Az Eszterházy Károly Tanárképző Főiskola Tudományos Közleményei. 2001. Sectio Mathematicae (Acta Academiae Paedagogicae Agriensis : Nova series ; Tom. 28)
NGUYEN CANH LUONG, The condition for generalizing invertible subspaces in Clifford algebras
90 Nguyen Canli Luong m (m — 1) = (-l) m(-l) = e A le A 3 .. .e Ai n + e A le A 2 .. .e Am = -e A le A 2 . . . e A m + e A le A 2 .. .e A m = 0. For k £ {1,2, ... , m} we have e A ké Am+ 1 + e Am+ 1e A k = e A ké Aj n .. .e A k . . .e A l + e A l .. .e A k .. .e Aj ne A k = {-l) m~ ké An i . ..e A ke A k . ..e A l + (-l) m~ ke A l . ..e A ke A k . ..e Am = (-l) m-^[(-l) m1e Ar a • . •e Ak+ 1e A k_ 1 ...e A l + e A l •. .e A k_ 1e Ak+ 1 ...e A j = (-l) mk [ - (-1) ( m" 1 ) 3 ( ma ) e A l ... e A k_ le Ak+ 1 ...e An • • .e^u^e,!,^ . ..e A m] = 0. TO+1 Take 0 ^ a = a 0e 0 + £ ciie A j £ L(e 0, e A l,. .. e A m,e Am+ 1 ). i = 1 Let a — 1 ( m+ 1 \ 1 = Jä\2 ( a° e° + E ai eA t\- Tlle n ^ / TO+l \ / m+1 \ 1 - ( ' a° e° + E ai eAr\ ( '°0e 0 + E J a • a — m+1 771 +1 771 + 1 aleo + a 0 [ ^ ai eA, + ai e Ai | + E al eA, eA, = i j= 1 i = 1 + y^aiaj(e A ie A j + e A je Ai i<j ^ /m + 1 E \ i = 0 a- e 0 = e 0. Similarly, one can check the equality a 1 • a = eoNow we prove the necessity. By Lemma 2 we have e A te A i + e A je A i = 0 for / ii i, J € {0,1,..., m] and by Lemma 3 we get that s = 1 and either e Am+ 1 = e A le A. 2 .. .e A m or e Am+ 1 = -e A le A s .. .e A We shall prove that m = 2 (mod 4). From e A ie A j + e A je A i = 0 for i ^ j; i, j G {0,1,..., 77?} one gets
