Timeline for Completely positive maps with commuting ranges can be extended to maximal tensor product
Current License: CC BY-SA 3.0
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| Nov 23 at 18:31 | comment | added | Martin Argerami | -1: This is false. It is not true that the product of positive operators with commuting ranges is max-bounded. An example is given by $A=B=B(H)$, and $C=B(H\otimes H)$, with $\varphi(a)=a^T\otimes I$ and $\psi(b)=I\otimes b$. The map $\varphi\times\psi$ is unbounded on $A\odot B$. The problem with the argument in this answer is that it is not true that the product is positive; the argument doesn't work because positive elements of $A\odot B$ are not necessarily of the form $x^*x$. $$ \ $$ I have posted a separate answer with a different argument. | |
| Oct 27, 2016 at 17:55 | history | edited | Sabrina Gemsa | CC BY-SA 3.0 |
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| Oct 27, 2016 at 17:44 | history | answered | Sabrina Gemsa | CC BY-SA 3.0 |