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Simple roots form a signed integral basis

Statement

Let Φ⊆E be a reduced crystallographic root system with positive system Φ+ and simple roots Δ⊆Φ+ (Positive systems and simple roots). Then Δ is a basis of E; more precisely:

  1. Δ is linearly independent and spans E, so ∣Δ∣=dim⁡E;
  2. every positive root is a sum of simple roots with nonnegative integer coefficients, and every negative root is a sum of simple roots with nonpositive integer coefficients.

Consequently every root is a unique integral combination ∑α∈Δnαα of the simple roots in which the nonzero coefficients all have the same sign, positive for positive roots and negative for negative roots.

Facts & Assumptions

Given: A reduced crystallographic root system Φ⊆E with a regular vector v∈E, the positive system Φ+={α:(v,α)>0}, and the set Δ of simple roots, a root being simple when it is not a sum of two positive roots.

[L1]

Φ is finite, spans E, 0∉Φ, sα(Φ)=Φ, all Cartan integers are integral, and Rα∩Φ={±α} (Reduced crystallographic Euclidean root system).

[L2]

Φ+={α:(v,α)>0} and Φ−=−Φ+ are disjoint and cover Φ; a positive root is simple when it is not a sum of two positive roots (Positive systems and simple roots).

[L3]

If α,β are nonproportional roots with (α,β)>0 then α−β∈Φ (Rank-two root-system classification).

Proof

technique · direct
1.1L1L2algebra

Every positive root is a nonnegative integral sum of simple roots: if α∈Φ+ is not simple, it is a sum α=β+γ of two positive roots, and (v,β),(v,γ) are positive and add up to (v,α); iterating this decomposition and always choosing a summand that is not simple cannot continue forever, since the finitely many values (v,δ), δ∈Φ+, strictly decrease along each branch, so the process terminates and exhibits α as a sum of simple roots.

1.2L2L3algebra

Distinct simple roots α,β satisfy (α,β)≤0. Indeed, if (α,β)>0 then α−β∈Φ by [L3]; this root is positive or negative, and if it is positive then α=(α−β)+β is a nontrivial sum of two positive roots, while if it is negative then β=(β−α)+α is a nontrivial sum of two positive roots, contradicting the simplicity of α or of β.

2.1L1L2step 1.2algebra

The simple roots are linearly independent. For coefficients ci≥0 with at least one ci>0, a combination ∑iciαi of simple roots satisfies (v,∑iciαi)=∑ici(v,αi)>0 by [L2], so it cannot vanish; negating also excludes a nonzero relation with all coefficients nonpositive. Thus any nontrivial real linear relation can be written ∑i∈Iciαi=∑j∈Jcjαj with disjoint nonempty index sets and all ci,cj>0. Its common vector γ=∑i∈Iciαi satisfies (v,γ)>0, hence γ≠0 and (γ,γ)>0. On the other hand, expanding one side against the other gives (γ,γ)=∑i∈I,j∈Jcicj(αi,αj)≤0, since distinct simple roots have nonpositive inner product by step 1.2. This contradiction proves linear independence.

3.1L1step 1.1step 2.1algebra∎

The simple roots span E: every root is a simple-root combination by step 1.1 or its negative, and Φ spans E. Together with step 2.1 the set Δ is a basis of E, and with step 1.1 every root is an integral combination whose coefficients all have the sign of the root. Uniqueness of the coefficients is basis uniqueness, and no choice-theoretic input is used.

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