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Generators of an orthogonal group over a local valuation domain
JOURNAL
OF ALGEBRA 55, 302-307
Generators
(1978)
of an Orthogonal Group Local Valuation Domain HIROYUKI
Department
of Mathematics,
ISHIBASHI
Josai University,
Communicated Received
over a
Sakadv, Saitama,
Japan
by J. Dieudonne’
September
20, 1977
I,et o be a local valuation domain, i.e., o is a commutative integral domain which has the unique maximal ideal R and for any a and 6 in o, either a divides b or b divides a. We assume 2 is a unit of o. V is an n-ary quadratic mod& over o with unit discriminant, O(V) or O,(V) is the orthogonal group on V, and S(V) is the set of symmetries on V. Since O(V) is generated by S(V) and any symmetry has order 2, any element CJin 0( 5’) is a product of finite number of elements of S(V). So we define Z(o) is the minimal number of factors in the expression of CJof O(V) as a product of symmetries on V. The case in which o is a field Z(a) was determined by Scherk [6] and DicudonnC [l]. DicudonnC also treated the other classical groups, not only O(V) but S’(V) or U(V). In [3] I have generalized the results of Scherk to an orthogonal group on a local principal ideal domain. In the present paper J generalize my results of [3] to an orthogonal group on a local valuation domain o. Let (Tbe in O,(V). WC denote the quotient field of o by F, the fixed module of u in V by V, , i.e., V, = {x E V / OX = X} and d = dim FV. Then the result of this note is Z(u) = rc (See Section 2, Statement
d
or
n---+2.
of the Theorem.)
1.
PRELIMINARY
LEMMA
Let o be a local valuation domain with 2 as a unit and R the unique maximal ideal of o, we use - or 7r to denote the canonical homomorphism from o onto o/R. F denotes the quotient field of o. If U is a direct summand of V then we call it a space over o and dim U means dim FU. We write V = &Xi oxi , xi E V. 302 0021~8693/78/0552-0302$02.00/O Copyright 0 1978 byAcademicPress,Inc. AU rights ofreproduction in any form reserved.
GENERATORS OF o,(v)
OVER LVZI
We use the same notation - or T, to denote the canonical map from V onto vp v. For sets A and B the set theoretic difference will be written A-B. Lmmh
1.7 Let U be a submodule of V. Then we have the following”
Sf r = 0 then V = U. im D < dimFU. (b) (c) 1j V = U @ T then P = u @ T and dim U = dim vi. (d) pS dim 0 = dim FU then U is a direct s~rnrn~~d of V. (a)
Proof.
We use -
for 27.
(a) Since T = 87 and V = @F=, oxi , we can take zli’s in U with z for i = 1, 2,..., n. Hence for 1 < i < n, xi - ui is contained in RV. xi = ui i Cj”=, aijxj i aij E R, and M = (a& Then we have *(zL~,.~., u,) = t(XI >.“, xlz) (E-&f). E is th e identity matrix. Since (1 - aii / 1 < i < B>are units in o? E-M is an invertible matrix, hence (x1 )..., ~3 C U. ‘I’herefore L7 =
v.
(b) Write P = u @ (@i=, GQ, wi E V. Put W = CL1 CIW~. Thea v = U + W. Hence by (a) we have V = U + IV. Therefore 1z< dim Site dim D = n - Y, we have (b). [By (b) if @b, F@$then &, 0’~‘~.] (c)
‘I’his part is clear by (b), since dim V = dim v.
(d) Write r = u @ (@T=, G@~), wi E V. I’ut = &, owi . since r = LJ + W, by (a) we have V = U + El’* Considering the dimensions we have b’ = U @ IV. QED.
2. STATEMENT
OF THE THEOREM
V is an n-ary nonsingular quadratic space over o. Nonsingular means the discriminant of V is a unit of o. O,(V) or O(V) is the orthogonal group on P’. We define canonically z + p = x + y, EZ = ax, and 57 =G for x, y in V and a in o. It can be shown that V/RV is an n-ary nonsingular quadratic space over o/R. If U is a nonempty subset of V, U* is its orthogonal complement (in V), i.e., U* = (X E I/ 1XU = 01. -4 submodule t’ of V is said to be an orthogonal component of V or pure in P’ if V = U _L W foor some submodule IV of ‘J. where U I @‘means U @ Wand UW = 0. It is easy to seethat if V’ = U 1 E7 t-hen w =
U”.
