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I am looking for a list of the irreducible representations of O(2). Could someone please provide a reference?

EDIT: I am particularly interested in the representations on IR^2 (irreducible or not)

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    $\begingroup$ Real or complex? $\endgroup$
    – S. Carnahan
    Commented Sep 2, 2010 at 12:00
  • $\begingroup$ For SO(n,C), see Theorem 2 here: books.google.com/books?id=Mi1fy70II8cC&pg=PA221 $\endgroup$ Commented Sep 2, 2010 at 13:02
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    $\begingroup$ If you are interested in complex irreps, then the book mentioned in the (deleted) answer by mathphysicist has the answer: it is Theorem 11.3, except that as stated the theorem is not right because it fails to mention that this is true for complex irreps. $\endgroup$ Commented Sep 2, 2010 at 13:51
  • $\begingroup$ @mathphysicist: I would undelete your answer and simply point out the above "caveat". $\endgroup$ Commented Sep 2, 2010 at 13:52
  • $\begingroup$ @Jose: I've just undeleted the answer. $\endgroup$ Commented Sep 2, 2010 at 15:14

2 Answers 2

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I just see now, that the issue is appearently real representations. I consider complex representations. I not experienced with real representations and whether my strategy works there as well.

You can induce from $SO(2)$. Define on $SO(2)$ the rep $\epsilon_n: \theta \mapsto e^{i \theta n}$. Let $\rho_n$ be the induced one, then $\rho_n$ is irreducible if $n \neq 0$. You have $\rho_n \cong \rho_{-n}$ and $\rho_{0} = 1 \oplus det$. These are up to isomorphism all irreducible representations.

Reference: Traces of Hecke operator by Knightly and Li.

A proof also is in my thesis: http://ediss.uni-goettingen.de/bitstream/handle/11858/00-1735-0000-000D-F074-7/palm.pdf?sequence=1 on pg 101.

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You can look into this section of the book Group theory in physics by Wu-Ki Tung to begin with.

EDIT: Theorem 11.3 from this book works for complex irreps only (see Jose's comments to the answer and to the question).

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    $\begingroup$ This books contains the assertion that all irreps of SO(2) are one-dimensional. So reader beware... $\endgroup$ Commented Sep 2, 2010 at 10:57
  • $\begingroup$ @Jose: Thanks for pointing this out. I guess I'd then better remove my answer. $\endgroup$ Commented Sep 2, 2010 at 12:49
  • $\begingroup$ @Jose: I hope the answer is OK now. If it isn't, please feel free to correct me. $\endgroup$ Commented Sep 2, 2010 at 15:11

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