I am going through the notes on QFT by Srednicki.
When describing fermions, from the very beginning he introduces the Lorentz Group and its algebra, and proves that it is equivalent to two copies of $SU(2)$, so that a representation is specified by two (half) integers, say $n,n'$ (see pages 213-214). He writes such a representation as $(2n+1,2n'+1)$.
For example, some important representations are $(1,1)$: scalar, $(2,1)$: left-handed spinor and so on. Some pages later (p. 217), he writes the relation $2\otimes 2=1\oplus 3$, which is just the usual result from addition of angular momentum. My problem is, some pages later (p. 219) he writes the following "group theoretic relation" $$(2,2)\otimes (2,2)=(1,1)\oplus(1,3)\oplus(3,1)\oplus(3,3).$$ I am having a hard time trying to understand this relation.
At first glance, it seems that we have to add four spin one-half momenta, that is, $$(2,2)\otimes(2,2)=2\otimes2\otimes2\otimes2.$$ If I go through the usual steps to construct such a sum, there is no way I get the expected result.
On the other hand, if I write $(2,2)=1\oplus 3$ and "distribute $\oplus$ over $\otimes$" as if they were actual products and sums, I get the result given by Srednicki, but I feel there is something wrong about that. Maybe I feel it's wrong just because there is something I'm missing or that I don't understand.
If this "distribute $\oplus$ over $\otimes$" is the right thing to do, I would really appreciate someone to explain why is that. If it's not the right thing, then I'd be glad if someone told me how should I deal with "group theoretic relations" like these, or where could I find some literature about this subject.
For example, on page 218, Srednicki writes $$(2,1)\otimes(1,2)\otimes(2,2)=(1,1)\oplus...$$
If my approach is the right one, then, as $$(2,1)\otimes(1,2)=1\oplus3=(2,2),$$ the full answer is $$(2,1)\otimes(1,2)\otimes(2,2)=(2,2)\otimes(2,2)=(1,1)\oplus(1,3)\oplus(3,1)\oplus(3,3).$$ Is this it?