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Jan 16, 2013 at 7:31 comment added Marc Palm I only read your second comment now. Okay, when you expand in that way, which is equivalent to expansion into matrix coefficients, then you can fully recover the function. So there was an ambiguity in what "Fourier transform" actually means :|
Jan 16, 2013 at 7:27 comment added Marc Palm @Yemon Choi: No there is no inverse in general, but only on the subalgebra of conjuation invariant functions. Trace of irreducible reps are conjugation invariant distributions, so they do not seperate the full algebra. Note that via Plancherel, you can only recover the value at the identity. Translation doesn't behave well here, so you should rather not think about 1 as an element, but as a conjugacy class. Btw, there is no difference here to the finite group situation and you might want to check his for a small non-abelian group first.
Jan 15, 2013 at 19:53 comment added Yemon Choi [deleted some earlier comments]
Jan 15, 2013 at 19:03 comment added Yemon Choi oh, wait, I see that you are defining f-hat of rho to be the trace of rho(f) whereas I would define it to be the matrix rho(f) itself. If you allow matrix-valued Fourier transform then it is injective on L^1(G)
Jan 15, 2013 at 18:45 comment added Yemon Choi I don't understand your comments about losing information because you take the trace. There is such a thing as the non-abelian Fourier transform and its inverse (via Plancherel)
Jan 15, 2013 at 15:52 comment added Marc Palm Indeed, then $f$ can be recovered as $f(k) = \sum\limits_\rho \hat{f}(\rho) \tr \rho(k)$:)
Jan 15, 2013 at 15:51 comment added Marc Palm Btw, you might now think to impose $G$-conjugation-invariance. I'd conjecture that then you can recover $f$.
Jan 15, 2013 at 15:29 history edited Marc Palm CC BY-SA 3.0
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Jan 15, 2013 at 15:24 history answered Marc Palm CC BY-SA 3.0