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 $B \cap B^- = T$.
Best Answer
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 $\mathfrak b$.