18.745: Lecture 21

Professor: Victor Kač Scribe: Ruochuan Liu, Zhongtao Wu

Let (V,△) be a root system. Define a reflection r_{α} ∈End(V) for every α ∈ △ by:

r_{α}(v) = v − \frac{2(v, α)}{(α, α)}α

r_{α} has properties:

(i) r_{α} fixes pointwise the hyperplane τ_{α} = {u ∈ V |(u, α) = 0}. (ii) r_{α} = −α. (iii) r_{α}^{2} = 1 and γ_{α} ∈ O(V, (·, ·)), i.e r_{α} is invertible and (γ_{α}(v), γ_{α}(v)) = (v, v). (iv) r_{α}(△) = △. (i), (ii) and (iii) are obvious. (iv) follows from the string property. Exercise 21.1 (optional). Show that property (iv) along with \frac{2(α,β)}{(α,α)} ∈ ℤ for all α, β ∈ Δ implies the string property. Proof. Denote < β, α >:= \frac{2(β,α)}{(α,α)}. Then < β, α >∈ ℤ for all α, β ∈ Δ, and r_{α}(β) = β− < β, α > α. Step I: Let α, β be nonproportional roots. If (α, β) > 0, then α − β is a root. If (α, β) < 0, then α + β is a

root. Proof: If (α, β) > 0, then both < α, β > and < β, α > are positive. By Cauchy’s inequality,

< α, β >< β, α >= \frac{4(α, β)^{2}}{(α, α)(β, β)} < 4,

where the inequality is strict since α and β are nonproportional roots. Hence, at least one of < α, β >, < β, α > equals 1. If < α, β >= 1, then r_{β}(α) = α − β ∈ Δ; similarly, if < β, α >= 1, then β − α ∈ Δ, hence r_{β−α}(β − α) = α − β ∈ Δ. The case (α, β) < 0 is similar.

Step II: The α-string through β is unbroken, i.e., if p, q ∈ ℤ_{+} are the largest integers for which β−pα, β+qα ∈

Δ, then β + iα ∈ Δ, ∀ − p ≤ i ≤ q. Proof: If not, we can find −p ≤ i < j ≤ q such that β + iα ∈ Δ, β + (i + 1)α ∉ Δ, β + (j − 1)α ∉ Δ,

β + jα ∈ Δ. But then the claim in Step I implies both (β + iα, α) ≥ 0, (β + jα, α) ≤ 0. Hence, (α, α) ≤ 0. Contradiction!

Step III: p, q as in Step II, then p − q =< β, α >. Proof: Since r_{α} just adds or subtracts a multiple of α to any root, the string is invariant under r_{α}. In

particular, r_{α}(β + qα) = β − pα. The left hand side is also equal to β− < β, α > α − qα. Hence,

p − q =< β, α >.

Thus, we proved the string property for Δ.

Definition 1. The Weyl group of a root system is a subgroup W of O(V, (·, ·)) generated by all reflections γ_{α} for α ∈ △. This is a finite group since it permutes elements of a finite set which spans V.

Example 2. △_{A_{r}} = {ε_{i} − ε_{j}|1 ≤ i, j ≤ r + 1, i ≠ j}. Let α = ε_{i} − ε_{j}, then

r_{α}(ε_{k}) = \{ \begin{matrix}ε_{k} : & k ≠ i, j \\ ε_{j} : & k = i \\ ε_{i} : & k = j \end{matrix} \}

Hence r_{α} = (ij), i.e., it just permutes ε_{i} and ε_{j}. Thus the Weyl group of A_{r} is S_{r+1}.

Exercise 21.2. Compute the Weyl group of the root systems B, C and D.

Proof.

(a) Root system Δ_{B_{r}} = {±ϵ_{i} ± ϵ_{j} (i ≠ j), ±ϵ_{i}}.

• When α = ϵ_{i} − ϵ_{j}, r_{α}(ϵ_{k}) = \begin{cases}ϵ_{k} & if k ≠ i, j; \\ ϵ_{j} & if k = i; \\ ϵ_{i} & if k = j. \end{cases}

• When α = ϵ_{i} + ϵ_{j}, r_{α}(ϵ_{k}) = \begin{cases}ϵ_{k} & if k ≠ i, j; \\ −ϵ_{j} & if k = i; \\ −ϵ_{i} & if k = j. \end{cases}

• When α = ϵ_{i}, r_{α}(ϵ_{k}) = \begin{cases}ϵ_{k} & if k ≠ i; \\ −ϵ_{i} & if k = i. \end{cases}

Hence, the Weyl group is generated by all permutations of the set {ϵ_{1}, ϵ_{2}, …, ϵ_{r}} and the operations of sign changes. In terms of group, it is isomorphic to S_{r} × ℤ_{2}^{r}.

