Root systems [expository notes]

11 Pages·2012·0.319 MB·English

by  John Dusel

Root systems
John Dusel
Contents
1. Axiomatics 2
1.1. Reflections in euclidean space 2
1.2. Root systems 2
1.3. Examples 4
1.4. Pairs of roots 4
2. Simple roots and the Weyl group 6
2.1. Bases and Weyl chambers 6
2.2. Lemmata on simple roots 7
2.3. The Weyl group 9
2.4. Irreducible root systems 10
References 11
1
This follows Chapter III of Humphreys [Hum72], with some additions from [Gre11].
1. Axiomatics
1.1. Reflections in euclidean space.
(cid:0) (cid:1)
1.1.1. Let E,(·,·) beaafinite-dimensionalrealvectorspacewithasymmetricpositive
definite bilinear form. Fix nonzero vectors α,β ∈ E. We can reflect β across the hyperplane
H of vectors in E orthogonal to α (figure) via s ∈ O(E). Write
α α
(cid:18) (cid:19)
(β,α) (β,α)
β = β − α + α
(α,α) (α,α)
and use s (α) = −α, s | = 1 to get
α α Hα
2(β,α)
s (β) = β − α.
α
(α,α)
Theseprojectioncoefficientsmakefrequentappearancesandaredenoted(cid:104)β,α(cid:105) = 2(β,α), soin
(α,α)
particular s (β) = β−(cid:104)β,α(cid:105)α. Because of the setup with a fraction, the map (cid:104)·,α(cid:105) : E → C
α
is linear but the maps (cid:104)·,·(cid:105) : E2 → C and (cid:104)β,·(cid:105) : E → C are non-linear.
1.1.2. Since s is just a (special kind of) reflection, it is a member of O(E). There is a
α
criterion for σ ∈ O(E) to be recognized as σ = s where α belongs to a suitable subset of E.
α
Lemma. Let Φ ⊂ E be a finite spanning set such that all s with α ∈ Φ leave Φ invariant.
α
If σ ∈ O(E)
(1) leaves Φ invariant,
(2) fixes pointwise some hyperplane P ⊂ E,
(3) sends some nonzero α ∈ Φ to its negative −α
then σ = s and P = H accordingly.
α α
1.2. Root systems.
1.2.1. We are interested in subsets Φ ⊂ E with the following properties (lifted from the
Lie theory):
(R1) Φ is a finite spanning set of E that does not contain 0.
(R2) For all α ∈ Φ, the only multiples of α that belong to Φ are ±α.
(R3) For all α ∈ Φ, s leaves Φ invariant.
α
(R4) For all α,β ∈ Φ, (cid:104)β,α(cid:105) is integral.
Such Φ are known as root systems in E. The important applications of the proposition
above occur when Φ is a root system.
1.2.2. Since we are really only interested in root systems, and each vector α in a root
system determines a reflection s ∈ O(E) it makes sense to consider the reflections {s : α ∈
α α
Φ}. The Weyl group W of a root system Φ in a Euclidean space E is the subgroup of O(E)
generated by {s : α ∈ Φ}. Denote Φ = {α ,··· ,α } and let W be its Weyl group. Define
α 1 r
a group homomorphism ι : W → S by extending s (cid:55)→ ι(s ), where ι(s )(j) = j(cid:48) if and
r αi αi αi
only if s (α ) = α . Observe σ ∈ kerι if and only if ι(σ) is the identity permutation. But
αi j j(cid:48)
then by definition σ(α ) = α for all j = 1,··· ,r, which means σ is the identity of O(E).
j j
Thus W identifies with a subgroup of S , meaning in particular W is a finite group. We can
r
thus regard W in two ways, W < O(E) as reflections or W < S as permutations of Φ.
r
2
1.2.3. There will be occasion to consider elements of GL(E) that aren’t members of the
Weyl group. The need to at least leave Φ invariant or there is no point. Conjugating the
Weyl group by such an element has a predictable effect:
Lemma. For a root system Φ ⊂ E with Weyl group W, if τ ∈ GL(E) leaves Φ invariant
then
τs τ−1 = s
α τ(α)
for all α ∈ Φ and
(cid:104)β,α(cid:105) = (cid:104)τ(β),τ(α)(cid:105)
for all α,β ∈ Φ.
