Timeline for A completely positive equivariant map $\varphi: A \to B$ induces a map on the full crossed products
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 29, 2021 at 20:23 | comment | added | Andromeda | In fact we can simply take the universal GNS-representation of $B \rtimes G.$ | |
| Dec 28, 2021 at 10:54 | history | bounty ended | Andromeda | ||
| Dec 28, 2021 at 10:54 | vote | accept | Andromeda | ||
| Dec 27, 2021 at 14:25 | comment | added | Andromeda | Ah yes you can take the direct sum! I didn't realise this. Thanks for your answer! | |
| Dec 27, 2021 at 13:50 | comment | added | Matthew Daws | Also, didn't your original question specify that we could find a faithful such $\tilde\pi$? (which is then an isometry thanks to the metrical natural of $C^*$-algebras). | |
| Dec 27, 2021 at 13:50 | comment | added | Matthew Daws | What's the definition of the "full" crossed product $B\rtimes G$? For me, it satisfies the universal property given by covariant representations. You can construct it by, for example, taking the direct sum over (all equivalence classes of not too large in a cardinality sense) covariant reps: this direct sum will then be the "universal" one you seek. | |
| Dec 27, 2021 at 13:46 | comment | added | Andromeda | But why can we pick $\widetilde{\pi}$ to be an isometry? Is this always possible? | |
| Dec 27, 2021 at 11:30 | history | answered | Matthew Daws | CC BY-SA 4.0 |