(b) Root system Δ_{C_{r}} = {±ϵ_{i} ± ϵ_{j} (i ≠ j), ±2ϵ_{i}}.

We can check that r_{α}(ϵ_{k}) has exactly the same form as the previous case, hence its Weyl group is also S_{r} ⋉ ℤ_{2}^{r}.

(c) Root system Δ_{D_{r}} = {±ϵ_{i} ± ϵ_{j}, i ≠ j}.

In this case, the Weyl group is generated by the first two types of reflections in (a). Also, notice that

r_{ϵ_{i}−ϵ_{j}} ∘ r_{ϵ_{i}+ϵ_{j}}(ϵ_{k}) = \begin{cases}ϵ_{k} & if k ≠ i, j; \\ −ϵ_{i} & if k = i; \\ −ϵ_{j} & if k = j. \end{cases}

This is the same as changing signs for a pair (ϵ_{i}, ϵ_{j}). Hence, each element in the Weyl group is a permutation of set {ϵ_{1}, ϵ_{2}, …, ϵ_{r}}, composed with an even number of sign changes. In terms of group, it is isomorphic to S_{r} ⋉ ℤ_{2}^{r−1}.

Exercise 21.3. If A ∈ O(V, (·, ·)), then Ar_{α}A^{−1} = r_{A(α)}.

Proof. Let AT_{α} := {A(u)| (α, u) = 0}. Since A ∈ O(V, (·, ·)), we have (A(u), A(α)) = (u, α) = 0 for all A(u) ∈ AT_{α}. Hence r_{A(α)}(AT_{α}) = AT_{α}.

Also, Ar_{α}A^{−1}(AT_{α}) = Ar_{α}(T_{α}) = AT_{α}. So both reflections Ar_{α}A^{−1} and r_{A(α)} fix the hyperplane AT_{α}, hence Ar_{α}A^{−1} = r_{A(α)}.

Recall that, given a choice of f ∈ V^{∗} such that f(α) ≠ 0 for all α ∈ △, we get a subset △_{+} = {α ∈ △|f(α) > 0} and Π = {α_{1}, α_{2}, …, α_{r}} of simple roots such that any α ∈ △_{+} is of the form α = Σ_{i}k_{i}α_{i}, k_{i} ∈ ℤ_{≥0}. The reflections s_{i} = r_{α_{i}} are called simple reflections. We use the notation height(α) := ∑_{i} k_{i}.

Theorem 3.

(a) △_{+}\{α_{i}} is s_{i}-invariant.

(b) If α ∈ △_{+}\Π, there is a s_{i} such that height(s_{i}(α)) <height(α).

(c) If α ∈ △_{+}\Π, we can choose a sequence of simple reflections s_{i_{1}}, …, s_{i_{k}} such that s_{i_{1}} … s_{i_{k}}(α) ∈ Π and s_{i_{j}} … s_{i_{k}}(α) ∈ △_{+} for each 1 ≤ j ≤ k.

(d) W is generated by simple reflections.

Proof.

(a) Applying simple reflection s_{i} changes sign of at most one coefficient k_{i} of α ∈ △_{+}. If k_{i} changes to negative, then s_{i}(α) wouldn’t be a root, hence s_{i}(α) ∈ △_{+}.

(b) If height(s_{i}(α)) doesn’t decrease for all s_{i}, then from s_{i}(α) = α − \frac{2(α,α_{i})}{(α_{i},α_{i})}α_{i} we get (α, α_{i}) ≤ 0 for all i. Hence (α, α) = Σ_{i}k_{i}(α, α_{i}) ≤ 0, a contradiction!

(c) follows from (b) and (a).

(d) Denote W′ the subgroup of W generated by simple reflections. By (c), for any α ∈ △_{+}, there exists w ∈ W′ such that w(α) = α_{i} ∈ Π for some i. By Ex21.3, wr_{α}w^{−1} = r_{w(α)}, hence r_{α} = w^{−1}s_{i}w ∈ W′.

Consider V \ ⋃_{α∈△_{+}} T_{α} = ∐_{j} C_{j}, where C_{j} are connected components of this set. C_{j}’s are called open chambers, \overline{C_{j}}’s are called (closed) chambers. Also, define the fundamental chamber: \overline{C} = \{v ∈ V |(α_{i}, v) ≥ 0 i = 1, …, r}.

Exercise 21.4. Show that T_{α} ∩ C = ∅, where C = {v ∈ V | (α_{i}, v) > 0, i = 1, …, r} is the open fundamental chamber. Hence the fundamental chamber is a chamber.

