The Big Book of Small Modules

Barbara Baumeister Ulrich Meierfrankenfeld

Gernot Stroth

April 2, 2004

Notes for the authors

U: strong steinberg tensor product theorem

U: the weights of a module are really weights on Φ^{∗}

U: Need to think about the notations (α, β) and <α, β >. For example <α, β > currently also denotes the root system generated by α and β.

U: I added the GLS reference. According the Parker/Rowler, it contains a proof ( Theorem 2.8.2) that the irreducible modules of the untwisted groups stay irreducible for the twisted group.

Chapter 1

Introduction

In this book we classify modules for finite groups fullfiling certain properties which forces the module to be ”small” in some sense or annother. The main motivation for the book is provide the information about modules necessary in the local classification of finite groups of local characterisic p [LGCP].

4 CHAPTER 1. INTRODUCTION

Some Group Theory

Lemma 2.0.1 [three subgroup lemma] Let A, B, C be subroups of G with [A, B, C] + [B, C, A] = 1. Then [C, A, B] + 1.

Proof:

Lemma 2.0.2 [nilpotent groups] Let M be a nilpotent group and A a proper subgroup of M. Then A is a proper subgroup of ℕ_{M}(A) and ⟨A^{M}⟩ is a proper subgroup of M.

6 CHAPTER 2. SOME GROUP THEORY

Chapter 3

Some elementary representation theory

Lemma 3.0.3 Let G be a finite group and V an irreducible 𝕂G-module. If char 𝕂 = p, p a prime and Ω^{p}(G) acts homogenously on V, Ω^{p}(G) acts irreducible on V. Proof: Comment: ref? any extra assumptions on 𝕂 ?

8 CHAPTER 3. SOME ELEMENTARY REPRESENTATION THEORY

Quadratic pairs in odd characteristic

The proof of the Glauberman-Thompson Theorem presented in

4.1 sec:glauberman thompson

is essentially due to Paul Flavell.

4.2 a,b and ab-quadratic

[sec:ab quadtratic] Lemma 4.2.1 [ab quadratic] Let G a group, R a ring, V a faithful RG-module and a, b ∈ G such that a, b and ab are quadratic in V and G = ⟨a, b⟩. Then

(a) [z] If G is abelian, then G is quadratic on 2V.

(b) [a] G is nilpotent of class at most two.

(c) [y] [V, G′] ≤ 2[V, G, G].

(d) [b] [V, G′, G] = [V, G, G′] = 0.

(e) [c] ⟨h⟩G′ is quadratic for all h ∈ G.

(f) [d] αβ = −βα and γ = 2αβ, where α = a − 1, β = b − 1 and γ = [a, b] − 1.

(g) [e] Put C_{δ} = {v ∈ V | vδ = 0}. Then C_{2β} ≤ V α + C_{2β} ≤ C_{γ} ≤ V and V γ ≅ V α + C_{2β}/C_{2β} ≅ V/C_{γ} as R-modules.

(h) [f] Suppose that R is field, then dim_{R}[V, [a, b]] ≤ \frac{1}{2} min\{dim_{R}[V, a], dim_{R}[V, b]\}.

10 CHAPTER 4. QUADRATIC PAIRS IN ODD CHARACTERISTIC

Proof: Let δ = ab−1. Then δ = (α+1)(β+1)−1 = αβ+α+β. Since α^{2} = β^{2} = δ^{2} = 0 we conclude that 1^{∘} [1] αβαβ + αβα + αβ + βαβ + βα = 0

Suppose that G is abelian, then αβ = βα and so (1^{∘}) implies 2αβ = 0. Thus (a) holds. Multiplying (1^{∘}) with β from the right we have

2^{∘} [2] αβαβ + βαβ = 0

Multiplying (2^{∘}) with α from the left we get 3^{∘} [3] αβαβ = 0.

Substituting (3^{∘}) into (2^{∘}) we have 4^{∘} [4] βαβ = 0

Multiplying (1^{∘}) with α from the right and using (3^{∘}) we have 5^{∘} [5] αβα = 0

From (1^{∘}), (3^{∘}),(4^{∘}) and (5^{∘}) we get 6^{∘} [6] αβ + βα = 0

Let g = [a, b]. Then g −1 = (1−α)(1−β)(1+α)(1+β)−1 = αβ −αβ −βα +αβ = 2αβ Thus (f) holds. (b),(c) and (d) are immediate consequences of (f). By (b) every element h ∈ G can be written as h = a^{k}b^{l}g^{m} with κ, l, m ∈ ℤ. Thus

h − 1 = (1 + kα)(1 + lβ)(1 + 2mαβ) − 1 = kα + lβ + (kl + 2m)αβ

and so

(h − 1)^{2} = kl(αβ + βα) = 0

Thus all elements in G are quadratic. Hence (e) follows from (d).

(g) follows from γ = 2αβ. Suppose R is a field. Then by (g),

dim V β = dim V/C_{β} ≥ dim V/C_{2β} ≥ dim V α + C_{2β}/C_{2β} + dim V/C_{γ} = 2 dim V_{γ} By symmetry in a and b we conclude that (g) holds. □

4.3 The Glauberman-Thompson Theorem

Let 𝔽 be a field with char 𝔽 ≠ 2, G a group and V a faithfully and finitary 𝔽G-module. ote here that we allow infinite fields and fields in characteristic 0. For a ∈ G let C_{a} = C_{V} (a), V_{a} = V (a − 1) and d_{a} = dim V_{a}. Let Q be the set on non trivial quadratic elements in G, that is Q = {1 ≠ a ∈ G | V_{a} ≤ C_{a}}. Put d = min_{a∈Q} d_{a} and QR = {a ∈ Q | d_{a} = d}. The elements of QR are called roots. Fix two roots a and b and let H = ⟨a, b⟩. Put α = a − 1, b = β − 1, g = [a, b] and γ = g − 1. Then V_{a} = V α and C_{α} = ker α. Suppose that H is niloptent, i.e that H acts unipotenly on V. In this section we prove the Glauberman-Thompson theorem which says that among other things H has class at most 2.

Lemma 4.3.1 [class two] Suppose that H has class two. Then

(a) [a] g is a root.

(b) [b] V_{a} ∩ V_{b} = 0.

(c) [c] V_{g} = V αβ ⊕ V βα and V αβ = V β ∩ C_{b} = V β ∩ V γ.

(d) [d] V = C_{a} + C_{b}.

(e) [e] V is a direct sum of indecomposable 𝔽H-submodules.

(f) [f] Let W be non-trivial indecomposable direct summand of the 𝔽H-module V. Then there exists a basis for V such that the matrices for α and β are ( in some order)

(\begin{matrix}0 & 0 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}) (\begin{matrix}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 1 & 0 \end{matrix})

(g) [g] Let 1 ≠ d ∈ H. Then d is quadratic on V iff q is a root and iff d ∈ ⟨a^{H}⟩ ∪ ⟨b^{H}⟩. Moreover, both ⟨a^{H}⟩ and ⟨b^{H}⟩ act quadratically on V.

Proof: Let A = ⟨a, a^{b}⟩ = ⟨a, g⟩. Since H has class two, A is abelian and A is normal in H. Moreover, a^{b^{2}} = ag^{2} = a^{2b}a^{−1}. Hence 4.2.1(b) implies that ⟨a^{2b}, a^{−1}⟩ is quadratic. Since char 𝔽 ≠ 2 and a is quadratic, we have a^{2} ≠ 1. Hence the minimality of d_{a} = d implies V_{a^{2}} = V_{a} and we conclude that A is quadratic. Then a centralizes V_{g} and by symmetry b centalizes V_{g}. Therefore

1^{∘} [0] V g ≤ C_{V} (H).

Note that

V α + V γ = [V, A] = V α + V α^{b} = V α + V αβ

In particular, C_{b} ∩ [V, A] = (V α ∩ C_{b}) + V αβ and so

V γ ≤ (V α ∩ C_{β}) + V αβ

By definition of d, dim V γ ≥ d. On the otherhand dim(V α∩C_{b})+dim V αβ = dim V α = d and we conclude that

2^{∘} [1] V γ = (V α ∩ C_{β}) ⊕ V αβ and g is a root.

In particular, V αβ ≤ V γ ≤ C_{V} (H). By symmetry V βα ≤ V γ ≤ C_{V} (H). In particular, V αβ +V βα is an 𝔽H-submodule in V and H acts quadtratically on V/(V αβ +V βα). Thus V γ ≤ V αβ + V βα. Since V βα ≤ V α ∩ V γ ≤ V α ∩ C_{b} we conclude from (2^{∘}) that

3^{∘} [2] V γ = V βα ⊕ V αβ ≤ C_{V} (H) and V α ∩ C_{β} = V βα = V α ∩ V γ

In particular, V α ∩ V β = (V α ∩ C_{b}) ∩ (V β ∩ C_{a}) = V βα ∩ V αβ = 0 and (a), (b) and (c) are proved. (d) follows from (b) applied to the dual module of V.

Let U_{a} be an 𝔽-complement to V α + C_{V} (H) in C_{a}. Then U_{a} ∩ C_{b} ≤ U_{a} ∩ C_{V} (H) = 0 and so dim U_{a} = dim U_{a}β. Let u ∈ U_{a} with uβα = 0. Then uβ ∈ V β ∩ C_{α} = V αβ and so uβ = vαβ for some v ∈ V. Hence u − vα ∈ C_{b} and u ∈ (C_{b} + V α) ∩ C_{α} = (C_{α} ∩ C_{β}) + V α = V α+C_{V} (H). Thus u = 0 and U_{a}β∩C_{a} = 0. Also U_{a}β∩U_{a}βα ≤ U_{a}β∩C_{a} = 0. Furthermore, U_{a}β + U_{a}βα ∩ C_{a} ≤ (V β ∩ C_{α}) + V α ≤ C_{V} (H) + V α and so U_{a} ∩ U_{a}b + U_{a}αβ = 0. Put W_{a} = U_{a} + U_{a}β + U_{a}βα. Then W_{a} = U_{a} ⊕ U_{a}β ⊕ U_{a}βα and dim U_{a} = dim U_{a}β = dim U_{α}β. Thus if u_{i}, 1 ≤ i ≤ m, is a basis for U_{α}, then u_{i}, u_{i}β, u_{i}βα, 1 ≤ i ≤ m is a basis for W_{a}. Since a centralizes U_{a} and U_{a}βα and b centralizes U_{a}β and U_{a}βα we see that 𝔽⟨u_{i}, u_{i}β, u_{i}βα is a 3-dimensional 𝔽H-submodule on which α and β as in (f).

Similarly define U_{b} and W_{b}. Suppose that W_{a}∩W_{b} ≠ 0. Then also W_{α}∩W_{β}∩C_{V} (H) ≠ 0. But W_{a} ∩ C_{V} (H) = U_{a}βα and W_{β} ∩ C_{V} (H) = U_{b}αβ and we obtain a contradiction to (b). Thus W_{a} ∩ W_{b} = 0.

Put \overline{V} = V/W_{a} + W_{b} + C_{V} (H). Since V = C_{a} + C_{b} = U_{a} + U_{b} + V α + V β + C_{V} (H) = W_{a} + W_{b} + C_{V} (H) we have \overline{V} = [\overline{V} , H]. Since H is nilpotent on \overline{V}, this implies \overline{V} = 0. Thus

V = W_{a} + W_{b} + C_{V} (H) = W_{a} ⊕ W_{b} ⊕ C

for some C ≤ C_{V} (H). Hence (e) holds and (f) follows from the above and the Krull-Schmidt Theorem. (g) follows easily from (f). □

Lemma 4.3.2 [va cap vb] Suppose that V α ∩ V β ≠ 0. Then H is abelian.

Proof: Note that V α ∩ V β ≤ C_{V} (H) and so V α ∩ V β ≤ V δ for all δ ∈ α^{H} ∩ β^{H}. Hence by induction on the maximum of the subnormal lengths of a and b in H, both ⟨a^{H}⟩ and ⟨b^{H}⟩ are abelian. Thus H has class at most two and the lemma follows from 4.3.1(b).

Lemma 4.3.3 [A abelian] H has class at most two.

4.4. THE SL_{2}(Q)-LEMMA 13

Proof: Since H is unipotent on V, there exists v ∈ V with [v, H, H] = 0 and [v, H] ≠ 0. Interchanging a and b if necessary we may assume that vα ≠ 0. Then vα ∈ V α ∩ C_{V} (H) ≤ V δ for all δ ∈ α^{H}. Thus by 4.3.2, A := ⟨a^{H}⟩ is abelian. By induction on the subnormal maximum of the subnormal lengths of a and b to H we may assume that B := ⟨b, b^{a}⟩ has class at most two. If B is abelian that g ∈ Z(B) ∩ A ≤ Z(H) and we are done. So suppose that B has class two. Note that H = AB. Since A is abelian, A∩B is normal in A. Since A is normal in H, A ∩ B is normal in B and so A ∩ B ⊴ H. Since g ∈ A ∩ B, H/A ∩ B is abelian. Hence [A, B] ≤ A ∩ B and B is normal in H. By 4.3.1(g), B has exactly two maximal quadratic subgroups, namely ⟨b^{B}⟩ and ⟨b^{aB}⟩. Thus a^{2} normalizes ⟨b^{B}⟩ and b^{a^{2}} ∈ ⟨b^{B}⟩. Since [b, a, a] ∈ [A, A] = 1 we conclude b^{2a} = (b[b, a])^{2} ∈ bb[b, a]^{2}B′ = bb^{a^{2}}B′ ⊆ < b^{B}⟩. Thus [B, b^{2a}, B] = 0. By minimality of d, [V, b^{2a}] = [V, b^{a}] and so [V, b^{a}, B] = 0. Since B = B^{a^{−1}} we also have [V, b, B] = 0, [V, B, B] = 0 and B is abelian, a contradiction. □

4.4 The SL_{2}(q)-Lemma

Let G be a finite group, 𝔽 a field with positive characteristic p ≠ 2, V a faithful, finite dimensional 𝔽G-module and a, b quadratic elemennts. Put H = ⟨a, b⟩. In this section we show (with some exceptions for p = 3), that if a and b are roots than either H is p-group or H ≅ SL_{2}(q) for some power q of p. Put δ = αβ + βα.

Lemma 4.4.1 [d commutes] δh = hδ for all h ∈ H.

Proof: αδ = αβα = δα.

Lemma 4.4.2 [d as scalar] Suppose that d acts as the scalar ξ on V.

(a) [a] If ξ = 0, then H is nilpotent of class at most two and all elements in H act quadratically on V.

(b) [b] Suppose ξ ≠ 0.

(a) [a] V is the direct sum of isomorphic 2-dimensional simple 𝔽H submodules.

(b) [b] For each simple 𝔽H-submodule in V there exists a basis such that the matrices of α and β are

(\begin{matrix}0 & 1 \\ 0 & 0 \end{matrix}) (\begin{matrix}0 & 0 \\ ξ & 0 \end{matrix})
(c) [c] H ≅ SL_{2}(𝔽_{p}[ξ]) or p = 3, |ξ| = 4 and H ≅ SL_{2}(5).

Proof: Suppose first that ξ = 0. Then αβ = −βα and so ab acts quadratic on V. Thus (a) follows from 4.2.1.

So suppose that ξ ≠ 0. Then V = ξV = V δ = V α + V β = C_{a} + C_{b} and C_{a} ∩ C_{b} ≤ C_{V} (H) ≤ ker δ = 0. Thus V α = C_{a}, V β = C_{b} and V = C_{b} ⊕ C_{a}. In particular the C_{b} → C_{a}, v → vα is an isomorphism. Let v ∈ C_{β}. Then vαβ = v(δ − βα) = ξv. Let v_{i}, 1 ≤ i ≤ m be a basis for C_{b}. Then v_{i}, v_{i}α is a basis for V and we see (b:a) and (b:b) holds. (b:c) now follows from Dickson’s Theorem ( see [?]) □

For ξ ∈ 𝔽 let V_{ξ} be the generalized ξ-eigenspace for δ on V. Note that by 4.4.1V_{ξ} is an H-submodule.

Lemma 4.4.3 [v xi] Let ξ ∈ 𝔽^{♯} and suppose that V = V_{ξ}. Then H/O_{p}(H) ≅ SL_{2}(𝔽_{p}[ξ]) respectively SL_{2}(5) if p = 3 and |ξ| = 4.

Proof: Note that ϵ := δ − ξ act nilpotenly on V and δ acts as the scalar ξ in each V ϵ^{n}/V ϵ^{n+1}. Thus the lemma follows from 4.4.2.

Lemma 4.4.4 [n central] Let N/O_{p}(H) = Z(H/O_{p}(H) and put Z = O^{p}(N). Then Z acts as a scalar±1 on each V_{ξ}. Inparticular, Z ≤ Z(H), Z is an elementary abelian 2-group and N = O_{p}(H) × Z.

Proof: Note that N acts as a scalar on each composition factor of H on V. In particular, N/O_{p}(H) is a p′-group and so N = O_{p}(H)Z. Let h be a p′ element in N and ξ ∈ 𝔽. If ξ = 0, then O^{p}(H) and so also h centralize V_{ξ}. So suppose that ξ ≠ 0. Since [h, H] centralizes all the composition factors for H on V_{ξ} we conclude from 4.4.3 that h either centalizes all the composition factors or h inversts all the composition factors of H on V_{ξ}. Since h is a p′ we conclude that h acts as ±1 on V_{ξ}. To prove the remaing assertions we may assume that 𝔽 is algebraicly closed. Then V is the direct sum of its eigenspaces and so h^{2} = 1 and [h, H] = 1. □

Lemma 4.4.5 [op central] Suppose that a is a root. Then [O_{p}(H), O^{p}(H)] = 1.

Let W be a non-trivial simple submodule for H in V. The W α ≠ 0. Moreover, N normalizes W α, A := ⟨a^{N}⟩ is a p-group and [W α ≤ V α^{n} for all n ∈ N. Thus by 4.3.2, A is abelian. Thus A act on N. Let X be a composition factor for H on O_{p}(H). Then by 4.4.4, N acts trivially on X. On the otherhand by 4.4.3 H/N is a subdirect product of L_{2}(q) for odd q′s and so H/N is p-stable. Thus a and so also O^{p}(H) centralizes X. □

With ring we mean a ring with one. Let M_{n}(R) be the ring of n × n matrices over the ring R.

Lemma 4.4.6 [ideals] Let R and S be a rings, ϕ : R → S an onto ringhomomorphism and I = ker ϕ. Then

(a) [a] ϕ_{1} : M_{m}(R) → M_{m}(S), (a_{ij}) ↦ (ϕ(a_{ij}) is an onto ring homomorphism with ker ϕ_{1} = M_{n}(I).

(b) [b] ϕ_{2} : GL_{m}(R) → GL_{m}(S), (a_{ij}) ↦ (ϕ(a_{ij}) is an group homomorphism with ker ϕ_{1} = 1 + M_{n}(I).

(c) [c] Let a, b ∈ GL_{m}(R). Then the following are equivalent: ab^{−1} ∈ 1 + M_{n}(R), a − b ∈ M_{n}(R) and ϕ_{2}(a) = ϕ_{2}(b).

Proof: Obvious. □

Lemma 4.4.7 [direct sum of rings] Let R_{1} and R_{2} be commuative rings. Then GL_{n}(R_{1}⊕ R_{2}) ≅ GL_{n}(R_{1}) × GL_{n}(R_{2}) and SL_{n}(R_{1} ⊕ R_{2}) ≅ SL_{n}(R_{1}) × SL_{n}(R_{2}).

Proof: Obvious. □

Lemma 4.4.8 [trace 0] Let k, l, m be postive integers, R a commutative ring with one, 𝔽 a subfield of R and μ ∈ R with R = 𝔽[μ] and μ^{k} = 0 for some k ∈ ℕ. Let M_{m}^{∘}(R) be the ring of trace 0, m × m matrices over 𝔽. Put

I_{l} := {1 + μ^{l}b | b ∈ M_{m}^{∘}(R)}

Nonsense, thsi is not even a subgroup. Just use determined to describe the correct subgroup.

(a) [a] I_{l} is a normal subgroup of GL_{m}(R)

(b) [b] [1 + μ^{r}a, 1 + μ^{s}b] ∈ 1 + μ^{r+s}(ba − ab)I_{r+s+1} for all a, b ∈ M_{m}^{∘}(R) and r, s ∈ N.

(c) [c] [I_{r}, I_{s}] = I_{r+s}. In particular, I_{2} = I′_{1} and I_{1} is nilpotent of class k − 1.

(d) [d] For all 1 ≤ l ≤ k, I_{l}/I_{l+1} is isomorphic to M_{m} ∘ (𝔽) as a module for GL_{m}(𝔽).

(e) [e] Let a ∈ GL_{m}(𝔽) and i ∈ I_{1}. Then a^{i} ∈ (a + μd)I_{2} for some uniquely determined d ∈ M_{m}(𝔽). Moreover, d has trace 0 and μ^{2} divides the trace of a^{i} − a.

Proof:

Let a, b ∈ M_{m}^{∘}(R) and d ∈ GL_{m}(R).

(a) (1 + μ^{l}a)(1 + μlb) = 1 + μ(a + b + μ^{l}ab) and so I_{l} is a subgroup of GL_{m}(R). Also (1 + μ^{l}a)^{d} = 1 + μ^{l}a^{d} and so I_{l} is a normal subgroup of GL_{m}(R).

(b)

Let x = 1 + μ^{r}a, y = 1 + μ^{s}b and z = 1 + μ^{r+s}(ba − ab). Then modulo I_{r+s+1}

xyc ≡ (1+μ^{r}a+μ^{s}b+μ^{r+s}ab)(1+μ^{r+s}ba−μ^{r+s}ab) ≡ (1+μ^{r}a+μ^{s}b+μ^{r+s}ab)+μ^{r+s}ba−μ^{r+s}ab ≡ 1+μ^{r}a+μ^{s}b+μ^{r+s}ba ≡ yx

Thus [x, y] = x^{−1}y^{−1}xy ≡ c modulo I_{r+s+1} and (b) holds.

(c) follows fro (b) and some straightforward calculations.

(d) The map M_{m} ∘ (𝔽) → I_{l}/I_{l+1}, a → (1 + μ^{l}a)I_{l+1} is GL_{m}(𝔽)-isomorphism.

(e) We may assume without loss that μ^{2} = 0. The I_{2} = 0. We first show the uniqueness of d. So suppose that a + μd = a + μe) with d, e ∈ M_{m}(𝔽). Then μ(d − e) = 0 and since d − e ∈ M_{m}(𝔽), d − e = 0 and d = e.

For the existence of d, note that i = 1 + μb with b ∈ M_{m}(𝔽) and so i^{−1} = 1 − μb. Thus

a^{i} ∈ (1 − μb)a(1 + μb) = a + μ(ab − ba)

So d = ab − ba works. Also ab and ba have the same trace and so ab − ba has trace 0.□

Let R be a commutative ring and O ≠ ξ ∈ R. Let S_{ξ} be the subgroup of SL_{2}(R) generated by

a := (\begin{matrix}1 & 1 \\ 0 & 1 \end{matrix}) and b_{ξ} : (\begin{matrix}1 & 0 \\ ξ & 1 \end{matrix})

Lemma 4.4.9 [sxi irreducible] Let 𝔽 be a locally finite field with 0 ≠ p := char F ≠ 2 and 0 ≠ ξ ∈ 𝔽. Then S_{ξ} acts irreducible on M_{2}^{∘}(𝔽_{p}[ξ])

Proof: Let b = b_{ξ}, S = S_{ξ} and 𝕂 = 𝔽_{p}[ξ]. Then S ≤ SL_{2}(𝕂). Put V = M_{2}^{∘}(𝕂) and let 0 ≠ U an 𝔽_{p}S submodule in V. We need to show that U = V. Note that V is an

𝕂S-module. Put x = (\begin{matrix}0 & 1 \\ 0 & 0 \end{matrix}), y := (\begin{matrix}1 & 0 \\ 0 & −1 \end{matrix}) and z := (\begin{matrix}0 & 0 \\ 1 & 0 \end{matrix}). Direct martix calculations

show that

1^{∘} [1] [x, a] = 0, [y, a] = 2x and [z, a] = y − x.

2^{∘} [2] [x, b] = ξy − ξ^{2}z, [y, b] = −2ξz and [z, b] = 0

In particular,

3^{∘} [3] [x, a, a] = [y, a, a] = 0 and [z, a, a] = 2x.

and

4^{∘} [4] [z, a, a] = [y, a, a] = 0 and [x, a, a] = −2ξ^{2}z.

