A Characterization of the Natural Module for some Classical Groups

U.Meierfrankenfeld

May 2, 2003

preliminary version

1 Introduction

Let K be a skew field, R a ring, G a ”classical” group defined over K, Z a ”long root” group contained in G and N a ”natural” KG-module, where N is allowed to be finite or infinite dimensional over K. Further, put L = C_{G}([N, Z]) and let V be an RG-module such that [V, Z, L] = 0 and V = [V, G]. The main goal of this paper is to prove that any such V has to be of the form M ⊗_{K} N , for some (R, K)-module M with G acting trivally on M. This is achieved in Theorems A and B. We became interested in this problem through the work of J.I.Hall and R.E.Phillips [Ha1], [Ha2], [Ph] on groups of finitary transformations. They classified certain classes of such groups. In Theorem C we are able to classify the corresponding modules. Theorems A, B and C partially generalize similar results found in [Cu],[LP], [Ha3], and [Tf].

In order to state exactly what we mean with ”classical” groups, ”natural ”module” and ”long root” groups we now introduce some notations and definitions which will be used throughout the paper:

K is a skew field.

R is a ring.

G is a group.

An (R, K)-module M is an abelian group , which is a left R-module and a right vector space over K such that (rm)k = r(mk) for all r ∈ R, k ∈ K and m ∈ M. M is called unitary if R has a unit 1 and 1m = m for all minM.

An RG-module M is an abelian group , which is a left R-module and a right ℤG-module such that (rm)k = r(mk) for all r ∈ R, k ∈ K and m ∈ M.

N is a left vector space over K. Ñ is a subspace of the dual space N^{∗} of N with C_{N}(Ñ) = 0, i.e for every 0 ≠ n ∈ N there exists ñ ∈ Ñ such that nñ ≠ 0. (Note that Ñ is a right vector space over K and so a left vector space over K_{op}, the opposite skew field of K.)

For n ∈ N and n^{∗} ∈ N^{∗} define t(ñ, n) ∈End_{K}(N) by

v.t(n^{∗}, n) = v + vn^{∗} · n

For a subspace X of N and a subspace X̃ of Ñ the following subgroup of GL_{K}(N) was introduced in [CH]

T (X̃, X) = ⟨t(x̃, x)|x ∈ X, x̃ ∈ X̃, xx̃ = 0⟩.

For the convenience of the reader we will recall the definition of a pseudo quadratic space (see [Ti] and [Gr] for basic properties of pseudo quadratic spaces).

(N, q, f) is a (σ, s)-pseudo quadratic space provided that

(PQ1) σ is an anti-automorphism of K and 0 ≠ s ∈ K such that for all x ∈ K

x^{σ^{2}} = x^{s} = s^{−1}xs and s^{σ} = s^{−1}.

(PQ2) f is a trace-valued (σ, s)-hermitian form on N, i.e

f : N × N → K is biadditive,

f(μv, λw) = μf(v, w)λ^{σ},

f(v, w) = sf(w, v)^{σ},
f(v, v) ∈ K_{+},

for all μ, λ ∈ K and v, w ∈ N. Here K_{+} is the additive subgroup of K defined by K_{+} = {k + sk^{σ}|k ∈ K}.

(PQ3) q is a map from N to \overline{K} = K/K_{−} so that

q(v, w) = q(v) + q(w) + \overline{f(v, w)},
q(λv) = λ ∗ q(v).

Here K_{−} = \{k − sk^{σ}|k ∈ K\} and k ∗ \overline{λ} = \overline{kλk^{σ}}.

(N, q, f) is called a quadratic space if σ = id_{K} and s = 1, that is if K_{−} = 0. Note that in this case K is necessarily commutative.

We mention some other special cases of pseudo-quadratic spaces. If σ = id_{K}, s = −1 and char K ≠ 2, then K = K_{−} and so q = 0. Hence q is redundant and (N, f) is a symplectic space. The symplectic spaces over fields of even characteristic are also included, since they can be written as N/N^{⊥}, where N is an appropriate quadratic space. More generally, any vector space with a trace-valued (σ, s)-hermitian form can be written as N/N^{⊥}, where N is an appropriate pseudo-quadratic space (see [Ti]). Finally, the case |σ| = 2 and s = 1 covers the unitary spaces over fields.

O = O(N, q, f) is the group of invertible K-linear transformations g of N such that f(v, w) = f(vg, wg) and q(v) = q(vg) for all v, w ∈ N.

If U is a subset of N, then U^{⊥} = {v ∈ N|f(v, u) = 0 for all u ∈ U} . A subspace U of N is called singular provided that the restrictions of q and f to U vanish.

rad N = {n ∈ N^{⊥}|q(n) = 0.

N is called degenerate if rad N ≠ 0 and defective if N^{⊥} ≠ 0.

S(i) is the set of i-dimensional singular subspaces of N.

The Witt index of (N, q, f) is the maximal dimension of a singular subspace in N. Note that the Witt index can be zero, any positive integer or infinite.

For U ∈ S(i) let P_{U} = N_{O}(U), Q_{U} = C_{O}(U^{⊥}/U) ∩ C_{O}(U), T_{U} = C_{O}(U^{⊥}) and Z_{U} = C_{O}(N/U). LetΣ be a set of subgroups of G. We will consider the following hypotheses:

Hypothesis (A) 1. Σ is the set of all {T (X̃, X) where X and X̃ are 1-dimensional subspaces of N and Ñ, respectively, with XX̃ = 0; and G = ⟨Σ⟩ = T (Ñ, N)

2. N is at least 3-dimensional over K.

Hypothesis (B) 1. Σ = {T_{U}|U ∈ S(1)} and G = ⟨Σ⟩ where (N, q, f) is a nondegenerate pseudo-quadratic space with Witt index at least two.

2. If (N, q, f) is quadratic, then (N, q, f) defective.

Hypothesis (C) Σ = {Z_{U}|U ∈ S(2)} and G = ⟨Σ⟩, where (N, q, f) is a nondegenerate quadratic space with Witt index at least two and dim N ≥ 5.

Hypothesis (C*) Hypothesis (C) holds and if char K = 2, then dim N/N^{⊥} ≥ 6, and if |K| = 2 and dim N = 6, then (N, q, f) has Witt index 2.

We define the following graph on Σ. Let Z_{1}, Z_{2} ∈ Σ. Then Z_{1} and Z_{2} are adjacent if, in case (A), ⟨Z_{1}, Z_{2}⟩ ≅ SL_{2}(K), or if, in case (B) or (C), [N, Z_{1}] ∩ [N, Z_{2}]^{⊥} = 0.

For convenience we view N under Hypothesis (A) as a singular pseudo quadratic space. In particular, O = GL_{K}(N) and Q_{X} = C_{GL_{K}(N)}(X) ∩ C_{GL_{K}(N)}(N/X).

We are now able to state our main results:

Theorem A Suppose that (G, Σ) fulfills Hypothesis (A),(B) or (C*) from above. Let Z ∈ Σ and put L = C_{G}([N, Z]). Let V be an RG-module with [V, Z, L] = 0. Then there exist an (R, K)-module M and an R-submodule C of M ⊗_{K} N with [C, G] = 0 such that one of the following holds:

1. [V, G] and (M ⊗_{K} N)/C are isomorphic as RG-modules.

2. G fulfills (B), |K| = 4, σ ≠ id, dim N = 4 and [V, G]/C_{V} (G) and M ⊗_{K} N are isomorphic as RG-modules.

Theorem B Assume that (G, Σ) fulfills Hypothesis (A). Put L_{0} = C_{G}([N, Z])∩C_{G}([Ñ, Z]) and let V_{0} be an RG-module with [V_{0}, Z, L_{0}] = 0. Then there exists an (R, K)-module M and an (R, K_{op})-module M̃ such that

[V_{0}, G] is isomorphic to M ⊗_{K} N ⊕ M̃ ⊗_{K_{op}} Ñ as RG-modules.

Theorem C Remark: Need Witt index assumption ? Suppose that (G, Σ) fulfills Hypthesis (A), (B) or (C) and that N/N^{⊥} is infinite dimensional over K. Let R be a divison ring and V be a non-trivial irreducible finitary RG-module. Then one of the following holds:

1. There exists an irreducible (R, K)-module M, which is finitely generated over R so that V is isomorphic to M ⊗_{K} N as RG-module.

2. Hypothesis (B) holds and there exists an irreducible (R, K_{op})-module M̃, which is finitely generated over R, so that V is isomorphic to M̃ ⊗_{K_{op}} Ñ as RG-module.

Some remarks on Hypothesis’ (B) and (C). In case (B) the quadratic space are assumed to be defective to ensure that T_{U} ≠ 1 for U ∈ S(1) (see 3.2). Note that Hypothesis (C) can be used to characterize the natural module of G for non-defective quadratic spaces. The assumption that N contains 2-dimensional singular subspaces is needed in the proof of 4.8. We do not know whether Theorem A holds also in the case where the maximal singular subspaces are 1-dimensional.

Under hypothesis (C) the assumption that (N, q, f) is quadratic rather then pseudoquadratic is redundant. Indeed, if N is pseudo-quadratic but not quadratic, it is easy to see that (G, V) fulfills the hypothesis (B). (C*) is needed since, if char K = 2 and dim N/N^{⊥} < 6, or if |K| = 2, dim N = 6 and N has Witt index three, the half-spin module for G fullfills Hypothesis (C) but not the conclusion of Theorem A.

We conclude this introduction with some remarks on the structure of M ⊗_{K} \overline{N}, where \overline{N} = N/N^{⊥}. Since M is a right vector space over K, we can decompose M into a direct sum of 1-dimensional K-subspaces. This leads to a ℤG-decomposition of M ⊗_{K} \overline{N} into a direct sum of copies of N. But as an RG-module M ⊗_{K} \overline{N} might still be irreducible or indecomposable. For example, if R = M = K, ϕ is an embedding of K into itself and M is regarded as an R-module by multiplication from the left and as a K-module by multiplication from the right by ϕ(k), then M ⊗_{K} \overline{N} is an irreducible RG-module. Now if ϕ is not onto, then M ⊗_{K} \overline{N} is not irreducible as a ℤG-module. In particular, M ⊗_{K} \overline{N} is not isomorphic to N. For another example, let K be a field with a non-trivial derivation δ, ( that is a map δ : K → K with δ(kl) = δ(k)l + kδ(l)), R = K and M = R^{2}, viewed as left

vector space over R. Embed K into Hom_{R}(M) by mapping k to ( \begin{matrix}k & 0 \\ δ(k) & k \end{matrix} ). Since δ is

a derivation this map is indeed a homomorphism. Via this homomorphism M becomes an (R, K)-module and M ⊗_{K} \overline{N} is as an RG-module a non-split extension of \overline{N} by \overline{N}.

On the otherhand, if R = K and K is an algebraic Galois extension of its ground field, M ⊗_{K} \overline{N} is as an RG-module the direct sum of modules algebraically conjugate to \overline{N}. Indeed, this follows easily from 2.1 below.

2 Preliminaries

P reliminaries
KML

Lemma 2.1 Let K, L be such that L is isomorphic to a subfield of K and M an indecomposable unitary (K, L)-module.If L is algebraic and Galois over its ground field, then there exist a ℤ-isomorphims α : M → K and a field monomorphism σ : L → K such that α(kml) = kα(m)l^{σ} for all k ∈ K, m ∈ M, l ∈ L.

