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Владикавказский математический журнал 2008, Том 10, Выпуск 3, С. 29-33
УДК 517.98
ON SOME PROPERTIES OF ORTHOSYMMETRIC BILINEAR OPERATORS
A. G. Kusraev
This note contains some properties of positive orthosymmetric bilinear operators on vector lattices which are well known for almost /-algebra multiplication but despite of their simplicity does not seem appeared in the literature.
Mathematics Subject Classification (2000): 46A40, 47A65.
Key words: vector lattice, square of a vector lattice, bilinear operator, orthosymmetry, lattice bimorphism, /-algebra multiplication.
The aim of this note is to present some properties of orthosymmetric bilinear operators which are well known for f -algebra multiplication but despite of their simplicity does not seem appeared in the literature. All unexplained terms can be found in [1] or [13]. All vector lattices under consideration are assumed to be Archimedean.
1. A bilinear operator b : E x E ^ G is called orthosymmetric if x Л y = 0 implies b(x, y) = 0 for all x, y £ E. This definition was introduced in [8]. Recall also that b is said to be symmetric if b(x, y) = b(y, x) for all x, y £ E and positively semidefinite if b(x, x) ^ 0 for every x £ E .In the special case that b is the multiplication of a commutative almost f-algebra the following proposition is presented in [2, Proposition 1.13].
Proposition 1. Let F and G be vector lattices. A positive bilinear operator b from E x E to G is orthosymmetric if and only if b(x, y) = b(x V y, x Л y) for all x, y £ E.
< Orthosymmetry implies b(x — x Л y, y — x Л y) =0. Since a positive orthosymmetric bilinear operastor is symmetric (see [8]), we deduce
b(x, y) = b(x, x Л y) + b(x Л y, y) — b(x Л y, x Л y)
= b(x + y — x Л y,x Л y) = b(x V y, x Л y).
Conversely, if x Л y = 0, then b(x, y) = b(x V y, 0) = 0. >
A bilinear operator b : E x F ^ G is said to be lattice bimorphism if the mappings y ^ b(e, y) (y £ F) and x ^ b(x, f) (x £ E) are lattice homomorphisms for all 0 ^ e £ E and 0 ^ f £ F, see [11]. Evidently, every lattice bimorphism is positive. The following characterization of lattice bimorphism was given in [15].
Proposition 2. For a positive bilinear operator b : E x E ^ G the following assertions are equivalent:
(1) b is a lattice bimorphism;
(2) |b(x, y)| = b(|x|, |y|) for all x £ E and y £ F;
© 2008 Kusraev A. G.
(3) if 0 ^ x, u £ E and 0 ^ y, v £ F satisfy xAu = 0 and y Av = 0, then b(x, y) Ab(u, v) = 0.
2. It was mentioned in [12] that an orthosymmetric positive bilinear operator is positively semidefinite. The converse is also true for lattice bimorphisms as was observed in [7, Proposition 1.7]. The following characterization of symmetric lattice bimorphisms is well known at least for d-algebra multiplication (see, for example, [2, Theorems 4.3, 4.4, 4.5] and [4, Proposition 3.6]).
Theorem 1. Let E and F be vector lattices and let b : E xE ^ F be a lattice bimorphism. Then the following assertions are equivalent:
(1) b is symmetric;
(2) b(x, x) — b(y, y) = b(x — y, x + y) for all x, y £ E;
(3) b(x, x) A b(y, y) ^ b(x, y) ^ b(x, x) V b(y, y) for all x, y £ E+;
(4) b(x A y, x A y) = b(x, x) A b(y, y) and b(x V y, x V y) = b(x, x) V b(y, y) for all x, y £ E+;
(5) x A y = 0 implies b(x, y) = b(y, x) for all x, y £ E;
(6) b(x, |x|) = b(x+, x+) — b(x-, x-) for all x £ E;
(7) b is orthosymmetric;
(8) b is positively semidefinite.
< (1) ^ (2): It is obviously true for every bilinear operator b.
(2) ^ (3): For any x, y £ E+ we deduce making use of (2):
b(x, x) Ab(y, y) — b(x, y) ^ b(x, x) A b(y, y) — b(x A y, x A y)
=[b(x, x) — b(x A y, x A y)] A [b(y, y) — b(x A y, x A y)]
=b(x — x A y, x + x A y) A b(y — x A y, y + x A y)
^b(x — x A y, x + y) A b(y — x A y, x + y)
=b((x — x A y) A (y — x A y), x + y) = 0.
The second inequality is deduced likewise.
(3) ^ (4): Using the first inequality in (3) we can write the following chain of equalities:
b(x, x) A b(y, y) = [b(x, x) A b(x, y)] A [b(y, x) A b(y, y)]
= b(x, x A y) A b(y, x A y) = b(x A y, x A y).