Thus C_{V} (a) = 𝕂x. Put E := {a ∈ K | lx ∈ U} and D := {d ∈ 𝕂 | dx ∈ U + 𝕂y + 𝕂z. Then E and D are 𝔽_{p} subspace of 𝕂 and E ≤ D. Since C_{U}(a) ≠ 0, E ≠ {0}. Let d ∈ D. Then [dx, b, b] ∈ U and so from (4^{∘}), −2ξ^{2}z ∈ U and so by (4^{∘}) [−2ξ^{2}z, a, a] = −4ξ^{2}x ∈ U. Thus ξ^{2}E ≤ ξ^{2}D ≤ E. Since multiplication by ξ^{2} is invertible we conclude that

5^{∘} [5] E = D = ξ^{2}D.

Let e ∈ E. Then [ex, b, a] = e[ξy − ξ^{2}z, a] = e((2ξ + ξ^{2})x − ξ^{2}y) ∈ U. Thus (2ξ + ξ^{2})e ∈ D = E. Since by (5^{∘}), ξ^{2}e ∈ E we get eξ ∈ E. Thus E is invaraint under multiplication by ξ. Since 𝕂 = 𝔽_{p}[ξ]], ξ acts irreducible on 𝕂 by left multipication. Hence E = 𝕂. It now follows from (2^{∘}) and (4^{∘}) that V = 𝕂x + 𝕂y + 𝕂z ≤ Ex + [Ex, b] + [Ex, b, b] ≤ U and so V = U. This completes the proof of the lemma.

□

Lemma 4.4.10 [a+y a root] Let 𝔽 be a field, f a polynomial over 𝔽, n be a non-negative integer and a a root of f in some extension field 𝕂 of 𝔽.

(a) [a] a has multiplicity at least n as a root of f if and only if a + y is a root of f in 𝕂[y]/(y^{n}).

(b) [b] The map ϕ : 𝔽[x]/(f^{n}) → 𝕂[y]/(y^{n}), g + (f^{n}) → f(a + y) + (y^{n}) is a well-defined ringhomomorphsim.

(c) [c] Suppose that 𝕂 = 𝔽(a)(≅ 𝔽[x]/(f)) and that f is irreducible and seperable. Then ϕ is an isomorphism.

Proof: (a) The proof is by induction on n. Write f = g · (x − a) + b with g ∈ 𝕂[x] and b = f(a) ∈ 𝕂. Then f(a + y) = g(a + y)y + b. Hence f(a + y) ∈ (y^{n}) if and only if b = 0 and g(a + y) ∈ (y^{n−1}). By induction this is true if and only if a is a root of f and a is has multiplicity at least n − 1 as a root of g. Thus (a) holds.

(b) Consider the ringhomorphism ψ : 𝔽[x] → 𝕂[y]/(y^{n}), g → g(a + y). Since a has multiplicity at least n as a root of f^{n} we get from (a) that ψ(f^{n}) = 0. Thus (f^{n}) ≤ ker ψ and (b) holds.

(c) Hence 𝔽[x] is a PID, ker ψ = (h) for some h ∈ 𝔽[x]. Since ψ(f^{n}) = 0, h divides f^{n}. Since f is irreducible we can choose h = f^{m} for some m ≤ n. From (a) we have that a has multiplicity at least n as a root of h = f^{m}. As f is seperable( that is f has no double roots) we concclude that m ≤ n and so m = n. Hence ker ψ = (f^{n}) and so ϕ is one to one. Let d = deg f. Then both 𝔽[x]/(f^{n}) and 𝕂[y]/(y^{n}) have dimension nd over 𝔽 and so ϕ is an isomorphism. □

Lemma 4.4.11 [Sf] Let p be a prime and f a non-constant polynomial over 𝔽_{p} with f(0) ≠ 0. Let ξ_{f} = x + (f) ∈ R_{f}, R_{f} = 𝔽_{p}[x]/(f) and S_{f} = S_{ξ_{f}}

(a) [a] Suppose that f is irreducible. Then R_{f} is a field and exactly one of the follwing holds.

1. [a] S_{ξ} = SL_{2}(R_{ξ}) and either p ≠ 3, or ξ^{2} = −1.

2. [b] S_{ξ} ≅ SL_{2}(5), p = 3 and ξ^{2} = −1.

(b) [b] Suppose that f = g^{n} for an irreducible polynomial g. Then (a) [a] R_{f} ≅ R_{g}[y]/(y^{n}).

(b) [b] According to (b:a), view R_{g} has a subfield of R_{f}. Then

S_{f} = (1 + M_{2}^{∘}(R_{f})S_{g}.

(c) [c] Supposse that f = ∏_{i=1}^{m} g_{i}, where g_{i} = f_{i}^{n_{i}} and the f_{i} are pairwise distinct irreducible polynomials in F_{p}[x]. Then

(a) [a] R_{f} ≅ ⊕_{i=1}^{m}R_{g_{i}} ≅ ⊕_{i=1}^{m}R_{f_{i}}[y]/(y^{n_{i}}).

(b) [b] SL_{2}(R_{f}) ≅ \/_{i=1}^{m} SL_{2}(R_{g_{i}}) ≅ \/_{i=1}^{m} SL_{2}(R_{f_{i}}[y]/(y^{n_{i}})).

(c) [c] S_{f} ≅ \/_{i=1}^{n} S_{g_{i}}.

Proof: (a) is Dickson’s Theorem ([?]).

(b:a) follows from 4.4.10(c).

(b:b) Let 𝔽 = R_{g} and R = 𝔽[y]/(y^{n}). Then ξ := ξ_{g} = x + (f) is root of f in 𝔽. Also put μ = y + (y^{n}). Then μ^{n} = 0 and R = 𝔽[μ]. By (b:a), SL_{2}(R_{f}) and SL_{2}(R) are isomorphic. Moreover, (see 4.4.10(b)) we can choose this isomorphism such that

(\begin{matrix}1 & 1 \\ 0 & 1 \end{matrix}) ↦ (\begin{matrix}1 & 1 \\ 0 & 1 \end{matrix})

and

(\begin{matrix}1 & 0 \\ ξ_{f} & 1 \end{matrix}) ↦ (\begin{matrix}1 & 0 \\ ξ + μ & 1 \end{matrix})

So we need to compute the subgroup S := S_{ξ+μ} of SL_{2}(R). Let I = (1 + M_{2}^{∘}(R), H = IS_{ξ} and ρ = ξ + μ. By 4.6.6, S_{g} normalizes I and so H is a subgroup of SL_{2}(R). b_{ξ+μ} = b_{μ}b_{ξ} and b_{μ} ∈ I we see that both a and b_{ξ+μ} are in H. Hence S ≤ H and H = SI. Note that we can choose ϵ ∈ ±1 such that ab_{ϵξ} is a p′ element. Suppose that S ∩ I ≤ Φ(I). Then S/S ∩I′ ≅ S_{ξ} and so abϵρI′ is a p′ element. Thus there exists i ∈ I with ab_{ϵρ} = (ab_{ϵξ})^{i} modulo I′. We will now apply 4.6.6(d). Note that with the notations from 4.6.6, I_{1} = I and by 4.6.6(c), I_{2} = I′. We conclude that μ^{2} divides the trace of ab_{ϵρ} − ab_{ϵξ} = a(b_{ϵμ} − 1). But the latter has trace ϵμ, a contradiction.

Thus S ∩ I ≰ I′. By 4.6.6(d) and 4.4.9 we have that S acts irreducible on I/I′. Thus I = (S ∩ I)I′. Since I is nilpotent this implies I ≤ S and so H = SI = S. Thus (b:b) is proved.

The first part of (c:a) follows from the Chinese Remainder Theorem. The second part follows from (b:a).

(c:b) follows from (c:a) and 4.4.7.

(c:c) Put S = S_{f}, S_{i} = S_{g_{i}} and b_{i} = b_{g_{i}}. We use the isomorphism in (c:b) to identify SL_{2}(R_{f}) with \/_{i=1}^{m} SL_{2}(R_{g_{i}}). Then S ≤ \/_{i=1}^{m} S_{i}. Let A = S_{1} and B = \/_{i=2}^{m} S_{i}. Then AB = SB and by induction on m, AB = SA. Hence A ∩ S ⊴ A, B ∩ S ⊴ B and A/A ∩ S ≅ 4.5. A SECOND PROOF FOR THE SL_{2}(Q)-LEMMA 19

S/(A ∩ S)(B ∩ S) ≅ B/B ∩ S. If A ≤ S we also get that B ≤ S and we are done. So we may assume that S_{i} ≰ S for all i.

Suppose first that n_{i} = 1 for all 1 ≤ i ≤ m.

If A is not perfect then by (a), R_{g_{i}} ≅ 𝔽_{3}. So p = 3 and g_{i} = x ± 1. Moreover,B∞ ≤ S^{∞} and S_{i} ≰ S implies that m = 2 and wthout loss g_{1} = x + 1 and g_{2} = x − 1. Since ab_{1} ∈ S_{1}′ but ab_{2} ∉ S_{2}′, we get that S/(AB′) = AB. Thus S contains a Sylow p-subgroup of AB. Hence A = (A ∩ S)A′ and as A ∩ S is normal in A, A ≤ S, contrary to our assumnptions.

So we may assume that A is perfect and by symmetry that all the S_{i}’s are perfect. Hence by (a) each of the S_{i} are quasisimple. In particular, A ∩ S_{i} ≤ Z(A) and A/A ∩ S is quasimple. Suppose that B ∩ S ≰ Z(B). Then B contains a component or B and thus S_{i} ≤ S for some i, contray to our assumptions.

Thus B ∩ S ≤ Z(B). Since B/B ∩ S ≅ A/A ∩ S is quasisimple we conclude that m = 2. Moreover, the exists an isomorphism ϕ : S_{1}/Z(S_{1}) → S_{2}/Z(S_{2}) which sends a_{1}Z(S_{1}) to a_{2}Z(S_{2}) and b_{1}Z(S_{1}) to b_{2}Z(S_{2}). Note that for p = 3 at most one of the S_{i} are isomorphic to SL_{2}(5) (since g_{i} = x^{2} ∗ 1 if this holds. We conclude that S_{i} = SL_{2}(R_{i}) and that ϕ is induced from a ismorphism of field σ : R_{1} → R_{2}. But then g_{1} = g_{2}, a contradiction.

This completes the analysis of the case n_{i} = 1 for all 1 ≤ i ≤ m. Let T = O_{p}(AB) = O_{p}(A)O_{p}(B). Put h = ∏_{i=1}^{m} f_{i}. Since AB/T = S_{h} we conclude from the preceeding case that AB = ST. Then AT = (S ∩ AT)T and so O^{p}(A) ≤ S. By 4.6.6(d) and 4.4.9, O_{p}(A) = [O_{p}(A), A] = [O_{p}(A), O^{p}(A)] ≤ O^{p}(A) and so O_{p}(A) ≤ S. By symmetry O_{p}(S_{i}) ≤ S for all i and so T ≤ S and AB = ST = S. □

4.5 A second proof for the SL_{2}(q)-lemma

.

Lemma 4.5.1 [ab semisimple] Let 𝔽 a field, V a finite dimensional vector space over 𝔽, and a, b ∈ GL_{𝔽}(V). Suppose that a, b are quadratic and put H = ⟨a, b⟩.

(a) [a] Let λ GL_{𝔽}(V) with [λ, H] = 1. Let R be a commutative of End_{𝔽H}(V) containing 𝔽, λ and λ^{−1}. Suppose that v be an eigenvector with eigenvalue λ for ab on V, that is v^{a}b = λv. Put W := Rv + Rv^{a} and w = v^{a}, then

(a) [a] v^{a} = w and w^{a} = −v + 2w.

(b) [b] v^{b} = 2v − λ^{−1}w and w^{b} = λv

(c) [c] v^{ab} = λv and w^{ab} = 2(λ − 1) + λ^{−1}w

(d) [d] W is H-invariant.

(e) [e] Suppose that λ + λ^{−1} is invertible in R. Put t := w − 2(λ + λ^{−1})^{−1}(λ − 1)v. Then t^{ab} = λ−1t and W = Rv ⊕ Rt.

(b) [b] Suppose that ab is semisimple.

(a) [a] V = C_{V} (ab) ⊕ [V, ab] and both [V, ab] and C_{V} (ab) are H invariant.

(b) [b] If 𝔽 is algebraicly closed then [V, ab] is the direct sun of simple 2-dimensional 𝔽H-submodules.

(c) [c] Suppose that −ab is quadratic and p = char 𝔽 ≠ 2.

(a) [a] If p ≠ 0, then H ≅ SL_{2}(p).

(b) [b] V is the direct sum of isomorphic simple 2-dimensional 𝔽H-module.

(c) [c] There exists basis v_{i}, w_{i}, 1 ≤ i ≤ m for V such that 𝔽v_{i} ⊕ 𝔽w_{i} is H invariant and the matrix for a and b with respect to v_{i}, w_{i} is

(\begin{matrix}1 & 2 \\ 0 & 1 \end{matrix}) (\begin{matrix}1 & 0 \\ −2 & 1 \end{matrix})

Proof: (a) Since a is quadratic, v^{a}−v = (v^{a}−v)^{a} and so w−v = w^{a}−w and w^{a} = −v+2w. Thus (a:a) holds.

Note that λv = v^{ab} = w^{b}. Hence v^{b^{−1}} = λ^{−1}w. Since b is quadratic, [v, b] = −[v, b^{−1}] = v − v^{b^{−1}} = v − λ^{−1}w amd so v^{b} = 2v − λ^{−1}w. Thus (a:b) holds. (a:d) follows immediately form (a:a) and (a:b).

Moreover, v^{ab} = λv and w^{ab} = (v^{a^{2}})^{b} = (−v + 2w)^{b} = (−2v + λ^{−1}w) + 2λv = 2(λ − 1)v + λ^{−1}w and so also (a:c) holds.

That t^{ab} = λ^{−1}t follows by direct calulation from (??). The definition of t implies Rv + Rt = Rv + Rw = W. Let u ∈ Rv ∩ Rt. Then λu = u^{ab} = λ−1u and so (λ − λ^{−1})u = 0. Since by assumption (λ − λ^{−1}) is invertible we have u = 0 and Rv ∩ Rt = 0. Thus (a:e) holds.

(b) We may assume that 𝔽 is algebraicly closed. Since ab is semsimple, V is the direct sum of the eigenspaces for ab on V. Let λ ∈ 𝔽, V_{λ} the corresponding eigenspaces and v ∈ V_{λ}. Since ab is semisimple, (a : c) implies that w ∈ 𝔽v + V_{λ^{−1}}. Thus V_{λ} + V_{λ^{−1}} is H invariant. Thus (b:a) holds.

Suppose now that λ ≠ 1 and v ≠ 0. If w ∈ 𝔽v, then the qudratic action of a and b imply that H centralizes v, a contradiction to v^{ab} = λv ≠ v. Thus 𝔽v + 𝔽w is 2-dimensional and we conclude that (b:b) holds.

(c) Suppose that −ab is quadratic. Then (ab + 1)^{2} = 0 and −1 is the only eigenvalue for ab on V. Let v be a nonzero eigenvector with eigenvalue −1 for ab on V. Then as we saw in the previous paragraph, 𝔽v + 𝔽w is 2-dimensional. Moreover, by (a:c), 𝔽v is the unique 1-dimesional ab-invariant subspace of 𝔽v+𝔽w and we conclude that U := ⟨ker ab+1^{H}⟩ is the direct sum of simple 2-dimesional 𝔽H-submodule. Thus dim V ≥ dim U = 2 dim ker ab + 1 and since ab + 1 = 0 we conclude that V = U. Thus (c:b) holds.

Finally we compute from (a) that the matrices of a and b with respect to the basis v + w, v − w of 𝔽v + 𝔽w is are as given in (c:c). Thus (c:c) and so also (c:a) holds. □

4.6. R-COMPOSTION RINGS 21

4.6 R-compostion rings

Definition 4.6.1 [def:composition ring] Let R be a commutative ring with 1. An Rcomposition ring is pair (A, \overline{·}) such that

(a) [a] A is a ring with R ≤ Z(A) ( and 1_{R} = 1_{A}).

(b) [b] \overline{·} is an R-linear anti-automorphism of A.

(c) [c] N(a) := a\overline{a} ∈ R for all a ∈ A.

(d) [d] tr(a) := a + \overline{a} ∈ R for all a ∈ A.

Lemma 4.6.2 [norm quadratic] Let (A, ·) be an R-composition ring and define f(a, b) = a\overline{b} + b\overline{a} = tr(a\overline{b}).

(a) [a] N : A → R is a multiplicative homomorphism.

(b) [b] N is a quadratic form on A over R with f as associate R-bilinear symmetric form.

Proof: (a) N(ab) = ab\overline{ab} = ab\overline{b}a = a(b\overline{b})α = (b\overline{b})(a\overline{a}) = N a N b.

(b) Note that f(·, ·) is a R-bilinear symmetric form. Also N(a + b) = (a + b)(\overline{a + b}) = a\overline{a} + a\overline{b} + b\overline{a} + b\overline{b} = N a + f(a, b) + N b. □

Lemma 4.6.3 [groups from nilpotent rings] Let A be a ring and N a nilpotent subring of A. Let G = 1 + N. Then G is a nipotent subgroup of A^{∗}. Let k, l ∈ ℤ^{+} and define G_{k} = 1 + N^{k}. Then [G_{k}, G_{l}] ≤ G_{k+l} and for all n ∈ N_{k}, m ∈ N_{l}

[1 + n, 1 + m] ≡ 1 + ⌈n, n⌉ mod G_{k+l+1},

where ⌈n, m⌉ = nm − mn.

Proof: Let n ∈ N. Since N is nilpotent, n^{k} = 0 for some k ∈ ℕ. Thus ∑_{i=0}^{∞}(−n)^{i} is well defined and is an inverse for 1 + n in 1 + N. Also (1 + n)(1 + m) = 1 + (n + m + nm) and so 1 + N is closed under multiplication. Thus G is a group under multiplication. Now let n ∈ N_{k} and m ∈ N_{l}. Put x = 1 + n, y = 1 + m and x = ⌈x, y⌉ = ⌈n, m⌉. Then

[x, y] = x^{−1}y^{−1}xy = x^{−1}y^{−1}(yx + z) = 1 + x^{−1}y^{−1}z

Since x^{−1}y^{−1} ∈ G, x^{−1}y^{−1} = 1 + r for some a ∈ N. Now z ∈ N^{k+l}, az ∈ N^{k+l+1} and [x, y] = 1 + z + az ∈ 1 + z + N^{k+l+1}. Hence (1 + z)^{−1}[x, y] ∈ 1 + N^{k+l+1} = G_{k+l+1}. □

Lemma 4.6.4 [rnf(n)] Let R be a commutative ring and N a nilpotent ideal in R. Let f ∈ R[x] sich that f(0) is invertible. Then the map N → N : n → nf(n) is a bijection.

Proof: Let k be minimal with N^{k} = 0 and put A = N^{k−1}. If k = 0, N = 0 and the lemma holds. Suppose k > 0 and let m ∈ N. By induction on k, there exists a unique n+A ∈ N/A with a := m − nf(n) ∈ A. Let b ∈ A. Since A is an ideal in R, f(n + b) = f(n) + d for some d ∈ A. Also f(n) = f(0) + e with e ∈ N. From NA = 0 we conclude

(n + b)f(n + b) = (n + b))(f(n) + d) = (n + b)f(n) = nf(n) + bf(n) = nf(n) + bf(0) = m − a + f(0)b. Thus n + f(0)^{−1}a is the unique solution of m = nf(n) in N. □

Lemma 4.6.5 [nilpotent and composition] Let (A, \overline{·}) be an R-composition ring and N be nilpotent subring of A with \overline{N} = N. Supppose that 2 is invertible in A. For S ⊆ A let S_{∘} = S ∩ ker tr. Let H = 1 + N and H^{∗} = {h ∈ H | N(h) = 1}.

(a) [a] For each a ∈ A there exist unique r_{a} ∈ R and t_{a} ∈ A^{∘} with a = r_{a} + t_{a}.

(b) [b] For any subring B of A with B = \overline{B}, B = (B ∩ R) ⊕ B_{∘}.

(c) [z] Let a, b ∈ A_{∘}. Then t_{ab} = \frac{1}{2}⌈a, b⌉.

(d) [c] For each n ∈ N_{∘} there exists a unique s_{n} ∈ R ∩ N with 1 + s_{n} + n ∈ H^{∗}. Moreover, s_{n} ∈ R ∩ N^{2}.

(e) [d] The map ϕ : N_{∘} → H^{∗}, n → 1 + s_{n} + n is a bijection with inverse h → t_{h}.

(a) r_{a} = \frac{1}{2} tr a = \frac{1}{2}(a + \overline{a}) and t_{a} = a − r_{a} = \frac{1}{2}(a − \overline{a}).

(b) Let b ∈ B. Since B = \overline{B}, tr b ∈ B. Thus also r_{b} and t_{b} ∈ B. Therefore B = (B ∩ R) + B^{∘}. If b ∈ R ∩ B^{∘}, then 2b = tr b = 0 and so b = 0 since 2 is invertible.

(c) Since a, b ∈ A_{∘} we have \overline{a} = −a and \overline{b} = −b. Thus

t_{ab} = \frac{1}{2}ab − \overline{ab} = \frac{1}{2}(ab − \overline{b}\overline{a}) = \frac{1}{2}(ab − ba).

(d) Let s ∈ R ∩ N and n ∈ N^{∘}. Then N(1 + s + n) = (1 + s)^{2} + (1 + s) tr n + N s = (1 + s)^{2} + N s. Thus 1 + s + n ∈ H^{∗} if and only if

N(s) = −2s(s − \frac{1}{2})

So by 4.6.4, there exists a unqique such s. Since N(s) ∈ N^{2}, we can alreay such an s in R ∩ N^{2}.

(e) Let n ∈ N^{∘}. Since 1 + s_{n} ∈ R, t_{1+s_{n}+n} = n. Let h ∈ H^{∗}. Then h = 1 + m for some m ∈ N. Also h = 1 + (r_{m} + t_{m}) = (1 + r_{m}) + t_{m}. Thus implies t_{h} = t_{m} ∈ N and ϕ(t_{m}) = h. □

Lemma 4.6.6 [trace 0] Let (A, \overline{·}) be an R-composition ring and N be nilpotent subring of A which is generated by N_{∘} as a ring. Suppose also 2 is invertible in A. Then N = \overline{N} and for all k ∈ ℤ^{+}:

(a) [a] N^{k} = (R ∩ N^{k}) ⊕ N_{∘}^{⌈k⌉} and

(b) [b] ϕ(N_{∘}^{⌈k⌉}) = (1 + N^{k}) ∩ H = H^{[k]}

(c) [c] ϕ_{k} : N_{∘}^{⌈k⌉}/N_{∘}^{⌈k+1⌉} → H^{[k]}/H^{[k+1]}, n + N_{∘}^{⌈k+1⌉} → ϕ(n)H^{[k+1]} is a well defined isomorphism.

Proof: Since \overline{a} = −a for all a ∈ N_{∘} we have N = \overline{N} and we can apply 4.6.5.

(a) For k = 1 this follows from 4.6.5(b). Suppose ?? holds for k, we will show that it also holds for k + 1. Let a ∈ N_{∘} and b ∈ N_{∘}^{⌈k⌉}. Thus by 4.6.5(a) and d

ab = r_{ab} + t_{ab} = r_{ab} + ⌈a, b⌉ ∈ D := (R ∩ N^{k+1}) + N_{∘}^{⌈k+1⌉}

So the set of all m ∈ A with mN_{∘}^{⌈k⌉} ≤ D is a subring containing N^{∘} and so NN_{∘}^{⌈k⌉} ≤ D. Similarly N_{∘}^{⌈k⌉}N ≤ D

This implies

N^{k+1} = ((R ∩ N) + N_{∘})((R ∩ N^{k}) + N_{∘}^{⌈k⌉})
= (R ∩ N)(R ∩ N^{k}) + NN_{∘}^{⌈k⌉} + N_{∘}^{⌈k⌉}N
≤ D

As D ≤ N^{k+1} we get N^{k+1} = D By 4.6.5(b) the sum defining D is a direct sum and (a) holds.

(b) Let n ∈ N_{∘}^{⌈k⌉} ≤ N^{k} and put H_{k} = (1 + N^{k}) ∩ H. Then t_{n} ∈ N^{k}, s_{n} ∈ N^{2k} and ϕ(n) ∈ H_{k}. Conversely, m ∈ N^{k} with 1 + m ∈ H. By (b) t_{1+m} = t_{m} ∈ N_{∘}^{⌈k⌉}. Since ϕ(t_{m})1 + m we have ϕ(N_{∘}^{⌈k⌉}) = H_{k}.