Proof: Let Q be the ground field of L and σ the isomorphism from Q to the ground field of K. View M as a left vector space over K. Since M is unitary, each q ∈ Q acts as scalar multiplication by q^{σ} on M. Let l ∈ L and f the minimal polynomial of l over Q. Then f^{σ}(l) = 0 as an element of End_{K}(M). Since L : Q is Galois, f splits over L and has no double roots. Since K contains a subfield isomorphic to L, f^{σ} splits over K and has no double roots. Thus as a left vector space over K, M decomposes into the direct sum of the eigenspaces for l on M. Since L is commutative, L normalizes each of the eigenspaces. But M is indecomposable as a (K, L)-module and so l acts as a scalar l^{σ} ∈ K on M. Since this is true for each l ∈ L, L normalizes all K-subspaces in M and M is 1-dimensional over K. The lemma is now readily verified.

BwB

Lemma 2.2 Let S be a group, T a subgroup of S, B = N_{S}(T), ω ∈ S\B, W an RS-module and Y an RB-submodule of C_{W} (T). Suppose that each of the following holds:

(i) S = B ∪ BωT and S = ⟨T^{S}⟩,

(ii) [W, T, T] = 0,

(iii) W = ⟨Y S⟩.

Then each of the following is true:

1. W = [W, S] + C_{W} (S) = Y + Y ω + C_{W} (S),

2. C_{W} (T) = Y + C_{W} (S) = Y + [W, T] = [W, T] + C_{W} (S),

3. Put \overline{W} = W/C_{W} (S). Then C\overline{W} (S) = 0.

4. If T = [T, H] and [Y, H] = 0, for some H ≤ B with H = H^{ω}, then [W, S] = 0 and W = Y.

Proof: Put Y_{0} = Y + [W, T]. By the assumptions T centralizes Y_{0}. By (a)

T^{S} = T^{B} ∪ T^{BωT} = T ∪ T^{ωT}

and hence

S = ⟨T^{S}⟩ = ⟨T, T^{ω}⟩.

This implies

W = ⟨Y S⟩ = Y + [W, S] = Y + [W, T] + [W, T^{ω}] = Y_{0} + [W, T]ω.

Similarly W = ⟨(Y ω)S⟩ = Y_{0}ω + [W, T]. Form the last two statements and the modular laws

Y_{0} + C_{[W,T]ω}(T) = C_{W} (T) = C_{Y_{0}ω}(T) + [W, T].

Moreover, C_{Y_{0}ω}(T) ≤ C_{W} (⟨T, T^{ω}⟩) = C_{W} (S) and C_{Y_{0}ω}(T) ≤ C_{W} (S) ∩ Y_{0}ω ≤ Y_{0}. So

C_{W} (T) = Y_{0} = [W, T] + C_{W} (S).

Since W = ⟨Y S⟩ we conclude from (a) that

W = Y + Y ωT = Y + Y ω + [Y ω, T] = Y ω + C_{W} (T).

Therefore

C_{W} (T^{ω}) = Y ω + (C_{W} (T) ∩ C_{W} (T^{ω})) = Y ω + C_{W} (S)

and

W = Y + C_{W} (T^{ω}) = C_{W} (T^{ω}) + C_{W} (T) = Y ω + C_{W} (S) + Y =
= [W, T] + [W, T]ω + C_{W} (S) = [W, S] + C_{W} (S).

This completes the proof of (1) and(2).

Let U be the inverse image of C\overline{W} (S) in W. By (2) applied to \overline{W} we have \overline{U} ≤ \overline{Y} [\overline{W} , T]. Thus U ≤ Y [W, T]C_{W} (S) ≤ C_{W} (T) and [U, T] = 0. Since S = ⟨T^{S}⟩ we get U ≤ C_{W} (S) and \overline{U} = 0, proving (3).

To prove (4) note that H centralizes W = Y + Y w + C_{W} (S) and hence H ≤ C_{S}(W). It follows that

T = [T, H] ≤ C_{S}(W) and S = ⟨T^{S}⟩ ≤ C_{S}(W).

So [W, S] = 0 and the lemma is proved.

SL2K

Lemma 2.3 Suppose that (G, Σ) fulfills (A)(1) with dim N = 2. Let Z ∈ Σ, L = C_{G}[N, Z] and V an RG modules with [V, Z, L] = 0, V = [V, G] and C_{V} (G) = 0, then there exists an (R, K)-module M such that V ≅ M ⊗_{K} N.

Proof: We start with defining some elements of G = SL_{2}(K). Let

a(k) = ( \begin{matrix}1 & 0 \\ k & 1 \end{matrix} ) , k ∈ K; h(λ) = ( \begin{matrix}λ & 0 \\ 0 & λ^{−1} \end{matrix} ) , 0 ≠ λ ∈ K; ω = ( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ) ,

Z = \{a(k)|k ∈ K\} and H = \{( \begin{matrix}1 & 0 \\ 0 & λ \end{matrix} ) |λ ∈ (K \ \{0\}′\}.

Note that Z as defined above is indeed an element of Σ. We reader might verify that the following relation holds for all t ∈ K^{#}. ( But the relation will follows from some computation below)

(∗)  ω^{−1}a(−t)ω = a(t^{−1})h(t)ωa(t^{−1}).

Since both a(t^{−1}) and a(t)^{ω} are in G, this implies h(t)ω ∈ G. Hence h(t)h(r^{−1}) ∈ G for all t, r ∈ K^{∗} and we conclude that all h(t) and ω ∈ G. Hence also h(t)h(r)h(r^{−1}t^{−1} ∈ H and so H ≤ G. In particular, ZH ≤ L (actually L=ZH, but we will not need that). Put M = [V, Z]. Since G = ⟨Z, Z^{ω}⟩ we get

V = [V, G] = M + Mω and M ∩ Mw ≤ C_{V} (G) = 0.

Hence,

V = M ⊕ Mω.

Let D = Hom_{R}(M, M) and M_{2}(D) the ring of 2 × 2-matrices with coefficients in D. We

define a ring isomorphism from M_{2}(D) to Hom_{R}(V, V) by sending ( \begin{matrix}a & b \\ c & d \end{matrix} ) to ϕ where

ϕ is defined by (u + v)ϕ = ua + vω^{−1}c + ubω + vω^{−1}dω for u ∈ M and v ∈ Mω. Direct computations show that this indeed defines a ring isomorphism. Furthermore we see that:

ω ↔ ( \begin{matrix}0 & 1 \\ ϵ & 0 \end{matrix} ) , ϵ^{2} = 1,

h(λ) ↔ ( \begin{matrix}y(λ) & 0 \\ 0 & z(λ) \end{matrix} ) ,

a(k) ↔ ( \begin{matrix}1 & 0 \\ x(a) & 1 \end{matrix} ) .

a(k)a(l) = a(k + l) implies

(1) x(k) + x(l) = x(k + l).

Since ω^{2} ∈ Z(G) we have ϵy(λ) = y(λ)ϵ. From ω^{−1}h(λ)ω = h(λ^{−1}) we get

(2) z(λ) = y(λ^{−1}).

Since h(λ)h(κ) ≡ h(λκ) (mod H) and [M, H] = 0, we have h(λ)h(κ) ≡ h(λκ) (mod C_{G}(M)). Thus

(3) y(λ)y(κ) = y(λκ).

Next we make another use of (*). a(t^{−1})h(t)ωa(t^{−1}) corresponds to

( \begin{matrix}1 & 0 \\ x(t^{−1}) & 1 \end{matrix} ) ( \begin{matrix}y(t) & 0 \\ 0 & y(t^{−1}) \end{matrix} ) ( \begin{matrix}0 & 1 \\ ϵ & 0 \end{matrix} ) ( \begin{matrix}1 & 0 \\ x(t^{−1}) & 1 \end{matrix} )

= ( \begin{matrix}y(t) & 0 \\ x(t^{−1})y(t) & y(t^{−1}) \end{matrix} ) ( \begin{matrix}x(t^{−1}) & 1 \\ ϵ & 0 \end{matrix} )

= ( \begin{matrix}y(t)x(t^{−1}) & y(t) \\ x(t^{−1})y(t)x(t^{−1}) + y(t^{−1})ϵ & x(t^{−1})y(t) \end{matrix} )

and ω^{−1}a(−t)ω to

( \begin{matrix}0 & ϵ \\ 1 & 0 \end{matrix} ) ( \begin{matrix}1 & 0 \\ x(t^{−1}) & 1 \end{matrix} ) ( \begin{matrix}0 & 1 \\ ϵ & 0 \end{matrix} ) = ( \begin{matrix}ϵx(−t) & ϵ \\ 1 & 0 \end{matrix} ) ( \begin{matrix}0 & 1 \\ ϵ & 0 \end{matrix} )

= ( \begin{matrix}1 & ϵx(−t) \\ 0 & 1 \end{matrix} )

So y(t)x(t^{−1}) = x(t^{−1})y(t) = 1, x(t^{−1}) = y(t)^{−1} = y(t^{−1}) and hence

(4) x(t) = y(t), for all 0 ≠ t ∈ K. Moreover, y(t) = ϵx(−t) = −ϵx(t) = −ϵy(t) and since y(1) = 1,

(5) ϵ = −1. By (1),(3) and (4) x : K → D is a ring homomorphism. In particular, M is an (R, K)-

module. Moreover,

ω ↔ ( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ) = ( \begin{matrix}x(0) & x(1) \\ x(−1) & x(0) \end{matrix} ) ,

a(k) ↔ ( \begin{matrix}1 & 0 \\ x(a) & 1 \end{matrix} ) = ( \begin{matrix}x(1) & x(0) \\ x(a) & x(1) \end{matrix} ) .

Since G = ⟨Z, Z^{ω}⟩ = ⟨Z, ω⟩ we conclude

( \begin{matrix}a & b \\ c & d \end{matrix} ) ↔ ( \begin{matrix}x(a) & x(b) \\ x(c) & x(d) \end{matrix} )

for all ( \begin{matrix}a & b \\ c & d \end{matrix} ) ∈ G.

On the other hand cleary M ⊗_{K} K ≅ M as R-modules. Since N ≅ K ⊕ K as Kmodules, we conclude that M ⊗_{K} N ≅ M ⊕ M as R-modules. Let e_{1} = (1, 0) ∈ K ⊕ K, e_{2} = (0, 1) ∈ K ⊕ Ka nd m_{1}, m_{2} ∈ M. Direct computations now show that

m_{1} + m_{2}ω → m_{1} ⊗ e_{1} + m_{2} ⊗ e_{2}

defines an RG isomorphism from V to M ⊗_{K} N .

CV G = 0

Lemma 2.4 Suppose that (G, Σ) fulfills part 1 of Hypothesis (A) with dim N = 2. Let Z ∈ Σ, L = C_{G}[N, Z] and V an RG modules with [V, Z, L] = 0 and V = [V, G]. If char K ≠ 2 or K is not commutative, then C_{V} (G) = 0.

Proof: Put \overline{V} = V/C_{V} (G). Since V = [V, G], V = [V, Z]^{G} and we can apply 2.2(c). Thus C\overline{V} (G) = 0 and by 2.3 \overline{V} ≅ M ⊗_{K} N, for some (R, K)-module M.

Assume first that char K ≠ 2 amd let t = −id ∈ Z(G). Then \overline{V} = [\overline{V} , t] and C\overline{V} (t) = 0. It follows that V = [V, t] ⊕ C_{V} (G) and V = [V, G] = [V, t], and so C_{V} (G) = 0.