The second equality is deduced likewise.
(4) ^ (5): Take x, y £ E with x A y = 0. By the first equality of (3) b(x, x) and b(y, y) are
disjoint. Using the second equality we have b(x,x) + b(y, y) = b(x V y,x V y) = b(x + y,x + y) = b(x,x) + b(x, y) + b(y, x) + b(y, y), so that b(x, y) = b(y,x) = 0.
(5) ^ (6): It is sufficient to observe that b(x, |x|) — b(x+,x+) + b(x-,x-) = b(x+,x-) —
b(x-, x+).
(6) ^ (7): If b obey (6), then b(x+,x-) and b(x-,x+) coincide, see (5) ^ (6). At the same time these elements are disjoint, since b(x+ ,x-) ^ b(x+, |x|), b(x-,x+) ^ b(x-, |x|) and b(x+, |x|) A b(x-, |x|) = 0. Thus, b(x+,x-) = b(x-,x+) = 0, from which (7) follows
(7) ^ (1): Follows from [8, Corollary 2].
(7) ^ (8): If b is ortosymmetric, then b(x, x) = b(x+,x+) — b(x+,x-) — b(x-,x+) +
b(x-,x-) = b(x+,x+) + b(x-,x-) ^ 0, see [12].
(8) ^ (7): Let b be a positively semidefinite lattice bimorphism. Take x, y £ E and put a := b(x, x), P := b(y, y), 7 := b(x, y) + b(y, x). Then a + ft — 7 = b(x — y, x — y) ^ 0. If x A y = 0, then b(x, y) ^ b(x, y) A b(y, y) = b(x A y, y) =0 and, since b(x, ■) and b(-, x) are lattice homomorphisms, we have a A b(x, y) = b(x, x A y) =0 and a A b(y, x) = b(x A y, x) = 0.
Thus, a ± y and analogously 3 ± 7. Therefore, (a + 3) ± 7, and taking into account the inequality a + 3 — 7 ^ 0 we derive 7 = 0, i.e. b(x, y) = b(y, x) = 0. >
3. Let E be a vector lattice. A pair (E0,0) is said to be a square of E if the following two conditions are fulfilled:
(1) E0 is a vector lattice and 0 is a symmetric lattice bimorphism from E x E to E0,
(2) if b is a symmetric lattice bimorphism from E x E to some vector lattice F, then there exists a unique lattice homomorphism $5 : E0 ^ F with b = $60.
For an arbitrary vector lattice E there exists the square (E0, 0) which is essentially unique, i. e. if some pair (E®, @) obeys (1) and (2) above, then there exists a lattice isomorphism i from E0 onto E® such that i0 = @ (and, of course, i-1@ = 0), see [10]. Moreover (see [10] and [7, Theorem 3.1]), for every positive bilinear orthoregular operator b : E x E ^ G there exists a unique linear regular operator $5 : E0 ^ G such that
b(x, y) = $b(x 0 y) (x,y £ E).
The symmetric lattice bimorphism 0 : E x E ^ E0 is called the canonical bimorphism of the square. The operator $5 is called the linearization of b via square. If E is a sublattice of a semiprime /-algebra A, then the canonical bimorphism 0 can be expressed in terms of the algebra multiplication, see [7, Proposition 2.5].
Proposition 3. Let A be a semiprime /-algebra with a multiplication • and E be a sublattice of A. Then there exists a sublattice F C A and an isomorphism 1 from E0 onto F such that i(x 0 y) = x • y for all x, y £ E. In other words, the pair (F, •) is a square of E.
4. A vector lattice E is called square-mean closed if the set {(cos 9)x + (sin 9)y : 0 ^ 9 < 2n} has a supremum s(x, y) in E for all x, y £ E .A vector lattice E is called geometric-mean closed if the set {(t/2)x + (1/2t)y : 0 < t < +<^} has an infimum g(x, y) in E for all x, y £ E+. The following result see in [5, Theorems 3.1 and 3.4].
Proposition 4. If A is a square-mean closed Archimedean /-algebra, then
s(x,y)2 = x2 + y2 (x, y £ A).
If A is a geometric-mean closed Archimedean /-algebra, then
g(x,y)2 = xy (x, y £ A+).
Every relatively uniformly complete vector lattice is square-mean closed and geometric-mean closed [5, Theorems 3.3]. However, neither a square-mean closed nor a geometric-mean closed Archimedean vector lattice need not be uniformly complete. But a geometric-mean closed Archimedean f-algebra is square-mean closed [5, Theorem 3.6]. The following result is a generalization of Proposition 4.
