I make my question more precisley :

Let $f$ be a conjugation invariant function on a compact semi-simple Lie group $G$. So $f$ can be regarded as a function on the maximal tours ( or Lie algebra of the maximal torus) but perhaps $f$ supported in a convex subset A of the lie algebra of the maximal torus. Assume that we have the fourier transform of this function (on the dual of the lie algebra of the maximal torus). Now from this data we want to describe the support $A$ of the function $f$.

From the fourier expansion (rather than fourier transform) I think is a hard problem to find the support $A$. Any comment/answer are welcome.

  • $\begingroup$ Could you be a bit more precise about what you mean by "recover"? are you looking for an algorithm? $\endgroup$ – Yemon Choi Jan 15 '13 at 18:50
  • $\begingroup$ Also, I assume that you are referring to the matrix-valued Fourier transform? $\endgroup$ – Yemon Choi Jan 15 '13 at 18:56
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    $\begingroup$ As reflected in the other comments and answers, the question itself is ambiguous. E.g., if "Fourier transform" means expansion in "characters" (=traces of irreds), then only conjugation-invariant functions are "represented" by "spectral expansions". If at the other end, "Fourier expansion" means expression as sum of functions from irreducibles, then there is a standard Sobolev-lemma that would assert that sufficiently smooth functions are represented (in various senses) by their spectral expansions. The subtler question of whether continuous functions are represented by their spectral... $\endgroup$ – paul garrett Jan 16 '13 at 1:12
  • $\begingroup$ ... expansions has a straightforward answer for compact Lie groups: "even worse" than for "ordinary" Fourier series, pointwise convergence typically fails, for Baire-category-corollary reasons. Nevertheless, the $L^2$ convergence determines an $L^2$ function, which, if it is a.e. a continuous function, unambiguously determines its support. $\endgroup$ – paul garrett Jan 16 '13 at 1:14
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    $\begingroup$ Perhaps, Jafar has something like the Paley-Wiener-Schwartz theorem in mind which describes the convex hull of the support (of a test function on $\mathbb R^n$ or a distribution with compact support) by growth conditions of the Fourier-Laplace transform. $\endgroup$ – Jochen Wengenroth Jan 16 '13 at 7:49

Certainly you can recover $f$ for a circle, because the group is abelian. In general, $f$ can be written as a linear combination of matrix coefficients, and for the Fourier transform of a general compact Lie group you take a trace, so you will loose information about $f$.

For example, there are nonzero functions with vanishing Fourier transform in the non-commutative situation, e.g. take two orthogonal vectors in an irreducible subspace $V_\rho$ and consider the function given by the matrix coefficient $f:k \mapsto \langle v_1 , \rho(k) v_2 \rangle$. We have $\hat{f}( \rho')=0$ for all irreducible representations $\rho'$. Sketch of proof: Use that $tr\rho'(f) = tr\rho'(f^G)$ where $$f^G (x) = \int\limits_G f(g^{-1}xg) d g$$ and use the orthogonality relations to show that $f^G=0$.

But you can always check the support at the identity with the Plancherel formula $$ f(1) = \sum\limits_{\rho} C_\rho \hat{f}(\rho).$$ The fudge factor $C_\rho$ depend only on the dimension. Unfortuantely, I don't recall how.

  • $\begingroup$ Btw, you might now think to impose $G$-conjugation-invariance. I'd conjecture that then you can recover $f$. $\endgroup$ – Marc Palm Jan 15 '13 at 15:51
  • $\begingroup$ Indeed, then $f$ can be recovered as $f(k) = \sum\limits_\rho \hat{f}(\rho) \tr \rho(k)$:) $\endgroup$ – Marc Palm Jan 15 '13 at 15:52
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    $\begingroup$ 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) $\endgroup$ – Yemon Choi Jan 15 '13 at 18:45
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    $\begingroup$ 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) $\endgroup$ – Yemon Choi Jan 15 '13 at 19:03
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    $\begingroup$ 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 :| $\endgroup$ – Marc Palm Jan 16 '13 at 7:31

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