Next assume that char K = 2 and K is not commutative. We claim that 2v = v + v = 0 for all v ∈ V. Indeed let X = \{v ∈ V |2v = 0\}. Since 2\overline{V} = 0, [2V, G] = 0. Moreover, V/X ≅ 2V and so V = [V, G] ≤ X, proving the claim.

Let a(t), ω and Z be as in 2.3. By 2.2b, C_{V} (Z) = [V, Z] + C_{V} (G). Pick λ ∈ (K \ 0)′

with λ ≠ 1 and put h = ( \begin{matrix}λ & 0 \\ 0 & 1 \end{matrix} ) . Note that h ∈ G. Then a(t)^{h} = a(tλ) and since

[N, h] ≤ [N, Z], [V, h] ≤ C_{V} (Z). So for v ∈ V and t ∈ K we get

[v, a(t)]h = [vh, a(t)^{h}] = [v + [v, h], a(tλ)] = [v, a(tλ)].

Hence

[v, a(t(1 + λ))] = [v, a(t)a(tλ)] = [v, a(t)] + [v, a(tλ)] =
= [v, a(t)] + [v, a(t)]h = [v, a(t), h],

where we used 2V = 0 in the last equality. So [v, a(t(l + 1))] ∈ [V, Z, h]. Since every s ∈ K is of the form t(λ + 1) for some t we conclude that [V, Z] ≤ [V, Z, h]. Now C_{[V̄ ,Z]}(h) = 0 and so [V, Z, h] ∩ C_{V} (G) = 0. Thus [V, Z] ∩ C_{V} (G) = 0. On the other hand by 2.2b applied to Y = [V, Z], C_{V} (G) ≤ [V, Z] and so C_{V} (G) = 0.

Remark: we should make some remark on the existence of universial central extensions

defUE

Definition 2.5 Let X be an ℤG-module. Then a ℤG-module X̂ is called a universial central ℤG-extension of M provided that

(a) There exists a 𝕏G-homomorphism ϕ : X̂ → X with [ker ϕ, G] = 0.

(b) Whenever W is a ℤG-module and ψ : W → X is a ℤG-homomorphism with [ker ψ, G] = 0, then there exist a unique ℤG-homomorphism α : X̂ → W with ϕ = ψα.

UE

Lemma 2.6 Let X be a KG-module and (X̂, ϕ) a universial central ℤG-extension of X. (a) X̂ is a KG-module and ϕ is KG-linear.

(b) Let M be an right vector space over K. Then M ⊗_{K} X̂ is a universial central ℤG-extension of M ⊗_{K} X.

Proof: (a) Let 0 ≠ k ∈ K. Then by part (b) of 2.5 applied W = X̂ and ψ = k^{−1}ϕ, there exists a ℤG-homomorphism α_{k} : M̂ → M̂ with ϕ = k^{−1}ϕα_{k}, i.e such that kϕ = ϕα_{k}. By the uniqueness of α_{k}, α_{1} = id,α_{k+l} = α_{k} + α_{l} and α_{kl} = α_{k}α_{l}. Defining kx̂ = α_{k}(x̂), X̂ becomes a vector space over K and (a) is proved.

(b) Let W be a ℤG-module and ψ : W → M ⊗ X be any ℤG-homomorphism with [ker ψ, G] = 0. For 0 ≠ m ∈ M, let W_{m} be the inverse image of m ⊗ X under ψ. The map x → m ⊗ x is a ℤG from X to m ⊗ X and we can define a ℤG-homomorphism ψ_{m} : W_{m} → X by ψ(w) = m ⊗ ψ_{m}(w). By the universial property of X̂ (2.5b) there exists a ℤG homomorphism α_{m} : X̂ → W_{m} with ϕ = ψ_{m}α_{m}. Also put α_{0} = 0. We obtain a map α : M ×X̂ → W, (m, x) → α_{m}(x). Clearly, α is additive in the second coordinate and by the uniqueness of α_{m} it is easy to check that α is additive in the first coordinate and is balanced ( that is α(m, kx) = α(mk, x) for all m ∈ M, x ∈ X, k ∈ K.) Thus by the universial property of a tensor product α can be extended to a ℤ-homomorphism α : M ⊗_{K} X̂ → W. Moreover, one readily verifies that α commmutes with G and ϕ = ψα.

rUE

Remark 2.7 Let C be a class of ℤG-modules. For X in C define a universial central C extension of X as in 2.5 except that X̂ and W are assumed to be in C. Then (with the same proof) 2.6 is still true for universial central C-extensions provided that M ⊗_{K} X̂ is in C.

3 Some properties of pseudo quadratic spaces

f(n, n)

Lemma 3.1 Let (N, q, f) be a pseudo quadratic space and put K^{+} = {k ∈ K|k + sk^{σ} = 0}. Then

(a) K_{−} ≤ K^{+}.

(b) For all a ∈ N and q ∈ q(a), f(a, a) = q + sq^{σ}.

(c) Suppose that K_{−} = K^{+}. Then q(a) is uniquely determine by f(a, a). In particular, f(a, a) = 0 implies q(a) = 0 and rad N = N^{⊥}.

(d) Suppose that charK ≠ 2 or that σ acts non trivially on Z(K). Then K_{−} = K^{+}.

Proof: (a) (k − sk^{σ})^{σ} = k^{σ} − k^{σ^{2}}s^{σ} = k^{σ} − s^{−1}kss^{−1} = k^{σ} − s^{−1}k and so k − sk^{σ} = −s(k − sk^{σ})^{σ}, k − sk^{σ} ∈ K^{+} and K_{−} ≤ K^{+}.

(b) Let k, l ∈ K. Then on the one hand

q(ka + la) = q(ka) + q(la) + \overline{f(ka, la)} = k ∗ q(a) + l ∗ q(a) + \overline{kf(a, a)l^{σ}}

= \overline{kqk^{σ} + lqlb^{σ} + kf(a, a)l^{σ}}

and on the other hand

q(ka + la) = q((k + l)a) = (k + l) ∗ q(a) = \overline{kqk^{σ} + kql^{σ} + lqk^{σ} + lql^{σ}}.

Thus

\overline{kf(a, a)l^{σ}} = \overline{kql^{σ} + lqk^{σ}}.

But lqk^{σ} − ksq^{σ}l^{σ} = lqk^{σ} − s(lqk^{σ})^{σ} ∈ K_{−} and so

\overline{kf(a, a)l^{σ}} = \overline{k(q + sq^{σ})l^{σ}}

for all k, l ∈ K. If \overline{K} ≠ 0 this cleary implies f(a, a) = q + sq^{σ}. If \overline{K} = 0, then K = K_{−} and so by (a) K = K^{+}. Thus K_{+} = 0 and q + sq^{σ} = 0. Moreover, by (PQ1), f is trace valued and so also f(a, a) = 0.

(c) By (b) q(a) is unique up to an element \overline{K^{+}}. Thus (c) holds

(d) If char K ≠ 2, put λ = 1 and if char K = 2 and σ acts non trivially on Z(K) pick λ ∈ Z(K) with λ ≠ λ^{σ}. Then in any case, λ ∈ Z(K) and λ + λ^{σ} ≠ 0. Put μ = λ(λ + λ^{σ})^{−1}. Since λ ∈ Z(K), λ^{σ^{2}} = λ. Thus μ + μ^{σ} = 1 and μ^{σ} ∈ Z(K). Let k ∈ K with k = −sk^{σ}. Then

kμ − s(kμ)^{σ} = kμ − sμ^{σ}k^{σ} = kμ − sk^{σ}μ^{σ} = k(μ + μ^{σ}) = k.

Hence k ∈ K_{−}, K^{+} ≤ K_{−} and (d) follows from (a).

transvection

Lemma 3.2 Let N be a left vector space over K, 0 ≠ n ∈ N and 0 ≠ ϕ ∈ N^{∗}.

(a) t(ϕ, n) is invertible if and only if nϕ ≠ −1, in which case the inverse is given by t(ϕ, −(1 + nϕ)^{−1}n).

(b) Let (q, f) be a (σ, s)-pseudo quadratic form on N. Then t(ϕ, n) ∈ O(V, q, f) if and only if one of the following holds:

(b1) n ∈ rad N and nϕ ≠ −1.

(b2) n ∉ N^{⊥} and there exists 0 ≠ k ∈ K with −k^{−1} ∈ q(n) and vϕ = f(v, n)k for all

v ∈ N.

Proof: Put t = t(n^{∗}, n).

(a) If nϕ = −1 then n.t = 0 and t is not invertible. If nϕ ≠ −1 it is trivial to verify that t(ϕ, −(1 + nϕ)^{−1}.n) is an inverse for t.

(b) Recall that t ∈ O(N, q, f) if and only if f(at, bt) = f(a, b) and q(at) = q(a) for all a, b ∈ N. Since f(at, bt) = f(a + aϕ.n, b + bϕ.n) weget

(1)  f(at, bt) = f(a, b) + aϕ · f(n, b) + f(a, n).(bϕ)^{σ} + aϕ · f(n, n) · (bϕ)^{σ}.

and since q(at) = q(aϕ.n + a)

(2)  q(at) = aϕ ∗ q(n) + \overline{aϕ.f(n, a)} + q(a).

In particular, if n ∈ radN and t is invertible, t ∈ O(N, q, f). Thus we may assume that n ∉ radN.

Assume now that t ∈ O(V, q, f). Suppose that n ∈ N^{⊥}. Since n ∉ rad V, q(n) ≠ 0. Hence by (2) aϕ = 0 for all a ∈ N, a contradiction.

Thus n ∉ N^{⊥}. Let b ∈ ker ϕ. Then by (1) aϕ.f(n, b) = 0 for all a in N and so b ∈ N^{⊥} and ker ϕ ≤ n^{⊥}. Since both ker ϕ and n^{⊥} are hyperplanes in N, ker ϕ = n^{⊥}. Therefore there exists k ∈ K with

(3)  aϕ = f(a, n)k for all a ∈ N.

Since f(n, a) = sf(a, n)^{σ}, (2) now implies f(a, n)k ∗ q(n) + \overline{f(a, n)ksf(a, n)^{σ}} = 0. Thus

f(a, n) ∗ (k ∗ q(n)) + f(a, n) ∗ \overline{ks} = 0

Hence k ∗ q(n) = −\overline{ks} and q(n) = k^{−1} ∗ \overline{ks} = −\overline{sk^{−σ}}. Now k^{−1} − sk^{−σ} ∈ K_{−} and so \overline{k^{−1}} = \overline{sk^{−σ}}. Thus q(n) = \overline{−k^{−1}} and (b2) holds in this case.

Suppose next that t fulfils (b2). Reading the above calculations backwards we see that q(a) = q(at) for all a ∈ N. Put q = −k^{−1}. By assumption q ∈ q(n) and so by 3.1 f(n, n) = q +sq^{σ}. Using (1) it is now readily verified that f(at, bt) = f(a, b) for all a, b ∈ N. It remains to show that t is invertible. Otherwise by (a), −1 = nϕ = f(n, n)k and so q = k^{−1} = f(n, n). Thus q = f(n, n) = q + sq^{σ}, sq^{σ} = 0 and q = 0, a contradiction to the definition of q.

