Presentation on Root Systems

Marcus and Katie

Let V be an finite dimensional inner product space over ℝ

Definition 1 Define

O (V) = {s ∈ End_{ℝ} (V) : ⟨sξ, sη⟩ = ⟨ξ, η⟩ for all v, w ∈ V } .

A reflection s is an element of O (V) such that s ≠ 1, and s fixes every element in some hyperplane ⟨α⟩^{⊥}

Proposition 1 Let H = ⟨α⟩^{⊥} ≤ V be a hyperplane. Then there exists a unique reflection s ∈ O (V) that fixes H. Moreover,

s (ξ) = ξ − \frac{2 ⟨ξ, α⟩}{⟨α, α⟩}α.

Proof. First assume that s ∈ O (V) is such a map. For any β ∈ H we have

⟨sα, β⟩ = ⟨sα, sβ⟩ = ⟨α, β⟩ = 0

so that sα ∈ ⟨α⟩. But since s preserves length and isn’t the identity, we have sα = −α. Let ξ ∈ V. Since V = H ⊕ ⟨α⟩, we have

ξ = η + tα for some η ∈ H, t ∈ ℝ.

Then

⟨ξ, α⟩ = t ⟨α, α⟩

so that

t = \frac{⟨ξ, α⟩}{⟨α, α⟩}

and

sξ = η − tα = ξ − 2tα = ξ − \frac{2 ⟨ξ, α⟩}{⟨α, α⟩}α.  (1)

We check that the map defined as in (1) satisfies s ≠ 1, s ∈ O (V) and s fixes H. This shows that s exists and is unique.

Always write s_{α} for the (unique) reflection in ⟨α⟩^{⊥}.

Proposition 2 Let g ∈ O (V). Then s_{gα} = gs_{α}g^{−1}.

Proof. Let ξ ∈ ⟨gα⟩^{⊥}. This means that ⟨gα, ξ⟩ = 0. But since g^{−1} ∈ O (V), we have that 0 = ⟨gα, ξ⟩ = ⟨α, g^{−1}ξ⟩ so that g^{−1}ξ ∈ ⟨α⟩^{⊥}. Then gs_{α}g^{−1} fixes ξ. But we clearly have that gs_{α}g^{−1} ≠ 1 and gs_{α}g^{−1} ∈ O (V), so we must have s_{gα} = gs_{α}g^{−1}.

Definition 2 A root system Δ is a finite set of vectors satisfying

1. 0 ∉ Δ and Δ spans V.

2. (Reduced Condition) If α ∈ Δ, then −α ∈ Δ. Moreover, if α and cα are both in Δ, then c = ±1.

3. For each α ∈ Δ, we have s_{α}Δ = Δ.

Let Δ be a fixed root system and let ξ be a fixed vector with ⟨ξ, α⟩ ≠ 0 for all α ∈ Δ. Write

Δ_{+} = {α ∈ Δ : ⟨α, ξ⟩ > 0}
Δ_{−} = {α ∈ Δ : ⟨α, ξ⟩ < 0} .

Example 1

Lemma 1 Suppose 0 = ∑_{j} a_{j}α_{j} for a_{j} ≥ 0 and α_{j} ∈ Δ_{+}. Then a_{j} = 0 for all j.

Proof. If 0 = ∑_{j} a_{j}α_{j}, then

0 = ⟨∑_{j} a_{j}α_{j}, ξ⟩ = ∑_{j} \underbrace{a_{j}}_{≥0} \overbrace{⟨α_{j}, ξ⟩}^{>0}

so that a_{j} = 0 for all j.

Definition 3 Define the positive span of a set of roots S

Sp (S) = {∑_{j} a_{j}α_{j} : a_{j} ≥ 0, α_{j} ∈ S, not all a_{j} = 0}

Definition 4 Let Π ⊂ Δ_{+} be such that Sp (Π) ∩ Δ = Δ_{+} and assume Π is a minimal subset of Δ_{+} with this property. Π is called a fundamental system and the roots in Π are called fundamental roots.

Example 2

Theorem 1 Π is a basis for V.

Write Π = {α_{1}, α_{2}, …, α_{m}}.

Lemma 2 For α_{i}, α_{j} ∈ Π with α_{i} ≠ α_{j}, it is impossible to have c_{i}α_{i} − c_{j}α_{j} ∈ Δ_{+} with both c_{i}, c_{j} > 0.

