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It is mentioned in Wiki that the spaces $\mathcal{H}_{k}$ of spherical harmonics of degree $k$ give ALL the irreducible representations of $SO(3)$. Could anyone tell me where can I find the proof? Thanks!

EDIT: I am seeking for an elementary proof that dose not require too many big machineries in representation theory.

Community
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Syang Chen
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    Bröcker-tom Dieck is a very accessible source for such things. – t.b. Nov 04 '11 at 03:35
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    Dear Xianghong, What is your background? If you know a little bit about how to pass between Lie group representations and Lie algebra representations, then this is easily verified, using highest weight theory for the Lie algebra ${\mathfrak sl}_2$. Regards, – Matt E Nov 04 '11 at 04:17
  • @t.b.: Thanks! It's helpful. – Syang Chen Nov 05 '11 at 03:45
  • @Matt E: But representations of the Lie algebra are not necessarily the representations of the group? And how are the representations of $SO(3)$ related to the representations of $sl_2$? – Syang Chen Nov 05 '11 at 03:46
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    @XianghongChen: Dear Xianghong, That's right; the representations of the Lie algebra correspond to representations of the simply connected cover, e.g. in the case of $SO(3)$, reps. of its Lie algebra corresond to reps. of its simply connected cover $SU(2)$. But it's not hard to figure out which of these reps. actually come from reps. of $SO(3)$; one just uses the fact that $SO(3) = SU(2)/\langle \pm 1\rangle$. And to figure out the reps. of $SU(2)$, one uses the fact that is complexified Lie algebra is equal to $\mathfrak{sl}_2$. I think this is all explained in Fulton and Harris, ... – Matt E Nov 05 '11 at 13:50
  • ... and in many other places too. Regards, – Matt E Nov 05 '11 at 13:51
  • @MattE: Thanks for your explanation. I think I will need to learn more to fully understand it. – Syang Chen Nov 05 '11 at 16:47
  • @MattE: Can this be shown by some physical arguments? –  Jun 05 '14 at 13:19
  • @ramanujan_dirac: Dear ramanujan_dirac, I don't know off the top of my head, and I'm not sure I'm the right person to really answer this kind of question. Regards, – Matt E Jun 06 '14 at 05:52

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The proof is rather simple, just calculate characters (for rotations around OZ, since the axis does not affect the character) , and, using orthogonality theorem, note that all Fourier series coefficient for any other character are zero. But functions $\cos (l*\phi) (l - n)$, (where $n$ is an integer) form a complete set on $<0, \pi>$, so there are no more irreducible, unequivocal representations of $SO(3)$.

Simon Hayward
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Krystian
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  • It makes sense. Why doesn't this work for $SO(4)$? – Syang Chen Aug 28 '13 at 17:02
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    Characters of these representations for SO(3) are dependent only on one rotation parameter. This is not true for SO(4). You can find a basis of Lie algebra A1..A3, B1..B3 such that: [Ai, Aj] = constAk, [Bi, Bj] = constBk, [Ai,Bj] = 0 for any i, j. So SO(4) is isomorphic to SO(3) x SO(3). That means each conjugancy class is defined by two parameters, which makes further calculations much harder. Moreover, this method with Fourier series works for SO(3), only because the characters are some simple trigonometric functions, dependent on dimmension. In the general case it may not work. – Krystian Aug 31 '13 at 11:38