# Jiang

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## Registered User

 Name Jiang Member for 2 years Seen May 9 at 2:34 Website Location Buffalo, NY Age
I am a first year gradute student in University at Buffalo, the State University of New York. I am fond of von Neumann algebras, especially $\mathrm{II}_1$ factors, ergodic theory, etc.
 Apr15 comment clarify a question in group cohomology@Misha, thanks! Apr12 asked clarify a question in group cohomology Apr5 asked Find a lower bound for a pre-invariant $Fol(L(F_m), X_m)$ Feb21 comment Not measure equivalent ICC groups $G$ and $H$, but $L(G)\cong L(H)$ We can further ask a weak version of the big one: Does $L(G)\cong L(H)$ and $G$ satisfies property $(T)$ imply that $G, H$ are measure equivalent? We can also replace the property $(T)$ by any other known invariant properties under Measure equivalent. Still, much work to be done on both sides, ME and von Neumann algebra equivalence. Feb21 comment Not measure equivalent ICC groups $G$ and $H$, but $L(G)\cong L(H)$ If this type question is sensitive to the separable argument, then a naive approach would be to find ICC groups $\{G_{\alpha}\}_{\alpha\in A}$, where $A$ is a uncountable set, and require $\{L(G_{\alpha})\}$ has at most countable different equivalent classes (maybe we can achieve this by considering the invariant associated to the $L(G_{\alpha})$, dimension, etc.) and at the same time require $G_{\alpha}$ has uncountable many different classes module the ME relation. Of course, these $\{G_{\alpha}\}$ should not have property $(T)$, and it may be not practical. Feb19 comment Not measure equivalent ICC groups $G$ and $H$, but $L(G)\cong L(H)$ sorry, I do not quite understand the meaning you say "so that L(G) has some restrictions on the fundamental group". The problem asked for constructing $G, H$ such that $L(G)\cong L(H)$ and $G, H$ are not measure equivalent or show such $G, H$ do not exist. Is the expression misleading or am I miss some point? Feb19 revised Not measure equivalent ICC groups $G$ and $H$, but $L(G)\cong L(H)$ fix some misleading expression? Feb19 asked Not measure equivalent ICC groups $G$ and $H$, but $L(G)\cong L(H)$ Jan10 comment Kadison-Singer problem in exotic Hilbert spacesMaybe the following links are helpful on this problem: aimath.org/WWN/kadisonsinger math.missouri.edu/~pete/index.html zentralblatt-math.org/portal/en/zmath/search/… Dec2 comment Fuglede-Kadison determinants in $L(\mathbb{F}_2)$@Andreas, the result of the spectral measure in the above paper looks complicated, it seems unclear whether we could get a similar result as Li's in this case. I guess one reason is that we could not apply finite approximation argument in $L(\mathbb{F}_2)$ as the case when the gorup is amenable? Nov30 comment Fuglede-Kadison determinants in $L(\mathbb{F}_2)$@ Andreas, thanks for bringing these papers into attention. Nov30 awarded ● Popular Question Nov30 asked Fuglede-Kadison determinants in $L(\mathbb{F}_2)$