I've read that one may prove that all irreducible representations of $SO(3)$ are tensor product representations of the fundamental representation (or tensor product representations of the spin 1/2 representation if we're allowing projective reps). How would one go about proving this?

I've only ever done representation theory from a physicist's viewpoint, so I feel like I'm missing the tools to rigorously show it! Many thanks.

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    $\begingroup$ Read the beginning of the book by Fulton and Harris on representation theory. This question is not quite adopted to a site such as this, because an answer is bound to be chapter-long... $\endgroup$ – Mariano Suárez-Álvarez Dec 21 '12 at 17:04
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    $\begingroup$ @MarianoSuárez-Alvarez Now you've done it. The challenge is to fit this in a single screen (1080 pixels). $\endgroup$ – David E Speyer Dec 21 '12 at 17:09
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    $\begingroup$ I honestly think that a «very terse summary» would not help much. Books are written for a reason. Your question suggests that you want a «rigorous» approach: the rigorous approach is to study the subject a bit in detail; I think that familiarity with the subject (which, in all likelyhood you alread have) would make reading the first few chapters of Fulton a Harris a breeze, for example. $\endgroup$ – Mariano Suárez-Álvarez Dec 21 '12 at 17:54
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    $\begingroup$ @EdwardHughes ... Maybe! Just wanted to be sure that less-informed readers were not misled. Yes, that terse phrase is often used, but it might give the wrong impression if someone didn't know. :) $\endgroup$ – paul garrett Dec 21 '12 at 19:19
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    $\begingroup$ Dear Edward, Take the symmetric square of the standard three dim'l rep. It contains an invariant line, spanned by the inner product on the three dim'l rep. (this is a symmetric two-tensor, which we can think of as an element of the symmetric square). The orthogonal complement of this line is five-dimensional, and irred. In general, there is one irrep. of $SO(3)$ of each odd dimension, and none of even dimension. (You probably know about spherical harmonics, and you can realize these irreps. in spaces of spherical harmonics via the action of $SO(3)$ on the sphere.) Regards, $\endgroup$ – Matt E Dec 22 '12 at 16:12

I think I can help with this one, too, though I'm only going to give a sketch. I think the details are worked out somewhere to the lecture notes I linked to in my comment on your last question.

First you should understand why the corresponding result (that every irrep is the tensor product of the fundamental representation) for $SU(2)$ is true. To begin, note that any representation of $SU(2)$ restricts to a representation of a $U(1)$ subgroup, so by decomposing a $SU(2)$ irrep into $U(1)$ irreps we see that every irrep of $SU(2)$ has the form $e^{i\theta k_1} \oplus \ldots \oplus e^{i \theta k_n}$. Here the $k_i$'s are integers, and conjugating with the matrix

$$\left(\begin{array}{cc} 0 & 1 \\ -1 & 0 \end{array}\right)$$

shows that $-k_i$ is an exponent whenever $k_i$ is. If the chosen $U(1)$ subgroup is generated by the basis vector $X_1$ for the Lie algebra of $SU(2)$, then it turns out that the eigenvalues of $X_1$ are simply $k_i/2$ (the $2$ comes from the fact that $SO(3)$ is a simply connected Lie group with the same Lie algebra as $SU(2)$ and $SU(2)$ double covers $SO(3)$). Using the other two directions in the Lie algebra of $SU(2)$, one can construct operators $X_+$ and $X_-$ (the "raising and lowering" or "creation and annihilation" operators) such that $X_+$ sends any $k/2$-eigenvector of $X_1$ to either $0$ or a $k/2 + 1$-eigenvector of $X_1$ and similarly for $X_-$.

This tells you exactly what irreps of $SU(2)$ have to look like: pick an eigenvector $v$ of $X_1$ whose eigenvalue $n/2$ is as large as possible (a "highest weight" vector), and repeatedly apply $X_-$. You'll generate a list of eigenvectors of $X_1$ with eigenvalues $-n/2, -n/2 + 1, \ldots, n/2-1, n/2$. None of this depended on which $X_1$ we started with, so in fact this sequence of numbers determines the irrep up to conjugation. Moreover any such sequence can be obtained by tensoring the fundamental representation (direct calculation), so we're done.

If you believe all of this works for $SU(2)$, then $SO(3)$ is easy. Every $SO(3)$ irrep gives rise to a $SU(2)$ irrep via the double cover $SU(2) \to SO(3)$, so we have already constrained the list of possible $SO(3)$ representations dramatically. The list can be restricted further by noting that a representation of $SU(2)$ descends to a representation of $SO(3)$ if and only if $-1$ goes to the identity, and this holds only when the largest eigenvalue $n/2$ is an integer, i.e. the dimension of the representation is odd.

  • $\begingroup$ Thank you - again that is marvellous. You're clearly my representation theory guru! $\endgroup$ – Edward Hughes Dec 21 '12 at 18:17
  • $\begingroup$ “The $2$ comes from the fact that $SO(3)$ is a simply connected Lie group with the same Lie algebra as $SU(2)$ ...” I think some of the $SU(2)$s in and around this sentence should be $SO(3)$s and vice versa, but not exactly sure which. $\endgroup$ – Alex Shpilkin Feb 12 '18 at 17:05
  • $\begingroup$ @AlexShpilkin :facepalm: Yes, $SO(3)$ and $SU(2)$ should be switched in the part you quoted. I think it's fine everywhere else. Not really sure if I should edit a 5 year old answer or just leave this comment. $\endgroup$ – Paul Siegel Feb 12 '18 at 21:04
  • $\begingroup$ @PaulSiegel Don’t see anything wrong with correcting the answer. I was doubting myself because the statement comes from the part of your answer where you (supposedly) only discuss $SU(2)$, but somehow the “Rodrigues’ $2$” still comes up. $\endgroup$ – Alex Shpilkin Feb 12 '18 at 22:18

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