Gelfand pair

Remarks

Example: $G = \mathrm{SL}_2(\R)$ and $K = \mathrm{SO}(2)$. Let $\theta(g) = g^{T}$, the transpose, which reverses products and preserves Haar measure. By the Cartan decomposition $g = k_1 a k_2$ with $a$ diagonal, so $g^{T} = k_2^{-1} a k_1^{-1}$ lies in $K g K$. So every bi-invariant $f$ satisfies $f \circ \theta = f$, and then $f_1 * f_2 = (f_2 * f_1) \circ \theta = f_2 * f_1$. This is Gelfand's trick.

Non-example: $G$ a finite non-abelian group and $K = \set{e}$. Then every function is bi-invariant, and the convolution of the indicators of $\set{a}$ and $\set{b}$ is the indicator of $\set{ab}$. So the algebra is not commutative when $ab \ne ba$.

Equivalent form: $(G, K)$ is a Gelfand pair exactly when every irreducible unitary representation of $G$ has a space of $K$-fixed vectors of dimension at most $1$. For finite $G$, this says that the permutation representation of $G$ on functions on $G/K$ is multiplicity-free. If $(G, K)$ is a Gelfand pair, then $G$ is unimodular.