We denote the element in O(V, q, f) of form t(ϕ, n) as in 3.2b2 by t(k, n), that is

v.t(k, n) = v + f(v, n)k · n where n ∈ N \ N^{⊥}, k ∈ K^{#}, −k^{−1} ∈ q(n), v ∈ N

t(k, n) is called a pseudo-transvection with axis Kn. We remark that for any given n ∈ N \ radN, there exists a pseudo-transvection with axis Kn, unless n is a singular vector in a quadratic space.

transitive

Lemma 3.3 Let (N, q, f) be a non-degenerate pseudo-quadratic space such that S(1) ≠Ø.

(a) N = ⟨S(1)⟩.

(b) Let i be a positive integer with S(i) ≠Ø and U ∈ S(i). Then Q_{U} acts transitively on the set of all U_{0} ∈ S(i) with U_{0} ∩ U^{⊥} = 0.

Proof: (a) Let M = ⟨S(1)⟩ and suppose N ≠ M. Let n ∈ N \ M. Then n is not singular and so there exists a pseudo-transvection t in O(N, q, f) with axis Kn. Since t normalizes M and n ∉ M, [M, t] ≤ M ∩ Kn = 0 and so n is perpendicular to M. It follows that M ∪ M^{⊥} = N, M^{⊥} = N, and M ≤ N^{⊥}. Since M is generated by singular vectors, M ≤ radN = 0, a contradiction.

(b) Let U_{1}, U_{2} ∈ S(i) with U^{⊥} ∩ U_{k} = 0, k = 1, 2. As N is non degenerate, N ∩ U = 0 and so dim N/U^{⊥} = i. Thus N = U^{⊥} + U_{k}. Let x_{1}, x_{2}, …, x_{i} be a basis for U over K and for k = 1, 2 let y_{1}(k), y_{2}(k), …, y_{i}(k) be the basis for U_{k} with f(x_{j}, y_{l}(k) = δ_{jl}. Note that U^{⊥} = U ⊕ (U + U_{k})^{⊥}. Hence for every x ∈ (U + U_{k})^{⊥} there exists a unique y ∈ (U + U_{2})^{⊥} with x + U = y + U, and the map x → y is an isometry. Extend this map to h ∈ GL_{K}(N) such that x_{j}h = x_{j} and y_{j}(1)h = y_{j}(2). Then it is easy to see that h is an isometry, h ∈ Q_{U} and U_{1}h = U_{2}.

parabolics

Lemma 3.4 Let (N, q, f) be a nondegenerate pseudo quadratic space, i a positive integer and U ∈ calS(i). Then

(a) P_{U}/Q_{U} ≅ GL_{K}(U) × O(U^{⊥}/U, q, f).

(b) Q_{U}/T_{U} ≅ Hom_{K}(U^{⊥}/U^{⊥⊥}, U) ≅ Hom_{K}(N/U^{⊥}, U^{p}erp/U^{⊥⊥}.

(c) T_{U}/Z_{U} ≅ Hom_{K}(N/U^{⊥}, N^{⊥})

(d) Q_{U}/Z_{U} ≅ Hom_{K}(N/U^{⊥}, U^{⊥}/U)

(e) Z_{U} is isomorphic to the additive group of all i × i-matrices M with M^{T} = −sM^{σ} and M_{jj} ∈ K_{−} for all 1 ≤ j ≤ i.

(f) T_{U} is isomorphic to the additive group o fall i × i-matrices M with M^{T} = −sM^{σ} and M_{jj} + K_{−} ∈ q(N^{⊥}) for all 1 ≤ j ≤ i.

(g) Z(Q_{U}) = T_{U} (unless i = 1 and K_{−} = 0, in which case Q_{U} is abelian)

(h) Q′_{U} = Z_{U} (unless Q_{U} = T_{U}, that is U^{⊥} = U^{⊥⊥})

Proof: It follows easily from 3.3a and induction that there exists a singular subspace E in N with N = U^{⊥} ⊕ E.

Put X = (U ⊕ E)^{⊥}. Since U is finite dimensional and U ∩ N^{⊥} = 0, N = U ⊕ X ⊕ E and U^{⊥⊥} = U ⊕ N^{⊥}.

(a) By 3.3b and a Frattini argument, P_{U} = N_{P_{U}} (E)Q_{U}. Since N_{P_{U}} (E) normalizes X, N_{P_{U}} (E) ∩ Q_{U} = 1 and it suffices to show that N_{P_{U}} (E) ≅ GL_{K}(U) × O(X, q, f). Let g ∈ GL_{k}(U) and h ∈ O(X, q, f). It is an easy exercise to show that there exists unique ĝ ∈ GL_{K}(E) with f(ug, eĝ) = f(u, e) for all u ∈ U, e ∈ E. Moreover, there exists a unique t ∈ GL_{K}(N) which acts as g on U, as ĝ on E and as h on X. Clearly, t ∈ O and so (a) holds.

(b) Let α ∈ Hom_{K}(X, U) with N^{⊥}α = 0. Put X_{α} = {x + x.α|x ∈ X} and note that U^{⊥} = U ⊕ X_{α}, Moreover, X_{α}^{⊥} = U ⊕ N^{⊥} ⊕ E_{α} for some i-dimensional singular subspace E_{α}. As in 3.3b, there exists t ∈ O such that t centralizes U and xt = x + xα for all x ∈ X. Clearly t ∈ Q_{U} and so the map

Q_{U} → Hom_{K}(X/N^{⊥}, U)
q → (x + N^{⊥} → [x, q])

is onto. Its kernel is obviously T_{U} and so the first equality in (b) holds. The second is just the dual version of the first.

Let β ∈ Hom_{K}(N, N^{⊥}) with U^{⊥}β = 0 and dim Nβ = 1. Then clearly (ker β)^{⊥} = Ku + N^{⊥} for some 0 ≠ u ∈ U. Let e ∈ N with f(e, u) = 1 and put r = eβ. Then 0 ≠ r ∈ N^{⊥}, q(r) ≠ 0 and so there exists 0 ≠ λ ∈ K with −λ^{−1} ∈ q(r). Pick μ ∈ K with μ^{σ} = λ^{−1} and put n = r + μ · e. Since e is singular and perpendicular to r, q(n) = q(e) and so we obtain the pseudo-transvection t(λ, n) ∈ O. Moreover,

(1)  [e, t(λ, n)] = f(e, n)λ · n = μ^{σ}λ · (r + μu) = r + μ · u ≡ rmodU.

Note that U^{⊥⊥} = U ⊕ N^{⊥} and [N, t] ∈ U^{⊥⊥} for all t ∈ T_{U} = C_{O}(U^{⊥}). Consider the homomorphism ρ;

T_{U} → Hom_{K}(N/U^{⊥}, U^{⊥⊥}/U)
t → (x + U^{⊥} → [x, t] + U)

Clearly ker ρ = Z_{U}. Let β̂ the the compostion of β with the natural isomorphism N^{⊥} → U^{⊥⊥}/U. Then by (1), ρ maps t(λ, n) onto β̂. As the holds for all β’s,ρ is onto and (c) holds.

(d) Cosider the map

Q_{U} → Hom_{K}(N/U^{⊥}, U^{⊥}/U)
t → (x + U^{⊥} → [x, t] + U)

By (b) and (c) this map is onto. Its kernel is Z_{U} and so (d) holds.

(e,f) Let u_{1}, u_{2}, …, u_{i} a basis for U and pick e_{1}, e_{2}, … e_{i} ∈ E with f(u_{l}, e_{j}) = δ_{lj}. Let z ∈ GL_{k}(N) with [N, z] ≤ U^{⊥⊥} and [U^{⊥}, z] = 0. Define a i × i matrix M = (m_{kl}) and r_{k} ∈ N^{⊥} by [e_{k}, z] = ∑_{l=1}^{i} m_{kl}u_{l} + r_{k}. Then z is uniquely determined by M and (r_{k}). We need to find necessary and sufficient conditions on M and (r_{k}) for z to be in O. So we compute

f(e_{j}z, e_{k}z) = f(e_{j} + ∑_{l=1}^{i} m_{jl}u_{l}, e_{k} + ∑_{l=1}^{i} m_{kl}u_{l}) =

= f(m_{jk}u_{k}, e_{k}) + f(e_{j}, m_{kj}u_{k}) = m_{jk} + sm_{kj}^{σ}

and

q(e_{k}z) = q(e_{k} + ∑_{l=1}^{i} m_{kl}u_{l} + r_{k}) = \overline{f(m_{kk}u_{k}, e_{k})} + q(r_{k}) = \overline{m_{kk}} + q(r_{k}).

Thus z ∈ O if and only if M^{T} = −sM^{σ} and m_{kk} + K_{−} ∈ −q(r_{k}) for all 1 ≤ k ≤ i. Note that q_{N^{⊥}} is injective. Hence for any M with M^{T} = −sM^{σ} and m_{kk} + K_{−} ∈ q(N^{⊥}) there exists unique r_{k} ∈ N^{⊥} with m_{kk} + K_{−} ∈ q(r_{k}). Thus (f) holds. Moreover, z ∈ T_{U} if and only if in addition r_{k} = 0. Thus also (e) holds.

(g,h) By definition of Q_{U}, [N, Q_{U}] ≤ U^{⊥} and [U^{⊥}, Q_{U}] ≤ U. Thus [N, Q_{U}, Q_{U}] ≤ U and the three subgroup lemma implies [N, Q′_{U}] ≤ U. Thus Q′_{U} ≤ Z_{U}. Note that [N, T_{U}] ≤ U^{⊥⊥} = U+ rad N. Thus [N, T_{U}, Q_{U}] = 0 and [N, Q_{U}, T_{U}] = 0. Hence [Q_{U}, T_{U}, N] = 0 and T_{U} ≤ Z(Q_{U}). So to prove (g) and (h) we need to show Z(Q_{U}) ≤ T_{U} and Z_{U} ≤ Q′_{U}.

Let a, b ∈ Q_{U}. We wish to compute [a, b]. For this put a_{i} = [e_{i}, a] and b_{i} = [e_{i}, b]. Then a_{i}, b_{i} ∈ U^{⊥}. Let x ∈ U^{⊥}. Then [x, a] ≤ U and so 0 = f(x, e_{i}) = f(x.a, e_{i}.a) = f(x + [x, a], e_{i} + a_{i}) = f(x, a_{i}) + f([x, a], e_{i}]. Hence

−[x, a] = ∑_{k=1}^{i} f(x, a_{k})u_{k}.

A similar formula holds for b. Furthermore, [e_{k}, a^{−1}] = −a_{k} + r_{k} and [e_{k}, b^{−1}] = −b_{k} + s_{k} for some r_{k}, s_{k} ∈ U^{⊥⊥}. Also [x, a^{−1}] = −[x, a] and so

(e_{k}.[a, b] = e_{k}.a^{−1}b^{−1}ab = (e_{k} − a_{k} + r_{k}).b^{−1}ab

= e_{k} − b_{k} + s_{k} − a_{k} − ∑_{j=1}^{i} f(a_{k}, b_{j})u_{j} + r_{k}).ab

Using (e_{k} − a_{k} + r_{k}).a = e_{k} we get

e_{k}.[a, b] = (e_{k} − b_{k} + ∑_{j=1}^{i} f(b_{k}, a_{j})u_{j} − ∑_{j=1}^{i} f(a_{k}, b_{j})u_{j} + s_{k}).b

= e_{k} + ∑_{j=1}^{i} f(b_{k}, a_{j})u_{j} − ∑_{j=1}^{i} f(a_{k}, b_{j})u_{j}.

