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

Dark_Knight
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    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. – Jyrki Lahtonen Jan 22 '17 at 12:45

3 Answers3

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

Dietrich Burde
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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.

anomaly
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    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 :-) – Jyrki Lahtonen Jan 22 '17 at 12:44
  • 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. – anomaly Jan 23 '17 at 04:49
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[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.

Bill Cook
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  • How to show that there is no problem if dim(Z(L))=dim(L)-2 ? – Infinity Aug 21 '23 at 14:31
  • No there is no problem. But in the question we need to prove the statement that Dim(Z(L))<=Dim(L)-2. How to prove this ? – Infinity Aug 21 '23 at 14:48
  • Since $Z(L)$ is a subalgebra, $\dim Z(L)\in {0,1,\ldots ,n}$ with $n=\dim(L)$. So it is enough to exclude $n$ and $n-1$ as possibilities. – Dietrich Burde Aug 21 '23 at 15:04
  • Non-abelian says $L \not= Z(L)$. This rules out $\dim(Z(L))=n$, so that only leaves us to rule out $n-1$ (see the second paragraph). – Bill Cook Aug 25 '23 at 12:19