5-reflectionality of anisotropic orthogonal groups over valuation rings

Abh. Math. Sem. Univ. Hamburg 61 (1991), 53-59

5-reflectionality of Anisotropic Orthogonal Groups over Valuation Rings By F. KNi~PPEL

In this article R is a commutative local ring with 1 and 1 and V is a free R-module of finite dimension n < oe. The ring R has a unique maximal ideal J. Let - denote the canonical homomorphisms R --, R / J , V --. V / J V and GL(V) ~ GL(V). A vector v E V is called unimodular if v~0 is a unit of R for some R-linear mapping ~0 9 V ~ R. This means that ~ ~ 0, or, equivalently, that v can be completed to a basis of V. A submodule U of the R-module F is called a subspace if V = U ~ W for some submodule W. Then U and W are free finite-dimensional R-modules with dim U + dim W = n. Let V carry a symmetric bilinear form f 9 V x V ~ R. Then we call V a metric or an orthogonal R-module. Clearly, f(~, ~) := f(v, w) is a well-defined symmetric bilinear form on the vector space V. Call V regular (anisotropic) if V is regular (anisotropic). Observe that V is regular if and only if the map V ~ V* (the R-module of linear functions of V into R), x ~-~ fx (where Yfx := f ( x , y)) is bijective. A commutative ring with 1 is called a valuation ring if ~ divides fl or/3 divides ~ for every pair ~,/3 c R. Every valuation ring is a local ring. If R is a valuation ring then every vector v E V admits some unimodular vector w E V such that v E Rw.

Theorem A. Let R be a valuation ring and V an anisotropic metric R-module. Let n E O(V) such that V(n 2 - 1) is a subspace. Then n is a product o f two involutions in O(V). Theorem B. Let R be a valuation ring and V an anisotropic metric R-module. I f dim V is even (odd) then every element o f O(V) is a product o f four (five) involutions in O(V). Remark. If V is a finite-dimensional regular metric vector space over a field then M.J.WONNENBURGER [7] and D.Z.DoKovIc [2] proved that O(V) is bireflectional. In the light of this result our above theorems seem to be quite unsatisfactory. Roughly said, WONNENBURGER and DOKOVIC consider for a given n E O(V) the decomposition of V into orthogonaUy indecomposable n-modules and make use of the particular structure of these n-modules. In the sequel we will prove theorems A and B. After that we shall give a short proof of LJUBIC theorem [6] that an isometry group of n-dimensional absolute geometry is bireflectional (Theorem C).

54

E Kniippel

Let V be an n-dimensional free module over a local ring R. First we collect some basic facts. 1. a) A tupel of vectors el . . . . . e. is a basis for V if and only if ~i-..... U. is a basis for the R-vector space V. b) If U is a subspace of V and W c U a submodule then W = U implies that W = U. c) For subspaces U and W of V one has V = U ~ W if and only if V=U@W. To every 7r E End(V) we assign five submodules: B(n) := V(Tr - 1),

F(~) := ker(rc - 1),

A(rc) := F(rc 2)

and

N(Tr) := ker(~ + 1),

D(n) = B(n2).

Observe that N(z) c B(z). If 7r is an involution then B(n) = N(n). 2. Let n c End(V). Then a) B(~) = B(r0, D(~) = D(r0, F(r 0 c F(~) and A(r0 = A(~). b) If B(z) is a subspace then F(r0 is a subspace and F(z) = F(~). If D(rt) is a subspace then A(n) is a subspace and A(z) = A(~). c) If Q is an involution then B(0) = N(Q) and V = B(0) @ F(Q). Indeed, if B(r0 = V(z - 1) is a subspace then F(z) = ker(z - 1) is a subspace and dim B(r~) + dim F(n) = n, hence dim B(n) + dim F(n) = n. Furthermore, dim B(~) + dim F(~) = n, B(~) = B(n) and F(n) c F(~). This yields that F(~) = F(~). In the sequel let V be a regular metric module. 3. a) For every rc 6 0 ( V ) one has F(rt) = B(n) • and A(rt) = D(Tt)• Suppose that V is anisotropic and ~ 6 0 ( V ) . Then V = B(~) Q F(~); if B(z) is a subspace then V = F(Tt) 9 B(Tz). b) If U is a regular subspace of V then V = U ~ U • Clearly there is a unique involution Q 6 0 ( V ) such that B(Q) = U, and then F(r = U • In particular, if a E V is anisotropic (i.e. f(a, a) is a unit) then V = Ra 9 a • and the symmetry oa c O(V) along Ra is the involution with B(oa) = Ra. If # E O(V) is an involution then # = ~ for some involution ~ 6 0 ( V ) . The last assertion in a) follows from B(~) = B(z) and F(~) = F(rt) (cf. 2a) and b)) and lc). Remark. In the last statement of 3a) we cannot delete the assumption that B(z) is a subspace. As an example take V = R 2 with the usual scalar product. For an arbitrary 2 ~ R with 1 + 2 2 a unit one has

rc := (1 +22)_1 ( 1 --2~ 22

--22) 1--22

~O(V).

