2
$\begingroup$

Let $L$ be a non-abelian Lie algebra. I need to show that $$\dim(Z(L)) \leq \dim(L) - 2$$

Now, if $\dim(L) = 2$ , then I know that this $L$ is a unique non-abelian Lie algebra such that its centre $Z(L) = 0$. Therefore, I'm done with the trivial case. But how do I prove the above inequality when $\dim(L) > n$ ,($n>2$)?

  • 1
    This could (should?) be seen as the Lie analogy of the known result about groups: If $G/Z(G)$ is cyclic then $G$ is abelian. – 2017-01-22

2 Answers 2

2

Let $x\in Z(L)^\perp$. Then $[x, z] = 0$ for all $z\in Z(L)$, and clearly $[x, x] = 0$. If $\dim Z(L) \geq \dim L - 1$, that forces $L$ to be abelian.

  • 1
    I guess you meant to take any $x$ from the *complement* of $Z(L)$ rather than the *orthogonal complement* (w.r.t. what inner product??): Anyway, this is the right idea :-) – 2017-01-22
  • 0
    Yes, I'm being lazy and assuming that the ground field is something like $\mathbb{R}$ or $\mathbb{C}$. Taking any $x$ such that $x$ and $Z(L)$ span $L$ (i.e., $x\not \in Z(L)$) is sufficient. – 2017-01-23
2

In fact, there is no $n$-dimensional Lie algebra $L$ with $\dim(Z(L))=n-1$. The argument is, as anomaly said, that in this case $L=K\oplus Z(L)$ with a $1$-dimensional Lie algebra $K$ with basis, say, $x$. Then $[x,z]=0$ for all $z\in Z(L)$ and $[x,x]=0$, so that $L=Z(L)$, a contradiction to $\dim(Z(L))=n-1$.