Representations of $SO_3(\mathbb{R})$ from $SU_2(\mathbb{C})$

Question

Define $V_n$ as the linear space of all homogeneous polynomials of degree $n$ in two variables $x$ and $y$. Define also the representation $\rho_n$ of $SL_2(\Bbb{C})$ on $V_n$ by: $$\rho_n(\begin{pmatrix}a&b\\c&d\end{pmatrix})f(x,y)=f(ax+cy,bx+dy)$$ This gives a representation of $SL_2(\mathbb{C})$. By restriction we get also a representation on $SU_2(\Bbb{C})$. I know that $SO_3(\mathbb{R})\cong SU_2(\mathbb{C})/\{\pm1\}$. There is a unique irreducible representation of $SU_2(\mathbb{C})$ in each (positive) dimension $m\in\Bbb{N}$. The fact is that only the odd dimensions are coming in the irreducible representations of $SO_3(\mathbb{R})$, but why?? Is this because of the isomorphism?

Thanks

Answer 1

The representations of $SO_3(\mathbb{R})$ are the representations of $SU_2(\mathbb{C})$ which are trivial on $\pm I$. One gets that $\rho_n(-I) = (-1)^n I$. Thus only the representations with $n$ even are representations of $SO_3(\mathbb{R})$.

Answer 2

Because of the isomorphism $SO_3(\mathbb{R}) \cong SU_2(\mathbb{C}) / \{\pm1\}$, in order for $\rho_n$ to give a representation on $SO_3(\mathbb{R})$, the element -1 must act trivially. But $\rho_n(-1) f(x,y) = f(-x, -y)$ so we must have $f(-x, -y) = f(x,y)$. This only happens when the degree $n$ is even.