Let $G$ an algebraic (reductive) group. $T$ a maximal torus, $B$ a Borel subgroup containing $T$, and $w\_0$ the longest element of the Weyl group.


I'm looking for a reference explaining why when you conjugate $B$ by $w\_0$, the result is the opposite Borel subgroup $B^-$.


Is there a proof involving roots of $G$ relative to $T$ ?


I've found a proof in a book of M. Geck, but this proof doesn't involve roots at all, but only the fact that $B^-$ is uniquely defined by the relation $B \cap B^- = T$. 



Longest element of a Weyl group 
The proof depends on how you're setting things up. In my opinion the cleanest approach is the Lie algebraic one, and it goes as follows. Your Borel subalgebra $\mathfrak b$ determines a choice of simple roots $\Delta$ and consequently a choice of positive roots $\Phi^+$: $\mathfrak b = \mathfrak t \oplus \bigoplus\_{\alpha \in \Phi^+} \mathfrak g\_\alpha$. The action of $w \in W$ takes $\mathfrak b$ to $\mathfrak b\_w = \mathfrak t \oplus \bigoplus\_{\alpha \in \Phi^+} \mathfrak g\_{w\alpha}$. With respect to the length function defined using $\Delta$, the longest element $w\_0$ of $W$ takes $\Phi^+$ to $-\Phi^+$. It follows that $b\_{w\_0}$ is the Borel subalgebra opposite to $\mathfrak b$.


