bio | website | mathnet.ru/php/… |
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visits | member for | 2 years, 8 months |
seen | 9 hours ago | |
stats | profile views | 1,397 |
Mar 21 |
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A variant of the Stone-Weierstrass theorem?
Of course, I am speaking about unital algebras (excuse me for not clarifying this from the very beginning). Douglas, how is this connected with what M.Dupre and R.Gillette write in "Banach bundles, Banach modules and automorphisms of C*-algebras"? |
Mar 21 |
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A variant of the Stone-Weierstrass theorem?
Ulrich, the Yemon Choi reference led me to a book by M.Dupre and R.Gillette "Banach bundles, Banach modules and automorphisms of C*-algebras". As far as I understand, from their Theorem 2.4 (at p.40, see also the discussion at pp.38-39) it follows that if C is a closed subalgebra in the center of a (unital) C*-algebra A, then A is isomorphic to the algebra of sections of a C*-algebra bundle over the spectrum T of C. I didn't understand, whether your words contradict to what they write... Just in case: everywhere I speak about unital algebras, of course. |
Dec 22 |
accepted | A variant of the Stone-Weierstrass theorem? |
Dec 22 |
comment |
A variant of the Stone-Weierstrass theorem?
Excuse me, I actually meant one question: 1) and 2) are just two conditions of one statement (I have now put "and" between them). But I would be satisfied if B is replaced by a C*-algebra bundle. |
Dec 22 |
revised |
A variant of the Stone-Weierstrass theorem?
added 4 characters in body |
Dec 21 |
asked | A variant of the Stone-Weierstrass theorem? |
Dec 4 |
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Algebraically simple Banach algebras
Besides this, do you mean algebras with identity? |
Dec 4 |
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Algebraically simple Banach algebras
Yes, the ideal $F(H)$ of all finite-dimensional operators in $K(H)$ is non-trivial. But as far as I understand, if you don't claim that $J$ is closed, then $K(H)$ is not a counterexample for you. I.e. $K(H)$ is not algebraically simple... |
Dec 4 |
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Algebraically simple Banach algebras
Will the algebra $K(H)$ of compact operators on a Hilbet space satisfy you? It has not non-trivial (closed) two-sided ideals. |
Nov 30 |
awarded | Enthusiast |
Nov 24 |
awarded | Supporter |
Nov 23 |
comment |
Weakened notion of extremal epimorphism?
Сердечное спасибо! (Thank you very much, по-ихнему.) |
Nov 23 |
accepted | Weakened notion of extremal epimorphism? |
Nov 22 |
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Complement of a subspace which is a cartesian product
Does $\oplus$ mean orthogonal sum of Hilbert spaces, or just direct sum of topological vector spaces? |
Nov 22 |
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Weakened notion of extremal epimorphism?
Of course, this must be the same in some categories. For example, in Abelian categories, or, more generally, in categories where every morphism $f$ can be represented as a composition $f=m\circ g\circ e$, with $m$ a strong monomorphism, $e$ a strong epimorphism, and $g$ a bimorphism (see details in arxiv.org/abs/1110.2013, Theorem 1.3; unfortunately, in Russian). But I do not know counterexamples at all. Perhaps, these definitions are equivalent in any category? |
Nov 22 |
revised |
Weakened notion of extremal epimorphism?
edited body |
Nov 22 |
asked | Weakened notion of extremal epimorphism? |
Nov 18 |
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Is the Mendeleev table explained in quantum mechanics?
"Any textbook will do" - this sends us back to the begining of my post. Antoine, you can just try to find a book on quantum mechanics, where there is 1) a system of axioms (or postulates), 2) a phrase like "by atom we mean..." (i.e., a definition of atom), and 3) a Proposition or a Theorem about atoms. Warning: in the Berezin-Shubin book they mention atoms in Proposition 2.2 (chapter 4), but these are atoms in the sense of measure theory, they have nothing in common with physical atoms. |
Nov 18 |
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Is the Mendeleev table explained in quantum mechanics?
I doubt that mathematical physicists will agree with this: "Physics is not a subset of mathematics, and cannot be understood as such". But anyway if you say that everything is so simple, then there must be a reference. What you say (your definition, proposition) - where is this written? |
Nov 18 |
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How to find a problem ???
@ colleagues: The idea of asking advisor, of course, couldn't occur without your help. Instead of whooping you could suggest something useful. @ kakalotte: My advice is that you should widen your outlook, and when doing this you will find many interesting questions which have no answers by now. |