Jiang
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Registered User
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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.
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Apr 15 |
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clarify a question in group cohomology @Misha, thanks! |
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Apr 12 |
asked | clarify a question in group cohomology |
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Apr 5 |
asked | Find a lower bound for a pre-invariant $Fol(L(F_m), X_m)$ |
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Feb 21 |
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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. |
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Feb 21 |
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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. |
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Feb 19 |
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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? |
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Feb 19 |
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Not measure equivalent ICC groups $G$ and $H$, but $L(G)\cong L(H)$ fix some misleading expression? |
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Feb 19 |
asked | Not measure equivalent ICC groups $G$ and $H$, but $L(G)\cong L(H)$ |
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Jan 10 |
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Kadison-Singer problem in exotic Hilbert spaces Maybe 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/… |
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Dec 2 |
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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? |
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Nov 30 |
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Fuglede-Kadison determinants in $L(\mathbb{F}_2)$ @ Andreas, thanks for bringing these papers into attention. |
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Nov 30 |
awarded | ● Popular Question |
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Nov 30 |
asked | Fuglede-Kadison determinants in $L(\mathbb{F}_2)$ |
