A parabolic subgroup of a connected algebraic group is one which contains a Borel subgroup. I'm trying to understand why parabolic subgroups are connected.
Let $P$ be a parabolic subgroup containing a Borel group $B$. Since $B$ is connected, $B\subset P^\circ$, the component of the identity in $P$. Also, $P^\circ$ is a connected algebraic group itself, so any other Borel subgroup of $G$ in $P^\circ$ is $P^\circ$-conjugate to $B$. Thus $$ N_G(P^\circ)=P^\circ N_G(B)=P^\circ B=P^\circ. $$ Since $P\subset N_G(P^\circ)$ since $P^\circ\unlhd P$, it follows $P=P^\circ$.
I'm missing the equality $N_G(P^\circ)=P^\circ N_G(B)$, it seems to follows from the fact that the Borel subgroups of $G$ in $P^\circ$ are still conjugate in $P^\circ$, but I don't see it.
If $g \in N_G(P^\circ)$, then $B^g$ is a Borel subgroup of $P^\circ$, so $B^g = B^p$ for some $p \in P^\circ$. Thus $gp^{-1} \in N_G(B)$, which gives $g \in P^\circ N_G(B)$.