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  "body": "\n## Definition\n\nLet $G$ be a [[locally compact group|AHV05ZBCYW]] with a [[Haar measure|VY0HXD3GC4]], and let $K$ be a [[compact|0R6EWFKRCF]] [[subgroup|0X1JXT47GA]] of $G$.\nA function $f$ on $G$ is *$K$-bi-invariant* if $f(k_1 g k_2) = f(g)$ for all $g \\in G$ and $k_1, k_2 \\in K$.\nLet $C_c(K \\backslash G / K)$ be the space of continuous, compactly supported, $K$-bi-invariant functions on $G$.\nIt is closed under [[convolution|2Z29JN3VES]].\nThe pair $(G, K)$ is a *Gelfand pair* if $f_1 * f_2 = f_2 * f_1$ for all $f_1, f_2 \\in C_c(K \\backslash G / K)$.\n\n## Remarks\n\nExample: $G = \\mathrm{SL}_2(\\R)$ and $K = \\mathrm{SO}(2)$.\nLet $\\theta(g) = g^{T}$, the transpose, which reverses products and preserves Haar measure.\nBy the Cartan decomposition $g = k_1 a k_2$ with $a$ diagonal, so $g^{T} = k_2^{-1} a k_1^{-1}$ lies in $K g K$.\nSo every bi-invariant $f$ satisfies $f \\circ \\theta = f$, and then $f_1 * f_2 = (f_2 * f_1) \\circ \\theta = f_2 * f_1$.\nThis is Gelfand's trick.\n\nNon-example: $G$ a finite non-abelian group and $K = \\set{e}$.\nThen every function is bi-invariant, and the convolution of the indicators of $\\set{a}$ and $\\set{b}$ is the indicator of $\\set{ab}$.\nSo the algebra is not commutative when $ab \\ne ba$.\n\nEquivalent 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 $1$.\nFor finite $G$, this says that the permutation representation of $G$ on functions on $G/K$ is multiplicity-free.\nIf $(G, K)$ is a Gelfand pair, then $G$ is unimodular.\n",
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