I have been given that there is a 2:1 surjective homomorphism from $SU(2)$ to $SO(3)$ with kernel $\{-I,+I\}$ and that $V_n$ are irreducible representations of $SU(2)$ where $V_n$ is the space of homogeneous polynomials in two variables over the complex numbers, degree $n$ with $SU(2)$ action given as follows:
If $F(z,w)$ is a polynomial in $V_n$ and $A$ is a matrix in $SU(2)$ then $A^{-1}$ acts on the column vector $(z,w)$ by multiplication and we obtain a new homogeneous polynomial $F(A^{-1}(z,w))$.
For $n=2m$ even, the action of $-I$ is trivial on $V_{2m}$ so that we obtain an $SO(3)$ action on $V_{2m}$. Moreover, $V_{2m}$ has a real structure induced by the map $T$ on $V_{2m}$ which acts by $T[F(z,w)] = \overline{F(w,-z)}$ so that we obtain a real, $SO(3)$ representation $W_m$, which is the subspace of $V_{2m}$ preserved by $T$.
My question is, from this how do I find the character of an element of $SO(3)$ on this representation?
I know that every element is in a maximal torus, and as the representation is the same as that of $SU(2)$ we may restrict our focus to the maximal torus of diagonal matrices (as characters are invariant under conjugation, and all maximal tori are mutually conjugate).
It is easy to find the character of $V_n$ as in: https://en.wikipedia.org/wiki/Representation_theory_of_SU(2)#The_characters
and the character of $W_m$ should be the same as in page 20 of: http://www.cmth.ph.ic.ac.uk/people/d.vvedensky/groups/Chapter8.pdf
But I cannot see how to relate the two.
The homomorphism $\mathrm{SU}(2)\to\mathrm{SO}(3)$ restricts to a map
$$ \begin{bmatrix} e^{i\theta} & 0 \\ 0 & e^{-i\theta} \end{bmatrix} \mapsto \begin{bmatrix} \cos2\theta & -\sin2\theta & 0 \\ \sin2\theta & \phantom{-}\cos2\theta & 0 \\ 0 & 0 & 1 \end{bmatrix} $$
of max tori, or in this case circle subgroups (with respect to some orthonormal basis of $\mathbb{R}^3$).
Set $\varphi=2\theta$. Thus, the character of a rotation by $\varphi$ will match the character of $\mathrm{diag}(e^{i\theta},e^{-i\theta})$ (whose real and complex characters match). The eigenvalues of the operator acting on $V_n$ are $e^{2n\theta i},e^{(2n-2)\theta i},\cdots$ so the trace is given by $\chi=e^{n\varphi i}+\cdots+e^{-n\varphi i}=\sin((n+\frac{1}{2})\varphi)/\sin(\frac{1}{2}\varphi)$.