Examples

Finite Reflection Groups

1.5 The Simplicial Complex of a Reflection Group

1.5.3 Examples

Example 1.79.Suppose that (W, V) is essential and of rank 2. One could simply give a direct analysis of this situation, but it will be instructive to see what Theorem 1.69 says about it. Let m := |H|. Then m ≥ 2, and the m lines in H divide the plane V into 2m chambers, each of which is a sector determined by two rays. The transitivity of W on the set of sectors implies that they are all congruent, so each sector must have angle 2π/2m= π/m.

In view of assertion (1) of Theorem 1.69, W is generated by two reflections in lines L1 and L2 that intersect at an angle of π/m. In other words,W is dihedral of order2mand(W, V)looks exactly like Example 1.9.

Let us also record, for future reference, the following fact about this ex- ample: Let L₁ and L₂ be the walls of one of the chambers C, and let e_i (i = 1,2) be the unit normal toL_i pointing to the side of L_i containingC;

see Figure 1.10. Then the inner product ofe₁ ande₂ is given by e1, e₂=−cos π

m .

[To understand the sign, note that the angle betweene1 and−e2isπ/m.]

Example 1.80.This is a trivial generalization of the previous example, but it will be useful to have it on record. Assume that (W, V) has rank 2 but is not necessarily essential. In other words, if we write V = V₀⊕V₁ as in Section 1.1, then dimV₁= 2. By the previous example applied to (W, V₁), we have W ∼= D_2m for some m ≥ 2. Moreover, if C₁ is a chamber in V₁ with wallsL_i and normalse_i as above, thenV₀×C₁is a chamber inV with walls V0⊕Liand the same normalsei. In particular, it is still true that a chamberC has two walls and that the corresponding unit normals (pointing toward the side containingC) satisfy

e1, e2=−cos π

m . (1.17)

L1

L₂

e1

e2

C

Fig. 1.10.The canonical unit normals associated with a chamber.

Example 1.81.(Type An−1) Let W be the symmetric group on n letters acting onRⁿ as in Example 1.10. Thus a permutationπacts byπ(vi) =v_π(i) for 1 ≤ i ≤ n, where v1, . . . , vn is the standard basis for Rⁿ. [These basis vectors were called ei in Example 1.10, but we now call them vi in order to avoid confusion with the canonical unit vectors associated to a chamber.] In terms of coordinates, the action is given byπ(x1, . . . , xn) = (y1, . . . , yn) with xi=y_π(i) for 1≤i≤n . (1.18) Indeed, we haveπ_n

i=1xivi

=_n

i=1xiπ(vi) =_n

i=1xiv_π(i), whence (1.18).

To analyze this example we can takeH to be the braid arrangement dis- cussed in Section 1.4.7; this set is clearlyW-invariant, and the corresponding reflections generateW. We already saw in Section 1.4.7 that the chambers are in 1–1 correspondence with elements ofW. The correspondence given there is identical with the one predicted by Theorem 1.69, if we take the fundamental chamber Cto be given by

x1> x2>· · ·> xn. (1.19) To see this, just observe that by (1.18),πC is defined by

x_π(1)> x_π(2)>· · ·> x_π(n). (1.20) The set of inequalities (1.19) is a minimal set of defining inequalities forC, so the latter hasn−1 walls, theith of which is the hyperplaneHi,i+1given by xi=xi+1 (i= 1, . . . , n−1). [Note:n−1 is the “right” number of walls, since this example has rank n−1.] The reflection with respect to the ith wall is the transpositionsi:=si,i+1that interchangesiandi+ 1, so assertion (1) of Theorem 1.69 reduces to the well-known fact that then−1 pairwise adjacent transpositions generate the symmetric group.

We remark in passing that equation (1.15) also reduces to a well-known fact about the symmetric group. Recall first that an inversion of a permutation

1.5 The Simplicial Complex of a Reflection Group 43 π ∈ W is a pair (i, j) with 1 ≤ i < j ≤n and π(i)> π(j). Then the well- known fact is that the lengthl_S(π) is equal to the number of inversions ofπ.

To derive this from (1.15), observe that by (1.20), a wall x_i =x_j withi < j separatesCfromπCif and only ifioccurs later thanj in the list of numbers π(1), . . . , π(n), i.e., if and only ifπ⁻¹(i)> π⁻¹(j). Thus (1.15) says thatl(π) is equal to the number of inversions of π⁻¹; since l(π) =l(π⁻¹), this proves our assertion.

Let’s compute, now, the canonical unit vectors e1, . . . , en−1 associated toC. If we letv1, . . . , vn be the standard basis vectors forV :=Rⁿ as above, then the ith inequality defining C can be written vi−vi+1, x> 0, so the unit vector ei perpendicular to the ith wall and pointing toward the side containingC is given by

ei =vi−√vi+1

2 .

In particular, we can calculate the inner product

ei, ej=

⎧⎪

⎨

⎪⎩

1 forj=i,

−1/2 forj=i+ 1, 0 forj > i+ 1.