1.2.4. Toreallyappreciatethislemmawemustelevatethediscourse. Givenrootsystems
(cid:0) (cid:1) (cid:0) (cid:1)
Φ ⊂ E,(·,·) and Φ(cid:48) ⊂ E(cid:48),(·,·)(cid:48) define a morphism of root systems to be a linear
transformation T : E → E(cid:48) (not necessarily an isometry) satisfying
(1) T(Φ) ⊂ Φ(cid:48),
(2) For all roots α,β ∈ Φ, (cid:104)T(β),T(α)(cid:105) = (cid:104)β,α(cid:105).
Thus we have the category of root systems. With this setup the Weyl group of Φ also
acts on Φ(cid:48). Indeed there is a map W → W(cid:48) produced by extending s (cid:55)→ s . In a precise
α T(α)
sense, T intertwines these actions:
s (T(β)) = T(β)−(cid:104)T(β),T(α)(cid:105)T(α)
T(α)
(cid:0) (cid:1)
= T β −(cid:104)T(β),T(α)(cid:105)α
(cid:0) (cid:1)
= T β −(cid:104)β,α(cid:105)α
(cid:0) (cid:1)
= T s (β) .
α
There is, of course, a notion of isomorphism of root systems Φ ∼= Φ(cid:48) entailing a vector
space isomorphism (again, not necessarily an isometry) T; in this case T(Φ) = Φ(cid:48). What
happens with the s for α ∈ Φ? The natural diagram to consider is
α
T (cid:47)(cid:47)
E E(cid:48)
sα sT(α)
(cid:15)(cid:15) (cid:15)(cid:15)
(cid:47)(cid:47)
E E(cid:48)
T
which indicates we need Ts = s T. Well, given β ∈ Φ ⊂ E we can map in two ways:
α T(α)
β (cid:55)−T→ T(β) s(cid:55)−T→(α) T(β)−(cid:104)T(β),T(α)(cid:105)T(α)
β (cid:55)−s→α β −(cid:104)β,α(cid:105)α (cid:55)−T→ T(β)−(cid:104)β,α(cid:105)T(α),
and since (cid:104)T(β),T(α)(cid:105) = (cid:104)β,α(cid:105), equality holds. Accordingly there is an isomorphism of
Weyl groups W → W(cid:48) given by extending s (cid:55)→ Ts T−1 = s . This is just another way of
α α T(α)
expressing the “intertwining” of the Weyl groups’ actions, this time in the case of isomorphic
root systems.
Specializing further, an automorphism of Φ is an isomorphism from Φ with itself. In
other words, an automorphism of Φ is a T ∈ O(E) that leaves Φ invariant. The set AutΦ of
automorphism of Φ is a group under composition, having the Weyl group of Φ as a subgroup.
The content of our Lemma is that W is a normal subgroup of AutΦ.
3
1.2.5. An natural example of a root system to which Φ is isomorphic: The dual root
system of Φ is the set of re-scaled roots
(cid:26) (cid:27)
2α
Φ= α∨ = : α ∈ Φ .
(α,α)
It’s not automatic that Φ∨ is a root system, but it turns out to be true. Note that (α,α∨) =
(α,2(α,α)−1α) = 2 for all α ∈ Φ, and also that (β,α∨) = (cid:104)β,α(cid:105) for all β ∈ Φ. The root
systems Φ and Φ∨ are canonically isomorphic under φ : E → E(cid:48) with φ(v) = 2(v,v)−1v,
which implies also that W ∼= W∨.
1.3. Examples. See [Hum72] §9.3. These make more sense after studying §9.4 and I
don’t want to draw them with a computer.
