Zero-divisors in a graded Lie algebra

Let $\mathfrak{g}$ be positively graded Lie algebra over $\mathbb{Q}$, concentrated in even degrees.

Question: If $\mathfrak{g}$ is not free, must there exist linearly independent elements $a,b\in\mathfrak{g}$ such that $[a,b]=0$?

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The Yang-Mills Lie algebra $\mathfrak{y}\mathfrak{m}(2)$ over $\mathbb{Q}$ has a grading concentrated in even degrees. It is isomorphic to the rational Heisenberg Lie algebra $h_1$ with basis $(e_1,e_2,e_3)$ and $[e_1,e_3]=[e_2,e_3]=0, [e_1,e_2]=e_3$. Did I misunderstand something ? – Dietrich Burde Jun 4 '13 at 11:38
@Dietrich: Perhaps. I changed one word to make the question clearer. – Mark Grant Jun 4 '13 at 11:43
@Mark, I see. If $\mathfrak{g}$ is finite-dimensional, then it is nilpotent by Jacobson's theorem, and we may take a nonzero element $a\in Z(\mathfrak{g})$ and any $b$ with $a$ and $b$ linearly independent to obtain $[a,b]=0$. – Dietrich Burde Jun 4 '13 at 11:53
@Dietrich: yes, thanks. I should have mentioned that we assume that $\mathfrak{g}$ is infinite-dimensional. – Mark Grant Jun 4 '13 at 12:32

Consider the complex Lie algebra $\mathfrak{h}=\mathbb{C}[t,t^{-1}]\partial_t$ with Lie bracket $$[t^m\partial_t,t^n\partial_t]=(n-m)t^{n+m-1}\partial_t$$ ie the natural bracket that arises considering the natural action of $\mathfrak{h}$ on $\mathbb{C}[t,t^{-1}]$ by derivations. This is known as the Witt algebra. If we give $t^n\partial_t$ degree $n-1$ then this is a $\mathbb{Z}$-graded Lie algebra.

I claim that the $\mathbb{Q}$-Lie-subalgebra $\mathfrak{g}$ of $\mathfrak{h}$ spanned by the elements $t^n\partial_t$ with $n\geq 3$ and odd does not contain linearly independent elements $a,b$ such that $[a,b]=0$. Indeed $\mathfrak{h}$ does not.

Suppose $a=\sum_{i\leq n}\lambda_it^i\partial_t$ and $b=\sum_{j\leq n}\mu_jt^j\partial_t$ are linearly independent with $[a,b]=0$. We can assume WLOG that $\lambda_n=1$. Suppose that $m\leq n$ is the largest number such that $\mu_m\neq 0$. Then the degree ${n+m-2}$ piece of $[a,b]$ is simultaneously $\mu_m (n-m)$ and $0$. Thus $m=n$ and by rescaling $b$ we may assume $\mu_n=1$.

Since $a\neq b$ we may now find $k$ maximal such that $\lambda_k\neq \mu_k$. Write $a=a_1+a_2$ and $b=b_1+b_2$ with $a_1,b_1$ a sum of homogenous terms of degree less than $k$ and $a_2=b_2$ a sum of homogenous terms at least $k$. Then $$0=[a,b]=[a_1,b_1]+[a_2,b_1]+[a_1,b_2]+[b_2,a_2]=[a_1,b_1]+[a_2,b_1-a_1].$$

The leading term of $[a_1,b_1]$ has degree at most $(2k-2)$ and the leading term of $[a_2,b_1-a_1]$ is $(\mu_k-\lambda_k)(n-k)t^{n+k-1}\partial_t$ which has degree $n+k-2$ and so we obtain the desired contradiction.

It is straightforward to verify that $\mathfrak{g}$ is not free.

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@Simon: Thanks for this. Are you missing a "not" from your second paragraph? In which case, is this giving an example of a non free Lie algebra for which all brackets of linearly independent elements are nonzero? – Mark Grant Jun 4 '13 at 12:54
In line $5$ it should be "I claim that ... does not contain", I think. – Dietrich Burde Jun 4 '13 at 12:57
Yes. That's right. I'll change it. – Simon Wadsley Jun 4 '13 at 13:03
If positively graded means non-negatively graded then $n\geq 3$ can be replaced by $n\geq 1$. – Simon Wadsley Jun 4 '13 at 13:09