Let $G$ be a compact and connected and simply connected Lie group and $\mathfrak{g}$ be its Lie algebra and $x\in\mathfrak{g}^\*$. How can we compute the fundamental group of $G/G\_x$ where $G\_x$ is the isotropy group of $G$. i.e. $\pi\_1(G/G\_x)=?$



Computing the fundamental group of a flag variety
The answer is that $\pi\_1(G/G\_x)=0$. Use the long-exact sequence of homotopy groups one obtains from a fibration. In this case, the fibration is $G\_x\rightarrow G\rightarrow G/G\_x$. We have $$\ldots\rightarrow\pi\_1(G)\rightarrow\pi\_1(G/G\_x)\rightarrow\pi\_0(G\_x)\rightarrow\pi\_0(G)\rightarrow\ldots.$$ Since $G$ is connected and simply-connected, $\pi\_1(G)=0=\pi\_0(G).$ By exactness, $\pi\_1(G/G\_x)=\pi\_0(G\_x)$. However, $G\_x$ is always connected, so $\pi\_1(G/G\_x)=0$. (A reference for the connectedness of $G\_x$ is Section 2.3 of "Symplectic Fibrations and Multiplicity Diagrams" by Guillemin.) 


Another approach would be to realize $G/G\_x$ as a homogeneous space of the complexification $G\_{\mathbb{C}}$. We would have $G/G\_x=G\_{\mathbb{C}}/P$, where $P\subseteq G\_{\mathbb{C}}$ is a parabolic subgroup. The Bruhat decomposition of $G\_{\mathbb{C}}/P$ involves cells in even-dimensions only, meaning that $\pi\_1(G\_{\mathbb{C}}/P)=0$.


