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Generally, Gamma matrices could be constructed based on the Clifford algebra. \begin{equation} \gamma^{i}\gamma^{j}+\gamma^{j}\gamma^{i}=2h^{ij}, \end{equation}

My question is how to generally construct the charge conjugation matrix to raise one spinor index in the gamma matrix.

In even dimensions (D=2m), consider complex Grassmann algebra $\Lambda_{m}[\alpha^{1},...,\alpha^{m}]$ with generators $\alpha^{1},...,\alpha^{m}.$) Namely, we define $\widehat{\alpha }^{i}$ and $\widehat{\beta}_{i}$ as multiplication and differentiation operators: \begin{equation} \widehat{\alpha}^{i}\psi=\alpha^{i}\psi, \end{equation} \begin{equation} \widehat{\beta}_{i}\psi=\frac{\partial}{\partial\alpha^{i}}\psi. \end{equation}

According to the Grassmann algebra, we have \begin{equation} \widehat{\alpha}^{i}\widehat{\alpha}^{j}+\widehat{\alpha}^{j}\widehat{\alpha }^{i}=0, \end{equation} \begin{equation} \widehat{\beta}_{i}\widehat{\beta}_{j}+\widehat{\beta}_{j}\widehat{\beta}% _{i}=0 \end{equation} \begin{equation} \widehat{\alpha}^{i}\widehat{\beta}_{j}+\widehat{\beta}_{j}\widehat{\alpha }^{i}=\delta_{j}^{i} \end{equation} This means that $\widehat{\alpha}^{1},...,\widehat{\alpha}^{m},\widehat{\beta }_{1},...,\widehat{\beta}_{m}$ specify a representation of Clifford algebra for some choice of $h$ (namely, for $h$ corresponding to quadratic form $\frac{1}{2}(x^{1}x^{m+1}+x^{2}x^{m+2}+...+x^{m}x^{2m})$). It follows that operators \begin{equation} \Gamma^{j}=\widehat{\alpha}^{j}+\widehat{\beta}_{j},1\leq j\leq m, \end{equation} \begin{equation} \Gamma^{j}=\widehat{\alpha}^{j-m}-\widehat{\beta}_{j-m},m<j\leq2m, \end{equation} determine a representation of $Cl(m,m,\mathbb{C})$

For example, in $D=4$, we can obtain $$\Gamma^{1}=\begin{pmatrix}0& 1& 0& 0\\ 1& 0& 0& 0\\ 0& 0& 0& 1\\ 0& 0& 1& 0\\ \end{pmatrix}$$, $$\Gamma^{2}=\begin{pmatrix}0& 0& 0& 1\\ 0& 0& {-1}& 0\\ 0& {-1}& 0& 0\\ 1& 0& 0& 0\\ \end{pmatrix}$$, $$\Gamma^{3}=\begin{pmatrix}0& {-1}& 0& 0\\ 1& 0& 0& 0\\ 0& 0& 0& 1\\ 0& 0& {-1}& 0\\ \end{pmatrix}$$, $$\Gamma^{4}=\begin{pmatrix}0& 0& 0& {-1}\\ 0& 0& 1& 0\\ 0& {-1}& 0& 0\\ 1& 0& 0& 0\\ \end{pmatrix}$$

My question is how to generally construct the charge conjugation matrix C, so that we could have $$C\Gamma C^{-1}=+/-\Gamma^T$$

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Lecture notes Supersymmetry summer term 2010 by Maximilian Kreuzer Here, in page 8 there a general answer to your question. Thanks for sharing that interesting representation in terms of Grassmann operators! – user77990 Apr 16 '15 at 20:27

Explicit expressions for the Euclidian signature are given in the following Hitoshi Murayama lecture notes (Section 1.3). The expressions are given in the Pauli matrix tensor product basis.

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Thank you. But they used a different method to construct the gamma matrices. Any idea how to construct the charge conjugation matrix for the gamma matrices given in my original post? – Osiris Xu Oct 4 '12 at 5:01
I can show you how to construct in this particular case, but can you please check your calculations because in your expressions $(\Gamma^1)^2=(\Gamma^2)^2=1$ while $(\Gamma^3)^2=(\Gamma^4)^2=-1$, i.e., you are working in a signature $(1, 1, -1, -1)$. Is this really the case you need. – David Bar Moshe Oct 4 '12 at 16:06
Dear David Yes, I am working with the general signatures of (1..1,-1..-1), where there are m +1s, and m -1s, for the 2m dimensional spacetime. – Osiris Xu Oct 4 '12 at 21:41

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