LEMM.~ 2.1. Let T be a submodule of V and p in O,(V); direct s~rnrna~d~ of V.
then T* and V,, are
304
HIROYUKI ISHIBASHI
Proof. Put U = V,, or T*. Suppose dim a < dimFU. We split v = 0 @L and write L = @I=, 6~~) wi E V. Put W = @L=, owi . By the assumption we have U A W # (0). Take x’ # 0 in U n W. Since o is a valuation ring we can write x’ = ax with a in o and x in V - RV. Since X’ belongs to U, hence so does x. Hence dim ox + W = Y + 1, but dim F(ox + W) = Y, which contradicts (b) of Lemma 1.1. Hence dim 0 = dimFU. By (d) of Lemma 1.1 we have the lemma. Q.E.D. LEMMA 2.2.
If U is a direct summand of V then (i7)* = (U*).
Proof.
By Lemmas 1.l and 2.1 it is obvious. For a submodule U of V, P( 77) denotes the set of all maximal pure submodules of V which contained in U. We regard the zero module as a pure one. Clearly -P(U) # @. We denote a typical one in P(U) byp( U). Since u = p( U) I (p(U)*) is a radical splitting with rad a = (p(U)*) and by Lemma 1.1 we know dimp( U) = dim p( U), so we see dimp( U) is unique for U, i.e., dim p( U) depends only on U itself and not the choice of p( U) in P(U). Now we state our theorem. We have put n = dim V. Further, for g in Q,(V), put d = dim V, , dl = dim p( V,), and d = dl + d,, . THEOREM. Let 1 f
0 E O,(V).
(i)
Ifn-d-dO#OthenZ(u)=n-d.
(ii)
i’j n -
d - d,, = 0 then Z(O)= n - d + 2.
3. PROOF FOR (i) OF THE THEOREM A pure submodule U of V is called a pure line or plane according to whether dim U = 1 or 2 (resp.). For a vector x in V if ox is a pure line then we say x is pure. If x is a pure vector of V, then the linear mapping 7, which carries x to --x and is identity on (ox)* is clearly in O(V); it is called the symmetry with respect to (ox)*. Let 0 # 1 be in O,(V). Extending u naturally to (T’, an isometry on FV, we know Z(o’) = n - d. (See [4] or [8].) So Z(a) 3 n - d. Hence it suffices to show
Z(u) < n - d. We now define K(o) = n - d - do and prove that if K(U) # 0 then Z(U) < n - d. First we show K(u) > 0. Split V, = p(V’) 1 U and V = p( V,) i V’. Then U C Y,. Then p is nonsingular and a is totally isotropic. Hence we have 0 < dim B’ - 2 dim a = dim V’ - 2 dim U = n - dl - 2d,, = n - d - do = K(o). Therefore K(u) > 0. Since K(u) f 0, we have K(u) > 0. Our proof will proceed by induction on n = dim V and n - d = codim V, = dim( V,)*.
GENERATORS
OF o,(v)
OVER
L-D
(a) Let n = I. Since 0 is an integral domain, Q = fl. o = -1 is a symmetry and d = 0. Hence I(O) = 1 = FZ- d.
305
Since D f I,
(b) Let V, contains a pure line, Call the line L. Split Y = h, J- %*. Then L* is nonsingular, dimL* = n - 1, p = (TIL* belongs to 0,-,(%*), (L”), = V0 n L* and so dim(L*), = d - 1. Therefore by the inductive hypothesis on n, we have I(p) < (TZ- 1) - (d - 1) = n - d. Hence I(U) < E - d. (c) By (a) and (b) we assume n 3 2 and V, has no pure line, i.e,, d = $ and k(o) = n - 2~7.Since k(o) > 0, we have d < n/2. ence dim(V,)* > nj2, Hence by Lemma 2.1 and (c) of Lemma 1.1 we have dim(VO)* > 42. Therefore (V,)* contains a pure vector y, which is fixed throughout this section. ~EMhK.4
3.1~
Let
v
be a pure vector in V.
