Tomek Kania
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Registered User
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5h |
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Why don’t more mathematicians improve Wikipedia articles? Hi Mark. It's nice to see people from our department here :-) |
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May 15 |
awarded | ● Yearling |
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May 8 |
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How long can it take to generate a $\sigma$-algebra? Do you mean Theorem 9.2 here: math.wisc.edu/~miller/res/dstfor.pdf ? Well, it does not say anything about $\sigma$-completeness. |
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Apr 19 |
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Can nuclearity be determined by tensoring with a single C*-algebra? Just out of curiosity, is there a known example of a non-nuclear $C*$-algebra $A$ with a unique C*-norm on $A\odot A$? |
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Apr 18 |
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isomophism, commutator I must admit I don't quite understand the question but perhaps this might be helpful to you. For $X$ being the James space or $X=C[0,\omega_1]$ there is a unique trace $\tau$ on $B(X)$. Hence $\tau$ is centre-valued and kills all the commutators. Obviously, the kernel of $\tau$ has codimension one in $B(X)$. Also, every Banach-algebra isomorphism $B(X)\to B(X)$ is implemented by an isomorphism $V$ on the Banach-space level. Thus, $\psi(T) = V^{-1}TV$. |
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Mar 14 |
answered | How identify bounded Borel measurable functions |
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Mar 13 |
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B(H) as a direct sum of a closed two sided ideal and a subalgebra @Bill, under Martin's axiom and failure of CH, we have $2^{\omega} = 2^{\omega_1}$, I am afraid. |
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Mar 12 |
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B(H) as a direct sum of a closed two sided ideal and a subalgebra @Bill, I agree that it requires some justification. To do this, one can tweak the argument for the fact that there is no 1-1 operator form $\ell_\infty(\omega_1)$ to $\ell_\infty^{\omega_1}(\kappa)$. I am sorry for just pointing a reference but today I have no time for going into more details. Of course, I am very curious about your proof! |
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Mar 12 |
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B(H) as a direct sum of a closed two sided ideal and a subalgebra modification |
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Mar 12 |
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B(H) as a direct sum of a closed two sided ideal and a subalgebra links |
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Mar 12 |
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B(H) as a direct sum of a closed two sided ideal and a subalgebra clarification |
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Mar 12 |
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B(H) as a direct sum of a closed two sided ideal and a subalgebra dr |
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Mar 12 |
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B(H) as a direct sum of a closed two sided ideal and a subalgebra presentation |
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Mar 12 |
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B(H) as a direct sum of a closed two sided ideal and a subalgebra Thank you, Bill. I hope my answer is now fairly complete. |
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Mar 12 |
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B(H) as a direct sum of a closed two sided ideal and a subalgebra +1 |
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Mar 12 |
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B(H) as a direct sum of a closed two sided ideal and a subalgebra type |
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Mar 12 |
answered | B(H) as a direct sum of a closed two sided ideal and a subalgebra |
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Mar 11 |
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zero-dimensional completely regular space with $\sigma$-complete clopen algebra Have you looked on M.H. Stone, Boundedness properties in function-lattices, Canadian Math. Soc. 1 (1949), 176-186? cms.math.ca/cjm/v1/cjm1949v01.0176-0186.pdf |
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Mar 9 |
awarded | ● Nice Answer |
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Mar 7 |
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Is there an editors checklist for mathematics? typo |
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Mar 7 |
answered | Is there an editors checklist for mathematics? |
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Jan 4 |
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Existence of model of ZF without AC, but with many choice function Thanks for this. |
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Jan 4 |
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Nuclear Space problem @Koushnik, these facts are probably due to Dixmier; they employ the Radon-Nikodym theorem in a very clever way. A good source of knowledge about $C(K)^{**}$ is the paper of H. G. Dales, A. T.-M. Lau and D. Strauss, Second duals of measure algebras, Dissertationes Math. 481 (2012). Of course, there are more C*-algebraic ways to prove nuclearity of $C(X)$, though this one is my favourite. |
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Jan 4 |
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Existence of model of ZF without AC, but with many choice function Just out of curiosity, assuming $ZF+AD$ is consistent, is $ZF+AD$ consistent with "there exists a well-orderable uncountable subset of $[0,1]$? |
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Jan 4 |
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Nuclear Space problem He is talking about en.wikipedia.org/wiki/Nuclear_C*-algebra |
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Jan 4 |
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Nuclear Space problem For $X$ compact and Hausdorff, $C(X)^{**}$ is *-isomorphic to some algebra of the form $L_\infty(\mu)$, which as a Banach space, is 1-complemented in any superspace. In particular, it is 1-complemented in the algebra $\mathscr{B}(L_2(\mu))$ (it is identified with the subalgebra of multiplication operators by elements from $L_\infty(\mu)$), hence it is an injective von Neumann algebra. This implies that $C(X)$ is nuclear. |
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Dec 28 |
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Extending complete filters So, might it happen that "there are no strongly comapct cardinals" is a theorem of ZFC? |
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Dec 28 |
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Extending complete filters Do we know whether Con(ZFC) implies Con(ZFC + there is a strongly compact cardinal)? |
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Dec 28 |
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Extending complete filters Actually, this counts as an answer since the filter I am dealing with is rather mysterious and, consequently, I have no chance for a "painless" extension, by the above. |
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Dec 28 |
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Extending complete filters +1 |
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Dec 28 |
asked | Extending complete filters |
