Assume you have a Lie algebra $G$ and a Clebsch-Gordan series $A\bigotimes{B}=C\bigoplus{D}\bigoplus...$
Assume you have a Lie algebra $g$ and a Clebsch-Gordan series $a\bigotimes{b}=c\bigoplus{d}\bigoplus...$
Now consider the direct sum $G+g$ and their irreps $(A,a)$ and $(B,b)$.
$(A,a)\bigotimes{(B,b)}=(C,c)\bigoplus{(C,d)}\bigoplus{(D,c)}\bigoplus...$ or what?
At least if $G=g$ it can't work with the distributive law that nicely since then some vector spaces coincide. (If you want to make me very happy, use the adjoint of $G=A_2$ as example - it costed me months to realize it's not $(11,11)$ but $(00,11)+(11,00)$. Or such. That would be my bonus question - is the adjoint of $G+g$ something like $(E,a)+(A,e)$ with $E,e$ the trivial and $A,a$ the adjoint irreps?)
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asked May 27, 2014 at 12:24
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$\begingroup$ I am pretty sure that the distributive law works. What do you mean by some spaces coinciding? The sum is direct, so the Lie subalgebras $G$ and $g$ are isomorphic but distinct, and a representation can distinguish the action of $G$ on it from the action of $g$ on it. (I am assuming that the Clebsch-Gordan series is the decomposition of a tensor product of two irreps in a direct sum of irreps.) $\endgroup$– darij grinbergCommented May 27, 2014 at 12:35
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$\begingroup$ And the answer to your bonus question is "yes" -- the action of $G+g$ on $G+g$ is componentwise Lie bracketing, so that the subspace $G$ is isomorphic to the representation $(A, e)$, and the subspace $g$ is isomorphic to the representation $(E, a)$. $\endgroup$– darij grinbergCommented May 27, 2014 at 12:37
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$\begingroup$ @Darij: THX. Here is an example the intuitive law can't work when G=g. SU2=A1 defining dimension already works. (I label by dimension, not J.) 2*2=1+3. (2,2)*(2,2)=1+3+3+9 (so far, so good). (2,2,2)*(2,2,2)=1+9+27+27. With simple distributive law, you would get eight terms! (But (1,1,3),(1,3,1),(3,1,1) somehow fall together, and so do (1,3,3) etc., which explains the dimensions. At least they do for the span of the clebsch - the R matrix fulfils a cubic equation and that's what relevant to me.) $\endgroup$– Hauke ReddmannCommented May 27, 2014 at 12:46
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$\begingroup$ How do you get (2,2,2)*(2,2,2)=1+9+27+27 ? $\endgroup$– darij grinbergCommented May 27, 2014 at 12:49
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$\begingroup$ By reading it somewhere :-) Myself, I of course would have used the naive distributive law. But here is an even more striking example from 3A1: Since it's on the Vogel plane and a member of the quarternion series, one just needs to look up the relevant (Westbury) paper to get adjoint^2: 9^2=1+9+2+15+27+27. Dissecting this: 9 is 113+131+311, 15 is 115+151+511, 1 is 111, 2 is the other two 111+111 (from 113*113), 27 is 133+313+331 (from 113*113), other 27 is 133+313+331 (from 113*131). The distributive law gives the building blocks, but "clustering" of irreps is completely wacky. $\endgroup$– Hauke ReddmannCommented May 28, 2014 at 8:53
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