Put B = (f(a_{k}, b_{j})). Then [a, b] ∈ Z_{U} corresponds to the matrix −B + sB^{T σ}.

Suppose there exists a ∈ Z(Q_{U}) \ T_{U}. Then a_{k} ≰ U^{⊥⊥} for some k. Let 1 ≤ k, j ≤ i and 0 ≠ λ ∈ K. By (b) we can choose b such that b_{l} ∈ U^{⊥⊥} if l ≠ j and f(a_{k}, b_{j}) = λ. Since a ∈ Z(Q_{U}) we get B = sB^{T σ}. Since all but the j’th colummns of B are zero and the k − j spot of B is λ and so not zero, we conclude that j = k and λ = sλ^{σ}. Since j and λ are arbitrary we conclude that i = 1 and K_{−} = 0. Also if i = 1 and K_{−} = 0, [a, b] = 1 for any a, b ∈ Q_{U} and so (g) holds.

To complete the proof for (h) we may assume that U^{⊥} ≠ U^{⊥⊥}. Fix 1 ≤ k, j ≤ i and λ ∈ K and choose a, b ∈ Q_{U} such that a_{l} ∈ U^{⊥⊥} if l ≠ k, b_{l} ∈ U^{⊥⊥} if l ≠ j and f(a_{k}, b_{j}) = λ. Then B is zero everywhere except in the k − j spot, where it is λ. Clearly every matrix M with M^{T} = −sM^{σ} and m_{ll} ∈ K_{−} for all l is the sum of matices of the form −B + B^{T σ}, 1 ≤ kj ≤ i, λ ∈ K and so (h) is proved.

transitive2

Lemma 3.5 Let (N, q, f) be a non-degenerate pseudo-quadratic space with S(1) ≠ ∅.

(a) Suppose that (N, q, f) is not a 2-dimensional quadratic space.

(a1) ⟨Q_{U}|U ∈ S(1)⟩ acts transitively on S(1).

(a2) If x, y ∈ S(1) with x ≠ y ,then S(1) ≠ (S(1) ∩ x^{⊥}) ∪ (S(1) ∩ y^{⊥}).

(a3) Define x, y ∈ S(1) to be adjacent if x ⊥̸ y. Then the corresponding graph on S(1) is connected.

(b) ⟨T_{U}|U ∈ S(1)⟩ acts transitively on S(1) unless Nis a non defective quadratic space.

Proof: (a1) Let U_{1}, U_{2} ∈ S(1) with U_{1} ≠ U_{2}.

Suppose first that U_{1} ⊥ U_{2}. Pick U_{3} ∈ S(1) with U_{3} ≤ U_{1} + U_{2} and U_{1} ≠ U_{3} ≠ U_{2}. By 3.4a, Q_{U_{3}} induces a full unipotent subgroup on U_{1} + U_{2} and so U_{2} and U_{1} are conjugate under Q_{U_{3}}.

Suppose next that U_{1} and U_{2} are not perpendicular. Assume that S(2) ≠ ∅. Then U_{1} is contained in a 2-dimensional singular subspace space W. Put U_{3} = W ∩ U_{2}^{⊥}. Then U_{1} and U_{2} are both conjugate to U_{3} under ⟨Q_{U}|U ∈ S(1). Thus we may assume that S(2) = ∅.

Suppose that there exists U_{3} ∈ S(1) with U_{1} ≠ U_{3} ≠ U_{2}. Then by 3.3b U_{1} and U_{2} are conjugate under Q_{U_{3}} and we may assume that no such U_{3} exists. Thus S(1) = {U_{1}, U_{2}}, Q_{U_{1}} centralizes U_{2} and so Q_{U_{1}} = 1. By 3.4 this implies U = U^{⊥} and K_{−} = 0, i.e. N is a 2-dimensional quadratic space.

(a2) Let x, y ∈ S(1) and suppose that S(1) = (S(1) ∩ x^{⊥}) ∪ (S(1) ∩ y^{⊥}).

If S(2) = ∅, then S(1) = (S(1) ∩ x^{⊥}) ∪ (S(1) ∩ y^{⊥}) = {x, y}. As seen in the proof of(a) this implies that (N, q, f) is a 2-dimensional quadratic space,.

So we may assume that S(2) ≠ ∅. By 3.3a (a) applied to y^{⊥}/y, y^{⊥} is generated by its singular subspaces. Thus there exists z ∈ S(1)∩y^{⊥} with z ≠ y and z ⊥̸ x. Let w ∈ S(1)∩z^{⊥} with w ∉ y^{⊥}. Then w ∈ x^{⊥}. So w + z ∈ S(2), (w + z) ∩ x^{⊥} = w and (w + z) ∩ y^{⊥} = z. But w + z contains more than two 1-dimensional subspaces contradicting the assumption that S(1) = (S(1) ∩ x^{⊥}) ∪ (S(1) ∩ y^{⊥}).

(a3) By (a2) we can choose z ∈ S(1) with z ⊥̸ x and z ⊥̸ y. Thus x is adjacent to z and z is adjacent to y, proving (a3).

(b) Without loss N is not a non defective quadratic space. Let U_{1}, U_{2} ∈ S(1) with U_{1} ≠ U_{2}. We need to show that U_{1} and U_{2} are conjugate under ⟨T_{U}|U ∈ S(1)⟩. By (a3) we may assume that U_{1} and U_{2} are not perpendicular. Put R = U_{1} +U_{2} +N^{⊥}. Then R itself is a non degenerate, pseudo quadratic space which is not non-defective quadratic. So we may assume N = R. By (a1) U_{1} and U_{2} are conjugated under ⟨Q_{U}|U ∈ S(1)⟩. Let U ∈ S(1). As N/N^{⊥} is 2-dimensional, U^{⊥} = U + N^{⊥} = U^{⊥⊥} and so T_{U} = Q_{U}. Thus (b) holds.

T U

Lemma 3.6 Let (N, q, f) be a nondegenerate pseudo quadratic space and 0 ≠ U a finite dimensional singular subspace of N. Then

(a) T_{U} = ⟨T_{U} ∩ Q_{E}|E ∈ S(1) ∩ U⟩

(b) Q_{U} = ⟨Q_{U} ∩ Q_{E}|E ∈ S(1) ∩ E⟩

(c) T_{U} = ⟨T_{E}|E ∈ S(1) ∩ E⟩, unless N is a non-defective quadratic space.

(d) Let 0 ≠ E ≤ U. Then C_{Q_{U}} (E^{⊥}/E) = Q_{U} ∩ Q_{E} and Q_{U}/Q_{U} ∩ Q_{E} ≅ C_{E^{⊥}/E}(U^{⊥}/U) ∩ C_{E^{⊥}/E}(U/E)

Proof: Let u_{1}, …, u_{i} be a basis for U. We use the correspondence between T_{U} and certain i × i matrices established in the proof of 3.4f without further reference.

(a) Let 1 ≤ j ≤ i. Then T_{U} ∩ Q_{Ku_{j}} corresponds to the set of matrices with m_{kl} = 0 for all 1 ≤ k, l ≤ i with k ≠ j ≠ l. This implies (a).

(b) Let E ∈ S(1) ∩ U and α ∈ Hom_{K}(U^{⊥}, U) with U^{⊥⊥}α = 0 and U^{⊥}α ≤ E. Extend α to an element β of Hom_{K}(E^{⊥}, E). By 3.4b, there exists q ∈ Q_{E} with β = q − 1|_{E^{⊥}}. Then clearly q ∈ Q_{U} ∩ Q_{E} and α = q − 1|_{U^{⊥}}. Thus

Q_{U} = ⟨Q_{U} ∩ Q_{E}|E ∈ S(1) ∩ U⟩T_{U}

and so (b) follows from (a).

(c) Without loss N is not a non-defective quadratic space. Thus K_{−} ≠ 0 or N^{⊥} ≠ 0. Thus there exists 0 ≠ ρ ∈ K with ρ̄ ∈ q(N^{⊥}). Since f vanishes on N^{⊥}, 3.1 implies ρ + sρ^{σ} = 0. Let λ be an arbitrary element in K and put μ = −λρ^{−1}sλ^{σ}. Then

μ = −λρ^{−1}(sρ^{σ})ρ^{−σ}λ^{σ} = (λρ^{−1})ρ((λρ^{−1})^{σ}

and so \overline{μ} = (λρ^{−1}) ∗ \overline{ρ} ∈ q(N^{⊥}).

Note that

( \begin{matrix}0 & −sλ^{σ} \\ λ & 0 \end{matrix} ) = ( \begin{matrix}ρ & 0 \\ 0 & 0 \end{matrix} ) + ( \begin{matrix}0 & 0 \\ 0 & μ \end{matrix} ) + ( \begin{matrix}−ρ & −sλ^{σ} \\ λ & μ \end{matrix} )

and −λρ^{−1} · (−ρ, −sλ^{σ}) = (λ, −μ). In particular all three matrices on the right side of the above equation correspond to transvections and it is easy to see that (c) holds.

(d) Let F be a complement to E^{⊥} and X = (E +F)^{⊥}. Then clearly C_{O(X)}(X ∩U^{⊥}/X ∩ U) ∩ C_{O(X)}(U ∩ X) ≅ C_{Q_{U}} (E + F) and (d) is easily verified.

connected1

Lemma 3.7 Let (N, q, f) be a nondegenerate pseudo quadratic space such that S(2) ≠ ∅, and define x, y ∈ S(2) to be adjacent if x ∩ y^{⊥} = 0. Then the corresponding graph on S(2) is connected, unless (N, q, f) is a four dimensional quadratic space.

Proof: Without loss (N, q, f) is not a four dimensional quadratic space. Let x ≠ y ∈ S(2).

Consider first the case that U = x ∩ y ≠ 0. By 3.5a2 applied to U^{⊥}/U in place of N there exists z ∈ S(2) with U ≤ z, z ⊥̸ x and z ⊥̸ y. Let F ∈ S(1) ∩ z with F ≠ U and choose S ∈ S(1) ∩ F^{⊥} with S ⊥̸ U. Then F + S ∈ S(2).

We claim that F + S is adjacent to x. Suppose not. Then T = x ∩ (F + S)^{⊥} ≠ 0. Since U ⊥̸ S, U ≰ T. Hence Z = F + U ⊥ T + U = x and x ⊥ z, contradicting our choice of z. Similarly F + S is adjacent to y and so x and y lie in the same connected component.

Next consider the case with x ∩ y = 0. Let P_{1} ∈ x ∩ S(1) and P_{2} ≤ y ∩ P^{⊥} ∩ S(1). Then P_{1} + P_{2} ∈ S(2). By the previous paragraph x, P_{1} + P_{2} and y lie in the same connected component and the lemma is proved.

connected2

Lemma 3.8 Suppose (G, Σ) fulfills Hypothesis’ (A),(B) or (C). Then graph on Σ is connected.

Proof: If (B) holds this is 3.5a3 and if (C) holds this is 3.7. So suppose (A) holds. Let Z_{1} and Z_{2} be in Σ. Then it is readily verified that there exists Z_{3} in Σ such that Z_{3} is adjacent to Z_{1} and Z_{2}.

Q_{U}inG

Lemma 3.9 Suppose (G, Σ) fulfills Hypothesis (B) or (C).

(a) Let 0 ≠ U be a finite dimensional singular subspace of N. Then Q_{U} ≤ G.

