Lorentz homogeneous group and observables For generators of the Lorentz group we have the following algebra:
$$
[\hat {R}_{i}, \hat {R}_{j} ] = -\varepsilon_{ijk}\hat {R}_{k}, \quad [\hat {R}_{i}, \hat {L}_{j} ] = -\varepsilon_{ijk}\hat {L}_{k}, \quad [\hat {L}_{i}, \hat {L}_{j} ] = \varepsilon_{ijk}\hat {R}_{k}. 
$$
For the splitting of algebra, we can introduce operators 
$$
\hat {J}_{k} = \hat {R}_{k} + i\hat {L}_{k}, \quad \hat {K}_{k} = \hat {R}_{k} - i\hat {L}_{k}.
$$ 
So
$$
[\hat {J}_{i}, \hat {J}_{j} ] = -\varepsilon_{ijk}\hat {J}_{k}, \quad [\hat {K}_{i}, \hat {K}_{j} ] = -\varepsilon_{ijk}\hat {K}_{k}, \quad [\hat {J}_{i}, \hat {K}_{j}]  = 0.
$$
So, each irreducible representation of Lie algebra is characterized by $(j_{1}, j_{2})$, where $j_{1}$ is max eigenvalue of $\hat {J}_{3}$ and $j_{2}$ is max eigenvalue of $\hat {K}_{3}$.
Then I can classify objects that transform through the matrices of the irreducible representations,
$$
\Psi_{\mu \nu}' = S^{j_{2}}_{\mu \alpha }S^{j_{2}}_{\nu \beta}\Psi_{\alpha \beta},
$$
where $S^{j_{i}}_{\gamma \delta}: (2j_{i} + 1)\times (2j_{i} + 1)$.
For $(0, 0)$ I have scalar field, for $\left(\frac{1}{2}, 0\right); \left(0; \frac{1}{2}\right)$ I have spinor, for $(1, 0); (0, 1)$ I have 3-vectors $\mathbf a, \mathbf b -> \mathbf a + i\mathbf b$ creating antisymmetrical tensor etc. 
Also, for scalar $j_{1} + j_{2} = 0$, for spinor - $\frac{1}{2}$, for tensor - $1$. So, the question: is sum $j_{1} + j_{2}$ experimentally observed? Is it connected with a spin?
 A: Yes, a representation labeled by $(j_1,j_2)$ corresponds to the total spin $j_1+j_2$, (rigourously speaking of spin needs that one of $j_1$ or $j_2$ is zero) and if $j_1=j_2$, this is a real representation, but you may have a representation which is a sum or irreductible representations ,  some examples:
$(\frac{1}{2},0)$ corresponds to a left-handed Weyl spinor
$(0,\frac{1}{2})$ corresponds to a right-handed Weyl spinor
$(\frac{1}{2},0) + (0,\frac{1}{2})$, is the  Dirac bi-spinor
$(\frac{1}{2},\frac{1}{2})$ corresponds to a  Lorentz vector.
$(1,0) + (0,1)$, is the  electromagnetic field representation
More generally, if the complex conjugate of a representation (interverting in all terms $j_1$ and $j_2$) is the same as the representation, then the representation is real.
For instance, the Dirac representation, or the electromagnetic field representation, are real representations.
A: Yes, the comination $j_1 + j_2$ determines the spin of the particle. Note however, that this is an addition of angular mementum which may be complicated.
Furthermore, you can count the degrees of freedom:
In $(j_1, j_2)$, each contribute $2j_1 + 1$ states and we construct a tensor product, so $(j_1, j_2)$ gives $(2j_1 + 1) * (2j_2 + 2)$ degrees of freedom. For the vector we have $(1/2, 1/2) \mapsto 2 * 2 = 4$ degrees of freedom. 
If the representation is reducible, i.e. of the form $(j_1, j_2) + (k_1, k_2)$, then you simply add the d.o.f. you get from each pair. The dirac spinor has $(1/2, 0) + (0, 1/2) \mapsto (2) + (2) = 4$ degrees of freedom, the field-strength tensor has $(1, 0) + (0, 1) \mapsto 3 + 3 = 6$ d.o.f.
As a sidenote: the representations $(1, 0)$ do not corresond to vectorlike degrees of freedom, but rather to antisymmetric self-dual tensors. $(0, 1)$ is the antisymmetric anti-self-dual tensor. The vector (and the only way to get a vector out of this) is $(1/2, 1/2)$!
