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Tensor products can sometimes be decomposed. For example $$2\otimes 2 = 3 \oplus 1$$ $$4\otimes 3 = 6\oplus 4\oplus 2$$

Where the numbers represent the dimension of a representation of $SU(2)$ (I think) and so are equal to $2j+1$. I don't understand how this was done, and most internet resources use mathematical notation that is too complicated for me. I can see that multiplying the two on the LHS gives the same number as adding on the RHS, but I don't know how to find the numbers for the RHS. Thanks for any help!

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    I'm sure it's me, but we are talking tensor products of *what*? – 2017-02-11
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    @AndreasCaranti Good question, I'm not entirely sure myself. The numbers are dimensions of representations of groups, I think. – 2017-02-11
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    Thanks. That would have been my first guess. Now which particular groups are we talking - one could guess from the numbers, but I don't think this is the point. – 2017-02-11
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    @AndreasCaranti Does the group affect the answer? In this case, I think it relates to some special unitary group, $SU(n)$ maybe, although I'm not sure of the dimension. Probably $SU(2)$. – 2017-02-11
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    Yes, it does. For instance if we are talking representations of $S_{3}$, we have $2 \otimes 2 = 1 \oplus 1 \oplus 2$, where the two ones are different. – 2017-02-11
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    I can understand the numbers in those tensors/direct sums are dimensions or whatever. What I can't still understand is what **ring**, module or whatever are they taken from. Tensor product can be taken for some kinds of algebraic structures, yet I'm not sure where those "numbers" fit in... – 2017-02-11
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    @AndreasCaranti Good point. Then there is *no* tensor product (at least not as I, and maybe also you, understood it at the beginning) . – 2017-02-11
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    Perhaps the left hand side has been quotiented by some equivalence classes which gives the dimensions on the right? Think I have seen such stuff somewhere. – 2017-02-11
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    @AndreasCaranti Ok, fairly sure it's $SU(2)$ being talked about here, because it relates to spins and particle physics. – 2017-02-11
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    Then I think you can find an excellent answer [in this MSE post](http://math.stackexchange.com/questions/95797/product-of-su2-representations). – 2017-02-11
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    @AndreasCaranti Don't quite understand how to apply the method. So instead of a $\rho_i,\rho_j$ I have just the dimension, and the dimension is $2j+1$ in my case not $i+1$ as in that question. Having made that comparison, then if I calculate my $i,j$ they are $=\frac{1}{2}$, and for $k = i+j$ I do get $k=1$. But then the progression apparently continues with $k=i+j-2$, which isn't 3. – 2017-02-11
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    Let us consider the second case. Write $4 = u +1$, $3 = v + 1$, so $u = 3$, $v = 2$. Then $c_{1} = u+v= 5$, $c_{2} = u + v - 2 = 3$, $c_{3} = u + v - 4 = 1 = u - v$. Then $4 \otimes 3 = (c_{1} + 1) \oplus (c_{2} + 1) \oplus (c_{3} + 1) = 6 \oplus 4 \oplus 2$. – 2017-02-11
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    @AndreasCaranti Oh I see! Also answers my next question, which would have been when to stop. Thank you! – 2017-02-11
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    You're welcome. I recommend you +1 the excellent answer in the post I linked to. I've just done that myself. – 2017-02-11

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