Real representations of Lie algebra $\mathfrak{so}(3)$

Question

How does one construct an $n$-dimensional, irreducible, real-valued and non-zero representation of the three generators of the Lie algebra $\mathfrak{so}(3)$ for a given value of $n$?

Answer 1

A google search lead me to this paper

V. M. Gordienko
Matrix entries of real representations of the group O(3) and SO(3)
Siberian Mathematical Journal (2002), 43(1):36-46
DOI: 10.1023/A:1013816403253

which describes the construction of the real representations of $SO(3)$ using homogeneous polynomials $$ \begin{align} P^{(N)}(x,y) =& a_1 x^{2N} + a_2 x^{2N-1} y + a_3 x^{2N-2} y^2 + \ldots \\ & + a_{N+1} x^{N} y^{N} + \ldots + a_{2N} x y^{2N-1} + a_{2N+1} y^{2N} \end{align} $$ in $x$ and $y$ with $x,y \in \mathbb{C}$. Gordienko chooses the following basis in the vector space of polynomials $P^{(N)}(x,y)$ $$ \begin{align} e^{(N)}_n(x,y) =& i^{N+1}\sqrt{\frac{(2N+1)!}{2(N-n)!(N+n)!}} \\ &\times(x^{N-n}y^{N+n} - (-1)^{-n}x^{N+n}y^{N-n}) \quad \mathrm{if}\; n \le -1 \notag\\ e^{(N)}_{n=0}(x,y) =& i^{N} \frac{\sqrt{(2N+1)!}}{N!} x^N y^N \quad \mathrm{if}\; n = 0 \notag\\ e^{(N)}_n(x,y) =& -i^{N}\sqrt{\frac{(2N+1)!}{2(N+n)!(N-n)!}} \notag\\ & \times(x^{N+n}y^{N-n} + (-1)^{n}x^{N-n}y^{N+n}) \quad \mathrm{if}\; n \ge 1 \notag \end{align} $$ with $n = -N,\ldots,N$. The point of this particular construction is, that $e^{(N)}_n(x,y) = (e^{(N)}_n(-y,x))^*$.

Now she/he takes a generic rotation matrix from the group $SU(2)$, parametrized by the three real numbers $\alpha$, $\beta$ and $\gamma$ and writes with $\delta\equiv\sqrt{1-\alpha^2-\beta^2-\gamma^2}$ $$ M = \begin{pmatrix} \delta - i\gamma & \beta -i\alpha\\ -\beta - i\alpha &\delta + i\gamma \end{pmatrix} \equiv \begin{pmatrix} A^* & B^*\\ -B & A \end{pmatrix} $$ The rotation matrix $M$ is then applied on the 2-dimensional complex vector $(x,y)$, i.e. $$ (x',y') \equiv (x,y) \begin{pmatrix} A^* & B^*\\ -B & A \end{pmatrix}=(A^*x-By, B^*x+Ay) $$ For the "rotated" basis $e^{(N)}_n(x',y')$ Gordienko obtains $$ \begin{align} e^{(N)}_n(x',y') =& e^{(N)}_n(A^*x-By, B^*x+Ay) \\ =& \sum_{m=-N}^{+N} R^{(N)}_{n,m}(A,B) \; e^{(N)}_m(x,y)\notag \end{align} $$ The $(2N+1)\times(2N+1)$ matrix $R^{(N)}_{m,n}(A,B)$ is the $(2N+1)$-dimensional representation of SO(3), I was looking for.

Why is the matrix $R^{(N)}_{m,n}(A,B)$ real-valued? Gordienko shows that for basis vectors with the property $$ e^{(N)}_m(x,y) = (e^{(N)}_m(-y,x))^* $$ the entries of the representation matrices $R^{(N)}_{n,m}(A,B)$ are indeed real. Specifically, $$ \begin{align} e^{(N)}_m(x',y') =& e^{(N)}_m(A^*x-By, B^*x+Ay) \\ =& (e^{(N)}_m(-B^*x-Ay,A^*x-By))^* \notag\\ =& (e^{(N)}_m(B^*(-x)+A(-y),A^*x-By))^* \notag\\ =& (e^{(N)}_m(-y',x'))^* \notag\\ =& (\sum_{n=-N}^N R^{(N)}_{m,n}(A,B)\;e^{(N)}_n(-y,x))^* \notag\\ =& \sum_{n=-N}^N (R^{(N)}_{m,n}(A,B))^*\;(e^{(N)}_n(-y,x))^* \notag\\ =& \sum_{n=-N}^N (R^{(N)}_{m,n}(A,B))^*\;e^{(N)}_n(x,y) \notag \end{align} $$ Thus, $R^{(N)}_{m,n}(A,B) = (R^{(N)}_{m,n}(A,B))^*$ since the $e^{(N)}_n(x,y)$ are linear independent; in other words, the matrix $R^{(N)}_{m,n}(A,B)$ is real.

In my blog the three generators of $\mathfrak{so}(3)$ in 3, 5 and 7 dimensions obtained by this technique are given explicitly.