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A relative arrangement need not be a reflection arrangement

PDF p. 2: §1.1.2; PDF p. 27: §7.1; Figure 1; PDF p. 28: §7.1, paragraph following Figure 1

Statement refuted

The relative arrangement of a Coxeter system need not be a finite real reflection arrangement. Specifically, for the D4 diagram with central vertex 2 and J={2}, the relative arrangement is three-dimensional and irreducible, with seven hyperplanes. An irreducible finite real reflection arrangement in dimension three has six, nine or fifteen hyperplanes, corresponding to types A3, B3 and H3 respectively.

Facts & Assumptions

[F1]

In the specified D4 example, the relative cone is the three-dimensional space E=ΘJ, and in coordinates x(f)=f(α1), y(f)=f(α3) and z(f)=f(α4) its seven distinct relative hyperplanes are the kernels of x, y, z, x+y, x+z, y+z and x+y+z. Seven relative D4 hyperplanes and thirty-two chambers

[F2]

The relative arrangement consists of the distinct nontrivial restrictions of root hyperplanes to ΘJ; the D4 diagram and its root-space notation are as in the specified data. Tits cone intersections, relative roots and arrangements D4 diagram and generator conventions

[F3]

A finite real reflection arrangement is the full arrangement of reflecting hyperplanes of a finite linear reflection group. An arrangement is irreducible if it has no product decomposition over a direct sum of two nonzero spaces, even if one factor has no hyperplanes. A product decomposition E=E1⊕E2 requires every hyperplane to be either H1⊕E2 or E1⊕H2, with Hi a hyperplane of Ei. Linear hyperplane arrangements and local finiteness Irreducible rank-three reflection-arrangement counts

[F4]

An irreducible finite real reflection arrangement in dimension three has 6, 9 or 15 hyperplanes, respectively for types A3, B3 and H3. Irreducible rank-three reflection-arrangement counts

Counterexample

technique · direct

Given: The D4 system with central vertex 2 and J={2}, with its relative arrangement as in [F1] and [F2].

Goal: Prove that this arrangement is irreducible and cannot be a finite real reflection arrangement.

1.1F1F3given

Suppose the arrangement has a product decomposition E=E1⊕E2 with both summands nonzero. The dual decomposes as E∗=E1∗⊕E2∗, where each summand is extended by zero on the other space. By the product condition, a nonzero defining form for each hyperplane belongs to one of these two dual summands: forms with the same kernel are nonzero scalar multiples. In particular x, y and z each lie in one summand. Since x+y also lies in one summand, uniqueness of the two components in the direct sum forces x and y to lie in the same summand. The same argument with x+z puts z there too. But x,y,z form a basis of E∗, so the other dual summand is zero, contrary to both E1 and E2 being nonzero. Thus the arrangement is irreducible.

2.1F1F4step 1.1∎

By [F1] the arrangement has seven hyperplanes in dimension three. If it were a finite real reflection arrangement, its irreducibility from 1.1 would make [F4] applicable, requiring its number of hyperplanes to be 6, 9 or 15. Since 7 is none of these, it is not a finite real reflection arrangement. This example therefore refutes the asserted universal property.

Source notes

The article reference attributes the example and the stated reflection counts. The displayed argument uses the mathematical suppliers and proves irreducibility locally.

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