Gelfand's trick

Remarks

Proof idea. The modular function $\Delta$ is trivial on the compact group $K$, so the hypothesis gives $\Delta(\theta(g)) = \Delta(g)^{-1}$; an involutive automorphism preserves Haar measure and $\Delta$, so $\Delta = 1$ and $G$ is unimodular. For a bi-$K$-invariant function $f$ write $f^\theta = f \circ \theta$ and $f^\vee(g) = f(g^{-1})$; the hypothesis gives $f^\theta = f^\vee$. Since $\theta$ is an automorphism, $(f_1 * f_2)^\theta = f_1^\theta * f_2^\theta$, and by unimodularity $(f_1 * f_2)^\vee = f_2^\vee * f_1^\vee$. Comparing these gives $f_1^\vee * f_2^\vee = f_2^\vee * f_1^\vee$ for all bi-invariant $f_1, f_2$, and $f \mapsto f^\vee$ is a bijection of $L^1(K \backslash G / K)$.

Examples. For $G$ abelian and $K = \set{e}$, take $\theta(g) = g^{-1}$. For $G = \mathrm{SL}(2, \R)$ and $K = \mathrm{SO}(2)$, take $\theta(g) = (g^{T})^{-1}$: writing $g = k_1 a k_2$ with $a$ diagonal, one finds $\theta(g) = k_1 a^{-1} k_2 \in K g^{-1} K$. The same argument with the Cartan decomposition $G = K A K$ shows that $(G, K)$ is a Gelfand pair for every connected semisimple Lie group $G$ with finite centre and maximal compact subgroup $K$.

Going deeper

For a student

The trick is the standard way to prove that the pairs coming from Riemannian symmetric spaces are Gelfand pairs, with $\theta$ the Cartan involution. Once $(G, K)$ is a Gelfand pair, the theory of spherical functions and the spherical transform applies. The same idea for a finite group $G$ and subgroup $K$ shows that the permutation representation on $G / K$ is multiplicity free.