Let ε 1 , . . . , ε n denote the standard basis of R n . For each σ ∈ S n , we define g σ ∈ O(R n ) by setting

Exercise 12. Let n ≥ 2 be an integer, and let S n denote the symmetric group of degree n.

Let ε ₁ , . . . , ε _n denote the standard basis of R ⁿ . For each σ ∈ S n , we define g _σ ∈ O(R ⁿ ) by setting

g _σ (

∑ n

i=1

c _i ε _i ) =

∑ n

i=1

c _i ε _σ(i) ,

and set G _n = { g _σ | σ ∈ S n } . Show that { s _ε

_{− ε}

| 1 ≤ i < j ≤ n } is precisely the set of reflections in G n . In other words, for σ ∈ S n , show that g σ is a reflection if and only if σ is a transposition.

Proof. We saw earlier that g _σ is a reflection if σ is a transposition. (See p. 11 of our lecture note.) Next assume g _σ is a reflection. Then g ² _σ = 1. Since the mapping S n → G _n defined by σ 7→ g _σ is an isomorphism, we have

σ ² = 1.

Therefore there exist 2m integers 1 ≤ k ₁ , . . . , k _2m ≤ n such that σ = (k ₁ k ₂ )(k ₃ k ₄ ) · · · (k _{2m − 1} k _2m ).

Without loss of generality, we may assume k i = i for 1 ≤ i ≤ 2m, so that σ = (1 2)(3 4) · · · (2m − 1 2m).

We need to show that m = 1. We give two independent proofs of this.

(1) Since g _σ is a reflection, there exists a nonzero vector α ∈ R ⁿ such that g _σ = s _α . For any 1 ≤ i ≤ m,

s _α (ε _{2i − 1} ) = ε _{2i − 1} − 2(ε _{2i − 1} , α) (α, α) α Also since σ = (1 2)(3 4) · · · (2m − 1 2m),

g _σ (ε _{2i − 1} ) = ε _2i .

Therefore we get

α ∈ R(ε _{2i − 1} − ε _2i ).

Since i was arbitrary, this holds for every 1 ≤ i ≤ m. But since α is nonzero and ε _{2i − 1} − ε _2i (1 ≤ i ≤ m) are linearly independent, m must be equal to 1.

(2) For 1 ≤ i ≤ m, by the definition of g _σ , we have

g _σ (ε _2i−1 − ε _2i ) = ε _2i − ε _2i−1 = − (ε _2i−1 − ε _2i ).

Since ε _{2i − 1} − ε _2i (1 ≤ i ≤ m) are linearly independent, g _σ has an eigenvalue − 1 with multiplicity at least m. On the other hand, since g _σ is a reflection, g _σ has an eigenvalue − 1 with multiplicity exactly 1. This proves m = 1 as desired.

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Exercise 13. With reference to Exercise 12, set Φ = {± (ε _i − ε _j ) | 1 ≤ i < j ≤ n } is a root system, with a positive system Π = { ε _i − ε _j | 1 ≤ i < j ≤ n } . For w ∈ W (Φ), setting n(w) = | Π ∩ w ^{− 1} ( − Π) | , show that

n(g _σ ) = |{ (i, j) | i, j ∈ { 1, 2, . . . , n } , i < j, σ(i) > σ(j) }| (σ ∈ S n ).

Proof. Fix σ ∈ S n . By definition, we have

g _σ Π = { g _σ (ε _i − ε _j ) | 1 ≤ i < j ≤ n }

= { ε _σ(i) − ε _σ(j) | 1 ≤ i < j ≤ n } . Since ε _σ(i) − ε _σ(j) ∈ − Π if and only if σ(i) > σ(j ),

g _σ Π ∩ ( − Π) = { ε _σ(i) − ε _σ(j) | 1 ≤ i < j ≤ n, σ(i) > σ(j) } . Therefore

n(g _σ ) = | Π ∩ g ⁻ _σ ¹ ( − Π) |

= | g σ Π ∩ ( − Π) |

= |{ ε _σ(i) − ε _σ(j) | 1 ≤ i < j ≤ n, σ(i) > σ(j ) }|

= |{ (i, j) | 1 ≤ i < j ≤ n, σ(i) > σ(j) }| . The result follows.

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