Associative Lie algebra

If $L$ is an associative Lie algebra, then for any $x,y,z\in L$, we have $$\big[x,[y,z]\big]=\big[[x,y],z\big].$$ By Jacobi's identity $$\big[x,[y,z]\big]=\big[[x,y],z\big]+\big[y,[x,z]\big].$$ This means $$\big[y,[x,z]\big]=0$$ for all $x,y,z\in L$. Consequently, $[L,L]$ is in the center $Z(L)$ of $L$.

It can be easily seen that if $[L,L]\subseteq Z(L)$, then $\big[L,[L,L]\big]=\{0\}$. Hence, we conclude that a Lie algebra is associative if and only if $[L,L]\subseteq Z(L)$. However, this means the lower central series of $L$ is of length at most $1$, i.e., if $L_0=L$ and $L_{k+1}=[L,L_{k}]$, then the lower central series $L_0\supseteq L_1\supseteq L_2\supseteq \ldots$ is stabilized with $L_2=L_3=\ldots=\{0\}$. This means $L$ is nilpotent of depth at most $1$.

The depth-$0$ case consists of abelian Lie algebras. The depth-$1$ case is more interesting and contains examples like the Heisenberg Lie algebras.

You can in fact construct an associative Lie algebra $L$ over a field $F$ such that $\dim L=n$ and $\dim Z(L)=k$, and every such $L$ arises from this construction. First let $Z(L)=\operatorname{span}\{z_1,z_2,\ldots,z_k\}$. Let $x_1,x_2,\ldots,x_m$ ($m=n-k$) be elements of $L$ such that $\{x_1,x_2,\ldots,x_m,z_1,z_2,\ldots,z_k\}$ is a basis of $L$. Obviously, $$[x_i,z_j]=0\ \ \ \ \ (1)$$ for all $i,j$. For $1\le i<j\le m$, we clearly have $$[x_p,x_q]=\sum_{r=1}^k \lambda^{r}_{p,q}z_r\ \ \ \ \ (2)$$ for some $\lambda^r_{p,q}\in F$. Since $\operatorname{span}\{x_1,x_2,\ldots,x_m\}\cap Z(L)=\{0\}$, for any $x=\sum_{i=1}^m\mu^ix_i$, if $[x,x_j]=0$ for all $j$, then $\mu^i=0$ for all $i$. Note that $$[x,x_j]=\sum_{i=1}^{j-1}\mu^i\sum_{r=1}^k\lambda^r_{i,j}z_r-\sum_{i=j+1}^m\mu^i\sum_{r=1}^k\lambda^r_{i,j}z_r=\sum_{r=1}^k\left(\sum_{i=1}^{j-1}\mu^i\lambda^r_{i,j}-\sum_{i=j+1}^m\mu^i\lambda^r_{i,j}\right)z_r.$$ Since $[x,x_j]=0$, $$\sum_{i=1}^{j-1}\mu^i\lambda^r_{i,j}-\sum_{i=j+1}^m\mu^i\lambda^r_{i,j}=0$$ for all $j=1,2,\ldots,m$ and $r=1,2,\ldots,k$. Therefore, $x=0$ is the only possible solution if and only if the matrix $A=[a_{u,v}]_{(mk)\times m}$ has rank $m$, where $$a_{(r-1)m+j,i}=\left\{\begin{array}{ll}\lambda^r_{i,j}&\text{if }i<j\\0&\text{if }i=j\\-\lambda^r_{i,j}&\text{if }i>j.\end{array}\right.$$ Note that $[L,L]$ is a subspace of $Z(L)$ of dimension $t$ if $t$ is the rank of the matrix $B$ given by $B=[b_{i,j}]_{\binom{m}{2}\times k}$ where $$b_{(q-p-1)m+p,r}=\lambda^r_{p,q}$$ for $1\le p<q\le m$ and $r=1,2,\ldots,m$.

For fixed integers $n>k\geq t>0$, there exists an associative non-abelian Lie algebra $L$ such that $n=\dim L$, $k=\dim Z(L)$, and $t=\dim [L,L]$ if and only if $t\leq \binom{n-k}{2}$. You can construct such a Lie algebra $L$ up to isomorphism by assuming that $$[L,L]=\operatorname{span}\{z_1,z_2,\ldots,z_t\},,$$ $$Z(L)=\operatorname{span}\{z_1,z_2,\ldots,z_k\},$$ and $$L=\operatorname{span}\{x_1,x_2,\ldots,x_{n-k},z_1,z_2,\ldots,z_k\}.$$ The Lie bracket relations are given by (1) and (2) with $\lambda^{r}_{p,q}=0$ if $r>t$, and the matrices $A$ and $B$ defined above must satisfy $\operatorname{rk}A=n-k$ and $\operatorname{rk}B=t$.