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Ordered roots and the mu-dot-root matrix in A3

Example

Work in the standard realization of A3=S4. Let E1,…,E4 be the coordinate vectors, put V0:={x∈R4:∑ixi=0}, and use B(x,y):=12∑ixiyi. Put σ1:=E1−E2, σ2:=E3−E4, σ3:=E2−E3 and let s1=(1 2), s2=(3 4), s3=(2 3) be their reflections. Thus J={s1,s2}, K={s3}, and c=s1s2s3. For i<j, write (i j) for the positive root Ei−Ej. Multiplication of permutations is right to left. Let ℓ be Coxeter word length and ℓT reflection length. Then:

(i) Ordered roots. One has c=(1 2 4 3), h=4, and ρ1=(1 2),ρ2=(3 4),ρ3=(1 4),ρ4=(2 4),ρ5=(1 3),ρ6=(2 3). In the simple-root basis these are σ1,σ2,σ1+σ2+σ3,σ2+σ3,σ1+σ3,σ3. Hence Φ+={ρ1,…,ρ6}, ∣Φ+∣=ℓ(w0)=nh/2=6, and ρi+3=CVρi for i≥1, where CV=ρ(c).

(ii) The μ-root matrix. The matrix (B(μi,ρj))1≤i,j≤6 is (101010011100001111−1001010−10011−1−1−1001). Its diagonal and upper-triangular entries are nonnegative, its strictly lower-triangular entries are nonpositive, and the first two subdiagonals vanish, as in The mu-dot-root identities, the cone separation, and the canonical simple systems of the subintervals [1, sigma] (2)(b)-(d).

(iii) The longest word and the second half. The longest element is w0=c2=(1 4)(2 3) and (s1s2s3)2 is a reduced expression of length 6=nh/2, with prefix roots ρ1,…,ρ6. The prefix roots of (s1s2s3)4 are ρ1,…,ρ12, and ρ7,…,ρ12=−ρ2,−ρ1,−ρ3,−ρ5,−ρ4,−ρ6, so the second half is −Φ+.

(iv) One factorization-criterion instance. For the pair (ρ1,ρ2), ℓT(R(ρ1)R(ρ2)c)=1=n−2⟺B(μ(ρ2),ρ1)=0.

No Choice is used.

Facts & Assumptions

Given: The type-A3 coordinate model and the bipartite data just specified, together with the conventions of the cited root and reflection items.

[F1]

The Coxeter form and reflection formula are those of The real Coxeter form, its radical, reflections, and form-preserving maps; the canonical representation preserves the form and identifies root reflections with their orthogonal actions by Descent of the reflection representation, unit root norms, and conjugation of reflections. The ordered simple generators (s1,s3,s2) are the standard type-A3 chain, so the group is S4 and Coxeter word length is permutation inversion number by Support, intrinsic parabolic presentations, minimal coset representatives and length additivity, with the type-A identification (4).

[F2]

The dual vectors βi, prefix-root and dual-vector recursions, and ρi+3=CVρi, μi+3=CVμi are as in The bipartite Coxeter element, its ordered prefix roots, and the conditional vector map mu(a) = -2(c-1)^{-1}a (2)-(4). For this irreducible rank-three system with h=4, The Coxeter plane, ordered-root enumeration, and invertibility of rho(c) - id (3)-(4) gives Φ+={ρ1,…,ρ6} and the longest-word formula.

[F3]

For each positive root a, μ(a)=μi when a=ρi, B(μi,ρi)=1, and the sign and zero-pairing rules in The mu-dot-root identities, the cone separation, and the canonical simple systems of the subintervals [1, sigma] (1)-(2) apply.

[F4]

Reflection length ℓT(w) is the minimum number of reflections in T whose product is w; the empty product represents the identity. Reflection length, the absolute order on a finite Coxeter group, and the moved and fixed spaces of an orthogonal operator (1)

[F5]

For an increasing tuple a1<⋯<ak, the length equality ℓT(R(a1)⋯R(ak)c)=n−k is equivalent to B(μ(ai),aj)=0 for every i>j. The factorization criterion, linear independence of the faces, and the geometric simplicial structure of X(sigma) (1)

Verification

technique · compute in the simple-root basis, with the standard coordinate action fixing all signs and permutation products
1.1F1

The Gram matrix of (σ1,σ2,σ3) is G=(10−1/201−1/2−1/2−1/21), with determinant 1/2>0; B is positive definite as it is half the Euclidean form on V0. Thus the map from the abstract simple-root basis to (σ1,σ2,σ3) is an isometry. In this basis CV=(01−110−111−1). For any i<j, the reflection with normal Ei−Ej sends x to x−(xi−xj)(Ei−Ej) and swaps coordinates i,j. Thus the generators act as the stated transpositions, c=(1 2 4 3), and its order is 4.

2.1F1F2step 1.1

The prefix convention gives ρ1=σ1, ρ2=σ2, and ρ3=R(σ1)R(σ2)σ3=E1−E4. Applying c to these roots gives E2−E4, E1−E3, E2−E3, respectively, which are ρ4,ρ5,ρ6. The roots of this coordinate realization are exactly ±(Ei−Ej); the six displayed vectors are exactly the ones with i<j, and their listed simple-root coordinates are nonnegative. This verifies the order, positivity, and full positive-root set in (i), while the cyclic recursion gives ρi+3=CVρi for all i≥1.

3.1F2F3step 2.1

Inverting G gives G−1=(3/21/211/23/21112), so β1=(3/2,1/2,1), β2=(1/2,3/2,1), and β3=(1,1,2) in the simple-root basis. Since B(β2,σ1)=0 and B(β3,σ1)=B(β3,σ2)=0, the prefix formula gives μ1=β1, μ2=β2, and μ3=β3; the cyclic recursion gives μ4=(−1/2,1/2,1), μ5=(1/2,−1/2,1), and μ6=(−1,−1,0). Taking B(μi,ρj)=μiTGρj yields exactly the displayed matrix. Reading its entries proves the diagonal, sign, and two-subdiagonal zero assertions.

3.2F1F2step 2.1

Since c=(1 2 4 3), c2=(1 4)(2 3) is the reverse permutation (4,3,2,1) and has six inversions, the maximum possible in S4. Thus w0=c2 and ℓ(w0)=6. The six-letter word c2=(s1s2s3)2 is therefore reduced. Applying CV2 to ρ1,…,ρ6 gives −ρ2,−ρ1,−ρ3,−ρ5,−ρ4,−ρ6; the period-three prefix recursion then gives the twelve prefix roots of c4, with its second half exactly −Φ+.

4.1F4F5step 3.1∎

The matrix in step 3.1 has B(μ2,ρ1)=0. Also, using R(ρ1)=(1 2) and R(ρ2)=(3 4), R(ρ1)R(ρ2)c=(1 2)(3 4)(1 2)(3 4)(2 3)=(2 3), a nonidentity reflection. Hence its reflection length is 1=n−2, so both sides of the stated equivalence hold for this pair.

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