Lemma 5

We have

$J(\mu$)(exp($z\gamma$ )) $=e^{z\delta(\mu)(\gamma)}\prod_{\alpha >0}(1-e^{-z\delta(\mu)(K_{\alpha})})$.

Thus, the function $J(\mu$)(exp($z\gamma$ )) is the product of a function that tends to 1 when $z$ tends to 0 and of

where N = Card $R_+$.

Assume first of all that $\rho \in X(T)$; applying Cor. 2 of Th. 2 we see that, as $z$ tends to $0,\chi_{\lambda}(z\gamma )$ tends to

hence the theorem in this case.

In the general case, it suffices to remark that, in proving Th. 3, we can always replace G by a suitable connected covering and thus reduce to the preceding case.

We now prove Lemma 5. Let $z\in \mathbf{C}$; denote by $\varphi_z$ the map from $\mathfrak{t}$ to the $\mathbf{C}$-algebra Map(X(T)$,\mathbf{C}$) of maps from X(T) to $\mathbf{C}$ that associates to $H\in \mathfrak{t}$ the map

$\varphi_z(H) :\mu \rightarrow \mu$(exp $zH$) $=e^{z\delta(\mu)(H)}$.

We have $\varphi_z(H+H') =\varphi_z(H)\varphi_z(H')$, so there exists a homomorphism of rings

$\psi_z:\mathbf{Z}[\mathfrak{t}]\rightarrow$ Map(X(T)$,\mathbf{C}$)

such that $\psi_z(e^H)(\mu) =e^{z\delta(\mu)(H)}$. On the other hand, by Chap. VI, §3, no. 3, Prop. 2, we have the following relation in $\mathbf{Z}[\mathfrak{t}]:$

Applying the homomorphism $\psi_z$, and using the equality

we deduce the stated formula.