How to show that there is a bijection from Weyl group to $\{B \subset G: T \subset B, \text{$B$ is a Borel subgroup of $G$}\}$?

Question

Let $G$ be a semisimple Lie group, $T$ a maximal torus of $G$, $W=N_G(T)$ the Weyl group of $G$. How to show that there is a bijection from $W$ to $\{B \subset G: T \subset B, \text{$B$ is a Borel subgroup of $G$}\}$? Thank you very much.

Answer 1

For the Weyl group $W$ you want $N_G(T)/Z_G(T)$, not $N_G(T)$. Actually $Z_G(T) = T$, so $W = N_G(T)/T$, but this is a difficult theorem.

Fix a Borel subgroup $B$ of $G$ containing $T$. The claim is that

$$nZ_G(T) \mapsto nBn^{-1} \tag{$n \in N_G(T)$}$$

is a bijection from $W$ onto the set of Borel subgroups of $G$ containing $T$.

Well defined/injective: If $n \in N_G(T)$, then $nBn^{-1}$ is obviously still a Borel subgroup of $G$ containing $T$. For $n_1, n_2 \in N_G(T)$, we have

$$n_1Bn_1^{-1} = n_2Bn_2^{-1} \iff n_1^{-1}n_2 \in N_G(B) \iff n_1^{-1}n_2 \in B$$

because $N_G(B) = B$ (Springer, Linear Algebraic Groups, 6.49). We also have $$N_G(T) \cap B = N_B(T) = Z_B(T) = Z_G(T) \cap B$$ where the middle equality is 6.36 (ii). Since is always in $n_1^{-1}n_2 \in N_G(T)$, we have $n_1^{-1}n_2 \in B \iff n_1^{-1}n_2 \in Z_G(T) \iff n_1Z_G(T) = n_2Z_G(T)$.

Surjective: Let $B'$ be a Borel subgroup of $G$ containing $T$. All Borel subgroups of $G$ are conjugate (6.27 (ii)), so there exists a $g \in G$ such that $gBg^{-1} = B'$. Then $gTg^{-1}$ is a maximal torus of $B'$. All maximal tori of $B'$ are conjugate (6.35 (iii)), so there exists a $b' \in B'$ such that $b'gTg^{-1}b'^{-1} = T$. Then $n = b'g \in N_G(T)$, with

$$nBn^{-1} = B'$$

Answer 2

It is well known that the group $W$ is in bijection with the set of all base of $\Phi$ where $\Phi$ is the root system associated to $\mathfrak h := \text{Lie}(T)$. In fact it is probably the hardest statement since you took a different definition of the Weyl group, first you should show that $W$ coincide with the reflection group of the root system.

Once you know it, you notice that if $\Delta$ is a base, that is a set of simple roots, then $\mathfrak b(\Delta) := \mathfrak h \oplus \bigoplus_{\alpha \in \Phi^+} \mathfrak g_{\alpha}$ is a Borel subalgebra, that is a maximal solvable algebra. It is surely solvable because its commutator is nilpotent, and by construction it's maximal.

For prove that any Borel $\mathfrak b \supset \mathfrak h$ is on this form, let $\mathfrak b = \mathfrak h \oplus \bigoplus_{\alpha} \mathfrak g_{\alpha}$ the weight decomposition. Again, as before the $\alpha$ should be a subset of $\Delta^+$ for some base $\Delta$. But we are done because it means $\mathfrak b \subset \mathfrak b(\Delta)$ and by maximality they are equal.