# 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?

• WP. What is your specific problem with $(a\otimes c, b\otimes d)$? Commented Apr 13, 2020 at 15:48
• For example $$(1/2,1/2)\otimes (1/2,1/2)\otimes (1/2,1/2)=(3/2,3/2)$$ Commented Apr 13, 2020 at 16:27
• I was reading also physics.stackexchange.com/questions/113798/… Commented Apr 13, 2020 at 16:36
• So you wish to compose symmetrically three 4-vectors to a 16-dimensional Lorentz rep by shedding 48 d.o.f's? Try working out directly what $J_x$ would look like in this rep. Commented Apr 13, 2020 at 16:36
• Where I can find that group (a,b)⊗(c,d) =(a⊗c,b⊗d) ? Commented Apr 13, 2020 at 16:46

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).