5-reflectionality of Anisotropic Orthogonal Groups Now suppose that if 2 @ 0 then v ~ numbers over the is anisotropic and

55

22 = 0. The vector v := (0, 2) satisfies v E F(n) N B(n), and 0. One can take R := R[x]/x2~,[x] (the ring of the dual reals) and 2 := x + x2R[x]. Then the usual scalar product 2 4: 0.

4. For r~ ~ O(V) one has A(n) = F(n) Q N(n) = D(n) • Suppose that V is anisotropic and 7~ ~ O(V). Then V = D(~) ~ A(~) ; if D(n) is a subspace then V = A(n)QD(n). Indeed, ifv E A(r0 then 2v = v ( n + l ) - v ( ~ t - 1 ) ~ F ( n ) ~ N ( z O. The other statements follow from 3 when ~2 replaces n. Since the next lemma is nice we include it though it will not be used.

Lemma 1. Suppose that rc~ O(V) and U is a regular subspace containing B(rc). Then BOzp) = U for some involution ~ ~ O(V). Proof Select anisotropic vectors ~ such that U = R~Q...

QR~

B(u).

This is possible since U is regular. Furthermore, we can assume that el . . . . . ek E U and the ei's are pairwise orthogonal (the usually orthogonalization process applies). Let a~ denote the symmetry along Rei. Then Q := a l ' . . . ' a k is an involution, and ~ is the symmetry along R~. It is well-known and easy to prove, that B(g~) = B ( ~ ) = U. Furthermore, B(rcQ) ~ B(rc) + B(al) + . . . + B(ak) ~ U. Since U is a subspace this implies that B(nQ) = U; cf. lb). w

L e m m a 2. Suppose that n is even and V is not a hyperbolic plane over the field GF3. Then D(rcr = V for some involutions ~ and a in O(V). Proof Using lemma 2.2 and 2.5 in [5] we construct some 09 ~ O(V) such that = D(og). Take involutions v,# 6 0 ( V ) such that ~ - 1 o = v#; see remark following theorem B. So D(~vl~) = V. Select any involutions q, a ~ O(V) such that ~ = v and ~ =/~. We obtain D(rcpa) = D ( ~ a ) = V, hence D ( ~ a ) = V. L e m m a 3. Suppose that n is odd and that V is not a 3-dimensional vectorspace o f index 1 over GF3. Let 7z E O(V). Then V = A(~zzQa) 9 OOzzOa) for some involutions z, O, tr E O(V). Proof. Select an arbitrary anisotropic vector a. Then b := a g - a or c := a ~ + a are anisotropic, and a~ab = a respectively agac = --a. Let z := ab respectively T := ae. We have V = Ra ~ a • Let ~o denote the restriction of ~z to a • Since a • is not a hyperbolic plane over GF3 the previous lemma yields that a • = D(~0r for some involutions q,a C O(a• Lift Q and a to O(V) such that a ~ F ( Q ) n F(a). Then A(~zQa) = Ra. We conclude that V = A(uzQa) Q D(uzQa).

56

E Knfippel

L e m m a 4. Suppose that W is a finite-dimensional f r e e anisotropic R-module, tp E O ( W ) and DOp) = W . L e t d E W be an unimodular vector. Then a := d~p - d~p -1 is anisotropic and q~ := ~paa satisfies (i) d E a • • A(~p);

(ii) W = A(q~) @ D(~o); (iii) D(~o)(~o 2 - 1) = D(~o); (iv) dim A(q~) = 2. P r o o f We have ~p, q~2 _ 1, ~o - 1, ~p + 1 E G L ( V ) . Let

r : = d + dq~ -1 = d t p -1 (~o + 1)

and

s :=d - dip -1 = dip-1 (lp - 1).

(1) The vectors d , a , r , s are anisotropic; d _L a; r .1_ s; rq~ = r a n d s~p = - s . P r o o f o f the statement r~o = r. Let b := dip -1 + dip. T h e n b E a • a n d 2dtp-l ~ra = (b - a)tTa = b + a = 2d~p ,

hence r~0 -- r. Similarly one obtains sq~ = - s . F r o m (1) follows i m m e d i a t e l y (2) W = R r ~ R s ~ T w h e r e

T :=(Rr@Rs)

•

We claim that (2') T(q~ 2 - 1) = T. Clearly, T(~o 2 - 1) c T, a n d since T is a subspace the assertion will follow once we have proved that ~-(~2 _ ]-) = ~-; of. lb). In o t h e r words, we need A(~) n T = 0. A n easy a r g u m e n t yields IdimF(N)-dimF(~)l