(a) Ij ‘o F (V,)* then V, C VT, aad dim V,, = d + 1. (b) If v $ (VW)* then VTgq C V, and dim VTsO= d - 1, .?p.ooJ The lemma holds in the orthogonal group over F (see [2]), and hence it holds over o. Suppose k(T,u) # 0. Since y is in (VU)*, by (a) of the Iemma and inductive hypothesis on codim V, , we have E(T,cJ)= n - (d + 1) = n - d - 1, hence l(a) < n - d. So we assume k(T,u) = 0. In the rest of this section, we show that there exists another pure vector v in (VO)* with k(a,o) # 0, which completes mr proof* We determine when we have k(T,o) # 0 for a pure z’ in (VU)*. LEMMA 3.2.
For
a pure
a, in (I/cr)*,
R(T,G) # 0 if ad
only ij
k(c) $
dim p(VTZID) > 3. So in the following cases we have k(r+~) # 0. (a)
k(0) > 3.
(b)
k(u) = 2 and VTu, contains a pure line.
(c)
k(o) = 1 and V,., contains a pure plane.
I’~oo,f~ K(T,u) = dim V - 2 dim VT*, + dim P(V,*~) = n - 2(d + 1) + im P(V~~,,.) = k(a) - 2 + dimp(V7,,UB).
y the lemma we may assume k(o) = 1 or 2. Now we list our assumptions to clearify the proof. (I) n >, 2. (2) k(0) = n - 2a = 1 or 2. (3) V, has no pure vector, i.e., TV is totally isotropic (4) y is a pure vector in ( VO)*, so y2 is a unit. (5) k(~~u) = 0, i.e., k(a) + dim p( V,*.J < 3.
306
HIROYUKI
ISHIBASHI
Under these assumptions all we have to do is to find a new pure vector v in (VU)* with K(u) + dimp(i?7,yU) > 3. By (a) of Lemma 3.1 we have V, C VTw,, and dim V, ~ = d + 1. Further, since by Lemma 2.1 fixed spaces are direct summands of “V, we have for some x in VTVO. Therefore
T+,(IX= x and so
OX = rl/x = x + ay
for some a in 0.
(In fact a # 0, for x # V, .) The above equation implies (ox)” = (x + ay)” and so 0 = 2axy + azyz. Since a # 0, we have 0 = 2xy + ay2. By assumption (4), ys is a unit. Therefore we conclude xy is a unit if and only if so is a. The following is a key lemma.
LEMMA 3.3. k(T,o) = 0 implies that a is a unit of o. Proof. We put V,, = U. Then dim U* e n - (d + 1) = (d + 1) + (n - 2d) - 2 = dim U + k(u) - 2. Since - preserves dimensions of direct summands, dim (U)* = dim u + k(u) - 2. Hence dim U* = dim U + (- 1 or 0) according to k(u) = 1 or 2. We know U* = (u)*
by Lemma 2.2. There-
fore if U* were not contained in g then u would contains a pure plane or line according to K(a) = 1 or 2, so K(a) + dimp( U’) 3 3, contradicts assumption (5). Thus p C n. Consequently U” = (u)* is totally isotropic. On the other hand, U = V, @ ox and y is contained in (Vu)*, i.e., yV, = 0, by assumption (4). Hence if xy were not a unit then y U = 0 [i.e., 7 would belong to the totally isotropic space (u)*], contradicts y2 is a unit. Hence xy is a unit and a is a unit. Q.E.D. It is easy to see that (u - 1) V C ( VO)*, since for‘s in V and t in V, we have ((u - 1) s) t = (us - s)t = (US) t - st = usut - st = 0. These lemmas now proved, we try to find the desired pure vector v. First we suppose d = 0. Hence K(o) = n - 2d = n. However, since we have k(a) = 1 or 2 and n > 2, we have k(u) = n = 2. Since o is a field, we can apply the well-known result in the orthogonal group over a field that if n < 4 then Z(i) = n dim r=On
dim VP for any jj in O,(r), (see [5, 43: 41). Hence Z(T+J) = 2 the other hand, d = 0 implies VTVU= V, @ ox = ox. It is easy
to see V7V0C F7yU , so 0 # x is contained in vcg . In particular,
dim VT70 # 0.