(2) Let U ∈ S(1). Then C_{G}(U) acts transitively on the 1-dimensional singular subspaces of U^{⊥}/U.

Proof: (a) Let U ∈ S(1). Pick x ∈ N \ U^{⊥}. By 3.4 the map

α : Q_{U}/Z_{U} → U^{⊥}/U, qZ_{U} → [x, q] + U

is an isomorphism. For V ∈ S(2) with U ≤ V define Z(U, V) = Z_{V} ∩ Q_{U}. Then under the isomorphism in 3.4e ( where we choose the first basis vector for V iin U), Z(U, V) corresponds to the 2 × 2 matrices M with M^{T} = −T^{σ}, m_{11} ∈ K_{−} and m_{22} = 0. It follows that α(Z(U, V)/Z_{U}) = V/U. By 3.3a, U^{⊥}/U is spanned by its 1-dimensional singular subspaces and so

Q_{U} = ⟨Z(U, V)|U ≤ V ∈ S(2)⟩.

We claim that Z_{V} ≤ G. Indeed, in case (C) Z_{V} ∈ Σ and in case (B) 3.6(c) yields Z_{V} ≤ T_{V} ≤ ⟨T_{E}|E ∈ S(1)⟩ ≤ ⟨Σ⟩. In particular, Z(U, V) ≤ G and so Q_{U} ≤ G. Thus (a) follows from 3.6b.

(b) Suppose first that Hypothesis (B) holds. Then by 3.5b applied to U^{⊥}/U, ⟨T_{E}|E ≤ S ∩ U^{⊥}⟩ acts transitively on the singular 1-spaces of U^{⊥}/U.

Under Hypothesis (C) U^{⊥}/U is at least three dimensional and so by 3.5a1 (applied to U^{⊥}/U) and 3.6d we get that ⟨Q_{V} |U ≤ V ∈ S(2)⟩ acts transitively on the singular 1-spaces of U^{⊥}/U. Thus (b) holds also in this case.

Qtransitive

Lemma 3.10 Suppose that (G, Σ) fulfills Hypothesis (A).

(a) Let 0 ≠ x ∈ N and X = Kx. Then Q_{X} ∩ G = {t(ϕ, x)|ϕ ∈ Ñ, xϕ = 0}, and C_{G}(x) acts transitively on Q_{X}^{#} ∩ G. Moreover, Q_{X} ∩ G acts transitively on the 1-dimensional subspaces of Ñ outside of CÑ(x).

(b) Let 0 ≠ ϕ ∈ Ñ and Φ = Kϕ. Then Q_{X} ∩ G = {t(ϕ, x)|x ∈ ker ϕ} and C_{G}(x) acts transitively on Q_{Φ}^{#} ∩ G. Moreover, Q_{Φ} ∩ G acts transitively on the 1-dimensional subspaces of N outside of ker ϕ.

(c) Let Z ∈ Σ and Q be the stabilizer in G of the series 0 ≤ [N, Z] ≤ C_{N}(Z) ≤ N. Then Q acts transitively on the set of all Z_{0} ∈ Σ adjacent to Z.

Proof: (a) Let 1 ≠ t ∈ Q_{X} ∩ G. The clearly x is a transvection with axis Kx and so t = t(ϕ, x) for some ϕ ∈ N^{∗} with xϕ = 0. Furthermore, ϕ ∈ [N^{∗}, t] ≤ [N^{∗}, G] ≤ Ñ and so first part of (a) holds. T (CÑ(x), N) acts transitively on the nonzero vectors of Ñ and so also on Q_{X}^{#} ∩ G. Finally, let ϕ_{1}, ϕ_{2} ∈ Ñ with xϕ_{1} = 1 = xϕ_{2}. Then ϕ_{1}.t(ϕ_{1} − ϕ_{2}, x) = ϕ_{2} and (a) is proved.

(b) Follows by a dual argument or by observing that T (Ñ, N) = T (N, Ñ) if we identify N with its copy in Ñ^{∗}.

(c) Let Z = T (Φ, X) and Z_{i} = T (Φ_{i}, X_{i}) ∈ Σ, i = 1, 2 with Z_{i} adjacent to Z. We have to show that Z_{1} and Z_{2} are conjugate under Q. Since Z_{i} is adjacent to Z, XΦ_{i} ≠ 0 and X_{i}Φ ≠ 0. Note that Q_{Φ} ≤ Q and so by (b) we may assume that X_{1} = X_{2}. Since X_{i}Φ_{i} = 0, the element of t(ϕ_{1} − ϕ_{2}, x) (found in (a)), conjugates Φ_{1} to Φ_{2} and fixes X_{1} = X_{2} . This proves (c).

Q = xCGU

Lemma 3.11 Suppose that (G, Σ) fullfils Hypothesis’ (A),(B) or (C). Let Z ∈ Σ, U a 1dimensional subspace of [N, Z] and Q = Q_{U} ∩G. Let x ∈ Q\Q′ such that [N, x] is singular. Then Q = ⟨x^{C_{G}(U)}⟩, unless dim N/N^{⊥} = 4 and q(N^{⊥}) ≠ K/K_{−}.

Proof: Under Hypothesis (A) this follows directly from 3.10a.

So suppose Hypotesis’ (B) or (C). Let 0 ≠ y ∈ [N, x] + U/UU and A the subgroup of U^{⊥}/U generated by y.C_{G}(U). In view of 3.4d,h Q = ⟨x^{C_{G}(U)}⟩ if and only if A = U^{⊥}/U. By 3.9 it suffices to show that A contains a singular 1-space. Also A spans U^{⊥}/U as a K-space and so f does not vanish on A. Let z ∈ A \ y^{⊥} and E = [N, x]U. By 3.6d, Q_{E} acts as Q_{E/U} on U^{⊥}/U.

Suppose that A ∩ y^{⊥} ≰ E + radN/U. Then by 3.4b, E/U ≤ [A ∩ y^{⊥}, Q_{E}] ≤ A and the lemma holds.

So we may assume that A ∩ y^{⊥} ≤ E + radN/U. In particular, [z, Q_{E}] ≤ E + radN/U and so by 3.4b, E^{⊥} = E + radN. Hence dim N/N^{⊥} = 4. If q(N^{⊥}) = K/K_{−} then by 3.4f, [z, Q_{E}] + radN + U/U = E + radN/U. Thus A + rad(U^{⊥}/U) = U^{⊥}/U. Hence [(U ⊥ /U, Q_{E}] ≤ A. If |K| = 2, {0, y} is a singular 1-space and we may assume that |K| ≠ 2. But then [(U^{⊥}/U, Q_{E}] = E + radN/U, and the lemma is proved.

perfect

Lemma 3.12 Suppose that Hypothesis (A), (B) or (C) holds, Then G is perfect.

Proof: Let

4 The structure of [V, G]/[V, G] ∩ C_{V} (G).

Throughout this section (G, Σ) fulfills Hypothesis (A),(B) or (C), Z ∈ Σ, L = C_{G}([N, Z] and V is an RG-modules with [V, Z, L] = 0. Let Z_{0} ∈ Σ be adjacent to Z and let U be a 1-dimensional singular subspace of [N, Z]. Put X = ⟨Z, Z_{0}⟩, and P = C_{G}(U).

o + 4

Lemma 4.1 Suppose that Hypothesis (C) holds. The there exists g_{1} ∈ P so that the following holds for X_{1} = X^{g_{1}} and L^{∗} = LX_{1}.

(a) [X, X_{1}] = 1 and [N, X] = [N, X_{1}].

(b) X_{1} nomalizes [N, Z] and L. In particular L^{∗} is a subgroup of N_{G}([N, Z]).

(c) [N, Z].g_{1} = U + [N, Z_{O}] ∩ U^{⊥}, [N, Z_{0}].g_{1} = [N, Z] ∩ [N, Z_{0}].g_{1} + [N, Z_{0}] ∩ [N, Z_{0}].g_{1}.

Proof: Let U_{1} be a singular 1-space in [N, Z] different from U. Put U_{2} = [N, Z_{0}]∩U^{⊥} and U_{3} = [N, Z_{0}] ∩ U_{1}^{⊥}. Then U + U_{1} + U_{2} + U_{3} = [N, Z] + [N, Z_{0}] = [N, X] is a 4 dimensional quadratic space ”+”-type. Let E = [N, Z] = U +U_{1}, E_{0} = [N, Z_{0}] = U_{2} +U_{3}, E_{1} = U +U_{2}, E_{2} = U_{1} + U_{3}, Z_{i} = T_{E_{i}}, i = 1, 2 and X_{1} = ⟨Z_{1}, Z_{2}⟩. Note that for z ∈ Z and n ∈ N, n.z is perpendicular to n. Hence [U_{1}, Z_{1}] ≤ U_{1}^{⊥} ∩ [N, Z_{1}] = U_{1}^{⊥} ∩ (U + U_{2}) = U. It follows that Z_{1} normalizes E and so centralizes Z. By symmetry, also Z_{2} normalises E and centralizes Z. So (b) holds. Morover, X_{1} centralizes Z and (by symmetry) Z_{O}. Thus (a) holds. To complete the proof of this lemma it now suffices to find g_{1} ∈ P with Z^{g_{1}} = Z_{1} and Z_{O}^{g_{1}} = Z_{2}, i.e with E.g_{1} = E_{1} and E_{O}.g_{1} = E_{2}.

By 3.9b there exists g ∈ P with E.g = E_{1}. Since E^{⊥} ∩ E_{0}^{=}0, E_{1}^{⊥} ∩ E_{0}.g = 0. Since also E_{1} ⊥ ∩E_{2} = 0 we conclude from 3.3b that E_{0}.gq = E_{2} for some q ∈ Q_{E_{1}}. Put g_{1} = gq. As g_{1} centralizes E_{1}, we get g_{1} ∈ P, E.g_{1} = E_{1} and E_{O}.g_{1} = E_{2}.

G = LZ0

Lemma 4.2 G = ⟨L, Z_{0}⟩ = ⟨L, X⟩.

Proof: Let Σ_{0} = Z^{⟨L,Z_{0}⟩}. By 3.3b, 3.9a and 3.10, Z_{0}^{L} contains all elements in Σ adjacent to Z. Further, if Z_{1} and Z_{2} are adjacent in Σ, then they are conjugated in ⟨Z_{1}, Z_{2}⟩. It follows that Σ_{0} contains the connected component of Σ which contains Z. So 3.8 implies Σ_{0} = Σ. Since G = ⟨Σ⟩, G = ⟨L, Z_{0}⟩.

{[W, Z] = \atop W 0}

Lemma 4.3 Let B = N_{X}(Z) and W_{0} an R(LB)-submodule of V with [W_{0}, L] = 0. Put W = ⟨W_{0}.X⟩. Then

C_{W} (X) ≤ C_{V} (G), [W, Z] + C_{W} (G) = W_{0} + C_{W} (G) and W = [W, X] + C_{W} (G).

Proof: By 2.2 we have C_{W} (X) ≤ W_{0} + [W, Z]. Since [V, Z, L] = 0 this implies [C_{W} (X), L] = 0. By 4.2, ⟨L, X⟩ = G and so C_{W} (X) ≤ C_{V} (G). The other assertions now follow from 2.2.

{[M, Z] = \atop M1}

Lemma 4.4 Let M_{0} be an RN_{G}(U)-submodule of V with [M_{0}, P] = 0. Put M_{1} = M_{0}, in cases (A) and (B), and M_{1} = ⟨M_{0}.L^{∗}⟩,in case (C). Put M = ⟨M_{0}.G⟩. Then [M, Z] + C_{M}(G) = M_{1} + C_{M}(G).