We will show by induction on k that H^{[k]}H_{k+1} = H_{k}. For k = 1, this is obvious. So suppose its true for n. By 4.6.3, H^{[k+1]} ≤ [H, H_{k}] ≤ H_{k+1}. Let h ∈ H_{k} and choose n ∈ N_{∘}^{⌈k+1⌉} with h = ϕ(n). Then there exists finitely many n_{i} ∈ N^{∘} and m_{i} ∈ N_{∘}^{⌈k}⌉ with n = ∑_{i}⌈n_{i}, m_{i}⌉. By the inductions assumption ϕ(m_{i}) ∈ h_{i}H_{k+1} for some h_{i} ∈ H^{[k]}. Put d = ∏_{i}[ϕ(n_{i}), h_{i}]. Then d ∈ H^{[k=1]}. By 4.6.3,

[ϕ(n_{i}), h_{i}] ≡ 1 + ⌈ϕ(n_{i}) − 1, h_{i}⌉modH^{k+2}

and so also

[ϕ(n_{i}), h_{i}] ≡ 1 + ⌈ϕ(n_{i}) − 1, h_{i} − 1⌉ mod N^{k+2}

sicne h_{i} − 1 − m_{i} ∈ R and ϕ(n_{i}) − 1 − n_{i} ∈ R. We get

[ϕ(n_{i}), h_{i}] ≡ 1 + ⌈n_{i}, m_{i}⌉ mod N^{k+2}

and so

d ≡ 1 + ∑_{i}[⌈n_{i}, m_{i}] ≡ 1 + n mod N^{k+2}

By 4.6.5(d), s_{n} ∈ N_{k+1}^{2} ∈ N_{k+2}.

h = ϕ(n) = 1 + s_{n} + n ≡ 1 + n ≡ d mod N^{k+2}

Hence also

h ≡ d mod H^{[k+2]}.

This completes the proof that H^{[k]}H_{k+1} = H_{k}. In particular if H^{[k+1]} = H_{k+1} then also H^{[k]} = H_{k}.. Let t ∈ ℕ with N^{t} = 0. Then H^{[t]} ≤ H_{t} = 1. Thus H^{[k]} = H_{k} for all k and (b) holds.

(c) Let n, m ∈ N_{∘}^{⌈k⌉}. Then ϕ(n) ≡ 1 + n mod N^{k+1}. So ϕ(n) ≡ ϕ(m) mod H_{k+1} if and only if n ≡ m mod N^{k+1}. By (a), N^{k+1} ∩ N_{∘}^{⌈k⌉} = N_{∘}^{⌈k+1⌉}. So ϕ(n)H_{k+1} = ϕ(m)H_{k+1} if and only if n + N_{∘}^{⌈k+1⌉} = m + N_{∘}^{⌈k+1⌉}. Thus ϕ_{k} is well defined and one to one.

Also ϕ(n)ϕ(m) ≡ 1 + n + m ≡ ϕ(n + m) mod N^{k+1} and so ϕ_{k}(n)ϕ_{k}(m) = ϕ_{k}(n + m). Thus ϕ_{k} is a homomorphism and (c) is proved. □

4.7 The ring M_{R}(δ)

Let R be a commutative ring and δ ∈ R. Define M = M_{R}(δ) to be the ring with R ≤ Z(M) and generated by R, α and β subject to the relation α^{2} = 0, β^{2} = 0 and δ = αβ + βα.

Lemma 4.7.1 [aba] Let n ∈ N. Then

(a) [a] αβα = δα.

(b) [b] βαβ = δβ

(c) [c] (αβ)^{2} = δαβ.

(d) [d] M is a free R-module with basis 1, α, β, αβ.

Proof: Since αβ = δ − βα and αα = 0, (a) holds. By symmetry (b) holds and (c) follows from (a). From (a)-(b) we conclude that M is spanned by 1, α, β and αβ as an R-module and it is easy to see that (d) holds. □

Since δ = (−β)(−α) + (−α)(b), the opposite ring of M is isomorphic to R and their exists an unique R-linear anti-automorphism \overline{·} : M → M, m → \overline{m} with \overline{α} = −α and \overline{β} = −β. Note that \overline{·} has order two. For m, x, y ∈ N define tr m = m + \overline{m}, N m = m\overline{m} and f(x, y) = x\overline{y} + y\overline{x}.

4.7. THE RING M_{R}(δ) 25

Lemma 4.7.2 [direct sums and m]

(a) [a] Suppose that R = \/_{i∈I} R_{i} and δ = (δ_{i})_{i∈I} ∈ R. Then M_{R}(δ) ≅ \/_{i∈I} M_{R_{i}}(δ_{i}).

(b) [b] Let I be an ideal in R. Then MI is an ideal in M, MI = I ⊕ Iα + Iβ + Iαβ, and M_{(}R/I)(δ + I) ≅ M/MI.

Proof: (a): Put M^{∗} = \/_{i∈i} M_{R_{i}}(δ_{i}), α^{∗} = (α_{i})_{i∈I} and β^{∗} = (β_{i})_{i∈I}. Then R ≤ Z(M^{∗}) and α^{∗}β^{∗} + β^{∗}α^{∗} = δ. Hence there exists an unique R-linear ring homomorphism ϕ : M_{R}(δ) → M^{∗} with ϕ(α) = α^{∗} and ϕ(β) = β^{∗}.

Lemma 4.7.3 [trace and norm]

(a) [a] \overline{1} + 1, \overline{α} = a, \overline{β} = −β and \overline{αβ} = βα = δ − αβ.

(b) [b] tr 1 = 1, tr α = tr β = 0 and tr(αβ) = δ.

(c) [c] N 1 = 1, N α = N b = N(αβ) = 0.

(d) [d] tr : M → R is R linear.

(e) [e] f : M → M → R is a symmetric and R-bilinear.

(f) [f] N : M → R is a quadratic form with f as its associate symmetric form, that is N(x + y) = N(x) + f(x, y) + N(y) for all x, y ∈ M.

(g) [g] N : M → R is a multiplicative homomorphisms.

Proof: (a),(b) and (c) are readily verified. Clearly tr is a Rlinear map from M to M. Since M is spanned by 1, α, β and αβ we conclude from (b) that tr(M) ≤ R and so (d) holds.

Clealry f is symmetic and R-bilinear. Since f(x, y) = tr(x\overline{y}), we conclude from (d) that f takes values in R and so (d) holds.

N(x + y) = (x + y)(\overline{x + y}) = (x + y)(\overline{x} + \overline{y}) = x\overline{x} + x\overline{y} + y\overline{x} + \overline{y}\overline{y} = N(x) + f(x, y) + N(y). Also for r ∈ R, N(rx) = r^{2} N(x). So (c) and (d) imply that N(M) ⊆ R and (e) is proved.

Since N(y) ∈ R ≤ Z(M) we compute N(xy) = (xy)(\overline{xy}) = xy\overline{y}x = x N(y)\overline{x} = (x\overline{x}) N(y) = N(x) N(y). So also (g) is proved. □

Define GL_{R}(δ) the set of invertible elements in M. Let SL_{R}(δ) = {m ∈ M | N(m) = 1}. Note that both GL_{R}(δ) form groups under multiplication. Let R^{∗} be set of invertibel elements in R.

Lemma 4.7.4 [glrd]

(a) [a] Let m ∈ M. Then m ∈ GL_{R}(δ) if and only if N(m) ∈ R^{∗}.

(b) [b] Let m ∈ GL_{R}(δ). Then m^{−1} = N m^{1}\overline{m}.

(c) [c] SL_{R}(δ) is a normal subgroup of GL_{R}(δ) and SL_{R}(δ) = \{m ∈ GL_{R}(δ) | \overline{m} = m^{−1}\}.

Proof: Let m be in M. If m is invertible than N(m) N(m^{−1}) = N(mm^{−1}) = N(1) = 1 and so N(m) is inverible. Suppose now that N(m) is invertible, then N(m^{−1}\overline{m})m = N(m^{−1}) N(m) = 1. Thus m is invertible and (a) and (b) hold. Since SL_{R}(δ) is the kernel of the group homorphism N : GL_{R}(δ) → R^{∗}, the first statement in (c) holds. The second is obvious. □

Lemma 4.7.5 [1+a in sl]

(a) [a] Let m ∈ M. Then N(m) = 1 + tr(m) + N(m). In particular, 1 + a ∈ SL_{R}(δ) if and only if tr(m) = − N(m).

(b) [b] Let x ∈ M with tr(α) = N(a) = 0. Then x^{2} = 0 and 1 + Rx is a subgroups of SL_{R}(δ) isomorphic to (R/ Ann_{R}(x), +). In particular, 1+Ra and 1+Rb are subgroups of SL_{R}(δ) isomorphic to R.

Proof: N(1 + x) = N(1) + tr(y) + N(y) and so (a) holds.

(b): Since tr(x) = 0 we have \overline{x} = −x and so N(x) = −x^{2}. Thus x^{2} = 0. Let r, s ∈ R. Then also N(rx) = 0 = tr(rx) and so by (a), 1 + rx ∈ SL_{R}(δ). Since x^{2} = 0, the map (R, +) → 1 + Rx, r ↦ 1 + rx is an onto groups homomorphism and so (b) holds. □

Put R_{δ} = {r ∈ R | rδ = 0}. For m ∈ M let ϕ_{m} be the ring homomorphism from M to End_{R}(mM) resulting from the right M module mM.

Lemma 4.7.6 [sld=sl2]

(a) [a] αM is a free R-module with basis α, αβ.

(b) [b] The matrices of ϕ_{(}α), ϕ_{a}(β), ϕ_{a}(αβ) and ϕ_{α}(βα) with respect to the basis α αβ are

(\begin{matrix}0 & 0 \\ δ & 0 \end{matrix}), (\begin{matrix}0 & 1 \\ 0 & 0 \end{matrix}) (\begin{matrix}0 & 0 \\ 0 & δ \end{matrix}) and (\begin{matrix}δ & 0 \\ 0 & 0 \end{matrix})

(c) [c] The image of M in End_{R}(aM) = M_{2}(R) consists of all the matrices (\begin{matrix}r & s \\ t & u \end{matrix}), with

r − s ∈ Rδ and t ∈ Rδ.

(d) [d] ker ϕ_{a} = R_{δ}a + R_{δ}βα.

(e) [e] M/αM + βM ≅ R/Rd and αM ∩ βM = R_{δ}αβ.

(f) [f] If δ is invertible, then ϕ_{α} is an isomorphism and M = aM ⊕ bM.

Proof: (a) follows easily from 4.7.1(d). (b) is readily verified. Let m = r_{1} + r_{α}α + r_{β}β + r_{αβ}αβ ∈ M. Then by (b), ϕ_{α}(M) has the matrix

(\begin{matrix}r_{1} & r_{α}δ \\ (_{,}β) & r_{1} + (_{,}αβ)δ \end{matrix})

Thus (c) holds. Moreover, ϕ(m) = 0 if and only if r_{1} = r_{β} = r_{α}δ = r_{αβ}δ = 0 and so (d) is proved.

From (a) and symmetry bM = Rβ + Rβα. Since βα = δ − αβ we get that aM + bM = Rd + Rα + Rβ + Rαβ. Thus M/aM + bM ≅ R/Rd. Also if m ∈ aM ∩ bM, then m = rαβ = sβα = sδ + sαβ for some r, <∈, R>. Thus r = s and sδ = 0 and (e) is proved.

(f) is an easy consequence of the previous statements. □

Let M^{∘} be the ideal in M generated by α and β.

Lemma 4.7.7 [mcirc] M∘ = ℝδ + Rα + Rβ + Rαβ, M/M^{∘} ≅ R/Rδ and M^{∘} is nilpotent if an only if δ is nilpotent.

Proof: This is easily verified. It might be also interesting to observe that the definition of M implies that M/M∘ is the quotient ring of R definition by setting δ = 0. □

Lemma 4.7.8 [ideals] Let I be an ideal in M_{R}(δ) with I ∩ R = 0.

(a) [a] Let m = r_{1} + r_{α}α + r_{β}β + r_{αβ}αβ ∈ I. Then δ^{2}r_{1} = 0, δ^{2}r_{α} = 0, δ^{2}r_{β} = 0 and δ^{3}r_{αβ} = 0. In particular, δ^{3}I = 0.

(b) [b] Suppose that R = 𝔽[δ] for some field 𝔽 ≤ R. Suppose also that there exists n ∈ ℕ with δ^{n+1} = 0 and that n is minimal with this property.

Proof:

1^{∘} [1] If r ∈ R with rα ∈ I or rβ ∈ I then δr = 0.

From rα ∈ I we get rαβ ∈ I and rβα = 0. Hence also rδ = r(αβ + βα) ∈ I. From R ∩ I = 0 we conclude that rδ = 0.

Let m = r_{1} + r_{α}α + r_{β}β + r_{αβ}αβ ∈ I.

2^{∘} [2] δ^{2}r_{α} = 0 = δ^{2}r_{β}.

αmα = r_{β}αβα = δr_{β}β. So by (1^{∘}), δ^{2}r_{β} = 0. Also βmβ = δr_{α}α and by (1^{∘}) δ^{2}r_{α} = 0.

3^{∘} [3] δ^{2}r_{1} = 0

αm = r_{1}α + r_{β}αβ ∈ I so (3^{∘}) follows from (2^{∘}) applied to αm in place of m.

4^{∘} [4] δ^{3}r_{αβ} = 0

mα = r_{1}α + (_{,}b)βα + r_{αβ}δα = r_{β}δ + (r_{1} + δr_{αβ})α − r_{β}αβ. So by (2^{∘}) applied to mα and using (3^{∘}) we have 0 = δ^{2}(r_{1} + δr_{αβ}) = δ^{3}r_{αβ}. □

What to do next: assume R = 𝔽[δ], zentral series of for M^{∘} ( maybe only if δ nilpotent). ideals in M, subgroup H generated by 1 + α, 1 + β. Prove some lemma if H is nilpotent. For example usually there exists 1 ≠ h ∈ Z(H) with [V, h, H] = 0.

Root Systems

5.1 Root Systems

Definition 5.1.1 [def:root system] A root system is a set Φ together with a vectorspace V_{Φ} over ℚ and a non-degenerate, positive definite, symmetric form ( , ) on V_{Φ} such that (a) [RS1] Φ is a finite set of non zero vectors in V_{Φ} and Φ spans V_{Φ}.

(b) [RS2] For all α, β ∈ Φ, <α, β > := 2\frac{(α,β)}{(β,β)} ∈ ℤ.

(c) [RS3] For all α, β ∈ Φ, ω_{α}(β) ∈ Φ, where

ω_{α} : V_{Φ} → V_{Φ}, v → v − <v, α>α

is the reflection associated to α.

(d) [RS4] If α, β ∈ Φ are linearly dependent over ℚ then α = ±β. Let Φ be a root system. The elements of Φ are called roots. Put W := ⟨ω_{α} | α ∈ Φ⟩ ≤

O(V_{ℚ}, ( , )). Note that (RS3) just says that Φ is invariant under W. Since Φ is finite and spans V_{ℚ}, W is finite. Lemma 5.1.2 [dual root system] Let Φ be a root system. For α ∈ Φ define α^{∗} := \frac{2}{(α,α)}α. Let Φ^{∗} = {α^{∗} | α ∈ Φ} Then for all α, β ∈ Φ.

(a) [a] <α, β > = (α, β^{∗}).

(b) [b] <α, β > = <β^{∗}, α^{∗} >.

(c) [c] ω_{α} = ω_{α^{∗}}

(d) [d] ω_{α^{∗}}(β^{∗}) = (ω_{α}(β))^{∗}

(e) [e] Φ^{∗} ( together with V_{Φ} and ( , )) is a root system.

Proof: (a)-(d) are readily verified and (e) follows from (c) and (d). □

Definition 5.1.3 [basis] Let Φ be a root system. A basis for Φ is a linearly independent subset Π of Φ such that Φ = Φ^{+} ∪. Φ^{−} where Φ^{+} = Φ ∩ ℚ^{+}Π and Φ^{−} = Φ ∩ ℚ^{−}Π = −Φ^{+}.

Lemma 5.1.4 [alpha string] Let Φ be a root system and α, β ∈ P hi with α ≠ ±β. Then

(a) [a] There exists non-negative integers p_{−}, p_{+} such that for i ∈ ℤ, β + iα ∈ Φ if and only if −p ≤ i ≤ q.

(b) [b] Put ϵ = sgn(α, β), then one of the following holds.

1. [a] p_{−ϵ} = |<β, α>| and p_{ϵ} = 0.

2. [b] p_{−ϵ} = |<β, α>| + 1 and p_{ϵ} = 1.

(c) [c] (b) holds iff α and β are not long and ℤ{α, β} ∩ Φ is a root system of type B_{2} or G_{2}.

Proof: See [?] □

Lemma 5.1.5 [linear combinations of roots] Let Φ be a root system, A ⊆ Φ and for a ∈ A let n_{a} ∈ ℚ. Put ϕ = ∑_{a∈A} n_{a}a and suppose that ϕ ∈ Φ. Then there exists b ∈ A with n_{b}(ϕ, b) > 0. Moreover, for any such b either ϕ = ±b or ϕ − sgn(n_{a})a ∈ Φ.

Proof: Note that 0 < (ϕϕ) = ∑_{a∈A} n_{a}(ϕ, a). Hence there exists b ∈ A with n_{b}(ϕ, b) > 0.

Given any such b with ϕ ≠ ±b. Then ϵ := sgn(ϕ, b) = sgn n_{n}. Also |<ϕ, a>| ≥ 1 and 5.1.4 implies that ϕ − ϵa ∈ Φ. □

Lemma 5.1.6 [existence of simple roots] Let Φ be a root system.

(a) [a] Φ has basis.

(b) [b] Any two basis are conjugate under W.

(c) [c] If Π is any basis, than Φ^{+} = Φ ∩ ℕ^{+}Π.

Definition 5.1.7 [def: long] A root α in a root system Φ is called long ( short) if (α, α) ≥ (β, β) ((α, α) ≤ (β, β)) for all β ∈ Φ.

Note here that if all roots in Φ have the same length, then all roots are long and short.

Lemma 5.1.8 [dual basis] Let Φ be a root system with basis Π. Then Π^{∗} := {α^{∗} | α ∈ Π} is a basis for Φ^{∗}.

5.2. ROOT SUBSYSTEMS 31 Proof: Since for all α ∈ Φ, α and α^{∗} only differ by a positive rational factor, ℚ^{+}Π = ℚ^{+}Π^{∗} and α ∈ Q^{+}Π if and only if α^{∗} ∈ ℚ^{+}Π^{∗}. Hence the lemma follows from the definition of a basis. □

Λ := {λ ∈ V_{Φ} | (λ, α^{∗}) ∈ ℤ, ∀α^{∗} ∈ Φ^{∗}}.

The elements in Λ are the integral weights.

The elements in

{λ ∈ V_{Φ} | (λ, α^{∗}) > 0, ∀α^{∗} ∈ Φ^{∗}}

are called dominant weights.

Note that by (RS2) Φ ⊆ λ. Let (λ_{α} | α ∈ Π) be the basis of V_{ℚ} dual to Π^{∗} so (λ_{α}, β^{∗}) = \begin{cases}1 & if α = β \\ 0 & if α ≠ β \end{cases}. Then (λ_{α} | α ∈ Π) is a ℤ basis for Λ.

For α, β ∈ Φ let r, s ∈ ℕ be maximal such that

β − rα, β − (r − 1)α, …, β − α, β, β + α, … β + sα)

all are roots. We call this sequence of roots the α-string through β. r will be denoted by r_{αβ} and s by s_{αβ}.

5.2 Root Subsystems

Definition 5.2.1 [def:root subsystem] Let Φ be a root system and Ψ ⊆ Φ.

(a) [a] Ψ is a root subsystem of Φ if (Ψ, ℚΨ) is a root system.

(b) [b] Let R be a subring of Q. Then Ψ is called R-closed if Ψ = Φ ∩ RΨ. Lemma 5.2.2 [root subsystems] Let Φ be a root system and Ψ ⊆ Φ. Then

(a) [a] Ψ is a root subsystem iff Φ is invariant under W (Ψ) := ⟨ω_{ψ} | ψ ∈ Ψ⟩.

(b) [b] Ψ is ℤ-closed iff −Ψ ⊆ Ψ and α + β ∈ Ψ for all α, β ∈ Ψ with α + β ∈ Φ.

(c) [c] If Ψ is ℤ-closed, then Ψ is a root subsystem. If Ψ is ℚ-closed then Ψ is ℤ closed.

(d) [d] Ψ is a root subsystem if and only if Ψ^{∗} is a root subsystem of Φ^{∗}. Ψ is ℚ-closed if and only if Ψ^{∗} is ℚ closed.

(e) [e] If Ψ is a root subsystem and all roots in Ψ are long, then Ψ is ℤ-closed.

Proof: (a): Note that (Ψ, ℚΨ) fulfills (RS1),(RS2) and (RS4). Hence Ψ is a root subsystem iff ω_{a}(b) ∈ Ψ for all α, β ∈ Ψ. This is the case iff Ψ is invariant W (Ψ).

(b) One direction is obvious. Suppose now that −α ∈ Ψ for all α ∈ Ψ, and α + β ∈ Ψ for all α, β ∈ Ψ with α + β ∈ Φ. Let ϕ = ∑_{ψ∈Ψ} n_{ψ}ψ ∈ ℤΨ ∩ Φ, where n_{ψ} ∈ ℤ. We show by induction on ∑ |n_{ψ}| that ϕ ∈ Ψ. By 5.1.5 we can choose ψ ∈ Ψ with n_{ψ}(ϕ, ψ) > 0. If ϕ = ±ψ, then ϕ ∈ Ψ. So suppose ϕ ≠ ±ψ. Then by 5.1.5 α := ϕ − sgn n_{ψ}ψ ∈ Φ. By induction α ∈ Ψ and so also ϕ = α + sgn n_{ψ}ψ ∈ Φ.

(c), (d) and (e) are readily verified. □

Given a connected root system Φ with two different root lengths. Then the short roots form a root subsystem which is not ℤ-closed. And the long roots form a subsystem which is ℤ-closed but not ℚ-closed.

From the affine diagram of E_{8} we see that E_{8} has a root subsystem D_{8}. Since D_{8} and E_{8} both have rank 8, the ℚ closure of D_{8} is E_{8}. On the otherhand D_{8} is ℤ-closed and contains the ℚ-closure of any two of its elements.

The ℤ closure of Φ_{long} × Φ_{short} in Φ × Φ is Φ_{long} × Φ. The ℚ closure is Φ × Φ.

Let n, m be integers with n ≥ 2 and m ≥ 1. Then B_{n+m} has a subsystem B_{n} × B_{m}. The long roots in B_{n} form a subsystem D_{n} and the short roots in B_{m} a subsystem A_{1}^{m}. Then ℤ-closure of D_{n} × A_{1}^{m} is D_{n} × B_{m}, while the ℚ-closure is B_{n_{m}}.

Now let Ψ be a connected root subsystem of Φ. We claim that either Ψ is ℤ closed or that the ℤ closure of Ψ is ℚ-closed. So suppose that Ψ is not ℤ closed. Then Ψ contains roots which are not long in Φ. Without loss Φ is the ℚ-closure of Ψ. Then Φ is connected. Since Ψ is connected ℚΨ_{short} = ℚΨ = ℚΦ. Thus Ψ_{short} has the same rank as Φ_{short}. Since Φ_{short} is of type A_{n}, D_{n} or A_{1}^{m} we conclude that Φ_{short} = Φ_{short}.

Thus Φ ⊆ ℤΦ_{short} ≤ ℤΨ and Φ is the ℤ closure of Ψ.

Lemma 5.2.3 [closure in rank 2] Let Φ be a root system and α, β ∈ Φ with (α, β) ≠ 0. Then

(a) [a] α ∈ ℚ⟨β, ω_{β}(α)⟩.

(b) [b] If α is not shorter than β then α ∈ ℤ⟨β, ω_{β}(α)⟩.

(c) [c] If α and β have the same length, the α ∈ ⟨β, ω_{β}(α)⟩.

Proof: Readily verified, for example by inspection of the rank 2 root system ℚ⟨α, β⟩ □

Lemma 5.2.4 [z closure of phishort] Let Φ be a connected root system. Then Φ is the ℤ closure of Φ_{short}.

Proof: Let α be a long root and choose a short root β with (α, β) ≠ 0. Then by 5.2.3(b), α is in the ℤ closure of β and ω_{α}(β). □

Lemma 5.2.5 [covering root systems] Let Φ be a root system.

(a) [a] Let Ψ be a root subsystem of Φ, α ∈ Ψ and β ∈ Φ \ Ψ. Then ω_{α}(β) ∉ Ψ.

