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Let $G$ denote a connected semisimple Lie group and $\pi$ an irreducible representation on $G$. Define $\pi':C^{\infty}_c(G) \to \mathbb{C}$ by:

$$\pi'(f) = \int_G f(x) \pi(x) dx $$

where $dx$ is the Haar measure on $G$. Then $\pi'$ is of trace class and the map $T_{\pi}$ defined by $f \mapsto tr(\pi (f))$ is a distribution on $G$ (the character of $\pi$).

Denote the regular set of $G$ by $G'$. Harish Chandra showed that $T_{\pi}$ coincides on $G'$ (an open dense subset of $G$) with an analytic function $F_{\pi}$.

How does one define an analytic function on a Lie group? As is shown at this link on stackexchange:

https://math.stackexchange.com/questions/2532282/analytic-functions-on-smooth-manifolds

the notion of an analytic function on a smooth manifold doesn't make sense a priori. I had read somewhere though that Lie groups have a unique analytic atlas making them into an analytic manifold, so perhaps is an analytic function on a Lie group defined to be analytic in analytic charts?

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    $\begingroup$ A Lie group has a canonical real analytic structure: locally around 1 one chart is given by the exponential map and elsewhere first left-translate to 1; then the product and inverse are both analytic. $\endgroup$
    – YCor
    Jun 2, 2019 at 6:17
  • $\begingroup$ I am sorry but I don't see what is your problem. Since you start with an algebraic object(semisimple Lie group), so as you mentioned, Harish-Chandra tried to find a well-defined 'trace' function(note that the fixed irreducible representation may not be finite dimensional). Indeed they are functions on conjugacy classes of (regular, strongly-.) semisimple elements of rational points of G, and these are widely studied objects nowadays. On the other hand, on geometric side, you have GAGA for underlying field being complex or p-adic. So you can really find 'analytic' functions, eg rigid geomrtry. $\endgroup$
    – Yuan Zhiri
    Jun 2, 2019 at 14:39

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