Let $G$ be a compact Lie group and $T$ be its maximal tours, and $a\in \mathfrak{g}^\*$. and $G\_a$ be the isotropy group of $G$ then $T\subset G\_a$ and we know that $\pi\_1(T)=\mathbb{Z}^n$. My question is how can we find a formula for $\pi\_1(G\_a)$ by using maximal tours. Is there any relation?



A formula for isotropy group $\pi_1(G_a)$
Assume that $G$ is connected and simply-connected. From the principal $G\_a$-bundle $G\_a\rightarrow G\rightarrow G/G\_a=\mathcal{O}\_a$, we obtain a long-exact sequence $$\ldots\rightarrow\pi\_i(G\_a)\rightarrow\pi\_i(G)\rightarrow\pi\_i(\mathcal{O}\_a)\rightarrow\pi\_{i-1}(G\_a)\rightarrow\ldots.$$ Since $\pi\_1(G)=0$ and $\pi\_2(G)=0$, exactness considerations imply that $\pi\_1(G\_a)=\pi\_2(\mathcal{O}\_a)$. Often, orbits $\mathcal{O}\_a$ are partial flag varieties of $G\_{\mathbb{C}}$, and their second homotopy groups are reasonably well-known. I am not sure if this helps.


