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I've learned a few things about harmonic analysis on semisimple Lie groups recently and the amount of effort that goes into the proof of the Plancherel formula seems overwhelming. Of course it has led to great discoveries in representation theory, but I was wondering whether there are some direct applications of the Plancherel formula. And now when I'm thinking about it, I don't even recall any applications of the classical Plancherel formula.

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Any time I want to integrate some product of matrix coefficients without looking up explicit Haar measure and messing about with changes of variable. Sorry if that sounds banal, but as stated the scope of your question is rather wide. (A popular proof that $\zeta(2)=\pi^2/6$ uses the Plancherel formula for the circle.) –  Yemon Choi Nov 20 '10 at 19:26
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There is a killer application of the Plancherel formula in the proofs of various known instances of Langlands Functoriality, if my understanding is correct. My understanding is certainly limited though. The application goes like this: we have two groups, say inner forms of each other, and we want to check the automorphic forms on them basically naturally biject with each other. We use the trace formula and check each term on one side matches up with a term on the other side. But for the terms corresponding to the identity element this is not formal: it is basically the Tamagawa Number Conj... –  Kevin Buzzard Nov 21 '10 at 8:34
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...ecture. However we're OK: if those terms don't match up then one gets a statement of the form "the trace of this wide class of functions $f$ on a space of $L^2$ functions is equal to some constant times $f(0)$". One now rules this out immediately as a consequence of Plancherel---it's so trivial that (in my experience) most experts just dismiss it as obvious, but when I was actually trying to read some proofs of these things recently I got very hung up about this and only unravelled it once I found the argument spelt out in Langlands' base change book. –  Kevin Buzzard Nov 21 '10 at 8:37
    
@Kevin Buzzard: A very good point. –  MBN Mar 10 '11 at 16:08
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4 Answers

The standard way to decompose $L^2(G)$ (where $G$ can be a wide variety of things) is to decompose a dense subspace (usually Schwartz or compactly supported), then prove the Plancherel formula, which implies that the decomposition extends to $L^2$. So it's foundational. If you already believe the result, the work you have to do might seem excessive, though.

People don't seem to do so much "applied" harmonic analysis on, say, semisimple Lie groups, so you (by which I mean "I") don't see many applications. But you can use the (noncommutative) Fourier transform to solve certain differential equations (e.g. when the operator comes from the center of the universal enveloping algebra), and you can talk about Sobolev spaces, so you can prove things about smoothness of solutions. The spectral decomposition of automorphic forms is used a lot (especially applied to analytic number theory).

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The key fact which underlies the Plancheral formula (say for real semi-simple Lie groups) is Harish--Chandra's theorem describing the discrete series (when they exist, what their infinitesimal characters are, what their Harish-Chandra characters are, and so on). (See my answer here for an elaboration on this.)

The results on the existence and basic properties of discrete series are fundamental in all of the theory of automorphic forms, Shimura varieties, the trace formula, and so on. It's impossible to overstate this.

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The Plancherel formula does almost all the work when you show that the continuous wavelet transform or the short-time Fourier transform are isometries (up to a constant).

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The Plancherel Transform does get complicated especially for non commutative groups, and semi simple Lie groups happen to be the worse ones. A fairly simple application of the Plancherel transform on $\mathbb R$ is that it takes complicated convolutions to simple pointwise multiplication on the frequency domain.

Also in wavelets analysis and representation theory, complicated operators get simplified by intertwining them with the Fourier transform. For example consider the coefficient operator $W_f:L^2(\mathbb R)\to L^2(\mathbb R)$ such that $W_{g}(f)(x) = \langle f,L(x)g \rangle $ if you try to prove that the Left regular representation $L$ of $\mathbb R$ is not admissible in term of continuous wavelets, you will need to take the Fourier (plancherel in this case) transform of $W_{g}f$ in order to prove that it is not square integrable.

Vignon S. Oussa

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