I am reading A. Zee's group theory book. In the chapter of spinor representation, page 416, he was trying to describe how to decompose of the product of spinor representation into irreducible representation of $SO(2n)$ group.

In particular, he showed that you can decompose $16^{+}\otimes 16^{+}$ into $10\oplus120 \oplus 126$ for $SO(10)$ by constructing the following thing $$\psi^{T}(\gamma_i\gamma_j\gamma_k)\psi$$ Where $\psi$ are spinors that have 16 independent components, corresponding to $16^{+}$. He then showed that this construction would transform like a vector for $SO(10)$, finally making the decomposition.

The thing I don't understand here is that this seems to be vastly different from the decomposition of direct product of representation, for example, the CG coefficient thing. There, for example, when we consider $10\otimes 10$, we form a two-index tensor via direct product of two vectors, and then consider how this tensor would fall into different pieces when considering representation.

The way we put together two $16^+$ seems so unnatural when you put additional $\gamma$ matrix in between. I initially thought that when you write $\otimes$, you should actually consider how $\psi_{\alpha}\psi_{\beta}$, which is a direct product, will decompose into irreducible representation. Why isn't this the case here?

  • 2
    $\begingroup$ Are not the $\gamma$ matrices playing the role of C-G coefficients as they do in $\psi^\dagger \gamma^0 \gamma^\mu\psi$ being a vector? $\endgroup$
    – mike stone
    Jan 30 at 14:21
  • $\begingroup$ What he's trying to do is explain the 'Fierz decomposition' of a product of spinors by noting that the answer has to be a sum of things involving spinors contracted against anti-symmetrized products of gamma matrices - here he is discussing what the coefficients in the Fierz expansion must be (without making it clear this is what he's doing) in a kind of direct way. If you go over inserting the 'Fierz Identity' between two spinors in a QFT book and compare to his discussion it will hopefully become clearer and look more like the usual approach with C-G coefficients. $\endgroup$
    – bolbteppa
    Jan 30 at 14:31


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