Symplectic $2$-step nilpotent Lie algebras 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).
 A: It may be interesting to note that the minimal dimension for a $2$-step nilpotent Lie algebra without symplectic structure is $6$. 
In fact, only the direct product $\mathfrak{h}_5(\mathbb{R})\times \mathbb{R}$ is not symplectic in dimension $6$, $2$-step nilpotent. Here $\mathfrak{h}_5(\mathbb{R})$ denotes the $5$-dimensional Heisenberg Lie algebra with basis $(e_1,\ldots ,e_6)$ and non-trivial brackets $[e_1,e_2]=[e_3,e_4]=e_5$. In fact, $\mathfrak{h}_5(\mathbb{R})$ admits a linear contact form and its Reeb vector field is in the kernel of any closed alternating bilinear form of $\mathfrak{h}_5(\mathbb{R})\times \mathbb{R}$. To satisfy the requirement of $2\dim Z(\mathfrak{g})=\dim (\mathfrak{g})$, one can add a suitable factor $\mathbb{R}^k$; see Tsemo's answer.
A: Consider the following Lie algebra:
${\cal G}=Vect\{a,a',b,b',c,c',d,d'\}$ $[a,a']=[b,b']=c$, 
$[a,b]=[a',b]=[a,b']=[a',b']=[c,a]=[c,b]=[c,a']=[c,b']=[c',a]=[c',a']=[c',b]=[c',b']=[c,c']=[d,a]=[d,a']=[d,b]=[d,b']=[d,c]=[d,c']=[d',a]=[d',a']=[d,b]=[d',b']=[d',c]=[d',c']=[d,d]=0.$
${\cal G}$ is a 2-nilpotent algebra and its center is $Vect\{c,c',d,d'\}$. Its derived ideal $[{\cal G},{\cal G}]=Vect\{c\}$.
Suppose that there exists a symplectic form $\omega$ on ${\cal G}$, you have:
$\omega(c,c')=\omega([a,a'],c')$, since the form is closed, you have:
$\omega([a,a'],c)+\omega([c,a],a')+\omega([a',c],a)=0=\omega([a,a'],c)=\omega(c,c')$ since $c$ is in the center.
$\omega(c,d)=\omega([a,a'],d)$.
$\omega([a,a'],d)+\omega([d,a],a')+\omega([a',d],a)=0=\omega([a,a'],d)=\omega(c,d)=0$ since $d$ is in the center of ${\cal G}$
$\omega(c,d')=\omega([a,a'],d')$.
$\omega([a,a'],d')+\omega([d',a],a')+\omega([a',d'],a)=0=\omega([a,a'],d')=\omega(c,d')=0$ since $d'$ is in the center of ${\cal G}$
$\omega(c,a)=\omega([b,b'],a)$, we have:
$\omega([b,b'],a)+\omega([a,b],b')+\omega([b',a],b)=0$, since $[a,b]=[b',a]=0$, we deduce that $\omega([b,b'],a)=\omega(c,a)=0$.
$\omega(c,a')=\omega([b,b'],a')$
$\omega([b,b'],a')+\omega([a',b],b')+\omega([b',a'],b)=0$, since $[a',b]=[b',a']=0$, we deduce that $\omega([b,b'],a')=\omega(c,a')=0$.
$\omega(c,b)=\omega([a,a'],b)$
we have:
$\omega([a,a'],b)+\omega([b,a],a')+\omega([a',b],a)=0$, since $[b,a]=[a',b]=0$, we deduce that $\omega([a,a'],b)=\omega(c,b)=0$.
$\omega(c,b')=\omega([a,a'],b')$
we have:
$\omega([a,a'],b')+\omega([b',a],a')+\omega([a',b'],a)=0$, since $[b',a]=[a',b']=0$, we deduce that $\omega([a,a'],b')=\omega(c,b')=0$.
Since $\omega(c,c)=0$, we deduce that for every $x\in {\cal G}, \omega(c,x)=0$. Contradiction since a symplectic form is not degenerated.
There does not exist a symplectic form on the 2-nilpotent Lie algebra ${\cal G}$.
A: 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. 
