Manifold of primitive idempotents in a Jordan-Hilbert algebra????????? 3? 11? 25????

Let V be a Jordan-Hilbert algebra (JH-algebra for short). By this we mean (1) V is a real Hilbert space with inner product h ¢ j ¢ i, (2) V is a real Jordan algebra, so that a bilinear product x; y 7! xy is de¯ned and one has for all x; y 2 V, (i) xy = yx; (ii) x2(xy) = x(x2y); (3) hxy j zi = hy jxzi for all x; y; z 2 V. Lemma. The Jordan product V £ V 3 (x; y) 7! xy 2 V is continuous. In what follows, we denote by L(x) the multiplication operator by x: L(x)y: = xy: Example. Let H be a real Hilbert space. We denote by Sym2(H) the set of all symmetric Hilbert-Schmidt operators on H. Then, Sym2(H) becomes a JH-algebra if we de¯ne a Jordan product A ¢B and an inner product hA jBi respectively by A ¢B: = 1 2 (AB +BA); hA jBi: = trace(AB): It should be noted here that if dim H = 1, then Sym2(H) does not have a unit element. Throughout this talk, the following basic assumptions are in force: Basic Assumptions. (i) dimV> 0, (ii) L(x) = 0 =) x = 0. The assumption (ii) is necessary in order to exclude the algebra in which xy = 0 for all x; y. 1 Proposition. V contains non-zero idempotents. Now, for every idempotent a 2 V, we have the Peirce decomposition of V relative to a: for k = 0, 1=2, 1, let us put Vk(a): = fx 2 V; ax = kxg, then (¤) V = V0(a) © V1=2(a) © V1(a) (orthogonal direct sum) with the multiplication rule: (here we write Vk instead of Vk(a) for simplicity)