# Spherical harmonics give all the irreducible representations of $SO(3)$?

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.

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Bröcker-tom Dieck is a very accessible source for such things. –  t.b. Nov 4 '11 at 3:35
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 4 '11 at 4:17
@t.b.: Thanks! It's helpful. –  Syang Chen Nov 5 '11 at 3:45
@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 5 '11 at 3:46
@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 5 '11 at 13:50

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 $SO(3)$.
It makes sense. Why doesn't this work for $SO(4)$? –  Syang Chen Aug 28 '13 at 17:02