This means Z(G) = 0 or 1. Further we shall show Z(z) # 1. By assumption (5) we have K(u) + dimp( V,,) < 3. Since k(u) = 2 and VTy, = ox, this means dimp(ox)
= 0, i.e., x is an isotropic
be a symmetry,
vector fixed by G. -
(note that n = dim Y = 2), i.e.,
Z(T,U)
Therefore, #
1. So Z(q)
G
cannot = 0, i.e.,
GENERATORS OF o,(Y)
OVER LVD
G is an identity map on V. Hence 67 = -r* Put u = y and Then u is contained in (VO)*, because of Im(u - 1) C (V,,)*. (c)z = I-27)” # 0, v is pure. Finally K(o) + dimp(VTeD) = K(a) 2 + 1 = 3. Thus 2: is a desired vector for d = 0. next let d f 0. We have V, ~ = V,, @ ox. Since V, is a direct
307
7j = (0 - 1) 21, Further, since $ dimp(oy) = sumrnand of V,
we have F0 # (0). y assumition (3), V, is totally isotropic. Hence if 5!(TG) were not zero then V,, would contain a pure plane, and so K(T~D) # 0, a contradiction. Therefore Z(K) = 0. Take z in Y with %(VU) + 0. If (U - 1) x is pure then we put u = z and TJ= (u - 1) U. If (5 - 1) x is not pure then we take 6 E C&l> such that y + S(u - 1) z is pure, put zk = x + n&z and av = (5 - I) u. In fact such 6 exists, since 2 is a unit in 0. Then whether (0 - 1) z is pure or not, u is pure in (VO)*~ Further, by Lemma 3.3 a is a unit. Hence uV, # 0. Therefore V, ~ = 7, @ ozk contains a pure plane, consequently K(U) + dimp(V7,210) > 3. ?&us we have completed our proof for (i) of the theorem.
4. PROOF FOR (ii) OF THE THEOREM In the same way as in the proof for (i), we may assume p(Vi) = (0) and k(a) = n. - 2d = 0. Take a pure y not in (VU)*. Then considering the orthogonal group over E, we find dim IfTsO =dimV*---l=d--1. Hence K(T,0)=n--2dimV7V,== n - 2d + 2 = 2 # 0. So, applying (i) of the theorem, we have ~(T~G.)= n - (d - 1) = n - d + 1. Therefore Z(o) < n - d + 2. The facts that E(o) f az- d and + n - d + 1 are clear by the results of O,cFFv). Thus we have completed the proof for (ii) of the theorem.
REFERENCES I. J. ~ImJDoNN~, Sur les generateurs des groupes classiques, Swnma Brad. Math. 3 (395§), 149-179. 2. H. IsH~BAS~-II, On some system of generators of the orthogonal groups, Sci. I&p. Tokyo Kyoiku Duigaku, Sect. A 11, No. 287 (1972), 24-33. 3. H. ISHIBASHI, The decomposition of isometries into symmetries in an orthogonal group ever a local principal ideal domain, Sci. Rep. Tokyo Kyoiku Daigaku Sect. A 12, No. 321 (1972), 49-58. 4. W. KLINGENBERG, Orthogonal gruppen iiber lokaien ringen, Amw. 1. Math. 83 (1961), 261-320. to Quadratic Forms,” Springer-Verlag, Berlin/ 5. 0. T. O'MEARA, “Introduction Giittingen/Heidelberg, 1963. 6. P. SCHERK, On the decomposition of orthogonalities into symmetries, PYOC. Amer. Math. Sot. 1 (1950), 481-491.
