Hi. In Georgi's book page 143, eqn. (10.29) he gives an example of decomposing a tensor product into irreps:(adsbygoogle = window.adsbygoogle || []).push({});

[tex]

u^iv_k^j=\frac{1}{2} \left( u^iv_k^j+u^jv_k^i-\frac{1}{4}\delta_k^iu^\ell v_\ell^j-\frac{1}{4}\delta_k^ju^\ell v_\ell^i \right)\\

+\frac{1}{4} \varepsilon^{ij\ell} \left( \varepsilon_{\ell m n}u^m v_k^n + \varepsilon_{kmn}u^m v_\ell ^n \right)\\

+\frac{1}{8} \left( 3 \delta_k^i u^\ell v_\ell^j - \delta_k^j u^\ell v_\ell^i \right)

[/tex]

I have a few novice questions about this.

1. In the first line, he symmetrizes and removes the trace. Looking at the second line, it can be written as [itex] \frac{1}{2} (u^i v_k^j-u^jv_k^i) - \frac{1}{4} \delta_k^i u^\ell v_\ell ^j + \frac{1}{4} \delta_k^j u^\ell v_\ell ^i [/itex] which corresponds to also removing the trace from the anti-symmetric part. Why do we need to also remove the trace from the anti-symmetric part? aren't the irreducible representations - traceless symmetric, anti-symmetric, and trace?

2. Why is the trace [itex] v_j^j [/itex] zero?

3. How do we get the dimensions of the irreps. For example, how do we figure out that the first line corresponds to [itex] \mathbf{\bar{15}} [/itex]?

Thanks.

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# A Decomposition of tensors into irreps (Georgi's book)

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