Proof. Suppose v ∈ T_{α} ∩ C, then (v, α) = 0 and (v, α_{i}) > 0 for i = 1, …, r. But α = ∑_{i=1}^{r} k_{i}α_{i}, where k_{i} ∈ ℤ_{+}. Hence, k_{i} = 0 for all i, and consequently v = 0.

So we proved T_{α} ∩ C = ∅. Since C is connected, C ⊂ C_{j} for some j. On the other hand, (α_{i}, v) ≠ 0, ∀i, v ∈ C_{j} by definition. And since the inner product is a continuous function of v, we conclude that (α_{i}, v) > 0 ∀v ∈ C_{j}. Hence C_{j} ⊂ C.

Thus, C̄ = C̄_{j}, i.e, the fundamental chamber is a chamber.

Theorem 4.

(a) W permutes all chambers transitively, i.e for any chamber \overline{C_{1}} and \overline{C_{2}}, there exists w ∈ W such that w\overline{C_{1}} = \overline{C_{2}}.

(b) Let △_{+} and △+^{′} be subsets of positive roots of △ defined by f and f′ respectively . Then there exists w ∈ W such that w(△_{+}) = △+^{′}. In particular, the Cartan matrix is independent of the choice of f.

Proof.

(a) Choose P_{i} ∈ C_{i} (i = 1, 2) such that the interval [P_{1}, P_{2}] doesn’t intersect any of τ_{α} ∩ τ_{β}, where α, β ∈ △_{+} and α ≠ β. Now move along the interval [P_{1}, P_{2}] until we hit a hyperplane τ_{α}. Apply reflection r_{α} to \overline{C_{1}}. Keep moving and applying reflections until we reach \overline{C_{2}}.

(b) △_{+} and △+^{′} define fundamental chamber \overline{C} and \overline{C′} respectively. By (a), there exists w ∈ W such that

w(\overline{C}) = \overline{C′}. Hence w(△_{+}) = w(△+^{′}), since \overline{C} = \{v ∈ V |(α_{i}, v) ≥ 0 i = 1, …, r\}.

Definition 5. Fix Π ⊂ △_{+} ⊂ △, then we have simple reflections s_{1}, …, s_{r} ∈ W. Given w ∈ W, an expression w = s_{i_{1}} … s_{i_{l}} is called reduced if l is minimal possible. We let l = l(w) called the length of w. Note that det w = (−1)^{l(w)} since det s_{i} = −1.

Example 6. l(s_{i}) = 1, l(s_{i}s_{j}) = 2 if i ≠ j, but l(s_{i}^{2}) = 0.

Lemma 7 (Exchange Lemma). Suppose that s_{i_{1}} … s_{i_{t−1}}(α_{i_{t}}) ∈ △_{−}, then the expression w = s_{i_{1}} … s_{i_{t}} is not reduced. Namely, there exists 1 ≤ r < t such that w = s_{i_{1}} … s_{i_{r−1}}s_{i_{r+1}} … s_{i_{t−1}}.

Proof. Consider the roots β_{k} = s_{i_{k+1}} … s_{i_{t−1}}(α_{i_{t}}) for 0 ≤ k ≤ t − 1. Then β_{0} ∈ △_{−} by assumption and β_{t−1} = α_{i_{t}} ∈ △_{+}. Hence there exists 0 ≤ r ≤ t − 1 such that β_{r−1} ∈ △_{−} and β_{r} ∈ △_{+}. But by definition β_{r} = s_{i_{r}}(β_{r}), hence β_{r} = α_{i_{r}}. Recall that, by definition, β_{r} = s_{i_{r+1}} … s_{i_{t−1}}(α_{i_{t}}) = α_{i_{r}}. Thus if we denote \overline{w} = s_{i_{r+1}} … s_{i_{t−1}}, using Ex21.3 we see that \overline{w}s_{i_{t}} \overline{w}^{−1} = s_{i_{r}}, thus \overline{w}s_{i_{t}} = s_{i_{r}} \overline{w}. Now multiplying both sides by s_{i_{1}} … s_{i_{r}}, we get the result.

Corollary 8. W acts simply transitive on chambers, i.e if w(\overline{C}) = \overline{C}, then w = 1. In particular if λ ∈ C(open chamber), w(λ) = λ, then w = 1.

Proof. If w ≠ 1, take its reduced expression: w = s_{i_{1}} … s_{i_{l}}. If w(\overline{C}) = \overline{C}, then w(Π) = Π, in particular w(α_{i_{l}}) ∈ △_{+}. On the other hand, w(α_{i_{l}}) = s_{i_{1}} … s_{i_{l−1}}(−α_{i_{l}}) ∈ △_{+} means s_{i_{1}} … s_{i_{l−1}}(α_{i_{l}}) ∈ △_{−}, hence by exchange lemma s_{i_{1}} … s_{i_{l}} is not a reduced expression. That’s a contradiction!