OF ALGEBRA 55, 302-307
Generators
(1978)
of an Orthogonal Group Local Valuation Domain HIROYUKI
Department
of Mathematics,
ISHIBASHI
Josai University,
Communicated Received
over a
Sakadv, Saitama,
Japan
by J. Dieudonne’
September
20, 1977
I,et o be a local valuation domain, i.e., o is a commutative integral domain which has the unique maximal ideal R and for any a and 6 in o, either a divides b or b divides a. We assume 2 is a unit of o. V is an n-ary quadratic mod& over o with unit discriminant, O(V) or O,(V) is the orthogonal group on V, and S(V) is the set of symmetries on V. Since O(V) is generated by S(V) and any symmetry has order 2, any element CJin 0( 5’) is a product of finite number of elements of S(V). So we define Z(o) is the minimal number of factors in the expression of CJof O(V) as a product of symmetries on V. The case in which o is a field Z(a) was determined by Scherk [6] and DicudonnC [l]. DicudonnC also treated the other classical groups, not only O(V) but S’(V) or U(V). In [3] I have generalized the results of Scherk to an orthogonal group on a local principal ideal domain. In the present paper J generalize my results of [3] to an orthogonal group on a local valuation domain o. Let (Tbe in O,(V). WC denote the quotient field of o by F, the fixed module of u in V by V, , i.e., V, = {x E V / OX = X} and d = dim FV. Then the result of this note is Z(u) = rc (See Section 2, Statement
d
or
n---+2.
of the Theorem.)
1.
PRELIMINARY
LEMMA
Let o be a local valuation domain with 2 as a unit and R the unique maximal ideal of o, we use - or 7r to denote the canonical homomorphism from o onto o/R. F denotes the quotient field of o. If U is a direct summand of V then we call it a space over o and dim U means dim FU. We write V = &Xi oxi , xi E V. 302 0021~8693/78/0552-0302$02.00/O Copyright 0 1978 byAcademicPress,Inc. AU rights ofreproduction in any form reserved.
GENERATORS OF o,(v)
OVER LVZI
We use the same notation - or T, to denote the canonical map from V onto vp v. For sets A and B the set theoretic difference will be written A-B. Lmmh
1.7 Let U be a submodule of V. Then we have the following”
Sf r = 0 then V = U. im D < dimFU. (b) (c) 1j V = U @ T then P = u @ T and dim U = dim vi. (d) pS dim 0 = dim FU then U is a direct s~rnrn~~d of V. (a)
Proof.
We use -
for 27.
(a) Since T = 87 and V = @F=, oxi , we can take zli’s in U with z for i = 1, 2,..., n. Hence for 1 < i < n, xi - ui is contained in RV. xi = ui i Cj”=, aijxj i aij E R, and M = (a& Then we have *(zL~,.~., u,) = t(XI >.“, xlz) (E-&f). E is th e identity matrix. Since (1 - aii / 1 < i < B>are units in o? E-M is an invertible matrix, hence (x1 )..., ~3 C U. ‘I’herefore L7 =
v.
(b) Write P = u @ (@i=, GQ, wi E V. Put W = CL1 CIW~. Thea v = U + W. Hence by (a) we have V = U + IV. Therefore 1z< dim Site dim D = n - Y, we have (b). [By (b) if @b, F@$then &, 0’~‘~.] (c)
‘I’his part is clear by (b), since dim V = dim v.
(d) Write r = u @ (@T=, G@~), wi E V. I’ut = &, owi . since r = LJ + W, by (a) we have V = U + El’* Considering the dimensions we have b’ = U @ IV. QED.
2. STATEMENT
OF THE THEOREM
V is an n-ary nonsingular quadratic space over o. Nonsingular means the discriminant of V is a unit of o. O,(V) or O(V) is the orthogonal group on P’. We define canonically z + p = x + y, EZ = ax, and 57 =G for x, y in V and a in o. It can be shown that V/RV is an n-ary nonsingular quadratic space over o/R. If U is a nonempty subset of V, U* is its orthogonal complement (in V), i.e., U* = (X E I/ 1XU = 01. -4 submodule t’ of V is said to be an orthogonal component of V or pure in P’ if V = U _L W foor some submodule IV of ‘J. where U I @‘means U @ Wand UW = 0. It is easy to seethat if V’ = U 1 E7 t-hen w =
U”.
