Parabolic subgroups of real semisimple Lie groups

Question

Let $G$ be a semisimple closed and connected subgroup of $\mathrm{GL}_n(\mathbb R)$ and let $B$ be a maximal solvable subgroup of $G$. Let $P$ be any subgroup that contains $B$.

1. Do we call $P$ a parabolic subgroup of $G$ as in the case of complex algebraic groups?
2. Is $N_G(P)=P$?
3. Is $P$ connected?

Answer 1

I would say that the answer to this question is no in all three cases.

Namely, let's assume that $G$ is something like $U(2)(\mathbb{R})$. Note then that by the standard equivalence between anisotropic $\mathbb{R}$-groups and compact Lie groups the maximal sovlable subgroup of $G$ is the $\mathbb{R}$-points of a maximal solvable subgroup of $U(2)$ which is a torus $T$. Note then that taking $B=T=P$ one easily sees that the answer to 2. is no since, as you can show, $N_G(T)$ contains $T$ as an index $2$ subgroup. Moreover, one then easily sees that the answer to 3. is no by taking $P=N_G(T)$ since it has the disconnected quotient $N_G(T)/T\cong \mathbb{Z}/2\mathbb{Z}$.

EDIT: I didn't do a semisimple example, but this changes nothing. I assume you can cook up a semisimple example yourself--all you need is compactness.