Remark

Consider the function $\lambda_{\mathfrak{g}}$ associated to a compact Lie group G; in view of §2, no. 1, Th. 1, to calculate $\lambda_{\mathfrak{g}}$ it suffices to know its values on $\mathfrak{t}$. But, for $x\in \mathfrak{t}$, the endomorphism ad $x$ of $\mathfrak{g}$ is semi-simple, and has eigenvalues 0 (with multiplicity $r$) and, for all $\alpha \in R(G,T),\delta (\alpha )(x)$ (with multiplicity 1). It follows immediately that

$\alpha \in R(G,T)$

with $\delta_{\mathfrak{g}}(x) =\delta_G$(exp $x$) and $\pi_{\mathfrak{g}}(x) =\prod_{\alpha\in R(G,T)}\delta (\alpha )(x) =$ det ad$_{\mathfrak{g}/\mathfrak{t}}(x)$.

Let $\omega_{G/T}$ be an invariant differential form of degree $n-r$ on $G/T$ and $\omega_{\mathfrak{t}}$ a translation-invariant differential form of degree $r$ on $\mathfrak{t}$. With the notation of no. 1, denote by $\omega_{G/T}\cap \omega_{\mathfrak{t}}$ the unique translation-invariant differential form $\omega_{\mathfrak{g}}$ of degree $n$ on $\mathfrak{g}$ such that $\omega_{\mathfrak{g}}(0) =\omega_{G/T}(\overline{e})\cap \omega_{\mathfrak{t}}(0)$.

Finally, denote by $\psi : (G/T)\times \mathfrak{t}\rightarrow \mathfrak{g}$ the morphism of manifolds induced by the map $(g, x) \rightarrow$ (Ad $g$)$(x)$ from $G\times \mathfrak{t}$ to $\mathfrak{g}$ by passage to the quotient.