It is known that the irreducible representations of $\operatorname{SO}(3)$ are of dimension $1$, $3$, $5$, etc. Can anyone give a proof that there is no two-dimensional irreducible representation of $\operatorname{SO}(3)$?
It is known that $\operatorname{SO}(3)$ and $\operatorname{SU}(2)$ share the same Lie algebra. The group $\operatorname{SU}(2)$ and its Lie algebra $\mathfrak{su}(2)$ have irreducible representations of dimension $1$, $2$, $3$, etc. So, why cannot the $2$-dim irreducible representation of $\operatorname{SU}(2)$ or $\mathfrak{su}(2)$ carries over to $\operatorname{SO}(3)$?
If $V$ is a $2$-dimensional irreducible representation of $SO(3)$, then it pulls back along the double cover $SU(2) \to SO(3)$ to a $2$-dimensional irreducible representation of $SU(2)$ on which the center $\{ \pm 1 \}$ acts trivially. But $SU(2)$ only has one $2$-dimensional irreducible representation, and the center does not act trivially on it. So there is no such $V$.