I'm trying to understand a statement on page 5 in the book "Analysis on Symmetric Cones" by Faraut and Koranyi.
The situation is as follows. The connected component $G$ of the identity in $G(\Omega)$ is acting transitively on the open cone $\Omega$. A stabilizer $G_a$ for $a\in \Omega$ apparently is connected because the quotient space $G/G_a$ is homeomorphic to $\Omega$ which is simply connected as it is convex.
I think I understand why $G/G_a$ is homeomorphic to $\Omega$, but would like to see why this implies that $G_a$ is connected. What is the simplest way to understand this claim, using the least amount of Lie theory as possible?
The Lie theory fact you need here is that for a closed subgroup $H\subset G$ the projection $G\to G/H$ defined by $g\mapsto gH$ is a locally trivial fiber bundle with typical fiber $H$. The rest is an issue of algebraic topology: Locally trivial fiber bundles (involving paracompact spaces) are fibrations in the sense of algebraic topology. Such a fibration induces a long exact sequence of homotopy groups which contains the part $\dots \to \pi_1(G)\to \pi_1(G/H)\to \pi_0(H)\to \pi_0(G)\to\dots $. You assume that $G$ is connected, so $\pi_0(G)$ consist of just one point, which means that the map $\pi_1(G/H)\to\pi_0(H)$ is surjective, so unless $H$ is connected, $G/H$ cannot be simply connected.