Show that $\dim(Z(L)) \leq \dim(L) - 2$

Question

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 $\dim(L) > n$ ,($n>2$)?

Answer 1

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.

Answer 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 $\dim(Z(L))=n-1$.

Answer 3

[Note: The accepted answer is not quite correct. To reference an orthogonal complements we require some kind of non-degenerate form.]

Let $L$ be a finite dimensional non-abelian Lie algebra. Non-abelian means $L \not= Z(L)$, so $\dim(Z(L))<\dim(L)$. It just remains to rule out $\dim(Z(L))=\dim(L)-1$. For sake of contradiction assume this is true.

Since $L$ is non-abelian, there is some $x,y \in L$ such that $[x,y] \not=0$. Consider $x+Z(L),y+Z(L) \in L/Z(L)$ (the quotient algebra). Since $\dim(L/Z(L)) = \dim(L)-\dim(Z(L)) = 1$, we must have that one of our cosets is a multiple of the other. Without loss of generality assume $x+Z(L) = c(y+Z(L))$ for some scalar $c$. Then $x+Z(L)=cy+Z(L)$ so $x-cy \in Z(L)$, say $x-cy=z \in Z(L)$. Thus $x=cy+z$. Then $[x,y] = [cy+z,y] = c[y,y]+[z,y]=0$ by alternation and the fact that $z$ is central. This contradicts our assumption that $[x,y]\not=0$. Therefore, $\dim(Z(L))=\dim(L)-1$ is impossible.