Timeline for What's the natural equivalence of subfactors in general?
Current License: CC BY-SA 3.0
11 events
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| Nov 19, 2013 at 16:59 | history | edited | Sebastien Palcoux | CC BY-SA 3.0 |
Minor edit : I've removed the last question
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| Sep 17, 2013 at 10:23 | vote | accept | Sebastien Palcoux | ||
| Sep 17, 2013 at 10:23 | |||||
| Jul 12, 2013 at 17:17 | history | edited | Sebastien Palcoux | CC BY-SA 3.0 |
I change the order of the attempts
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| Jul 12, 2013 at 15:44 | history | edited | Sebastien Palcoux | CC BY-SA 3.0 |
I add two attempts.
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| Jul 12, 2013 at 13:52 | comment | added | Owen Sizemore | $P^t\subset Q^t=pM_n(P)p\subset pM_n(Q)p$ for an appropriate projection $p\in P$. Similarly for $P^t\otimes\mathcal{R}\subset Q^t\otimes\mathcal{R}$, but $p\in P\otimes\mathcal{R}$ in the last case. The isomorphism type only depends on the trace of $p$ and since $P\otimes\mathcal{R}$ is a factor we get a projection with correct trace in $\mathcal{R}$. So we can assume $p\in 1\otimes\mathcal{R}$. Then we just view $M_n(P)\otimes\mathcal{R}$ as $M_n(\mathbb{C})\otimes P \otimes\mathcal{R}= P\otimes M_n(\mathcal{R})$. Since $p\in P'$ we get the amplification is happening on $\mathcal{R}$. | |
| Jul 12, 2013 at 13:43 | comment | added | Sebastien Palcoux | I'm not sure to well understand the first isomorphism, but anyway, it seems imply that the uncountably many non-isomorphic subfactors in BNP are in fact equivalent in this sense. This is good news ! | |
| Jul 12, 2013 at 13:07 | comment | added | Owen Sizemore | However, this notion kills any information from the fundamental group. Specifically, consider the BNP examples $P\subset Q$. Then $P$ and $Q$ are the hyperfinite $II_1$, which we call $\mathcal{R}$, which is absorbing for itself. Then we have $P^t\otimes\mathcal{R} \subset Q^t\otimes\mathcal{R}\simeq P\otimes\mathcal{R}^t\subset Q\otimes\mathcal{R}^t\simeq P\otimes\mathcal{R}\subset Q\otimes \mathcal{R}$. This will always happen if the absorbing factor ($M$ above) has full fundamental group, and anything that absorbs $\mathcal{R}$ is a McDuff factor and has full fundamental group. | |
| Jul 12, 2013 at 7:53 | history | edited | Sebastien Palcoux | CC BY-SA 3.0 |
I invite to read Owen's comment about "existence".
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| Jul 11, 2013 at 22:12 | comment | added | Owen Sizemore | About existence, yes given any finite collection of $II_1$ factors $P_1, ..., P_k$. Consider the the infinite tensor product factor $\bigotimes (P_1\otimes\cdots\otimes P_k)$. This will be absorbing for all the $P_i$. | |
| Jul 11, 2013 at 18:50 | history | edited | Sebastien Palcoux | CC BY-SA 3.0 |
I replace "isomorphism" by "equivalence".
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| Jul 11, 2013 at 16:20 | history | answered | Sebastien Palcoux | CC BY-SA 3.0 |