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Irreducible reflection arrangements give connected Coxeter diagrams

Statement

Suppose $A (W)$ is an irreducible finite real reflection arrangement on $E$ , with $\dim E > 1$ , in the sense of Invariant metric, product arrangements and essentiality. Its chamber-wall Coxeter system from Chamber walls of an essential finite reflection group give full Coxeter rank has connected Coxeter diagram in the convention of Coxeter-system, roots, length and weak-order conventions. In particular, if $\dim E = 3$ , this system has connected rank three.

Facts & Assumptions

[F1]

The wall normals form a basis, and $S$ generates $W$ (Chamber walls of an essential finite reflection group give full Coxeter rank).

[F2]

An absent diagram edge means $m (s, t) = 2$ , hence $s$ and $t$ commute (Coxeter-system, roots, length and weak-order conventions).

[F3]

Reflections are orthogonal and uniquely determined by their hyperplanes; a nontrivial product decomposition contradicts irreducibility (Invariant metric, product arrangements and essentiality).

Proof

Given: The irreducible arrangement, invariant metric and essentiality of Invariant metric, product arrangements and essentiality, the wall-normal basis and Coxeter generators from Chamber walls of an essential finite reflection group give full Coxeter rank, and the diagram convention of Coxeter-system, roots, length and weak-order conventions. Goal: Exclude a disconnected diagram.

Proof technique: direct.

1.1F1F2F3algebra

If the diagram is disconnected, split its vertices as nonempty unions of components $S_{1} ⊔ S_{2}$ . For $s \in S_{1}$ , $t \in S_{2}$ , [F2] gives $s t = t s$ . The $- 1$ eigenline of $s$ is therefore $t$ -invariant. An orthogonal reflection has invariant lines either in its hyperplane or equal to its normal line: its restriction to an invariant line is $+ 1$ or $- 1$ . The normal lines of distinct wall reflections cannot coincide, by [F3]. Their normals are consequently perpendicular. Thus the spans $E_{1}, E_{2}$ of the two sets of wall normals are orthogonal and, by [F1], $E = E_{1} \oplus E_{2}$ with both factors nonzero.

2.1step 1.1F1F3algebra∎

The generators in $S_{i}$ act trivially on the other factor, so $W = W_{1} \times W_{2}$ on this orthogonal direct sum, where $W_{i}$ is generated by $S_{i}$ . For $(w_{1}, w_{2})$ the codimension of the fixed subspace is the sum of the codimensions for $w_{i}$ . If it is a reflection, exactly one summand is one and the other zero; faithfulness of each factor's action makes the zero-codimension factor the identity. Hence every reflection hyperplane is $H_{1} \oplus E_{2}$ or $E_{1} \oplus H_{2}$ , and all such factor reflection hyperplanes occur. This is a product arrangement, contradicting [F3]. The diagram is connected. When $\dim E = 3$ , [F1] also gives $∣ S ∣ = 3$ .

Depends on

Used by · 1

Irreducible rank-three reflection-arrangement counts

Dependency tree · 2 levels

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