Timeline for Morita equivalence for compact groups
Current License: CC BY-SA 3.0
4 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 19, 2012 at 10:00 | vote | accept | Marc Palm | ||
| Mar 17, 2012 at 19:48 | comment | added | Marc Palm | Okay, thank you for this counter example. So you consider $A_3 = <d>$ inside $S_3$. The conjugation orbits are $1$, $<d> - 1$ and $S_3 -<d>$, but that is how far I can go. @Reimundo Heluani: No, I was not confused with this example. My proof goes as follows. I identify invariant functionals with irreducible representation,and observe that the invariant functionals on $C(K)$ are isomorphic to invariant functionals on $C(K)^H$. No I guess, I am simply mistaken that modules over $C(K)^H$ are not associated to representations anymore;( | |
| Mar 17, 2012 at 18:54 | comment | added | Reimundo Heluani | I think pm might be confused with the algebra of bi-invariant functions which also happens to be a commutant in the spirit of Morita's theorem: $Hom_K(L(X),L(X))=L(H\K/H)$ where $X=K/H$. | |
| Mar 17, 2012 at 17:16 | history | answered | Tom Goodwillie | CC BY-SA 3.0 |