I ask this question basically because I need help, just an hint, for a computation with bispinors. The background is string theory / supergravity, where many data of a supersymmetric vacuum can be encoded in a compatibile pair of pure spinors. Consider a compactification down to 4-dimension, $N=1$, where $\eta_{1+}$, $\eta_{2+}$ are two spinors on the internal manifold $M_6$, $(\eta_{1, 2+})^*=\eta_{1,2-}$. I can construct a compatible pair of pure spinors $\Phi_{\pm} = \eta_{1+} \otimes \eta_{2 \pm}^{\dagger}$. Using Fierz identities and Clifford map I can write: \begin{equation} \Phi_{\pm} = \eta_{1+} \otimes \eta_{2 \pm}^{\dagger} = \frac{1}{8} \sum_{k=0}^6 \frac{1}{k!} \left( \eta_{2 \pm}^{\dagger} \gamma_{m_k...m_1}\eta_{1+} \right)dx^{m_1} \wedge ... \wedge dx^{m_k}. \end{equation} I am looking for an explicit expression of this pair, using a parametrization for the two spinors: \begin{equation} \eta_{1+} = cos(\psi) \eta_+ + sin(\psi) \chi_+ \qquad \eta_{2+} = icos(\psi) \eta_+ - isin(\psi) \chi_+, \end{equation} where $\eta_+$ and $\chi_+= z\eta_-$ form an $SU(2)$ structure $(j, \omega, z)$ since they are orthogonal ($z$ is complex vector which is meant to act by Clifford moltiplication $z = z_m \gamma^m$).

Question: At the end of the day, I want to write $\Phi_{\pm}$ in term of $(j, \omega, z)$, the result is $(3.17)$ in (https://arxiv.org/abs/0708.1032). Consider for example $\Phi_+$: there are only terms with $k = even$. For the first one, with $k=2$, I can easily write it in terms of $(j, \omega, z)$; BUT for a term like $k=4$: $ \eta_{2+}^{\dagger} \gamma_{m n p q}\eta_{1+} $, how can I write it in terms of $(j, \omega, z)$, since $j,\omega$ are bilinears with 2 gamma matrices and $z$ has only one? I don't know how to expand a term with 4 gamma matrices (or more) in ones with 2 or 1. Can I write $ \eta_{2+}^{\dagger} \gamma_{m n p q}\eta_{1+} $ as something like $(\eta_{2+}^{\dagger} \gamma_{m n}\eta_{1+})^2 $ or something like that?


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.