Timeline for Convolution theorem on a non-abelian Lie group
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11 events
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| Aug 7, 2018 at 0:49 | comment | added | Yemon Choi | @paulgarrett True, but (as you know!) tensoring irreps doesn't usually give an irrep, and so defining the dual convolution in terms of the Fourier coefficients (as one defines convolution of two functions in terms of the function values) is going to be tricky | |
| Aug 6, 2018 at 22:36 | comment | added | paul garrett | At least figuratively, "pointwise multiplication" is "tensor product". The point is that "tensor products" maybe have more apparent structural properties than pointwise multiplication, so may suggest what features you'd want/require... | |
| Aug 6, 2018 at 20:36 | comment | added | Yemon Choi | To the OP: if you are just starting out, and you are interested mainly in SU(2) for now, then a good way to gain intuition for what this dual convolution looks like is to take an $f$ for which you know its non-abelian Fourier coefficients, and consider the Fourier coefficients of $f^n$ (pointwise product of functions). For $f(p) = {\rm Tr}(p)$ this basically ends up as the problem of writing powers of cosine in terms of cosines of various frequencies, a.k.a. Chebyshev polynomials | |
| Aug 6, 2018 at 20:26 | comment | added | Yemon Choi | @RW if $f$ and $g$ are not class functions then I think one needs a bit more than just the hypergroup structure (of course when they are class functions then you are just getting the hypergroup on ${\bf N}_0$ given by the usual Clebsch-Gordan rules). For general (smooth) functions on SU(2) we have a kind of operator-valued hypergroup, I think | |
| Aug 6, 2018 at 20:24 | comment | added | Yemon Choi | I forget the precise conditions needed on the group, but when $G$ is unimodular there is an old paper of Stinespring that discusses an abstract approach to defining "dual convolution" in the sense that you describe. When $G$ is a compact group this is more or less what @RW describes (although you need to know more than merely the isomorphism classes if the irreducible constituents, you need the actual intertwining maps). For a semi-explicit formula for what happens when G is the real Heisenberg group, see Section 5 of arxiv.org/abs/1405.6403 | |
| Aug 6, 2018 at 20:05 | comment | added | R W | In order to define a convolution it is enough to have a hypergroup structure (so that the "product" of two points is a measure). In your setup such a structure is provided by the decomposition of tensor products of irreducible representations. | |
| Aug 6, 2018 at 18:32 | history | edited | onamoonlessnight | CC BY-SA 4.0 |
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| Aug 6, 2018 at 18:11 | comment | added | onamoonlessnight | @LSpice That's exactly what I would like to know. I wouldn't mind a result in terms of sums over irreducible representations, either. Perhaps there's a naturally defined convolution product on $\hat{\mathrm{G}}$? | |
| Aug 6, 2018 at 17:57 | comment | added | LSpice | By the way, the Fourier transform of a character is a delta distribution, so I guess that uniquely pins down what convolution must be if it is to satisfy your desired identity. | |
| Aug 6, 2018 at 17:50 | comment | added | LSpice | What would the convolution mean on the dual side? (Maybe this is part of your question.) There's a measure on that side (the Plancherel measure), but the lack of algebraic structure means that I have a hard time guessing how convolution should behave. | |
| Aug 6, 2018 at 17:17 | history | asked | onamoonlessnight | CC BY-SA 4.0 |