1.4. Pairs of roots.
1.4.1. Two nonzero vectors α,β ∈ E have an angle θ between them according to the
formula (α,β) = (cid:107)α(cid:107)(cid:107)β(cid:107)cosθ. Let’s see what can be said about θ. Axiom (R4) im-
plies that (cid:104)β,α(cid:105) = 2(β,α) is integral, but (cid:104)β,α(cid:105) = 2(cid:107)β(cid:107)(cid:107)α(cid:107)−1cosθ and, similarly, (cid:104)α,β(cid:105) =
(α,α)
2(cid:107)α(cid:107)(cid:107)β(cid:107)−1cosθ. Now
(cid:104)β,α(cid:105)(cid:104)α,β(cid:105) = 4cos2θ ∈ {0,1,2,3,4},
sincesgn(cid:104)β,α(cid:105) = sgn(cid:104)α,β(cid:105). (Thelaterfollowsfromthebilinearlyandpositive-definitenessof
(·,·)). Evidently cos2θ = 1 iff θ ∈ {0,π/2} iff α = ±β and cos2θ = 0 iff θ = π/2. Excluding
these possibilities we can have (cid:104)β,α(cid:105)(cid:104)α,β(cid:105) ∈ {1,2,3}. For the sake of concreteness take
(cid:107)α(cid:107) ≤ (cid:107)β(cid:107).
Case # (cid:104)β,α(cid:105)(cid:104)α,β(cid:105) (cid:104)α,β(cid:105) (cid:104)β,α(cid:105) θ (cid:107)β(cid:107)2/(cid:107)α(cid:107)2
1 1 1 1 π/3 1
2 1 -1 -1 2π/3 1
3 2 1 2 π/4 2
4 2 -1 -2 3π/4 2
5 3 1 3 π/6 3
6 3 -1 -3 5π/6 3
To fill in the table, first observe that (cid:104)α,β(cid:105) ≤ (cid:104)β,α(cid:105) since (cid:107)α(cid:107) ≤ (cid:107)β(cid:107). (Square and look at
reciprocals.)
In case 1 we have
2(cid:107)β(cid:107)(cid:107)α(cid:107)−1cosθ = 1 = 2(cid:107)α(cid:107)(cid:107)β(cid:107)−1cosθ
which implies (cid:107)β(cid:107)2(cid:107)α(cid:107)−2 = 1 and cosθ = 1/2, so θ = π/3. Similarly, in case 2 we have
(cid:107)β(cid:107)2(cid:107)α(cid:107)−2 = 1 and cosθ = −1/2 so that θ = 2π/3.
In case 3 we have
2(cid:107)α(cid:107)(cid:107)β(cid:107)−1cosθ = (cid:107)β(cid:107)(cid:107)α(cid:107)−1cosθ,
√
which implies (cid:107)β(cid:107)2(cid:107)α(cid:107)−2 = 2 and cosθ = 1/ 2, so θ = π/4. Similarly in case 4 we have
√
(cid:107)β(cid:107)2(cid:107)α(cid:107)−2 = 2 and cosθ = −1/ 2 so that θ = 3π/4.
The other two cases are similar. The interested reader can fill them in for exercise.
What’s interesting to note here is how everything follows from (R4).
4
1.4.2. Given α,β ∈ E it is natural to ask if any linear combination of α,β belongs to
Φ. Axiom (R2) sort of says we need only consider α±b since
−α+β = −(α−β) ∈ Φ iff α−β ∈ Φ,
−α−β = −(α+β) ∈ Φ iff α+β ∈ Φ.
So if α,β ∈ Φ we would like sufficient conditions to know either of α ±β ∈ Φ. Of course,
(R2) also says we should only consider non-proportional α,β.
Lemma. If α,β ∈ Φ are non-proportional then
(1) If the angle between α and β is strictly acute, (α,β) > 0, then α−β ∈ Φ.
(2) If the angle between α and β is strictly obtuse, (α,β) < 0, then α+β ∈ Φ.
No information is given when α ⊥ β. In the rank 2 system A ×A we have α ⊥ β and
1 1
neither α±β ∈ Φ (figure). Whereas in B both of α±γ ∈ Φ (figure).
