Timeline for Why all irreducible representations of compact groups are finite-dimensional ? [EDIT: Subtleties: AC,etc]
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
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| Jan 22, 2013 at 12:51 | comment | added | Alexander Chervov | Oops, stupid :) too much thinking is not good :) | |
| Jan 21, 2013 at 13:38 | comment | added | Benjamin Steinberg | Or said differently every simple module for a finite dimensional algebra is cyclic and hence a quotient of the regular module and hence finite dimensional. | |
| Jan 21, 2013 at 13:25 | comment | added | Eric Wofsey | If it's finite, any vector in any representation generates a finite-dimensional subrepresentation, so any irreducible representation must be finite-dimensional. | |
| Jan 21, 2013 at 6:37 | comment | added | Alexander Chervov | Thank you very much ! By the way if semi-group is finite, is it also false ? (Your argument breaks, since set is discrete and it can be mapped to discrete set of idempotents). | |
| Jan 20, 2013 at 22:03 | comment | added | Benjamin Steinberg | This actually just shows you can't separate points with finite dimensional representations, which means no Peter-Weyl theorem. For a compact idempotent commutative semigroup there are no infinite dimensional irreps. This is because the only central idempotents can be 0 or 1 in an irreducible rep of a compact semigroup because the image of a central idempotent is invariant. | |
| Jan 20, 2013 at 21:32 | comment | added | Todd Trimble | That's actually pretty cute... | |
| Jan 20, 2013 at 21:24 | history | answered | Benjamin Steinberg | CC BY-SA 3.0 |