(b) [b] Suppose that Φ ⊆ X ∪ Y where X and Y are proper root subsytems of Φ. If X is ℚ closed or both X and Y are ℤ-closed, then Φ is disconnected.

(c) [c] Suppose that Φ is connected and α, β ∈ Φ. Then there exists γ ∈ Φ such that γ is neither perpendicular to α nor to β. In particular α and β are contained in a connected subroot system of rank at most 3.

Proof: (a) If ω_{α}(β) ∈ Ψ, then β = ω_{α}(ω_{α}(β)) ∈ Ψ a contradiction.

(b) Choose X and Y as in (b) with |X ∩ Y | minimal. Let A = Φ \ Y, B = Φ \ X and C = Φ∩X∩Y. Let a ∈ A and b ∈ B. Suppose for contradiction that (a, b) ≠ 0. Then by (a), ω_{b}(a) ∉ Y and so ω_{b}(a) ∈ X. If X is ℚ closed, then by 5.2.3(a), b ∈ ℚ⟨a, ω_{b}(a)⟩ ∩ Φ ⊆ X, a contradiction. Thus X is not ℚ closed and so by assumption, X and Y are ℤ-closed. Hence we may assume that b is not shorter than a. Thus by 5.2.3(b) b ∈ ℤ⟨a, ω_{b}(a)⟩ ∩ Φ ⊆ X, again a contradiction.

Thus A ⊥ B. Let X̃ = B^{⊥} ∩ X and Ỹ = A^{⊥} ∩ Y. Then X̃ and Ỹ are subsystems. Moreover, either X̃ is ℚ-closed or both X̃ and Ỹ are ℤ-closed. Also A ⊆ X and B ⊆ Y.

We claim that Φ = X̃ ∪ Ỹ, that is that C ⊆ X̃ ∪ Ỹ. Let c ∈ C and suppose that c ∉ X̃. Then (c, a) ≠ 0 for some a ∈ A. Since c ∈ Y and a is not, (a) implies ω_{c}(a) = a− < a, c > c ∈ A. Thus ω_{c}(a) and a both perpendicular to B. Hence c ⊥ B and c ∈ Ỹ.

Thus C = X̃ ∪ Ỹ. The minimal choice of X ∩ Y implies X ∩ Y = X̃ ∩ Ỹ. Hence C ⊆ X̃ ∩ Ỹ ≤ A^{⊥} ∩ B^{⊥}. Since also A ⊥ B, A ∪ B ∪ C is a decompostion of Φ into pairwise orthorgonal subsets. Thus Φ is disconnected and (b) is proved.

(c) By (a) there exists γ ∈ Φ \ (α^{⊥} ∪ β^{⊥}). Also Φ ∩ ℚ⟨α, β, γ⟩ is connected root system of rank at most 3. Thus (c) holds. □

Lemma 5.2.6 [generation by non perpendicular roots] Let Φ be a connected root system, and α a short root.

(a) [a] Then ℚΦ = ℚΦ_{long} = ℚΦ_{Short}.

(b) [b] Let Ψ be the root subsystem generated by α and the long roots, then Ψ = Φ. Comment:false for F_{4}

(c) [c] Let Ψ be the root subsystem generated by α and the long roots which are not perpendicular to α. If Φ is not of type C_{n}, n ≥ 3 or F_{4}, then Ψ = Φ. Comment:maybe false for F_{4} – indeed, it is false: if α = e_{1}, then we obtain a subsystem of type B_{4}

Proof:

(a) Let {i, j} = {long, short}. Since Φ is connected there exists αΦ_{i} and β ∈ Φ_{j} with <α, β > ≠ 0. If b ∉ ℚΦ_{i} then 5.2.5(a) implies ω_{β}(α) ∉ ℚΦ_{i} a contradiction. Thus β ∈ ℚΦ_{i} and the transitivity of W_{Φ} on Φ_{j} implies ℚΦ_{j} ⊆ ℚΦ_{i}.

For (b) and (c) note that if Φ has rank two, then every subsystem containing a long and a short system equals Φ (Comment:false for G_{2}, it contains a A_{1}(long) × A_{1}(short) Also ℚΦ_{long} = ℚΦ and so Ψ contains a long root. So we may assume that Φ has rank at least two. Let Σ be the subsystem generated by the long root.

(b) Without loss α is the highest short root. Let β be any short root. By (a) there exists a long root δ with <δ, β > < 0. Then ω_{δ}(β) has larger height than β Comment:this is false if β is negative and so by induction ω_{δ}(β) ∈ Ψ. Hence also β ∈ Ψ.

(c) We may assume that Φ is not of type C_{n} or F_{4}. Thus Σ is connected. By definition of Ψ, Σ = (Σ ∩ Ψ) ∪ (Σ ∩ α^{⊥}). Since Σ ∩ α^{⊥} is closed in Σ, 5.2.5(b) implies that Σ ⊆ Ψ. So (c) follows from (b). □

Lemma 5.2.7 [height induction] Let Φ be a root system with simple roots Π and α ∈ Φ^{+} \ Π. Then there exists β ∈ Π and γ, δ ∈ Φ^{+} with α = ω_{β}(γ) = β + δ and (β, γ) < 0. Comment:maybe combine with 5.1.5

Proof: Since α is a positive linear combination of Π and since (α, α) > 0 there exists β ∈ Π with (α, β) > 0. Put γ = ω_{β}(α) and δ = α−β. Then (β, γ) = (ω_{β}(β), ω_{β}(γ)) = (−β, α) < 0. Since γ = α − <α, β >β and α is not a multiple of β, γ is positive. Also δ is on the β-string from γ to α. So δ is a root and δ is positive. □

Lemma 5.2.8 [perp of weight] Let Φ be a root system with simple roots Π and λ a dominant integral weight for Π. Then Π ∩ λ^{⊥} is a system of simple roots for Φ ∩ λ^{⊥}.

Proof: Let Ψ be the root subsytem generated by Π ∩ λ^{⊥}. It suffices to show that Ψ = Φ ∩ λ^{⊥}. Let α ∈ Φ^{+} with λ(α) = 0 (that is α ∈ λ^{⊥}). We show by induction on ht α that α ∈ Ψ. If α has height 1, then α ∈ Π and so α ∈ Ψ. If α ahs height larger than 1 then α ∉ Π By 5.2.7, there exists β, γ ∈ Φ^{+} with α = ω_{β}(γ) and m := <γ, β > > 0. Then α = mβ + γ. Since λ is dominant, both λ(β) and λ(γ) are non-negative. Thus λ(β) = 0 = λ(γ). By induction, both β and γ are in Ψ and so also α ∈ Ψ. □

5.3 Quadratic weights

Definition 5.3.1 [def: quadratic weight] An integral weight λ on the root system Φ is called quadratic provided that 1 ≤ (α, λ) ≤ 1 for all short roots α ∈ Φ_{short}.

Theorem 5.3.2 [quadratic weights] Let Φ be a connected root system and λ a non-zero dominant integral weight on Φ. Let t ∈ {long, short} and α_{t} the heighest t-root in Φ. Then the following are equivalent.

(a) [d] (α_{t}, λ) ≤ 1.

(b) [e] λ = λ_{β} for some root β ∈ Π with n_{β}^{t} = 1, where n_{γ}^{t} for γ ∈ Π is defined by α_{t} = ∑_{γ∈Π} n_{γ}^{t}γ.

(c) [d+] λ = λ_{β} for some root β ∈ Π such that β is long if t = long and such that Φ_{t} is contained in the root subsystem generated by Φ ∩ λ^{⊥} and α

(d) [g] One of the following holds: Comment:labeling of roots needs to be introduced Comment:needs to be updated, inparticular

1. [1] Φ = A_{n} and λ = λ_{i} for some 1 ≤ i ≤ n.

2. [2] Φ = B_{n} and λ = λ_{1} or λ_{n}.

3. [3] Φ = C_{n} and λ = λ_{i} for some 1 ≤ i ≤ n.

4. [4] Φ = D_{n} and λ = λ_{1}, λ_{n−1} or λ_{n}.

5. [5] Φ = E_{6} and λ = λ_{1} or λ_{6}.

6. [6] Φ = E_{7} and λ = λ_{1}.

7. [7] Φ = E_{8}: No such module.

8. [8] Φ = G_{2} and λ = λ_{1}.

9. [9] Φ = F_{4} and λ = λ_{1}

(e) [f] λ is the (unique ) minimal (with respect to ≺) dominant weight in λ^{W (Φ)} + Φ_{t}^{∗}. Comment:needs some work, maybe make extra lemma

Proof: Put α = α_{t}.

(a)⟺ (b): Let λ = ∑_{β∈Π} m_{β}λ_{β}. Then each m_{β} is a non-negative integer and each n_{γ}^{t}

is a positive integer. Also (α, λ) = ∑_{β∈Π} m_{β}n_{β}^{t} and so (a) and (b) are equivalent. (b)⟹ (c): Let Ψ be the root subsystem generated by Φ ∩ λ^{⊥} and α.

Let δ ∈ Φ_{t}^{+}. We need to show that δ ∈ Ψ. Since (b) implies (a), (α, λ) = 1 and so (δ, λ) = 0 or (δ, λ) = 1.

Suppose that (δ, λ) = 0, then δ ∈ Φ ∩ λ^{⊥} and so δ ∈ Ψ.

Suppose next that (δ, λ) = 1 and that δ is not perpendicular to α. Since (α, δ) ≥ 0 and α_{t} and δ have the same length we conclude that <α, δ > = 1 and so ω_{α}(δ) = δ − α. Also (δ, λ) = (α, λ) and hence δ − α ∈ Φ ∩ λ^{⊥} ⊆ Ψ. Thus δ = ω_{α}(δ − α) ∈ ψ.

Suppose finally that (δ, λ) = 1 and δ is perpendicular to α. By 5.2.5(c), there exists γ ∈ Φ such that γ is neither perpendicular to α nor to δ. If γ ∈ Φ_{t} then by the previous paragraph both γ and ω_{γ}(δ) are in Ψ, so also δ ∈ Ψ. So suppose that γ ∉ Φ_{t}. Since the diagram of (α, γ, δ) is not sperical, we see that α, γ, δ are not linear independent. Let Δ be the root sytem generated by α, γ and δ. Then Δ has rank two, is connected and has a pair of perpendicular roots of the same length. Δ is of type B_{2}. Put r = |<γ, α>| and μ = α−rδ. In δ we see that μ ∈ Φ, and δ = ω_{μ}(α). Since (δ, λ) = (μ, λ) we have μ ∈ Φ∩λ^{⊥}. Hence δ = ω_{μ}(α) ∈ Ψ.

It remains to show that β is long if t is not short. So suppose that Φ has two distinct roots length, t = long and β is a short. The there exists δ ∈ P hi_{long}^{+} with (δ, β) < 0. Put then −<α, β > ≥ 2 and so (ω_{β}(δ), λ) = (δ, λ) − <α, β > ≥ 2, a contradiction.

(c)⟹ (a): Let δ ∈ Φ_{t}^{+} with (δ, λ) ≠ 0. By (c), δ = γ + nα for some n ∈ ℤ and γ ∈ λ^{⊥}.

Since (δ, λ) ≥ 0, n > 0. Since α is the highest t- root, the β-coefficent of δ is not larger than the β coefficent of α. Thus n ≤ 1 and so n = 1. Hence (δ, λ) = (α, δ). It remains to show that there exists δ ∈ Φ^{t} with (δ, λ) = 1. If β ∈ Φ_{t} we can choose δ = β. So we may assume that β ∉ Φ_{t}. The assumptions of (c) imply that t = short. Thus β is long. Let Ψ be the ℚ closure of Φ_{short} ∩ λ^{⊥}. Then Π_{short} ∈ Ψ and all roots in λ^{⊥} \ Ψ are perpendicular to Ψ. Since Φ is connected we conclude that β is not perpendicular to Ψ. Thus there exists a γ ∈ Φ_{short} ∩ λ^{p}erp with (γ, β) < 0. Put δ = ω_{β}(γ). Since β is long, δ = γ + β. Thus (δ, λ) = 1 and we are done.

(b)⟺ (d): Follows from a glance at the highest t- root of Φ (??).

(a) ⟹ (e): Suppose that μ is a dominant weight with μ ≺ λ and μ ∈ λ + ℤΦ_{t}^{∗}. Put

δ = λ − μ. Then δ ∈ ℕΦ^{∗} ∩ ℤΦ_{t}^{∗}. In particular (α, δ) ≥ 0. Since (α, λ) = 1 we conclude that (α, μ) = 1 and (α, δ) = 0. Hence for all ϕ ∈ Φ_{t}^{+} we have (ϕ, μ) ∈ {0, 1}. It follows that |(ϕ, δ)| ≤ 1. Therefore there exists w ∈ W (Φ) such that ρ := δ^{ϕ} is an dominant integral weight with (α, ρ) = 1. Also ρ ∈ ℤΦ_{t}^{∗}. Using (d) we can express the restriction of ρ to ℤΦ_{t} as rational linear combination of a basis for Φ_{t}^{∗}. Since not all the coefficents are integers we obtain a contradiction. Comment:make an explicit list of the quadratic weights as linear combination of Π^{∗}. Or find a better proof

(e)⟹ (a): Let λ be a dominant weight such that λ is minimal under the dominant

weights in λ + ℤΦ_{t}^{∗}. We will show that (α, λ) ≤ 1 and that λ is unique in λ + ℤΦ_{t}^{∗}. Consider first the case where all roots in Φ have the same length. Suppose that (α, λ) ≥ 2 and choose δ ∈ Φ_{t} = Φ of minimal height with respect to

(δ, λ) ≥ 2. By minimality of λ, λ − δ^{∗} is not dominant and so there exists β ∈ Π such that (β, λ − δ^{∗}) < 0. Thus (β, λ) < (β, δ^{∗}). Since β and δ^{∗} have equal length we conclude that (β, δ^{∗}) + 1 and (β, λ) = 0. Thus δ − β is a root and (δ − β, λ) = (δ, λ) ≥ 2, contradicting the minimal height of δ.

Hence (α, λ) ≤ 1.

Suppose next that μ ∈ λ + ℤΦ^{∗} is also minimal with respect to being dominant. Then also (α, μ) ≤ 1. Put δ = λ − μ. Then −1 ≤ (ϕ, δ) ≤ 1 for all ϕ ∈ Φ^{=} and so ϕ is a quadratic weight. Since δ ∈ ℤΦ^{∗} we conclude from the “ (a) ⟹ (e):” step that δ = 0. Thus λ = μ

and the one root length case is completed.

Now consider the case where Φ has roots of two different lengths. Let {r, t} = {long, short}. Put Σ = ⋃ Π_{t}^{W (Π_{r})}. Then Σ is a basis for Φ_{t}. and Σ is invariant under W (Π_{r})). Note that W (Φ_{t})) acts trivial on Λ(Φ)/ℤΦ_{t}^{∗} and W (Φ) = W (Φ_{t})W (Π_{r})). So λ^{W (Φ)} + ℤΦ_{t}^{∗} = λ^{W (Π_{r})} + ℤΦ_{t}^{∗}. For μ ∈ Λ(Φ) let \overline{μ} the restriction of μ to ℤΦ_{τ}. Then \overline{l} is a minimal dominant integral weight in \overline{l} + ℤ\overline{Φ_{t}^{∗}}. Thus by the one root length case (α, λ) = 1. Let μ be any minimal dominant weight in λ^{W (Φ)} + ℤΦ_{t}^{∗} and pick w ∈ W (Π_{r}) with μ ∈ λ^{w} + ℤΦ_{t}^{∗}. Since Σ is invariant under w, \overline{l}^{w} and \overline{μ} are minimal dominant weights in \overline{μ} + ℤ\overline{Φ}_{t}^{∗}. Thus by the one root length case, \overline{μ} = \overline{l}^{w}. Thus μ − λ^{w} ∈ Φ_{t}^{⊥} and so μ = λ^{w}. Since every W (Φ) orbit on Λ(Φ) contains a unique dominant weight we conclude that μ = λ. □

5.4 Subdiagrams

Definition 5.4.1 [def:diagram] Let \underline{α} = (α_{1}, …, α_{n}) be a tuple of roots. Then the diagram of \underline{α} is the matrix (<α_{i}, α_{j} >). If Φ is a connected root system and \underline{α} = (α_{1}, …, α_{n}) is basis for Φ in the standard order, then the diagram of \underline{α} is called an Φ-diagram.

Note that the Φ-diagram is just the Cartan matrix of Φ.

Lemma 5.4.2 [conjugation to pi] Let Φ be a root system with simple roots Π and λ a dominant integral weight for Φ. Let ϕ ∈ Φ with λ(ϕ) = 1. Let ω ∈ W_{Φ∩λ^{⊥}} with ht ϕ^{ω} minimal. Then there exist β_{1}, β_{2}, …, β_{k} ∈ Π such that

(a) [a] ϕ^{ω} = β_{1} + β_{2} + … + β_{k}.

(b) [b] If k > 1 then (β_{1}, β_{2}, …, β_{k}) has B_{k} or G_{2} diagram.

(c) [c] λ(β_{1}) = 1 and λ(β_{i}) = 0 for 2 ≤ i ≤ k.

Proof: Put Ψ = Φ ∩ λ^{⊥} and α := ϕ^{ω}. If α ∈ Π, then the lemma holds with k = 1 and β_{1} = α. So suppose that α ∉ Φ. Let β and γ be as in 5.2.7 and put m = (α, β^{∗}). Then α = mβ + γ. Suppose that λ(β) = 0. Then ω_{β} ∈ W_{Ψ} and γ = ω_{β}(α) has smaller height than α, a contradiction to the choice of α. Thus λ(β) ≠ 0. Since λ is dominant integral and λ(α) = 1 we conclude that m = 1, λ(β) = 1 and λ(γ) = 0. Since m = 1, (α, α) ≤ (β, β) and α = β + γ.

Suppose that (α, α) = (β, β). Then ω_{γ}(α) = β and β has smaller height than α, contradiction the choice of α.

Suppose that 2(α, α) = (β, β). Since β ∈ Π \ Ψ, β^{∗} is a dominant integral weight on Ψ. Moreover by 5.2.8, Π ∩ Ψ is basis for Ψ. Also (γ, β^{∗}) = m = 1. Let w ∈ W_{Ψ∩β^{⊥}}. Then w ∈ W_{Ψ} and α^{w} = β + γ^{w}. Thus the choice of α implies that ht(γ) ≤ ht(γ^{w}). So by induction on Π there exists b_{2}, …, b_{k} ∈ Π ∩ Ψ such that

• [e] γ = β_{2} + … + β_{k}.

• [f] (β_{2}, β_{2}, …, β_{k}) has B_{k−1} or G_{2} diagram.

• [g] (β_{2}, β) = 1 and (β_{i}, β) = 0 for 3 ≤ i ≤ k

Since no connected component of Φhas roots of three different lengths, (β_{2}, …, β_{k}) cannot have G_{2}-diagram. Put β_{1} = β. Then clearly (a) and (c) holds. If k ≤ 3, then both β_{1} and β_{2} are long and so (β_{1}, …, β_{k}) has B_{k}-diagram. If k = 2, then β_{2} = γ is short and (β_{1}, β_{2}) has B_{2}-diagram. Thus in any case (b) holds.

Suppose finally that 3(α, α) = (β, β). Then Φ is of type G_{2} and Π = {β, γ}. Thus the lemma holds with β_{1} = β and β_{2} = γ. □

Lemma 5.4.3 [an strings] Let Φ be a connected root system and (α_{0}, α_{1}, …, α_{k}) an ordered tuple of roots in Φ with diagram X_{k+1} where X ∈ {A, B, G}. Suppose that α_{0} is long. Then there exists an integer m ≥ k, roots β_{0}, β_{1}, …, β_{m} in Φ and w ∈ W (Φ) such that

(a) [z] α_{i}^{w} = β_{i} for all 0 ≤ i < k and a_{k}^{w} = β_{k} + … + β_{m}.

(b) [a] β_{0} = −α_{long}.

(c) [b] β_{i} ∈ Π for all 1 ≤ i ≤ m.

(d) [e] One of the following holds;

1. [a] X = A, k = m, a_{k}^{w} = b_{k} and (β_{0}, β_{1}, …, β_{k}) has diagram A_{k+1}.

2. [b] X ≠ A and (β_{0}, β_{1}, …, β_{m}) has diagram X_{m+1}.

(e) [c] Put Ψ_{i} = (Φ ∩ α_{0}^{⊥} ∩ … ∩ α_{i}^{⊥})^{w}. Then for all 0 ≤ i ≤ k, Ψ_{i} ∩ Π is a system of simple roots for Ψ_{i}.

Proof: By induction on k. Suppose first that k = 0. Since 𝔸_{o} is long and Φ is connected α_{0}^{w} = −α_{long} for some w ∈ W. Put m = 0 and β_{0} = −α_{long}. Then clearly (b) to (d) holds. Note also that α_{long}^{∗} induces a dominant integral weight on Π and so (e) follows from 5.2.8.

Suppose now that k ≤ 1 and that the statement has been proved for k − 1. Since (α_{0}, …, a_{k−1}) has A_{k} diagram we conclude that exists v ∈ W_{Φ} and β_{0}, …, β_{k−1} in Φ such that α_{i}^{v} = β_{i} for all 0 ≤ i < k, β_{0} = −α_{long} and (e) holds for all i < k. Put Ψ = Ψ_{k−2} if k ≥ 2 and Ψ = Φ if k = 1. Also put α = a_{k}^{v} and β = β_{k−1}. Note that β ∉ Ψ and ϕ ∈ Ψ. Also since (e) holds for k − 2, Π ∩ Ψ is a system of simple roots for Ψ. Thus −β^{∗} induces a dominant integral weight λ on Ψ. Note also that λ(α) = (ϕ, −β^{∗}) = −(α_{k−1}, α_{k}) = 1.

Thus by 5.4.2 there exists ω ∈ W_{Ψ∩λ^{⊥}} and β_{k}, …, β_{m} ∈ Ψ such that

1^{∘} [1]

(a) [1:a] α^{ω} = β_{k} + … + β_{m}.

(b) [1:b] (β_{k}, …, β_{m}) has B_{m−k} or G_{m−k} diagram.

(c) [1:c] λ(β_{k}) = 1 and λ(β_{i}) for k ≤ i ≤ m.

Note that ω fixes β_{0}, β_{1}, … b_{k−1}. Put w = vω. Then (a) to (c) holds. Also since (e) holds for i = k − 1, Ψ_{k−1} ∩ Π is a system of simple roots for Ψ_{k−1}.

Suppose that k = m. Then b_{κ} = a_{k}^{w} and so (d) holds. Also −β_{k}^{∗} induces a dominant integral weight on Ψ_{k−1} ∩ Π. Also Ψ_{k} = Ψ_{k−1} ∩ b_{k}^{⊥} we conclude from 5.2.8 that (e) holds.

Suppose next that k ≠ m. Then α is not long and so X ≠ A. Let Y be the diagram type of (β_{k}, …, β_{m}). From (1^{∘})(a) we conclude that Y = B or Y = G. Thus (1^{∘})(c) implies that (d) holds. Put δ = α_{k}^{w} = α^{ω}. Then Ψ_{k} = Ψ_{k−1} ∩ δ^{⊥}. From (1^{∘})(a) we conclude that −δ^{∗} is a dominant integral weight on Ψ_{k−1} ∩ Π \ {β_{k}, …, β_{m}}.

Suppose Y = B. Then δ is perpendicular to b_{k+1}, …, b_{m}. Now b_{κ} ∉ Ψ_{k} and thus −δ^{∗} is a dominant integral weight on Ψ_{k}. Thus (e) follows from 5.2.8.

Suppose Y = G. Then X = G, Ψ = Φ, k = 1 and α_{0}, α_{1} generate Φ. Hence Ψ_{k} = ∅ and again (e) holds. □

40 CHAPTER 5. ROOT SYSTEMS

Same Characteristic Representations

This chapter is devoted to 𝕂G(𝔽) modules, where 𝕂 and 𝔽 are fields in the same characteristic and G(𝔽) is a group of Lie type over a field 𝕂.

6.1 Lie Algebras

Let Φ be a root system. We continue to use the notation introduced in 5. Definition 6.1.1 [chevalley basis] Let 𝕂 be a field and 𝔤 a Lie-algebra over 𝕂. A Chevalley basis for 𝔤 is a basis

(𝔊_{α}, α ∈ Φ; ℌ_{γ}, γ ∈ Π^{∗})

such that for all α, β ∈ Φ, γ, δ ∈ Π^{∗}:

(a) [CB1] [ℌ_{γ}, ℌ_{δ}] = 0.

