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Let $\chi$ be the character of some representation $\rho:G \to GL(M)$ over $\mathbb C$.

Suppose $G$ is a group, then $\forall g \in G$ of finite order $n$, $ \chi(g^{-1})=\overline{\chi (g)}$

Proof: $\rho(g)^n=\rho(g^n)=\rho(e)=\textrm{id}$. Hence the characteristic polynomial of $\rho(g)$ divides $x^n-1$ and so the characteristic polynomial of $\rho(g)$ has distinct roots. Thus there is a basis $\mathcal B$ of $M$ composed of eigenvectors of $\rho(g)$ and so $[\rho(g)]_\mathcal B$ is diagonal. Then $\chi(g)=\sum_i \lambda_i$ where the $\lambda_i$ are the eigenvalues of $\rho(g)$. Now $[\rho(g^{-1})]_\mathcal B=[\rho(g)]_\mathcal B ^{-1}$ which has the $\lambda_i ^{-1}$ on the diagonal. So $\chi(g^{-1})=\sum_i \lambda_i^{-1}.$ But since $\rho(g)^n=\textrm{id}, \, [\rho(g)]_\mathcal B^n=I $ so for any eigenvector $v_i$ with eigenvalue $\lambda_i$, $\rho(g)^n(v_i)=\lambda_i^n v=v_i$. Hence $\lambda_i$ is an $n$-th root of unity and has $\lambda_i^{-1}=\overline{\lambda_i}$. Therefore $\chi(g^{-1})=\sum_i \lambda_i^{-1}=\sum_i \overline{ \lambda_i}=\overline{\sum_i \lambda _i}=\overline{\chi(g)}.$

You can see that my proof of this relies heavily on the existence of this finite $n$. I am therefore wondering:

For an element $g\in G$ of infinite order, is it still true that $ \chi(g^{-1})=\overline{\chi (g)}$?

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    $\begingroup$ What is $\overline{u}$ when $u$ is in an algebraically closed field that isn't necessarily $\mathbb{C}$ ? $\endgroup$ – darij grinberg May 5 '15 at 23:52
  • $\begingroup$ Yes. I'll edit my post. The course on group rep that I'm studying only covers $\mathbb C$, so I'm inadvertently restricting myself to that case. :( It also only covers finite groups, which is why I'm struggling with this kind of question in the first place $\endgroup$ – James May 5 '15 at 23:54
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As a rule of thumb, look at simple examples first. Consider any nonunitary one-dimensional complex representation of your favorite locally compact noncompact group, $\Bbb R$...

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  • $\begingroup$ What is meant by a non-unitary rep? $\endgroup$ – James May 5 '15 at 23:57
  • $\begingroup$ @James: Image not lying in the unit circle. $\endgroup$ – user98602 May 5 '15 at 23:58
  • $\begingroup$ @James for a one-dimensional representation, nonunitary means the image is not contained in the unit circle, in which case it must be a spiral in the context of my hint. orangeskid has the same hint, but with $\Bbb Z$ instead of $\Bbb R$, which is equally good. In general, unitary means a representation by unitary operators on a Hilbert space (you can google the terms if you wish to know more). $\endgroup$ – whacka May 5 '15 at 23:58
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    $\begingroup$ I am saying $\rho(G)$ is not contained within $\{z\in\Bbb C:|z|=1\}$. A representation is a group homomorphism, so when I say image, I mean image of that group homomorphism. Note that ${\rm GL}_1(\Bbb C)\cong\Bbb C^\times$ so the representation is essentially a complex-number-valued function. $\endgroup$ – whacka May 6 '15 at 0:14
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    $\begingroup$ @James Right. Although being 1-dim is not necessary for a rep to be unitary; I mentioned that "unitary" is more general, and provided some keywords to google if you're interested. Also you seem to be thinking more along the lines of orange's discrete example, where you have a linear map and apply it an integer-number of times, whereas my example draws out a continuous spiral as the argument from $\Bbb R$ is allowed to vary. The former (infinite abstract group) may be more suited to your thinking, but the latter (locally compact group) is more what representation theory looks like out there. $\endgroup$ – whacka May 6 '15 at 4:08
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Hint: Take $G = (\mathbb{Z}, +)$. Consider the representation $G\to GL(1, \mathbb{C}) = \mathbb{C}^{\times}$, $n \mapsto a^n$. If $|a| \ne 1$, then you will not have the equality.

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