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

  • $\begingroup$ What's your question? $\endgroup$ – user38268 Jun 12 '13 at 1:22
  • $\begingroup$ Thanks, I improved the question. $\endgroup$ – Vasco Jun 12 '13 at 13:10
  • $\begingroup$ It is unclear to me why the WP page says the action on the dual space is the one that is required; the rest of the page seems to make no such assertion. The Weyl group acts both on the Cartan subalgebra and (as usual by inverse transposes) on its dual; giving one or the other is equivalent. $\endgroup$ – Marc van Leeuwen Jun 12 '13 at 13:24
  • $\begingroup$ @MarcvanLeeuwen It seems to me that coming from the Lie algebra side one tends to think of the Weyl group as a kind of symmetry group of the root system. So the action on the dual space is "the one required" (and the action on the subalgebra is merely induced by the action on the root system). However, those coming from a Lie group background tend to think of the Weyl group as the quotient of the normalizer of a maximal torus (by the torus) and so the Weyl group more properly belongs to the group/algebra. So maybe the WP author is more of an algebra guy. That's just my guess anyway. $\endgroup$ – Bill Cook Jun 14 '13 at 2:09

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 $P\cdot e_i = e_{\sigma(i)}$.

  • $\begingroup$ I'm not getting this induced action relate to the Weyl group. $\endgroup$ – Vasco Jun 13 '13 at 17:31
  • $\begingroup$ There is something strange about setting $P\cdot x = P^{-1}xP$, as it implies that $PQ\cdot x=Q\cdot P\cdot x$. The only reason I can think of is that the unspecified correspondence between permutations and permutation matrices is such that it is an anti-homomorphism (reversing the order of multiplication). so that one does get a left action of permutations. Unfortunately the involution taken as example does not make this clear. It would seem more natural to choose the correspondence to be a group homomorphism, wouldn't it? $\endgroup$ – Marc van Leeuwen Jun 14 '13 at 8:37
  • $\begingroup$ Oops! Thanks @MarcvanLeeuwen ! I wrote down a right action instead of a left action. $\endgroup$ – Bill Cook Jun 14 '13 at 14:05
  • $\begingroup$ @MarceloSilva I added a little more about what such an action does in terms of a dual basis. I hope that helps. $\endgroup$ – Bill Cook Jun 14 '13 at 14:05

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 $(g\alpha)(gv)=\alpha(v)$.

  • 1
    $\begingroup$ If you haven't seen this before, it may be a useful exercise to check that the use of $g^{-1}$ rather than just $g$ in the definition is needed in order to get $g(h\alpha)=(gh)\alpha$. If you omit the inverse, you get $g(h\alpha)=(hg)\alpha$. $\endgroup$ – Andreas Blass Jun 12 '13 at 13:23

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