LEMM.~ 2.1. Let T be a submodule of V and p in O,(V); direct s~rnrna~d~ of V.
then T* and V,, are
304
HIROYUKI ISHIBASHI
Proof. Put U = V,, or T*. Suppose dim a < dimFU. We split v = 0 @L and write L = @I=, 6~~) wi E V. Put W = @L=, owi . By the assumption we have U A W # (0). Take x’ # 0 in U n W. Since o is a valuation ring we can write x’ = ax with a in o and x in V - RV. Since X’ belongs to U, hence so does x. Hence dim ox + W = Y + 1, but dim F(ox + W) = Y, which contradicts (b) of Lemma 1.1. Hence dim 0 = dimFU. By (d) of Lemma 1.1 we have the lemma. Q.E.D. LEMMA 2.2.
If U is a direct summand of V then (i7)* = (U*).
Proof.
By Lemmas 1.l and 2.1 it is obvious. For a submodule U of V, P( 77) denotes the set of all maximal pure submodules of V which contained in U. We regard the zero module as a pure one. Clearly -P(U) # @. We denote a typical one in P(U) byp( U). Since u = p( U) I (p(U)*) is a radical splitting with rad a = (p(U)*) and by Lemma 1.1 we know dimp( U) = dim p( U), so we see dimp( U) is unique for U, i.e., dim p( U) depends only on U itself and not the choice of p( U) in P(U). Now we state our theorem. We have put n = dim V. Further, for g in Q,(V), put d = dim V, , dl = dim p( V,), and d = dl + d,, . THEOREM. Let 1 f
0 E O,(V).
(i)
Ifn-d-dO#OthenZ(u)=n-d.
(ii)
i’j n -
d - d,, = 0 then Z(O)= n - d + 2.
3. PROOF FOR (i) OF THE THEOREM A pure submodule U of V is called a pure line or plane according to whether dim U = 1 or 2 (resp.). For a vector x in V if ox is a pure line then we say x is pure. If x is a pure vector of V, then the linear mapping 7, which carries x to --x and is identity on (ox)* is clearly in O(V); it is called the symmetry with respect to (ox)*. Let 0 # 1 be in O,(V). Extending u naturally to (T’, an isometry on FV, we know Z(o’) = n - d. (See [4] or [8].) So Z(a) 3 n - d. Hence it suffices to show
Z(u) < n - d. We now define K(o) = n - d - do and prove that if K(U) # 0 then Z(U) < n - d. First we show K(u) > 0. Split V, = p(V’) 1 U and V = p( V,) i V’. Then U C Y,. Then p is nonsingular and a is totally isotropic. Hence we have 0 < dim B’ - 2 dim a = dim V’ - 2 dim U = n - dl - 2d,, = n - d - do = K(o). Therefore K(u) > 0. Since K(u) f 0, we have K(u) > 0. Our proof will proceed by induction on n = dim V and n - d = codim V, = dim( V,)*.
GENERATORS
OF o,(v)
OVER
L-D
(a) Let n = I. Since 0 is an integral domain, Q = fl. o = -1 is a symmetry and d = 0. Hence I(O) = 1 = FZ- d.
305
Since D f I,
(b) Let V, contains a pure line, Call the line L. Split Y = h, J- %*. Then L* is nonsingular, dimL* = n - 1, p = (TIL* belongs to 0,-,(%*), (L”), = V0 n L* and so dim(L*), = d - 1. Therefore by the inductive hypothesis on n, we have I(p) < (TZ- 1) - (d - 1) = n - d. Hence I(U) < E - d. (c) By (a) and (b) we assume n 3 2 and V, has no pure line, i.e,, d = $ and k(o) = n - 2~7.Since k(o) > 0, we have d < n/2. ence dim(V,)* > nj2, Hence by Lemma 2.1 and (c) of Lemma 1.1 we have dim(VO)* > 42. Therefore (V,)* contains a pure vector y, which is fixed throughout this section. ~EMhK.4
3.1~
Let
v
be a pure vector in V.
