie Algebra is abelian then the Lie Group of the same dimension is also abelian.

I thought that seemed a little basic and I wonder if I have missed anything else?

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    $\begingroup$ By definition a rank one Lie algebra is a one dimensional vector space $V$ with a bilinear mapping satisfying some conditions. What you want to see is that $[x,y]=0$ for every $x,y\in V$. The condition that $[x,x]=0$ for every $x$ is always satisfied by definition. Since $V$ is one dimensional, $V=\langle v\rangle $ for some $v$, and thus you can write $x=\lambda v,y=\mu v$ and obtain $[x,y]=\mu\lambda [v,v]=0$. $\endgroup$
    – Pedro ♦
    Commented Oct 20, 2015 at 2:09
  • $\begingroup$ Thanks, I didn't like my definition as you have both pointed out I'm proving something by it's own axiom! $\endgroup$
    – Alexander McFarlane
    Commented Oct 20, 2015 at 2:11

1 Answer 1

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Let $\{e_1\}$ be a basis of your $1$-dimensional Lie algebra $L$ over a field $K$. The Lie bracket $[x,y]$ is determined by the brackets of the basis elements. So the only bracket to consider is $[e_1,e_1]$, which is zero by the definition of a Lie algebra.

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