Proof. If c_{i}α_{i} − c_{j}α_{j} ∈ Δ_{+} with c_{i}, c_{j} > 0, write

c_{i}α_{i} − c_{j}α_{j} = ∑_{k} a_{k}α_{k}

for a_{k} ≥ 0 and α_{k} ∈ Π. If a_{i} − c_{i} ≥ 0, we then have

0 = ∑_{k≠i,j} a_{k}α_{k} + (a_{i} − c_{i}) α_{i} + (a_{j} + b_{j}) α_{j}

with all the coefficients ≥ 0 and a_{j} + b_{j} > 0. This contradicts Lemma 1. If, on the other hand we have a_{i} − c_{i} < 0, then we have

(c_{i} − a_{i}) α_{i} = ∑_{k≠i,j} a_{k}α_{k} + (a_{j} + b_{j}) α_{j}

We have that c_{i} −a_{i} > 0, so dividing through by c_{i} −a_{i}, we have expressed α_{i} as a positive linear combination of Π\α_{i}, which contradicts the minimality of Π.

Lemma 3 For α_{i}, α_{j} ∈ Π with α_{i} ≠ α_{j}, ⟨α_{i}, α_{j}⟩ ≤ 0.

Proof. We have by Proposition 1 that

s_{α_{i}} (α_{j}) = α_{j} − \frac{2 ⟨α_{i}, α_{j}⟩}{⟨α_{i}, α_{i}⟩}α_{i}

so that since Δ is a root system, we have that one or the other of

α_{j} − \frac{2 ⟨α_{i}, α_{j}⟩}{⟨α_{i}, α_{i}⟩}α_{i} or  \frac{2 ⟨α_{i}, α_{j}⟩}{⟨α_{i}, α_{i}⟩}α_{i} − α_{j}  (2)

is in Δ_{+}. In both cases, we have by Lemma 2 that ⟨α_{i}, α_{j}⟩ ≤ 0.

Remark 1 The second case in (2) above is not actually possible by the Main Lemma below, but we don’t know this yet.

Proof of Theorem 1. We need to show that Π is linearly independent. We prove the stronger result that any set {α_{1}, α_{2}, …, α_{m}} ⊂ Δ_{+} with ⟨α_{i}, α_{j}⟩ ≤ 0 for all i, j with i ≠ j is independent. If not, we have a dependence relation

ρ = ∑_{i∈I} a_{i}α_{i} = ∑_{j∈J} a_{j}α_{j}

for disjoint, nonempty I, J ⊂ {1, 2, …, m} and a_{i}, a_{j} ≥ 0. On the one hand, we have ⟨ρ, ρ⟩ ≥ 0. On the other hand, we have

⟨ρ, ρ⟩ = ∑_{i,j} a_{i}a_{j} ⟨α_{i}, α_{j}⟩ ≤ 0,

by Lemma 3. Thus, we must have

0 = ρ = ∑_{i∈I} a_{i}α_{i} = ∑_{j∈J} a_{j}α_{j}

Then by Lemma 1, we have that all the a_{i} and all the a_{j} are 0.

Corollary 1 The set Π is unique given ξ and Δ_{+}. Conversely, given a fundamental system Π, we have a unique positive set of roots Δ_{+} for some ξ.

Proof. We show that

Π = {α ∈ Δ_{+} : α is not a positive linear combo of two or more β ∈ Δ_{+}}

Write

Δ_{+} = {β_{1} = α_{1}, β_{2} = α_{2}, … β_{m} = α_{m}, β_{m+1}, …, β_{n}} .

Let β_{j} = ∑_{k=1}^{m} d_{j,k}α_{k}. Observe that d_{j,k} ≥ 0. Suppose

α_{1} = ∑ c_{j}β_{j}  (3)

with c_{j} ≥ 0. Then

α_{1} = ∑_{j,k} c_{j}d_{j,k}α_{k}

so that ∑_{j} c_{j}d_{j,1} = 1 and ∑_{j} c_{j}d_{j,k} = 0 for all k ≠ 1. We know that d_{1,k} = 0 for all k ≠ 1, so we must have c_{1} = 1. Similarly, we have d_{j,k} = δ_{j,k} for 2 ≤ k ≤ m so we must have c_{j} = 0 for these k. Finally, inserting these values into into (3), we have

α_{1} = α_{1} + ∑_{j=m+1}^{n} c_{j}α_{j},

but this by Lemma (1) means that c_{j} = 0 for m + 1 ≤ j ≤ n. This shows that α_{1} is not a positive linear combination of positive roots.

Conversely, if β ∉ Π, then since Π is a basis for V, we must have that β is a linear combination of at least 2 elements of Π ⊂ Δ_{+}.