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P-Adic Clifford Algebras Annales Mathématiques Blaise Pascal, Tome 1, No 1 (1994), P

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P-Adic Clifford Algebras Annales Mathématiques Blaise Pascal, Tome 1, No 1 (1994), P ANNALES MATHÉMATIQUES BLAISE PASCAL BERTIN DIARRA p-adic Clifford algebras Annales mathématiques Blaise Pascal, tome 1, no 1 (1994), p. 85-103 <http://www.numdam.org/item?id=AMBP_1994__1_1_85_0> © Annales mathématiques Blaise Pascal, 1994, tous droits réservés. L’accès aux archives de la revue « Annales mathématiques Blaise Pascal » (http:// math.univ-bpclermont.fr/ambp/) implique l’accord avec les conditions générales d’utilisation (http://www.numdam.org/conditions). Toute utilisation commerciale ou impression systématique est constitutive d’une infraction pénale. Toute copie ou impression de ce fichier doit contenir la présente mention de copyright. Article numérisé dans le cadre du programme Numérisation de documents anciens mathématiques http://www.numdam.org/ Ann. Math. Blaise Pascal, Vol. I N°1, 1994, pp. 85-103 P-ADIC CLIFFORD ALGEBRAS Bertin DIARRA In a previous paper [2], we gave the. index of the standard quadratic form of rank n over the field of p-adic numbers. Here, we recover, as a consequence, the structure ,of the associated Clifford ’ algebra. " ,’, _ ’. ~~ The classification of all ( equivalence classes of) quadratic forms over a p-adic field is well known ( cf.[5]), with this classification, one is able to classify all p-adic Clifford algebras. I - INTRODUCTION . , Let K be a field of characteristic = 2 and E a vector space over K of finite dimension n. A mapping q : E --~ K is a quadratic f orm over E if there exists a bilinear symmetric ’ form f : E x E -; K such that . We assume that q is regular, that is f is non-degenerated. An element x E E is isotropic if = 0. Let Y be a subspace of E the orthogonal subspace of V is the set V-~- = {y E~ = 0 for all x E Y}. The subspace V is called totally isotropic if V C It is well known ( cf. for example that any totally isotropic subspace is contained in a maximal totally isotropic subspace. The maximal totally isotropic subspaces have the same dimension v, called the indez of q and 2v n. If 2v = n, then (E, q) is called a hyperbolic space and for the case n = 2, one says hyberbolic plane. The index v = 0 iff ~(a?) ~ 0 for a? ~ 0 i.e. ~E, q) is anisotropic. 86 Bertin DIARRA Let E = ~f" and B = (e~, , . , , en) be the canonical basis of E; the standard quadratic form qo is the quadratic form associated to the bilinear form n hence qg(x) = x,x > = £ x2. j=1 Let (E, q) be a quadratic space, possibly non regular ; an algebra C == C(E, q) over K, with unit 1, is said to be a Clifford algebra for (E, q) if (i) There exists a one-to-one linear mapping p : E -~ C such that p(x)~ = q(x) .1. (ii) For every algebra A with unit 1 and linear mapping ~ :: E -~ A satisfying ~(x)2 = q(x) ~ 1, there exists an algebra homomorphism ~ : C --> A such that ~ o p = ~. Clifford algebra exists and is unique up algebra isomorphism (cf. for instance [I] or For [3]). example,, let K X~, ~ ~ ,Xn > be the free algebra with free system of generators XI, ~ ~ ~ , .Xn and .I be the two-sided ideal of K Xl, ~ ~ ~ , Xn > generated by XX J + X’jX= - 2 f (e;, e~) . I,1 i, j r~, where (e~, . ~ ~ , en) is an orthogonal basis of then = (E, q); C(E, q) K X’z, ... ~ Xn > , II - THE P-ADIC STANDARD QUADRATIC FORM qo II -1 .. The index of qo Let p be a prime nurnber and Qp be the p-adic field i.e. the completion of the field of rational numbers Q for the p-adic absolute value. We denote by ~a~ the integral part of the real number a. Proposition 1 ~2~ n The standard quadratic form q4(x) = £ xa over E = C~p has indez j=1 P-Adic Clifford Algebras 87 Proof : 1°) If p ~ 1 (mod.4), it is well known that ~ = E Qp. Let u = 2 and E? = v _ = of i e2_1 + e2j, I ~ j v, then V is a totally isotropic subspace 2°) p = 3 (mod.4) Therefore i ~ Qp and if n = 2 the index of qo is 0. If n = 3 , applying Chevalley’s theorem and Newton’s method to qo(z) = x1 +x2+x3 we such that az bz 1 ~ 0. Therefore e~ = a ei + b e2 + e3 find a, b E ~p,, a ~ 0, b ~ 0, + + is isotropic in and v = ~ ~ == 1’ (a) For n = 4m, put = a e4J_~ + b e4~.