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I recently thought about representations of the orthogonal group $O(2)$ and found the one dimensional representations a bit confusing.

We have that $SO(2)$ is a normal subgroup of $O(2)$. In fact, we can identify $SO(2)$ with the unit circle $S^{1}$, which is an abelian group and consequentially only has one-dimensional representations. We can say more about these representations since any continuous group homomorphism

$\rho: S^{1} \rightarrow \mathbb{C}^{\times}$

will have an image that is compact and consequentially lives inside of $S^{1}$, thus we are left what we really have is a continuous group homomorphism $\rho: S^{1} \rightarrow S^{1}$, which can be thought of as a map from $\mathbb{R}/\mathbb{Z} \rightarrow S^{1}$ and hence constitutes a map $ \rho \circ \pi : \mathbb{R} \rightarrow S^{1}$. We have that $\rho \circ \pi$ will then be a map that sends $ x \mapsto e^{iax}$ See here for proof. We then have that $\rho \circ \pi$ contains the integers in its kernel and thereforer $e^{ia}=1$ which means $a=2\pi{k}$, it follows that $\rho$ will be a map that sends $z \mapsto z^{k}$. For each $k$, we obtain a different one-dimensional representation.

I am essentially wondering how the ideas above extend to finding one dimensional representations of $O(2)$

We have that since $SO(2)$ is a subgroup and in fact a maximal tori, we must have that a continuous one-dimensional representations of $O(2)$ must restrict to a power map on $SO(2)$. More generally, since $O(2)$ is compact, the image of such a map must live in the unit circle.

We have that in $O(2)$, that a rotation by $\theta$ and a rotation by $-\theta$ are conjugate to each other. This point confuses me as far as computing trace.

Any help on this is appreciated. Thanks for the help.

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Hint: any homomorphism from $O(2)$ to an abelian group must have kernel containing the commutator subgroup of $O(2)$.

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