Let $\mathfrak{n}$ be a $2k$ dimensional $2$-step nilpotent Lie algebra and suppose that its center is $k$ dimensional. Does $\mathfrak{n}$ admit symplectic structure?


Let $\{f\_1,\dots,f\_k\}$ be a basis of the center of $\mathfrak{n}$ and complete it to a basis of $\mathfrak{n}$ $\{e\_1,\dots,e\_k,f\_1,\dots,f\_k\}$. Consider de dual basis $\{e^1,\dots,e^k,f^1,\dots,f^k\}$. Notice that the $2$-step nilpotency implies that $df^i\in \wedge^2\langle e^1,\dots,e^k\rangle$ and $de^j=0$.


A somewhat natural way to try to build a symplectic form is to pick a permutation $\sigma \in S\_k$ and define the $2$-form $$\omega=e^1 \wedge f^{\sigma(1)} + \dots + e^k \wedge f^{\sigma(k)} $$ which is non-degenerate. But I don't know how to choose $\sigma$ such that $d\omega=0$ (or even if it is possible to make such choice).



Symplectic $2$-step nilpotent Lie algebras
Consider the Lie algebra with basis $(e\_1,\dots,e\_7,z\_1,\dots,z\_7)$, with nonzero brackets (up to skew-symmetry):
$$z\_1=[e\_1,e\_2]=[e\_3,e\_4]=[e\_1,e\_6]=[e\_5,e\_7];$$
$$z\_2=[e\_2,e\_5],z\_{3}=[e\_2,e\_6],z\_{4}=[e\_2,e\_7],z\_5=[e\_3,e\_5],z\_{6}=[e\_3,e\_6],z\_{7}=[e\_3,e\_7].$$


Note that it's 2-step nilpotent with center being equal to the derived subalgebra, namely with basis $(z\_1,\dots,z\_7)$.


A general immediate fact is that $b([\mathfrak{g},\mathfrak{g}],\mathfrak{z}(\mathfrak{g}))=0$ for every closed alternating 2-form $b$.


Now we check that for every such $b$, we have $b(z\_1,e\_i)=0$ for all $i$ (and thus $b$ is degenerate). The reason is that for every $i$ we can find $j,k\neq i$ such that $z\_1=[e\_j,e\_k]$ and $[e\_j,e\_i]=[e\_k,e\_i]=0$. Indeed writing $(j,k)=t(i)$, we can choose $t(1)=t(2)=(3,4)$, $t(3)=t(4)=(1,2)$ $t(5)=t(7)=(1,6)$, and $t(6)=(5,7)$. Writing that $b$ is closed on the triple $(i,j,k)$ then yields $b(z\_1,e\_i)=0$ since both other terms vanish. 


