Remark 4

Denote by $\mathfrak{g}_{reg}$ the set of regular elements of $\mathfrak{g}$ (Chap. VII, §2, no. 2, Def. 2) and put $\mathfrak{t}_{reg}=\mathfrak{t}\cap \mathfrak{g}_{reg}$. For $x\in \mathfrak{t}$, we have

det(X $-$ ad$_{\mathfrak{g}}x$) $= X^{dim}\prod_{\alpha\in R(G,T)}^{\mathfrak{t}}(X-\delta (\alpha )(x))$,

and hence $\mathfrak{t}_{reg}$ is the set of elements $x$ of $\mathfrak{t}$ such that $\delta (\alpha )(x)\not= 0$ for all $\alpha \in R(G,T)$, that is the union of the chambers of $\mathfrak{t}$ (so $\mathfrak{t}_r\subset \mathfrak{t}_{reg}$). Consequently $C\cap \mathfrak{t}_{reg}= C$, so we have homeomorphisms

$C\rightarrow \mathfrak{t}_{reg}/W\rightarrow \mathfrak{g}_{reg}/$Ad(G).