Kirillov orbit method

Remarks

A polarization always exists for nilpotent $\mathfrak g$, and its dimension is $\tfrac12 (\dim \mathfrak g + \dim \mathfrak g_\ell)$, where $\mathfrak g_\ell = \set{X : \ell([X, \mathfrak g]) = 0}$.

Example: let $\mathfrak g$ have basis $X, Y, Z$ with $[X, Y] = Z$ and $Z$ central, the Lie algebra of the Heisenberg group. For $\ell$ with $\ell(Z) = \lambda \ne 0$, the coadjoint orbit is the plane $\set{\ell' : \ell'(Z) = \lambda}$. The subalgebra $\mathfrak h = \operatorname{span}(Y, Z)$ is abelian, so $\ell([\mathfrak h, \mathfrak h]) = 0$, and it is maximal with this property since $\ell([X, Y]) = \lambda \ne 0$. The induced representation $\pi_\ell$ is the Schrödinger representation with central character $e^{i \lambda z}$. For $\ell(Z) = 0$ the orbit is the single point $\ell$, and $\pi_\ell$ is the one-dimensional character $\exp(aX + bY + cZ) \mapsto e^{i(a \ell(X) + b \ell(Y))}$.

Non-example: for $\ell(Z) = \lambda \ne 0$, the center $\mathfrak h = \operatorname{span}(Z)$ satisfies $\ell([\mathfrak h, \mathfrak h]) = 0$ but is not maximal. The induced representation $\mathrm{Ind}_H^G \chi_\ell$ acts on $L^2(\R^2)$ and is a direct sum of infinitely many copies of the Schrödinger representation, so it is reducible.