Timeline for Is there a non-Kac complex finite dimensional semisimple Hopf algebra?
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
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| Nov 2, 2022 at 17:18 | comment | added | dm82424 | I opened a similar question here: mathoverflow.net/questions/433176/hopf-algebras-vs-kac-algebras | |
| Mar 9, 2021 at 13:46 | comment | added | Daniel | Oh, yes sure, I meant C*-algebra such that $\Delta$ is a $*$-homomorphism, but that's exactly what you write in the question, I just did not notice, sorry. You can ignore my comment. | |
| Mar 9, 2021 at 13:34 | comment | added | Sebastien Palcoux | @Daniel: how does you sentence show that the comultiplication $\Delta$ is a $*$-homomorphism? | |
| Mar 9, 2021 at 13:29 | comment | added | Sebastien Palcoux | @Daniel: a finite-dimensional semisimple algebra over $\mathbb{C}$ is always a C*-algebra. Proof: by the Artin–Wedderburn theorem a finite dimensional semisimple algebra over a field $k$ is a finite product of matrix algebras over division algebras over $k$. Now over an algebraically closed field $k$ (for example the complex numbers $\mathbb{C}$), there are no finite-dimensional (associative) division algebras, except $k$ itself. | |
| Mar 9, 2021 at 12:01 | comment | added | Daniel | I am not sure if I am not missing something, but I guess that equivalently you need to show that $H$ is a C*-algebra, right? Because it is known that for a semisimple Hopf algebra we have $S^2={\rm id}$. And if $H$ is a C*-algebra, this would mean that it defines a finite (compact) quantum group and for CQGs we have the Haar state and we know that $S^2={\rm id}$ if and only if the Haar state is tracial. | |
| Feb 26, 2019 at 12:57 | history | edited | Sebastien Palcoux | CC BY-SA 4.0 |
minor edit: title edit
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| Feb 25, 2019 at 18:02 | history | asked | Sebastien Palcoux | CC BY-SA 4.0 |