# Why are invariant tensors also Clebsch-Gordan coefficients?

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1. Mar 19, 2015

### Primroses

On one hand, in reading Georgi's book in group theory, I comprehend the invariant tensor as a special "tensor", which is unchanged under the action of any generators. On the other hand, CG decomposition is to decompose the product of two irreps into different irreps.

Now it is claimed that invariant tensors are Clebsch-Gordan coupling constants for the product of two irreps. Why?

Because they are invariants, we can use them to isolate invariant sub-spaces in the space of tensor product. Consider $SU(3)$ and, for example, decompose the tensor product $\psi^{i} \phi^{j}$ (of the fundamental representations, i.e. $[3] \otimes [3]$) into symmetric and antisymmetric tensors $$\psi^{i} \phi^{j} = S^{i j} + A^{i j} ,$$ where $2 S^{i j} = \psi^{i} \phi^{j} + \psi^{j} \phi^{i}$ is symmetric and, therefore, has 6 independent components. And $2 A^{i j} = \psi^{i} \phi^{j} - \psi^{j} \phi^{i}$ is antisymmetric and, therefore, has 3 independent components. However, it is easy to see that $A^{i j}$ transforms exactly like $\epsilon^{i j k} \chi_{k}$, where $\chi_{k}$ belongs to the conjugate (anti-fundamental) representation $[\bar{3}]$. Therefore, for some numerical constant $a$, you can rewrite your decomposition as $$\psi^{i} \phi^{j} = S^{i j} + a \epsilon^{i j k} \chi_{k} ,$$ which is basically $$[3] \otimes [3] = [6] \oplus [\bar{3}] .$$