Opposite Borel subgroup of a conjugated Borel subgroup Let $G$ be a connected split reductive group over some field $k$, with split maximal torus $T$ and given Borel $B \supset T$. Further $W=N_G(T)/T$ is the Weyl group of the split pair $(G,T)$ with longest element $w_0$. Then for $w \in W$,  $B'=wBw^{-1}$ is another Borel subgroup of $G$ containing $T$.
The opposite Borel subgroup of $B$ is the unique Borel subgroup $B^-$ of $G$ such that $$B \cap B^- =T.$$ Similar for $B'$. It is well known that for the opposite Borel subgroup of $B$ we have $B^-=w_0Bw_0$ resp. for the one of $B'$ we have $(B')^{-}=w_0B'w_0$.
On the other hand $$B' \cap wB^{-}w^{-1}=wBw^{-1} \cap wB^{-}w^{-1}=w(B \cap B^{-1})w^{-1}=wTw^{-1}=T.$$
This implies that $wB^{-}w^{-1}$ is the opposite Borel subgroup of $B'$.
In examples with $SL_3$ it seems that in general $$ww_0Bw_0w^{-1} \neq w_0wBw^{-1}w_0,$$
especially $(B')^- \cap B'=w_0wBw^{-1}w_0 \cap B' \neq T$. That confused me. Can someone tell what my mistake is?
 A: Let us forget about algebraic groups for a second and think only about Coxeter groups. The longest element $ w_0 $ is not intrinsic to the Weyl group $ W $, it depends on the choice of a positive subsystem $ \Phi^+ $ of the root system $ \Phi $: it is the unique element of $ w_0 $ which maps $ \Phi^+ $ to $ -\Phi^+ $.
Going back to reductive groups, the choice of a Borel subgroup $ B $ determines a positive system
$$ \Phi^+(B) = \{ \alpha \in \Phi | \operatorname{Lie}(G)_\alpha \subset \operatorname{Lie}(B) \} = \{ \alpha \in \Phi | U_\alpha \subset B \} $$
(see e.g. 21.35 in Milne's book (the book, not the online notes)), and the symmetry with respect to $ B $ is defined to be the longest element with respect to $ \Phi^+(B) $ (see Milne's book, 21.72). Then your statement "$ w_0 B w_0 $ is the opposite Borel subgroup of $ B $" should, in a more precise way, be formulated as "$ w_0 B w_0 $ is the opposite Borel subgroup of $ B $ where $ w_0 $ is the symmetry with respect to $ B $" (see the paragraphs after 21.85 in Milne's book).
Now let $ B $ be any Borel subgroup, let $ w \in W $, let $ w_0 $ be the symmetry with respect to $ B $ and set $ B' := wBw^{-1} $. By the second description of $ \Phi^+(B) $ and the fact that $ U_{w(\alpha)} = wU_\alpha w^{-1} $, it is easy to see that $ \Phi^+(B') = w \Phi^+(B) w^{-1} $ (because $ U_\alpha \subset wBw^{-1} $ if and only if $ U_{w^{-1}(\alpha)} \subset B $). It follows that $ w_0' := ww_0w^{-1} $ satisfies
$$ w_0' \Phi^+(wBw^{-1}) w_0' = ww_0 w^{-1} w \Phi^+(B) w^{-1} ww_0 w^{-1} = -w\Phi^+(B) w^{-1} = -\Phi^+(wBw^{-1}), $$
so $ w_0' $ is the symmetry with respect to $ B' $. Thus the Borel subgroup opposite to $ B' $ is
$$ w_0' B' w_0' = ww_0w^{-1} wB w^{-1} ww_0w^{-1} = ww_0 B w_0 w^{-1} = wB^-w^{-1}, $$
which is consistent with your second computation.
