# Plancherel formula for compact groups from Peter-Weyl Theorem

I'm trying to derive the following Plancherel formula:

$$\|f\|^{2}=\sum_{\xi\in\widehat{G}}{\dim(V_{\xi})\|\widehat{f}(\xi)\|^{2}}$$

from the statement of the Peter-Weyl Theorem as given by Terence Tao here:

Let $G$ be a compact group. Then the regular representation $\tau\colon G\rightarrow U(L^{2}(G))$ is isomorphic to the direct sum of irreducible representations. In fact, one has $\tau\cong\bigoplus_{\xi\in\widehat{G}}{\rho_{\xi}^{\bigoplus \dim(V_{\xi})}}$, where $(\rho_{\xi})_{\xi\in\widehat{G}}$ is an enumeration of the irreducible finite-dimensional unitary representations $\rho_{\xi}\colon G\rightarrow U(V_{\xi})$, up to isomorphism.

I managed to prove Fourier inversion from this without any difficulty at all, but I'm really struggling to see how the Plancherel formula follows from it. I'm pretty sure that the fact that we have $\|\operatorname{Proj}_{\xi}f\|=\dim(V_{\xi})^{1/2}\|\widehat{f}(\xi)\|$ is the main ingredient of the proof (here, I've used $\|\operatorname{Proj}_{\xi}f\|$ to denote the orthogonal projection of $f$ to $L^{2}(G)_{\xi}$), and most of my attempts have boiled down to trying to show the equality by proving inequality in both directions - showing $\sum_{\xi\in\widehat{G}}{\dim(V_{\xi})\|\widehat{f}(\xi)\|^{2}}\le \|f\|^{2}$ is rather trivial, although if this is the correct approach, I simply can't make the inequalities work in the other direction, despite having tried pretty much every possible way of looking at it - my main issue is that I always end up at the point where I could prove the result if I could prove that a square of a sum of norms is bounded by the sum of the squares of norms, which clearly isn't true in general (i.e. if all of the norms are 1 and the sum is nontrivial), and I can see no reason why they should be equal in this case. Any help or suggestions would be much appreciated!

Edit: Alternatively, I can see how this is from Tao's observation that we may write $L^{2}(G)\cong\bigoplus_{\xi\in\widehat{G}}{\dim(V_{\xi})\cdot HS(V_{\xi})}$, this is immediate - $\|f\|^{2}=\langle f,f\rangle=\sum_{\xi\in\widehat{G}}{\dim(V_{\xi})\cdot\langle\widehat{f},\widehat{f}\rangle}=\sum_{\xi\in\widehat{G}}{\dim(V_{\xi})\|\widehat{f}(\xi)\|^{2}}$, so I'd be fine with an explanation of how one obtains this isomorphism (if it can be obtained by any means other than combining the statements of Fourier inversion and the Plancherel formula).

• I don't have the book at hand but I once saw this proof in Folland's A Course in Abstract Harmonic Analysis. It should be in chapter 4. – Hui Yu Oct 19 '12 at 9:11
• Fantastic, thanks! I'll have to see if I can get my hands on a copy. edit: if anyone reading this is interested, I found the book online here: xwww.uni-math.gwdg.de/graboluk/seminar-2010/… – user123123 Oct 19 '12 at 9:13

I don't have the book at hand but I once saw this proof in Folland's A Course in Abstract Harmonic Analysis. It should be in chapter 4. ''