Timeline for Connes' correspondences of two $L^\infty$-algebras
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 16, 2015 at 16:32 | comment | added | Nik Weaver | The result is not obtained for pairwise disjoint rectangles, it is obtained for sequences $\{A_n\}$ and $\{B_k\}$ each of which is pairwise disjoint. | |
| Feb 16, 2015 at 15:07 | comment | added | Danila Zaev | It is very likely that I am wrong, but I can't see the mistake. After we obtain the result for pairwise disjoint rectangles, we check that $\gamma(\bigcup D_n)=\sum \gamma(D_n)$ for arbitrary sequence of step-sets $\{D_n\}$. To show that, we split $\bigcup D_n$ into a union of disjoint rectangles $\{D_{n,i,j}\}$ and apply the result about pairwise disjoint sequences to it. | |
| Feb 15, 2015 at 16:44 | comment | added | Nik Weaver | I really don't think this proof is right. The result about $\gamma(\bigcup A_n \times B_n)$ is only proven when the sequences $\{A_n\}$ and $\{B_k\}$ are pairwise disjoint. | |
| Feb 15, 2015 at 9:06 | comment | added | Danila Zaev | $D_n$ is an element of the algebra of step-sets, hence it can be represented as a finite union of disjoint rectangles. I agree that a shorter proof for the lemma is definitely possible, and probably it already exists. | |
| Feb 14, 2015 at 19:29 | comment | added | Nik Weaver | I don't follow the last part of the argument --- where does the decomposition $D_n = \bigcup_{i=1}^{M_n} D_{n,i}$ come from? But it seems to me that the usual proof of this lemma for product measures, where you integrate characteristic functions, should work. | |
| Feb 14, 2015 at 16:37 | history | answered | Danila Zaev | CC BY-SA 3.0 |