bio | website | maths.lancs.ac.uk/~kania |
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location | Toronto, Canada | |
age | ||
visits | member for | 2 years, 11 months |
seen | 10 hours ago | |
stats | profile views | 1,479 |
Apr 14 |
comment |
Do all $L^{\infty}(\mu)$ spaces have the Grothendieck property?
user46855, you should post that as the answer. |
Apr 14 |
comment |
Decomposable Banach Spaces
Koszmider proved that consistently you can have an indecomposable $C(K)$-space of density $2^{\mathfrak{c}}=\aleph_2$. There are however certain limitations of the proof so that it is unlikely that it can be generalised to higher cardinals. On large indecomposable Banach spaces; J. Funct. Anal. 264 (2013), no. 8, 1779–1805 |
Apr 13 |
awarded | Revival |
Mar 29 |
accepted | Introducing a dual space structure |
Mar 29 |
asked | Introducing a dual space structure |
Mar 19 |
revised |
Banach spaces with no reflexive complemented subspaces
a more appropriate link |
Mar 15 |
revised |
Banach spaces with no reflexive complemented subspaces
banach-spaces tag added |
Mar 15 |
suggested | suggested edit on Banach spaces with no reflexive complemented subspaces |
Mar 15 |
comment |
Banach spaces with no reflexive complemented subspaces
The example with the Schreier space actually shows that a space failing DPP can have no reflexive subspaces at all. |
Mar 15 |
revised |
Banach spaces with no reflexive complemented subspaces
link |
Mar 15 |
answered | Banach spaces with no reflexive complemented subspaces |
Mar 15 |
comment |
Is the space of trace class operators finitely representable in an $L^1$-space?
On a tangential note, do we know what are the reflexive subspaces of $S_1$? At least do we know if they are super-reflexive? |
Mar 15 |
comment |
When is it $C(X)$?
To the best of my knowledge Dales et al. are going to publish (rather soon) a survey article on this and related problems. I will keep you posted. |
Mar 15 |
comment |
When is it $C(X)$?
Yemon, Haydon in his proof uses Pełczyński's decomposition method hence his result is only isomorphic not isometric, so the trick with uniqueness of preduals doesn't work. This situation is analogous to the fact that there exists a Banach-space between apparently very different Banach algebras $\ell_\infty$ and $L_\infty$. I am not sure if the OP is happy with Banach-space isomorphisms, though. |
Mar 12 |
revised |
Extension of bounded linear operators
typo |
Mar 12 |
revised |
Extension of bounded linear operators
examples |
Mar 12 |
answered | Extension of bounded linear operators |
Mar 10 |
revised |
States and extremal states of quantum SU(2) and the Podleś sphere
spelling of *Podleś* |
Mar 10 |
suggested | suggested edit on States and extremal states of quantum SU(2) and the Podleś sphere |
Mar 1 |
comment |
Noncommutative geometry and category theory
That is right. In the realm of C*-algebras it is however more natural to work with the so-called strong Morita equivalence introduced by Rieffel (see for instance his paper Morita equivalence for $C^\ast$-algebras and $W^\ast$-algebras). |