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).

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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

<

1

[dimN(~)--dimN(~)l

<

1.

and

U s i n g A(N) = F ( ~ ) (9 N ( ~ ) we c o n c l u d e that dim A(~) < 2, hence A(-~) = Rr r R s .

N o w (2) implies A(~) r3 T = 0. (3) W = Rr@Rs@D(~o), D(q~)(q~2-1) = D(~o), A(q~) = R r @ R s , A(~) = A(q)) a n d dim A(~o) = 2. P r o o f o f (3). (1), (2) a n d (2') yield that

D(q~)

=

W(~0

2 -

1) = T(~o 2 - 1) = T ,

hence the first a n d the second equality are true. F u r t h e r m o r e , R r + Rs = A(~o) c D(q9) •

T • = R r + Rs ;

cf. 3. T h u s we proved that A(~o) = R r @ Rs. This implies the other assertions.

5-reflectionality of Anisotropic Orthogonal Groups

57

L e m m a 5. Let R be a valuation ring and V an anisotropic R-module. Let n E O(V) such that V = V(n 2 - 1) and al E V unimodular. Then n = dim V is even, and every k with 1 < k < ~ admits a2 .... , ak ~ V satisfying the following set o f properties called (Ek). (lg) The vectors al .... ,ak are anisotropie and pairwise orthogonal. Let qgk := hal "..." ak where cri denotes the symmetry along Ra~. (2k) al . . . . . ag-1 ~ A(~Ok). (3k) V = A(~pk) Q D(qgk), dim A(qgk) = 2k and D(rpk)(~o~ -- 1) = D(Cpk). P r o o f The minimum polynomial of ~ has even degree since it is symmetric and neither 1- nor ---]-is an eigenvalue of~. Hence n is even. For k = 1 lemma 4 yields the assertion immediately. Now suppose that the tupel al . . . . . ak fulfills (Ek) and k < _ ~. n We want to construct ak+l such that al . . . . . ak, ak+l is a tupel which satisfies (Ek+l). Let W :--- D(~0k) and ~p the restriction of~ok to W. Then W :~ 0. Since R is a valuation ring we can choose some unimodular d ~ W such that the projection of a~ to D(~Ok) (with respect to the decomposition V = A(q)k) (~ D(Cpk)) is contained in Rd. Let a := d~ - dip -1. Then a • ak, and W, ~p, d, a satisfy the requirements of the previous lemma. Let ak+l := a. Property (3k) and (ii) of the previous lemma yield that V = A(~Ok+l)O)D(~Ok+l). Also the other statements in (Ek+1) are easy conclusions from (Ek) and the assertions of the previous lemma. Corollary 1. Suppose that V, R and n satisfy the assumptions o f the previous lemma. Furthermore, let a E V. Then n = Or f o r some involutions O, z E O(V) with a E B(z). P r o o f Choose some unimodular vector al such that a E Ral and apply the previous lemma with k = ~. n We get involutions hal" . .'ak and al "...'ak where the mappings ai are pairwise commuting symmetries and al is the symmetry along Ral. Hence n = (hal "..." ak)(al "..." ak) is a product of two involutions O, r and Ral c B(z). P r o o f o f Theorem A. D(n) = V ( n 2 - 1 ) is a subspace, so is A(n) = k e r ( n 2 - 1 ) . Furthermore, A(n) l D(n), and V = A(~) @ D(~) since V is anisotropic. This yields that V = A(n) 9 D(n). The restriction of 7r to A(n) is an involution. The corollary implies that the restriction of n to D(n) is a product of two involutions. Hence the assertion is true. Theorem B follows immediately from lemma 2, lemma 3 and theorem A. Finally, we will give a simple proof that n-dimensional absolute geometry is bireflectional. Let V be a vectorspace of dimension n + 1 (n _> 2) over a field K of charK ~ 2 and f " V • V ~ K a symmetric bilinear form. Suppose that V contains an n-dimensional anisotropic subspace W. Hence dim radV = 1

Preliminaries.

58

E Kniippel

and indV = 0, or dim radV = 0 and indV < 1. For an anisotropic vector a let aa denote the symmetry along Ka. If a, b, c are linearly dependent anisotropic vectors then aaahac = cra and Q(d) = Q(a) . Q(b) . Q(e) (where Q(v) := f(v,v)) for some vector d. Put a • b • c := d. For a set T of anisotropic vectors of V we introduce the following properties. (T0 W \ 0 ~

T.

(T2) If a, b, c E T and a, b, c are linearly dependent then a • b • c E T. (T3) If a, b E T are orthogonal then (Ka + Kb) \ 0 c T. (T4) T @ W \ 0 a n d

T@ V\0.

