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
The condition for generalizing invertible subspaces in Clifford algebras 89 We shall need the following lemmas. Lemma 1. (see Lemma 1 [3]) If L(e A l , e A 2,..., e A k), where e A i G E, e A t ^ e A j for i ^ j, i,j G {1, 2,..., k], is invertible if and only if L(e A le A k, e A 2e A k ,. .., e A k e A k ) is invertible. By Lemma 1 we shall study subspaces of A in the form L(e 0, e A l,..., e A l). Lemma 2. (see Lemma 3 [3]) L(eo, e A l,..., e A l), e A i G E, e A i / e A . for i ^ j, is invertible if and only if e A ie A j + e A je A i = 0 for i / j , i,j G {0,1, 2,..., where e A o = e 0. Lemma 3. (see Theorem [3]) // L(eo, e A l , e A 2,..., e A l), e A t G E,e A i / e A j for i zfi j ) ij £ {1,2,... ./} is invertible, then (i) / < m + 1. (ii) If I = m + 1, either e A l = e A l ... e A,_ 1 or e A l = -e A l e A., ... e A l_ 1 . The purpose of this paper is to prove the following. Theorem. L(eo, e A l,. . ., e A m , e A m, x, • • •, e Am+ s ) is invertible if and, only if the following conditions simultaneously hold: (1) e A ie A j + e A jé A i = 0 for i ± j i,j G {0,1,2,...m}, where e A o = e 0, (2) rn = 2 (mod 4), (3) S = 1, (4) Either e Am+ l = e A le A 2 .. .e A m or e Am+ 1 = -e A le A 2 . . .e A m . Proof. First we prove the sufficiency. From e A ie A j +e A je A t = 0 for i. ^ j, i,j G {0,1,... ?7? } we have éA,é A j + e A ]e A i = 0 and e A i + e A i = 0 for i / j, i,j G {1,...m}. We shall prove that e A ke Ar n+ l + e Am+ 1e A k = 0 for k G {0,1,... m}. For k = 0, by ab = bei and by m = 2 (mod 4), we get that eoé Am+ 1 + e,i m+ 1e 0 = e A le A., . . . e A m + e A le A 2 .. .e Aj n = e A me A m_ l .. .e A l + e A le A :, .. .e Am = (-1 ) me A me A m_ 1 .. . e A l + e A le A,, . . .e A m
