Lemma 4

Let $g\in G,t\in T$, and let $\overline{g}$ be the image of $g$ in $G/T$. Identify the tangent space of $G/T$ (resp. T, resp. G) at $\overline{g}$ (resp. $t$, resp. $gtg^{-1}$) with $\mathfrak{g}/\mathfrak{t}$ (resp. $\mathfrak{t}$, resp. $\mathfrak{g}$) by means of the left translation $\gamma (g)$ by $g$ (resp. $t$, resp. $gtg^{-1}$). The tangent linear map of $f$ at $(\overline{g}, t)$ is then identified with the linear map $f': (\mathfrak{g}/\mathfrak{t})\times \mathfrak{t}\rightarrow \mathfrak{g}$ defined as follows: if $z\in \mathfrak{g}, x\in \mathfrak{t}$, and if $\overline{z}$ denotes the image of $z$ in $\mathfrak{g}/\mathfrak{t}$, then

$f'(\overline{z}, x) =$ (Ad $gt^{-1}$)$(z-$(Ad $t$)$z+x)$.

Let F be the map from $G\times T$ to T such that $F(g, t) =gtg^{-1}$. Since $F\circ (\gamma (g)$, Id$_T) =$ Int $g\circ F$, we have $T_{(g,t)}(F)(gz, tx) = T_t$(Int $g$)$\circ T_{(e,t)}(F)(z, tx)$; by Chap. III, §3, no. 12, Prop. 46,

$T_{(e,t)}(F)(z, tx) =t$((Ad $t^{-1}$)$z-z$) $+tx=t$((Ad $t^{-1}$)$(z-$ (Ad $t$)$z+x$))

and consequently

$T_{(g,t)}(F)(gz, tx) =gtg^{-1}$((Ad $gt^{-1}$)$(z-$ (Ad $t$)$z+x$)).

The lemma follows immediately from this formula by passage to the quotient.