(a) Ij ‘o F (V,)* then V, C VT, aad dim V,, = d + 1. (b) If v $ (VW)* then VTgq C V, and dim VTsO= d - 1, .?p.ooJ The lemma holds in the orthogonal group over F (see [2]), and hence it holds over o. Suppose k(T,u) # 0. Since y is in (VU)*, by (a) of the Iemma and inductive hypothesis on codim V, , we have E(T,cJ)= n - (d + 1) = n - d - 1, hence l(a) < n - d. So we assume k(T,u) = 0. In the rest of this section, we show that there exists another pure vector v in (VO)* with k(a,o) # 0, which completes mr proof* We determine when we have k(T,o) # 0 for a pure z’ in (VU)*. LEMMA 3.2.
For
a pure
a, in (I/cr)*,
R(T,G) # 0 if ad
only ij
k(c) $
dim p(VTZID) > 3. So in the following cases we have k(r+~) # 0. (a)
k(0) > 3.
(b)
k(u) = 2 and VTu, contains a pure line.
(c)
k(o) = 1 and V,., contains a pure plane.
I’~oo,f~ K(T,u) = dim V - 2 dim VT*, + dim P(V,*~) = n - 2(d + 1) + im P(V~~,,.) = k(a) - 2 + dimp(V7,,UB).
y the lemma we may assume k(o) = 1 or 2. Now we list our assumptions to clearify the proof. (I) n >, 2. (2) k(0) = n - 2a = 1 or 2. (3) V, has no pure vector, i.e., TV is totally isotropic (4) y is a pure vector in ( VO)*, so y2 is a unit. (5) k(~~u) = 0, i.e., k(a) + dim p( V,*.J < 3.
306
HIROYUKI
ISHIBASHI
Under these assumptions all we have to do is to find a new pure vector v in (VU)* with K(u) + dimp(i?7,yU) > 3. By (a) of Lemma 3.1 we have V, C VTw,, and dim V, ~ = d + 1. Further, since by Lemma 2.1 fixed spaces are direct summands of “V, we have for some x in VTVO. Therefore
T+,(IX= x and so
OX = rl/x = x + ay
for some a in 0.
(In fact a # 0, for x # V, .) The above equation implies (ox)” = (x + ay)” and so 0 = 2axy + azyz. Since a # 0, we have 0 = 2xy + ay2. By assumption (4), ys is a unit. Therefore we conclude xy is a unit if and only if so is a. The following is a key lemma.
LEMMA 3.3. k(T,o) = 0 implies that a is a unit of o. Proof. We put V,, = U. Then dim U* e n - (d + 1) = (d + 1) + (n - 2d) - 2 = dim U + k(u) - 2. Since - preserves dimensions of direct summands, dim (U)* = dim u + k(u) - 2. Hence dim U* = dim U + (- 1 or 0) according to k(u) = 1 or 2. We know U* = (u)*
by Lemma 2.2. There-
fore if U* were not contained in g then u would contains a pure plane or line according to K(a) = 1 or 2, so K(a) + dimp( U’) 3 3, contradicts assumption (5). Thus p C n. Consequently U” = (u)* is totally isotropic. On the other hand, U = V, @ ox and y is contained in (Vu)*, i.e., yV, = 0, by assumption (4). Hence if xy were not a unit then y U = 0 [i.e., 7 would belong to the totally isotropic space (u)*], contradicts y2 is a unit. Hence xy is a unit and a is a unit. Q.E.D. It is easy to see that (u - 1) V C ( VO)*, since for‘s in V and t in V, we have ((u - 1) s) t = (us - s)t = (US) t - st = usut - st = 0. These lemmas now proved, we try to find the desired pure vector v. First we suppose d = 0. Hence K(o) = n - 2d = n. However, since we have k(a) = 1 or 2 and n > 2, we have k(u) = n = 2. Since o is a field, we can apply the well-known result in the orthogonal group over a field that if n < 4 then Z(i) = n dim r=On
dim VP for any jj in O,(r), (see [5, 43: 41). Hence Z(T+J) = 2 the other hand, d = 0 implies VTVU= V, @ ox = ox. It is easy
to see V7V0C F7yU , so 0 # x is contained in vcg . In particular,
dim VT70 # 0.
