Does dimension of irreducible representations of the double cover $SU(2)$ of the 3D rotation group define spin of particle? In quantum field theory, does dimension of irreducible representations of  the double cover $SU(2)$ of the 3D rotation group conclusively define spin? In other words, Is spin 1 particle only thing that vector field in 4D spacetime can generate? Does spinor field always generate spin 1/2 particle?
 A: *

*In relativistic theories, an irreducible representation with spins $(j_L,j_R)\in\frac{1}{2}\mathbb{N}_0\times\frac{1}{2}\mathbb{N}_0$ for the [double cover $\tilde{G}_1=SL(2,\mathbb{C})\times SL(2,\mathbb{C})$ of the] complexified proper Lorentz group $G_1=SO(1,3;\mathbb{C})$ decomposes into a direct sum $$(j_L,j_R)~=~\bigoplus_{j=|j_L-j_R|}^{j_L+j_R} j$$ of irreducible representations with spins $j\in\frac{1}{2}\mathbb{N}_0$ for the [double cover $\tilde{G}_2=SL(2,\mathbb{C})$ of the] restricted Lorentz group $G_2=SO^+(1,3;\mathbb{R})$.

*In turn, an irreducible representation with spin $j\in\frac{1}{2}\mathbb{N}_0$ for the [double cover $\tilde{G}_2=SL(2,\mathbb{C})$ of the] restricted Lorentz group $G_2=SO^+(1,3;\mathbb{R})$ is also an irreducible representation with the same spin $j\in\frac{1}{2}\mathbb{N}_0$ for the [double cover $\tilde{G}_3=SU(2)$ of the] 3D rotation group $G_3=SO(3,\mathbb{R})$.

*Examples. A Lorentz vector is a $(1/2,1/2)$ representation. Left & right Weyl spinors are $(1/2,0)$ & $(0,1/2)$ representations, cf. e.g. this related Phys.SE post. 

*The particles corresponding to a field are usually classified in terms of the little group, cf. e.g. this & this related Phys.SE posts and links therein. 
