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A vector with mixed signs is not a root, while every root has a sign

Example

Let S={s,t} with m(s,t)≥3, V=RS, B the Coxeter form and Φ the root system of the canonical reflection representation (The real Coxeter form, its radical, reflections, and form-preserving maps, The canonical reflection homomorphism, roots, reflections, and the positive cone). Then:

(i) the vector v:=es−et has coefficient −1 at et, so v∉V+∪(−V+): it has mixed signs;

(ii) consequently v∉Φ, although it is nonzero and B-non-isotropic, because by Root sign coherence and the action of simple reflections on positive roots (2) every root lies in V+∖{0} or in −V+∖{0}; explicitly B(v,v)=2+2c(s,t)>0 while every root has B-norm 1 (Descent of the reflection representation, unit root norms, and conjugation of reflections (3)), so v fails both the sign test and the norm test;

(iii) for m(s,t)=3 one has B(v,v)=3 and the comparison case es+et has B(es+et,es+et)=1 and lies in Φ+; for m(s,t)=∞ one has B(v,v)=4 and v still fails to be a root.

Facts & Assumptions

Given: a two-element set S={s,t} with m:=m(s,t)≥3, the space V=RS with basis (es,et), the Coxeter form B, the canonical reflection homomorphism ρ, the root system Φ=Φ+⊔Φ− and the positive cone V+.

[F1]

B(es,es)=B(et,et)=1 and B(es,et)=−c with c:=c(s,t) equal to cos⁡(π/m) when m<∞ and to 1 when m=∞; the reflection with normal et is rt(u)=u−2B(u,et)et, so ρ(t)es=rtes=es+2cet; and V+={λes+μet:λ≥0, μ≥0} with −V+={u:−u∈V+} (The real Coxeter form, its radical, reflections, and form-preserving maps, The canonical reflection homomorphism, roots, reflections, and the positive cone).

[F2]

Every root lies in V+∖{0} or in −V+∖{0}, and not in both; every root α satisfies B(α,α)=1; Φ+=Φ∩V+ (Root sign coherence and the action of simple reflections on positive roots, Descent of the reflection representation, unit root norms, and conjugation of reflections).

[F3]

For every x∈W and every generator s one has xC∘⊆Bs or xC∘⊆sBs, and xC∘⊆sBs if and only if ℓ(sx)<ℓ(x); moreover ρ(y)es∈Φ+  ⟺  y−1C∘⊆Bs for the root ρ(y)es (The rank-two half-space alternative and the chamber-length induction (Pn), (Qn), Root sign coherence and the action of simple reflections on positive roots).

[F5]

The addition formulas sin⁡(x+y)=sin⁡xcos⁡y+cos⁡xsin⁡y, cos⁡(x+y)=cos⁡xcos⁡y−sin⁡xsin⁡y hold for all real x,y; sin⁡(−x)=−sin⁡x, cos⁡(−x)=cos⁡x and sin⁡2x+cos⁡2x=1; cos⁡(π/2)=0 and cos⁡π=−1; cosine is strictly decreasing on [0,π]; and π>0 (The addition formulas for sine and cosine, Parity and the Pythagorean identity for sine and cosine, Quarter-turn values and shifts by pi/2 and pi, Signs, monotonicity intervals, and ranges of sine and cosine, Pi as twice the smallest positive zero of cosine).

[F6]

Every x∈V=R{s,t} equals x(s)es+x(t)et; evaluating at s and t shows that these coordinates are unique (Reflections: involutivity, form invariance, fixed hyperplane, and exact rank-two order (1)).

Verification

technique · direct
1.1F1F5algebra

Set-up. By [F1] the vector v=es−et has coordinates v(s)=1 and v(t)=−1 in the basis (es,et), and B(v,v)=B(es,es)−2B(es,et)+B(et,et)=2+2c. The constant c is positive: c=1>0 for m=∞, while for finite m≥3 one has 0<π/m≤π/3<π/2 and cosine is strictly decreasing on [0,π] with cos⁡(π/2)=0 by [F5], so c=cos⁡(π/m)>cos⁡(π/2)=0. Hence B(v,v)>2>0, so v≠0 and v is B-non-isotropic.

1.2F1F6algebra

The vector has mixed signs. By [F6] the coordinates of a vector in the basis (es,et) are unique, so u∈V+ exactly when both coordinates of u are ≥0 and u∈−V+ exactly when both coordinates are ≤0. Since v has the coordinates 1 and −1, it lies in neither cone: v∉V+∪(−V+). This is (i).

2.1F1F5step 1.1algebra

The value c=cos⁡(π/3). For x with cos⁡x=:q and sin⁡x=:r, the addition formulas and sin⁡2x+cos⁡2x=1 of [F5] give cos⁡(2x)=q2−r2=2q2−1 and sin⁡(2x)=2qr, hence cos⁡(3x)=cos⁡(2x)cos⁡x−sin⁡(2x)sin⁡x=(2q2−1)q−2qr2=2q3−q−2q(1−q2)=4q3−3q. Applied to x=π/3 and combined with cos⁡π=−1 this gives 4c3−3c+1=0, that is (c+1)(2c−1)2=0. Since c>0 by 1.1, the factor c+1 does not vanish, so (2c−1)2=0 and c=1/2. Consequently 2c=1, B(v,v)=2+1=3, and ρ(t)es=es+2cet=es+et; moreover B(es+et,es+et)=1+1−2c=2−1=1.

2.2F2step 1.1step 1.2

v is not a root. By [F2] every root lies in V+∖{0} or in −V+∖{0}, and every root has B-norm 1. By 1.2 the vector v lies in neither cone, so it is not a root; independently, by 1.1 its B-norm is 2+2c>2≠1, so it also fails the norm test for roots. This is (ii).

3.1F3F4step 2.1

The comparison vector es+et is positive. By [F3] applied to x=t either tC∘⊆Bs or tC∘⊆sBs. In the second case ℓ(st)<ℓ(t)=1 by [F3], so ℓ(st)=0 and hence st=1 by [F4], which gives s=t−1=t and contradicts s≠t; therefore tC∘⊆Bs, and the equivalence of [F3] with y=t gives ρ(t)es∈Φ+. By 2.1, ρ(t)es=es+et when m=3, so es+et∈Φ+ with B(es+et,es+et)=1.

4.1step 2.1step 2.2step 3.1F1∎

The cases m=3 and m=∞. If m=3 then c=1/2 by 2.1, so B(v,v)=3, and es+et=ρ(t)es is a positive root of B-norm 1 by 3.1. If m=∞ then c=1 by [F1], so B(v,v)=4, and v∉Φ by 2.2. With (i) from 1.2 and (ii) from 2.2, all three clauses are verified: the sign theorem applies to the W-orbit of the simple roots, not to arbitrary vectors of V.

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