Plancherel theorem for compact groups

Remarks

Fix orthonormal bases and write $\pi_{ij}$ for the matrix entries of $\pi$. By the Peter–Weyl theorem the functions $\sqrt{d_\pi} \, \pi_{ij}$ form an orthonormal basis of $L^2(G)$. The $(j, i)$ entry of $\hat f(\pi)$ is $\int_G f \, \overline{\pi_{ij}} \, dg = d_\pi^{-1/2} \ip{f}{\sqrt{d_\pi} \pi_{ij}}$, so Parseval's identity for this basis is exactly the formula in 1, and surjectivity in 2 is the completeness of the basis. In the language of the abstract Plancherel theorem for groups, the Plancherel measure of $G$ gives the point $\pi$ the mass $d_\pi$.

Example: for the circle group $\R / 2\pi\Z$, all $d_\pi = 1$ and the formula is Parseval's identity $\frac{1}{2\pi} \int_0^{2\pi} \abs{f}^2 \, d\theta = \sum_{n \in \Z} \abs{c_n}^2$ for the Fourier coefficients $c_n$. For a finite group $G$ with normalized counting measure, applying 1 to the indicator function of $\set{e}$ gives $\abs{G} = \sum_\pi d_\pi^2$.

Going deeper

For a student

This is the Peter–Weyl theorem read as a statement about the Fourier transform. It shows that $L^2(G)$ is the Hilbert direct sum of the spaces of matrix coefficients, each of dimension $d_\pi^2$. It is the model case of the Plancherel theorem for locally compact groups, where $\widehat{G}$ need not be discrete and the sum becomes an integral against the Plancherel measure.