Jesse Peterson
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Registered User
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May 1 |
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Metrics and completions on the direct limit of matrices of all sizes over arbitrary fields For $A \in M_n$, the two norm is $\| A \|_2 = (\frac{1}{n} {\rm Tr}(A^*A) )^{1/2}$. This is different than the uniform norm/maximum singular value $\| A \|$. Although, one has the inequality $\| A \|_2 \leq \| A \|$. One way to construct the hyperfinite II$_1$ factor is to complete each uniform ball in the two norm and then take the union. But if you don't restrict to the uniform balls then you will not get an algebra since multiplication is not jointly continuous in the two norm |
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Apr 30 |
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Metrics and completions on the direct limit of matrices of all sizes over arbitrary fields To construct the hyperfinite II$_1$ factor in this way you also need to use the uniform norm. You need to restrict to uniformly bounded subsets when you take the completion so that multiplication is continuous in the two norm. |
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Apr 28 |
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Weights on Von Neuman factors In either case, if $H = \mathbb C^2$, and $A = \mathbb M_2(\mathbb C)$. Then requiring $A$ to be isomorphic to $\mathbb C_{H_2} \otimes B(H_1)$ means that $H_1$ is two dimensional. Since $H$ is isomorphic to $H_2 \otimes H_1$ you then know that $H_2$ is one dimensional. Thus $P_{H_2} \not= 1$. |
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Apr 28 |
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Weights on Von Neuman factors Perhaps I am still confused about your question. When you ask for $H$ to be isomorphic to $H_2 \otimes H_2$ such that $A$ is isomorphic to $\mathbb C_{H_2} \otimes B(H_1)$, am I correct that you mean for the second isomorphism to be implemented by the first? |
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Apr 25 |
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Weights on Von Neuman factors In that case a simple counter example is to take $A = B(H)$. If the dimension of $H$ is at least 2 then the right hand side of the formula cannot be faithful. Just consider $T = 1 - P_{H_2}$. |
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Apr 25 |
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Weights on Von Neuman factors It's a bit unclear to me how you are making sense of the formula. Does $P_{H_2}$ denote the orthogonal projection onto $H_2$? If so, what is the domain, and where does $T$ live? |
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Mar 27 |
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Stone–Čech Compactification of $\mathbb{Z}$ with Fürstenberg Topology I'm just joining the conversation here and I must say that I am very confused since it appears that the notation $\beta \mathbb Z$ is being used by some to denote the Stone-Cech compactification of $\mathbb Z$ with the discrete topology, and by others to denote the Stone-Cech compactification with respect to the topology $\mathcal T$. Perhaps writing $\beta (\mathbb Z, \mathcal T)$ to denote the latter would help? |
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Mar 21 |
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Which von Neumann algebras have inner permutation of tensor factors? Fixed typo. |
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Mar 21 |
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Which von Neumann algebras have inner permutation of tensor factors? Added reference and outline of Sakai's paper |
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Mar 19 |
awarded | ● Nice Question |
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Feb 21 |
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Not measure equivalent ICC groups $G$ and $H$, but $L(G)\cong L(H)$ This problem is still open as far as I know. |
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Feb 13 |
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Compact subgroups of the unitary group of operators in a hilbert space @Andras Batkai: Actually, the unitaries with the weak topology do not form a compact group. In fact, the weak and strong topologies give the same relative topology on the space of unitaries. |
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Dec 15 |
awarded | ● Nice Answer |
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Dec 10 |
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Which von Neumann algebras have inner permutation of tensor factors? In fact, many of the examples of ${\rm II}_1$ factors without ourter automorphisms can also be shown to be prime, i.e., they are not isomorphic to any tensor product of other ${\rm II}_1$ factors. |
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Dec 10 |
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Which von Neumann algebras have inner permutation of tensor factors? For the hyperfinte ${\rm II}_1$ factor the flip is approximately inner which can be seen by restricting to finite dimensional subalgebras, but the unitaries you get won't converge. |
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Dec 10 |
answered | Which von Neumann algebras have inner permutation of tensor factors? |
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Nov 29 |
awarded | ● Nice Answer |