(b) [CB2] [ℌ_{γ}, 𝔊_{α}] = (α, γ)𝔊_{α}

(c) [CB3] [𝔊_{α}, 𝔊_{−α}] = ℌ_{α^{∗}} where ℌ_{ρ} for ρ = ∑_{γ∈Π^{∗}} m_{γ}γ ∈ Φ^{∗} is define by ℌ_{ρ} := ∑_{γ∈Π^{∗}} m_{γ}ℌ_{γ}.

(d) [CB4] [𝔊_{α}, 𝔊_{β}] = ±(r_{αβ} + 1)𝔊_{α+β} if α + β ∈ Φ.

(e) [CB5] [𝔊_{α}, 𝔊_{β}] = 0 if 0 ≠ α + β ∉ Φ. Lemma 6.1.2 [nilpotent action for lie algebras] Let 𝔤 be a Lie algebra over 𝕂 and V be a finite dimensional 𝔤-module.

(a) [a] Then there exists unique maximal ideal 𝔲_{v}(𝔤) which acts nilpotently on V.

(b) [b] Let 𝔡 be an ideal in 𝔤, X a 𝔡-submodule of V and 𝔊 ∈ 𝔤.

(a) [a] Define T : X → V/X, x → 𝔊x + X. Then T is a 𝔡-equivariant. In particular 𝔊X + X is a 𝔡-submodule of V.

(b) [b] If V is irreducible for 𝔤 then all composition factors for 𝔡 on V are isomorphic.

(c) [c] If X is irreducible for 𝔡 and 𝔊X ≰ X then 𝔊X ∩ X = 0 and Ann_{X}(𝔊) = 0.

Proof: (a) 𝔲_{V} (𝔤) is just the intersection of the annhilators of the composition factors of 𝔤 on V.

(b) Let 𝔇 ∈ 𝔡 and x ∈ X. Then [𝔊, 𝔇]x ∈ 𝔡x ≤ X and so

T (𝔇x) = 𝔊𝔇x + X = (𝔇𝔊 + [𝔊, 𝔇])x + X = 𝔇(𝔊x + X) = 𝔇(T (x))

So (b:a) holds.

For (b:b) let Y be a 𝔡-submodule maximal such that all composition factors for 𝔡 on Y are isomorphic. By (b:a) applied to Y, all composition factors of 𝔡 on 𝔊Y + Y/Y are isomorphic to a composition factor of Y. Hence by maximality of Y, 𝔊Y ≤ Y. Since 𝔊 ∈ 𝔤 was arbitray and 𝔤 acts irreducibly, V = Y.

For (b:c) note that the irreducibilty of X and (b:a) imply ker T = 0. □

We remark that under the assumption of part (b:b) of the preceeding lemma, V does not need be completely reducible for 𝔡. For example let 𝔤 = 𝔰𝔩_{2}(𝕂) with char 𝕂 = 2 and V the natural 2-dimensional module. Then 𝔡 := 𝕂⟨𝔊_{α}, ℌ_{α}⟩ is an ideal in 𝔰𝔩_{2}(𝕂) and has a unique proper submodule ( namely 𝔊_{α}V). This example also shows that an ideal does not need to act faithfully on its proper submodules.

Lemma 6.1.3 [X+bX] Let 𝔤 be a Lie algebra, 𝔞 and 𝔟 subspaces of 𝔤 with 𝔤 = 𝔞 + 𝔟. Let X be an 𝔞 invariant subspace of V.

(a) [a] For all n ∈ ℕ, ∑_{i=0}^{n} 𝔟^{i}X is 𝔞-invariant.

(b) [b] ∑_{i=0}^{∞} 𝔟^{i}X is 𝔤-invariant.

(c) [c] If X ≠ 0 and V is irreducible as 𝔤-module, then V = ∑_{i=0}^{∞} 𝔟^{i}X.

Proof: (a) By induction on i it suffices to show that X + 𝔟X is 𝔞 invariant. Note that 𝔤X = (𝔞 + 𝔟)X ≤ X + 𝔟X. Let 𝔄 ∈ 𝔞 and 𝔅 ∈ 𝔟. Then

(𝔄𝔅)X = (𝔅𝔄 + [𝔄, 𝔅])X ≤ 𝔅(𝔄X) + 𝔤X ≤ X + 𝔟X.

So (a) holds.

(b) By (a)

∑_{i=0}^{∞} 𝔟^{i}X = ⋃_{n=1}^{∞}(∑_{i=0}^{n} 𝔟^{i}X)

is 𝔞 invariant. Clearly it is also 𝔟 invariant and so (b) follows from 𝔤 = 𝔞 + 𝔟. (c) Follows from (b). □

6.1. LIE ALGEBRAS 43

Proposition 6.1.4 [smith’s lemma] Let 𝔤 be a Lie algebra, 𝔩, 𝔮_{+} and 𝔮_{−} sub algebras and V an irreducible 𝔤 module. Suppose that

(a) [i] 𝔤 = 𝔮_{+} + 𝔩 + 𝔮_{−}

(b) [ii] [𝔩, 𝔮_{+}] ≤ 𝔮_{+} and [𝔩, 𝔮_{−}] ≤ 𝔮_{−}.

(c) [iii] 𝔮_{+} and 𝔮_{−} both act nilpotently on V. Then

(a) [a] 𝔩 acts irreducible on Ann_{V} (q_{+}).

(b) [b] V = Ann_{V} (𝔮_{+}) ⊕ Ann_{V}^{∗} (𝔮_{−}), where Ann_{V}^{∗} (𝔮_{−}) is smallest q_{−} submodule of V containing 𝔮_{−}V.

(c) [c] V = ∑_{i=0}^{∞} 𝔮_{−}^{i} Ann_{V} (𝔮_{+}). Proof: Since 𝔮_{+} acts nilpotently on V, Ann_{V} (𝔮_{+}) ≠ 0. By (ii) Ann_{V} (𝔮_{+}) is an 𝔩 submodule. Let X be a non-zero 𝔩-submodule of Ann_{V} (𝔮_{+}) and Y = ∑_{i=1}^{∞} 𝔮_{−}^{i}X. Then X is an 𝔮_{+} + 𝔩 submodule of Ann_{V} (𝔮_{+}) and Y ≤ Ann_{V}^{∗} (𝔮_{−}). By 6.1.3,

(*) V = X + Y

Suppose that X̃ := Ann_{V} (𝔮_{+}) ∩ Ann_{V}^{∗} (𝔮_{−}) ≠ 0. Since X̃ is 𝔩 invariant, (*) applied to X yields V = X̃ + Ann_{V}^{∗} (𝔮_{−}) ≤ Ann_{V}^{∗} (𝔮_{−}). Since 𝔮_{−} acts nilpotently this implies Ann_{V}^{∗} (𝔮_{−})) = 0, a contradiction to X̃ ≠ 0.

Thus X̃ = 0. Hence also Ann_{V} (𝔮_{+}) ∩ Y = 0 and so using (*)

Ann_{V} (𝔮_{+}) = X + (Ann_{V} (𝔮^{+}) ∩ Y) = X

Since X was an arbitray 𝔩 submodule of Ann_{V} (𝔮_{+}) the lemma is proved. □

Lemma 6.1.5 [q- quadratic] Let 𝔤 be a Lie algebra, 𝔩, 𝔮_{+} and 𝔮_{−} subalgebras and V an irreducible 𝔤-module. Suppose that

(a) [i] 𝔤 = 𝔮_{+} + 𝔩 + 𝔮_{+}

(b) [ii] [𝔩, 𝔮_{+}] ≤ 𝔮_{+} and [𝔩, 𝔮_{−}] ≤ 𝔮_{−}.

(c) [iii] 𝔮_{−}^{2}V = 0 and 𝔮_{−}V ≠ 0.

(d) [iv] 𝔮^{+} acts nilpotently on V. Then

(a) [a] V = Ann_{V} (𝔮_{+}) ⊕ Ann_{V} (𝔮_{−}).

(b) [b] 𝔩 acts irreducibly on both Ann_{V} (𝔮_{+}) and Ann_{V} (𝔮_{−}).

(c) [c] 𝔮_{+}^{2}V = 0 and 𝔮_{+}V ≠ 0.

(d) [d] Ann_{V} (𝔮_{+}) = 𝔮_{+}V and Ann_{V} (𝔮_{−}) = 𝔮_{−}V.

Proof: Note that 𝔮_{−}V ≤ Ann_{V} (𝔮^{−}). By 6.1.4(a) ( applied with the roles of + and − interchanged, Ann_{V} (𝔮_{−}) is an irreducible 𝔩 module. Thus

𝔮_{−}V = Ann_{V} (𝔮_{−}) = Ann_{V}^{∗} (𝔮_{−})

Thus by 6.1.4(b) implies that (a) holds. In particular 𝔮_{+}+𝔩 acts irreducibly on V/ Ann_{V} (𝔮_{+})). Hence 𝔮_{+} annhilates V/ Ann(𝔮_{+}) and the remaiing parts of the lemma are readily verified. □

Comment:The preceeding lemma could be also used to in some later places to avoid the use of the graph automorphism for A_{n}

Definition 6.1.6 [root faithful] Let V be a 𝔤_{Φ}(K)-module. We say that 𝔤_{Φ}(K) acts root faithful on V if 𝔊_{α}V ≠ 0 for all α ∈ Φ.

Lemma 6.1.7 [ideal] Let λ ≠ 0 be a p-restricted dominant weight and Φ a connected root system with basis Π, 𝔤_{Φ}(𝕂) the Lie algebra of type Φ over 𝕂 and V a 𝔤_{Φ}(𝕂)-module of highest weight λ. Let α ∈ Φ and suppose that 𝔊_{α}V = 0. Then the following holds.

(a) [a] Φ has two different root lengths and α is short.

(b) [b] (λ, β^{∗}) = 0, for all short roots β ∈ Π.

(c) [c] char 𝕂 = p_{Φ}.

6.2 Groups of Lie Type and Irreducible Rational Representations

Let Φ be a connected root system, 𝕂 a field, 𝔼 the algebraic closure of 𝕂 and G_{Φ}(K) the corresponding universial group of Lie type. Then G_{Φ}(K) is generated by elements χ_{α}(t), α ∈ Φ, t ∈ 𝕂 fulfilling the Steinberg Relations: For t ∈ 𝕂^{#} define ω_{α}(t) = χ_{α}(t)χ_{α}(t^{−1})χ_{α}(t) and h_{α}(t) := ω_{α}(t)ω_{α}(1)^{−1}.

(a) [St1] χ_{α}(t)χ_{α}(s) = χ_{α}(t + s)

(b) [St2] h_{α}(u)h_{α}(v) = h_{α}(uv)

(c) [St3] If α^{∗} = ∑_{i=1}^{n} n_{i}β_{i}^{∗} for some n_{i} ∈ ℤ, β_{i} ∈ Φ then h_{α}(u) = ∏_{i=1}^{n} h_{β_{i}}(u^{n_{i}}).

(d) [St4] h_{α}(u)χ_{β}(t)h_{α}(u)^{−1} = χ_{α}(u^{(β,α^{∗})}t)

6.2. GROUPS OF LIE TYPE AND IRREDUCIBLE RATIONAL REPRESENTATIONS45

(e) [ST5] ω_{α}(1)χ_{β}(t)ω_{α}(1)^{−1} = χ_{ω_{α}(β)}(ϵ_{αβ}t) for some ϵ_{αβ} = ±1.

(f) [ST6] If α + β is not a root, and α ≠ −β then [χ_{α}(t), χ_{β}(s)] = 1.

(g) [ST7] If α + β is a root then

[χ_{α}(t), χ_{β}(s)] = χ_{α+β}(N_{αβ}ts) ∏_{i,j>1} χ_{iα+jβ}(C_{αβij}t^{i}s^{j})

Let H_{α} = {h_{α}(u) | u ∈ 𝕂^{#}}, X_{α} = {χ_{α}(t) | t ∈ 𝕂^{#}}, U = ∏_{α∈Φ^{+}} X_{α}, H = ∏_{α∈P} H_{α} and B = HU.

Let V be a finite dimensional rational 𝔼G_{Φ}(𝔼) module. For g ∈ G_{Φ}(𝔼) denote by g^{V} the image of g ∈ End_{𝔼}(V). Slighty abusing notation we will often just write g for g^{V}. Since V is rational and finite dimensional we have

χ_{α}(t) = ∑_{i=0}^{d_{α}} t^{i}𝔊_{α,i}

for some d_{α} ∈ ℕ and some 𝔊_{α,i} ∈ End_{𝔼}(V). Note that 𝔊_{α,0} = χ_{α}(0) = 1.

(We remark that, if V is obtained from a module in characteristic zero via an admissible lattice and taking tensor products, then 𝔊_{α,i} = (\frac{1}{i!}𝔊_{α}^{i}) ⊗ 1.)

Comment:It might be interesting to figure out what (ST1) means for the 𝔊_{α,i}

Since 𝔼 is infinite ( and so |E| > d_{α}) it is easy to see that the subalgebra of End_{𝔼}(V) generated by X_{α} contains all of the 𝔊_{α,i}. Let 𝔊_{α}^{V} = 𝔊_{α,1} and 𝔤^{V} the Lie subalgebra of 𝔤𝔩(V) generated by the G_{α}^{V}. Let A^{V} be the subalgebra of End_{𝔼}(V) generated by all the 𝔊_{α,i} (As usual we will ommit the superscript V). Then every G_{Φ}(𝔼) submodule of V is also an 𝔤 submodule and 𝔊_{Φ}(E) and A have the same submodules. Comment: Maybe One should define ℌ_{α}^{V} and verify the remaining relation for the Lie algebra )).

From (ST6)

[𝔊_{α}, 𝔊_{β}] = 0

if α + β is not a root and from (ST7)

[𝔊_{α}, 𝔊_{β}] = N_{αβ}𝔊_{α+β}

if α + β is a root. By (ST4)

h_{α}(u)𝔊_{β,i}h_{α}(u)^{−1} = u^{i(α^{∗},β)}𝔊_{β}

Let μ ∈ Λ and v ∈ V. We say that v is a weight vector for μ if

h_{α}(u)v = u^{(α^{∗},μ)}v

for all u ∈ 𝕂^{#} and α ∈ Φ. Since 𝔼 is infinite and every polynomial as at most finitely many roots, two weights with a common non zero weight vector are equal. Let V_{μ} be the set of all weight vectors for μ.

Lemma 6.2.1 [Gb Vmu] Let β ∈ Φ, i ∈ ℕ and μ ∈ Λ(Φ^{∗}) and v ∈ V_{μ}

(a) [a] 𝔊_{β,i}V_{μ} ≤ V_{μ+iβ}.

(b) [b] ℌ_{β}v = (β, μ)v.

Proof: (a): Let α ∈ Φ and u ∈ 𝕂^{♯}. Then

h_{α}(u)𝔊_{β,i}v = u^{i(α^{∗},β)}𝔊_{β,i}h_{α}(u)v = u^{i(α^{∗},β)}𝔊_{β,i}u^{(α^{∗},μ)}v = u^{(α^{∗},μ+iβ)}𝔊_{β,i}v

(b) ???????? □

Since the different weight spaces are linear independent ( that is the sum of the weight spaces is a direct sum) for a weight vector v that X_{α} fixes v if and only if 𝔊_{α,i}v = 0 for all 1 ≤ i ≤ ∞.

A weight vector is called a highest weight vector if uv = v for all u ∈ U. In the view of the preceeding this means 𝔊_{α,i}v = 0 for all α ∈ Φ^{+}. If V is irreducible there exists a non-zero weight vector. Indeed, since U acts unipotenly Comment:why? C_{V} (U) ≠ 0. Since H is abelian and 𝔼 is algebraicly closed, there exists a one dimensional 𝔼H submodule 𝕂v in C_{V} (U). Since V is rational it is easy to see that v is a weight vector for some weight λ ∈ Λ. Now V = Av and so (∗∗) implies that V is the direct sum of its weight spaces.

A way to obtain the group G_{Φ}(𝕂) is to start with a faithful representation

π : 𝔤_{Φ}(ℂ) → gl(V)

of the Lie algebra 𝔤_{Φ}(C) and to identify the complex vector space V with ℂ^{n} via the admissable ℤ-lattice Λ (in fact, Λ consists of all the weights of all the rational representations of 𝔤_{Φ}(ℂ)). Then

G_{Φ}(𝕂) = ⟨x_{a}(t) | t ∈ 𝕂⟩.

We obtain the quotients of G_{Φ}(𝕂) by following the same procedure but replacing Λ by any admissable ℤ-lattice. Then G_{Φ}(𝕂) is the group of rational points of the algebraic group G = G_{Φ}(𝔼). There is a well know method to relate a Lie-algebra L(G) to an algebraic group G: see for instance Humphreys. According to Borel [Bo, 3.3] the Lie algebras L(G_{Φ}(𝔼)) and 𝔤_{Φ} ⊗_{ℤ} 𝔼 = 𝔤_{Φ}(𝔼) are isomorphic as well as the Lie algebras L(G_{Φ}(𝕂)) = L(G_{Φ})(𝕂) and 𝔤_{Φ} ⊗_{ℤ} 𝕂 = g_{Φ}(𝕂).

Let π : G_{Φ}(𝕂) → GL(W) be an irreducible and faithful 𝔽_{p}-representation for G_{Φ}(𝕂), where p = char𝕂. Then π induces an irreducible, faithful and rational representation π : G_{Φ}(𝔼) → GL(V) for G = G_{Φ}(𝔼) on V = W ⊗_{End_{𝔽pG}(V)} 𝔼 (rational means that all the weights λ of π are in the lattice Λ) and the differential dπ defines a representation

dπ : L(G) → gl(V)

of the to G related Lie algebra 𝔤 ≅ L(g).

Definition 6.2.2 [L]et λ be the highest weight of π. We say that π is p-restricted, if λ = ∑ c_{i}λ_{i} with 0 ≤ c_{i} < p.

According to [Bo, 6.4] the following holds.

Theorem 6.2.3 (Curtis, Borel) If π is a p-restricted irreducible representation of G, then dπ is an irreducible representation of 𝔤.

If V is an irreducible 𝔽_{p}-module for ^{2}G_{Φ}(𝕂), then V is an irreducible 𝔽_{p}-module for G_{Φ}(𝕂), as well, and therefore the following fact holds.

Theorem 6.2.4 [trans] If V (λ) is a p-restricted irreducible 𝔽_{p}^{σ}G_{Φ}(𝕂)-module, then V (λ)⊗_{End_{𝔽pG}(V)} 𝔼 is an irreducible module for 𝔤_{Φ}(𝔼).

Later we will need some more information about the elements of a unipotent subgroup U^{1} of ^{2}F_{4}(K). Here we follow the description given in the book of Carter [Ca, 13.6].

Lemma 6.2.5 [system F4] Let Φ be a root system of type F_{4} and Π = {α, β, γ, δ} a fundamental system of Φ where α, β are long and γ, δ short and where α and δ are perdendicular. Let τ be a mapping of V_{Φ}(𝕂) into V_{Φ}(𝕂) defined by

τ(r) = f(r)σ(r), where

r is a root and σ(r) the permutation of Φ induced by the graph automorphism and

f(r) = \sqrt{\frac{1}{2}} if r is short and f(r) = \sqrt{2} if r is long.

Then τ is an isometrie of V_{Φ}(𝕂). Let W be the related Weyl group (the group generated by the reflections on the hyperplanes perpendicular to the roots) and let W^{1} = C_{W} (τ). Then

W^{1} = ⟨w_{α}w_{δ}, (w_{β}w_{γ})^{2}⟩ ≅ D_{16},

where for r a root, w_{r} is the reflection on the hyperplane perpendicular to r. The orbits of W^{1} on Φ^{+} partition Φ^{+}. These orbits are

S_{1} = {α, δ},
S_{2} = {α + 2β + 2γ, β + 2γ + δ},
S_{3} = {α + 2β + 2γ + 2δ, α + β + 2γ + δ},
S_{4} = {a + 2β + 4γ + 2δ, α + 2β + 2γ + δ},
S_{5} = {β, γ, β + γ, β + 2γ},
S_{6} = {α + β, γ + δ, α + β + 2γ + 2δ, α + β + γ + δ},

S_{7} = {α + β + 2γ, β + γ + δ, α + 2β + 3γ + δ, α + 3β + 4γ + 2δ},
S_{8} = {β + γ + 2δ, α + β + γ, α + 2β + 3γ + 2δ, 2α + 3β + 4γ + 2δ}.

Let S = {S_{i} | 1 ≤ i ≤ 8}. Notice that these orbits are either of type A_{1} × A_{1} or B_{2}. If S_{i} = {r, s} is of type A_{1} × A_{2} with r long and s short, then define

x_{S_{i}}(t) = x_{s}(t^{θ})x_{r}(t),

where t ∈ 𝕂 and θ a field automorphism of 𝕂. In this case set

X_{S_{i}} = ⟨x_{S_{i}}(t) | t ∈ 𝕂⟩.

If S_{i} is of type B_{2} and {r, s} is a fundamental system with r long and s short, then define

x(t, u) = x_{r}(t^{θ})x_{s}(t)x_{r+s}(t^{θ+1} + u^{θ})x_{2s+r}(u^{2θ}),

where t, u ∈ 𝕂 and θ a field automorphism of 𝕂. In this case set

X_{S_{i}} = ⟨x(t, u) | t, u ∈ 𝕂⟩.

Then

U^{1} = ∏_{T ∈S} X_{S}^{1}.

6.3 Translation form the group to the Lie algebra

Comment:This is taking from Tim’s file, needs to be adapted

Lemma 6.3.1 [splitting field] Let K ⊆ k be a subfield of k and λ a dominant weight with λ(α) < |K|, for all α ∈ Σ. Then A(λ) is irreducible as a kG(K)-module.

Proof: □

Let λ be a dominant p-restricted integral weight and V = V (λ) an irreducible GF (p)G_{Φ}(K)- module with highest weight λ. Then by 6.2.4 V ⊗ E is an irreducible module for 𝔤_{Φ}(E) with E the algebraic closure of K.

Order Π in some way and then order the set of weights lexicographically. Comment: mention positive, by carter we can choose the order to be compatible with the height function

Definition 6.3.2 [u+]

(a) [a] U_{α}^{+} = ⟨X_{β} | β ≥ α⟩

(b) [b] U_{α}^{−} = ⟨X_{β} | β > α⟩. Note that U_{α}^{+} = X_{α}U_{α}^{−}.

(c) [c] V_{μ} a weight space (as usual)

(d) [d] V_{μ}^{+} = ⨁_{γ≥μ} V_{γ}

(e) [e] V_{μ}^{−} = ⨁_{γ>μ} V_{γ}

Let P be a subgroup of a unipotent group U of G_{Φ}(K) and let

Φ_{P} = {α ∈ Σ^{+} | P ∩ U_{α}^{+} ≰ U_{α}^{−}}.

For α ∈ Φ_{P}, pick g_{α} ∈ (P ∩ U_{α}^{+}) \ U_{α}^{−}. Then g_{α} = x_{α}(t)u_{α} for some u_{α} ∈ U_{α}^{−} and t ≠ 0.

Lemma 6.3.3 [special order] Let P be a subgroup of a unipotent group U of ^{2}G_{2}(q),^{B}_{2}(q) or ^{2}F_{4}(q). Then there is an ordering of Π such that Φ_{P} consists only of short roots.

Proof: Assume first that Φ is of type B_{2}. Then the elements of a unipotent subgroup U^{1} of ^{2}B_{2}(𝕂) are

x(y, u) = x_{α}(t^{θ})x_{β}(t)x_{α+β}(t^{θ+1} + u)x_{2α+β}(u^{2θ}),

where Π = {α, β} with α short and β long and u, t ∈ 𝕂 and θ a field automorphism, see for instance [Ca, 13.6.1]. Choose the ordering on Φ such that α < β. Then Φ_{P} is a subset of {α, α + β}, which is a set of short roots, the assertion.

Now assume that Φ is of type G_{2}. The elements of a unipotent subgroup U^{1} of ^{2}G_{2}(𝕂) have the form

x(t, u, v) = x_{α}(t^{θ})x_{β}(t)x_{α+β}(t^{θ+1} + u^{θ})x_{2α+β}(t^{2θ+1} + v^{θ})x_{3α+β}(u)x_{3α+2β}(v),

where Π = {α, β} with α short and β long, t, u, v ∈ 𝕂 and θ an automorphism of 𝕂, see for instance [Ca, 13.6.1]. We choose again the ordering on Φ such that α < β. Then Φ_{P} is a subset of {α, α + β, 2α + β}, which is again a set of short roots, the assertion.

Finally assume that Φ is of type F_{4}. The elements of a unipotent subgroup U^{1} of ^{2}F_{4}(𝕂) are described in 6.2.5. We are going to use the same notation as in 6.2.5. We order Φ such that β > γ > α > δ. Then Φ_{P} is again a subset of a set of short roots. □

𝔤 = ∑_{α∈Φ_{P}} 𝔼𝔊_{α} + ∑_{α∈Φ_{P}} 𝔼ℌ_{α}.

