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). What is the maximal possible dimension of an abelian Lie subalgebra of $\mathfrak{h}\_n$? I'm sure this result is well-known, but it's not well-known to me.



Maximal possible dimension of abelian Lie subalgebra of Heisenberg Lie algebra of dimension $2n+1$?
As already answered [here on MathSE](https://math.stackexchange.com/questions/1414661/), the answer is $n+1$, and furthermore all maximal abelian subalgebra have this dimension. This is a basic exercise, following the standard fact that in an $n$-dimensional symplectic vector space, maximal isotropic subspaces have dimension $n$. 