2
1.4.3. Now let α,β ∈ Φ be non-proportional and consider the α-string through β:
{β +iα : i ∈ Z}. This is a finite set so there exist natural numbers r,q such that
r = max{j ∈ N : b−jα ∈ Φ}
q = max{j ∈ N : β +jα ∈ Φ}.
These are like the beginning and end of the α-string through β, and a natural question to
ask is whether or not this string is “unbroken”, meaning {β +tα : −r ≤ t ≤ q} ⊂ Φ.
If there is a break in the string, say β + tα ∈/ Φ for −r < t < q, then we can find p,s
with −r < p < s < q satisfying
(1.1) β +pα ∈ Φ and β +(p+1)α ∈/ Φ
(1.2) β +(s−1)α ∈/ Φ and β +sα ∈ Φ.
To find p ask if (1.1) holds for p = t − 1. If so then OK, if not ask if (1.1) holds for
p = t − 2. Since β − (r − 1)α ∈ Φ this process halts at p = −(r − 1) at the latest. Why
does β −(r −1)α ∈ Φ? Well β −(r −1)α = β −rα+α ∈ Φ if (β −rα,r) < 0 by Lemma
1.4. But by the definition of r, β − (r − 1)α = β − rα −α ∈/ Φ implies (α,β) ≤ 0 and so
(β −rα,α) = (β,α)−r(α,α) < 0 which implies β −(r−1)α ∈ Φ. Similarly we can find s.
But then 1.1 implies (β+pα,α) ≥ 0 and 1.2 implies (β+sα,α) ≤ 0 by the contrapositive
of Lemma 1.4. And now: since (·,·) is positive definite we have
(β,α)+s(α,α) ≤ 0 ≤ (β,α)+p(α,α)
whence s ≤ p. No. So the α-string through β is “unbroken”.
How long can this string be? Observe s (β) = β − (cid:104)β,α(cid:105)α and for −r < t < q we get
α
s (β) = β−((cid:104)β,α(cid:105)−t)α. According to Humphreys it is geometrically clear that s reverses
α α
the α-string through β. Taking this statement as face value, we see that
s (β −rα) = β −((cid:104)β,α(cid:105)+r)α = β +qα
α
s (β +qα) = β −((cid:104)β,α(cid:105)−q)α = β −rα.
α
It follows that r = (cid:104)β,α(cid:105) − q and, accordingly, the length of the α-string through β is
r+q = (cid:104)β,α(cid:105) ≤ 4.
5
2. Simple roots and the Weyl group
2.1. Bases and Weyl chambers. Let Φ ⊂ E be a rank r root system with Weyl group
W. This is in particular a spanning set, and so contains a basis of E. Not just any basis of
E will do, though; for our purposes we need to work with integral combinations of roots.
2.1.1. A subset Π ⊂ Φ is a base provided that
(B1) Π is a basis of E.
(cid:80)
(B2) Each root β can be written as β = k α with all α ∈ Π, each k integral with
α∈Π α α
either all k non-negative or all k non-positive.
α α
Roots α ∈ Π are called simple, of course there are r of these.
(cid:80) (cid:80)
The height of a root β = k α ∈ Φ is the integer htβ = k . When k ≥ 0
α∈Π α α∈Π α α
(respectivelyk ≤ 0)forallα ∈ Πtherootβ iscalledpositive(respectivelynegative). The
α
sets of positive, respectively negative roots of Φ is denoted Φ+,Φ−; certainly Φ− = −Φ+.
Trivial observations: Π ⊂ Φ+; when the sum α + β of two positive roots is a root, it is
positive.
A base Π ⊂ Φ ⊂ E defines a partial order of E compatible with the notion of positivity.
Given µ,λ ∈ E define µ ≺ λ to mean either λ = µ or λ−µ is a sum of positive (simple)
roots.
2.1.2. Recall the highly restricted nature of the possible angles between roots. What
angles can occur between simple roots? Not surprisingly, the answer to this question follows
from (B2).
