5
$\begingroup$

Let $\mathrm{G}$ be a compact (simple, if it helps) non-abelian Lie group and let $\hat{\mathrm{G}}$ be its unitary dual of (equivalence classes) of irreducible unitary representations. Defining the Fourier transform by $$ \hat{f}(\pi) = \int_{\mathrm{G}} f(x) \pi(x^{-1}) \, \mathrm{d} \mu(x),$$ where $\mu$ is a normalized Haar measure on $\mathrm{G}$, the Peter--Weyl theorem says that for $f \in L^2(\mathrm{G})$ one has the Fourier series expansion $$ f(x) = \sum_{\pi \in \hat{\mathrm{G}}} d_\pi \operatorname{Tr}(\hat{f}(\pi) \pi(x)), $$ where $d_\pi$ is the dimension of $\pi$. Furthermore, if we define the convolution product $*$ by $$ (f* g)(x) = \int_{\mathrm{G}} f(y) g(x y^{-1}) \, \mathrm{d} \mu(y), $$ then a version of the convolution theorem holds, namely that $$ (\widehat{f * g})(\pi) = \hat{f}(\pi) \hat{g}(\pi). $$ Question. I am interested in whether there is a sense in which an ``inverse" convolution theorem might hold, something along the lines of $$ (\hat{f} * \hat{g})(\pi) = (\widehat{fg})(\pi). $$ Edit. On reflection this might be asking for a somewhat unnatural identity. Perhaps a more natural question to ask would be whether there exists a convolution product on $\hat{\mathrm{G}}$ that makes a similar identity hold?

If it helps, I am more specifically interested in the case when $\mathrm{G} = \mathrm{SU}(2)$. I am fairly new to this so would appreciate any pointers to standard references if this is quite well-known!

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – LSpice Aug 6 '18 at 17:50
  • 1
    $\begingroup$ 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. $\endgroup$ – LSpice Aug 6 '18 at 17:57
  • 3
    $\begingroup$ 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. $\endgroup$ – R W Aug 6 '18 at 20:05
  • 2
    $\begingroup$ 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 $\endgroup$ – Yemon Choi Aug 6 '18 at 20:24
  • 3
    $\begingroup$ 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 $\endgroup$ – Yemon Choi Aug 6 '18 at 20:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.