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  "body": "\n## Definition\n\nLet $G$ be a [[unimodular|CXG6JZGMCK]] [[locally compact group|AHV05ZBCYW]] with a [[Haar measure|VY0HXD3GC4]] $dg$.\nLet $\\lambda$ be the left regular representation of $G$ on $L^2(G)$, given by $(\\lambda(g) f)(x) = f(g^{-1} x)$.\nLet $\\pi$ be an [[irreducible|FX3NEFYFFN]] [[unitary representation|74BPYJNZTM]] of $G$ on a Hilbert space $H$.\nThen $\\pi$ is a *discrete series representation* if it is unitarily equivalent to a subrepresentation of $\\lambda$.\n\n## Remarks\n\nFor $u, v \\in H$, the matrix coefficient of $\\pi$ is the function $g \\mapsto \\ip{\\pi(g) u}{v}$.\nEquivalently, $\\pi$ is in the discrete series if and only if some nonzero matrix coefficient lies in $L^2(G)$.\nThen every matrix coefficient lies in $L^2(G)$.\n\nExample: let $G$ be a compact group, with Haar measure of total mass $1$.\nEach matrix coefficient of a unitary representation is continuous and bounded by $\\norm{u} \\norm{v}$, so it lies in $L^2(G)$.\nSo every irreducible unitary representation of $G$ is in the discrete series.\n\nNon-example: the irreducible unitary representations of $G = \\R$ are the characters $x \\mapsto e^{ixt}$ with $t \\in \\R$.\nEach matrix coefficient has constant absolute value, so it is not square-integrable on $\\R$.\nSo $\\R$ has no discrete series, and $L^2(\\R)$ decomposes as a continuous direct integral by Fourier analysis.\n\nFor groups with non-compact center, such as $\\mathrm{GL}_n(\\R)$, one asks instead that the matrix coefficients be square-integrable modulo the center.\nFor connected semisimple $G$ with finite center and maximal compact subgroup $K$, discrete series exist if and only if $\\rank G = \\rank K$, by a theorem of Harish-Chandra.\nFor example, $\\mathrm{SL}_2(\\R)$ has discrete series, while $\\mathrm{SL}_2(\\C)$ has none.\n",
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