Lemma. If Π is a base of Φ then (α,β) ≤ 0 for distinct α,β ∈ Φ and α−β is not a root.
Proof. By way of contradiction suppose (α,β) > 0. Since α,β are distinct, Lemma 1.4
says α−β ∈ Φ. But then we have a root that is not expressible in the manner indicated by
(B2), meaning Π is not a base.. (cid:3)
2.1.3. The definition of a base Π ⊂ Φ does not stipulate existence, although every root
system does in fact admit a base. The proof of this fact is deferred, but we’ll need a couple
of definitions that originate from it.
Given a vector γ ∈ E define the set of all roots lying on the “positive” side of the
hyperplane orthogonal to γ (figure)
Φ+(γ) = {α ∈ Φ : (γ,α) > 0}.
(When the angle between α and γ is strictly acute, α is said to be on the positive side of
(cid:83)
said hyperplane.) It is a fact from geometry that the finite union of hyperplanes H
α∈Π α
does not coincide with E. The compliment E (cid:114)(cid:83) H is the set of regular elements of
α∈Π α
E. When γ ∈ E is regular, by virtue of the setup we have Φ = Φ+(γ) ∪ Φ−(γ). A root
α ∈ Φ+(γ) is decomposable when it can be expressed as a sum α = β +β for some roots
1 2
β ∈ Φ+(γ), otherwise α is indecomposable. It turns out that
i
Theorem. For any regular γ ∈ E the set Π(γ) of all indecomposable roots in Φ+(γ) is a
base of Φ. Every base of Φ is of the form Π(γ) for some regular γ ∈ E.
Accordingly we’d like to better understand regular elements of E. The (finitely many)
connected components of E (cid:114)(cid:83) H are called the (open) Weyl chambers of E. These
α∈Π α
are the places where regular elements live, since each regular element γ of E belongs to
exactly one Weyl chamber C(γ) of E. One way of interpreting the statement C(γ) = C(γ(cid:48)) is
6
γ and γ(cid:48) lie on the same side of all H , α ∈ Φ. In particular Φ+(γ) = Φ+(γ(cid:48)) because for all
α
α ∈ Φ, (γ,α) > 0 if and only if (γ(cid:48),α) > 0. In particular particular Π(γ) = Π(γ(cid:48)), meaning
bases of Φ are in one-to-one correspondence with Weyl chambers.
For a base Π = Π(γ) of Φ one writes C(Π) = C(γ) and calls it the fundamental Weyl
chamber relative to Π. By definition C(Π) is the open set comprising all γ ∈ E for which
(γ,α) > 0 for all α ∈ Π.
Example. The root system A has E = R2 and base Π = {α,β}. Draw the figure. To find
2
Weylchamberslookatthehyperplanesorthogonaltotherootsα,β,α+β. Putthemtogether
to get a nice picture. To be in C(Π) a regular vector γ must satisfy (γ,α) > 0,(γ,β) > 0,
meaning γ ∈ Φ+(α)∩Φ+(β). Of course one such vector is α +β, and so the fundamental
Weyl chamber relative to Π can be expressed as C(Π) = C(α+β).
There are six Weyl chambers in total, each being fundamental for one of the six possible
bases of A . Label them on the figure.
2
2.1.4. Let’s talk about the Weyl group. It is geometrically obvious that an element of
W permutes the Weyl chambers. Indeed, if α ∈ Φ then s (C(γ)) = C(s (γ)). In particular if
α α
α ∈ C(γ) then s (C(γ)) = C(γ). Given a base Π ⊂ Φ and any generator s of the Weyl group
α α
it is clear that s (Π) is a basis of E. Now since for all β ∈ E there exist k ∈ Z ((cid:15) ∈ Π) all
α (cid:15)
(cid:80) (cid:80)
non-negative xor all non-positive such that β = k (cid:15) it follows that s (β) = k s ((cid:15)).
