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The Weyl group of $A_{n}$ is the symmetric group

Example

For $n \geq 1$ , $W (A_{n}) ≅ S_{n + 1}$ , acting on the sum-zero subspace of $R^{n + 1}$ by permuting coordinates.

Facts & Assumptions

Given: An integer $n \geq 1$ , the sum-zero subspace $E \subseteq R^{n + 1}$ , and the model $A_{n} = {ε_{i} - ε_{j} : i \neq j} \subseteq E$ .

[L1]

The coordinate model of $A_{n}$ consists of the roots $ε_{i} - ε_{j}$ in the sum-zero subspace (Classical root systems in coordinates).

[L2]

The Weyl group is generated by the root reflections (Weyl group).

Verification

technique · direct

1.1L1L2algebra

Let $ρ : S_{n + 1} \to O (E)$ be the homomorphism obtained by restricting coordinate permutations to $E$ . For $x \in E$ , the reflection formula gives $s_{ε_{i} - ε_{j}} (x) = x - (x_{i} - x_{j}) (ε_{i} - ε_{j}) = ρ ((i j)) x$ . Hence [L2] and the fact that the transpositions generate $S_{n + 1}$ give $W (A_{n}) = ρ (S_{n + 1})$ .

2.1L1step 1.1algebra∎

The homomorphism $ρ$ is injective. Indeed, if $ρ (σ)$ is the identity on $E$ , then for every $i \neq j$ it fixes $ε_{i} - ε_{j}$ , so $ε_{σ (i)} - ε_{σ (j)} = ε_{i} - ε_{j}$ ; uniqueness of the positive and negative coordinate positions gives $σ (i) = i$ and $σ (j) = j$ . Thus $σ = 1$ , and step 1.1 yields $W (A_{n}) ≅ S_{n + 1}$ with the asserted action.

Depends on

Used by

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Sources

Pavel Etingof, MIT 18.745 Lie Groups and Lie Algebras I, Lectures 19-24 (standard reference, not scraped)

The Weyl group of AnAn​ is the symmetric group