Skip to main content
9 events
when toggle format what by license comment
Jul 15, 2021 at 19:51 comment added Benjamin Steinberg Since he said irreducible you can automatically assume finite dimensional if the representation is on a Hilbert space.
Jul 15, 2021 at 19:49 comment added Alex B. @BenjaminSteinberg Oh yeah, sorry, I was not disputing that. Now corrected -- thanks!
Jul 15, 2021 at 19:48 history edited Alex B. CC BY-SA 4.0
inserted the hypothesis that the representations be finite dimensional
Jul 15, 2021 at 16:34 comment added Benjamin Steinberg Yes. The OP should have been clearer. As you point out even for finite groups you need a splitting field. But it is important that reps are finite dimensional to get it to factor through a finite quotient
Jul 15, 2021 at 15:45 vote accept Martin Skilleter
Jul 15, 2021 at 14:55 comment added Alex B. @BenjaminSteinberg Yes, we posted more or less simultaneously. I did not take the "topological vector space" part of the question to implicitly assume that we are working over the complex numbers. For example number theorists routinely consider continuous representations of profinite groups (e.g. of absolute Galois groups) over all sorts of rings.
Jul 15, 2021 at 14:44 comment added Benjamin Steinberg Anyway +1. The question could be more specific on the hypotheses
Jul 15, 2021 at 14:02 comment added Benjamin Steinberg The complex part is essentially my answer in the comments except you need to use that all irreducible representations are finite dimensional since the regular representation of a compact group is faithful. I think it is implicit from the question mentioning a counting argument for finite groups and talking about topological vector space and unitary representations that the OP was working over the complex numbers
Jul 15, 2021 at 13:16 history answered Alex B. CC BY-SA 4.0