Let $G$ be a compex Lie group with semisimple Lie algebra $\mathfrak{g}$, the connected subgroup $B$ of $G$ with Lie algebra $\mathfrak{b} := \mathfrak{b} \oplus \bigoplus_{\alpha \in R^{+}} \mathfrak{g}_{\alpha} $, ( here $R^{+}$ is the set of positive roots) is called a Borel subgroup.
In Fulton-Harris' Representation Theory on page 383 there is a
Claim 23.45. B is a closed subgroup of G, and the quotient G/B is compact.
PROOF. Consider the adjoint representation of $G$ on $\mathfrak{g}$. The action of the Borel subalgebra $b$ (? I think that the authors meant here the action by $B$) obviously preserves the subspace $\mathfrak{b} \subset \mathfrak{g}$, and, in fact, $\mathfrak{b}$ is just the inverse image of the subalgebra of $\mathfrak{gl} (\mathfrak{g})$ preserving this subspace: if $X = \sum X_{\alpha}$ is any element of $\mathfrak{g}$ with $ X_{\alpha} \subset \mathfrak{g}_{\alpha} $ for some $\alpha \in R^{-} $ , we could find an element $H$ of $\mathfrak{b} \subset \mathfrak{g}$ with $ ad(X)(H) \not \in \mathfrak{g}$ - any $H$ not in the annihilator of $\alpha \in \mathfrak{h}^* $ would do. $ B $ is thus (the connected component of the identity in) the inverse image in $ G $ of the subgroup of $GL(\mathfrak{g}) $ carrying b into itself. It follows that $ B$ is closed; and the quotient $G/B$ is contained in a Grassmannian and hence compact.
Moreover an example: in the case of the classical groups, it is easy (?) to describe the Borel subgroups and the corresponding quotients. For $G = SL_{n+1} \mathbb{C}$ , then $B $ is the group of all upper-triangular matrices in $G$, i.e., those automorphisms preserving the standard flag. It follows that $G/B$ is the usual (complete) flag manifold, i.e., the variety of all flags.
Questions:
I not understand the part in the proof where the authors conclude in the above that the quotient $G/B$ should be contained in a Grassmannian. I found several different longer proofs in the web on this issue but here seemingly the authors deduce it directly from previous arguments & assumptions, otherwise I not see any reason why at that point the authors not going into detail how this embedding of $G/B$ in certain Grassmanian manifold should be done.
Secondly, I not see how to deduce for $G = SL_{n+1} \mathbb{C}$ , that the Borel group $B $ is the group of all upper-triangular matrices in $G$. Could somebody sketch which ideas are involved? Is here really only the (a bit tedious) writing out of the exponantiation of $\mathfrak{b}$ considered as restriction of exponentiation of $\mathfrak{g}$ to $G$ involved, or can it be seen more directly that $B$ consists of upper-triangular matrices in $G$?