Let S := {a~ : a E T} and G the group generated by S. In [1] and [4] absolute geometry is defined in group-theoretic terms by a system of axioms. When a pair (15, ~) where 15 is a group and ~ a set of involutions generating 15 satisfies the axioms then (6i, ~) is called an isometry group (of n-dimensional absolute geometry). We will only take into account so-called non-elliptic isometry groups; if (15, ~) is elliptic then 15 is isomorphic to PO(V) for some n+l-dimensional anisotropic vectorspace, hence the remark following theorem B implies that 15 is bireflectional. A main theorem states: Every (non-elliptic) isometry group (15, ~) is isomorphic to a pair (G,S) as in 5 which satisfies (T1) to (T4). In [6] D.LJuBIC proved for every isometric group (15, ~) that 15 is bireflectional, i.e. that every element of ffi is a product of two involutions of 15. Avoiding vector space arguments LJUBIC gives a synthetic proof. Our theorem C is a slightly more general statement. Lemma 6. Let W be an n-dimensional anisotropic vectorspace over a field of charK :# 2. Let n E O ( W ) and a E W. Then ~ = c~fl and a E B(fl) for some involutions c~,fl E O(W). Proof We have W = A ( ~ ) Q D ( n ) and A(n) = F(~)(DN(~). Let a =b+c+d where b E F(n), c E N(r0 and d E O(Tr). Clearly zIA(~) = /~v and b + c E N(v) for some involutions #, v E O(A(rt)). Corollary 1 supplies involutions 0,T E O(D(n)) such that n[Ol,) = 0r and d E B(r). Clearly, := ~t ~ @and fl := # ~ r fulfill the assertion.

Theorem C. (D.LJuBIC). Let V be a vectorspace of dimension n+ 1 over afield of charK :~ 2 and f : V • V ~ K a symmetric bilinear form. Let W be an n-dimensional anisotropic subspace of V and T a set of anisotropic vectors of V such that (T1) and (T2) of 5 hold. Let G be the group generated by the set S of symmetries along elements o f T. Then G is bireflectional.

5-reflectionality of Anisotropic Orthogonal Groups

59

P r o o f F r o m (TI) and (T2) follows immediately: (i) Given a, b ~ T. Then a~ab

= tTdtTe

for some d E W \ 0 and e 6 T.

Every element o f G is a product of symmetries along elements of T ; hence (i) yields (ii) Given n ~ G. Then n = ~otr for some product ~o of symmetries along vectors o f W and a 6 S. (iii) Given a ~ W and ~0 6 0 ( W ) . involutions ~,fl c O(W),

Then ~0 = ctfl and a c F(fl) for some

P r o o f o f (iii). Apply l e m m a 6 and observe that B(fl) = F ( - f l ) involution ft.

for every

N o w let rc E G. Using (ii) we have n = q~r where ~o is a product o f symmetries along vectors o f W and tr = trc for some c E T. Since V = W 9 W • we can assign to every ~p E O ( W ) the element ~p ~ lw• E O(V) and thus consider O ( W ) as a subgroup of O(V). Clearly O(W) c G. Let c = a + b where a E W and b ~ W • Then (iii) supplies involutions ~, fl c O(W) such that q~ = ~fl and a ~ F(fl). Since b ~ W l c F(fl) we have c E F(fl). Therefore fla is an involution, and n = c~(fla) is a product of two involutions in G.

References

[1] EBACHMANN, Aufbau der Geometrie aus dem Spiegelungsbegriff,

2 nd

edition,

Springer 1973. [2] D.Z.DoKovIc, The Product of Two Involutions in the Unitary Group of a Hermitian Form, Indiana Univ. Math. J. 21 (1971), 449-456. [3] E.W.ELLERS and R.FRANK, Products of Quasireflections and Transvections over Local Rings, J. Geom. 31 (1988), 69-78. [4] H.KINDER, Begriindung der n-dimensionalen absoluten Geometrie aus dem Spiegelungsprinzip, Dissertation, Kiel 1965. [5] EKNi3PPEL and K.NIELSEN, Products of Involutions in O+(V), Lin. Alg. and Appl. 94 (1987), 217-222. [6] D.LuBIb, Bireflectionality in Absolute Geometry, Canad. Math. Bull. 32 (1989), 54-63. [7] M.J.WoNNENBURGER, Transformations Which Are Products of Two Involutions, J. Math. Mech. 16 (1966), 327-338.

Eingegangen am: 05.07.1990

Author's address: Frieder Kniippel, Mathematisches Seminar, OlshausenstraBe 40, W-2300 Kiel, Bundesrepublik Deutschland.