Lemma 6.3.4 [L1]

1. 𝔤 is a subalgebra of 𝔤_{Φ}(𝔼).

2. If P has nilpotent class m, then 𝔤 has nilpotent class at most m.

3. If [P [P [ \underbrace{…}_{n−times}[P [P, A]] …]]] = 0, then D^{n}A = 0.

4. dim(Ann_{A}(D)) ≥ dim(C_{A}(P)).

5. Suppose that there are α, β ∈ Φ_{D} and k ∈ 𝕂 such that α < β and (g_{α} −1) ≡ k(g_{β} −1). Then 𝔊_{α} ≡ 0.

Proof: Notice that [g_{α}, g_{β}]U_{α+β}^{−} = [x_{α}(t_{α}), x_{β}(t_{β})]U_{α+β}^{−} = x_{α+β}(N_{αβ}t_{α}t_{β}]U_{α+β}^{−}, where [X_{α}, X_{β}] = N_{α+β}X_{α+β} in 𝔤_{Φ}(E). If N_{α+β} ≠ 0, then [g_{α}g_{β}] ∈ U_{α+β}^{+} \ U_{α+β}^{−}. Hence, α + β ∈ Φ_{P}, so D is a subalgebra of 𝔤_{Φ}(E), proving (1).

Now [g_{α_{1}}, g_{α_{2}}, …, g_{α_{n}}]U_{α_{1}+α_{2}+···+α_{n}}^{−} = x_{α_{1}+α_{2}+···+α_{n}}(rt_{α_{1}}t_{α_{2}} … t_{α_{n}})U_{α_{1}+α_{2}+···+α_{n}}^{−}. So if [g_{α_{1}}, g_{α_{2}}, …, g_{α_{n}}] = 1, then r = 0 and so [X_{r_{1}}, X_{r_{2}}, …, X_{r_{n}}] = rX_{r_{1}+r_{2}+···+r_{n}} = 0.

Now let a ∈ A_{μ}^{+} with a = a_{μ} + a_{μ}^{−} where a_{μ} ∈ A_{μ} and a_{μ} ∈ A_{μ}^{−}. Then

[x_{α}(t_{α}), a] = ∑_{n=1}^{∞} frac1n!t_{α}^{n}X_{α}^{n})a ∈ t_{α}X_{α}a_{μ} + A_{μ+α}^{−}

So [g_{α}, a] ∈ t_{α}X_{α}a_{μ} + A_{μ+α}^{−}, and in particular,

[g_{α_{1}}[g_{α_{2}}[… [g_{α_{n}}, a] …]] ∈ t_{α_{1}}t_{α_{2}} … t_{α_{n}}X_{α_{1}}X_{α_{2}} … X_{α_{n}}a_{μ} + A_{μ+α_{1}+α_{2}+···+α_{n}}^{−}.

So, if [P [P [… [P, A] …]] = 0, then X_{α_{1}}X_{α_{2}} … X_{α_{n}}a_{μ} ∈ A_{μ+α_{1}+α_{2}+···+α_{n}}^{−}∩A_{μ+α_{1}+α_{2}+···+α_{n}} = 0.

Hence X_{α_{1}}X_{α_{2}} … X_{α_{n}}A = 0. That is, D^{n}A = 0, proving (2).

Choose E_{μ} ≤ A_{μ} so that C_{A_{μ}^{+}}(P) + A_{μ}^{−} ≥ E_{μ} + A_{μ}^{−} (E_{μ} = A_{μ} ∩ (C_{A_{μ}^{+}}(P) + A_{μ}^{−})). Let E = ⨁_{μ} E_{μ}. Then dim_{k}(E) = dim_{k}(C_{A}(P)).

Now, if a ∈ C_{A_{μ}^{+}}(P), then a = a_{μ}+a_{μ}^{−}, so [g_{α}, a] ∈ t_{α}X_{α}a_{μ}+A_{μ+α}^{−} implies that x_{α}a_{μ} = 0. Hence, X_{α}E = 0 and so DE = 0, proving (3). It remains to show (5). Let a ∈ A_{μ}. By what was proved before (g_{α} − 1)a = [g_{α}, a] ∈ t_{α}𝔊_{α}a_{μ} + A_{μ+α}^{+} and (g_{β} − 1)a[= g_{β}, a] ∈ t_{β}𝔊_{β}a_{μ} + A_{μ+β}^{+} ∈ A_{μ+α}^{+} Since (g_{α} − 1) ≡ k(g_{β} − 1) we conclude that t_{α}𝔊_{α}a_{μ} = 0 and so 𝔊_{α}a_{μ} = 0 and 𝔊_{α} ≡ 0,

hence (5).

□

Quadratic Modules

7.1 Quadratic modules for 𝔤

For a root system Φ let p_{Φ} := \frac{(α,α)}{(β,β)} where α is a long and β is a short root in Φ. Note that if Φ is connected then p_{Φ} ∈ {1, 2, 3}. If 𝔤 = 𝔤_{Φ}(𝕂) and p_{Φ} = char 𝕂, then 𝔤_{short} (the subalgebra of 𝔤 generated by {𝔊_{α} |, α ∈ Φ_{short}}) is an ideal in 𝔤. Note that this happens for p = 2 and Φ of type B_{n}, C_{n} and F_{4} and for p = 3 and Φ of type G_{2}. These cases will require special attention throughout this section.

Definition 7.1.1 [def:quadratic] A module V for 𝔤_{Φ}(𝕂) is called quadratic if (ℌ_{α^{∗}} − 1)𝔊_{α}V = 0 for all long roots α ∈ Φ.

The definition of a quadratic module is motivated by the following lemma:

Lemma 7.1.2 [quadratic in odd characteristic] Let V be a 𝔤_{Φ}(𝕂)-module and α ∈ Φ.

(a) [a] If (ℌ_{α^{∗}} − 1)𝔊_{α}V = 0 then 𝔊_{α}^{2} = 0.

(b) [b] If char 𝕂 ≠ 2, then 𝔊_{α}^{2}V = 0 iff (ℌ_{α^{∗}} − 1)𝔊_{α}V = 0.

(c) [c] Suppose that V comes from a module for U_{Φ}(ℤ) and that \frac{𝔊_{α}^{2}}{2} V = 0, then (ℌ_{α^{∗}} − 1)𝔊_{α}V = 0.

Proof: (a) Since (ℌ_{α^{∗}} − 1)𝔊_{α}V = 0 we have ℌ_{α^{∗}}𝔊_{α} ≡ 𝔊_{α} and so

0 = [𝔊_{α}, 𝔊_{α}] ≡ [ℌ_{α^{∗}}𝔊_{α}, 𝔊_{α}] = [ℌ_{α^{∗}}, 𝔊_{α}]𝔊_{α} = 𝔊_{α}^{2}

(b) We compute

[𝔊_{α}^{2}, 𝔊_{−α}] = [𝔊_{α}, 𝔊_{−α}]𝔊_{α} +𝔊_{α}[𝔊_{α}, 𝔊_{−α}] = = ℌ_{α^{∗}}𝔊_{α} +𝔊_{α}ℌ_{α^{∗}} = [𝔊_{α}, ℌ_{α^{∗}}]+2ℌ_{α^{∗}}𝔊_{α} = −2𝔊_{α} + 2ℌ_{α^{∗}}𝔊_{α} = 2(ℌ_{α^{∗}} − 1)𝔊_{α}. Thus

(∗) [𝔊_{α}^{2}, 𝔊_{−α}] = 2(ℌ_{α^{∗}} − 1)𝔊_{α}

So if 𝔊_{α}^{2} ≡ 0 and char 𝕂 ≠ 2 we conclude that (ℌ_{α^{∗}} − 1)𝔊_{α} ≡ 0.

(c) Note that the (*) is also a valid equation in U_{Φ}(ℤ). Thus in U_{Φ}(ℤ) we have [\frac{𝔊_{α}^{2}}{2} , 𝔊_{−α}] = (ℌ_{α^{∗}} − 1)𝔊_{α}. Thus (??) holds. □

The irreducible quadratic modules for 𝔤_{Φ}𝕂 are fairly easily classified (see the next theorem). The remainder of the section will be devoted to show that some weaker conditions already imply that a module is quadratic. If V is a module for 𝔤 and 𝔊_{1}, 𝔊_{2} ∈ 𝔤 we write 𝔊_{1} ≡ 𝔊_{2} if (𝔊_{1} − 𝔊_{2})V = 0, that is if the image of 𝔊_{1} and 𝔊_{2} in End(V) are equal.

Theorem 7.1.3 [classification of quadratic modules for Lie algebras] Let 𝕂 be a field, Φ a root system and 𝔤 = 𝔤_{Φ}(𝕂) the corresponding algebra. Let V = V (λ) be the irreducible restricted 𝔤-module of heighest weight λ ≠ 0. Let α = α_{long} be the heighest long root of Φ. Then the following are equivalent:

(a) [a] V is quadratic.

(b) [b] (ℌ_{α^{∗}} − 1)𝔊_{α}V = 0

(c) [c] 𝔊_{β}𝔊_{α}V = 0 for all β ∈ Φ with (β, α) > 0.

(d) [d] (λ, α^{∗}) = 1.

(e) [e] −1 ≤ (ρ, α^{∗}) ≤ 1 for all weights ρ for 𝔤 on V.

Proof: We assume without loss that 𝕂 is algebraicly closed.

(a)⟹ (b): Obvious.

(b)⟹ (c): Let β ∈ Φ with (β, α) > 0. If β = α then 𝔊_{α}^{2} ≡ 0 by 7.1.2(a). Suppose that

β ≠ α. Then (α^{∗}, β) = 1 and so [ℌ_{α^{∗}}, 𝔊_{β}] = 𝔊_{β}. Note that β is positive, so β + α ∉ Φ be the maximality of α. Thus 𝔊_{α}𝔊_{β} = 𝔊_{β}𝔊_{α}. Also by assumption (ℌ_{α^{∗}} − 1)𝔊_{α} ≡ 0 and so ℌ_{α^{∗}}𝔊_{α} ≡ 𝔊_{α}. We compute:

𝔊_{β}𝔊_{α} = [ℌ_{α^{∗}}, 𝔊_{β}]𝔊_{α} = ℌ_{α^{∗}}𝔊_{β}𝔊_{α} − 𝔊_{β}ℌ_{α^{∗}}𝔊_{α} =
= ℌ_{α^{∗}}𝔊_{α}𝔊_{β} − 𝔊_{β}ℌ_{α^{∗}}𝔊_{α} ≡ 𝔊_{α}𝔊_{β} − 𝔊_{β}𝔊_{α} = 0.

(c)⟹ (d): Let v_{−} be a lowest weight vector. Let 𝔲_{+} = 𝔤_{Φ^{+}}(𝕂) and let ω_{0} ∈ W (Φ)

with ω_{0}(Π) = −Π. Then v_{−} has weight ω_{0}(λ). The 𝔲v_{−} = V (??). Since [𝔲_{+}, α] = 0 we conclude that v^{−} ∉ Ann(𝔊_{α}). Hence v := 𝔊_{α}v_{−} ≠ 0 is a non zero weight vector with weight ω_{0}(λ) + α. Let

𝔮_{α} = 𝕂⟨G_{β} | β ∈ ϕ, (α, β) > 0⟩

and

𝔩_{α} = 𝕂⟨G_{β} | β ∈ ϕ, (α, β) = 0⟩ + 𝔥

7.1. QUADRATIC MODULES FOR 𝔊 53

By Smith’s Lemma 6.1.4 Ann(𝔮_{α}) is an irreducible module for 𝔩_{α}. Since v_{+} is a highest weight vector in Ann(𝔮_{α}) we conclude from ?? that all weights in Ann(𝔮_{α}) are of the form λ + μ for some μ ∈ ℕ(Φ^{−} ∩ α^{⊥}).

Recall that with weight vectors we mean weight vectors for the Cartan subgroup H of G_{𝕂}(Φ). In particular two weights in Λ which share a non-zero weight vector are equal. Thus

ω_{0}(λ) + α = λ + μ

for some μ ∈ Λ with (α, μ) = 0. Note also that ω_{0} has order two, peserves (·, ·) and ω_{0}(α) = −α. So we compute

(ω_{0}(λ) + α, α^{∗}) = (ω_{0}(λ), α^{∗}) + (α, α^{∗}) = (λ, ω_{0}(α^{∗})) + 2 = −(λ, α^{∗}) + 2

On the other hand

(λ + μ, α^{∗}) = (λ, α^{∗}) + (μ, α^{∗}) = (λ, α^{∗})

The last three displayed equations imply 2(λ, α^{∗}) = 2. Since this is a statement in ℤ we conclude (λ, α^{∗}) = 1.

(d)⟹ (e):

(d)⟹ (a): Suppose that (λ, α^{∗}) = 1. Note that ρ = λ − ϕ for some ϕ ∈ ℕΦ^{∗}. Also

(ϕ, α^{∗}) ≥ 0 and so (ρ, α^{∗}) ≤ (λ, α′^{∗}) ≤ 1. Similarly as ρ = ω_{0}(λ) ∗ ψ for some ψ ∈ ℕΦ^{∗} we have (ρ, α^{∗}) ≥ (ω_{0}(λ), α^{∗}) = −1 and so (e) holds.

(e)⟹ (a): It suffices to show that (ℌ_{α}d − 1)𝔊_{α}V_{μ}0 for all weights μ on V. If 𝔊_{α}V_{μ} = 0

this is obvious. So suppose that 𝔊_{α}V_{μ} ≠ 0. Thus both μ and μ + α are weights on V. Thus

(μ + α, α^{∗}) ≤ 1

On the other hand

(μ + α, α^{∗}) = (μ, α^{∗}) + (α, α^{∗}) ≥ −1 + 2 = 2

and we conclude that (μ+α, α^{∗}) = 1. Hence 6.2.1(b) implies that (ℌ_{α^{∗}} −1)V_{μ+α} = 0. Thus (a) holds. □

Definition 7.1.4 [def:quadratic tuple] A quadratic tuple is tuple (Φ, p, λ, α, β) where Φ is a connected root system, λ is a non-zero dominant integral p-restricted weight, α and β are roots, and V = V_{𝕂}(λ) for some field 𝕂 with char 𝕂 = p such that

(a) [a] 𝔊_{β}𝔊_{α}V = 0.

(b) [b] 𝔊_{α}V ≠ 0 ≠ 𝔊_{β}V.

(c) [c] If α = β then p ≠ 2.

In the next few lemmas we will determine all the quadratic tuples. Comment:We should once and for all introduce weight vectors for arbitrary fields: For the algebraicly closed case define it by the action of H, in general v ∈ V (λ) is called a weight vector if 1 ⊗_{𝕂} v is a weight vector in K̄ ⊗ 𝕂V. Note that for p-restricted weights, V will be the direct sum of the weight spaces.( just start with the lowest weight vector and take images under the 𝔊_{α}’s

Lemma 7.1.5 [quadratic tuple for a=b long] Let (Φ, p, λ, α, β) be a quadratic tuple with α = β and α long. Then V is a quadratic module.

Proof: By assumption p ≠ 2. So the lemma follows from 7.1.2 □

Lemma 7.1.6 [quadratic tuples for (a,b) positive and a long] Let (Φ, p, λ, α, β) be a quadratic tuple with α long, α ≠ β and (α, β) > 0. Then V is a quadratic module.

Proof: Without loss α is the highest long root. Then β is positive. Let Ψ = ⟨α, β⟩, the root subsystem generated by α and β. Then Ψ is of type A_{2}, B_{2} or G_{2}. In any case δ = α−β is a root, α = δ + β, α + β is not a root, 𝔊_{α}𝔊_{β} = 𝔊_{β}𝔊_{α} ≡ 0 and r_{βδ} + 1 = p_{Ψ}.

Suppose first that p ≠ 2 and p ≠ p_{Ψ}.

Since 𝔊_{β}𝔊_{α} ≡ 0 taking the Lie bracket with 𝔊_{δ} gives ±p_{Ψ}𝔊_{α}^{2} ≡ 0. Thus 𝔊_{α}^{2} = 0 and we are done by 7.1.5.

Suppose next that p = p_{Ψ}. Then Ψ is of type B_{2} or G_{2}, p = p_{Φ} and β is short. Let X be an irreducible 𝔤_{short}-submodule in V. If 𝔊_{β}X = 0 then also ℌ_{β} = [𝔊_{β}, 𝔊_{−β}] annihilates X. Thus by ??(bb), ℌ_{α} acts nilpotently on V. But ℌ_{β} is semisimple on V and so ℌ_{β}V = 0. Hence by ?? 𝔊_{β}V = 0, a contrdiction to the definition of a quadratic tuple.

Thus 𝔊_{β}X ≠ 0. Since 𝔊_{α}𝔊_{β}X = 0 we conclude Ann_{X}(𝔊_{α}) ≠ 0 and so by ??(bc), 𝔊_{α}X ≤ X. By symmetry the same holds for any long root subalgebra of 𝔤 and so 𝔤X ≤ X and V = X. Thus 𝔤_{short} acts irreducibly on V. Let 𝔮 = 𝕂⟨𝔊_{μ} | μ ∈ Φ_{short}, (μ, α) > 0⟩ and 𝔩 = 𝕂⟨𝔊_{μ} | μ ∈ Φ_{short} ∩ α^{⊥}⟩. Then 𝔮 + 𝔩 + 𝔥_{short} is a parabolic subalgebra and so by 6.1.4 Ann_{V} (𝔮) is an irreducible 𝔩-module. Note that 𝔮 is an ideal in 𝔮_{α} + 𝔩_{α} and so Ann_{V} (𝔮) is an irreducible module for 𝔮_{α} + 𝔩_{α}. It follows that 𝔮_{α} annihilates Ann_{V} (𝔮). On the other hand W (Φ ∩ α^{⊥}) acts transitively on {μ ∈ Φ_{short}, (μ, α) > 0} and thus 𝔮𝔊_{α}V = 0 and so also 𝔮_{α}G_{α}V = 0. Thus V is quadratic by 7.1.3.

Suppose now that Ψ is of type A_{2}. We claim that 𝔊_{μ}𝔊_{α} ≡ 0 for all μ ∈ Φ with (μ, α) > 0. This is obvious if μ = α or if (α, μ) is conjugate to (α, β) under W (Φ). If neither of this holds then Φ is of type A_{n}. Let V^{∗} be 𝔤-module dual to V. Then 𝔊_{α}𝔊_{β}V^{∗} = 0. Since 𝔊_{α} and 𝔊_{β} commute, 𝔊_{β}𝔊_{α}V^{∗} = 0. Now V^{∗} ≅ V^{σ} where σ is the graph automorphism of 𝔤. Thus 𝔊_{σ(β)}𝔊_{σ(α)}V = 0. Now (α, μ) is conjugate under W (Φ) to (σ(α), σ(β) and we again conclude that 𝔊_{μ}𝔊_{α} ≡ 0. Thus V is quadratic by 7.1.3.

Suppose finally that p = 2 and Ψ is of type G_{2}. Then β is short. Let γ = β − δ. Then γ is a root, r_{δβ} = 3, and α + γ is not a root.

0 ≡ [𝔊_{β}𝔊_{α}, 𝔊_{γ}] = ±3𝔊_{β+γ}𝔊_{α}

Thus 𝔊_{β+γ}𝔊_{α} ≡ 0. Using the action of W (Φ ∩ α^{⊥}) we conclude that 𝔮_{α}𝔊_{α} ≡ 0 and V is quadratic. □

Lemma 7.1.7 [a long implies quadratic] Let (Φ, p, λ, α, β) be a quadratic tuple with α long. Then V is quadratic.

Proof: Without loss α is the highest long root. If β = α we are done by 7.1.5. So we may choose β ∈ Φ maximal with β ≠ a, 𝔊_{β}V ≠ 0 and 𝔊_{β}𝔊_{α}V = 0. If (β, α) > 0 we are done by 7.1.6. So we may assume that (α, β) ≤ 0.

Suppose first that β is long. If Φ is of type A_{1} then β = −α and so 2𝔊_{α}^{2} = [𝔊_{β}𝔊_{α}, 𝔊_{α}, 𝔊_{α}] = iv0. Thus 𝔊_{α}^{2} ≡= 0 and V is quadratic by 7.1.3 ( Actually a moments thought even gives a contradiction).

So assume that Φ ≠ A_{1}. If Φ_{long} is connected there exists γ ∈ Π(Φ_{long}) with β + γ ∈ Φ_{long}. Then N_{βγ} ≠ 0 and so 𝔊_{β+γ}𝔊_{a} = 0. The maximal choice of β implies β + γ = α. But then (α, β) > 0.

So Φ_{l}o is disconnected, α ⊥ β,Φ is of type C_{n} and γ := \frac{1}{2}(α − β) ∈ Φ_{short}. Then N_{βγ} ≠ 0 and 𝔊_{γ+α}𝔊_{α} ≡ 0. The maximal choice of γ implies 𝔊_{γ+α}V = 0. In particular p = 2, 𝔤_{short}V = 0 and [ℌ_{β}, 𝔤]V = 0. Thus ℌ_{β} acts as a scalar on V. Since α ⊥ b, ℌ_{β}𝔊_{α} = [𝔊_{β}𝔊_{α}, G_{−β} =≡ 0 and so ℌ_{β}V = 0 But then 𝔤 acts nilpotent on V a contradiction.

Suppose next that β is not long. Note that the highest short root has positive inner product with α. So β is not the highest short root. Assume Φ_{short} is connect. Then we can choose γ ∈ Π(Φ_{short}) with β + γ ∈ Φ_{short} and we get a contradiction to the maximal choice of β. Hence Φ_{short} is disconnected and Φ is of type B_{n}. If β is not perpendicular to α then ((b, a) < 0, N_{βα} ≠ 0 and we get G_{α+β}𝔊_{α} = 0, contradiction the maximality of β. So β ⊥ α and as above ℌ_{β}𝔊_{α} = 0. Let γ ∈ Π with β + γ ∈ Φ. If N_{βγ} ≠ 0, we get a contradiction to the maximality of β. Thus p = 2 and so [Hb, 𝔤] = 0 and ℌ_{β} centralizes V. But then 𝔤_{short}V = 0, a contradiction as β is short and 𝔊_{β}V ≠ 0.

This settles the last case and the lemma is proved. □

Lemma 7.1.8 [quadratic tuples with GaGb not 0] Let (Φ, p, λ, α, β) be a quadratic tuple with 𝔊_{α}𝔊_{β}V (λ) ≠ 0. The up to conjugacy under W Φ = A_{n}, α = e_{0} − e_{n} and either β = −e_{0} + e_{1} and λ = λ_{n} or β = −e_{2} + e_{n} and λ = λ_{1}.

Proof: Let V^{∗} the dual of V. So V^{∗} = V (ω_{0}(λ)). Then 𝔊_{α}𝔊_{β}V^{∗} = 0 and we conclude that λ ≠ −ω(λ). Thus Φ = A_{n} or n ≥ 5, n is odd and Φ = D_{n} Also [G_{α}, G_{b}] ≠ 0 and so (α, β) < 0.

But in D_{n} for n > 3, W has a unique orbits on pairs of roots (γ, δ) with (γ, δ) < 0, namely all are conjugate to (e_{1} + e_{2}, −e_{1} + e_{3}). Thus (α β) is conjugate to (β, α) contradicting the assumptions.

Thus Φ is of type A_{n}. By 7.1.7 V is quadratic and so by 7.1.3 λ = λ_{i} for some 1 ≤ i ≤ n. Up to conjugation under W, we may assume α = e_{0} − e_{n} and either β = −e_{0} + e_{1} or

β = −e_{1}+e_{n}. In view of the graph automorphismus it suffices to treat the case β = −e_{0}+e_{1}. Let

Σ = ⟨β, Φ ∩ α^{⊥}⟩ = {±(e_{i} − e_{j}) | 0 ≤ i < j ≤ n − 1}.

Then Σ is a closed root subsystem of type A_{n−1}. Also 𝔊_{α}V is invariant under 𝔩_{α} and 𝔊_{β} and so under 𝔤_{Σ}. Since 𝔊_{β} annihilates 𝔊_{α}V and W (Σ) is tranisitive on Σ, g_{Σ} annihilates 𝔊_{α}V. As v_{+} ∈ 𝔊_{α}V we conclude that λ = λ_{n} and the lemma is proved. □

Lemma 7.1.9 [quadratic tuples for (a,b) not positive and a long] Let (Φ, p, λ, α, β) be a quadratic tuple with α long, α ≠ β and (α, β) ≤ 0. Then one of the following holds:

(a) [a] Φ = A_{n}, α = e_{0} − e_{n} and either

(a) [aa] λ = λ_{1} and β = e_{1} − e_{2} or −e_{2} + e_{n} or

(b) [ab] λ = λ_{n} and β = e_{1} − e_{2} or −e_{0} + e_{1}.