(cid:15) (cid:15) α (cid:15) (cid:15) α
Since s (β) runs through all Φ as β does it follows that s (Π) is a base of Φ. Accordingly,
α α
the Weyl group permutes the bases of Φ.
These actions of the Weyl group on Weyl chambers and bases are compatible with the
correspondence between bases and Weyl chambers: given Π corresponding to C(Π) = C(γ)
wehaveσ(Π)correspondingtoC(σ(Π)) = C(σ(γ)). (Thisfollowsfrom(σ(α),σ(β)) = (α,β))
for all α,β ∈ Φ.
2.2. Lemmata on simple roots. Select and fix a base Π of a root system Φ ⊂ E as
usual.
2.2.1. The first Lemma sez that given a positive root there is always a way to produce
from it another positive root.
Lemma. If α ∈ Φ is positive and not simple then α−β is a positive root for some simple
root β ∈ Π.
Corollary. Each positive root β can be expressed as β = α + ··· + α with α ∈ Π not
1 n i
necessarily distinct in such a way that each partial sum α +···+α is a root (1 ≤ j < n).
1 j
Proof. Induct on htβ. If htβ = 1 then β = α is a simple root and the result is true, so
(cid:80)
induction begins. Assume the result for height n and suppose htβ = n+1: β = k α
α∈Π α
(cid:80)
and k = n + 1. Certainly β is positive and not simple, so there exists α ∈ Π such
α α 0
that β − α is a positive root (previous Lemma). Now ht(β − α ) = n and by induction
0 0
β − α = α + ··· + α for α ∈ Π not necessarily distinct such that each partial sum is
0 1 n i
a root. But then β = α + ··· + α + α , and each partial sum is a root by virtue of the
1 n 0
setup. (cid:3)
2.2.2. The next result extends the discussion of the Weyl group from Bases and Weyl
chambers. Rather than describing the action of (the generators of) the Weyl group on a
base, we can instead look at its action on the set of positive roots.
7
Lemma. If α is a simple root then s permutes the positive roots other than α.
α
Proof. Given a positive root β different from α, we claim 0 ≺ s (β). Certainly in
α
(cid:80)
the expression β = k γ (all k ≥ 0) for some γ different from α the coefficient k is
γ∈Π γ γ γ
positive. By definition
(cid:88)
s (β) = k (γ −(cid:104)β,α(cid:105)α)
α γ
γ
so the coefficient of γ in s (β) is still k . Now axiom (B2) forces all the coefficients of s (β)
α γ α
to be positive. (cid:3)
Accordingly there is the following special element which is almost fixed by s for simple
α
α.
Corollary. For δ = 1 (cid:80) β the formula s (δ) = δ −α holds for all simple roots α.
2 0≺β α
2.2.3. An arbitrary element of the Weyl group is the product of finitely many generators
(since each generator has order 2). The next result is a criterion indicating when such an
expression may be shortened.
Lemma. Let α ,··· ,α be not necessarily distinct simple roots. For brevity write s for s .
1 t i αi
If the root s ···s (α ) is negative then for some index 1 ≤ s < t
1 t−1 t
s ···s = s ···s s ···s .
1 t 1 s−1 s+1 t−1
Proof. Let β = α and β = s ···s (α ) for each j = 0,··· ,t−2. Observe that
t−1 t j j+1 t−1 t
β is positive, β is negative (hypothesis), and s (β ) = β .
t−1 0 j j j−1
Let s be the minimal index such that β is positive (so 1 ≤ s ≤ t − 2). By design
j
s β = β is negative. Since s permutes Φ+ (cid:114){α } and maps α → −α it must be the
s s s−1 s s s s
case that β = α .
s s
For all simple roots α and all τ ∈ W ⊂ O(E) recall τs τ−1 = s . By virtue of the
α τ(α)
setup τ = s ···s : α → α so that
s+1 t−1 t s
s = (s ···s )s (s ···s ).
s s+1 t−1 t t−1 s+1
Re-arrange to get
s (s ···s ) = (s ···s )s
s s+1 t−1 s+1 t−1 t
s ···s = s s ···s s
s+1 t−1 s s+1 t−1 t
now compose on the left with s ···s to conclude. (cid:3)
1 s−1
Turning this result around yields a fact about the behavior of the so-called reduced
expression of an element of the Weyl group, more on this in The Weyl group.
