Maximal possible dimension of an abelian Lie subalgebra of Heisenberg Lie algebra of dimension $2n+1$.

Fix $n \in \mathbb{N}$, and let $\mathfrak{h}_n$ denote the Heisenberg Lie algebra of dimension $2n+1$ (over any given field $k$). Namely, $\mathfrak{h}_n$ is the Lie algebra with basis $x_1, \dots, x_n, y_1, \dots, y_n, c$ and with the Lie bracket defined by$$[x_i, y_j] = \delta_{ij}c,\text{ }[x_i, x_j] = [y_i, y_j] = [x_i, c] = [y_j, c] = 0$$(where $1 \le i, j \le n$ and $\delta_{ij}$ is the Kronecker delta). My question is, what is the maximal possible dimension of an abelian Lie subalgebra of $\mathfrak{h}_n$?

The maximal dimension of an abelian subalgebra of $\mathfrak{h}_n$ is equal to $n+1$. One direction is easy. Obviously $\langle x_1,\ldots ,x_n,c\rangle$ is an abelian subalgebra of dimension $n+1$. The non-trivial part is to show that there is no abelian subalgebra of dimension $n+2$. This can be found in the thesis of M. Ceballos here, in particular reference [78]. There is also an argument in the paper on a refinement of Ado's theorem here, for Heisenberg Lie algebras. Note that the maximal dimension of an abelian subalgebra coincides with the maximal dimension of an abelian ideal for nilpotent Lie algebras.
• That there is no abelian subalgebra of dimension $\ge n+2$ is trivial: it would project to an isotropic subspace of dimension $\ge n+1$ in a $2n$-dimensional symplectic vector space and it's clear that this does not exist. – YCor Oct 28 '15 at 23:19