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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).

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  • $\begingroup$ I don't know what you mean by putting a symplectic structure on a Lie algebra. What compatibility do you want between the symplectic structure and the Lie bracket? Without any compatibility, of course every even-dimensional vector space admits a symplectic structure. $\endgroup$ – Qiaochu Yuan Sep 2 '16 at 22:47
  • $\begingroup$ @QiaochuYuan The compatibility is the following: the 2-form must be closed for the Lie algebra cohomology of $\mathfrak{n}$. Like in this paper $\endgroup$ – 115465 Sep 3 '16 at 0:28
  • $\begingroup$ Explicitly, that a 2-form $b$ is closed means that $b(x,[y,z])+b(y,[z,x])+b(z,[x,y])=0$ for all $x,y,z$. $\endgroup$ – YCor Sep 3 '16 at 7:29
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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.

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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.

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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 ${\cal G}$.

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  • $\begingroup$ There was the requirement that the center is half of the dimension. Here the dimension is 6 and the dimension of the center is 2. $\endgroup$ – YCor Sep 3 '16 at 8:09
  • $\begingroup$ OK it works now. I don't know why you write all cases. You can just write them to show that $\omega(c,a)=0$ and $\omega(c,d)=0$, since all other cases follow by the same arguments (the second because the center and the derived subalgebra are orthogonal). In my post I forced the derived subalgebra to be half-dimensional as well, which forced me to increase the dimension. $\endgroup$ – YCor Sep 3 '16 at 9:05

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