Suppose we are given a vector space $V$ equipped with a bilinear form $[,]:V\times V\to V$ such that
$$[x,y]=h, \quad \text{and} \quad [x,h]=0=[y,h]$$ for any $x,y$ and $h$ in $V$.
How can we show that this bilinear form defines a Lie bracket on $V$?
A Lie bracket has to be antisymmetric meaning that $[x,y]=-[y,x]$ and satisfies in the Jacobi identity $ [x, [y, z]] + [y, [z, x]] + [z, [x, y]] = 0$.
Many thanks!
The Jacobi identity here is trivial, because each term in it is zero. We have $[x,[y,z]]=0$ for all $x,y,z$, since the Heisenberg Lie algebra is $2$-step nilpotent - see your previous question and the answers to it:
Associative Lie algebra
So we have $$ [x, [y, z]] + [y, [z, x]] + [z, [x, y]] = 0+0+0.$$
By convention, only the Lie bracket $[x,y]=h$ is given, with the implicit understanding that $[y,x]=-h$. Here $(x,y,h)$ is a basis of the Heisenberg Lie algebra.