How can the trivial representation not be in some $R\otimes R$?

Where do I foul up?

In (e.g.) $SU(3)=A_2$ the tensor product $(1,0)\otimes (1,0)=(2,0)+(0,1)$. No $(0,0)$. But on the other hand still $(1,0)\otimes (0,0)=(1,0)$. This immediately begs the question

1. what happened to Schur's lemma and
2. imagine some 6j symbol $(ABC|DEF)$ where $A=B=(1,0)$, $C=(0,0)$. Is $ABC$ no admissible triangle and the 6j vanishes?

My only idea is that the Dynkin diagram is symmetric and you first must symmetrize/antisymmetrize over the reps to get the correct ones, i.e. the proper irreps in this case would be $((1,0)+(0,1))/V2$ and $((1,0)-(0,1))/V2$. (And what does this do to the dimensions? Same as $(1,0)$ and $(0,1)$, I think.)

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I'm not familiar with the notation, but in general there's no reason that, for a representation $V$, $V \otimes V$ will contain a copy of the trivial representation. (It is true that $V \otimes V^{\vee} \simeq V \otimes \overline{V}$ will, though.) – Akhil Mathew Aug 29 '12 at 15:02
I don't think your question is comprehensible. Could you at least cite Schur's lemma and explain why it has any bearing on this question. – Marc van Leeuwen Aug 30 '12 at 9:01