The Weyl reflection in ${\mathfrak{s}\mathfrak{l}}_2$

Since $\mathfrak{h}=\mathbb{C}h$ is one-dimensional, so is ${\mathfrak{h}}^{\ast }$; it is spanned by $\alpha$ with $\alpha (h)=2$. The reflection formula gives $s_{\alpha }(\alpha )=\alpha -\alpha (h_{\alpha })\alpha =\alpha -2\alpha =-\alpha$, so $s_{\alpha }=-\mathrm{id}$ on the whole line.

The element $W=e^e{e}^{-f}{e}^e$ is the product of the three matrix exponentials $e^e=\left(\begin{array}{cc}1& 1\\ 0& 1\end{array}\right)$, $e^{-f}=\left(\begin{array}{cc}1& 0\\ -1& 1\end{array}\right)$, $e^e=\left(\begin{array}{cc}1& 1\\ 0& 1\end{array}\right)$, which multiplies to $\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)$; it lies in $S{L}_2(\mathbb{C})$ and satisfies $W^2=-I$, $Wh{W}^{-1}=-h$.

Define ${\tau }_{\alpha }={\mathrm{Ad}}_W$ on the displayed matrix algebra. Since ${\mathrm{Ad}}_A{\mathrm{Ad}}_B={\mathrm{Ad}}_{AB}$ for invertible matrices, step 1.2 gives ${\tau }_{\alpha }={\mathrm{Ad}}_{e^e}{\mathrm{Ad}}_{e^{-f}}{\mathrm{Ad}}_{e^e}$ for the explicitly chosen standard triple. Direct conjugation acts on its basis by $h\mapsto -h$, $e\mapsto -f$, $f\mapsto -e$: indeed $We{W}^{-1}=\left(\begin{array}{cc}0& 0\\ -1& 0\end{array}\right)=-f$ and $Wf{W}^{-1}=\left(\begin{array}{cc}0& -1\\ 0& 0\end{array}\right)=-e$. Hence ${\tau }_{\alpha }$ maps ${\mathfrak{g}}_{\alpha }=\mathbb{C}e$ to ${\mathfrak{g}}_{-\alpha }=\mathbb{C}f$ and back, and its action on $\mathfrak{h}$ is $-\mathrm{id}$.

Since ${\tau }_{\alpha }$ acts on $\mathfrak{h}$ as $-1$, its induced action on ${\mathfrak{h}}^{\ast }$ is also $-1$: for $\lambda \in {\mathfrak{h}}^{\ast }$ the induced functional is $\lambda \circ {\tau }_{\alpha }^{-1}=\lambda \circ (-\mathrm{id})=-\lambda$. This equals $s_{\alpha }$ by step 1.1, so the Weyl reflection of the root $\alpha$ is realized by the inner automorphism ${\tau }_{\alpha }$ and swaps the roots $\pm \alpha$.