# From $\gamma$ to $\sigma$, finding proper basis

For $d$ dimensional case, usual gamma matrices have basis $\Gamma^A = \{1, \gamma^a, \cdots \gamma^{a_1 \cdots a_d}\}$ (Let's think about even case only for simplicity, I know for odd case only up to rank $m$ gamma matrices plays role as a basis, where $d=2m+1$, After figuring out even case i think i can handle for odd case too).

Then it naturally follows, from $Tr[\Gamma^A \Gamma^B]=2^{\frac{d}{2}}\eta^{AB}$, from expansion of any matrix $M$ as $M= m_A \Gamma^A$, one can have completness relation like \begin{align} \delta_{\alpha}{}^\beta \delta_\gamma{}^\delta = 2^{-\frac{d}{2}} \sum_A (\Gamma_A)_\alpha{}^\delta (\Gamma^A)_{\gamma}{}^\beta \end{align}

What i want to do is decompose $2^{\frac{d}{2}}\times 2^{\frac{d}{2}}$ matrix $\gamma$ as $2^{\frac{d}{2}-1} \times 2^{\frac{d}{2}-1}$ matrix $\sigma$ and $\bar{\sigma}$, such that \begin{align} \gamma^a = \begin{pmatrix} 0 & \sigma^a \\ \bar{\sigma}^a & 0 \end{pmatrix} \end{align} and find the completeness relation for $\sigma$ and $\bar{\sigma}$.

1. What is the basis for this case? I guess $\Sigma^A=\{1, \sigma^a, \cdots \}$ or $\Sigma^A = \{1, \bar{\sigma}^a , \cdots \}$, but considering $\gamma^{12}=\gamma^1 \gamma^2= \begin{pmatrix} \sigma^1 \bar{\sigma}^2 & 0 \\ 0 & \bar{\sigma}^1 \sigma^2 \end{pmatrix}$, i need cross terms like that, but how? Just define it?

2. I can pretty sure from usual clifford algebra, $\sigma^a \bar{\sigma}^b + \bar{\sigma}^a \sigma^b = 2 \eta^{ab}$ and since the dimension for this matrix is $2^{\frac{d}{2}-1}$ the trace formula become $Tr[\sigma^a \bar{\sigma}^b]= 2^{\frac{d}{2}-1} \eta^{ab}$, am I right?

• One correction with the notation. $\sigma^a$ already contains $1$. Therefore there is no need to specify $1$ separately in $\Sigma^A$. As for your second question, yes. Both the relations you have written are correct. See Supergravity by Dan Freedman and A Van Proeyen Chapter 3, Exercise 3.12 – BoundaryGraviton Jul 22 '16 at 9:24
• @SubramanyaHegde, I see, Then back to question 1, How one can make basis with $\sigma$ not $\gamma$? – phy_math Jul 22 '16 at 13:17
• It is not clear to me what your first question is. What are you trying to do? – BoundaryGraviton Jul 22 '16 at 17:10
• @SubramanyaHegde, I want to construct basis with $\sigma$ not in terms of $\gamma$ – phy_math Jul 22 '16 at 23:35
• You can read up about product basis of $\gamma$ matrices from the same book – BoundaryGraviton Jul 25 '16 at 5:26