Is there a relation between the irreducible representations of SO(2) and SO(3)? For instance, consider an n-by-n matrix representation of SO(3), $G$, if I restrict all of my rotations to have a common axis then the expressions for the matrix elements of $G$ simplify, but it is still a representative of SO(3) explicitly. However, since all of the rotations have a common axis this is equivalent to SO(2). Is there a straightforward way to relate the matrix elements of $G$ to those of an irrep (possibly m-by-m where $n\ne m$) of SO(2), call it $F$?

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    $\begingroup$ There are many copies of SO(2) inside SO(3) (essentially, they correspond to picking an axis): you can restrict a represention of the latter to one of them and obtain a representation of SO(2) But I really do not understand what it is you are asking... $\endgroup$ Commented Oct 19, 2012 at 0:00

1 Answer 1


$SO(3)$ has a unique finite dimensional irreducible rep. of each odd dimension $2l+1$. If we understand that this representation is taken with complex coefficients (i.e. is by matrices with complex entries) then we can diagonalize the action of $SO(2)$, and it acts by the characters $e^{im\theta}$, where $m$ ranges between $-l$ and $l$. (As Mariano notes in a comment above, there are lots of different choices of $SO(2)$ in $SO(3)$, according to which axis you chosse, but since any two axes can be rotated one into the other, any two copies of $SO(2)$ are conjugate, and so it is no real loss of generality just to fix one.)

Since any representation of $SO(3)$ can be decomposed as a sum of these irreducible ones, this gives a precise description of how the representations of $SO(3)$ decompose when restricted to $SO(2)$.


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