~~ + and E2~ ’= -b e~~.:3 + a m. It is clear that = = b~ + I ~~= 0 and e4~_2 + , 1 j a2 ~- > = -ab+ab = 0. Therefore V = ~ is a totally isotropic ~=1.., = = subspace of t~p and v 2m r~ ’. If n = 4m + 1, with the same notations as above the subspace V is totally isotropic ’ ’ in ~p and v = 2m = ~ . m On the other hand if n = 4m + 3 the V = and subspaces ’ QpE2j) ;=1 where a e4m+1 -~- b e4~+~ + e4m+3, are totally isotropic and orthogonal. Therefore Vo = Qp E~m+1 is totally isotropic and v = 2m + 1 == 2 . m . = to (b) If n = 4m + 2, let V be as obove. It, is easy verify that if x e G~p is isotropic and x is orthogonal to V then x E V. Therefore Y is a maximal totally isotropic subspace, of C~p and v = 2m = r~ ~ 3.. .... 88 Bertin DIARRA Proposition 2 : Let p = 2. , 0 ~ ~ ~ 7. ~ ’ n The standard quadratic form = ~ x’ over ,~ ~ t~~ has index " .." ~..1 (i) II = 4m if 0_s_4 (ii) II = 4m + t if s=4~-t ,, 1t3 Proof : ~ ° ) If 1 n 4 , then the index of qo is 0. Indeed, this is clear when n = 1. If n = = = 2, let x + E ~Z be isotropic and different from 0 i.e. qo( x) xi + a?~ = 0 and say a?2 7~ 0. Therefore 1 + a~ = 0 with a = and v2(a) = 0 i.e. a == 1 = + , p.’ 1 .s 0. Then 1 + a2 = 2 + + 2~~‘a~ = 0 or 1 + + = 0 ; in other words 1 ~ 0 (mod.2) ; a contradiction. In the same way, one shows that if n = 3 or 4 , the index of qo is o. 2~) __...._ Let x~ = 2ei + e2 + e~ + e4 + e5 E G~~, then qo(xo) = 8 and = 2 ~ 0 (mod.4), ax~ 2j5 By Newton’s method there exists x = ~~l1 E G~~ such that = 0 with xx = 2 (mod.8) .~ 3 (mod.8), 2 ’5 Put a = b = c ~. = then a~ b~ c2 d~ 3 = 0. xix51,, x2x5 ~,, x~x51,, d x4x~ 1,, + -~ ~- -~ The two following elements of , = are isotropic with El, >= 2. Hence H is a hyperbolic plane in Q~.. Let = U be the orthogonal subspace of H in . The following three elements of Q~ :: P-Adic Clifford Algebras 89 are elements of ~ with Furthermore M,, ~ >= and (~1,~2? is a basis of !7. For every we have " " ,’.,. ’ c2~2~~~2_!~2.~21 2 a - and = 0 IS u = 0 because the standard form of c2 ’~ quadratic rank 4 is anisotropic. In other words is anisotropic and Q~ == is a Witt decomposition of (Q~, Hence the index of ~o is 1. 3~) n = 8m +s , 0 ~ ~ 4. Put, for and A straightforward computation shows that >= 0 = >, Q _ i, j rra -1 ;; 1 _, k, I 4 and >- 2 ;; U _ ~ m " 1’;; 1 t 4. Furthermore ~i,k~ >"~’ ~ if ~i, k~ ~ (~~ t~~ m-1 m-1 = = are with Hence the subspaces V ~ and W fl~ 1 isotropic ~=a ~~o Therefore H = is a hyperbolic subspace of E = with dim V = dim W = 4m. 90 Bertin DIARRA 8m ~ = But E = ( orthogonal sum ) where Em = ~ and ~ ~ k=l Q2. = with index If s = 0, we have E = Em = V ~ W ~l and (E, qo~ is a hyperbolic space 4m. If 1 s 4 ;; E = with ~m = V ~ W = H. Since 1 ~ dimE. = s ~ 4, the standard quadratic space is anisotropic. Consequently E = is a Witt decomposition of E and the index of qo is 4m. 4°) n = 8m 4. 4 -~t ,, 1 ~ ~ ~ 3. a) n == 8m + 5 5 With the same notations as above, we have E = where Es = k=1 ‘"~2.. Let us write , as for n = 5, and = The subspace Us = (~ of ~ is anisotropic. On the other hand, h~1 ~ = * >= 2 and are orthogonal to . Therefore E = vo == V ® and Wo = W ~ are isotropic subspaces of E and = = dim . is a Witt decomposotion of E. Hence the index of q~ is dim Vo W~ 4m+1. (b) n == 8 m+6. 6 = and As before, we have E = Em..LE6 where E6 hence , k=1 E6. P-Adic Clifford Algebras 91 Let us put and The subspace U6 = of E6 is anisotropic. Moreover, = >= 2 and are to Therefore 0 ~~‘E4m+2l ~; E4m+2~ E4m+2 ~m+~,, E~m.~2 orthogonal U6. Vi = Vo (B and Wi = are isotropic subspaces of E and E = is a Witt decomposition of E. Hence the index of qo is dim V’~ = dim 4~n + 2. (c) n= 8m 7. ’ 7 We have E = where E~ _ E6. k=1 - Let us write .. and. The subspace U7 - of E7 is anisotropic. Furthermore = 0 = = >= 2 and are to . Therefore * E4m+3~ ~4m+g E4m+3,, orthogonal U~. vZ and W2 = are isotropic subspaces of E and E = is a Witt decomposition of E. Hence the index of qo is dim V2 = dim W2 = 4rn + 3. Remark Let, K be a non formally real field. The level of K is the least integer s such that ~ .
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