Timeline for The functional equation $T(x\otimes y)=T(x)\otimes T(y)$ on certain $C^{*}$ algebras
Current License: CC BY-SA 3.0
11 events
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| May 2, 2016 at 19:53 | comment | added | Johannes Hahn | @AliTaghavi Well, yes for simple tensors this is a valid definition. But then you have only defined $t\otimes T$ on the algebraic tensor product which is a dense subspace of the/any C*-algebra-tensor product. You need some form of continuity to extend this maps from the subspace to the whole space. | |
| May 2, 2016 at 15:59 | comment | added | Ali Taghavi | @UlrichPennig thanks very much for your very helpful comment. | |
| May 2, 2016 at 15:37 | comment | added | Ali Taghavi | @WillSawin Thank you for this very interesting point. Your comment is interesting and is a motivation to ask: How many isomorphisms between $\mathcal{K} $ and $\mathcal{K} \otimes \mathcal{K}$ are there?Some isomorphisms diffrent from "matrix tensor product? | |
| May 2, 2016 at 15:09 | comment | added | Ali Taghavi | @JohannesHahn Thanks for your comment. For two element $a,b$ $T(a)\otimes T(b)$ is defined simply as a "simple tensor", so I think that the definition does mot need bounded ness of $T$.Am I right? | |
| Apr 30, 2016 at 20:11 | comment | added | Johannes Hahn | How do you define $T\otimes T$ on $A\otimes A$ without using boundedness in the first place? Remember that $A\otimes A$ is a completion of the algebraic tensor product and therefore you need some form of continuity to extend operators from the algebraic to the spatial tensor product. | |
| Apr 30, 2016 at 20:07 | comment | added | Ulrich Pennig | btw. the only scalars that satisfy your condition are $0$ and $1$, since the first equation gives you $\lambda = \lambda^2$. | |
| Apr 30, 2016 at 20:04 | comment | added | Ulrich Pennig | The property $A \cong A \otimes A$ is called self-absorbing. So-called strongly self-absorbing $C^*$-algebras have played a cornerstone role in the classification program. | |
| Apr 30, 2016 at 18:25 | comment | added | Will Sawin | Don't you think this could depend on the choice of isomorphism between $A$ and $A \otimes A$? | |
| Apr 30, 2016 at 16:02 | history | edited | Ali Taghavi | CC BY-SA 3.0 |
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| Apr 29, 2016 at 15:35 | history | edited | Ali Taghavi | CC BY-SA 3.0 |
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| Apr 29, 2016 at 15:29 | history | asked | Ali Taghavi | CC BY-SA 3.0 |