Note that 1 =−cos(π/1),−1/2 =−cos(π/3), and 0 =−cos(π/2). Hence the inner product calculation can be written in the more concise form

ei, e_j=−cos π m_ij ,

wherem_ijis the order ofs_is_j(or, equivalently, 2m_ijis the order of the dihedral subgroup generated bys_i ands_j). This formula should not be surprising, in view of (1.17).

Example 1.82.(Type Cn) LetW be the signed permutation group acting on Rⁿ as in Example 1.11. Then Hconsists of the hyperplanes x_i−x_j = 0 (i=j),x_i+x_j= 0 (i=j), andx_i= 0. To describe a chamber, one has to say which coordinates are positive and which are negative, and one has to specify an ordering of the absolute values of the coordinates. It follows that there are 2ⁿn! chambers, each defined byninequalities of the form

1x_π(1)> 2x_π(2)>· · ·> nx_π(n)>0

withi∈ {±1} andπ∈Sn. As fundamental chamber we can take x1> x2>· · ·> xn>0.

The interested reader can work out the fundamental reflections, the canonical unit vectors, and so on, as in Example 1.81. The reader might further want to work out the poset/semigroup of cells, as we did for type A_n−1in Section 1.4.7.

Example 1.83.(Type Dn) LetWbe the subgroup of the signed permutation group consisting of elements that change an even number of signs (Example 1.13). ThenHconsists of the hyperplanesx_i−x_j = 0 andx_i+x_j= 0 (i=j).

To figure out what the chambers look like, consider two coordinates, say x₁ andx₂. From the fact thatx₁is comparable to bothx₂and−x₂on any given chamber C, one can deduce that one of the coordinates is bigger than the other in absolute value and that this coordinate has a constant sign. In other words, we have an inequality of the formx1>|x2|orx2>|x1|onC, where =±1. It follows that there are 2ⁿ⁻¹n! chambers, each defined by inequalities of the form

1xπ(1)> 2xπ(2) >· · ·> _n−1xπ(n−1)>|xπ(n)| (1.21) with _i ∈ {±1} and π ∈ S_n. Note that the last inequality is equivalent to two linear inequalities,_n−1x_π(n−1)> x_π(n)and_n−1x_π(n−1)>−x_π(n), so we have nlinear inequalities in all.

As fundamental chamber we take

x1> x2>· · ·> xn−1>|xn|,

with walls x1 = x2, x2 = x3, . . . , x_n−1 = xn, and x_n−1 = −xn. Further analysis is left to the interested reader.

Example 1.84.This final example is intended to provide some geometric intuition. Several statements will be made without proof, and the reader is advised not to worry too much about this.

Let W be the reflection group of type H3, i.e., the group of symmetries of a regular dodecahedron in V :=R³. It is convenient to restrict the action of W to the unit sphere S² and to think of W as a group of isometries of this sphere. As such, it is the group of symmetries of the regular tessellation of the sphere obtained by radially projecting the faces of the dodecahedron onto the sphere. LetP be one of the 12 spherical pentagons that occur in this tessellation. It has interior angles 2π/3, since there are 3 pentagons at each vertex.

The circles of symmetry of this tessellation (corresponding to the planes of symmetry of the dodecahedron) barycentrically subdivideP, thereby cutting it into 10 spherical triangles. A typical such triangleT has angles π/2,π/3, and π/5. The angle π/5 = 2π/10 occurs at the center of P; the angle π/3, which is half of the interior angle 2π/3 of P, occurs at a vertex of P; and the angle π/2 occurs at the midpoint of an edge ofP, where the line from the center ofP perpendicularly bisects that edge. See Figure 1.11.^∗ Finally, a typical chamber C inV is simply the cone over such a triangleT. There are 12·10 = 120 such chambers, so|W|= 120. Thus the dodecahedral groupW is

∗Figure 1.11 first appeared in Klein–Fricke [145, p. 106] and is reprinted from a digital image provided by the Cornell University Library’s Historic Monograph Collection.

1.5 The Simplicial Complex of a Reflection Group 45

Fig. 1.11.The dodecahedral tessellation, barycentrically subdivided.

a group of order 120 generated by 3 reflections. The calculation of the angles of T above makes it easy to compute the orders of the pairwise products of the generating reflections. One has, for a suitable numberings₁, s₂, s₃of these reflections,

(s₁s₂)³= (s₂s₃)⁵= (s₁s₃)²= 1. Exercises

1.85.Recall from the discussion near the end of Section 1.5.1 that we can distinguish various types of adjacency. Spell out what that means in Examples 1.81, 1.82, and 1.83. (For the A_n−1case, see Example 1.51.)

1.86.Find w₀ and the induced involution of S in Examples 1.81, 1.82, and 1.83, wherew₀ is the element of maximal length (Section 1.5.2).

1.87.In Example 1.84, W is a familiar group of order 120. Which one is it?