(b) [b] Φ = C_{n}, λ = λ_{1}, α = 2e_{1} and either β = 2e_{2} or p ≠ 2, n > 2 and β = e_{2} − e_{3}.

(c) [c] Φ = B_{n}, n ≥ 3, α = e_{1} + e_{2} and either

(a) [ca] λ = λ_{n} and β = e_{1} − e_{2} or

(b) [cb] λ = λ_{1}, β = e_{2} − e_{3} or p ≠ 2 and β = e_{2}.

(d) [d] Φ = D_{4} α = e_{1} + e_{2} and one of the following holds:

(a) [da] λ = λ_{1} and β = e_{3} − e_{4} or e_{3} + e_{4}.

(b) [db] λ = λ_{3} and β = e_{1} − e_{2} or e_{3} + e_{4}.

(c) [dc] λ = λ_{4} and β = e_{1} − e_{2} or e_{3} − e_{4}.

(e) [e] Φ = D_{n}, n ≥ 5,α = e_{1} + e_{2} and either

(a) [ea] β = e_{3} − e_{4} and λ = λ_{1} or

(b) [eb] β = e_{1} − e_{2} and λ = λ_{n−1} or λ_{n}.

Proof: Without loss α is the highest root. Let Ψ be the closed root subsystem generated by α and β. By 7.1.7 that V is quadratic and so by 7.1.3 λ = λ_{μ} for some δ ∈ Π with n_{μ^{∗}}^{∗} = 1. Moreover, 𝔊_{α}V = Ann(𝔮_{α}) and so 𝔊_{β}𝔊_{α}V = 0 just means that 𝔊_{β} annihilates V_{α} := Ann(𝔮_{α}).

Supose first that (β, α) = 0. Then 𝔊_{b} ≤ Ann_{𝔩_{α}}(V_{α}). If (μ, α) ≠ 0 then all of 𝔩_{α} annihilates V_{α} and (a) or (b) holds. So suppose that (μ, α) = 0. Asume that Φ ⊥ α^{⊥} is connected. Then 𝔊_{β}V_{α} = 0 implies that β is short and p = p_{Φ}. On the otherhand G_{β}V ≠ 0 implies that μ is short. But the μ is conjugate to β in W (Φ∩α^{⊥}) and so 𝔊_{μ}V_{α} = 0, a contradiction.

Thus Φ ⊥ α^{⊥} is not connected and so Φ is of type B_{n}, n > 2 or D_{n}, n ≥ 4. It is now easu to see that one of (c), (d) or (e) holds, the assumption that p ≠ 2 in some cases is to make sure that 𝔊_{β}V ≠ 0.)

Suppose next that (β, α) < 0. If 𝔊_{α}𝔊_{β}V ≠ 0, then (a) holds by 7.1.8 So we may assume that 𝔊_{α}𝔊_{β} ≡ 0. Then also [𝔊_{α}, 𝔊_{β}] ≡ 0. Since (β, α) < 0, α + β is a root and since α is long N_{αβ} ≠ 0. It follows that G_{α+β} ≡ 0. Thus p = p_{Φ} and α + β is short. Since 𝔊_{β} ≢ 0, β is long. But the sum of two long roots always long , a contradiction to α + β short. □

Lemma 7.1.10 [p=pphi and a and b short] Let (Φ, p, λ, α, β) be a quadratic tuple and suppose that p = p_{Φ} and both α and β are short. Then Φ = C_{n}, p = 2 and λ = λ_{1} or λ_{1} + λ_{n}. Moreover, V (λ) is as a module for 𝔤_{short} isomorphic to a direct sum of natural modules.

Proof: Note that Φ is B_{n}, C_{n}, G_{2} or F_{4} and Φ_{short} is of type A_{1}^{n}, D_{n}, A_{2} and D_{4} respectively. Moreover W/W (Φ_{short}) induces the full group of graph automorphisms on Φ_{short}.

Let μ be the restriction of λ to Φ_{short}^{∗}. Then all composition factors for 𝔤_{short} on V are isomorphic to V (μ). Moreover (Φ_{short}, μ, α, β) is a quadratic tuple. This easily rules out the case Φ_{short} = A_{1}^{n}. Hence Φ_{short} is connected and so by 7.1.7 V (μ) is quadratic for 𝔤_{short}. Since μ is invariant under all graph automorphism, 7.1.3 implies that Φ_{short} = D_{n} and μ = ”μ_{1}”. Then λ = λ_{1} or λ = λ_{1} + λ_{n}. Note that V (λ_{1} + λ_{n}) ≅ V (λ_{1}) ⊗ V (λ_{n}) and g_{short} acts trivially on V (λ_{n}). So also the last statement of the lemma is proved. □

It remains to look at quadratic tuples where Φ has two root lengths, α and β are short and p ≠ p_{ϕ},

Lemma 7.1.11 [a=b short] Let (Φ, p, λ, α, β) be a quadratic tuple with α = β short and p ≠ p_{Φ} ≠ 1. Then V is minuscule. That is one of the following holds

(a) [a] Φ = B_{n} and λ = λ_{n}.

(b) [b] Φ = C_{n} and λ = λ_{1}

Proof: Without loss α is the highest short root. Since α is not the highest long, there exists γ ∈ Π with α + γ ∈ Φ. Since α is the highest short root, α + γ is long, N_{αγ} = ±p_{Φ} and neither α + 2γ nor 2α + γ are roots Thus

0 ≡ [𝔊_{α}^{2}, 𝔊_{γ}] = ±2p_{Φ}𝔊_{α+γ}𝔊_{α}

Since α = β, p ≠ 2. By assumtion p ≠ p_{Φ} and so 𝔊_{α+γ}𝔊_{α} ≡ 0. Thus by 7.1.7 V is quadratic. So λ = λ_{δ} for some δ ∈ Π so that δ^{∗} appears once in the highest short root of Φ^{∗}. A glance at the highest long root of Φ^{∗} shows that δ appears once or twice in α^{∗}. Thus (λ, α^{∗}) ∈ {1, 2}. Note that there exists a composition factor for 𝕂⟨Gα, ℌ_{α}𝔊_{−α}⟩ with heighest weight the restriction of λ. Since 𝔊_{a}^{2} annihilates this composition factor (λ, α^{∗}) = 1. So λ is minuscule. □

Lemma 7.1.12 [a,b short, (a,b) not negative] Let (Φ, p, λ, α, β) be a quadratic tuple with both α and β short, α ≠ β, (α, β) ≥ 0 and p ≠ p_{Φ} ≠ 1. Then up to conjugacy under W,

Φ = C_{n}, λ = λ_{1}, α = e_{1} + e_{2} and β = e_{2} + e_{3} or β = e_{3} + e_{4}.

Proof: Suppose that α + β is a long root. Then N_{αβ} = p_{Φ} ≠ p. By 7.1.8 𝔊_{β}𝔊_{α} ≡ 0 and so N_{αβ}G_{α+β} ≡ 0. Thus G_{α+β} ≡ 0 a contradiction.

Thus α + β is not a long root. This rules out the case Φ = B_{n} and Φ = G_{2}. It also shows that (α, β) > 0 for F_{4}. Also p ≠ p_{ϕ} = 2 and in view of 7.1.11 we will be done if we can show that 𝔊_{α}^{2} ≡ 0.

Suppose that (α, β) > 0. Then ⟨α, β⟩ is of type A_{2}. So γ = β − α is a short root, α + γ is not a root and N_{βγ} = ±1 ≠ 0. Hence

0 ≡ [𝔊_{β}𝔊_{α}, 𝔊_{γ}] = N_{βγ}G_{α}^{2}

and so 𝔊_{α}^{2} ≡ 0.

Suppose next that (α, β) = 0. Then Φ = C_{n}, n ≥ 4 and without loss α = e_{1} + e_{2} and β = e_{3} + e_{4}. Let γ = e_{2} − e_{3}. Then β + γ = e_{2} + e_{4} is a root, N_{βγ} = ±1 ≠ 0 and α + γ is not a root and so

0 ≡ [𝔊_{β}𝔊_{α}, 𝔊_{γ}] = N_{βγ}𝔊_{β+γ}𝔊_{α}

and so 𝔊_{β+γ}𝔊_{α} ≡ 0. Since (β + γ, α) > 0, we are done by the previous case. □

Lemma 7.1.13 [a,b short, (a,b) negative] Let (Φ, p, λ, α, β) be a quadratic tuple with both α and β short, α ≠ β, (α, β) < 0 and p ≠ Φ_{p} ≠ 1. Then up to conjugcay under W,

Φ is of type G_{2}, λ = λ_{1}, p = 2, α = α_{1} + 2α_{2}, β = α_{1} + α_{2}

Proof: By 7.1.8 𝔊_{α}𝔊_{β} ≡ 0 and so [𝔊_{α}, 𝔊_{β}] ≡ 0.

Suppose that β = −α then [𝔊_{α}, 𝔊_{β}] = ℌ_{α}. By ?? ℌ_{α} ≡ 0 implies 𝔊_{α} ≡ 0, a contradicion. Thus β ≠ −α and (α, β) ≠ 0 implies that α + β is a root. Hence N_{αβ}G_{αβ} ≡= 0 and as

p ≠ p_{Φ} we conclude N_{αβ} = 0. p ≠ p_{ϕ} implies N_{αβ} = ±2, p_{ϕ} ≠ 2 and so Φ = G_{2} and p = 2. Let Π = {α_{1}, α_{2}} with α_{1} short. Define

Σ_{+} = {α_{1}, α_{1} + α_{2}, −2α_{1} − α_{2}}

and

Σ^{−} = −Σ^{+}

Then Φ_{short} = Σ_{+} ∪ Σ_{−} and W (Φ_{long}) acts transitively on Φ_{long}, Σ_{+} and Σ_{−}. Let ϵ ∈ {+, −} and δ, μ ∈ Σ_{ϵ} with δ ≠ μ. Then (δ, μ) is conjugate under W (Φ) to (α, β) and so 𝔊_{δ}𝔊_{μ} ≡ 0. Since p = 2 also 𝔊_{δ}^{2} ≡ 0. Moreover [G_{δ}, G_{μ}] = ±2G_{δ+μ} = 0. Put

𝔮_{ϵ} = 𝕂⟨G_{δ} | δ ∈ Σ^{ϵ}⟩

We conclude that 𝔮_{ϵ} is an commuative subalgebra of 𝔤 and that

q_{ϵ}^{2} ≡ 0

Also 𝔊_{α_{2}} commutes with 𝔊_{α_{1}+α_{2}} and with 𝔊_{−2α_{1}−α_{2}} and [𝔊_{α_{2}}, 𝔊_{α_{1}}] = ±𝔊_{α_{1}+α_{2}}. Thus [𝔊_{α_{2}}, 𝔮_{+}] ≤ 𝔮_{+}. Let 𝔩 = 𝔤_{long}. The action of W (Φ_{long}) implies [𝔩, 𝔮^{+}] ≤ q_{+}. Since W (Φ) interchanges Σ^{+} and Σ^{−} we also have [𝔩, 𝔮^{−}] ≤ 𝔮^{−}. Thus we can apply ?? conclude that

V = V_{+} ⊕ V_{−}

where V_{ϵ} = Ann_{V} (q_{ϵ}).

Since V_{ϵ} is H invariant, v_{+} ∈ V_{ϵ} for some ϵ ∈ {+, −}. Hence v_{+} is annihilated by q_{ϵ} and 𝔲 = 𝕂⟨𝔊_{δ} | δ ∈ Φ^{+}⟩. It is easy to see that 𝔤 is (as a Lie algebra) generated by 𝔮_{−} and 𝔲. Thus v_{+} =∈ V_{+} and v+ is annihilated by 𝔮_{+} and 𝔲. In particular 𝔊_{±(2α_{1}+α_{2})}v_{+} = 0 and so ℌ_{2α_{1}+α_{2}}v_{+} = 0. Since (2α_{1} + α_{2})^{∗} = 2α_{1}^{∗} + 3α_{2}^{∗} and p = 2 we have ℌ_{2α_{1}+α_{2}} = ℌ_{α_{2}}. Thus ℌ_{α_{2}}v_{+} = 0 and so λ = λ_{1}. □

Comment:there probably exists more direct proof for the preceeding lemma, but I like the proof since it treats G_{2} for p = 2 like an A_{3}

Theorem 7.1.14 [all quadratic tuples] The following table lists all quadratic tuples:

Φ  p  λ  α, β
any  any  quadratic  α long, β ≠ α, (α, β) > 0
any  odd  quadratic  α = β long
classical any  natural  \begin{matrix}α long, β ≠ α, (α, β) = 0 \\ not α = ±e_{i} ± e_{j}, β = ±e_{i} ∓ e_{j} \end{matrix}
A_{n}  any  λ_{1} (natural)  α = e_{i} − e_{j}, β = e_{j} − e_{k}
A_{n}  any  λ_{n} (natural)  α = e_{j} − e_{k}, α = e_{i} − e_{j}
C_{n}  odd  natural  α = β short

C_{n} any λ_{1} or (for p=2)λ_{1} + λ_{n} α, β short, (α, β) ≥ 0 not α = ±e_{i} ± e_{j}, β = ±e_{i} ∓ e_{j}

B_{n}, D_{n} any  spin  α = ±e_{i} ± e_{j}, β = ±e_{i} ∓ e_{j}
B_{n}  odd  spin  α = β short
D^{4}

any

λ_{m}, m ∈ {3, 4} (spin)

α = ±e_{i} ± e_{j}, β = ±e_{k} ± e_{l} number of− = m − 1 mod 2

G_{2}  2  λ_{1}  α, β long, <α, β > = −1

Proof: This is just a summary of the results of this section □

Theorem 7.1.15 [quadratic subalgebras] Let 𝕂 be a field of characteristic p ≥ 0, Φ a connected root system and 𝔤 = 𝔤_{Φ}(𝕂) the corresponding algebra. Let V = V (λ) be the irreducible restricted 𝔤-module of heighest weight λ ≠ 0. Let ∅ ≠ Ψ ⊆ Σ such that 𝔊_{α}V ≠ 0 for all α ∈ Ψ. Suppose that 𝔤_{Ψ} is quadratic on V, that is 𝔤_{Ψ}^{2}V = 0. Then

(a) [a] If V is quadratic but neither natural nor spin, then one of the follwing holds:

1. [a] There exists a tuple (α_{0}, α_{1}, …, α_{k}) of roots with diagram Δ such that Ψ = {α_{0}, α_{0} + α_{1}, …, α_{0} + … α_{k}}. Moreover, either

1. [a] All roots in Ψ are long, and Δ = A_{k+1}.

2. [b] Ψ contains a unique short root, p = 2, Δ = B_{k+1} or G_{k+1} and Φ = C_{n}, F_{4} or G_{2}

2. [b] All roots in Ψ are short, Φ = G_{2}, p = 2, |Ψ| = 2 or 3 and <α, β > = −1 for all α ≠ β ∈ Ψ.

Proof: (a): Let β_{0}, …, β_{m} be the long roots in Ψ. Then by 7.1.14 <b_{i}, b_{j} > = 1 for all 1 ≤ i < j ≤ m. Put α_{0} = β_{0} and α_{i} = β_{i} − β_{i−1} for all 1 ≤ i ≤ m. Then clearly (α_{0}, α_{1}, …, α_{m}) has diagram A_{m+1}. If Ψ contains only long roots we conclude that (a:1:1) holds. Suppose next that Ψ contains a unique root β which is not long. Then by 7.1.14 <β_{i}, β > = 1 for all 1 ≤ i ≤ m. Put α_{m+1} = β −β_{m}. Then <b_{i}, a_{m+1} > = 0 for all 0 ≤ i < m and <β, β_{m} > = −1. Thus <a_{m+1}, a_{i} > = 0 for all 0 ≤ i < m and <a_{m+1}, a_{m} > = −1. Note also that a_{m+1} is short and so (a:1:2) holds.

Suppose finally that Ψ contains two distinct roots α and β which are not long. Then by 7.1.14, p = 2, Φ = G_{2} and <α, β > = −1. If |Ψ| = 2, then (a:2) holds. So assume δ ∈ Ψ \ {a, β}. Suppose that δ is long, then by 7.1.14, <α, δ > = <β, δ > = 1 and so <α + β, δ > = 2. But α + β is a short root and we obtain a contradiction.

So δ is short. Thus by 7.1.14 we get <α, δ > = <β, δ > = −1 and so <α + β, δ > = −1. Hence δ = −(α + β), δ is unique, |Ψ| = 3 and (a:2) holds.

(??): □

7.2 Quadratic modules for Groups of Lie Type

Definition 7.2.1 [A] quadratic system is a tuple (M, V, A, D, p) such that

(a) [a] M is a finite group.

(b) [b] p is a prime and V an irreducible faithful GF (p)M-module.

(c) [c] D is a p-subgroup of M with A ≤ Z(D) and |D| > 2.

(d) [d] M = ⟨A^{M}⟩D.

(e) [e] [V, A, D] = 0.

The purpose of this section is to study and (under some extra assumptions) classify quadratic system.

Lemma 7.2.2 [[V,D,A]=0] Let (V, M, A, D, p) be a quadratic system. Then

(a) [a] [V, D, A] = 0.

(b) [b] M = O^{p}(M)D.

Proof: (a) By the definition of a quadratic system [V, A, D] = 0 and A ≤ Z(D). Thus [A, D, V] = 0 and the Three Subgroup Lemma 2.0.1 implies [D, V, A] = 0.

(b) Since M = ⟨A^{M}⟩D, M = ⟨D^{M}⟩. So (b) follows from 2.0.2 applied to M/O^{p}(M). □

Lemma 7.2.3 [imprimitive quadratic systems] Let (M, V, A, D, p) be a quadratic system and suppose that Δ is a system of imprimitivity for M on V. Then

(a) [a] p = 2 and A acts non-trivially on Δ.

(b) [b] |D/ℂ_{D}(W)| = 2 = |W^{Q}| for all W ∈ Δ with A ≰ ℕ_{M}(W).

(c) [c] O^{p}(M) acts transitively on Δ.

Proof:

Since V is faithful and V = ∑ Δ, there exists W ∈ Δ with [W, A] ≠ 0. Suppose first that A acts trivially on Δ. Then 0 ≠ [W, A] ≤ C_{W} (D) and so D normalizes W. Since M = ⟨A^{M}⟩D = C_{G}(Δ)D we conclude that M normalizes W, a contradiction to the irreducibility of V.

So A acts non-trivially on Δ. Let W with A ≰ N_{M}(W). [W, A, D] = 0 implies |W^{A}| = W^{D}| = p = 2. Also [W, ℕ_{D}(W)] ≤ C_{W} (A) and so [W, ℕ_{D}(W)] = 0. Therefore |D/C_{D}(W)| = 2.

Suppose that O^{p}(M) does not act transitively on Δ. Replacing Δ by {∑ W^{O^{p}(M)} | W ∈ Δ} we may assume that O^{p}(M) acts trivially on Δ. Thus by 7.2.2(b) M = C_{M}(Δ)D. Hence Δ = W^{M} = W^{D}, |Δ| = 2, C_{D}(Δ) = C_{D}(W) ≤ C_{M}(V) = 1 and so |D| = 2 a contradiction to the assumption. □

Lemma 7.2.4 [e-linear] Let (M, V, A, D, p) be a quadratic system and suppose that there exists a field 𝔼 such that V is a vector space over 𝔼 and M acts 𝔼-semilinear on V. Then M is 𝔼-linear on V.

Proof: Let 1 ≠ a ∈ A and let σ be the (maybe trivial) field automorphism induced by a on 𝔼. Let 𝔼_{σ} be the fixed field of σ in 𝔼. As a is quadratic on V, e − e^{σ} ∈ 𝔼_{σ} for all e ∈ 𝔼. It easy to see that this implies that 𝔼 = 𝔼_{σ} or p = 2 and 𝔼 has dimension 2 over 𝔼_{σ} has index two in F. Moreover, [V, a] is an E_{σ}-subspace centralized by D. So D is E_{σ}-linear and we may assume that 𝔼_{σ} ≠ 𝔼. Since [V, C_{D}(𝔼)] is an 𝔼-space centralized by a, C_{D}(𝔼) = 1. Thus D is isomorphic to a subgroup of Aut_{𝔼_{σ}}(𝔼) ≅ C_{2}, a contradiction to |D| ≥ 2. □

Lemma 7.2.5 [OpM irreducible in quadratic system] Let (M, V, A, D, p) be a quadratic system. Then O^{p}(M) acts irreducibly on V.

Proof: By 7.2.3 O^{p}(M) is homogenous on V. So the lemma follows from the facts that V contains precisely q^{n} − 1 irreducible O^{p}(M)-submodules and that M/O^{p}(M) is ap-group, see also ??. □

Definition 7.2.6 [dtendec] Let 𝕂 be a field, H a group and V a 𝕂H-module. Then a tensor decomposition of V for H is a tuple (𝔽, V_{i}, i ∈ I) such that

(a) [a] F ≤ End_{K}(V) is a field with K ≤ F.

(b) [b] H acts F -semilinear on V.

(c) [c] Put E = C_{H}(F) ( the largest subgroup of H acting F -linear on V). Then V_{i} is an F E-promodule.

(d) [d] As F E-modules, V and ⨂_{F} {V_{i} | i ∈ I} are isomorphic.

Lemma 7.2.7 [qtp] Comment:need to allow the case that D acts on I, giving O_{4}^{+}(q) Let D be a group with |D| ≥ 3. 1 ≠ A ≤ Z(D), K a field with char K = p, p a prime, V a faithful KD-module with [V, A, D] = 0 and (F, V_{i}, i ∈ I) a tensor decomposition of V for D. Then D acts F -linear and one of the following holds:

1. [1] There exists i ∈ I so that [V_{i}, A, D] = 0 and D acts trivially on all other V_{j}’s.

2. [2] p = 2, D is F -linear and there exist i, j ∈ I, a_{k} ∈ End_{F} (V_{k}) with a_{k}^{2} = 0 (k=i,j)and a monomorphism λ : D → (F, +) so for q ∈ D,

(a) [a] For k = i, j, q acts on V_{k} as 1 + λ(q)a_{i}.

(b) [b] D centralizes all V_{s}’s with s ≠ i, j.

Proof: Note first that as A acts quadratically on V, A is an elementary abelian p-group. Also [V, A, D] = 0 and [D, A] = 1. So the three subgroup lemma implies that [V, D, A] = 1.

By 7.2.4 M acts 𝔽-linear on V. Since A is a p-group, we may assume that the V_{i}’s are actually F A-modules and not only promodules. If D acts trivially on some V_{k}, V is a direct sum of copies of the F D-module ⨂_{F} {V_{i} | i ∈ I − k}. So the latter has the same properties as V. Thus we may assume fom now on that D acts non-trivially on each V_{i}. If |I| = 1, then 1. holds

Suppose next that |I| = 2 and say I = {1, 2}. Note that

[C_{V_{1}}(A) ⊗ V_{2}, A] = C_{V_{1}}(A) ⊗ [V_{2}, D].

D acts as scalars on [V_{2}, A] and [V_{1}, A]. Hence we may choose the promodules V_{1} and V_{2} so that [V_{i}, A, D] = 0 for i = 1, 2. For q ∈ D let q_{i} be the endomorposim q − 1 of V_{i}. Then z_{i}q_{i} = 0. Moreover, in End_{F} (V_{1} ⊗ V),

z − 1 = (1 + z_{1}) ⊗ (1 + z_{2}) − 1⊗ = z_{1} ⊗ 1 + 1 ⊗ z_{2} + z_{1} ⊗ z_{2}.

Thus [V, z, q] = 0 implies

z_{1} ⊗ q_{2} = −q_{1} ⊗ z_{2}

If z_{1} = 0 then as V is faithful, z_{2} ≠ 0. Thus the previuos equation implies q_{2} = 0 for q, a contradcition to the assumption that D does not centalize V_{2}. Hence both z_{1} and z_{2} are not zero. Choosing q = z we see that p = 2. Hence for arbitray q, q_{1} = λ(q)z_{1} and q_{2} = λ(q)z_{2} for some λ(q) ∈ F. Thus 2. holds in this case.

Suppose now that |I| ≥ 3. Say 1, 2 ∈ I and but W = ⨂_{F} {V_{i} | i ∈ I \ {1, 2}. Then V ≅ (V_{1} ⊗ V_{2}) × W. Then by the prviuos case D acts faithfully on V_{1} ⊗ V_{2} z − 1 and q − 1 are linear dependent on V_{1} ⊗ V_{2}. Let λ = λ(q) be as above. Then on v_{1} ⊗ v_{2}

q − 1 = (1 + λz_{1}) ⊗ (1 + λz_{2}) − 1 ⊗ 1 = λ(z_{1} ⊗ 1 + 1 ⊗ z_{2} + λz_{1} ⊗ z_{2}).