Proof: Let g ∈ G. If [U.g, Z] = 0, then Z ≤ P^{g} and [M_{0}g, Z] = 0.

If [Ug, Z] ≠ 0, we claim that there exists ω ∈ N_{X}(Z) and h ∈ L ∩ C_{G}(Z) (or, in case (C), h ∈ L^{∗}) with Ug = Uωh. Indeed, in case (A) this follows from 3.10b. In case (B) U.g and U.ω both are not perpendicular to U and the claim follows from 3.3b. In case (C) we first choose h_{1} ∈ L_{∗} with Ugh_{1} ⊥ U. Then both U.gh_{1} and U.ω are perpendicular to U and neither Ugh_{1} nor Uω are perpendicular to [N, Z]. Thus by 3.6d and 3.3b (the latter applied to U^{⊥}/U) we get U.gh_{1}q = U.ω for some q ∈ Q_{[N,Z]}. This proves the claim. In particular,

[M_{0}.g, Z] = [M_{0}.ωh, Z] = [M_{0}.w, Z]h.

By 4.3 we have

[M_{0}.ω, Z] + C_{M}(G) = M_{0} + C_{M}(G).

Thus [M_{0}g, Z] + C_{M}(G) = M_{0}h + C_{M}(G) and M_{0}h + C_{M}(G) ≤ [M, Z] + C_{M}(G) ≤ M_{1} + C_{M}(G). Since L^{∗} normalises [M, Z] the lemma is established.

CV LG

Lemma 4.5 (a) ⟨C_{V} (L).X⟩ = [C_{V} (L), X] + C_{V} (G).

(b) ⟨C_{V} (L).G⟩ = [V, G] + C_{V} (G).

(c) If V = ⟨C_{V} (L).G⟩, then [V, G] = [V, G, G].

(d) [V, G, G, G] = [V, G, G].

(e) Let D be maximal in V with [V, G, G] = 0, then C_{V/D}(G) = 0.

Proof: (a) follows immediately from 4.3 applied to W_{0} = C_{V} (L).

(b) By (a) ⟨C_{V} (L).G⟩ ≤ [V, G] + C_{V} (L). Now [V, G] = ⟨[V, Z].G⟩ ≤ ⟨C_{V} (L).G⟩ and (b) is proved.

(c) follows from (b). (d) from (c) applied to [V, G] in place of V and (e) from (d) applied the inverse image of C_{V/D}(G) = 0 in V.

V QQ

Lemma 4.6 Suppose that (C) holds. Then (a) [C_{V} (L), Q_{U}, P] = 0

(b) [V, Q_{U}, Q_{U}, P] = 0 and [V, Q, Q] = [V, Q, Z].

(c) [V, Z, L^{∗}] ≤ ⟨C_{V} (P)G⟩

Proof: Let g ∈ P with [N, Z]g ⊥̸ [N, Z]. Then by ?? P = ⟨L, L^{g}⟩ and Q_{U} = Z^{g}(Q_{U} ∩ L). Thus

[V, Z, Q_{U}] ≤ [C_{V} (L), Q_{U}] = [C_{V} (L), Z^{g}] ≤ C_{V} (⟨L, L^{g}⟩ ≤ C_{V} (P)

. Thus (a) is proved. Now (b) follows from Q_{U} = ⟨Z^{P} ⟩ and (c) from L^{∗} = ⟨Q_{U}^{L^{∗}}⟩L.

V ZL∗

Lemma 4.7 Suppose that (C*) holds and [V, Z, L^{∗}] = 0. Then [V, G] = 0

Proof: By 4.5 we may assume that C_{V} (G) = 0. Let D = [V, Z]. Since G = ⟨Z^{G}⟩ we need to show that D = 0. Suppose not.

Assume first that char K ≠ 2. Then Z(X) = Z(X_{1}) ≤ L_{∗}. Thus [D, Z(X) = 0 and so we conclude from 2.3 that [D, X] = 0. By 2.3, D ≤ C_{V} (G) = 0 and we are done in this case.

Assume next that char K = 2. We may assume without loss that V = ⟨DG⟩. We will first prove (1) V, Q_{U}, Q_{U}] = 0.

Indeed by 4.6, [V, Q, Q] = [V, Q, Z] is centralized by P and by L^{∗} and so by G.

(2) If N has Witt index at least three, then N is nondefective of dimension 6 and |K| = 2.

Pick E ∈ calS(2) ∩ [N, Z]^{⊥} with E ∩ U = 0. Then Z_{E} ≤ P and so [C_{V} (Q_{U}), Z_{E}] is centralized by Y = ⟨Q_{U}, L^{∗}E. Hence if dim N > 6 we conclude from ?? that Y = G and C_{V} (G) = 0i mplies [C_{V} (Q_{U}), Z_{E}=0. Since [V, Z] ≤ C_{V} (Q_{U}), [V, Z] is centralized by ⟨Z_{E}^{P} ⟩ and L^{∗} and so by G. To show that K=2,pick Z_{2} ∈ P ∩ Σ adjacent to Z_{E}. Put S = ⟨Z_{E}, Z_{2}. Note that N_{S}(Z_{E}) ≤ L^{∗} and so centalizes D. Suppose that |K| ≠ 2. Then Z_{E} = [Z_{E}, N_{S}(Z_{E})] and so by 2.2d, S centralise D. But ⟨S, L^{∗}⟩ = G and (2) holds.

(3) N has Witt index at least three.

Suppose that N has Witt index 2. Since Q_{[N,Z} = (Q_{[N,Z} ∩ Q_{U})(Q_{[N,Z]} ∩ Q_{U})^{g} for ginL^{∗} \ N_{G}(U) we conclude from (1) that [C_{V} (Q_{U}), Q_{[N,Z]}, Q_{[N,Z]}] = 0 and we can apply 2.2 to ⟨D^{P} ⟩. Since L cenralizes D and since by ?? Q_{E}Q_{U} = [Q_{E}, L]Q_{U} we conclude from 2.2d that ⟨Q_{U}^{P}⟩ centralizes [D]. So again [D, G] = 0.

CV P G

Lemma 4.8 Suppose that (A),(B) or (C*) holds. Then ⟨C_{V} (P)G⟩ = [V, G] + C_{V} (G).

Proof: If (A) or (B) holds this is 4.5. So suppose that (C*) holds. Then by the same reference, C_{V} (P) ≤ C_{V} (L) ≤ [V, G] + C_{V} (G). Moreover, by 4.6c and 4.7 (applied to V/⟨C_{V} (P)G⟩), [V, G] ≤ ⟨C_{V} (P)G⟩.

For the case (B) we need to define a few more subgroups of G. Let U_{1} ∈ (2) with U ≤ U_{1}. Let U2 ∈ S(2) with U_{1} ∩ U_{2}^{⊥} = 0 and put F = U + (U_{2} ∩ U^{⊥}. Define X̃ = C_{O}((U1+U2)^{)}∩N_{O}(U_{1})∩N_{O}(U_{2}), Z^{∗} = CX̃(F), and X^{∗} = ⟨Z∗X̃⟩. Note that X^{∗} ≅ SL_{2}(K) and Z^{∗} is a maximal unipotent subgroup of X^{∗}. Moreover, Z^{∗} ≤ Z_{F} ≤ Q_{F} and so by 3.9 both Z^{∗} and X^{∗} are contained in G.

V Z ∗ Z∗

Lemma 4.9 Suppose that (B) holds.

(a) [V, T_{F} , C_{G}(F)) = 0. Inparticular, [V, Z^{∗}, Z^{∗}] = 0.

(b) G = ⟨X^{∗}, C_{G}(F)⟩.

(c) 4.3 still holdsif X, Z and L are replaced by X^{∗}, Z^{∗} and C_{G}(F), respectively.

Proof: By 3.6, T_{F} = ⟨T_{E}|E ∈ S(1) ∩ F ⟩. Futhermore, C_{G}(F) ≤ C_{G}(E) ≤ C_{G}([V, T_{E}] for all E in S(1) ∩ F. Thus (a) holds.

(b) G_{0} = ⟨X^{∗}, C_{G}(F)⟩ and Λ = F G_{0} ⊆ S(2). Since F X^{∗} contains elements adjacent to F (with respect to the graph defined in 3.7 )3.3b implies that Λ containes all elements in calS(2) adjacent to F. Since G_{0} acts transitive on Λ, Λ is a connected component of calS(2) and so by 3.7 Λ = calS(2). Since T_{F} ≤ G_{0} and G = ⟨Σ⟩, G = G_{0}.

(c) Using (a) and (b) the proof for 4.3 goes through.

main

Theorem 4.10 Suppose that V = ⟨C_{V} (P)^{G}⟩. Then there an (R, K)-module M such that V/V_{V} (G) ≅ M ⊗_{K} N/N^{⊥} as an RG -module.

Proof: By4.5 we may assume that C_{V} (G) = 0. Let P_{0} = NG(U). In case (B) define X^{∗} and Z^{∗} as above. In cases (A) and (C) let X^{∗} = X and Z^{∗} = Z. Then X^{∗} ∩ P_{0} acts transitively on U^{#} and therefore P_{0} = (X ∩ P_{0})P.

Let W = C_{V} (P) and W_{1} = ⟨W X^{⟩}. It is easily checked that X ≤ L, X^{∗}⟩ and so by 4.2, G = ⟨L, X^{∗}. Note that by 4.9b Z^{∗} acts quadratically on W_{1}. By 4.3 and 4.9c C_{W_{1}}(X^{∗}) ≤ C_{V} (G) = 0, W_{1} = [W_{1}, X∗] and W = [W_{1}, Z^{∗}]. Let U_{1} = ⟨UX∗⟩. Then U_{1} is a natural module for X∗ ≅ SL_{2}(K),U = [U_{1}, Z^{∗}] and C_{X^{∗}}(U) = P ∩ X∗. Hence [W, C_{X^{∗}}(U)] = 0 and [W_{1}, Z∗, C_{X^{∗}}(U)] = 0. Therefore we can apply 2.3 to X^{∗} and W_{1} and find an (R, K)-module M such that

(1) W_{1} ≅ M ⊗_{K} U_{1} as an RX^{∗}- module.

Let N_{0} = M ⊗K(N/N^{⊥}). We will show that N_{0} and V are isomorphic as RG- modules. P_{0} = (X^{∗} ∩ P_{0})P and (1) imply that M ⊗_{K} U and W are isomorphic RP_{0}-modules. Let M be the RG-module induced from the RP_{0}-module W. Let

M̃ = M/⟨[M, Z, L]^{G}⟩ and M̄ = M̃/CM̃(G).

By assumption V = ⟨W G⟩. Furthermore, N_{0} = ⟨M ⊗_{K} G⟩ and so by the universial property of induced modules there exist RG-epimorphisms:

ϕ_{1}M → V and ϕ_{2}M → N_{0}.

Since [V, , Z, L] = 0 and [N_{0}, Z, L] = 0 we get that ⟨[M, Z, L]⟩ ≤ ker ϕ_{i} for i=1,2. Further, C_{V} (G) = 0 and C_{N}0(G) = 0, the latter being true since N_{0} as a ℤG module is the direct sum of copies of N/N^{⊥}. Thus ϕ_{1} and ϕ_{2} induce RG-epimorphism

\overline{ϕ_{1}}M → V and ϕ̄_{2}M → N_{0}.

