1978
DOI: 10.1090/s0002-9947-1978-0515547-x
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A characterization and sum decomposition for operator ideals

Abstract: Abstract.Let L(H) be the ring of bounded operators on a separable Hubert space. Assuming the continuum hypothesis, we prove that in L(H) every two-sided ideal that contains an operator of infinite rank is the sum of two smaller two-sided ideals. The proof involves a new combinatorial description of ideals of L(H). This description is also used to deduce some related results about decompositions of ideals. Finally, we discuss the possibility of proving our main theorem under weaker assumptions than the continuu… Show more

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Cited by 11 publications
(5 citation statements)
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References 10 publications
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“…and this last set is forced to be a subset of k: by ps, hence also by the extension (r, s). Since x = AcG"), we have verified requirement (2). It remains to check (5): r IF "For every n E o, there is an extension t of s in (Po/Gk)/H forcing 'B* n n .x(m) ".…”
Section: B*n(k+l)~b*nmg{i<m(q(i)=l}mentioning
confidence: 87%
See 3 more Smart Citations
“…and this last set is forced to be a subset of k: by ps, hence also by the extension (r, s). Since x = AcG"), we have verified requirement (2). It remains to check (5): r IF "For every n E o, there is an extension t of s in (Po/Gk)/H forcing 'B* n n .x(m) ".…”
Section: B*n(k+l)~b*nmg{i<m(q(i)=l}mentioning
confidence: 87%
“…Indeed, (1) to ( 3) are trivial for k = 0 (as PO is trivial), (4) says p E N which is true by our choice of N, and ( 5) is exactly fact 4.5 above. Third, (1) implies that, for each k, pn 1 k is independent of n once n 2 k. Since P, is an inverse limit, we can define 4 E p, bY then 4 extends every pk, in particular p, and it forces "B* s A" by (2). Thus, CJ will be as desired, so the proof will be complete once we construct the pk's.…”
Section: ) Pk R K If "Pk 1 [K O) E N[gk]" and (5) Pk 1 K It-"for Every N E O There Is An Extension T Of Pk 1 [K O) In P/g Forcing 'B* N Nmentioning
confidence: 89%
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“…As a ( ) is am-closed, this yields: Further consequences of Theorem 10 are obtained by exploiting lattice properties of ideals, in particular, of some classes of principal ideals. Blass and Weiss (19) proved that K(H) is the sum of two proper ideals (neither of which can be countably generated) and in general, every ideal that properly contains F is the sum of two proper ideals. Here we obtain that with respect to the inclusion order, the lattice of principal ideals has no ''gaps'', that is, between any two principal ideals lies another one.…”
Section: Dimension Of I͞[i B(h)]mentioning
confidence: 99%