2003
DOI: 10.1090/s0002-9947-03-03152-0
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Twisted sums with đ¶(đŸ) spaces

Abstract: Abstract. If X is a separable Banach space, we consider the existence of non-trivial twisted sums 0For the case K = [0, 1] we show that there exists a twisted sum whose quotient map is strictly singular if and only if X contains no copy of 1 . If K = ω ω we prove an analogue of a theorem of Johnson and Zippin (for K = [0, 1]) by showing that all such twisted sums are trivial if X is the dual of a space with summable Szlenk index (e.g., X could be Tsirelson's space); a converse is established under the assumpti… Show more

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Cited by 36 publications
(18 citation statements)
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“…It therefore follows that K-splitting and K-lifting are not equivalent notions. It is not hard to see that K-lifting and W-lifting are also non-equivalent, even in the presence of the BAP: indeed, it is shown in [30] (see also [9,Thm. 2.3.…”
Section: Proposition 1 a Separable Banach Space Z Has The Bap If And ...mentioning
confidence: 99%
See 1 more Smart Citation
“…It therefore follows that K-splitting and K-lifting are not equivalent notions. It is not hard to see that K-lifting and W-lifting are also non-equivalent, even in the presence of the BAP: indeed, it is shown in [30] (see also [9,Thm. 2.3.…”
Section: Proposition 1 a Separable Banach Space Z Has The Bap If And ...mentioning
confidence: 99%
“…If, moreover, Z has the BAP, they are also equivalent to (8) The sequence P K -lifts. (9) The sequence H K -lifts.…”
Section: Extension and Lifting Of Polynomials And Holomorphic Mappingsmentioning
confidence: 99%
“…The problem here is that no current method is known to obtain a twisted sum 1 ⊕ ℩ c 0 , apart from its existence. Indeed, those obtained in [5,6,7] are actually non-constructive. Moreover, as we have already said, no current method is known to decide when two twisted sum…”
Section: Subspaces Of 1 In Different Non-isomorphic Positionsmentioning
confidence: 99%
“…In [CCKY,Theorem 4.1], the following crucial characterization of the spaces X for which Ext(X, C(ω ω )) = 0 is presented:…”
Section: While Theorem 43 Becomesmentioning
confidence: 99%
“…The parameter π n (X), introduced in [CCKY,Section 3], is easily checked to be π n (X) = z(X, C(ω n )). Now, since c 0 (C(ω n )) is isomorphic to a hyperplane of C(ω ω ) one has z(X, C(ω ω )) ≀ 2 sup n z(X, C(ω n )).…”
Section: While Theorem 43 Becomesmentioning
confidence: 99%