We now prove that

(2) ker \overline{phi_{i}} ∩ [\overline{M}, Z] = 0.

For this note first that by 4.4 applied to [\overline{M}, Z] = ⟨\overline{W} L^{∗}⟩, where we identified W with its canonical image in M. Obviously \overline{ϕ_{i}} restricted to W is one to one. So if [I, Z] = W, (2) is proved. Otherwise (C) holds. Recall the definition of X − 1 and g_{1} at the beginning of this section.Then

[\overline{I}, Z] = ⟨W L^{∗}⟩ = ⟨W X_{1}⟩⟨W X⟩g_{1} = ⟨W_{1}⟩g_{1}

Now [W_{1}g_{1}, Z_{1}^{g}] = W g_{1} = W and so [W_{1}g_{1}, Z] ∩ ker \overline{ϕ_{i}} = 0, [ker \overline{ϕ_{i}} ∩ W_{1}g_{1}, Z] = 0 and ker \overline{ϕ_{i}} ∩ W_{1}g_{1} ≤ C_{W_{1}g_{1}}(X_{1}) = 0. This proves (2).

By (2) we get that [ker \overline{ϕ_{i}}, Z] = 0 and so ker \overline{ϕ_{i}} ≤ C_{barM}(G). By 4.5c, CM̄(G) = 0 and thus \overline{ϕ_{i}} is one to one. Hence

V ≅ \overline{M} ≅ N_{0} as RG-modules,

Theorem 4.10 is established.

5 Determination of [V, G] ∩ C_{V} (G)

Retain the assumptions and notation from the previous section. This section is entirely devoted to the proof of

CE

Proposition 5.1 Suppose that V/C_{V} (G) ≅ M ⊗_{K} N/N^{⊥} for some (R, K)-module M and that V = [V, G]. Then one of the follwing holds

(a) There exists an R-submodule C ≤ M ⊗ N^{⊥} such that V ≅ M ⊗_{K} N/C as RGmodules.

(b) |K| = 4, σ ≠ id and dim N = 4.

Proof: Recall the notations introduced in 2.5 and 2.7. Then we are trying to proof that M ⊗_{K} N is a universial central C extension, where C is the class of RG-modules W with [W, Z, L] = 0. In view of 2.7 we may assume without loss that V/C_{V} (G) ≅ N/N^{⊥}. We first prove

(1) There exists an RG-module W and RG-submodules C_{1} and C_{2} of C_{W} (G) such that W/C_{1} ≅ V,W/C_{2} ≅ N and [W, Z, L] = 0.

Let μ be an RG-isomorphism from V/C_{V} (G) onto N/N^{⊥}. Put W = {(v, n)|v ∈ V, n ∈ N, μ(v + C_{V} (G)) = n + N^{⊥}}, Let C_{1} = {0} × N^{⊥} and C_{2} = C_{V} (G) × {0}. Then C_{i} is the kernel of the projection of W onto the i’th coordinate and so W/C_{1} ≅ V and W/C_{2} ≅ N.

In view of (1) we may assume that C_{V} (G) has a submodule C such that V/C ≅ N. Pick x ∈ Q_{U} \ T_{U} such that [N, x] is singular. Put V̄ = V/C_{V} (G) and A = [V, P, Q_{U}]. Note that V̄ ≅ N/N^{⊥}.

(2) CV̄ (x) = \overline{C_{V} (x)}.

Let R be maximal in G with respect to acting trivially on [V̄ , x], CV̄ (x)/[V̄ , x] and V̄ /CV̄ (x). Then by 3.4b and3.10a, [V̄ , R] = CV̄ (x). Note that [V, x, R, Z]=0. Indeed, if x is contained in an element of Σ, then this follows from [V, Z, L] = 0 and if not, (B) holds and it follows from 4.9a. Moreover, [R, Z] = 0 anf the three subgroup lemma implies [V, R, x] = 0.

(3) Asssume that (B) holds and |K| ≠ 2, 4. Then [V, Z^{∗}] ∩ C_{V} (G) = 0.

Recall the definitions of X^{∗}, X̃ and Z^{∗} (see before ??). It is enough to prove that [V, X^{∗}] ∩ C_{V} (G) = 0. By (2), V = [V, X^{∗}]C_{V} (X^{∗}) since the same statement holds for V̄ in plcae of V. Thus if K is not commutative or char K ≠ 2 we are done by 2.4 So assume that K is commutative. Since |K| ≠ 2, 4 there exist λμ ∈ K\{0} so that λμ ≠ 1 and λ^{2} = μ^{−1}μ^{σ}. This in turn yields an element 1 ≠ h ∈ X^{∗} with [h, X^{∗}] = 1 and acting as λμ on one of the 2dimensional singular subspaces of [N, X^{∗}] normalised by X^{∗}. Now [[V, X^{∗}], h] ∩ C_{V} (G) = 0, [V, X^{∗}, X^{∗}] ≤ [[V, X^{∗}], h] and since X^{∗} is perfect [V, X^{∗}] ∩ C_{V} (G) = 0. So (3) is proved.

(4) A ∩ C_{V} (G) = 0, except possibly in the case (B), |K| = 4, dim N = 4 and σ ≠ id.

Suppose first that Q = ⟨x^{P} ⟩. Since [V, P, x, P] = 0 we get that [V, P, Q] = ⟨x^{P} ⟩ = ⟨[V, P, x]⟩ = [V, P, x]. Now by (1), [V, x] ∩ C_{V} (G) = 0 and (4) holds in this case. Suppose next that the hypothesis of (3) holds. Note that by ??, Q = ⟨Z^{∗P} ⟩. So replacing x by Z^{∗} in the preceeding argument shows that (4) holds also in this case. Now by ?? we have covered all cases but the one excluded in (4).

(5) If A ∩ C_{V} (G) = 0, then C = 0.

Let v ∈ V \ ([V, P] + C_{V} (G)). We claim that [v, Q] ∩ (C_{V} (G) + A) ≤ A. Suppose not and put Ṽ = V/A. Since [Ṽ , Q, Q] = 0, we have [V, Q′] = 0 and [ṽ, Q] = {[ṽ, q]|q ∈ Q]}. Hence there exists q ∈ Q with [ṽ, q] ≠ 0 and [v, q] ∈ C_{V} (G) + A. Reading this equation modulo C and applying the ”Q′ = T ”-statement of ?? we get that q ∈ Q′, a contradiction to [ṽ, q] ≠ 0.

Therefore ([v, Q] + A) ∩ C_{V} (G) = 0. Since P normalizes [v, Q] + A, we conclude that [v, Q] + A = [V, P, P] = [V, Q] and [V, Q] ∩ C_{V} (G) = 0. Let g ∈ G with U^{g} ≰ U^{⊥}. Then V = [V, Q] ⊕ C_{V} (G) ⊕ A^{g}. P = (P ∩ P^{g})Q and [A^{g}, P ∩ P^{g}] = 0 imply that [V, P] = [V, Q]. Moreover, [V, Q] = A ⊕ [V, P ∩ P^{g}] and so V = [V, G] = [V, P] + [V, P]^{g} = A ⊕ [V, P ∩ P^{g}] ⊕ A^{g}. Finally, this direct sum remains a direct sum modulo C and C intersects each of the summands trivally. This implies that C = 0, proving (5).

Theorem A is now a direct consequence of (4) and (5).

6 Proof of Theorem B

P roofOfT heoremB
Reduction

Theorem 6.1 Suppose (G, Σ) fulfills the hypothesis of Theorem B. Let L_{1} = C_{G}([N, Z]), L_{2} = C_{G}([Ñ, Z]), F = [V_{0}, Z], F_{i} = C_{F} (L_{i}) and V_{i} = ⟨F_{i}^{G}⟩, (i = 1, 2). Then for i = 1, 2 [V_{i}, Z, L_{i}] = 0 and [V_{0}, G] = W_{1} + W_{2}.

Proof: Since dim N > 2 we can choose 1-dimensional subspaces U and Ũ of N and Ñ, respectively, so that Ũ([N, Z]) = 0, Ũ(U) = 0, [Ñ, Z](U) ≠ 0 and Ũ ≠ [Ñ, Z]. Let Z_{0} = T (Ũ, [N, Z]). Then [V_{0}, Z_{0}] is centralized by C_{G}(Ũ) ∩ C_{G}([N, Z]) and, in particular, by Q := C_{G}(Ũ^{⊥}) ∩ CG(˜). Note that Z centralizesŨ and therefore normalizes Q. Now [V_{0}, Z_{0}] ≤ C_{V_{0}}(Q) and so

[V_{0}, Z_{0}, Z] ≤ C_{V_{0}}(Q).

By ?? applied to L_{1}/C_{L_{1}}(CÑ([N, Z]) in place of G we have ⟨L_{0}, Z_{0}⟩ = L_{1}. Hence L_{1} = ⟨L_{0}, Q⟩ and

[V_{0}, Z_{0}, Z] ≤ C_{V_{0}}(L_{0}) ∩ C_{V_{0}}(Q) ≤ C_{V_{0}}(L1).

Since ⟨Z_{0}^{L_{0}}⟩ is normalized by ⟨L_{0}, Z_{0}⟩ = L_{1}, we have L_{1} = ⟨Z_{0}^{L_{0}}⟩ and thus

[V_{0}, ⟨Z_{0}^{L_{0}}⟩, Z] = [V_{0}, L_{1}, Z] ≤ C_{V_{0}}(L_{1})

By a symmetric argument [V_{0}, L_{2}, Z] ≤ C_{V_{0}}(L_{2}). Furthermore,

[V_{0}, G] = [V_{0}, ⟨L_{1}, L_{2}⟩] = [V_{0}, L_{1}] + [V_{0}, L_{2}]

and thus It follows that [V_{0}, G] = [V_{0}, G, G] = V_{1} + V_{2}.

To complete the proof of 6.1 it is enough to show that [V_{i}, Z] ≤ F_{i} + C_{V_{i}}(G). A glance at the proofs of ??, ?? and ?? shows that these lemmas hold with V replaced by V_{i} and L replaced by L_{0}. It follows that [V_{i}, Z] ≤ F_{i}+C_{V} (G). So [V_{i}, Z, L_{i}] = 0 and 6.1 is established.

To prove Theorem B we now merely have to apply Theorem A to the modules V_{1} and V_{2} of 6.1. Note here that we can view N as a subspace of the dual space of Ñ and that then T (˜,N) = T (N, Ñ).

7 Finitary modules for Classical Groups

finitary

Remark: we need to be more precise

Suppose G is one of the groups in the introduction and that N is infinite dimensional over K. Furthermore, let W be a G-module over the integers such that [W, g] has finite rank for all g ∈ G. In case (A) let L_{0} be defined as in Theorem B, otherwise let L_{0} = L. Then in any case [Z, L_{0}] = 0. Now it is well-known(?) that L_{0} has no non-central Z-module of finite rang. Hence [W, z, L_{0}] = 0, for all z ∈ Z, and so [W, Z, L_{0}] = 0. Therefore we can apply our main theorems with R the ring of integers to see that [W, G]/[W, G] ∩ C_{W} (G) is a direct sum of natural modules. Thus Theorem C holds.

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Ulrich Meierfrankenfeld Department of Mathematics Michigan State University East Lansing, Mi 48824 meier@math.msu.edu http://www.math.msu.edu/^{∼}meier/