# Maximal possible dimension of abelian Lie subalgebra of Heisenberg Lie algebra of dimension $2n+1$? [closed]

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.

## closed as off-topic by YCor, abx, Wolfgang, user1688, Vladimir DotsenkoJan 21 '16 at 10:21

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• It's equivalent to the question of determining the maximal dimension of an isotropic subspace in a $2n$-dimensional symplectic vector space; the latter is $n$ and hence the answer to the question is $n+1$. It's an exercise rather than a research question; I think I remember it already occurred in MathSE (I checked: it's here math.stackexchange.com/questions/1414661/…). Anyway Dietrich refers to a thesis but it's really an exercise. – YCor Oct 28 '15 at 23:17
• I think this is a legitimate MO question: helping with "I'm sure this result is well-known, but it's not well-known to me" was originally one of the roles MO hoped to play, IIRC. – Yemon Choi Oct 29 '15 at 0:15
• This is a straightforward linear algebra question and doesn't belong here. Project your subalgebra to a subspace in the span of $x_1,\ldots,x_n, y_1,\ldots,y_n$ and read en.wikipedia.org/wiki/Symplectic_vector_space#Subspaces – S. Carnahan Oct 29 '15 at 13:37
• @YemonChoi anyway the question seems to be essentially a copy-paste of the original question on MathSE, which already had a complete answer. Second, unlike what you claim, the question was not "the maximal dimension of ... is $n+1$, is this result well-known?", but "what is the dimension of..." without the answer. – YCor Oct 29 '15 at 15:43
• I am not at all happy with the votes to delete this question. It strikes me that some people believe all algebra questions they can easily do are therefore infra dig on MathOverflow. While I'm prepared to believe these people easily teach themselves all the functional analysis they might run into, I'm not convinced this is obviously the case – Yemon Choi Oct 9 '16 at 2:12

As already answered here on MathSE, 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 $n$.
• (Typo: I mean: $2n$-dimensional symplectic vector space) – YCor Jan 21 '16 at 10:34