Theorem 2. Let E and F be vector lattices and b : E x E ^ F a positive orthosymmetric bilinear operator. If E is square-mean closed, then
s(x, y) 0 s(x, y) = x 0 x + y 0 y, b(s(x,y),s(x,y)) = b(x,x) + b(y, y)
for all x, y £ E. If E is geometric-mean closed, then for all x, y £ E+ we have
fl(x,y) 0 g(x,y) = x 0 y, b(g(x,y), g(x,y)) = b(x, y).
< In each of two cases under consideration the second equality follows from the first one by applying $b, the linearization via square of b. Let A denotes the universal completion of E endowed with a semiprime /-algebra multiplication. Then by Proposition 3 there is a lattice isomorphism i of E0 onto a sublattice F C A. At the same time, according to Proposition 4, the following equalities are true in A:
s(x,y) • s(x,y)= x • x + y • y (x, y £ E), g(x,y) • g(x,y)= x • y (x,y £ E+).
Now, the first equalities are immediate by applying i-1, since s(x, y) £ E and g(x,y) £ E under the stated hypotheses and i-1(x • y) = x 0 y. >
5. In conclusion we present some corollaries to Theorem 2.
Corollary 1. Let E and F be vector lattices with E square-mean closed and b : E x E ^ F be a positive orthosymmetric bilinear operator. Then E+ := {b(x,x) : x £ E} is a convex pointed cone and E(5) := b(E x E) is a vector subspace of F ordered by a positive cone E+ such that E(5) = E+ — E^. If, in addition, b is a lattice bimorphism, then E(5) is a vector sublattice of F.
< The first part of Theorem 2 implies that E+ C F+ is a pointed cone. The equalities b(x, y) = (1/4)[b(x + y, x + y) — b(x — y, x — y)]) and b(x, x) — b(y, y) = b(x + y, x — y) show that E(5) = E+5) — E+5). Thus, (E(5), E+5)) is an ordered vector space. If b is a lattice bimorphism,
then E+5) is a sublattice of F+ in virtue of Theorem 1 (2). >
For an almost /-algebra multiplication this result was obtained in [4, Prposition 3.3, Corollary 3.7]. The first statement of the following corollary was proved in [9, Lemma 8] in case of uniformly complete E.
Corollary 2. Let E be a square-mean closed vector lattice. The the assertions hold:
(1) E0 = {x 0 y : x, y £ E} and E+0 = {x 0 x : x £ E};
(2) If F = h(E), then F0 = h0 (E0) for any vector lattice F and lattice homomorphism h : E ^ F;
(3) If J is a uniformly closed order ideal of E, then J0 := {x 0 y : |x| A |y| £ J} is a uniformly closed order ideal of E0 and the map x 0 y + J0 ^ (x + J) 0 (y + J) implements a lattice isomorphism of E0/J0 onto (E/J)0.
< (1) Put b:= 0 in Corollary 1 and observe that E0 = E( 5), since E0 coincides with the sublattice generated by b(E x E) = {x 0 y : x, y £ E}.
(2) If h : F ^ E is a lattice homomorphism then by [7, Proposition 2.4] there exists a lattice homomorphism h0 : F0 ^ E0 such that h0(x 0 y) = h(x) 0 h(y) (x, y £ F). Assume that T(E) = F. Then making use of by (1) we deduce
E0 = {h(x) 0 h(y) : x, y £ F} = {h0(x 0 y) : x, y £ F) C h0(F0) C E0.
(3): If 0 : E ^ E/J is a quotient homomorphism, then 00 is a surjective map from E0 to (E/J)0 by (2). According to (1) any u £ E0 have the representation u = x 0 y for some x,y £ E and 0 = 00(u) = 0(x) 0 0(y) implies 0(x) ± 0(y) by [7, Theorem 2.1 (3)]. But the latter is equivalent to |x| A |y| £ J, since 0 is a lattice homomorphism. Thus, J0 = ker(00) and the proof is complete. >
Corollary 3. Let E and F be vector lattices with E square-mean closed and let b : E x E ^ F be an order bounded orthosymmetric bilinear operator. Then for any finite collections x1, yi,..., xn, yN £ E there exist u, v £ E such that N=1 b(x&, y&) = b(u, v).
< According to Corollary 1(1) there exist u, v £ E such that u 0 v = ^^=1 x^ 0 y&. Now, if b = $ 0 for a linear operator $ from E0 to F, then
/ N \ N
b(u, v) = $b(u 0 v) = $5 I 0 yfc = ^ b(xfc,yk)
V ^=1 / ^=1
which is the desired representation. >
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Received June 5, 2008.
Anatoly G. Kusraev
Institute of Applied Mathematics and Informatics Vladikavkaz Science Center of the RAS Vladikavkaz, 362040, RUSSIA E-mail: [email protected]
