Timeline for Unitary representations of discrete (locally compact) groups
Current License: CC BY-SA 4.0
14 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 27 at 8:44 | answer | added | Omar Mohsen | timeline score: 1 | |
| Nov 20 at 2:31 | comment | added | David Gao | @YemonChoi Thank you for the information! That was interesting. I wasn’t aware of that. I usually only deal with discrete groups, where being type I is such a rare occurrence. | |
| Nov 20 at 1:43 | comment | added | Yemon Choi | @DavidGao Belatedly coming back to this: as soon as you allow G to be non-discrete, Type I examples are plentiful! all semisimple Lie groups, all nilpotent Lie groups, the Euclidean motion groups, various p-adic version ... Indeed, this is what allows one to do noncommutative harmonic analysis. | |
| Nov 15 at 2:35 | comment | added | user82261 | @DavidGao Thanks! I'll look into all this. | |
| Nov 15 at 2:09 | comment | added | David Gao | (I don’t know if an equivalent condition is known for general locally compact groups, but this is also fine if, say, $G$ is compact, in which case you do get $C^\ast(G)$ is type I. Maybe the equivalent condition is for $G$ to have a cocompact abelian subgroup? I’m not sure.) | |
| Nov 15 at 2:05 | comment | added | David Gao | @user82261 Though, it is fine if the Hilbert space your group is represented on is separable, as then the von Neumann algebra generated by the representation is separable, so a direct integral decomposition into factorial representations is possible. Again, you can only decompose into factorial representations instead of irreducible ones in general. As Yemon already mentioned, if you actually want decompositions into irreducible representations instead of just factorial ones, $C^\ast(G)$ must be type I, so in the discrete case this can only happen for virtually abelian groups. | |
| Nov 15 at 2:00 | comment | added | David Gao | @user82261 There is no direct integral decomposition for general $C^\ast$-algebras/von Neumann algebras which are not separable, more or less because everything can only de done up to almost everywhere, but an uncountable union of null sets needs not be null. As a representation of a $C^\ast$-algebra/von Neumann algebra induces a representation of any of its generating subgroup of the unitary group, this will basically say uncountable discrete groups cannot in general be expected to have direct integral decompositions for their representations. | |
| Nov 14 at 22:41 | review | Close votes | |||
| Nov 19 at 3:01 | |||||
| Nov 14 at 22:36 | comment | added | user82261 | @YemonChoi Agreed. I was looking at Krillov's "Elements of the Theory of Representations" and the analogous result there is only stated locally compact groups with a countable basis. Hence, it only applies to countable discrete groups. Do you know if this assumption can be relaxed? I will continue digging into the literature. I understand I'll have to narrow down the type of discrete group I'm considering. | |
| Nov 14 at 22:24 | comment | added | Yemon Choi | Dixmier's book only achieves this for Type I Cstar algebras (at least if one is looking for a disintegration into a direct integral of irreducible representations). The only discrete groups that are Type I are those which are virtually abelian; this is a theorem of Thoma. I think you need to narrow down what you are hoping to achieve | |
| Nov 14 at 14:32 | history | edited | LSpice | CC BY-SA 4.0 |
Typo
|
| Nov 14 at 5:02 | comment | added | David Gao | It’s not possible to decompose into irreducible representations in general. For example, for a discrete icc group $\Gamma$, its left regular representation generates a factor, so it cannot be decomposed as a direct integral, but the representation is not already irreducible. In general, what you can get is a direct integral decomposition into factorial representations (which, in the finite-dimensional case, does reduce to a direct sum of irreducible representations). This can be found in many books on $C^\ast$-algebras and von Neumann algebras. | |
| Nov 14 at 4:57 | comment | converted from answer | Omar Mohsen | I don't have enough reputation to comment but this is in Dixmier's book C* algebras (chapter integration and disintegration I think) | |
| Nov 14 at 4:42 | history | asked | user82261 | CC BY-SA 4.0 |