Let \[ S(\omega) \equiv e^{\frac{i}{4}\omega_{ij}\sigma_{ij}}, \qquad \Gamma_{ijk} \equiv \gamma_i\gamma_j\gamma_k. \] Then the right-handed spinor transforms as \[ \psi_R \to S\,\psi_R, \] so its transpose transforms as \[ \psi_R^T \to \psi_R^T S^T. \]
Start from the bilinear \[ B \equiv \psi_R^T\,C\,\Gamma_{ijk}\,\psi_R. \] Under an \(SO(10)\) transformation, \[ B \to B' = \psi_R^T S^T\, C\, \Gamma_{ijk}\, S\psi_R. \] So the main task is to simplify the factor \(S^T C\).
The charge-conjugation matrix satisfies \[ C^{-1}\gamma_i^T C = -\gamma_i. \] From this one obtains \[ C^{-1}\sigma_{ij}^T C = -\sigma_{ij}, \] because \[ \sigma_{ij}=\frac{i}{2}[\gamma_i,\gamma_j]. \]
Indeed, \[ \begin{aligned} C^{-1}\sigma_{ij}^T C &= \frac{i}{2}C^{-1}[\gamma_i,\gamma_j]^T C \\ &= \frac{i}{2}C^{-1}[\gamma_j^T,\gamma_i^T]C \\ &= \frac{i}{2}[C^{-1}\gamma_j^T C,\, C^{-1}\gamma_i^T C] \\ &= \frac{i}{2}[-\gamma_j,\, -\gamma_i] \\ &= \frac{i}{2}[\gamma_j,\gamma_i] \\ &= -\sigma_{ij}. \end{aligned} \]
Exponentiating this relation gives \[ C^{-1}S^T C = C^{-1} e^{\frac{i}{4}\omega_{ij}\sigma_{ij}^T} C = e^{\frac{i}{4}\omega_{ij}(C^{-1}\sigma_{ij}^T C)} = e^{-\frac{i}{4}\omega_{ij}\sigma_{ij}} = S^{-1}. \] Equivalently, \[ S^T C = C S^{-1}. \]
Therefore, \[ \begin{aligned} B' &= \psi_R^T S^T C \Gamma_{ijk} S \psi_R \\ &= \psi_R^T C S^{-1}\Gamma_{ijk}S \psi_R. \end{aligned} \]
Now use \(\psi_R=P_+\psi\), together with \(P_+^T=P_+\). Then \[ \psi_R^T = \psi^T P_+, \] and so \[ B' = \psi^T P_+\, C\, \bigl(S^{-1}\Gamma_{ijk}S\bigr)\, P_+\psi. \]
Hence \[ \boxed{ \psi_R^T C \gamma_i\gamma_j\gamma_k \psi_R \;\to\; \psi^T P_+\, C \left\{ e^{-\frac{i}{4}\omega\sigma} (\gamma_i\gamma_j\gamma_k) e^{\frac{i}{4}\omega\sigma} \right\} P_+\psi } \] which is the transformation law used in the text.
The important point is that the gamma-product transforms by the adjoint action \[ \Gamma_{ijk} \to S^{-1}\Gamma_{ijk}S. \] Since \[ S^{-1}\gamma_i S = R_{ii'}\gamma_{i'}, \] it follows that \[ S^{-1}(\gamma_i\gamma_j\gamma_k)S = R_{ii'}R_{jj'}R_{kk'}\gamma_{i'}\gamma_{j'}\gamma_{k'}. \] So this object transforms as a rank-3 antisymmetric tensor of \(SO(10)\). It corresponds to the \(120\)-dimensional representation.
Let \[ T_{ijk}=\psi^T C\,\gamma_i\gamma_j\gamma_k\,\psi . \] From the earlier argument, this transforms as a rank-3 \(SO(10)\) tensor. The question is why, for representation-theory purposes, we may replace it by a totally antisymmetric tensor.
The key point is that the Clifford algebra lets us decompose any product of gamma matrices into a totally antisymmetric part plus terms with fewer free indices.
Start from \[ \{\gamma_i,\gamma_j\}=2\delta_{ij}. \] Therefore, \[ \gamma_i\gamma_j =\tfrac12\{\gamma_i,\gamma_j\}+\tfrac12[\gamma_i,\gamma_j] =\delta_{ij}\mathbf 1+\tfrac12[\gamma_i,\gamma_j]. \] If we define \[ \sigma_{ij}=\frac{1}{2i}[\gamma_i,\gamma_j], \] then equivalently \[ \gamma_i\gamma_j=\delta_{ij}\mathbf 1+i\sigma_{ij}. \]
Now multiply by \(\gamma_k\): \[ \gamma_i\gamma_j\gamma_k =\delta_{ij}\gamma_k+\tfrac12[\gamma_i,\gamma_j]\gamma_k. \] When inserted into \(T_{ijk}\), the first term gives \[ \delta_{ij}\,\psi^T C\gamma_k\psi, \] which is not an independent irreducible rank-3 object. It is simply the metric \(\delta_{ij}\) multiplying the vector \[ T_k^{(1)}=\psi^T C\gamma_k\psi. \] So this part belongs to a lower-rank sector.
To isolate the genuine 3-index content, fully antisymmetrize. A standard identity is \[ \gamma_i\gamma_j\gamma_k = \gamma_{[i}\gamma_j\gamma_{k]} +\delta_{ij}\gamma_k-\delta_{ik}\gamma_j+\delta_{jk}\gamma_i, \] where \[ \gamma_{[i}\gamma_j\gamma_{k]} \equiv \frac{1}{3!}\sum_{\pi\in S_3}\operatorname{sgn}(\pi) \,\gamma_{\pi(i)}\gamma_{\pi(j)}\gamma_{\pi(k)}. \]
This identity follows by repeatedly using \[ \gamma_a\gamma_b=-\gamma_b\gamma_a+2\delta_{ab}. \] Every time two gamma matrices are interchanged, one gets the antisymmetric sign together with a \(\delta\)-term. After full antisymmetrization, the leftover non-antisymmetric terms are always expressible as one \(\delta\) times a single gamma matrix.
Hence \[ T_{ijk} = \psi^T C\,\gamma_{[i}\gamma_j\gamma_{k]}\psi +\delta_{ij}\,\psi^T C\gamma_k\psi -\delta_{ik}\,\psi^T C\gamma_j\psi +\delta_{jk}\,\psi^T C\gamma_i\psi. \]
So the rank-3 tensor decomposes into:
This is why we may “take \(T_{ijk}\) to be totally antisymmetric”: the non-antisymmetric part does not contain a new independent 3-index structure. It reduces to lower-rank tensors that can be handled separately.
In representation-theory language, \[ (\text{rank-3 tensor from }\gamma_i\gamma_j\gamma_k) \longrightarrow (\text{3-form})\oplus(\text{vector trace pieces}). \] So once the vector part is dealt with separately, the genuinely new rank-3 content is just the totally antisymmetric piece.
Concretely, one may define \[ T_{[ijk]}=\psi^T C\,\gamma_{[i}\gamma_j\gamma_{k]}\psi, \] and this contains the irreducible 3-index information.
This is additional material to complement page 416 of Group Theory in a Nutshell by Anthony Zee.
Created with assistance from ChatGPT.