Tensor product of representations of Lorentz group Where is the rules for tensor product of representations of Lorentz group $(a,b)\otimes (c,d)$ without decomposition of one of these in orthogonal sum?
 A: Your  answer is there for the  appreciating in WP 3.2.1.2, and all you need is understand the notation. (Several people attach to Ch. 5, of v I of Weinberg's QFT text.) There may be a confusion of the tensor product ⊗ used in both linking the Lorentz Group left ideal A to the right one B but also, again with common rotation angles!, to the Kronecker multiplication of different su(2) representations involved ("adding spins"). So in Kronecker multiplying two Lorentz representations, you consider "synchronized swimming" rotations in both the spaces of A and B, as well as the tensor space of the spins added. There may be an elegant notation around stressing the distinction, but it might not be worth the trouble, as essentially all operators, see below, commute with each other and act on different subspaces.
Consider a rotation around the x axis, so the generator $J_x$, in the (m,n) representation, of dimensionality N=(2m+1)(2n+1), so (reducible) N×N matrices, 
$$
  \pi_{(m,n)}(J_x)  =1\!\! 1_{(2m+1)} \otimes J^{(n)}_x + J^{(m)}_x \otimes 1\!\!1_{(2n+1)}   .
$$ 
The N×N matrix is the sum of two such: the first that leaves the (2m+1) dimensional space alone and rotates the  (2n+1)-vectors, and, symmetrically,  the one that rotates (2m+1) -vectors leaving  (2m+1) -vectors alone. The angle of rotation is the same for $\pi_{m,n} ; J_x^{(m)}; J_x^{(n)}$. I chose a rotation, since the Lorentz ideals treat it identically, unlike boosts, for instance.
Now, adding two reps, spin m and n would yield a composite representation coproduct  Δ of dimensionality N again, via the very same formula (!). For example, if you added two spin 1/2 (doublets), as proposed in your comment, you'd get a coproduct
$$
J^{(1/2)}_x=\frac{1}{2} \sigma_x  ~~~~~~ \leadsto ~~~~~\Delta_{(1/2,1/2)}(J_x)= \frac{1}{2}
 \begin{pmatrix} 0&1&0&1\\  1& 0&1&0\\     0&1&0&1\\  1& 0&1&0                   \end{pmatrix}.
$$
It might not show it, but, by a change of basis similarity transformation,  this generator is reducible to a spin 0 singlet rep, and a spin 1 triplet rep,
$$
 J_x^{(1)}= \frac{1}{\sqrt 2}
 \begin{pmatrix} 0&1&0 \\  1& 0&1 \\     0&1&0                     \end{pmatrix}.
$$
Adding yet another spin 1/2 doublet to this triplet, as described above,  yields a 6×6 matrix, which, Clebsching out a doublet, nets you a spin 3/2 quartet, 
$$
 J_x^{(3/2)}= \frac{1}{  2}
 \begin{pmatrix} 0&\sqrt 3 &0 &0 \\   \sqrt 3 & 0&2 & 0  \\ 0&2&0&  \sqrt 3  \\ 0&0 &  \sqrt 3   &0                             \end{pmatrix}.
$$
So, for your (1/2,1/2)⊗(1/2,1/2)⊗(1/2,1/2)=(3/2,3/2)+ ... comment example, the triple Kronecker product Lorentz rep would be a 16-dimensional one, as detailed,
$$
  \pi_{(3/2,3/2)}(J_x)  =1\!\! 1_{(4)} \otimes J^{(3/2)}_x + J^{(3/2)}_x \otimes 1\!\!1_{( 4)}   .
$$
(a,b)⊗(c,d) =(a⊗c,b⊗d) has been applied, so 
(1/2,1/2)⊗(1/2,1/2)⊗(1/2,1/2)=(1/2⊗1/2⊗1/2,1/2⊗1/2⊗1/2).  
Observe how the reducible dross of 48 states has beed discarded in Clebsching: the ... ellipsis reps. 
You started out by composing three 4-vectors, 64 states, and you ended up with a mere 16-tuplet! neat, huh?
