# Jesse Peterson

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 Name Jesse Peterson Member for 2 years Seen 13 hours ago Website Location Vanderbilt University Age
 May1 comment Metrics and completions on the direct limit of matrices of all sizes over arbitrary fieldsFor $A \in M_n$, 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 Apr30 comment Metrics and completions on the direct limit of matrices of all sizes over arbitrary fieldsTo 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. Apr28 comment Weights on Von Neuman factorsIn 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 $P_{H_2} \not= 1$. Apr28 comment Weights on Von Neuman factorsPerhaps 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? Apr25 comment Weights on Von Neuman factorsIn 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 $T = 1 - P_{H_2}$. Apr25 comment Weights on Von Neuman factorsIt'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? Mar27 comment Stone–Čech Compactification of $\mathbb{Z}$ with Fürstenberg TopologyI'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? Mar21 revised Which von Neumann algebras have inner permutation of tensor factors?Fixed typo. Mar21 revised Which von Neumann algebras have inner permutation of tensor factors?Added reference and outline of Sakai's paper Mar19 awarded ● Nice Question Feb21 comment Not measure equivalent ICC groups $G$ and $H$, but $L(G)\cong L(H)$ This problem is still open as far as I know. Feb13 comment 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. Dec15 awarded ● Nice Answer Dec10 comment 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. Dec10 comment 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. Dec10 answered Which von Neumann algebras have inner permutation of tensor factors? Nov29 awarded ● Nice Answer