Let $G$ be a complex lie group with semisimple lie algebra $g$.
Put $b$ the borel subalgebra (so the sum of $h$ and all the positive roots) for some choices.
One can there is the borel subgroup $B$ with lie algebra $b$ in $G$; given as a the stabilizer of $b$ in the adjoint action. My book claims now that $G/B$ is compact, since it's isomorphic to its image in the Grassmanian (i.e where the adjoint sends $b$) and the Grassmanian is compact. However, how do we know that the image is closed?
