Parabolic subalgebra

Remarks

Example: in the general linear Lie algebra $\mathfrak{gl}_3(k)$, let $\mathfrak p$ be the set of matrices whose $(3, 1)$ and $(3, 2)$ entries vanish. It is closed under the commutator, since it is the set of matrices that map the plane $k e_1 + k e_2$ into itself. It contains the upper triangular matrices, which form a Borel subalgebra, so $\mathfrak p$ is parabolic.

Non-example: in the special linear Lie algebra $\mathfrak{sl}_2(k)$, the line $kh$ spanned by $h = \operatorname{diag}(1, -1)$ is a subalgebra, but it is not parabolic. Every Borel subalgebra of $\mathfrak{sl}_2(k)$ has dimension $2$, so none fits inside the one-dimensional $kh$.

Every Borel subalgebra and the whole algebra $\mathfrak g$ are parabolic. For a semisimple $\mathfrak g$ with a fixed Borel subalgebra $\mathfrak b$ and base $\Delta$ of simple roots, the parabolic subalgebras containing $\mathfrak b$ correspond bijectively to the subsets of $\Delta$.