Weyl group of the Lie algebra $\mathfrak{sl}_n$

Question

The Weyl group of the Lie algebra $\mathfrak{sl}_n$ is just the symmetric group on $n$ elements, $S_n$. The action can be realized as follows. If $\mathfrak{h}$ is the Cartan subalgebra of all diagonal matrices with trace zero, then $S_n$ acts on $\mathfrak{h}$ via conjugation by permutation matrices. This action induces an action on the dual space $\mathfrak{h}^\ast$, which is the required Weyl group action. Weyl group - Wiki

question 1: "$S_n$ acts on $\mathfrak{h}$ via conjugation by permutation matrices", what would that action? question 2: "This action induces an action on the dual space", how would you describe this induced action?

Thank you very much

Answer 1

For question #1. The action is exactly what it says it is.

For example: $\mathfrak{sl}_2$'s Cartan subalgebra is $\mathfrak{h}=\mathrm{span} \left\{ H \right\}$ where $H = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}$.

$S_2 = \left\{ I_2, P \right\}$ where $I_2$ is the $2 \times 2$ identity and $P = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$. Notice that $I_2 H I_2^{-1} = H$ and $PHP^{-1} = -H$. So $S_2$ acts on $\mathfrak{h}$ via conjugation (specifically $I_2$ acts as the identity and $P$ -- the transposition interchanging 1 and 2 -- sends $x \in \mathfrak{h}$ to $-x \in\mathfrak{h}$). Or another way to look at it, $P$ interchanges the $(1,1)$-entry in $x \in \mathfrak{h}$ with the $(2,2)$-entry.

In general if $P$ is a permutation matrix associated with the permutation $\sigma \in S_n$ (thinking of $S_n$ as bijections from $\{1,2,\dots n\}$ to itself), then for $x=\mathrm{diag}(x_1,x_2,\dots,x_n) \in \mathfrak{h}$ we have $P\cdot x = PxP^{-1} = \mathrm{diag}(x_{\sigma(1)},x_{\sigma(2)},\dots,x_{\sigma(n)}) \in \mathfrak{h}$.

For question #2. As Andreas points out, we get an induced action on the dual space, $\alpha \in \mathfrak{h}^*$, by $P \cdot \alpha (x) = \alpha(P^{-1}\cdot x) = \alpha(P^{-1}xP)$ for all $x \in \mathfrak{h}$ (I denoted the group actions using a "$\cdot$").

A little more concretely: Let $\varepsilon_i$ be dual to $e_i=\mathrm{diag}(0,\dots,0,1,0,\dots,0)$ (1 in the $(i,i)$-position), so $\varepsilon_i(e_j)=\delta_{ij}$ (Kronecker delta). Let $\sigma$ be the permutation associated with the permutation matrix $P$. Then $(P \cdot \varepsilon_i)(e_j) = \varepsilon_i(P^{-1} \cdot e_j) = \varepsilon_i(Pe_jP^{-1}) = \varepsilon_i(e_{\sigma^{-1}(j)})=\delta_{i\sigma^{-1}(j)}=\delta_{\sigma(i)j}$. Thus $P\cdot \varepsilon_i = \varepsilon_{\sigma(i)}$ just like $P\cdot e_i = e_{\sigma(i)}$.

Answer 2

For question 2, whenever a group $G$ acts on a vector space $V$, it also acts on the dual space $V^*$ as follows. If $g\in G$ and $\alpha\in V^*$ then $g\alpha$ is the element of $V^*$ that (regarded as a function from $V$ to scalars) sends each $v\in V$ to $\alpha(g^{-1}v)$.

Perhaps a nicer way to remember the formula is the equivalent form $(g\alpha)(gv)=\alpha(v)$.