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Let $G$ be a connected compact semisimple Lie group. Let $V$ be a faithful representation of $G$, with character $\chi \colon G \to \mathbb{C}$.

Let $\mu_G$ be the normalized left Haar measure. (So $\mu_G(G) = 1$.) We can consider the pushforward of $\mu_G$ along $\chi$. This gives a measure $\chi_* \mu_G$ on $\mathbb{C}$ whose support is the compact set $\chi(G)$.

Question. Does the measure $\chi_*\mu_G$ uniquely determine the character $\chi$?

Remarks. (i) Note that this is false if $G = S^1$. If $\chi_n$ is the character $z \mapsto z^n$ ($n \ne 0$), then $\chi_{n,*}\mu_{S^1}$ is the same as $\chi_{1,*}\mu_{S^1}$. So all non-trivial characters give rise to the same measure. In particular, it seems important to require that $G$ is semisimple.

(ii) I am not sure if the hypothesis "connected" is necessary. But my gut feeling says that things might go horribly wrong if $G$ is a non-trivial finite group.

(iii) Similarly, I don't know if the hypothesis "compact" is necessary. (Of course, if one removes this hypothesis, then it the claim that $\chi_*\mu_G$ has compact support is no longer true.)

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    $\begingroup$ You're right that things don't work out if $G$ is a finite group. Consider the case of an abelian group. Then $\chi(G)$ is simply a finite subgroup and therefore the measure is the uniform measure on those finitely many points. So the only thing it determines is $|\chi(G)|$ which is not enough to distinguish all the characters from each other if $G$ is non-cyclic. $\endgroup$ Jan 31, 2019 at 14:46
  • $\begingroup$ Right. That is similar to my $S^1$ counterexample. But maybe $G$ finite and simple might still work? $\endgroup$
    – jmc
    Jan 31, 2019 at 14:49
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    $\begingroup$ Even finite and simple does not work. If $G$ is cyclic of order $p$, then the measure distinguishes the trivial character form the $p-1$ non-trivial characters, but nothing else. More generally, $\chi_\ast(\mu_G)$ can only give you the character values (=the set of points where the measure is concentrated) and the size of each fibre (=the measure of a point multiplied by $|G|$). For example it isn't possible to distinguish the two characters of degree three of $G=A_5$, because both take different values on all five conjugacy classes, but the set of values is the same. $\endgroup$ Jan 31, 2019 at 14:58
  • $\begingroup$ On the other hand, it does work for $G=SU_2(\mathbb{C})$, because the character of highest weight $m\in\mathbb{N}$ is uniquely determined by $\chi_m(diag(e^{i\theta},e^{-i\theta})) = e^{-im\theta} + e^{i(m-2)\theta} + \cdots + e^{-i(m-2)\theta} + e^{-im\theta} = \frac{\sin((m+1)\theta)}{\sin(\theta)}$. Thus $\chi_m(G) \subseteq [-(m+1),m+1]$ with $\chi_m(1) = m+1$. Thus $\chi_m$ is uniquely determined by $\chi_m(G)=supp((\chi_m)_\ast(\mu_G))$. $\endgroup$ Jan 31, 2019 at 15:48

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