On the otherhand z − 1 = z_{1} ⊗ 1 + 1 × z_{2} + z_{1} ⊗ z_{2} and we conclude that λ = 0, 1 and so |D| = 2, a contradiction. □

Lemma 7.2.8 [quadratic on exterior powers] Let 𝔽 be a field with p := char 𝔽 ≥ 0, A a group, V a faithful, finite dimensional 𝔽A-module. Put n = dim_{𝔽} V, let 2 ≤ m ≤ n−2 and suppose that A acts unipotenly on V and quadratically on ⋀^{m} V. Then A is an elementary abelian p-group and one of the follwing holds.

1. [a] dim[V, A] = 1.

2. [b] dim V/C_{V} (A) = 1.

3. [c] p = 2, m ∈ {2, dim V − 2} and A ⊆ 1 + 𝔽t for some t ∈ End_{𝔽}(V) with t^{2} = 0.

4. [d] p = 2, V = X ⊕ Y, where X and Y are 𝔽A-submodules of V with dim X = 4 and [Y, A] = 0. Moreover, put U = C_{X}(A). Then U = [V, A] is 2-dimensional and A is contained in an isotropic subspace of End_{𝔽}(X/U, U).

5. [e] p = 2, m = 2, dim V = 4 and A acts cubic but not quadratic on V.

6. [f] p = 2 = |A|.

Proof: Suppose first that all elements in A are transvections. Then (1) or (2) holds. So we may assume that there exists d ∈ A with dim[V, d] ≥ 2.

Suppose A does not act quadratically on V. Then there exists D ≤ A and an 𝔽D submodule X in V such that D is not quadratic on V and if we put k = dim U, then either p = 2, |D| = 4 and k = 4 or p ≠ 2, D is cylic and k = 3. Let l = k −1 if m ≥ k −1 and l = 1 if m < k. Then D does not act quadratically on ⋀^{l} X. Suppose that m − l ≤ n − k. Then ⋀^{l} X ⊗ ⋀^{m−1} V/X is isomorphic to an 𝔽D section of ⋀^{m} V and we obtain a contradiction to 7.2.7. Thus n−3−l ≥ m−l−1 ≥ n−k ≥ n−4. Thus l = 1, n−2 = m and k = 4. Sy by choice of l, m ≤ k − 2 = 2. Hence n = 4. Let D = ⟨a, b⟩. Then C_{V} (D) is 1-dimensional and so C_{V} (D) = C_{V} (A). Moreover, [V, D] = C_{V} (a)+C_{V} (b) and C_{V} (a)/C_{V} (A) is 1-dimensional. Thus [V, D, A] ≤ C_{V} (A), A is cubic on V and (5) holds.

So we may assume from now on that A acts quadratically on V.

Suppose that p ≠ 2. Let X be 2-dimensional non-trivial 𝔽⟨d⟩-submodule in V. Then d acts quadratically on X ⨂ ⋀^{m−1} V/X and we conlude from 7.2.7 that d centralizes ⋀^{m−1} V/X and so also V/X. Thus [V, d] ≤ C_{X}(d) and so [V, d] is 1-dimensional, a contradiction to the choice of d.

Thus p = 2 and we may assume that |A| ≥ 4. Let 1 ≠ a, b ∈ A and put D = ⟨a, b⟩. Suppose that C_{V} (a) ≰ C_{V} (b). Then there exists a non-trivial 2-dimensional 𝔽D-subspace X in V with [X, a] = 1. Since D acts quadratically on X ⨂ ⋀^{m−1} V/X we conclude from 7.2.7 that a centralizes V/X. Thus [V, a] ≤ C_{X}(A) = [X, b]. Since this hold for all such X we get [V, a] = [C_{V} (a), b], a is a transvection and dim[V, b] ≤ 2. In particular, a ≠ d and so C_{V} (d) = C_{V} (A). By a dual argument, [V, d] = [V, D]. If [V, d] is 2-dimensional it is easy to verify that (4) holds.

So assume that dim[V, d] ≥ 3. Hence d ≠ a. If C_{V} (a) ≠ C_{V} (d) we conclude that C_{V} (a) ≰ C_{V} (d), a contradiction. Thus C_{V} (a) = C_{V} (d) for all 1 ≠ a ∈ A.

Replacing V by V^{∗} and m by n − m if necessary we may assume that n ≥ 2m.

Let t = d − 1 ∈ End_{𝔽}(V).

Suppose that A ⊆ 𝔽d. Then V = V_{0}⊕V_{1}⊕V_{k}, where A centralizes V_{0}, k ≥ 3 and V_{1}, … V_{k} are pairwise isomorphic non-trivial 2-dimensional 𝔽A-submodules. If m ∈ {2, n − 2}, then (3) holds. So suppose for a contradiction that 3 ≤ m ≤ n − 3. Let Y = V_{0} + V_{3} + … V_{k}. Then dim Y = n − 6 ≥ 2m − 6 ≥ m − 3. Thus ⋀^{m} V has a section isomorphic to V_{1} ⊗ V_{2} ⊗ V_{3} ⊗ ⋀^{m−3} Y and we obtain a contradiction to 7.2.7.

So we may assume that there exists a ∈ A with a ∉ 𝔽d. Thus there exists x ∈ V with 𝔽[x, a] ≠ 𝔽[x, d]. Put D = ⟨a, d⟩. Since C_{V} (a) = C_{V} (d) we conclude that X := 𝔽⟨x^{D}⟩ is 3dimensional. Since n ≥ 2m we have dim V/X = n−3 ≥ 2m−3 ≥ m−1. Moreover, equality holds only for n = 2m and m = 2. But n ≥ 2 dim[V, d] ≥ 6 and so m − 1 < dim V/X. Since X ⊗ ⋀^{m−1} V/X is a section of ⋀^{m} V we conclude from 7.2.7 that D centralizes V/X. Thus [V, D] ≤ C_{X}(D) and [V, D] is at most 2-dimensional. This contradiction completes the proof of 7.2.8. □

Definition 7.2.9 [strong quadratic] Let M ∈ Lie_{p} and V a faithful 𝔽_{p}M-module. Then V is called strongly quadratic if there exists A ≤ D ≤ M such that (i) [a] (M, V, A, D, p) is a quadratic system.

(ii) [b] If p = 2 then |Φ_{A^{g}}| ≥ 2 for some g ∈ M̂ with A^{g} ∈ U.

Let (M, V, A, D, p) be a quadratic system, where M is a group of Lie type. Then we may assume that D is a subgroup of the unipotent group U. As in ?? for X = D, A let

Φ_{X} = {α ∈ Φ^{+} | X ∩ U_{α}^{+} ≰ U_{α}^{−}}

and

T_{X} = ∑_{α∈Φ_{X}} 𝕂𝔊_{α} ≤ 𝔤_{Φ}(𝕂).

Theorem 7.2.10 [same characteristic quadratic systems ] Let M ∈ Lie_{p} and V an irreducible, strongly quadratic 𝔽_{p}M-module. Then M and V are as listed below.

1. [1] M is a quotient of SL_{n}(𝕂) or SU_{n}(𝕂) and λ = λ_{i} for some 1 ≤ i ≤ n.

2. [2] M ≅ Ω_{2n+1}(𝕂), p ≠ 2 and λ = λ_{1} or λ_{n} and M ≅ Spin_{2n+1}(𝕂).

3. [3] M ≅ Sp_{2n}(𝕂) and λ = λ_{i} for some 1 ≤ i ≤ n.

4. [4] M ≅ Ω_{2n}^{±}(𝕂) and λ = λ_{1}, λ_{n−1} or λ_{n}.

5. [5] M ≅ ^{3}D_{4}(𝕂) and λ = λ_{1}, λ_{n−1} or λ_{n}.Comment:U: just λ_{1}

6. [6] M ≅ E_{6}(𝕂) or ^{2}E_{6}(𝕂) and λ = λ_{1} or λ_{6}. Comment:U: justλ_{1}

7. [7] M ≅ E_{7}(𝕂) and λ = λ_{1}.

8. [8] M ≅ G_{2}(𝕂) and λ = λ_{1} or p = 3 and λ = λ_{1} or λ_{2}.

9. [9] M ≅ F_{4}(𝕂) and λ = λ_{1} or p = 2 and λ = λ_{1} or λ_{4}.

Proof: The strategy of the proof is to translate the fact that we have a quadratic system into the fact that we have a quadratic tuple for a Lie algebra related to M and then to use the classification of quadratic tuples.

Let 𝔽 = End_{𝔽_{p}M}(V) and set W = V ⊗_{𝔽} 𝔼 where 𝔼 is the algebraic closure of 𝔽. Then W is an irreduzible and faithful 𝔼M-module, therefore W = W (λ) for some weight λ and

W ≅ ⊗_{i=0}^{a−1}W (μ_{i})^{(p^{i})},

where |𝕂| = q = p^{a} and λ = ∑_{i=0}^{a−1} p^{i}μ_{i} with μ_{i} p-restricted weights for 0 ≤ i ≤ a − 1, see ...

By 6.2.4 the modules W (μ_{i})^{(p^{i})} are irreducible modules for 𝔤_{Φ}(𝔼), for 0 ≤ i ≤ a − 1. We claim that we may assume that W_{1} := W (μ_{1}) is a root faithful module for 𝔤_{Φ}(𝔼). Assume that W_{1} is not a root faithful module for 𝔤_{Φ}(𝔼). Then Φ is of type B_{n}, C_{n}, F_{4} and p = 2 or Φ is of type G_{2} and p = 3, see 6.1.7. In all these cases there exists an automorphism of M which induces a bijection from Φ onto Φ^{∗}. The Lie algebra 𝔤_{Φ^{∗}}(𝔼) acts faithfully on W_{1}, see 6.1.7. Therefore we may assume that 𝔤_{ϕ}(𝔼) acts root faithfully on W_{1}.

Now 7.2.7 implies that either (a) [1] V ≅ V (μ_{i})^{(p^{i})} for some p-restricted weight μ_{i} or

(b) [2] p = 2 and V ≅ V (μ_{i})^{(p^{i})} ⊗ V (μ_{j})^{(p^{j})}.

Assume that (2) holds. Then, as p = 2, in fact |Φ_{D}| ≥ 2. Let α and β be two different elements in Φ_{D} such that α < β and let g_{α} and g_{β} be elements in D such that g_{α} ∈ U_{α}^{+} \U_{α}^{−} and g_{β} ∈ U_{β}^{+} \ U_{β}^{−}. Then by 7.2.7 there exists k_{α}, k_{β} ∈ 𝕂 and a_{l} ∈ End_{𝕂}(V_{l}) such that g_{x}

acts on V_{l} as 1 + k_{x}a_{l} for x ∈ {α, β} and l ∈ {i, j}. Hence g_{α} acts on V_{i}^{(p^{i})} and on V_{j}^{(p^{j})} as

\frac{k_{α}}{k_{β}} (g_{β} − 1).

Now 6.3.4(5) implies that 𝔊_{α} acts trivially on both, V_{i} and V_{j}. As we saw above we may assume that 𝔊_{α} does not act trivially on both, V_{i} and V_{j}, which is a contradiction. Hence (1) holds.

Thus W is a p-restricted 𝔼M-module and by 6.2.4 it is a root faithful and irreducible 𝔤_{Φ_{D}}(𝔼)-module, as well, which admits by 6.3.4a quadratic tuple. If Φ_{D} contains a long root, then by 7.1.7 W is quadratic. Hence if Φ_{D} contains a long root, then by 7.1.3 we have the possibilities (1.) – (9.) of the assertion.

If M is a twisted group and of type B_{2}, F_{4} or G_{2}, then by our choice of the ordering on Φ, the set Φ_{D} consists only of short roots, see 6.3.3.

Hence we may assume that Φ_{D} consists only of short roots. Then by 7.1.10, 7.1.11, 7.1.12 and 7.1.13 p = p_{Φ}, Φ is of type C_{n}, n ≥ 3 and λ = λ_{1} or λ = λ_{1} + λ_{n} or p ≠ p_{Φ}, Φ is of type B_{n}, C_{n} or G_{2} and λ = λ_{n}, λ_{1}, λ_{1}, respectively. If Φ is of type C_{n}, ≠ 3 and λ = λ_{1} + λ_{n}, then by ?? V (Λ) ≅ V (λ_{1}) ⊗ V (λ_{n}), which is not possible as we saw above.

These cases are listed in (2.), (3.) and (8.) of the assertion. □

Theorem 7.2.11 [same characteristic quadratic systems with outer automorphism]

Let (M, V, D, A, p) be a quadratic system such that

(a) [a] F^{∗}(M) is a quotient of ^{σ}G_{Φ}(𝕂) and char K = p.

(b) [b] D ≰ F^{∗}(M).

Then p = 2, M = O_{2n}^{ϵ}(𝕂_{σ}) and V is the corresponding natural module.

Proof:

Let H = F^{∗}(M). By 7.2.5 we have that H acts irreducibly on V. Let 𝔽 = End_{H}(V). Then by 7.2.4, M acts 𝔽-linear on V. Let 𝔼 be the algebraic closure of 𝔽 and W = V ⊗_{𝔽} 𝔼. Then W is a simple 𝔽H-module.

By the Strong Steinberg Tensor Product Theorem ?? W is as an 𝔼H-module isomorphic to W = ⨂_{σ∈I} V (λ_{σ})^{σ}, where I is a set of Frobenius automorphisms of L and λ_{σ} is a non-zero p-restricted weight. Moreover, the extension M̂ of M preserves this tensor decomposition and I is invariant under D via right multiplication.

Suppose first that D acts non-trivial on I. Then by 7.2.7 we conclude that |I| = 2 and dim V (λ_{σ}) = 2 for all σ ∈ I. Thus Φ = A_{1}, N := V (λ_{σ}) is the natural module for Ĥ and 𝔽 = 𝕂. Since C_{D}(I) acts 𝔼-linear on N we have that C_{D}(I) ≤ H. So the outer automorphism group induced by D on H is generated by a field automorphism τ of order 2. Since I = {σ, στ} we conclude that W = (N ⊗ N^{τ})^{σ} and so M ≅ O_{4}^{−}(𝕂_{τ}) and V is the natural O_{4}^{−}(𝕂_{τ}))-module.

Suppose next that D acts trivially on I but |I| ≥ 2. Then by ?? |I| = 2 and D acts linearly dependently on V_{σ} for all σ ∈ I. By induction on |I| we conclude that Φ = D_{n}, p = 2 and V (λ_{σ}) is the natural module. It follows, if a ∈ D, then [V (λ), a] is even dimensional if and only if a ∈ H. As |DH/H| = 2 and |D| ≥ 3 we have D ∩ H ≠ 1. And we obtain a contradiction to the linear dependency of D.

Suppose finally that |I| = 1. It follows from ?? the outer automorphism group induced by H is a standard graph automorphism of order p. Also Φ is of type A_{n}, D_{n} or E_{n}.In particular, |DH/H| = p and p ∈ {2, 3}. Moreover, the extensions of the 𝔼H-module W to G = G_{ϕ}(𝔼) and to the Lie-algebra 𝔤_{Φ}(𝔼) are invariant under M.

Since D is a p-group we may assume that D ≤ U⟨g⟩, where U = ∏_{α∈Φ^{+}} X_{α} is the standard maximal unipotent subgroup of G and g is the standard graph automorphism.

1^{∘} [1] Suppose that W is a strong quadratic module for G. Then Φ = D_{n}, n ≥ 3 and W is natural or Φ = A_{2m−1} and λ = λ_{m}.

Since λ = λ^{g} this follows immediately from ??.

2^{∘} [2] If p = 3, then A ≤ H.

Note that A acts quadratically on W. As p is odd, it follows that A centralizes the abelian group N_{G}(U)/U. Since g acts non-trvial on N_{G}(U)/U we conclude that A ≤ H.

3^{∘} [3] D ∩ H ≠ 1.

This follows immediately from |D| > 2 and (2^{∘}).

In particular, there exists 1 ≠ b ∈ D ∩ H with |b| = p and a d ∈ D \ H. If A ≤ H we choose b ∈ A and if A ≰ H we choose d ∈ A. So in any case [W, b, d] = 0. For 1 ≤ i ≤ |Π|, let H_{i} be the maximal parabolic subgroup of H corresponding to Π \ {α_{i}}. Also let U_{i} be the unipotent radical of H_{i}. We now split the analysis into four different cases:

Case 1 [d4] The case p = 3 does not occur.

Suppose that p = 3 and so Φ = D_{4}. Suppose that b ∉ U_{2}. Then 7.2.7 applied to some chief factor for M_{2}/U_{2}⟨d⟩ on W gives a contradiction. Thus b ∈ U_{2}. It is easy to [U_{2}/Z(U_{2}), d] is at least 4 dimensional over 𝔼. On the other hand, C_{U_{2}}(b)/Z(U_{2}) has codimension at most 1 in U_{2}/Z(U_{2}) and thus [C_{U_{2}}(b), d] ≰ 𝔼dZ(U_{2}). Since [W, b, [C_{U_{2}}(b), d]] = 0 we conclude that W is strongly quadratic.

Case 2 [dn] Suppose that p = 2 and Φ = D_{n} with n ≥ 3. Then V is the natural module.

Suppose that b ∈ U_{1}. If [C_{H_{1}}(b), d] ∩ U_{1} ≰ 𝕂b, then W is strongly quadratic and we are done by (1^{∘}). So we may assume that [C_{H_{1}}(b), d] ∩ U_{1} ≤ 𝕂b. In particular, [U_{1}, d] ≤ 𝕂b, d induces a transvection with center 𝕂b on the orthorgonal space U_{1}, b is non singular in U_{1} and [d, C_{H_{1}}(b)] ≤ U_{1}. Thus [C_{H_{1}}(b), d] ≤ 𝕂b. Let B/𝕂b = Z(C_{H_{1}}(b)/𝕂b). Then either (n, q) ≠ (3, 2) and |B| = 2|𝕂|, or (n, q) = (3, 2) and B ≅ D_{8}. In either case all involutions in B \ 𝕂b are transvections on the natural module N for H. If |d| = 2 we conclude that ⟨b^{C_{H}(d)}⟩ contains a long root element c with c ∈ U_{1}. Then [V, c, d] = 0 and the preceeding argument applied with c in place of b and realizing that c is not singular in U_{1} we get that W is strongly quadratic.

If d has order four, then (n, q) = (3, 2) and |C_{N}(d)| ≥ 8. Since d is not an involution, C_{N}(d) is not isotropic and so d centralizes non-degenerate 2-space in N. Thus d is contained in a subgroup L of M isomorphic to O_{4}^{ϵ}(2). Let T be a subgroup of L such that d normalizes T and T hasof order 3^{2} and 5 if ϵ = + or −, respectively). Let W_{1} be an faithful irreducible T ⟨d⟩-submodule of W. Then W_{1} has four different T -eigenspaces which are permuted transitively by d. Therefore, [W, b, d] ≠ 0.

Suppose next that b ∉ U_{1}. By ?? we have that [U_{1}, b] is even dimensional, while [U_{1}, d] is odd dimensional. Thus C_{U_{1}}(b) ≠ C_{U_{1}}(d). Suppose first that C_{U_{1}}(b) ≰ C_{U_{1}}(d). As [W, [C_{U_{1}}(b), d], b] = 0 we conclude that W is strongly quadratic, and we are done. Thus suppose C_{U_{1}}(b) ≤ C_{U_{1}}(d). Then there exists an element t ∈ C_{U_{1}}(d) \ C_{U_{1}}(b) and we can replace b by [t, b]. Since [t, b] ≤ U_{1} we are done by the preceeding paragraph.

Case 3 [an] Suppose that p = 2 and Φ = A_{n} with n ≥ 2. Then Φ = A_{3} = D_{3}

Put Z = U_{1} ∩ U_{2}. Suppose that n = 2. Then all the involutions in U are contained in U_{1} ∪ U_{2}. Since b is an involution centralized by d and U_{1}^{d} = U_{2} we get that b ∈ Z. Thus [W, [U_{1}, b], b] = 0 and W is strongly quadratic, a contradiction.

Thus n ≠ 2 and we may assume for contradiction that n ≥ 4. Suppose that b ∉ U_{1}U_{n}. Then by induction we see that every non-central chief factor for H_{1} ∩ H_{n}⟨g⟩ is a natural orthogonal module and n = 5. Moreover as b has order two, 2 dim C_{U_{1}}(b) ≥ dim U_{1} = n ≥ 4 and so C_{U_{1}}(b) ≰ Z. Thus [C_{U_{1}}(b), d] ≠ 1 and W is strongly quadratic. Thus by (1^{∘}), λ = λ_{3}. But then M_{1} ∩ M_{5} has a 4-dimensional composition factor.

Thus b ∈ U_{1}U_{n}.

Suppose that b ∈ Z. Then [W, [U_{1}, g], b] = 0 and so W is strongly quadratic. Thus by ?? λ = λ_{m} where n = 2m − 1. But then d acts non-trivially on C_{W} (U_{1}U_{n})) and C_{W} (U_{1}U_{n}) = [W, b], a contradiction.

Thus b ∉ Z and so also b ∉ U_{1}. Thus C_{U_{1}}(b) has codimension 1 in U_{1} and since n ≥ 4, T := [C_{U_{1}}(b), d] ≰ 𝔽bZ. Since [W, T, b] = 0, W is strongly quadratic and again λ = λ_{m}, where n = 2m − 1. In particular, n is odd and so n ≥ 5 and m = n + 1 − m ≥ 3. Let N be the natural module for H. Then dim[N, b] = 2 and 7.2.8 shows that C_{N}(b) ≤ C_{T} (b) and [N, T] ≤ [N, b]. It follows that T U_{1}/U_{1} is contained in a 1-dimensional subspace of U_{1}U_{n}/U_{1}, a contradiction to n ≥ 4. □

7.3 Some random results

Lemma 7.3.1 [half quadratic] Let 𝔽 be a field with char F = p > 0 and p ≠, let A be a finite abelian group, F an 𝔽A-module D the set of non-trivial quadratically acting elements in A. Suppose that |D| ≥ \frac{|A^{#}|}{2}. The one of the following holds:

1. [1] A acts quadratically on V.

2. [2] p = 3 and |A/B| = 9 where B = C_{A}([V, A]).

Let E be a maximal quadratic subgroup of A. If E = A then (1) holds. So suppose A ≠ E. Let |A/E| = p^{n}. For a ∈ D \ E and put E_{a} = {e ∈ E | ea ∈ D}. Let e ∈ E_{a}. Then by ?? ⟨e, a⟩ is quadratic and we conclude that E_{a} = C_{E}([V, a]). In particular, E_{a} is a subgroup of E_{a}. Note also that E_{a}⟨a⟩ is quadratic and contains all the quadratic elements in E⟨a⟩ not contained in E. In particular, by maximality of E, E_{a} ≠ E. Thus E_{a}a contains at most \frac{1}{p}|E| quadratic elements.

Hence

|D| ≤ |E| − 1 + \frac{p^{n} − 1}{p}|E|

On the otherhand

|D| ≥ \frac{1}{2}|A^{#}| = \frac{1}{2}(p^{n}|E| − 1)

Hence

\frac{1}{2}(p^{n}|E| − 1) ≤ |E| − 1 + \frac{p^{n} − 1}{p}|E|

(p^{n+1} − 2p^{n} − 2 − 2p) ≤ −\frac{p}{|E|} ≤ 0

(p − 2)(p^{n} − 2) ≤ 6

Thus p = 3 and n = 1. So A = E⟨a⟩ and E_{a} centralizes both [V, E] and [V, a]. Thus E_{a} ≤ B. If E_{a} < B, then A = EB or A = B⟨a⟩ and in both cases A acts quadratically, contradicting the maximal choice of E. Thus B = E_{a} and (2) holds. □

70 CHAPTER 7. QUADRATIC MODULES

Bibliography

[Bo] A.Borel, Springer Lecture Notes 131.

[Ca] R.Carter, Simple groups of Lie type

[LGCP] The Local Structure Of Finite Groups of Local Characteristic p. (2062)

[GLS] D. Gorenstein, R. Lyons and R. Solomon, The Classification of the Finite Simple Groups, Mathematical Surveys and Monographs, 40.3. American Mathematical Society, Providence, RI, 1998.

[St] R. Steinberg, Lectures On Chevalley Groups (1967) Yale University

Index

basis, 30 Chevalley basis, 39 quadratic, 49 root subsystem, 31 roots, 29