Fourier transform on a nonabelian group

Remarks

The operator norm satisfies $\norm{\hat f(\pi)} \le \norm{f}_1$, since each $\pi(g)^*$ is unitary. Many authors use $\pi(f) = \int_G f(g) \pi(g) \, dg$ instead; the two differ by the substitution $\pi(g)^* = \pi(g^{-1})$. Changing $\pi$ to an equivalent representation $U \pi U^{-1}$ changes $\hat f(\pi)$ to $U \hat f(\pi) U^{-1}$, so norms and traces of $\hat f(\pi)$ depend only on the class of $\pi$.

Example: let $G = \R$ with Lebesgue measure. The irreducible unitary representations are the characters $\pi_\xi(x) = e^{2\pi i \xi x}$ on $\C$, so $\hat f(\pi_\xi) = \int_\R f(x) e^{-2\pi i \xi x} \, dx$, the usual Fourier transform.

Example: let $G$ be a finite group with counting measure. Then $\hat f(\pi) = \sum_{g \in G} f(g) \pi(g)^*$ is a $d_\pi \times d_\pi$ matrix, where $d_\pi = \dim \mathcal{H}_\pi$.

If $f_1 * f_2$ is the convolution $\int_G f_1(h) f_2(h^{-1} g) \, dh$, then $\widehat{f_1 * f_2}(\pi) = \hat f_2(\pi) \hat f_1(\pi)$, with the order reversed because of the adjoint. A non-example: the family of the numbers $\int_G f(g) \chi_\pi(g) \, dg$, with $\chi_\pi$ the character, is not the Fourier transform. It records only the traces of the operators and loses information when some $d_\pi > 1$.