Corollary. If σ = s ···s is an expression for σ ∈ W in terms of reflections corresponding
1 t
to simple roots, with t as small as possible, then σ(α ) is negative.
t
Proof. The lemma says that if s ···s (α ) is negative then σ can be shortened. So
1 t−1 t
if σ cannot be shortened then s ···s (α ) is positive, denote this root β (cid:31) 0. Now α =
1 t−1 t t
s ···s β implies −α = s (α ) = s s ···s β ≺ 0. But then s ···s s (−α ) = β (cid:31) 0,
t−1 1 t t t t t−1 1 1 t−1 t t
and so s ···s (α ) = −β ≺ 0 (cid:3)
1 t t
8
2.2.4. The image of a simple root under an element of the Weyl group is a priori just
some other root. The following result helps the user decide whether said image is positive
or negative.
Proposition ([Gre11] 11.28). Let α be a simple root and w an element of the Weyl group.
(1) The image w(α) is a negative root if and only if (cid:96)(ws ) = (cid:96)(w)−1.
α
(2) The image w(α) is a positive root if and only if (cid:96)(ws ) = (cid:96)(w)+1.
α
2.3. The Weyl group.
2.3.1. Recall from group theory that an action of a group G on a set X is called tran-
sitive when for all x,∈ X there exists g ∈ G such that gx = y. The action is simply
transitive when g is uniquely determined by x and y.
2.3.2. Let Π ⊂ Φ ⊂ E and W be as usual. In Bases and Weyl chambers we discussed
the action of the Weyl group on the Weyl chambers of Φ (relative to Π) and on the bases of
Φ.
Recapitulation: a regular element γ ∈ E is a vector belonging to one of E’s Weyl
chambers relative to some base Π ⊂ Φ. The set of all vectors in E lying on the positive side
of the hyperplane orthogonal to γ is Φ+(γ) = {α ∈ Φ : (γ,α) > 0}. An indecomposable
root is a root α ∈ Φ+(γ) that cannot be expressed as a sum of two roots from Φ+(γ). The
set of all indecomposable roots is denoted Π(γ). Theorem 2.1 says that every base of Φ is
of the form Π(γ) for some regular γ ∈ E. When Π = Π(γ) one writes C(Π) = C(γ) for the
fundamental Weyl chamber
C(Π) = {γ ∈ E : (γ,α) > 0 for all α ∈ Π}.
The Weyl group acts on bases via Π = Π(γ) (cid:55)→ σ(Π) = Π(σ(γ)) and acts on Weyl chambers
via C(γ) (cid:55)→ C(σ(γ)). This action turns out to be particularly nice:
Theorem. Let Π be a base of the root system Φ.
(1) If γ ∈ E is regular then there exists an element σ of the Weyl group such that
(σ(γ),α) > 0 for all simple roots α.
(2) If Π(cid:48) is another base of Φ then σ(Π) = Π(cid:48) for some element σ of the Weyl group.
(3) If α is any root then there exists an element σ of the Weyl group such that σ(α) ∈ Π.
(4) The Weyl group is generated by the set {s : α ∈ Π}.
α
(5) If σ ∈ W satisfies σ(Π) = Π then σ = 1.
Remark. Interpretations of each statement:
(1) The Weyl group acts transitively on the Weyl chambers: the γ belongs to some
Weyl chamber, and σ takes γ (along with its Weyl chamber) into the fundamental
Weyl chamber. To get from the Weyl chamber of γ to the Weyl chamber of some
other γ(cid:48) find σ : C(γ) → C(Π) and σ(cid:48) : C(γ(cid:48)) → C(Π). By design σ(cid:48)σ : C(γ) → C(γ(cid:48)).
(2) The Weyl group acts transitively on the bases of Φ.
