Timeline for Irreducible group representation(algebraic and topological irreducibility)
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 28, 2022 at 19:26 | comment | added | Ruy | @Narutaka, thanks. This is a nice result. As I indicated above, I can only prove it for countable groups. | |
| Mar 28, 2022 at 6:49 | comment | added | Narutaka OZAWA | @Ruy: I think it's true that if a Banach space $X$ admits an algebraically cyclic continuous action of a second countable locally compact group, then it is finite-dimensional. | |
| Mar 28, 2022 at 2:06 | comment | added | Ruy | If the dimension of the space of the representation is infinite then $\text{span}\ G\zeta$ is always a proper invariant subspace: it is either finite dimensional, or a non-closed infinite dimensional subspace. | |
| Mar 27, 2022 at 20:08 | comment | added | Ali Taghavi | @Ruy No I do not mean all representation. I mean the representation is fixed but for every $\zeta \in H$, the span of $G\zeta$ isa finite dim. s[ace. So in this case I think that maybe the argument in your first comment does not work. | |
| Mar 27, 2022 at 16:32 | comment | added | Ruy | @Ali, I didn't quite understand your question. Could you please expand it. Do you mean $G\zeta$ finite for all $\zeta$ and all representations? | |
| Mar 27, 2022 at 12:53 | comment | added | Ali Taghavi | @Ruy if $G\zeta$ is a finite set for every $\zeta$ is it obvious that $G$ is necessarilly a finite group? | |
| Mar 26, 2022 at 15:56 | comment | added | Ruy | There is no algebraically irreducible representation of any countable group $G$ on an infinite dimensional Hilbert space! Any vector $\xi$ lies in the (countably generated, hence) proper subspace $\text{span}\ G\xi$. On the other hand, there are surely many examples of topologically irreducible representations! | |
| Mar 25, 2022 at 14:35 | comment | added | Benjamin Steinberg | The comments to math.stackexchange.com/questions/4384903/… give an example | |
| Mar 25, 2022 at 12:08 | comment | added | Benjamin Steinberg | Algebraic irreducibility for the group is the same as algebraic irreduciblility for the complex group algebra which is much smaller than the C*-algebra if you ignore all topology. So it shouldn't be that surprising if there are algebraically irreducible reps if the C*-algebra which are not algebraically irreducible for the group. | |
| Mar 25, 2022 at 10:49 | history | asked | Ali Taghavi | CC BY-SA 4.0 |