This means Z(G) = 0 or 1. Further we shall show Z(z) # 1. By assumption (5) we have K(u) + dimp( V,,) < 3. Since k(u) = 2 and VTy, = ox, this means dimp(ox)
= 0, i.e., x is an isotropic
be a symmetry,
vector fixed by G. -
(note that n = dim Y = 2), i.e.,
Z(T,U)
Therefore, #
1. So Z(q)
G
cannot = 0, i.e.,
GENERATORS OF o,(Y)
OVER LVD
G is an identity map on V. Hence 67 = -r* Put u = y and Then u is contained in (VO)*, because of Im(u - 1) C (V,,)*. (c)z = I-27)” # 0, v is pure. Finally K(o) + dimp(VTeD) = K(a) 2 + 1 = 3. Thus 2: is a desired vector for d = 0. next let d f 0. We have V, ~ = V,, @ ox. Since V, is a direct
307
7j = (0 - 1) 21, Further, since $ dimp(oy) = sumrnand of V,
we have F0 # (0). y assumition (3), V, is totally isotropic. Hence if 5!(TG) were not zero then V,, would contain a pure plane, and so K(T~D) # 0, a contradiction. Therefore Z(K) = 0. Take z in Y with %(VU) + 0. If (U - 1) x is pure then we put u = z and TJ= (u - 1) U. If (5 - 1) x is not pure then we take 6 E C&l> such that y + S(u - 1) z is pure, put zk = x + n&z and av = (5 - I) u. In fact such 6 exists, since 2 is a unit in 0. Then whether (0 - 1) z is pure or not, u is pure in (VO)*~ Further, by Lemma 3.3 a is a unit. Hence uV, # 0. Therefore V, ~ = 7, @ ozk contains a pure plane, consequently K(U) + dimp(V7,210) > 3. ?&us we have completed our proof for (i) of the theorem.
4. PROOF FOR (ii) OF THE THEOREM In the same way as in the proof for (i), we may assume p(Vi) = (0) and k(a) = n. - 2d = 0. Take a pure y not in (VU)*. Then considering the orthogonal group over E, we find dim IfTsO =dimV*---l=d--1. Hence K(T,0)=n--2dimV7V,== n - 2d + 2 = 2 # 0. So, applying (i) of the theorem, we have ~(T~G.)= n - (d - 1) = n - d + 1. Therefore Z(o) < n - d + 2. The facts that E(o) f az- d and + n - d + 1 are clear by the results of O,cFFv). Thus we have completed the proof for (ii) of the theorem.
REFERENCES I. J. ~ImJDoNN~, Sur les generateurs des groupes classiques, Swnma Brad. Math. 3 (395§), 149-179. 2. H. IsH~BAS~-II, On some system of generators of the orthogonal groups, Sci. I&p. Tokyo Kyoiku Duigaku, Sect. A 11, No. 287 (1972), 24-33. 3. H. ISHIBASHI, The decomposition of isometries into symmetries in an orthogonal group ever a local principal ideal domain, Sci. Rep. Tokyo Kyoiku Daigaku Sect. A 12, No. 321 (1972), 49-58. 4. W. KLINGENBERG, Orthogonal gruppen iiber lokaien ringen, Amw. 1. Math. 83 (1961), 261-320. to Quadratic Forms,” Springer-Verlag, Berlin/ 5. 0. T. O'MEARA, “Introduction Giittingen/Heidelberg, 1963. 6. P. SCHERK, On the decomposition of orthogonalities into symmetries, PYOC. Amer. Math. Sot. 1 (1950), 481-491.