(3) Every root is conjugate by W to a simple root. (Humphreys uses “conjugate by W”
to mean “connected by the W-action”.)
(4) A priori the Weyl group is generated by {s : α ∈ Φ}.
α
(5) The Weyl group acts simply transitively on bases.
9
2.3.3. It is significant that the Weyl group is generated by the simple reflections. An
expression σ = s ···s with α simple and t minimal is called a reduced expression,
α1 αt i
and the length of σ (relative to Π) is (cid:96)(σ) = t. Another characterization of length uses
n(σ), the number of positive roots α for which σ(α) ≺ 0.
Lemma. Each element σ of the Weyl group satisfies (cid:96)(σ) = n(σ).
2.3.4. The closure of the fundamental Weyl chamber relative to Π is a fundamental
domain for the action of the Weyl group on E, meaning each vector in E is conjugate by
W to exactly one vector in C(Π).
Lemma. Let λ,µ ∈ C(Π). If σλ = µ for some element σ of the Weyl group then σ is a
product of simple reflections which fix λ. In particular λ = µ.
2.4. Irreducible root systems.
2.4.1. In a precise sense, irreducible root systems are the building blocks of general root
systems. A root system Φ is irreducible if it cannot be partitioned into a union of two
mutually orthogonal proper subsets; otherwise Φ is reducible. This is the natural definition
because our universe of discourse is an inner-product space.
Forexample, therank1rootsystemA istriviallyirreducibleandtherank2rootsystems
1
A ,B ,G are irreducible while A ×A is reducible.
2 2 2 1 1
2.4.2. To determine the reducibility of a root system it is enough to consider a base:
Proposition. A root system Φ with base Π is irreducible if and only if Π cannot be partitioned
into a union of two mutually orthogonal proper subsets.
Proof. To prove the “if” direction we resort to proving the contrapositive: “if Φ is
reducible then Π is reducible.” To this end suppose Φ = Φ ∪Φ with Φ proper subsets of
1 2 i
Φ satisfying (Φ ,Φ ) = {(α ,α ) : α ∈ Φ } = 0. If Π ⊂ Φ (for instance) then (Π,Φ ) = 0
1 2 1 2 i i 1 2
but then since Π is a basis of E, γ = (cid:80) c α with c ∈ C for all γ ∈ E. Now (γ,Φ ) =
α∈Π α α 2
(cid:80) c (α,Φ ) = 0, which shows (E,Φ ) = 0 and hence Φ = ∅, contradiction. Since Π is
α α 2 2 2
not contained in either Φ there is a partition Π = Π ∪ Π with Π contained in Φ and
i 1 2 i i
(Π ,Π ) = 0.
1 2
To prove the “only if” direction suppose Φ is irreducible and attempt to partition Π =
Π ∪Π with (Π ,Π ) = 0. Using Theorem 2.3 part (3), for i = 1,2 let Φ denote the set of
1 2 1 2 i
all roots that are conjugated into Π by some element of the Weyl group (we do not stipulate
i
disjointjess at this step). Turning this around, any root in Φ can be obtained from a simple
i
root in Π by applying some element of the Weyl group. The formula s (γ) = γ −(cid:104)γ,Π(cid:105)Π
i Π
implies s and s commute when1 (α,β) = 0:
α β
(cid:18) (cid:19)
2(cid:104)γ,β(cid:105) 2(cid:104)γ,α(cid:105)
(s s −s s )(γ) = (α,β) α− β .
α β β α
(α,α) (α,α)
Since the Weyl group is generated by the simple reflections, applying the formula for the
latter shows that every root in Φ is obtained from a simple root in Π by adding and
i i
subtracting simple roots in Π . In short, Φ is contained in the span of Π . Now it is clear
i i i
that (Φ ,Φ ) = 0, and since Φ is irreducible some Φ = ∅. This fact forces the corresponding
1 2 i
Π = ∅, so our attempt at partitioning Π fails. (cid:3)
i
1in fact if γ ∈/ H ∪H this is an ‘if and only if’ statement
α β
10