I am following this statement / answer given in https://physics.stackexchange.com/a/230266/42982:
About degrees of freedom (dof):
"There are two spinor representation of SO(3,1) (1/2,0) and (0,1/2) denoted by dot and no-dot indices. Vector (spin 1) representation in spinor indices is (1/2,0) X (0,1/2) = (1/2,1/2) which has 4 dofs. If the theory is massless, the theory is gauge invariant, there is one gauge dof, which can be eliminated by the gauge fixing condition. One dof is time-like one, which is not physical. So 4-1-1 =2 physical dofs remain. For symmetric tensor we have (1/2,1/2)X (1/2,1/2) = (1,0)+ (0,1)+ (1,1)+(0,0) The irreducible symmetric representation is just 9-dimensional. The general coordinate transformation eliminate 4 dofs (which essentially is 4 space-time coordinates). So the massive spin 2 has 9-4=5 dofs. The massless spin 2 theory will be gauge invariant, where we can eliminate 1 dof. Two dofs are time-like which do not propagate. So only 5-1-2 dofs remain. We can extend the argument to arbitrary dimension. "
Which says (1/2,1/2) $\times$ (1/2,1/2) = (1,0)+ (0,1)+ (1,1)+(0,0) The irreducible symmetric representation is just 9-dimensional. Is this because the irreducible symmetric representation is simply that of (1,1), which is 3 $\times$ 3 =9-dimensional representation?
Is there some mathematical precise way to explain:
Why spin-1 boson has 4-1-1 =2 physical dofs remain?
Why spin-2 boson has 9-4-2-1 =2 physical dofs remain?