Finite Reflection Groups
1.5 The Simplicial Complex of a Reflection Group
1.5.10 Roots and Half-Spaces
1.5 The Simplicial Complex of a Reflection Group 55 Σ∼= (standard cosets)op
that is compatible with theW-action, whereW acts on the cosets by left trans-
lation.
We can express the theorem more briefly by saying thatΣisW-isomorphic to the poset of standard cosets inW, ordered by reverse inclusion.
Exercise 1.112.Let W be the reflection group of type An−1 (symmetric group on nletters), with the standard choice of fundamental chamber. Thus S is the set{s1, . . . , sn−1}of basic transpositions, wheresiinterchangesiand i+ 1. We stated without proof in the discussion of Example 1.10 that the complexΣ associated toW is the barycentric subdivision of the boundary of an (n−1)-simplex. (See also Exercise 1.109.) Prove this rigorously.
Lemma 1.115.Let αbe a positive root with corresponding hyperplane H :=
α⊥. For any w ∈ W, wα is a negative root if and only if H separates C from w−1C.
Proof. LetU be the half-space corresponding toα. Then wαis negative ⇐⇒ wU is negative
⇐⇒ wU does not containC
⇐⇒ U does not containw−1C
⇐⇒ H separatesC fromw−1C.
Let’s specialize to the case thatwis a fundamental reflections∈S. Then the wallHsfixed bysis the only wall that separatesCfromsC. So the lemma in this case says that sαis negative if and only if α⊥ =Hs. This gives the following interpretation of Φas a set with W-action:
Proposition 1.116.There is aW-equivariant bijectionΦ↔ H×{±1}, where the action of a generator sonH × {±1} is given by
(H, )→
(sH, ) if H=Hs, (H,−) if H=Hs.
Propositions 1.113 and 1.116 have a purely group-theoretic consequence, whose significance will become clear in the next chapter:
Corollary 1.117.LetT be the set of conjugates of elements ofS. Then there is an action of W on T× {±1} such that a generators∈S acts by
(t, )→
(sts, ) ift=s, (s,−) ift=s.
Finally, we will describe algebraically the half-spaces corresponding to the fundamental reflections. Here we identify a half-space with the set of chambers it contains, and we use the bijection wC↔wof Theorem 1.69 to relate this to a subset of W. Our task, then, is to describe the set ofw∈W such that wC⊆U+(s), whereU+(s) is the positive half-space bounded byHs.
Proposition 1.118.For alls∈S and w∈ W, wC ⊆U+(s)if and only if l(sw)> l(w).
1.5 The Simplicial Complex of a Reflection Group 57 Proof. We have
wC⊆U+(s) ⇐⇒ Hs does not separateC fromwC
⇐⇒ d(sC, wC)> d(C, wC)
⇐⇒ d(C, swC)> d(C, wC)
⇐⇒ l(sw)> l(w),
where we have used the W-invariance of d(−,−) to write d(sC, wC) =
d(C, swC).
Example 1.119.We illustrate the concepts in this section by applying them to the case that W is the symmetric group onn letters acting on Rn as in Examples 1.10 and 1.81. Recall that a permutationπacts onRnbyπei=eπ(i), wheree1, . . . , en is the standard basis forRn.
(a)Walls and reflections. There is one wallHij for eachunordered pairi, jof integers with 1 ≤i, j ≤n and i=j; it is the hyperplane given by xi =xj. The corresponding reflection is the transpositionsij ∈W that interchangesi andj.
(b)Roots and half-spaces. We can take our root system Φ to be the set of vectorsαij :=ei−ej. Thus there is one root for eachorderedpair of integersi, j with 1≤i, j≤nandi=j; the corresponding half-space is given byxi > xj. Recalling that the fundamental chamberCis defined byx1>· · ·> xn, we see thatαij is a positive root if and only ifi < j. So the bijectionΦ↔ H × {±1}
is given byαij↔(Hij, ), where= +1 ⇐⇒ i < j.
Let’s take the analysis one step further and identify the roots with subsets of W. Here ifα is a root andU is the associated half-space α,−>0, the corresponding subset ofW is{w∈W |wC⊆U}. Thinking of elements ofW as permutationsπ, we claim that
αij ↔
π∈W |π−1(i)< π−1(j)
. (1.25)
The condition π−1(i)< π−1(j) has a concrete interpretation if we represent a permutation π by its list of values π(1)π(2)· · ·π(n) as in Section 1.4.7.
Namely, it says thatioccurs beforej in the list. To prove the claim, we need only recall that πC is the chamber given byxπ(1) >· · ·> xπ(n). Clearly this chamber is contained in the half-space xi > xj if and only ifi precedesj in the list representingπ.
The reader might find it instructive to verify that the sets of permutations corresponding toα14andα41in Figure 1.7 do indeed form a pair of opposite hemispheres.
(c) The action ofW on roots. It is immediate from the definitions thatwαij = αw(i)w(j) for anyi, j and any permutationw ∈W. One can easily verify by direct calculation that this is consistent with the correspondence in (1.25), i.e., that left multiplication by w maps the set on the right side of (1.25) to π∈W |π−1(w(i))< π−1(w(j))
.
(d)The simple roots. Finally, we illustrate Proposition 1.118 in this example.
The fundamental reflections are the transpositions si := si,i+1, where 1 ≤ i≤n−1. The corresponding root isei−ei+1, and the corresponding subset of W, according to (b), is the set of permutations π such that π−1(i) <
π−1(i+ 1). Recall now that the length of an elementw∈W is the number of inversions (see Example 1.81). The interpretation of Proposition 1.118, then, is the following assertion, which one can easily check directly:iprecedesi+ 1 in the list π(1)· · ·π(n) if and only if interchangingi andi+ 1 increases the number of inversions.
Exercises
1.120.Show that every positive root is a nonnegative linear combination of simple roots.
1.121.Show that every root isW-equivalent to a simple root.
1.122.Write down the simple roots for the root systems of type An, Bn, Cn, and Dn, based on the fundamental chambers given in Section 1.5.3.
1.123.Givens∈S, letαsbe the corresponding simple root. For anyw∈W, show that wαs is a positive root if and only ifl(ws)> l(w).
1.124.A restatement of Proposition 1.118, in view of Exercise 1.76, is that wC U+(s) if and only if l(sw) < l(w), i.e., if and only w admits a re- duced decomposition starting withs. Interpret this geometrically in terms of galleries.
1.125.For any w ∈ W, show that l(w) is the number of positive roots α such thatwα is negative. [Note that this proves, again, that the length of a permutation is the number of inversions.] Deduce that the longest element w0 ∈ W (Section 1.5.2) is characterized by the property that it takes every positive root to a negative root.
1.126.Withw0as in the previous exercise, show that the action ofw0on the simple roots αs (s∈S) is given by
w0αs=−ασ0(s),
whereσ0 is the involution ofS introduced at the end of Section 1.5.2.